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Astronomy_381

如图所示, 质量分别为 $M$ 和 $m$ 的两个星球 $\mathrm{A}$ 和 $\mathrm{B}$ (均视为质点) 在它们之间的引力作用下都绕 $O$ 点做匀速圆周运动, 星球 $\mathrm{A}$ 和 $\mathrm{B}$ 之间的距离为 $L$ 。已知星球 $\mathrm{A}$ 和 $\mathrm{B}$ 和 $O$三点始终共线, $\mathrm{A}$ 和 $\mathrm{B}$ 分别在 $O$ 点的两侧。引力常量为 $G$ 。 在地月系统中, 若忽略其他星球的影响, 可以将月球和地球看成上述星球 $\mathrm{A}$ 和 $\mathrm{B}$,月球绕其轨道中心运行的周期记为 $T_{1}$ 。但在处理近似问题时, 常常认为月球是绕地心做圆周运动的, 这样算得的运行周期为 $T_{2}$ 。已知地球和月球的质量分别为 $m_{\text {地 }}=6 \times 10^{24} \mathrm{~kg}$和 $m_{\text {月 }}=7 \times 10^{22} \mathrm{~kg}$ 。求 $\frac{T_{2}^{2}}{T_{1}^{2}}$ 。(结果保留三位有效数字) [图1]

你正在参加一个国际天文竞赛，并需要解决以下问题。 这个问题的答案是一个数值。 问题: 如图所示, 质量分别为 $M$ 和 $m$ 的两个星球 $\mathrm{A}$ 和 $\mathrm{B}$ (均视为质点) 在它们之间的引力作用下都绕 $O$ 点做匀速圆周运动, 星球 $\mathrm{A}$ 和 $\mathrm{B}$ 之间的距离为 $L$ 。已知星球 $\mathrm{A}$ 和 $\mathrm{B}$ 和 $O$三点始终共线, $\mathrm{A}$ 和 $\mathrm{B}$ 分别在 $O$ 点的两侧。引力常量为 $G$ 。 在地月系统中, 若忽略其他星球的影响, 可以将月球和地球看成上述星球 $\mathrm{A}$ 和 $\mathrm{B}$,月球绕其轨道中心运行的周期记为 $T_{1}$ 。但在处理近似问题时, 常常认为月球是绕地心做圆周运动的, 这样算得的运行周期为 $T_{2}$ 。已知地球和月球的质量分别为 $m_{\text {地 }}=6 \times 10^{24} \mathrm{~kg}$和 $m_{\text {月 }}=7 \times 10^{22} \mathrm{~kg}$ 。求 $\frac{T_{2}^{2}}{T_{1}^{2}}$ 。(结果保留三位有效数字) [图1] 你输出的所有数学公式和符号应该使用LaTeX表示！ 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER是数值。

NV

Astronomy

ZH

multi-modal

Astronomy_1096

The Parker Solar Probe (PSP) is part of a mission to learn more about the Sun, named after the scientist that first proposed the existence of the solar wind, and was launched on $12^{\text {th }}$ August 2018. Over the course of the 7 year mission it will orbit the Sun 24 times, and through 7 flybys of Venus it will lose some energy in order to get into an ever tighter orbit (see Figure 1). In its final 3 orbits it will have a perihelion (closest approach to the Sun) of only $r_{\text {peri }}=9.86 R_{\odot}$, about 7 times closer than any previous probe, the first of which is due on $24^{\text {th }}$ December 2024. In this extreme environment the probe will not only face extreme brightness and temperatures but also will break the record for the fastest ever spacecraft. [figure1] Figure 1: Left: The journey PSP will take to get from the Earth to the final orbit around the Sun. Right: The probe just after assembly in the John Hopkins University Applied Physics Laboratory. Credit: NASA / John Hopkins APL / Ed Whitman. $$ v^{2}=G M\left(\frac{2}{r}-\frac{1}{a}\right) $$ Given that in its final orbit PSP has a orbital period of 88 days, calculate the speed of the probe as it passes through the minimum perihelion. Give your answer in $\mathrm{km} \mathrm{s}^{-1}$. Close to the Sun the communications equipment is very sensitive to the extreme environment, so the mission is planned for the probe to take all of its primary science measurements whilst within 0.25 au of the Sun, and then to spend the rest of the orbit beaming that data back to Earth, as shown in Figure 2. [figure2] Figure 2: The way PSP is planned to split each orbit into taking measurements and sending data back. Credit: NASA / Johns Hopkins APL. When considering the position of an object in an elliptical orbit as a function of time, there are two important angles (called 'anomalies') necessary to do the calculation, and they are defined in Figure 3. By constructing a circular orbit centred on the same point as the ellipse and with the same orbital period, the eccentric anomaly, $E$, is then the angle between the major axis and the perpendicular projection of the object (some time $t$ after perihelion) onto the circle as measured from the centre of the ellipse ( $\angle x c z$ in the figure). The mean anomaly, $M$, is the angle between the major axis and where the object would have been at time $t$ if it was indeed on the circular orbit ( $\angle y c z$ in the figure, such that the shaded areas are the same). [figure3] Figure 3: The definitions of the anomalies needed to get the position of an object in an ellipse as a function of time. The Sun (located at the focus) is labeled $S$ and the probe $P . M$ and $E$ are the mean and eccentric anomalies respectively. The angle $\theta$ is called the true anomaly and is not needed for this question. Credit: Wikipedia. The eccentric anomaly can be related to the mean anomaly through Kepler's Equation, $$ M=E-e \sin E \text {. } $$a. When the probe is at its closest perihelion: ii. Calculate the temperature the heat shield must be able to survive. Assume that the heat shield of the probe absorbs all of the incident radiation, radiates as a perfect black body, and that only one side of the probe ever faces the Sun (to protect the instruments) such that the emitting (surface) area is double the absorbing (cross-sectional) area.

You are participating in an international Astronomy competition and need to solve the following question. The answer to this question is a numerical value. Here is some context information for this question, which might assist you in solving it: The Parker Solar Probe (PSP) is part of a mission to learn more about the Sun, named after the scientist that first proposed the existence of the solar wind, and was launched on $12^{\text {th }}$ August 2018. Over the course of the 7 year mission it will orbit the Sun 24 times, and through 7 flybys of Venus it will lose some energy in order to get into an ever tighter orbit (see Figure 1). In its final 3 orbits it will have a perihelion (closest approach to the Sun) of only $r_{\text {peri }}=9.86 R_{\odot}$, about 7 times closer than any previous probe, the first of which is due on $24^{\text {th }}$ December 2024. In this extreme environment the probe will not only face extreme brightness and temperatures but also will break the record for the fastest ever spacecraft. [figure1] Figure 1: Left: The journey PSP will take to get from the Earth to the final orbit around the Sun. Right: The probe just after assembly in the John Hopkins University Applied Physics Laboratory. Credit: NASA / John Hopkins APL / Ed Whitman. $$ v^{2}=G M\left(\frac{2}{r}-\frac{1}{a}\right) $$ Given that in its final orbit PSP has a orbital period of 88 days, calculate the speed of the probe as it passes through the minimum perihelion. Give your answer in $\mathrm{km} \mathrm{s}^{-1}$. Close to the Sun the communications equipment is very sensitive to the extreme environment, so the mission is planned for the probe to take all of its primary science measurements whilst within 0.25 au of the Sun, and then to spend the rest of the orbit beaming that data back to Earth, as shown in Figure 2. [figure2] Figure 2: The way PSP is planned to split each orbit into taking measurements and sending data back. Credit: NASA / Johns Hopkins APL. When considering the position of an object in an elliptical orbit as a function of time, there are two important angles (called 'anomalies') necessary to do the calculation, and they are defined in Figure 3. By constructing a circular orbit centred on the same point as the ellipse and with the same orbital period, the eccentric anomaly, $E$, is then the angle between the major axis and the perpendicular projection of the object (some time $t$ after perihelion) onto the circle as measured from the centre of the ellipse ( $\angle x c z$ in the figure). The mean anomaly, $M$, is the angle between the major axis and where the object would have been at time $t$ if it was indeed on the circular orbit ( $\angle y c z$ in the figure, such that the shaded areas are the same). [figure3] Figure 3: The definitions of the anomalies needed to get the position of an object in an ellipse as a function of time. The Sun (located at the focus) is labeled $S$ and the probe $P . M$ and $E$ are the mean and eccentric anomalies respectively. The angle $\theta$ is called the true anomaly and is not needed for this question. Credit: Wikipedia. The eccentric anomaly can be related to the mean anomaly through Kepler's Equation, $$ M=E-e \sin E \text {. } $$ problem: a. When the probe is at its closest perihelion: ii. Calculate the temperature the heat shield must be able to survive. Assume that the heat shield of the probe absorbs all of the incident radiation, radiates as a perfect black body, and that only one side of the probe ever faces the Sun (to protect the instruments) such that the emitting (surface) area is double the absorbing (cross-sectional) area. All mathematical formulas and symbols you output should be represented with LaTeX! You can solve it step by step. Remember, your answer should be calculated in the unit of \mathrm{~K}, but when concluding your final answer, do not include the unit. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER is the numerical value without any units.

NV

Astronomy

EN

multi-modal

Astronomy_444

2020 年左右我国将进行第一次火星探测, 美国已发射了“凤凰号”着陆器降落在 火星北极勘察是否有水的存在。如图所示为“凤凰号”着陆器经过多次变轨后登陆火 星的轨迹图, 轨道上的 $P 、 S 、 Q$ 三点与火星中心在同一直线上, $P 、 Q$ 两点分别是粗 圆轨道的远火星点和近火星点, 且 $P Q=2 Q S$, (已知轨道II为圆轨道) 下列说法正确的 是 [图1] A: 着陆器在 $P$ 点由轨道I进入轨道II需要点火加速 B: 着陆器在轨道II上 $S$ 点的速度小于在轨道III上 $Q$ 点的速度 C: 着陆器在轨道II上 $S$ 点的加速度大小大于在轨道III上 $P$ 点的加速度大小 D: 着陆器在轨道II上由 $P$ 点运动到 $S$ 点的时间是着陆器在轨道III上由 $P$ 点运动到 $Q$ 点的时间的 2 倍

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个单选题（只有一个正确答案）。 问题: 2020 年左右我国将进行第一次火星探测, 美国已发射了“凤凰号”着陆器降落在 火星北极勘察是否有水的存在。如图所示为“凤凰号”着陆器经过多次变轨后登陆火 星的轨迹图, 轨道上的 $P 、 S 、 Q$ 三点与火星中心在同一直线上, $P 、 Q$ 两点分别是粗 圆轨道的远火星点和近火星点, 且 $P Q=2 Q S$, (已知轨道II为圆轨道) 下列说法正确的 是 [图1] A: 着陆器在 $P$ 点由轨道I进入轨道II需要点火加速 B: 着陆器在轨道II上 $S$ 点的速度小于在轨道III上 $Q$ 点的速度 C: 着陆器在轨道II上 $S$ 点的加速度大小大于在轨道III上 $P$ 点的加速度大小 D: 着陆器在轨道II上由 $P$ 点运动到 $S$ 点的时间是着陆器在轨道III上由 $P$ 点运动到 $Q$ 点的时间的 2 倍 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]

SC

Astronomy

ZH

multi-modal

Astronomy_561

地球赤道上的物体随地球自转的向心加速度为 $a$, 地球的同步卫星绕地球做匀速圆周运动的轨道半径为 $r_{1}$, 向心加速度为 $a_{1}$ 。已知万有引力常量为 $G$, 地球半径为 $R$, 地球赤道表面的加速度为 $g$ 。下列说法正确的是 ( ) A: 地球质量 $M=\frac{a R^{2}}{G}$ B: 地球质量 $M=\frac{a_{1} r_{1}^{2}}{G}$ C: 加速度之比 $\frac{a_{1}}{a}=\frac{R^{2}}{r_{1}^{2}}$ D: $a 、 a_{1} 、 g$ 的关系是 $a<a_{1}<g$

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个多选题（有多个正确答案）。 问题: 地球赤道上的物体随地球自转的向心加速度为 $a$, 地球的同步卫星绕地球做匀速圆周运动的轨道半径为 $r_{1}$, 向心加速度为 $a_{1}$ 。已知万有引力常量为 $G$, 地球半径为 $R$, 地球赤道表面的加速度为 $g$ 。下列说法正确的是 ( ) A: 地球质量 $M=\frac{a R^{2}}{G}$ B: 地球质量 $M=\frac{a_{1} r_{1}^{2}}{G}$ C: 加速度之比 $\frac{a_{1}}{a}=\frac{R^{2}}{r_{1}^{2}}$ D: $a 、 a_{1} 、 g$ 的关系是 $a<a_{1}<g$ 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]

MC

Astronomy

ZH

text-only

Astronomy_417

如图甲所示, 河外星系中两黑洞 $A 、 B$ 的质量分别为 $M_{1}$ 和 $M_{2}$, 它们以两者连线上的某一点为圆心做匀速圆周运动. 为研究方便可简化为如图乙所示示意图, 黑洞 $A$ 和黑洞 $B$ 均可看成匀质球体, 天文学家测得 $O A>O B$, 且黑洞 $A$ 的半径大于黑洞 $B$ 的半径. 根据你所学的知识, 下列说法正确的是 [图1] 图甲 [图2] 图乙 A: 两黑洞质量之间的关系为 $M_{1}>M_{2}$ B: 黑洞 $A$ 的第一宇宙速度小于黑洞 $B$ 的第一宇宙速度 C: 若两黑洞间距离不变, 设法将黑洞 $A$ 上的一部分物质移到黑洞 $B$ 上, 则它们间的万有引力将变大 D: 人类要将宇航器发射到距离黑洞 $A$ 或黑洞 $B$ 较近的区域进行探索, 发射速度一定大于 $16.7 \mathrm{~km} / \mathrm{s}$

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个多选题（有多个正确答案）。 问题: 如图甲所示, 河外星系中两黑洞 $A 、 B$ 的质量分别为 $M_{1}$ 和 $M_{2}$, 它们以两者连线上的某一点为圆心做匀速圆周运动. 为研究方便可简化为如图乙所示示意图, 黑洞 $A$ 和黑洞 $B$ 均可看成匀质球体, 天文学家测得 $O A>O B$, 且黑洞 $A$ 的半径大于黑洞 $B$ 的半径. 根据你所学的知识, 下列说法正确的是 [图1] 图甲 [图2] 图乙 A: 两黑洞质量之间的关系为 $M_{1}>M_{2}$ B: 黑洞 $A$ 的第一宇宙速度小于黑洞 $B$ 的第一宇宙速度 C: 若两黑洞间距离不变, 设法将黑洞 $A$ 上的一部分物质移到黑洞 $B$ 上, 则它们间的万有引力将变大 D: 人类要将宇航器发射到距离黑洞 $A$ 或黑洞 $B$ 较近的区域进行探索, 发射速度一定大于 $16.7 \mathrm{~km} / \mathrm{s}$ 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]

MC

Astronomy

ZH

multi-modal

Astronomy_851

Jupiter's deep atmosphere is very warm due to convection leading to an adiabatic temperature profile that increases with increasing pressure. Assuming (for simplicity) that this outer layer of Jupiter has a temperature of $500 \mathrm{~K}$, perform a back-of-the-envelope estimate of the characteristic thickness (or e-folding scale) of the envelope of Jupiter (you may find that this is independent of pressure level). You may further use that the specific gas constant in Jupiter's atmosphere is $3600 \mathrm{~J} \mathrm{~kg}^{-1} \mathrm{~K}^{-1}$. A: $20 \mathrm{~km}$ B: $73 \mathrm{~km}$ C: $568 \mathrm{~km}$ D: $3,120 \mathrm{~km}$ E: $10,233 \mathrm{~km}$

You are participating in an international Astronomy competition and need to solve the following question. This is a multiple choice question (only one correct answer). problem: Jupiter's deep atmosphere is very warm due to convection leading to an adiabatic temperature profile that increases with increasing pressure. Assuming (for simplicity) that this outer layer of Jupiter has a temperature of $500 \mathrm{~K}$, perform a back-of-the-envelope estimate of the characteristic thickness (or e-folding scale) of the envelope of Jupiter (you may find that this is independent of pressure level). You may further use that the specific gas constant in Jupiter's atmosphere is $3600 \mathrm{~J} \mathrm{~kg}^{-1} \mathrm{~K}^{-1}$. A: $20 \mathrm{~km}$ B: $73 \mathrm{~km}$ C: $568 \mathrm{~km}$ D: $3,120 \mathrm{~km}$ E: $10,233 \mathrm{~km}$ You can solve it step by step. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER should be one of the options: [A, B, C, D, E].

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Astronomy

EN

text-only

Astronomy_377

两颗人造卫星绕地球运动周期相同, 轨道如图所示, 分别为圆轨道和椭圆轨道, $\mathrm{AB}$为粗圆的长轴, $\mathrm{C} 、 \mathrm{D}$ 为两轨道交点. 已知椭圆轨道上的卫星到 $\mathrm{C}$ 点时速度方向与 $\mathrm{AB}$平行, 则下列说法中正确的是( ) [图1] A: 卫星在圆轨道的速率为 $v_{0}$, 卫星椭圆轨道 $\mathrm{A}$ 点的速率为 $v_{A}$, 则 $v_{0}>v_{A}$ B: 卫星在圆轨道的速率为 $v_{0}$, 卫星在椭圆轨道 $\mathrm{B}$ 点的速率为 $v_{B}$, 则 $v_{B}>v_{0}$ C: 两个轨道上的卫星运动到 $\mathrm{C}$ 点时的加速度相同 D: 两个轨道上的卫星运动到 $\mathrm{C}$ 点时的向心加速度大小相等

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个单选题（只有一个正确答案）。 问题: 两颗人造卫星绕地球运动周期相同, 轨道如图所示, 分别为圆轨道和椭圆轨道, $\mathrm{AB}$为粗圆的长轴, $\mathrm{C} 、 \mathrm{D}$ 为两轨道交点. 已知椭圆轨道上的卫星到 $\mathrm{C}$ 点时速度方向与 $\mathrm{AB}$平行, 则下列说法中正确的是( ) [图1] A: 卫星在圆轨道的速率为 $v_{0}$, 卫星椭圆轨道 $\mathrm{A}$ 点的速率为 $v_{A}$, 则 $v_{0}>v_{A}$ B: 卫星在圆轨道的速率为 $v_{0}$, 卫星在椭圆轨道 $\mathrm{B}$ 点的速率为 $v_{B}$, 则 $v_{B}>v_{0}$ C: 两个轨道上的卫星运动到 $\mathrm{C}$ 点时的加速度相同 D: 两个轨道上的卫星运动到 $\mathrm{C}$ 点时的向心加速度大小相等 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]

SC

Astronomy

ZH

multi-modal

Astronomy_154

宇航员在地球表面一斜坡上 $P$ 点, 沿水平方向以初速度 $v_{0}$ 抛出一个小球, 测得小球经时间 $t$ 落到斜坡另一点 $Q$ 上现宇航员站在某质量分布均匀的星球表面相同的斜坡上 $P$点, 沿水平方向以相同的初速度 $v_{0}$ 抛出一个小球, 小球落在 $P Q$ 的中点. 已知该星球的半径为 $R$, 地球表面重力加速度为 $g$, 引力常量为 $G$, 球的体积公式是 $V=\frac{4}{3} \pi R^{3}$ 。求: 该星球的质量 $M$; [图1]

你正在参加一个国际天文竞赛，并需要解决以下问题。 这个问题的答案是一个表达式。 问题: 宇航员在地球表面一斜坡上 $P$ 点, 沿水平方向以初速度 $v_{0}$ 抛出一个小球, 测得小球经时间 $t$ 落到斜坡另一点 $Q$ 上现宇航员站在某质量分布均匀的星球表面相同的斜坡上 $P$点, 沿水平方向以相同的初速度 $v_{0}$ 抛出一个小球, 小球落在 $P Q$ 的中点. 已知该星球的半径为 $R$, 地球表面重力加速度为 $g$, 引力常量为 $G$, 球的体积公式是 $V=\frac{4}{3} \pi R^{3}$ 。求: 该星球的质量 $M$; [图1] 你输出的所有数学公式和符号应该使用LaTeX表示！ 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER是一个不含等号的表达式，例如ANSWER=\frac{1}{2} g t^2

EX

Astronomy

ZH

multi-modal

Astronomy_896

Where on the Moon did astronauts walk in July 1969 ? A: Mare Serenitatis B: Mare Tranquillitatis C: Mare Imbrium D: Mare Nubium

You are participating in an international Astronomy competition and need to solve the following question. This is a multiple choice question (only one correct answer). problem: Where on the Moon did astronauts walk in July 1969 ? A: Mare Serenitatis B: Mare Tranquillitatis C: Mare Imbrium D: Mare Nubium You can solve it step by step. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER should be one of the options: [A, B, C, D].

SC

Astronomy

EN

text-only

Astronomy_194

为了探知未知天体, 假如宇航员乘坐宇宙飞船到达某星球, 测得飞船在该星球表面附近做圆周运动的周期为 $T$ 。飞船降落到该星球表面后, 宇航员将小球从 $H$ 高处以初速度 $v_{0}$ 水平抛出, 落地时水平位移为 $x$, 忽略空气阻力和该星球的自转, 已知引力常量为 $G$ ，将该星球视为质量分布均匀的球体，则以下说法正确的是（） A: 该星球的半径为 $\frac{H v_{0}^{2} T^{2}}{2 \pi^{2} x^{2}}$ B: 该星球的半径为 $\frac{H v_{0}^{2} T^{2}}{4 \pi^{2} x^{2}}$ C: 该星球的质量为 $\frac{H^{3} v_{0}^{6} T^{4}}{2 G \pi^{4} x^{6}}$ D: 该星球的质量为 $\frac{H^{3} v_{0}^{6} T^{4}}{G \pi^{4} x^{6}}$

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个多选题（有多个正确答案）。 问题: 为了探知未知天体, 假如宇航员乘坐宇宙飞船到达某星球, 测得飞船在该星球表面附近做圆周运动的周期为 $T$ 。飞船降落到该星球表面后, 宇航员将小球从 $H$ 高处以初速度 $v_{0}$ 水平抛出, 落地时水平位移为 $x$, 忽略空气阻力和该星球的自转, 已知引力常量为 $G$ ，将该星球视为质量分布均匀的球体，则以下说法正确的是（） A: 该星球的半径为 $\frac{H v_{0}^{2} T^{2}}{2 \pi^{2} x^{2}}$ B: 该星球的半径为 $\frac{H v_{0}^{2} T^{2}}{4 \pi^{2} x^{2}}$ C: 该星球的质量为 $\frac{H^{3} v_{0}^{6} T^{4}}{2 G \pi^{4} x^{6}}$ D: 该星球的质量为 $\frac{H^{3} v_{0}^{6} T^{4}}{G \pi^{4} x^{6}}$ 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]

MC

Astronomy

ZH

text-only

Astronomy_1219

In November 2020 the Aricebo Telescope at the National Astronomy and Ionosphere Centre (NAIC) in Puerto Rico was decommissioned due to safety concerns after extensive storm damage. First opened in November 1963, this brought an end to an illustrious contribution to radio astronomy where, with a dish diameter of $304.8 \mathrm{~m}$ (1000 ft), it was the largest radio telescope in the world until 2016. Its important discoveries range from detection of the first extrasolar planets around a pulsar to fast radio bursts, as well as a pivotal role in the search for extraterrestrial intelligence (SETI), however in this question we will explore its earliest major revelation: that Mercury was not tidally locked. [figure1] Figure 1: Left: The Aricebo telescope before it was damaged. Credit: NAIC. Right: When transmitting a pulse from a radio telescope, diffraction prevents the beam from staying perfectly parallel and so the width of the beam increases by $2 \theta$. Credit: OpenStax, College Physics. Mercury had already been studied with optical and infrared telescopes, however the advantage of a radio telescope was that you could send pulses and receive their reflections. This radar-ranging technique had already been used with Venus to measure the distance to it and hence provide the data necessary for a definitive measurement of an astronomical unit in metres. The radar echo from Mercury is much harder to detect due to the extra distance travelled and its smaller cross-sectional area (its radius is $2440 \mathrm{~km}$ ). In April 1965, Pettengill and Dyce sent a series of $500 \mu \mathrm{s}$ pulses at $430 \mathrm{MHz}$ with a transmitted power of $2.0 \mathrm{MW}$ towards Mercury whilst it was at its closest point in its orbit to Earth. In ideal circumstances the beam would stay parallel, however diffraction widens the beam as shown on the right in Fig 1. The signal-to-noise ratio of this echo was high enough that Doppler broadening of the received signal was reliably detected, allowing a determination of the rotation rate of Mercury. In August 1965 the same scientists sent $100 \mu$ s pulses and sampled the echo on short timescales as it returned. The strongest echo (received first) came from the point of the planet closest to the Earth (called the sub-radar point), with later echos coming from other parts of the surface in an annulus of increasing radius (see Fig 2). Photons from the approaching side would be blueshifted to a higher frequency, whilst those from the receding side would be redshifted to a lower frequency. Hence, by measuring the Doppler shift and the time delay, you can map the rotational velocity as a function of apparent longitude and so can calculate the apparent rotation rate (as well as the direction of rotation and co-ordinates of the pole). [figure2] Figure 2: Left: Snapshots of the reflections of a single $100 \mu$ sulse. The strength of the echo weakens as you move to later delays (as represented by the scale factor in the top right of each snapshot) and hence you need to use an annulus rather than detections from the horizon. The small arrows indicate the Doppler shifted frequency associated with intersection of the annulus with the apparent equator for each delay. The horizontal axis is in cycles per second (and so $1 \mathrm{c} / \mathrm{s}=1 \mathrm{~Hz}$ ). Credit: Dyce, Pettengill and Shapiro (1967). Top right: The key principles of the delay-Doppler technique, looking at a cross-section of the planet. At the very centre is the sub-radar point (the point on the planet's surface closest to the Earth). As you move away from the sub-radar point the light has to travel further before it can be reflected, and hence the echo from those regions arrives later. The brightest point of any given annulus is where it intersects the apparent equator (due to the largest reflecting area), and so in each of the snapshots that is why the extreme Doppler shifts are boosted relative to the middle. Credit: Shapiro (1967). Bottom right: The same as the snapshots, but this time summed over the first $500 \mu$ of reflections. Here the difference between the extreme left and right frequencies reliably detected is $\sim 5 \mathrm{~Hz}$, but when corrected for relative motion of the Earth and Mercury it becomes the value given in part c. Credit: Pettengill, Dyce and Campbell (1967). The Doppler shift with light is given as $$ \frac{\Delta f}{f}=\frac{v}{c} $$ where $\Delta f$ is the shift in frequency $f, v$ is the line-of-sight velocity of the emitting object and $c$ is the speed of light. Ever since the first maps of Mercury's surface by Schiaparelli in the late 1880s, many in the scientific community believed that Mercury would be tidally locked and so always present the same hemisphere to the Sun. The reason they expected the rotational period to be the same as its orbital period (i.e. a $1: 1$ ratio), rather like the Moon, is because it is so close to the Sun and the tidal torques causing this synchronicity are proportional to $r^{-6}$ where $r$ is the distance from the massive body. Given it is the closest planet to the Sun, it receives by far the largest torques, so the discovery it was in a different ratio was a complete surprise to many of the scientists at the time. [figure3] Figure 3: The orientation of Mercury's axis of minimum moment of inertia (the axis the tidal torque acts upon) displayed at six points in its orbit (equally spaced in time) if the ratio had been $1: 1$. Credit: Colombo and Shapiro (1966).c. The eccentricity of the planet's orbit became the prime suspect as to why its actual ratio would be stable over long time periods. ii. Assuming the tidal torque at perihelion is the dominating factor in setting Mercury's rotation rate, predict the rotational period of Mercury if it were to behave as though it was tidally locked when passing through perihelion. Compare this to the measured value and comment on validity of the assumption.

You are participating in an international Astronomy competition and need to solve the following question. The answer to this question is a numerical value. Here is some context information for this question, which might assist you in solving it: In November 2020 the Aricebo Telescope at the National Astronomy and Ionosphere Centre (NAIC) in Puerto Rico was decommissioned due to safety concerns after extensive storm damage. First opened in November 1963, this brought an end to an illustrious contribution to radio astronomy where, with a dish diameter of $304.8 \mathrm{~m}$ (1000 ft), it was the largest radio telescope in the world until 2016. Its important discoveries range from detection of the first extrasolar planets around a pulsar to fast radio bursts, as well as a pivotal role in the search for extraterrestrial intelligence (SETI), however in this question we will explore its earliest major revelation: that Mercury was not tidally locked. [figure1] Figure 1: Left: The Aricebo telescope before it was damaged. Credit: NAIC. Right: When transmitting a pulse from a radio telescope, diffraction prevents the beam from staying perfectly parallel and so the width of the beam increases by $2 \theta$. Credit: OpenStax, College Physics. Mercury had already been studied with optical and infrared telescopes, however the advantage of a radio telescope was that you could send pulses and receive their reflections. This radar-ranging technique had already been used with Venus to measure the distance to it and hence provide the data necessary for a definitive measurement of an astronomical unit in metres. The radar echo from Mercury is much harder to detect due to the extra distance travelled and its smaller cross-sectional area (its radius is $2440 \mathrm{~km}$ ). In April 1965, Pettengill and Dyce sent a series of $500 \mu \mathrm{s}$ pulses at $430 \mathrm{MHz}$ with a transmitted power of $2.0 \mathrm{MW}$ towards Mercury whilst it was at its closest point in its orbit to Earth. In ideal circumstances the beam would stay parallel, however diffraction widens the beam as shown on the right in Fig 1. The signal-to-noise ratio of this echo was high enough that Doppler broadening of the received signal was reliably detected, allowing a determination of the rotation rate of Mercury. In August 1965 the same scientists sent $100 \mu$ s pulses and sampled the echo on short timescales as it returned. The strongest echo (received first) came from the point of the planet closest to the Earth (called the sub-radar point), with later echos coming from other parts of the surface in an annulus of increasing radius (see Fig 2). Photons from the approaching side would be blueshifted to a higher frequency, whilst those from the receding side would be redshifted to a lower frequency. Hence, by measuring the Doppler shift and the time delay, you can map the rotational velocity as a function of apparent longitude and so can calculate the apparent rotation rate (as well as the direction of rotation and co-ordinates of the pole). [figure2] Figure 2: Left: Snapshots of the reflections of a single $100 \mu$ sulse. The strength of the echo weakens as you move to later delays (as represented by the scale factor in the top right of each snapshot) and hence you need to use an annulus rather than detections from the horizon. The small arrows indicate the Doppler shifted frequency associated with intersection of the annulus with the apparent equator for each delay. The horizontal axis is in cycles per second (and so $1 \mathrm{c} / \mathrm{s}=1 \mathrm{~Hz}$ ). Credit: Dyce, Pettengill and Shapiro (1967). Top right: The key principles of the delay-Doppler technique, looking at a cross-section of the planet. At the very centre is the sub-radar point (the point on the planet's surface closest to the Earth). As you move away from the sub-radar point the light has to travel further before it can be reflected, and hence the echo from those regions arrives later. The brightest point of any given annulus is where it intersects the apparent equator (due to the largest reflecting area), and so in each of the snapshots that is why the extreme Doppler shifts are boosted relative to the middle. Credit: Shapiro (1967). Bottom right: The same as the snapshots, but this time summed over the first $500 \mu$ of reflections. Here the difference between the extreme left and right frequencies reliably detected is $\sim 5 \mathrm{~Hz}$, but when corrected for relative motion of the Earth and Mercury it becomes the value given in part c. Credit: Pettengill, Dyce and Campbell (1967). The Doppler shift with light is given as $$ \frac{\Delta f}{f}=\frac{v}{c} $$ where $\Delta f$ is the shift in frequency $f, v$ is the line-of-sight velocity of the emitting object and $c$ is the speed of light. Ever since the first maps of Mercury's surface by Schiaparelli in the late 1880s, many in the scientific community believed that Mercury would be tidally locked and so always present the same hemisphere to the Sun. The reason they expected the rotational period to be the same as its orbital period (i.e. a $1: 1$ ratio), rather like the Moon, is because it is so close to the Sun and the tidal torques causing this synchronicity are proportional to $r^{-6}$ where $r$ is the distance from the massive body. Given it is the closest planet to the Sun, it receives by far the largest torques, so the discovery it was in a different ratio was a complete surprise to many of the scientists at the time. [figure3] Figure 3: The orientation of Mercury's axis of minimum moment of inertia (the axis the tidal torque acts upon) displayed at six points in its orbit (equally spaced in time) if the ratio had been $1: 1$. Credit: Colombo and Shapiro (1966). problem: c. The eccentricity of the planet's orbit became the prime suspect as to why its actual ratio would be stable over long time periods. ii. Assuming the tidal torque at perihelion is the dominating factor in setting Mercury's rotation rate, predict the rotational period of Mercury if it were to behave as though it was tidally locked when passing through perihelion. Compare this to the measured value and comment on validity of the assumption. All mathematical formulas and symbols you output should be represented with LaTeX! You can solve it step by step. Remember, your answer should be calculated in the unit of \text { days }, but when concluding your final answer, do not include the unit. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER is the numerical value without any units.

NV

Astronomy

EN

multi-modal

Astronomy_563

建造一条能通向太空的电梯 (如图甲所示), 是人们长期的梦想。材料的力学强度是材料众多性能中被人类极为看重的一种性能, 目前已发现的高强度材料碳纳米管的抗拉强度是钢的 100 倍, 密度是其 $\frac{1}{6}$ ，这使得人们有望在赤道上建造垂直于水平面的“太空电梯”。图乙中 $r$ 为航天员到地心的距离, $R$ 为地球半径, $a-r$ 图像中的图线 $A$ 表示地球引力对航天员产生的加速度大小与 $r$ 的关系, 图线 $B$ 表示航天员由于地球自转而产生的向心加速度大小与 $r$ 的关系, 关于相对地面静止在不同高度的航天员, 地面附近重力加速度 $g$ 取 $10 \mathrm{~m} / \mathrm{s}^{2}$, 地球自转角速度 $\omega=7.3 \times 10^{-5} \mathrm{rad} / \mathrm{s}$, 地球半径 $R=6.4 \times 10^{3} \mathrm{~km}$ 。下列说法正确的有 ( ) [图1] 图甲 [图2] 图乙 A: 随着 $r$ 增大, 航天员受到电梯舱的弹力减小 B: 航天员在 $r=R$ 处的线速度等于第一宇宙速度 C: 图中 $r_{0}$ 为地球同步卫星的轨道半径 D: 电梯舱停在距地面高度为 $6.6 R$ 的站点时, 舱内质量 $60 \mathrm{~kg}$ 的航天员对水平地板的压力为零

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个单选题（只有一个正确答案）。 问题: 建造一条能通向太空的电梯 (如图甲所示), 是人们长期的梦想。材料的力学强度是材料众多性能中被人类极为看重的一种性能, 目前已发现的高强度材料碳纳米管的抗拉强度是钢的 100 倍, 密度是其 $\frac{1}{6}$ ，这使得人们有望在赤道上建造垂直于水平面的“太空电梯”。图乙中 $r$ 为航天员到地心的距离, $R$ 为地球半径, $a-r$ 图像中的图线 $A$ 表示地球引力对航天员产生的加速度大小与 $r$ 的关系, 图线 $B$ 表示航天员由于地球自转而产生的向心加速度大小与 $r$ 的关系, 关于相对地面静止在不同高度的航天员, 地面附近重力加速度 $g$ 取 $10 \mathrm{~m} / \mathrm{s}^{2}$, 地球自转角速度 $\omega=7.3 \times 10^{-5} \mathrm{rad} / \mathrm{s}$, 地球半径 $R=6.4 \times 10^{3} \mathrm{~km}$ 。下列说法正确的有 ( ) [图1] 图甲 [图2] 图乙 A: 随着 $r$ 增大, 航天员受到电梯舱的弹力减小 B: 航天员在 $r=R$ 处的线速度等于第一宇宙速度 C: 图中 $r_{0}$ 为地球同步卫星的轨道半径 D: 电梯舱停在距地面高度为 $6.6 R$ 的站点时, 舱内质量 $60 \mathrm{~kg}$ 的航天员对水平地板的压力为零 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]

SC

Astronomy

ZH

multi-modal

Astronomy_604

木星有众多卫星, 木卫三是其中最大的一颗, 其直径大于行星中的水星。假设木卫三绕木星做匀速圆周运动的轨道半径为 $r_{1}$, 运行周期为 $T_{1}$, 木星半径为 $R$ 。已知行星与卫星间引力势能的表达式为 $E_{P}=-\frac{G M_{0} m_{0}}{r}, r$ 为行星与卫星的中心距离, 则木星的第二宇宙速度为 ( ) A: $\frac{\pi r_{1}}{T_{1}} \sqrt{\frac{2 r_{1}}{R}}$ B: $\frac{\pi r_{1}}{T_{1}} \sqrt{\frac{r_{1}}{R}}$ C: $\frac{2 \pi r_{1}}{R T_{1}} \sqrt{\frac{2 r_{1}}{R}}$ D: $\frac{2 \pi r_{1}}{T_{1}} \sqrt{\frac{2 r_{1}}{R}}$

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个单选题（只有一个正确答案）。 问题: 木星有众多卫星, 木卫三是其中最大的一颗, 其直径大于行星中的水星。假设木卫三绕木星做匀速圆周运动的轨道半径为 $r_{1}$, 运行周期为 $T_{1}$, 木星半径为 $R$ 。已知行星与卫星间引力势能的表达式为 $E_{P}=-\frac{G M_{0} m_{0}}{r}, r$ 为行星与卫星的中心距离, 则木星的第二宇宙速度为 ( ) A: $\frac{\pi r_{1}}{T_{1}} \sqrt{\frac{2 r_{1}}{R}}$ B: $\frac{\pi r_{1}}{T_{1}} \sqrt{\frac{r_{1}}{R}}$ C: $\frac{2 \pi r_{1}}{R T_{1}} \sqrt{\frac{2 r_{1}}{R}}$ D: $\frac{2 \pi r_{1}}{T_{1}} \sqrt{\frac{2 r_{1}}{R}}$ 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]

SC

Astronomy

ZH

text-only

Astronomy_268

据央视报道, 5 月 31 日搭载两名美国宇航员的美国太空探索技术公司 SpaceX 龙飞船发射成功, 乘“猎鹰 9 号”火箭飞往国际空间站。并在 31 日成功与国际空间站对接,把两位航天员送入国际空间站。龙飞船发射过程可简化为: 先让飞船进入一个近地的圆轨道 I, 然后在 $P$ 点变轨, 进入椭圆形转移轨道 II (该椭圆轨道的近地点与近地圆轨道 I 相切于 $P$ 点, 远地点与最终运行轨道 III 相切于 $Q$ 点), 到达远地点 $Q$ 时再次变轨, 进入圆形轨道 III。关于飞船的运动, 下列说法正确的是（ ） [图1] A: 飞船沿轨道 $\mathrm{I}$, 经过 $P$ 点时需要的向心力小于沿轨道 II 经过 $P$ 点时需要的向心力 B: 飞船沿转移轨道 II 运行到远地点 $Q$ 点时的速率 $v_{3}$ 等于在轨道 III 上运行的速率 $v_{4}$ C: 飞船沿轨道 I 经过 $P$ 点时受到的万有引力小于沿轨道 II 经过 $P$ 点时受到的万有引力 D: 飞船在轨道 I 上运行的速率 $v_{1}$ 等于在轨道 II 运行经过 $P$ 点时的速率 $v_{2}$

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个单选题（只有一个正确答案）。 问题: 据央视报道, 5 月 31 日搭载两名美国宇航员的美国太空探索技术公司 SpaceX 龙飞船发射成功, 乘“猎鹰 9 号”火箭飞往国际空间站。并在 31 日成功与国际空间站对接,把两位航天员送入国际空间站。龙飞船发射过程可简化为: 先让飞船进入一个近地的圆轨道 I, 然后在 $P$ 点变轨, 进入椭圆形转移轨道 II (该椭圆轨道的近地点与近地圆轨道 I 相切于 $P$ 点, 远地点与最终运行轨道 III 相切于 $Q$ 点), 到达远地点 $Q$ 时再次变轨, 进入圆形轨道 III。关于飞船的运动, 下列说法正确的是（ ） [图1] A: 飞船沿轨道 $\mathrm{I}$, 经过 $P$ 点时需要的向心力小于沿轨道 II 经过 $P$ 点时需要的向心力 B: 飞船沿转移轨道 II 运行到远地点 $Q$ 点时的速率 $v_{3}$ 等于在轨道 III 上运行的速率 $v_{4}$ C: 飞船沿轨道 I 经过 $P$ 点时受到的万有引力小于沿轨道 II 经过 $P$ 点时受到的万有引力 D: 飞船在轨道 I 上运行的速率 $v_{1}$ 等于在轨道 II 运行经过 $P$ 点时的速率 $v_{2}$ 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]

SC

Astronomy

ZH

multi-modal

Astronomy_202

北斗卫星导航系统组网成功后会有 5 颗静止轨道卫星。已知地球赤道上的重力加速 度大小为 $g$, 将地球视为半径为 $R_{0}$ 的球体, 地球的自转周期为 $T_{0}$, 则地球静止轨道卫星（同步卫星）的运行半径为（） A: $\sqrt{R_{0}{ }^{3}+\frac{g R_{0}{ }^{2} T_{0}{ }^{2}}{4 \pi^{2}}}$ B: $\sqrt[3]{\frac{g R_{0}{ }^{2} T_{0}{ }^{2}-R_{0}{ }^{3}}{4 \pi^{2}}}$ C: $\sqrt[3]{R_{0}{ }^{3}+\frac{g R_{0}{ }^{2} T_{0}{ }^{2}}{4 \pi^{2}}}$ D: $\sqrt{\frac{g R_{0} T_{0}}{2 \pi}+R_{0}^{2}}$

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个单选题（只有一个正确答案）。 问题: 北斗卫星导航系统组网成功后会有 5 颗静止轨道卫星。已知地球赤道上的重力加速 度大小为 $g$, 将地球视为半径为 $R_{0}$ 的球体, 地球的自转周期为 $T_{0}$, 则地球静止轨道卫星（同步卫星）的运行半径为（） A: $\sqrt{R_{0}{ }^{3}+\frac{g R_{0}{ }^{2} T_{0}{ }^{2}}{4 \pi^{2}}}$ B: $\sqrt[3]{\frac{g R_{0}{ }^{2} T_{0}{ }^{2}-R_{0}{ }^{3}}{4 \pi^{2}}}$ C: $\sqrt[3]{R_{0}{ }^{3}+\frac{g R_{0}{ }^{2} T_{0}{ }^{2}}{4 \pi^{2}}}$ D: $\sqrt{\frac{g R_{0} T_{0}}{2 \pi}+R_{0}^{2}}$ 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]

SC

Astronomy

ZH

text-only

Astronomy_1204

In order to move a spacecraft between orbits we must apply a thrust using rockets, which changes the velocity of the spacecraft by $\Delta v$. In this question we will ignore changes in the mass of the spacecraft due to the burning of fuel. For an object of mass $m$ in a circular orbit of radius $r$ around an object with mass $M$ (where $m \ll M$ ) the orbital velocity, $v_{\text {orb }}$, is given by the formula $v_{\text {orb }}=\sqrt{\frac{G M}As part of their plan to rule the galaxy the First Order has created the Starkiller Base. Built within an ice planet and with a superweapon capable of destroying entire star systems, it is charged using the power of stars. The Starkiller Base has moved into the solar system and seeks to use the Sun to power its weapon to destroy the Earth. [figure1] Figure 3: The Starkiller Base charging its superweapon by draining energy from the local star. Credit: Star Wars: The Force Awakens, Lucasfilm. For this question you will need that the gravitational binding energy, $U$, of a uniform density spherical object with mass $M$ and radius $R$ is given by $$ U=\frac{3 G M^{2}}{5 R} $$ and that the mass-luminosity relation of low-mass main sequence stars is given by $L \propto M^{4}$.{r}}$.d. Taking the Starkiller Base's ice planet to have a diameter of $660 \mathrm{~km}$, show that the Sun can be safely contained, even if it was fully drained.

You are participating in an international Astronomy competition and need to solve the following question. The answer to this question is a numerical value. Here is some context information for this question, which might assist you in solving it: In order to move a spacecraft between orbits we must apply a thrust using rockets, which changes the velocity of the spacecraft by $\Delta v$. In this question we will ignore changes in the mass of the spacecraft due to the burning of fuel. For an object of mass $m$ in a circular orbit of radius $r$ around an object with mass $M$ (where $m \ll M$ ) the orbital velocity, $v_{\text {orb }}$, is given by the formula $v_{\text {orb }}=\sqrt{\frac{G M}As part of their plan to rule the galaxy the First Order has created the Starkiller Base. Built within an ice planet and with a superweapon capable of destroying entire star systems, it is charged using the power of stars. The Starkiller Base has moved into the solar system and seeks to use the Sun to power its weapon to destroy the Earth. [figure1] Figure 3: The Starkiller Base charging its superweapon by draining energy from the local star. Credit: Star Wars: The Force Awakens, Lucasfilm. For this question you will need that the gravitational binding energy, $U$, of a uniform density spherical object with mass $M$ and radius $R$ is given by $$ U=\frac{3 G M^{2}}{5 R} $$ and that the mass-luminosity relation of low-mass main sequence stars is given by $L \propto M^{4}$.{r}}$. problem: d. Taking the Starkiller Base's ice planet to have a diameter of $660 \mathrm{~km}$, show that the Sun can be safely contained, even if it was fully drained. All mathematical formulas and symbols you output should be represented with LaTeX! You can solve it step by step. Remember, your answer should be calculated in the unit of m, but when concluding your final answer, do not include the unit. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER is the numerical value without any units.

NV

Astronomy

EN

multi-modal

Astronomy_597

靠近地面运行的近地卫星的加速度大小为 $a_{1}$, 地球同步轨道上的卫星的加速度大小为 $a_{2}$, 赤道上随地球一同运转 (相对地面静止) 的物体的加速度大小为 $a_{3}$, 则 ( ) A: $a_{1}=a_{3}>a_{2}$ B: $a_{1}>a_{2}>a_{3}$ C: $a_{1}>a_{3}>a_{2}$ D: $a_{3}>a_{2}>a_{1}$

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个单选题（只有一个正确答案）。 问题: 靠近地面运行的近地卫星的加速度大小为 $a_{1}$, 地球同步轨道上的卫星的加速度大小为 $a_{2}$, 赤道上随地球一同运转 (相对地面静止) 的物体的加速度大小为 $a_{3}$, 则 ( ) A: $a_{1}=a_{3}>a_{2}$ B: $a_{1}>a_{2}>a_{3}$ C: $a_{1}>a_{3}>a_{2}$ D: $a_{3}>a_{2}>a_{1}$ 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]

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Astronomy

ZH

text-only

Astronomy_354

2020 年 6 月 23 日, 我国北斗卫星导航系统最后一颗组网卫星成功发射, 这是一颗同步卫星。发射此类卫星时, 通常先将卫星发送到一个椭圆轨道上, 其近地点 $M$ 距地面高 $h_{1}$, 远地点 $N$ 距地面高 $h_{2}$, 进入该轨道正常运行时, 其周期为 $T_{1}$, 机械能为 $E_{1}$, 通过 $M 、 N$ 两点时的速率分别是 $v_{1} 、 v_{2}$, 加速度大小分别是 $a_{1} 、 a_{2}$ 。当某次飞船通过 $N$点时, 地面指挥部发出指令, 点燃飞船上的发动机, 使飞船在短时间内加速后进入离地面高 $h_{2}$ 的圆形轨道, 开始绕地球做匀速圆周运动, 这时飞船的运动周期为 $T_{2}$, 速率为 $v_{3}$,加速度大小为 $a_{3}$, 机械能为 $E_{2}$ 。下列结论正确的是 ( ) (1) $v_{1}>v_{3}$ (2) $E_{2}>E_{1}$ (3) $a_{2}>a_{3}$ (4) $T_{1}>T_{2}$ [图1] A: (1)(2)(3) B: (2)(3) C: (1)(2) D: (3)(4)

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个单选题（只有一个正确答案）。 问题: 2020 年 6 月 23 日, 我国北斗卫星导航系统最后一颗组网卫星成功发射, 这是一颗同步卫星。发射此类卫星时, 通常先将卫星发送到一个椭圆轨道上, 其近地点 $M$ 距地面高 $h_{1}$, 远地点 $N$ 距地面高 $h_{2}$, 进入该轨道正常运行时, 其周期为 $T_{1}$, 机械能为 $E_{1}$, 通过 $M 、 N$ 两点时的速率分别是 $v_{1} 、 v_{2}$, 加速度大小分别是 $a_{1} 、 a_{2}$ 。当某次飞船通过 $N$点时, 地面指挥部发出指令, 点燃飞船上的发动机, 使飞船在短时间内加速后进入离地面高 $h_{2}$ 的圆形轨道, 开始绕地球做匀速圆周运动, 这时飞船的运动周期为 $T_{2}$, 速率为 $v_{3}$,加速度大小为 $a_{3}$, 机械能为 $E_{2}$ 。下列结论正确的是 ( ) (1) $v_{1}>v_{3}$ (2) $E_{2}>E_{1}$ (3) $a_{2}>a_{3}$ (4) $T_{1}>T_{2}$ [图1] A: (1)(2)(3) B: (2)(3) C: (1)(2) D: (3)(4) 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]

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Astronomy

ZH

multi-modal

Astronomy_631

中国新闻网宣布: 在摩洛哥坠落的陨石被证实来自火星。某同学想根据平时收集的部分火星资料 (如图所示) 计算出火星的密度, 再与这颗陨石的密度进行比较。下列计 | 火星-Mars | | :--- | | 火星的小档案 | | 直径 $d=6794 \mathrm{~km}$ | | 质量 $M=6.4219 \times 10^{23} \mathrm{~kg}$ | | 表面重力加速度 $g_{0}=3.7 \mathrm{~m} / \mathrm{s}^{2}$ | | 近地卫星周期 $T=3.4 \mathrm{~h}$ | A: $\rho=\frac{3 g_{0}}{2 \pi G d}$ B: $\rho=\frac{g_{0} T^{2}}{3 \pi d}$ C: $\rho=\frac{3 \pi}{G T^{2}}$ D: $\rho=\frac{6 M}{\pi d^{3}}$

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个多选题（有多个正确答案）。 问题: 中国新闻网宣布: 在摩洛哥坠落的陨石被证实来自火星。某同学想根据平时收集的部分火星资料 (如图所示) 计算出火星的密度, 再与这颗陨石的密度进行比较。下列计 | 火星-Mars | | :--- | | 火星的小档案 | | 直径 $d=6794 \mathrm{~km}$ | | 质量 $M=6.4219 \times 10^{23} \mathrm{~kg}$ | | 表面重力加速度 $g_{0}=3.7 \mathrm{~m} / \mathrm{s}^{2}$ | | 近地卫星周期 $T=3.4 \mathrm{~h}$ | A: $\rho=\frac{3 g_{0}}{2 \pi G d}$ B: $\rho=\frac{g_{0} T^{2}}{3 \pi d}$ C: $\rho=\frac{3 \pi}{G T^{2}}$ D: $\rho=\frac{6 M}{\pi d^{3}}$ 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]

MC

Astronomy

ZH

text-only

Astronomy_672

宇宙空间有两颗相距较远、中心距离为 $d$ 的星球 $\mathrm{A}$ 和星球 $\mathrm{B}$ 。在星球 $\mathrm{A}$ 上将一轻弹簧坚直固定在水平桌面上, 把物体 P 轻放在弹簧上端, 如图 (a) 所示, P 由静止向下运动, 其加速度 $a$ 与弹簧的压缩量 $x$ 间的关系如图 (b) 中实线所示。在星球 $\mathrm{B}$ 上用完全相同的弹簧和物体 $\mathrm{P}$ 完成同样的过程, 其 $a-x$ 关系如图 (b) 中虚线所示（图中 $a_{0}$未知）。已知两星球密度相等。星球 $\mathrm{A}$ 的质量为 $m_{0}$, 引力常量为 $G$ 。假设两星球均为质量均匀分布的球体。 求星球 B 的质量 $M$; [图1] 图(a) [图2] 图(b)

你正在参加一个国际天文竞赛，并需要解决以下问题。 这个问题的答案是一个表达式。 问题: 宇宙空间有两颗相距较远、中心距离为 $d$ 的星球 $\mathrm{A}$ 和星球 $\mathrm{B}$ 。在星球 $\mathrm{A}$ 上将一轻弹簧坚直固定在水平桌面上, 把物体 P 轻放在弹簧上端, 如图 (a) 所示, P 由静止向下运动, 其加速度 $a$ 与弹簧的压缩量 $x$ 间的关系如图 (b) 中实线所示。在星球 $\mathrm{B}$ 上用完全相同的弹簧和物体 $\mathrm{P}$ 完成同样的过程, 其 $a-x$ 关系如图 (b) 中虚线所示（图中 $a_{0}$未知）。已知两星球密度相等。星球 $\mathrm{A}$ 的质量为 $m_{0}$, 引力常量为 $G$ 。假设两星球均为质量均匀分布的球体。 求星球 B 的质量 $M$; [图1] 图(a) [图2] 图(b) 你输出的所有数学公式和符号应该使用LaTeX表示！ 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER是一个不含等号的表达式，例如ANSWER=\frac{1}{2} g t^2

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Astronomy

ZH

multi-modal

Astronomy_774

What does the astronomical term ecliptic describe? A: The path of the Sun in the sky throughout a year. B: The axial tilt of the Earth throughout a year. C: The movement of the stars due to Earth's rotation. D: The central line through the axis of rotation.

You are participating in an international Astronomy competition and need to solve the following question. This is a multiple choice question (only one correct answer). problem: What does the astronomical term ecliptic describe? A: The path of the Sun in the sky throughout a year. B: The axial tilt of the Earth throughout a year. C: The movement of the stars due to Earth's rotation. D: The central line through the axis of rotation. You can solve it step by step. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER should be one of the options: [A, B, C, D].

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Astronomy

EN

text-only

Astronomy_909

NASA's Mars 2020 mission involves a large rover called Perseverance, and an audacious experiment to see if motor driven flight is possible on the red planet. The helicopter drone is called Ingenuity, shown in Figure 2, and is the first attempt to fly on another world. [figure1] Figure 2: The Ingenuity drone with its helicopter rotors ready to take off, with the solar panel used to charge its battery on the very top. In the background you can see a part of the Perseverance rover. Credit: NASA/JPL-Caltech. Although gravity is weaker on Mars, the Martian atmosphere is only about $1 \%$ the density of that on Earth, which makes it very hard to get lift. To overcome this, the probe has two counterrotating blades (directly above each other) with a tip-to-tip length of $1.2 \mathrm{~m}$ that will spin at 2400 rpm (about 5 times faster than helicopters on Earth), and is incredibly light with a mass of $1.8 \mathrm{~kg}$. A consequence of the weight restriction is that the battery is small (and charged by a solar panel on top of the drone, above the blades). Since flying in such a thin atmosphere is a very high power activity the maximum flight duration is therefore short, and given the time needed to recharge, they will be limited to only one flight per day. If the active region of the solar panel has an area of $544 \mathrm{~cm}^{2}$ and Mars takes 687 days to orbit the Sun, determine the minimum amount of time (in minutes) the battery needs to charge for between flights. Assume the panel is $30 \%$ efficient and the Sun is directly overhead throughout.

You are participating in an international Astronomy competition and need to solve the following question. The answer to this question is a numerical value. problem: NASA's Mars 2020 mission involves a large rover called Perseverance, and an audacious experiment to see if motor driven flight is possible on the red planet. The helicopter drone is called Ingenuity, shown in Figure 2, and is the first attempt to fly on another world. [figure1] Figure 2: The Ingenuity drone with its helicopter rotors ready to take off, with the solar panel used to charge its battery on the very top. In the background you can see a part of the Perseverance rover. Credit: NASA/JPL-Caltech. Although gravity is weaker on Mars, the Martian atmosphere is only about $1 \%$ the density of that on Earth, which makes it very hard to get lift. To overcome this, the probe has two counterrotating blades (directly above each other) with a tip-to-tip length of $1.2 \mathrm{~m}$ that will spin at 2400 rpm (about 5 times faster than helicopters on Earth), and is incredibly light with a mass of $1.8 \mathrm{~kg}$. A consequence of the weight restriction is that the battery is small (and charged by a solar panel on top of the drone, above the blades). Since flying in such a thin atmosphere is a very high power activity the maximum flight duration is therefore short, and given the time needed to recharge, they will be limited to only one flight per day. If the active region of the solar panel has an area of $544 \mathrm{~cm}^{2}$ and Mars takes 687 days to orbit the Sun, determine the minimum amount of time (in minutes) the battery needs to charge for between flights. Assume the panel is $30 \%$ efficient and the Sun is directly overhead throughout. All mathematical formulas and symbols you output should be represented with LaTeX! You can solve it step by step. Remember, your answer should be calculated in the unit of mins, but when concluding your final answer, do not include the unit. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER is the numerical value without any units.

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Astronomy

EN

multi-modal

Astronomy_775

Neptune's diameter is similar to ... A: Venus B: Jupiter C: Saturn D: Uranus

You are participating in an international Astronomy competition and need to solve the following question. This is a multiple choice question (only one correct answer). problem: Neptune's diameter is similar to ... A: Venus B: Jupiter C: Saturn D: Uranus You can solve it step by step. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER should be one of the options: [A, B, C, D].

SC

Astronomy

EN

text-only

Astronomy_347

2020 年 7 月 23 日, 我国的“天问一号”火星探测器, 搭乘着长征五号遥四运载火箭,成功从地球飞向了火星。如图所示为“天问一号”发射过程的示意图, 从地球上发射之后经过地火转移轨道被火星捕获, 进入环火星圆轨道, 经变轨调整后, 进入着陆准备轨道。已知“天问一号” 火星探测器在轨道半径为 $r$ 的环火星圆轨道上运动时, 周期为 $T_{1}$, 在半长轴为 $a$ 的着陆准备轨道上运动时, 周期为 $T_{2}$, 则下列判断正确的是 ( ) 地火转移轨道地球 [图1] A: 火星的平均密度一定大于 $\frac{3 \pi}{G T_{1}^{2}}$ B: “天问一号”从环火星圆轨道进入着陆准备轨道时需减速 C: “天问一号”在环火星圆轨道和着陆准备轨道上运动时满足 $\frac{r^{2}}{T_{1}^{3}}=\frac{a^{2}}{T_{2}^{3}}$ D: “天问一号”在环火星圆轨道上的机械能大于其在着陆准备轨道上的机械能

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个多选题（有多个正确答案）。 问题: 2020 年 7 月 23 日, 我国的“天问一号”火星探测器, 搭乘着长征五号遥四运载火箭,成功从地球飞向了火星。如图所示为“天问一号”发射过程的示意图, 从地球上发射之后经过地火转移轨道被火星捕获, 进入环火星圆轨道, 经变轨调整后, 进入着陆准备轨道。已知“天问一号” 火星探测器在轨道半径为 $r$ 的环火星圆轨道上运动时, 周期为 $T_{1}$, 在半长轴为 $a$ 的着陆准备轨道上运动时, 周期为 $T_{2}$, 则下列判断正确的是 ( ) 地火转移轨道地球 [图1] A: 火星的平均密度一定大于 $\frac{3 \pi}{G T_{1}^{2}}$ B: “天问一号”从环火星圆轨道进入着陆准备轨道时需减速 C: “天问一号”在环火星圆轨道和着陆准备轨道上运动时满足 $\frac{r^{2}}{T_{1}^{3}}=\frac{a^{2}}{T_{2}^{3}}$ D: “天问一号”在环火星圆轨道上的机械能大于其在着陆准备轨道上的机械能 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]

MC

Astronomy

ZH

multi-modal

Astronomy_301

由三颗星体构成的系统, 忽略其他星体对它们的作用, 存在着一种运动形式: 三颗星体在相互之间的万有引力作用下, 分别位于等边三角形的三个顶点上, 绕某一共同的圆心在三角形所在的平面内做角速度相等的圆周运动, 如图所示。已知星体 $\mathrm{A}$ 的质量为 $4 m$, 星体 $\mathrm{B} 、 \mathrm{C}$ 的质量均为 $m$, 三角形边长为 $d$ 。求: 星体 $\mathrm{A}$ 所受的合力大小 $F_{\mathrm{A}}$; [图1]

你正在参加一个国际天文竞赛，并需要解决以下问题。 这个问题的答案是一个表达式。 问题: 由三颗星体构成的系统, 忽略其他星体对它们的作用, 存在着一种运动形式: 三颗星体在相互之间的万有引力作用下, 分别位于等边三角形的三个顶点上, 绕某一共同的圆心在三角形所在的平面内做角速度相等的圆周运动, 如图所示。已知星体 $\mathrm{A}$ 的质量为 $4 m$, 星体 $\mathrm{B} 、 \mathrm{C}$ 的质量均为 $m$, 三角形边长为 $d$ 。求: 星体 $\mathrm{A}$ 所受的合力大小 $F_{\mathrm{A}}$; [图1] 你输出的所有数学公式和符号应该使用LaTeX表示！ 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER是一个不含等号的表达式，例如ANSWER=\frac{1}{2} g t^2

EX

Astronomy

ZH

multi-modal

Astronomy_723

如图所示, 有一质量为 $M$ 、半径为 $R$ 、密度均匀的球体, 在距离球心 $O$ 为 $2 R$ 的 $P$点有一质量为 $m$ 的质点, 从 $M$ 中挖去一个半径为 $\frac{1}{2} R$ 的球体, 设大球剩余部分对 $m$ 的万有引力为 $F_{1}$ 。若把质点 $m$ 移放在空腔中心 $O^{\prime}$ 点, 设大球的剩余部分对该质点的万有引力为 $F_{2}$ 。已知质量分布均匀的球壳对壳内物体的引力为 0 , 万有引力常量为 $G, O$ 、 $O^{\prime} 、 P$ 三点共线。下列说法正确的是（） [图1] A: $F_{1}$ 的大小为 $\frac{7 G M m}{36 R^{2}}$ B: $F_{2}$ 的大小为 $\frac{G M m}{4 R^{2}}$ C: 若把质点 $m$ 移放在 $O$ 点右侧, 距 $O$ 点 $\frac{3}{4} R$ 处, 大球的剩余部分对该质点的万有引力与 $F_{2}$ 相同 D: 若把质点 $m$ 移放在 $O$ 点右侧, 距 $O$ 点 $\frac{3}{4} R$ 处, 大球的剩余部分对该质点的万有引力与 $F_{2}$ 不同

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个多选题（有多个正确答案）。 问题: 如图所示, 有一质量为 $M$ 、半径为 $R$ 、密度均匀的球体, 在距离球心 $O$ 为 $2 R$ 的 $P$点有一质量为 $m$ 的质点, 从 $M$ 中挖去一个半径为 $\frac{1}{2} R$ 的球体, 设大球剩余部分对 $m$ 的万有引力为 $F_{1}$ 。若把质点 $m$ 移放在空腔中心 $O^{\prime}$ 点, 设大球的剩余部分对该质点的万有引力为 $F_{2}$ 。已知质量分布均匀的球壳对壳内物体的引力为 0 , 万有引力常量为 $G, O$ 、 $O^{\prime} 、 P$ 三点共线。下列说法正确的是（） [图1] A: $F_{1}$ 的大小为 $\frac{7 G M m}{36 R^{2}}$ B: $F_{2}$ 的大小为 $\frac{G M m}{4 R^{2}}$ C: 若把质点 $m$ 移放在 $O$ 点右侧, 距 $O$ 点 $\frac{3}{4} R$ 处, 大球的剩余部分对该质点的万有引力与 $F_{2}$ 相同 D: 若把质点 $m$ 移放在 $O$ 点右侧, 距 $O$ 点 $\frac{3}{4} R$ 处, 大球的剩余部分对该质点的万有引力与 $F_{2}$ 不同 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]

MC

Astronomy

ZH

multi-modal

Astronomy_99

宇宙中有一星球, 其半径为 $R$, 自转周期为 $T$, 引力常量为 $G$, 该天体的质量为 $M$ 。若一宇航员来到位于赤道的一斜坡前, 将一小球自斜坡底端正上方的 $O$ 点以初速度 $v_{0}$ 水平抛出, 如图所示, 小球垂直击中了斜坡, 落点为 $P$ 点, 求 该星球赤道地面处的重力加速度 $g_{1}$; [图1]

你正在参加一个国际天文竞赛，并需要解决以下问题。 这个问题的答案是一个表达式。 问题: 宇宙中有一星球, 其半径为 $R$, 自转周期为 $T$, 引力常量为 $G$, 该天体的质量为 $M$ 。若一宇航员来到位于赤道的一斜坡前, 将一小球自斜坡底端正上方的 $O$ 点以初速度 $v_{0}$ 水平抛出, 如图所示, 小球垂直击中了斜坡, 落点为 $P$ 点, 求 该星球赤道地面处的重力加速度 $g_{1}$; [图1] 你输出的所有数学公式和符号应该使用LaTeX表示！ 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER是一个不含等号的表达式，例如ANSWER=\frac{1}{2} g t^2

EX

Astronomy

ZH

multi-modal

Astronomy_344

行星 $\mathrm{A}$ 和行星 $\mathrm{B}$ 是两个均匀球体, 行星 $\mathrm{A}$ 的卫星沿圆轨道运行的周期为 $T_{A}$, 行星 $B$ 的卫星沿圆轨道运行的周期为 $T_{B}$, 两卫星绕各自行星的近表面轨道运行, 已知 $T_{A}: T_{B}=1: 4$, 行星 $\mathrm{A} 、 \mathrm{~B}$ 的半径之比为 $R_{\mathrm{A}}: R_{\mathrm{B}}=1: 2$, 则 () A: 这两颗行星的质量之比 $m_{\mathrm{A}}: m_{\mathrm{B}}=2: 1$ B: 这两颗行星表面的重力加速度之比 $g_{A}: g_{B}=2: 1$ C: 这两颗行星的密度之比 $\rho_{\mathrm{A}}: \rho_{\mathrm{B}}=16: 1$ D: 这两颗行星的同步卫星周期之比 $T_{\mathrm{A}}: T_{\mathrm{B}}=1: \sqrt{2}$

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个多选题（有多个正确答案）。 问题: 行星 $\mathrm{A}$ 和行星 $\mathrm{B}$ 是两个均匀球体, 行星 $\mathrm{A}$ 的卫星沿圆轨道运行的周期为 $T_{A}$, 行星 $B$ 的卫星沿圆轨道运行的周期为 $T_{B}$, 两卫星绕各自行星的近表面轨道运行, 已知 $T_{A}: T_{B}=1: 4$, 行星 $\mathrm{A} 、 \mathrm{~B}$ 的半径之比为 $R_{\mathrm{A}}: R_{\mathrm{B}}=1: 2$, 则 () A: 这两颗行星的质量之比 $m_{\mathrm{A}}: m_{\mathrm{B}}=2: 1$ B: 这两颗行星表面的重力加速度之比 $g_{A}: g_{B}=2: 1$ C: 这两颗行星的密度之比 $\rho_{\mathrm{A}}: \rho_{\mathrm{B}}=16: 1$ D: 这两颗行星的同步卫星周期之比 $T_{\mathrm{A}}: T_{\mathrm{B}}=1: \sqrt{2}$ 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]

MC

Astronomy

ZH

text-only

Astronomy_858

David is walking down MIT's infinite corridor (latitude $42^{\circ} 21^{\prime} 33^{\prime \prime}$ ) when he suddenly sees the sun aligning with the window at the end of the corridor. Being the observational master he is, David immediately pulls out his compass and measures the Sun to be at an azimuth of $245.81^{\circ}$. Forgetting to bring his jacket, he is painfully reminded as he walks outside that it has been less than 6 months since the previous winter solstice. Which of the following choices is closest to the current date? Assume the corridor is parallel to the surface of the Earth. A: January 15 B: January 30 C: February 15 D: March 20 E: April 1

You are participating in an international Astronomy competition and need to solve the following question. This is a multiple choice question (only one correct answer). problem: David is walking down MIT's infinite corridor (latitude $42^{\circ} 21^{\prime} 33^{\prime \prime}$ ) when he suddenly sees the sun aligning with the window at the end of the corridor. Being the observational master he is, David immediately pulls out his compass and measures the Sun to be at an azimuth of $245.81^{\circ}$. Forgetting to bring his jacket, he is painfully reminded as he walks outside that it has been less than 6 months since the previous winter solstice. Which of the following choices is closest to the current date? Assume the corridor is parallel to the surface of the Earth. A: January 15 B: January 30 C: February 15 D: March 20 E: April 1 You can solve it step by step. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER should be one of the options: [A, B, C, D, E].

SC

Astronomy

EN

text-only

Astronomy_1181

On $21^{\text {st }}$ December 2020, Jupiter and Saturn formed a true spectacle in the southwestern sky just after sunset in the UK, separated by only $0.102^{\circ}$. This is close enough that when viewed through a telescope both planets and their moons could be seen in the same field of view (see Fig 4). When two planets occupy the same piece of sky it is known as a conjunction, and when it is Jupiter and Saturn it is known as a great conjunction (so named because they are the rarest of the naked-eye planet conjunctions). The reason it happens is because Jupiter's orbital velocity is higher than Saturn's and so as time goes on Jupiter catches up with and overtakes Saturn (at least as viewed from Earth), with the moment of overtaking corresponding to the conjunction. The process of the two getting closer and closer together has been seen in sky throughout the year (see Fig 5). [figure1] Figure 4: Left: The view of the planets the day before their closest approach, as captured by the 16 " telescope at the Institute of Astronomy, Cambridge. Credit: Robin Catchpole. Right: The view from Arizona on the day of the closest approach as viewed with the Lowell Discovery Telescope. Credit: Levine / Elbert / Bosh / Lowell Observatory. [figure2] Figure 5: Left: Demonstrating how Jupiter and Saturn have been getting closer and closer together over the last few months. Separations are given in degrees and arcminutes $\left(1 / 60^{\text {th }}\right.$ of a degree). Credit: Pete Lawrence. Right: The positions of the Earth, Jupiter and Saturn that were responsible for the 2020 great conjunction. The precise timing of the apparent alignment is clearly sensitive to where Earth is in its orbit. Credit: timeanddate.com. The time between conjunctions is known as the synodic period. Although this period will change slightly from conjunction to conjunction due to different lines of perspective as viewed from Earth (see Fig 5), we can work out the average time between great conjunctions by considering both planets travelling on circular coplanar orbits and ignoring the position of the Earth. For circular coplanar orbits the centre of Jupiter's disc would pass in front of the centre of Saturn's disc every conjunction, and hence have an angular separation of $\theta=0^{\circ}$ (it is measured from the centre of each disc). In practice, the planets follow elliptical orbits that are in planes inclined at different angles to each other. Fig 6 shows how this affects the real values for over 8000 years' worth of data, which along with different synodic periods between conjunctions makes it a difficult problem to solve precisely without a computer. However, after one synodic period Saturn has moved about $2 / 3$ of the way around its orbit, and so roughly every 3 synodic periods it is in a similar part of the sky. Consequently every third great conjunction follows a reasonably regular pattern which can be fit with a sinusoidal function. [figure3] Figure 6: Top: All great conjunctions from 1800 to 2300, calculated for the real celestial mechanics of the Solar System. We can see that each great conjunction belongs to one of three different series or tracks, with Track A indicated with orange circles, Track B with green squares, and Track C with blue triangles. Bottom: The same idea but extended over a much larger date range, up to $10000 \mathrm{AD}$. It is clear the distinct series form broadly sinusoidal patterns which can be used with the average synodic period to give rough predictions for the separations of great conjunctions. The opacity of points is related to each conjunction's angular separation from the Sun (where low opacity means close to the Sun, so it is harder for any observers to see). Credit: Nick Koukoufilippas, but inspired by the work of Steffen Thorsen and Graham Jones / Sky \& Telescope.c. By empirically fitting a sinusoidal function (which is assumed to be the same for each track, just with a fixed phase difference between them) and assuming all conjunctions are separated by the average synodic period, we can give rough estimations for the separations of any given great conjunction. Note: be careful as your calculations will be very sensitive to rounding errors. ii. Without having to read anything else off the graph, write down the equations for Tracks B and $C$, given the same restrictions on $\phi_{B}$ and $\phi_{C}$.

You are participating in an international Astronomy competition and need to solve the following question. The answer to this question is an expression. Here is some context information for this question, which might assist you in solving it: On $21^{\text {st }}$ December 2020, Jupiter and Saturn formed a true spectacle in the southwestern sky just after sunset in the UK, separated by only $0.102^{\circ}$. This is close enough that when viewed through a telescope both planets and their moons could be seen in the same field of view (see Fig 4). When two planets occupy the same piece of sky it is known as a conjunction, and when it is Jupiter and Saturn it is known as a great conjunction (so named because they are the rarest of the naked-eye planet conjunctions). The reason it happens is because Jupiter's orbital velocity is higher than Saturn's and so as time goes on Jupiter catches up with and overtakes Saturn (at least as viewed from Earth), with the moment of overtaking corresponding to the conjunction. The process of the two getting closer and closer together has been seen in sky throughout the year (see Fig 5). [figure1] Figure 4: Left: The view of the planets the day before their closest approach, as captured by the 16 " telescope at the Institute of Astronomy, Cambridge. Credit: Robin Catchpole. Right: The view from Arizona on the day of the closest approach as viewed with the Lowell Discovery Telescope. Credit: Levine / Elbert / Bosh / Lowell Observatory. [figure2] Figure 5: Left: Demonstrating how Jupiter and Saturn have been getting closer and closer together over the last few months. Separations are given in degrees and arcminutes $\left(1 / 60^{\text {th }}\right.$ of a degree). Credit: Pete Lawrence. Right: The positions of the Earth, Jupiter and Saturn that were responsible for the 2020 great conjunction. The precise timing of the apparent alignment is clearly sensitive to where Earth is in its orbit. Credit: timeanddate.com. The time between conjunctions is known as the synodic period. Although this period will change slightly from conjunction to conjunction due to different lines of perspective as viewed from Earth (see Fig 5), we can work out the average time between great conjunctions by considering both planets travelling on circular coplanar orbits and ignoring the position of the Earth. For circular coplanar orbits the centre of Jupiter's disc would pass in front of the centre of Saturn's disc every conjunction, and hence have an angular separation of $\theta=0^{\circ}$ (it is measured from the centre of each disc). In practice, the planets follow elliptical orbits that are in planes inclined at different angles to each other. Fig 6 shows how this affects the real values for over 8000 years' worth of data, which along with different synodic periods between conjunctions makes it a difficult problem to solve precisely without a computer. However, after one synodic period Saturn has moved about $2 / 3$ of the way around its orbit, and so roughly every 3 synodic periods it is in a similar part of the sky. Consequently every third great conjunction follows a reasonably regular pattern which can be fit with a sinusoidal function. [figure3] Figure 6: Top: All great conjunctions from 1800 to 2300, calculated for the real celestial mechanics of the Solar System. We can see that each great conjunction belongs to one of three different series or tracks, with Track A indicated with orange circles, Track B with green squares, and Track C with blue triangles. Bottom: The same idea but extended over a much larger date range, up to $10000 \mathrm{AD}$. It is clear the distinct series form broadly sinusoidal patterns which can be used with the average synodic period to give rough predictions for the separations of great conjunctions. The opacity of points is related to each conjunction's angular separation from the Sun (where low opacity means close to the Sun, so it is harder for any observers to see). Credit: Nick Koukoufilippas, but inspired by the work of Steffen Thorsen and Graham Jones / Sky \& Telescope. problem: c. By empirically fitting a sinusoidal function (which is assumed to be the same for each track, just with a fixed phase difference between them) and assuming all conjunctions are separated by the average synodic period, we can give rough estimations for the separations of any given great conjunction. Note: be careful as your calculations will be very sensitive to rounding errors. ii. Without having to read anything else off the graph, write down the equations for Tracks B and $C$, given the same restrictions on $\phi_{B}$ and $\phi_{C}$. All mathematical formulas and symbols you output should be represented with LaTeX! You can solve it step by step. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER is an expression without equals signs, e.g. ANSWER=\frac{1}{2} g t^2

EX

Astronomy

EN

multi-modal

Astronomy_988

There are approximately 450 geostationary satellites currently in orbit. They have an orbital period of 1 day and orbit in the plane of the equator, so are therefore always directly above the same spot. Assuming circular orbits and an equidistant arrangement, what is the distance between two geostationary satellites? A: $560 \mathrm{~km}$ B: $570 \mathrm{~km}$ C: $580 \mathrm{~km}$ D: $590 \mathrm{~km}$

You are participating in an international Astronomy competition and need to solve the following question. This is a multiple choice question (only one correct answer). problem: There are approximately 450 geostationary satellites currently in orbit. They have an orbital period of 1 day and orbit in the plane of the equator, so are therefore always directly above the same spot. Assuming circular orbits and an equidistant arrangement, what is the distance between two geostationary satellites? A: $560 \mathrm{~km}$ B: $570 \mathrm{~km}$ C: $580 \mathrm{~km}$ D: $590 \mathrm{~km}$ You can solve it step by step. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER should be one of the options: [A, B, C, D].

SC

Astronomy

EN

text-only

Astronomy_1119

In November 2020 the Aricebo Telescope at the National Astronomy and Ionosphere Centre (NAIC) in Puerto Rico was decommissioned due to safety concerns after extensive storm damage. First opened in November 1963, this brought an end to an illustrious contribution to radio astronomy where, with a dish diameter of $304.8 \mathrm{~m}$ (1000 ft), it was the largest radio telescope in the world until 2016. Its important discoveries range from detection of the first extrasolar planets around a pulsar to fast radio bursts, as well as a pivotal role in the search for extraterrestrial intelligence (SETI), however in this question we will explore its earliest major revelation: that Mercury was not tidally locked. [figure1] Figure 1: Left: The Aricebo telescope before it was damaged. Credit: NAIC. Right: When transmitting a pulse from a radio telescope, diffraction prevents the beam from staying perfectly parallel and so the width of the beam increases by $2 \theta$. Credit: OpenStax, College Physics. Mercury had already been studied with optical and infrared telescopes, however the advantage of a radio telescope was that you could send pulses and receive their reflections. This radar-ranging technique had already been used with Venus to measure the distance to it and hence provide the data necessary for a definitive measurement of an astronomical unit in metres. The radar echo from Mercury is much harder to detect due to the extra distance travelled and its smaller cross-sectional area (its radius is $2440 \mathrm{~km}$ ). In April 1965, Pettengill and Dyce sent a series of $500 \mu \mathrm{s}$ pulses at $430 \mathrm{MHz}$ with a transmitted power of $2.0 \mathrm{MW}$ towards Mercury whilst it was at its closest point in its orbit to Earth. In ideal circumstances the beam would stay parallel, however diffraction widens the beam as shown on the right in Fig 1. The signal-to-noise ratio of this echo was high enough that Doppler broadening of the received signal was reliably detected, allowing a determination of the rotation rate of Mercury. In August 1965 the same scientists sent $100 \mu$ s pulses and sampled the echo on short timescales as it returned. The strongest echo (received first) came from the point of the planet closest to the Earth (called the sub-radar point), with later echos coming from other parts of the surface in an annulus of increasing radius (see Fig 2). Photons from the approaching side would be blueshifted to a higher frequency, whilst those from the receding side would be redshifted to a lower frequency. Hence, by measuring the Doppler shift and the time delay, you can map the rotational velocity as a function of apparent longitude and so can calculate the apparent rotation rate (as well as the direction of rotation and co-ordinates of the pole). [figure2] Figure 2: Left: Snapshots of the reflections of a single $100 \mu$ sulse. The strength of the echo weakens as you move to later delays (as represented by the scale factor in the top right of each snapshot) and hence you need to use an annulus rather than detections from the horizon. The small arrows indicate the Doppler shifted frequency associated with intersection of the annulus with the apparent equator for each delay. The horizontal axis is in cycles per second (and so $1 \mathrm{c} / \mathrm{s}=1 \mathrm{~Hz}$ ). Credit: Dyce, Pettengill and Shapiro (1967). Top right: The key principles of the delay-Doppler technique, looking at a cross-section of the planet. At the very centre is the sub-radar point (the point on the planet's surface closest to the Earth). As you move away from the sub-radar point the light has to travel further before it can be reflected, and hence the echo from those regions arrives later. The brightest point of any given annulus is where it intersects the apparent equator (due to the largest reflecting area), and so in each of the snapshots that is why the extreme Doppler shifts are boosted relative to the middle. Credit: Shapiro (1967). Bottom right: The same as the snapshots, but this time summed over the first $500 \mu$ of reflections. Here the difference between the extreme left and right frequencies reliably detected is $\sim 5 \mathrm{~Hz}$, but when corrected for relative motion of the Earth and Mercury it becomes the value given in part c. Credit: Pettengill, Dyce and Campbell (1967). The Doppler shift with light is given as $$ \frac{\Delta f}{f}=\frac{v}{c} $$ where $\Delta f$ is the shift in frequency $f, v$ is the line-of-sight velocity of the emitting object and $c$ is the speed of light. Ever since the first maps of Mercury's surface by Schiaparelli in the late 1880s, many in the scientific community believed that Mercury would be tidally locked and so always present the same hemisphere to the Sun. The reason they expected the rotational period to be the same as its orbital period (i.e. a $1: 1$ ratio), rather like the Moon, is because it is so close to the Sun and the tidal torques causing this synchronicity are proportional to $r^{-6}$ where $r$ is the distance from the massive body. Given it is the closest planet to the Sun, it receives by far the largest torques, so the discovery it was in a different ratio was a complete surprise to many of the scientists at the time. [figure3] Figure 3: The orientation of Mercury's axis of minimum moment of inertia (the axis the tidal torque acts upon) displayed at six points in its orbit (equally spaced in time) if the ratio had been $1: 1$. Credit: Colombo and Shapiro (1966).b. Averaging over a series of pulses from August 1965, after correcting for the relative motion of the Earth and Mercury and the rotation rate of the Earth during the observations, the difference between the frequencies of photons from the extreme left and right parts of an annulus received $500 \mu$ s after the initial echo was $\Delta f_{\text {total }}=4.27 \mathrm{~Hz}$. i. Given that the pulse was Doppler shifted twice (once when reflected, once again when received back at Aricebo) show that the rotational period of Mercury is $\approx 60$ days. Assume that the axis of rotation is normal to the plane of observations. [ 1 day $=24$ hours]

You are participating in an international Astronomy competition and need to solve the following question. The answer to this question is a numerical value. Here is some context information for this question, which might assist you in solving it: In November 2020 the Aricebo Telescope at the National Astronomy and Ionosphere Centre (NAIC) in Puerto Rico was decommissioned due to safety concerns after extensive storm damage. First opened in November 1963, this brought an end to an illustrious contribution to radio astronomy where, with a dish diameter of $304.8 \mathrm{~m}$ (1000 ft), it was the largest radio telescope in the world until 2016. Its important discoveries range from detection of the first extrasolar planets around a pulsar to fast radio bursts, as well as a pivotal role in the search for extraterrestrial intelligence (SETI), however in this question we will explore its earliest major revelation: that Mercury was not tidally locked. [figure1] Figure 1: Left: The Aricebo telescope before it was damaged. Credit: NAIC. Right: When transmitting a pulse from a radio telescope, diffraction prevents the beam from staying perfectly parallel and so the width of the beam increases by $2 \theta$. Credit: OpenStax, College Physics. Mercury had already been studied with optical and infrared telescopes, however the advantage of a radio telescope was that you could send pulses and receive their reflections. This radar-ranging technique had already been used with Venus to measure the distance to it and hence provide the data necessary for a definitive measurement of an astronomical unit in metres. The radar echo from Mercury is much harder to detect due to the extra distance travelled and its smaller cross-sectional area (its radius is $2440 \mathrm{~km}$ ). In April 1965, Pettengill and Dyce sent a series of $500 \mu \mathrm{s}$ pulses at $430 \mathrm{MHz}$ with a transmitted power of $2.0 \mathrm{MW}$ towards Mercury whilst it was at its closest point in its orbit to Earth. In ideal circumstances the beam would stay parallel, however diffraction widens the beam as shown on the right in Fig 1. The signal-to-noise ratio of this echo was high enough that Doppler broadening of the received signal was reliably detected, allowing a determination of the rotation rate of Mercury. In August 1965 the same scientists sent $100 \mu$ s pulses and sampled the echo on short timescales as it returned. The strongest echo (received first) came from the point of the planet closest to the Earth (called the sub-radar point), with later echos coming from other parts of the surface in an annulus of increasing radius (see Fig 2). Photons from the approaching side would be blueshifted to a higher frequency, whilst those from the receding side would be redshifted to a lower frequency. Hence, by measuring the Doppler shift and the time delay, you can map the rotational velocity as a function of apparent longitude and so can calculate the apparent rotation rate (as well as the direction of rotation and co-ordinates of the pole). [figure2] Figure 2: Left: Snapshots of the reflections of a single $100 \mu$ sulse. The strength of the echo weakens as you move to later delays (as represented by the scale factor in the top right of each snapshot) and hence you need to use an annulus rather than detections from the horizon. The small arrows indicate the Doppler shifted frequency associated with intersection of the annulus with the apparent equator for each delay. The horizontal axis is in cycles per second (and so $1 \mathrm{c} / \mathrm{s}=1 \mathrm{~Hz}$ ). Credit: Dyce, Pettengill and Shapiro (1967). Top right: The key principles of the delay-Doppler technique, looking at a cross-section of the planet. At the very centre is the sub-radar point (the point on the planet's surface closest to the Earth). As you move away from the sub-radar point the light has to travel further before it can be reflected, and hence the echo from those regions arrives later. The brightest point of any given annulus is where it intersects the apparent equator (due to the largest reflecting area), and so in each of the snapshots that is why the extreme Doppler shifts are boosted relative to the middle. Credit: Shapiro (1967). Bottom right: The same as the snapshots, but this time summed over the first $500 \mu$ of reflections. Here the difference between the extreme left and right frequencies reliably detected is $\sim 5 \mathrm{~Hz}$, but when corrected for relative motion of the Earth and Mercury it becomes the value given in part c. Credit: Pettengill, Dyce and Campbell (1967). The Doppler shift with light is given as $$ \frac{\Delta f}{f}=\frac{v}{c} $$ where $\Delta f$ is the shift in frequency $f, v$ is the line-of-sight velocity of the emitting object and $c$ is the speed of light. Ever since the first maps of Mercury's surface by Schiaparelli in the late 1880s, many in the scientific community believed that Mercury would be tidally locked and so always present the same hemisphere to the Sun. The reason they expected the rotational period to be the same as its orbital period (i.e. a $1: 1$ ratio), rather like the Moon, is because it is so close to the Sun and the tidal torques causing this synchronicity are proportional to $r^{-6}$ where $r$ is the distance from the massive body. Given it is the closest planet to the Sun, it receives by far the largest torques, so the discovery it was in a different ratio was a complete surprise to many of the scientists at the time. [figure3] Figure 3: The orientation of Mercury's axis of minimum moment of inertia (the axis the tidal torque acts upon) displayed at six points in its orbit (equally spaced in time) if the ratio had been $1: 1$. Credit: Colombo and Shapiro (1966). problem: b. Averaging over a series of pulses from August 1965, after correcting for the relative motion of the Earth and Mercury and the rotation rate of the Earth during the observations, the difference between the frequencies of photons from the extreme left and right parts of an annulus received $500 \mu$ s after the initial echo was $\Delta f_{\text {total }}=4.27 \mathrm{~Hz}$. i. Given that the pulse was Doppler shifted twice (once when reflected, once again when received back at Aricebo) show that the rotational period of Mercury is $\approx 60$ days. Assume that the axis of rotation is normal to the plane of observations. [ 1 day $=24$ hours] All mathematical formulas and symbols you output should be represented with LaTeX! You can solve it step by step. Remember, your answer should be calculated in the unit of days, but when concluding your final answer, do not include the unit. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER is the numerical value without any units.

NV

Astronomy

EN

multi-modal

Astronomy_521

为研究太阳系内行星的运动, 需要知道太阳的质量, 已知地球半径为 $R$, 地球质量为 $m$ ，太阳与地球中心间距为 $r$, 地球表面的重力加速度为 $g$, 地球绕太阳公转的周期为 $T$. 则太阳的质量为 $(\quad)$ A: $\frac{4 \pi^{2} r^{3}}{T^{2} R^{2} g}$ B: $\frac{T^{2} R^{2} g}{4 \pi^{2} m r^{3}}$ C: $\frac{4 \pi^{2} m g r^{2}}{r^{3} T^{2}}$ D: $\frac{4 \pi^{2} m r^{3}}{T^{2} R^{2} g}$

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个单选题（只有一个正确答案）。 问题: 为研究太阳系内行星的运动, 需要知道太阳的质量, 已知地球半径为 $R$, 地球质量为 $m$ ，太阳与地球中心间距为 $r$, 地球表面的重力加速度为 $g$, 地球绕太阳公转的周期为 $T$. 则太阳的质量为 $(\quad)$ A: $\frac{4 \pi^{2} r^{3}}{T^{2} R^{2} g}$ B: $\frac{T^{2} R^{2} g}{4 \pi^{2} m r^{3}}$ C: $\frac{4 \pi^{2} m g r^{2}}{r^{3} T^{2}}$ D: $\frac{4 \pi^{2} m r^{3}}{T^{2} R^{2} g}$ 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]

SC

Astronomy

ZH

text-only

Astronomy_869

Two planets A and B orbit a star with coplanar orbital paths that don't intersect. The major axes of the orbits are perfectly aligned, but the major axis of A is larger than that of B. A and $\mathrm{B}$ are observed to have eccentricities 0.5 and 0.75 , respectively. What is the minimal possible ratio of semi-major axes of $\mathrm{A}$ to $\mathrm{B}$ ? A: 1 B: $\frac{7}{6}$ C: $\frac{8}{3}$ D: $\frac{7}{2}$ E: 6

You are participating in an international Astronomy competition and need to solve the following question. This is a multiple choice question (only one correct answer). problem: Two planets A and B orbit a star with coplanar orbital paths that don't intersect. The major axes of the orbits are perfectly aligned, but the major axis of A is larger than that of B. A and $\mathrm{B}$ are observed to have eccentricities 0.5 and 0.75 , respectively. What is the minimal possible ratio of semi-major axes of $\mathrm{A}$ to $\mathrm{B}$ ? A: 1 B: $\frac{7}{6}$ C: $\frac{8}{3}$ D: $\frac{7}{2}$ E: 6 You can solve it step by step. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER should be one of the options: [A, B, C, D, E].

SC

Astronomy

EN

text-only

Astronomy_623

宇宙空间有两颗相距较远、中心距离为 $d$ 的星球 $\mathrm{A}$ 和星球 $\mathrm{B}$ 。在星球 $\mathrm{A}$ 上将一轻弹簧坚直固定在水平桌面上, 把物体 $\mathrm{P}$ 轻放在弹簧上端, 如图 (a) 所示, $\mathrm{P}$ 由静止向下运动, 其加速度 $a$ 与弹簧的压缩量 $x$ 间的关系如图 (b) 中实线所示。在星球 $\mathrm{B}$ 上用完全相同的弹簧和物体 P 完成同样的过程, 其 $a-x$ 关系如图 (b) 中虚线所示。已知两星球密度相等。星球 $\mathrm{A}$ 的质量为 $m_{0}$, 引力常量为 $G$ 。假设两星球均为质量均匀分布的球体。 若将星球 $\mathrm{A}$ 看成是以星球 $\mathrm{B}$ 为中心天体的一颗卫星, 求星球 $\mathrm{A}$ 的运行周期 $T_{1}$; [图1] 图(a) [图2] 图(b)

你正在参加一个国际天文竞赛，并需要解决以下问题。 这个问题的答案是一个表达式。 问题: 宇宙空间有两颗相距较远、中心距离为 $d$ 的星球 $\mathrm{A}$ 和星球 $\mathrm{B}$ 。在星球 $\mathrm{A}$ 上将一轻弹簧坚直固定在水平桌面上, 把物体 $\mathrm{P}$ 轻放在弹簧上端, 如图 (a) 所示, $\mathrm{P}$ 由静止向下运动, 其加速度 $a$ 与弹簧的压缩量 $x$ 间的关系如图 (b) 中实线所示。在星球 $\mathrm{B}$ 上用完全相同的弹簧和物体 P 完成同样的过程, 其 $a-x$ 关系如图 (b) 中虚线所示。已知两星球密度相等。星球 $\mathrm{A}$ 的质量为 $m_{0}$, 引力常量为 $G$ 。假设两星球均为质量均匀分布的球体。 若将星球 $\mathrm{A}$ 看成是以星球 $\mathrm{B}$ 为中心天体的一颗卫星, 求星球 $\mathrm{A}$ 的运行周期 $T_{1}$; [图1] 图(a) [图2] 图(b) 你输出的所有数学公式和符号应该使用LaTeX表示！ 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER是一个不含等号的表达式，例如ANSWER=\frac{1}{2} g t^2

EX

Astronomy

ZH

multi-modal

Astronomy_130

如图所示, 宇宙空间有一种由三颗星体 $A 、 B 、 C$ 组成的三星体系, 它们分别位于等边三角形 $A B C$ 的三个顶点上, 绕一个固定且共同的圆心 $O$ 做匀速圆周运动, 轨道如 图中实线所示, 其轨道半径 $r_{\mathrm{A}}<r_{\mathrm{B}}<r_{\mathrm{C}}$ 。忽略其他星体对它们的作用, 关于这三颗星体,下列说法正确的是（ ） [图1] A: 角速度大小关系是 $\omega_{\mathrm{A}}<\omega_{\mathrm{B}}<\omega_{\mathrm{C}}$ B: 线速度大小关系是 $v_{\mathrm{A}}<v_{\mathrm{B}}<v_{\mathrm{C}}$ C: 加速度大小关系是 $a_{\mathrm{A}}<a_{\mathrm{B}}<a_{\mathrm{C}}$ D: 质量大小关系是 $m_{\mathrm{A}}<m_{\mathrm{B}}<m_{\mathrm{C}}$

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个多选题（有多个正确答案）。 问题: 如图所示, 宇宙空间有一种由三颗星体 $A 、 B 、 C$ 组成的三星体系, 它们分别位于等边三角形 $A B C$ 的三个顶点上, 绕一个固定且共同的圆心 $O$ 做匀速圆周运动, 轨道如 图中实线所示, 其轨道半径 $r_{\mathrm{A}}<r_{\mathrm{B}}<r_{\mathrm{C}}$ 。忽略其他星体对它们的作用, 关于这三颗星体,下列说法正确的是（ ） [图1] A: 角速度大小关系是 $\omega_{\mathrm{A}}<\omega_{\mathrm{B}}<\omega_{\mathrm{C}}$ B: 线速度大小关系是 $v_{\mathrm{A}}<v_{\mathrm{B}}<v_{\mathrm{C}}$ C: 加速度大小关系是 $a_{\mathrm{A}}<a_{\mathrm{B}}<a_{\mathrm{C}}$ D: 质量大小关系是 $m_{\mathrm{A}}<m_{\mathrm{B}}<m_{\mathrm{C}}$ 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]

MC

Astronomy

ZH

multi-modal

Astronomy_376

一宇航员到达半径为 $R$ 、密度均匀的某星球表面, 做如下实验: 用不可伸长的轻绳拴一质量为 $m$ 的小球, 上端固定在 $O$ 点, 如图甲所示, 在最低点给小球某一初速度,使其绕 $O$ 点的坚直面内做圆周运动, 测得绳的拉力 $F$ 大小随时间 $t$ 的变化规律如图乙所示. $F_{1}=7 F_{2}$, 设 $R 、 m$ 、引力常量 $G$ 以及 $F_{1}$ 为已知量, 忽略各种阻力. 以下说法正确的是 [图1] A: 该星球表面的重力加速度为 $\frac{\sqrt{7} F_{1}}{7 m}$ B: 卫星绕该星球的第一宇宙速度为 $\sqrt{\frac{G m}{R}}$ C: 星球的密度为 $\frac{3 F_{1}}{28 \pi G m R}$ D: 小球过最高点的最小速度为 0

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个单选题（只有一个正确答案）。 问题: 一宇航员到达半径为 $R$ 、密度均匀的某星球表面, 做如下实验: 用不可伸长的轻绳拴一质量为 $m$ 的小球, 上端固定在 $O$ 点, 如图甲所示, 在最低点给小球某一初速度,使其绕 $O$ 点的坚直面内做圆周运动, 测得绳的拉力 $F$ 大小随时间 $t$ 的变化规律如图乙所示. $F_{1}=7 F_{2}$, 设 $R 、 m$ 、引力常量 $G$ 以及 $F_{1}$ 为已知量, 忽略各种阻力. 以下说法正确的是 [图1] A: 该星球表面的重力加速度为 $\frac{\sqrt{7} F_{1}}{7 m}$ B: 卫星绕该星球的第一宇宙速度为 $\sqrt{\frac{G m}{R}}$ C: 星球的密度为 $\frac{3 F_{1}}{28 \pi G m R}$ D: 小球过最高点的最小速度为 0 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]

SC

Astronomy

ZH

multi-modal

Astronomy_390

我国“嫦娥一号”探月卫星发射后，先在“ 24 小时轨道”上绕地球运行（即绕地球一圈需要 24 小时)然后, 经过两次变轨依次到达“48 小时轨道”和“72 小时轨道”, 最后奔向月球。如果按圆形轨道计算, 并忽略卫星质量的变化, 则在每次变轨完成后与变轨前相比 A: 卫星速度增大，引力势能和动能之和减小 B: 卫星速度减小, 引力势能和动能之和增大 C: 卫星速度增大, 引力势能和动能之和增大 D: 卫星速度减小, 引力势能和动能之和减小

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个单选题（只有一个正确答案）。 问题: 我国“嫦娥一号”探月卫星发射后，先在“ 24 小时轨道”上绕地球运行（即绕地球一圈需要 24 小时)然后, 经过两次变轨依次到达“48 小时轨道”和“72 小时轨道”, 最后奔向月球。如果按圆形轨道计算, 并忽略卫星质量的变化, 则在每次变轨完成后与变轨前相比 A: 卫星速度增大，引力势能和动能之和减小 B: 卫星速度减小, 引力势能和动能之和增大 C: 卫星速度增大, 引力势能和动能之和增大 D: 卫星速度减小, 引力势能和动能之和减小 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]

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Astronomy

ZH

text-only

Astronomy_792

Consider an eclipsing binary star system observed (in some fixed band) to have a combined apparent magnitude of 5.67. During the secondary transit, the second star is totally eclipsed by the first star, and the apparent magnitude dims to 6.28. What percent of the combined flux is produced by the second star? A: $10.8 \%$ B: $43.0 \%$ C: $57.0 \%$ D: $89.2 \%$

You are participating in an international Astronomy competition and need to solve the following question. This is a multiple choice question (only one correct answer). problem: Consider an eclipsing binary star system observed (in some fixed band) to have a combined apparent magnitude of 5.67. During the secondary transit, the second star is totally eclipsed by the first star, and the apparent magnitude dims to 6.28. What percent of the combined flux is produced by the second star? A: $10.8 \%$ B: $43.0 \%$ C: $57.0 \%$ D: $89.2 \%$ You can solve it step by step. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER should be one of the options: [A, B, C, D].

SC

Astronomy

EN

text-only

Astronomy_873

An exoplanet discovered by the radial velocity method is found to have an orbital period of 2.45 days around a Sun-like star. Assuming the planet has zero albedo (i.e., absorbs all incoming starlight) and perfectly transports heat across its surface, estimate the temperature at the photosphere of the planet. A: $395 \mathrm{~K}$ B: $954 \mathrm{~K}$ C: $1231 \mathrm{~K}$ D: $1476 \mathrm{~K}$ E: $2071 \mathrm{~K}$

You are participating in an international Astronomy competition and need to solve the following question. This is a multiple choice question (only one correct answer). problem: An exoplanet discovered by the radial velocity method is found to have an orbital period of 2.45 days around a Sun-like star. Assuming the planet has zero albedo (i.e., absorbs all incoming starlight) and perfectly transports heat across its surface, estimate the temperature at the photosphere of the planet. A: $395 \mathrm{~K}$ B: $954 \mathrm{~K}$ C: $1231 \mathrm{~K}$ D: $1476 \mathrm{~K}$ E: $2071 \mathrm{~K}$ You can solve it step by step. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER should be one of the options: [A, B, C, D, E].

SC

Astronomy

EN

text-only

Astronomy_1143

Plotting the position of the Sun in the sky at the same time every day, you get an interesting figure-ofeight shape known as an analemma (see Figure 1). For observers in the Northern hemisphere, you might expect to always see the Sun due South at midday, however on some days the Sun has already passed through that bearing and on others it needs a few more minutes before it gets there. This is due to two effects: the axial tilt of the Earth, and the fact the Earth's orbit is not perfectly circular [figure1] Figure 1: The analemma above was composed from images taken every few days at noon near the village of Callanish in the Outer Hebrides in Scotland. In the foreground are the Callanish Stones and the main photo was taken on the winter solstice (when the maximum angle the Sun reaches above the horizon is the lowest of the year, so is at the bottom of the analemma). Credit: Giuseppe Petricca. The vertical co-ordinate of a point in the analemma is entirely determined by the Earth's axial tilt. This is known as the solar declination, $\delta$, and varies sinusoidally throughout the year. The horizontal coordinate of a point in the analemma is determined by a combination of the Earth's axial tilt and the eccentricity of the Earth's orbit. Both of these individually vary sinusoidally, but the superposition of the two is no longer sinusoidal. We will define $\alpha$ as the angle between due South and the Sun at local midday as seen from Oxford, where a positive value means the Sun has already passed through due South (so is on the right of the figure above) whilst a negative value means the Sun has yet to pass through due South. If $\alpha_{\text {tilt }}$ is the contribution due to the axial tilt and $\alpha_{\text {ecc }}$ is the contribution due to the Earth's orbital eccentricity, then $$ \alpha=\alpha_{\text {tilt }}+\alpha_{\text {ecc }} $$ If the angle of the axial tilt is $\varepsilon$ and the eccentricity of the Earth's orbit is $e$, and we assume that both are small enough that the sinusoidal approximation of $\delta, \alpha_{\text {tilt }}$, and $\alpha_{\text {ecc }}$ apply, then we find the following boundary conditions: - $\delta$ has a period of 1 year, an amplitude of $\varepsilon$, is maximum at the summer solstice (21 $21^{\text {st }}$ June) and minimum at the winter solstice $\left(21^{\text {st }}\right.$ December $)$ - $\alpha_{\text {tilt }}$ has a period of 0.5 years, an amplitude (in radians) of $\tan ^{2}(\varepsilon / 2)$, is zero at the solstices and the equinoxes (vernal equinox $=21^{\text {st }}$ March, autumnal equinox $=21^{\text {st }}$ September), and (using our sign convention) positive just after the vernal equinox - $\alpha_{\text {ecc }}$ has a period of 1 year, an amplitude (in radians) of $2 e$, is zero at the perihelion (4 $4^{\text {th }}$ January) and the aphelion ( $6^{\text {th }}$ July), and (using our sign convention) negative just after the perihelion Given the $n^{\text {th }}$ day of the year, a value can be calculated for $\delta$ and $\alpha$, and these are the co-ordinates for the analemma (it is drawn by these parametric equations). For the Earth, $\varepsilon=23.44^{\circ}$ and $e=0.0167$. Consider an alternative version of Earth, known as Earth 2.0. On this planet, the year is unchanged and the perihelion and aphelion are at the same time, but it has a different axial tilt, a different orbital eccentricity, and a different month for the vernal equinox (although it is still on the $21^{\text {st }}$ day of that month). The analemma as viewed from Earth 2.0 is show in Figure 2 below. [figure2] Figure 2: The analemma of the Sun at midday as seen by an observer on Earth 2.0. In this situation, $\alpha$ ranges from -26 mins 47 secs to 18 mins 56 secs. The circled letters correspond to the same (unknown) day of each month (for example $5^{\text {th }}$ Jan, $5^{\text {th }}$ Feb, $5^{\text {th }}$ March etc.). Credit: Bob Urschel.c. Using your graph to guide you to the relevant point of the year, but using your precise algebraic expressions (rather than reading off the graph): i. What are the dates and values of the maximum and minimum values of $\alpha$ ?

You are participating in an international Astronomy competition and need to solve the following question. The answer to this question is a numerical value. Here is some context information for this question, which might assist you in solving it: Plotting the position of the Sun in the sky at the same time every day, you get an interesting figure-ofeight shape known as an analemma (see Figure 1). For observers in the Northern hemisphere, you might expect to always see the Sun due South at midday, however on some days the Sun has already passed through that bearing and on others it needs a few more minutes before it gets there. This is due to two effects: the axial tilt of the Earth, and the fact the Earth's orbit is not perfectly circular [figure1] Figure 1: The analemma above was composed from images taken every few days at noon near the village of Callanish in the Outer Hebrides in Scotland. In the foreground are the Callanish Stones and the main photo was taken on the winter solstice (when the maximum angle the Sun reaches above the horizon is the lowest of the year, so is at the bottom of the analemma). Credit: Giuseppe Petricca. The vertical co-ordinate of a point in the analemma is entirely determined by the Earth's axial tilt. This is known as the solar declination, $\delta$, and varies sinusoidally throughout the year. The horizontal coordinate of a point in the analemma is determined by a combination of the Earth's axial tilt and the eccentricity of the Earth's orbit. Both of these individually vary sinusoidally, but the superposition of the two is no longer sinusoidal. We will define $\alpha$ as the angle between due South and the Sun at local midday as seen from Oxford, where a positive value means the Sun has already passed through due South (so is on the right of the figure above) whilst a negative value means the Sun has yet to pass through due South. If $\alpha_{\text {tilt }}$ is the contribution due to the axial tilt and $\alpha_{\text {ecc }}$ is the contribution due to the Earth's orbital eccentricity, then $$ \alpha=\alpha_{\text {tilt }}+\alpha_{\text {ecc }} $$ If the angle of the axial tilt is $\varepsilon$ and the eccentricity of the Earth's orbit is $e$, and we assume that both are small enough that the sinusoidal approximation of $\delta, \alpha_{\text {tilt }}$, and $\alpha_{\text {ecc }}$ apply, then we find the following boundary conditions: - $\delta$ has a period of 1 year, an amplitude of $\varepsilon$, is maximum at the summer solstice (21 $21^{\text {st }}$ June) and minimum at the winter solstice $\left(21^{\text {st }}\right.$ December $)$ - $\alpha_{\text {tilt }}$ has a period of 0.5 years, an amplitude (in radians) of $\tan ^{2}(\varepsilon / 2)$, is zero at the solstices and the equinoxes (vernal equinox $=21^{\text {st }}$ March, autumnal equinox $=21^{\text {st }}$ September), and (using our sign convention) positive just after the vernal equinox - $\alpha_{\text {ecc }}$ has a period of 1 year, an amplitude (in radians) of $2 e$, is zero at the perihelion (4 $4^{\text {th }}$ January) and the aphelion ( $6^{\text {th }}$ July), and (using our sign convention) negative just after the perihelion Given the $n^{\text {th }}$ day of the year, a value can be calculated for $\delta$ and $\alpha$, and these are the co-ordinates for the analemma (it is drawn by these parametric equations). For the Earth, $\varepsilon=23.44^{\circ}$ and $e=0.0167$. Consider an alternative version of Earth, known as Earth 2.0. On this planet, the year is unchanged and the perihelion and aphelion are at the same time, but it has a different axial tilt, a different orbital eccentricity, and a different month for the vernal equinox (although it is still on the $21^{\text {st }}$ day of that month). The analemma as viewed from Earth 2.0 is show in Figure 2 below. [figure2] Figure 2: The analemma of the Sun at midday as seen by an observer on Earth 2.0. In this situation, $\alpha$ ranges from -26 mins 47 secs to 18 mins 56 secs. The circled letters correspond to the same (unknown) day of each month (for example $5^{\text {th }}$ Jan, $5^{\text {th }}$ Feb, $5^{\text {th }}$ March etc.). Credit: Bob Urschel. problem: c. Using your graph to guide you to the relevant point of the year, but using your precise algebraic expressions (rather than reading off the graph): i. What are the dates and values of the maximum and minimum values of $\alpha$ ? All mathematical formulas and symbols you output should be represented with LaTeX! You can solve it step by step. Remember, your answer should be calculated in the unit of \mathrm{mins}, but when concluding your final answer, do not include the unit. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER is the numerical value without any units.

NV

Astronomy

EN

multi-modal

Astronomy_118

随着我国航天技术的发展, 国人的登月梦想终将实现。若宇航员者陆月球后在其表面以一定的初速度坚直上抛一小球 (可视为质点), 经时间 $t$ 小球落回抛出点; 然后宇航员又在离月面高度为 $h$ 处, 以相同的速度大小沿水平方向抛出一小球，一段时间后小球落到月球表面, 抛出点与落地点之间的水平距离为 $L$ 。已知月球的质量为 $M$, 引力常量为 $G$ ，月球可看做质量分布均匀的球体，下列判断正确的是（） A: 月球表面的重力加速度大小为 $\frac{2 L^{2}}{h t^{2}}$ B: 小球上拖的初速度大小为 $\frac{L^{2}}{2 h t}$ C: 月球的半径为 $\frac{t}{L} \sqrt{2 G M h}$ D: 月球的第一宇宙速度大小为 $\sqrt{\frac{L}{h t} \sqrt{2 G M h}}$

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个多选题（有多个正确答案）。 问题: 随着我国航天技术的发展, 国人的登月梦想终将实现。若宇航员者陆月球后在其表面以一定的初速度坚直上抛一小球 (可视为质点), 经时间 $t$ 小球落回抛出点; 然后宇航员又在离月面高度为 $h$ 处, 以相同的速度大小沿水平方向抛出一小球，一段时间后小球落到月球表面, 抛出点与落地点之间的水平距离为 $L$ 。已知月球的质量为 $M$, 引力常量为 $G$ ，月球可看做质量分布均匀的球体，下列判断正确的是（） A: 月球表面的重力加速度大小为 $\frac{2 L^{2}}{h t^{2}}$ B: 小球上拖的初速度大小为 $\frac{L^{2}}{2 h t}$ C: 月球的半径为 $\frac{t}{L} \sqrt{2 G M h}$ D: 月球的第一宇宙速度大小为 $\sqrt{\frac{L}{h t} \sqrt{2 G M h}}$ 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]

MC

Astronomy

ZH

text-only

Astronomy_579

嫦娥三号探测器欲成功软着陆月球表面, 首先由地月轨道进入环月椭圆轨道I, 远月点 $A$ 距离月球表面为 $h$, 近月点 $B$ 距离月球表面高度可以忽略, 运行稳定后再次变轨进入近月轨道II。已知嫦娥三号探测器在环月椭圆轨道周期为 $T$ 、月球半径为 $R$ 和引力常量为 $G$ ，根据上述条件可以求得（） [图1] A: 探测器在近月轨道II运行周期 B: 探测器在环月椭圆轨道I经过 $B$ 点的加速度 C: 月球的质量 D: 探测器在月球表面的重力

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个多选题（有多个正确答案）。 问题: 嫦娥三号探测器欲成功软着陆月球表面, 首先由地月轨道进入环月椭圆轨道I, 远月点 $A$ 距离月球表面为 $h$, 近月点 $B$ 距离月球表面高度可以忽略, 运行稳定后再次变轨进入近月轨道II。已知嫦娥三号探测器在环月椭圆轨道周期为 $T$ 、月球半径为 $R$ 和引力常量为 $G$ ，根据上述条件可以求得（） [图1] A: 探测器在近月轨道II运行周期 B: 探测器在环月椭圆轨道I经过 $B$ 点的加速度 C: 月球的质量 D: 探测器在月球表面的重力 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]

MC

Astronomy

ZH

multi-modal

Astronomy_197

2021 年 6 月 17 日神舟十二号飞船搭载长征二号 $\mathrm{F}$ 火箭成功升空, 进入预定圆轨道 (离地约 $200 \mathrm{~km}$ )。此时, 天宫空间站天和核心舱位于离地约 $400 \mathrm{~km}$ 的圆轨道上。神舟十二号飞船只需绕地球飞行 4.3 圈, 连续进行 6 次变轨 (模拟如图所示), 到达空间站后下方的位置, 之后绕飞到空间站的正前方, 适当调整后就能跟核心舱前端接口成功对接, 形成组合体。已知地球半径 $R=6400 \mathrm{~km}$ ，则下列说法正确的是（） [图1] A: 神舟十二号飞船搭载长征二号 $\mathrm{F}$ 火箭的发射速度可能小于 $7.9 \mathrm{~km} / \mathrm{s}$ B: 对接前, 在各自预定圆轨道飞行时, 神舟十二号飞船的环绕速度比天和核心舱小 C: 神舟十二号飞船从低轨道变轨与天和号核心舱对接时, 需制动减速 D: 神舟十二号飞船进入离地约 $200 \mathrm{~km}$ 预定轨道后, 最快大约需 $6.5 \mathrm{~h}$ 就可与天和核心舱对接

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个单选题（只有一个正确答案）。 问题: 2021 年 6 月 17 日神舟十二号飞船搭载长征二号 $\mathrm{F}$ 火箭成功升空, 进入预定圆轨道 (离地约 $200 \mathrm{~km}$ )。此时, 天宫空间站天和核心舱位于离地约 $400 \mathrm{~km}$ 的圆轨道上。神舟十二号飞船只需绕地球飞行 4.3 圈, 连续进行 6 次变轨 (模拟如图所示), 到达空间站后下方的位置, 之后绕飞到空间站的正前方, 适当调整后就能跟核心舱前端接口成功对接, 形成组合体。已知地球半径 $R=6400 \mathrm{~km}$ ，则下列说法正确的是（） [图1] A: 神舟十二号飞船搭载长征二号 $\mathrm{F}$ 火箭的发射速度可能小于 $7.9 \mathrm{~km} / \mathrm{s}$ B: 对接前, 在各自预定圆轨道飞行时, 神舟十二号飞船的环绕速度比天和核心舱小 C: 神舟十二号飞船从低轨道变轨与天和号核心舱对接时, 需制动减速 D: 神舟十二号飞船进入离地约 $200 \mathrm{~km}$ 预定轨道后, 最快大约需 $6.5 \mathrm{~h}$ 就可与天和核心舱对接 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]

SC

Astronomy

ZH

multi-modal

Astronomy_422

2020 年 7 月 21 日将发生土星冲日现象, 如图所示, 土星冲日是指土星、地球和太阳几乎排列成一线, 地球位于太阳与土星之间。此时土星被太阳照亮的一面完全朝向地球, 所以明亮而易于观察。地球和土星绕太阳公转的方向相同, 轨迹都可近似为圆, 地球一年绕太阳一周，土星约 29.5 年绕太阳一周。则（） [图1] A: 地球绕太阳运转的向心加速度大于土星绕太阳运转的向心加速度 B: 地球绕太阳运转的运行速度比土星绕太阳运转的运行速度小 C: 2019 年没有出现土星冲日现象 D: 土星冲日现象下一次出现的时间是 2021 年

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个多选题（有多个正确答案）。 问题: 2020 年 7 月 21 日将发生土星冲日现象, 如图所示, 土星冲日是指土星、地球和太阳几乎排列成一线, 地球位于太阳与土星之间。此时土星被太阳照亮的一面完全朝向地球, 所以明亮而易于观察。地球和土星绕太阳公转的方向相同, 轨迹都可近似为圆, 地球一年绕太阳一周，土星约 29.5 年绕太阳一周。则（） [图1] A: 地球绕太阳运转的向心加速度大于土星绕太阳运转的向心加速度 B: 地球绕太阳运转的运行速度比土星绕太阳运转的运行速度小 C: 2019 年没有出现土星冲日现象 D: 土星冲日现象下一次出现的时间是 2021 年 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]

MC

Astronomy

ZH

multi-modal

Astronomy_489

将空间站“天和”核心舱绕地球的运动视为匀速圆周运动, 分别对其运行周期 $T$ 以及对应的轨道半径 $r$ 取对数, 得到 $\lg T-\lg r$ 图像如图所示, 其中 $a$ 为已知量。引力常量为 $G$ 。下列说法正确的是 ( ) [图1] A: $\lg T-\lg r$ 图像的斜率为 $\frac{2}{3}$ B: $\lg T-\lg r$ 图像的斜率为 $\frac{3}{2}$ C: 地球的质量为 $\frac{4 \pi^{2}}{G} \times 10^{3 a}$ D: 地球的质量为 $\frac{4 \pi^{2}}{G} \times 10^{2 a}$

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个多选题（有多个正确答案）。 问题: 将空间站“天和”核心舱绕地球的运动视为匀速圆周运动, 分别对其运行周期 $T$ 以及对应的轨道半径 $r$ 取对数, 得到 $\lg T-\lg r$ 图像如图所示, 其中 $a$ 为已知量。引力常量为 $G$ 。下列说法正确的是 ( ) [图1] A: $\lg T-\lg r$ 图像的斜率为 $\frac{2}{3}$ B: $\lg T-\lg r$ 图像的斜率为 $\frac{3}{2}$ C: 地球的质量为 $\frac{4 \pi^{2}}{G} \times 10^{3 a}$ D: 地球的质量为 $\frac{4 \pi^{2}}{G} \times 10^{2 a}$ 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]

MC

Astronomy

ZH

multi-modal

Astronomy_962

Below is one of the first images taken with the James Webb Space Telescope (JWST) released to the public on $12^{\text {th }}$ July 2022 . Which object is it from? [figure1] A: Carina Nebula B: Stephan's Quintet C: Southern Ring Nebula D: WASP-96 b

You are participating in an international Astronomy competition and need to solve the following question. This is a multiple choice question (only one correct answer). problem: Below is one of the first images taken with the James Webb Space Telescope (JWST) released to the public on $12^{\text {th }}$ July 2022 . Which object is it from? [figure1] A: Carina Nebula B: Stephan's Quintet C: Southern Ring Nebula D: WASP-96 b You can solve it step by step. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER should be one of the options: [A, B, C, D].

SC

Astronomy

EN

multi-modal

Astronomy_155

2022 年 11 月 29 日, 神舟十五号飞行任务是中国空间站建造阶段的最后一棒, 也是空间站应用与发展阶段的第一棒. 已知空间站 $\mathrm{Q}$ 和同步卫星 $\mathrm{P}$ 环绕地球运行的轨道均可视为匀速圆周运动。如图所示, 已知 $\mathrm{P}, \mathrm{Q}$ 运动方向均沿逆时针方向, $P Q$ 与 $O P$连线的夹角最大值为 $\alpha$ 。求: 由 $P Q$ 与 $O P$ 连线夹角最大时开始计时 (图示状态), 直到空间站 $\mathrm{Q}$ 和同步卫星 $\mathrm{P}$的间距第一次最近时所用的时间与空间站 $\mathrm{Q}$ 的周期之比。 [图1]

你正在参加一个国际天文竞赛，并需要解决以下问题。 这个问题的答案是一个表达式。 问题: 2022 年 11 月 29 日, 神舟十五号飞行任务是中国空间站建造阶段的最后一棒, 也是空间站应用与发展阶段的第一棒. 已知空间站 $\mathrm{Q}$ 和同步卫星 $\mathrm{P}$ 环绕地球运行的轨道均可视为匀速圆周运动。如图所示, 已知 $\mathrm{P}, \mathrm{Q}$ 运动方向均沿逆时针方向, $P Q$ 与 $O P$连线的夹角最大值为 $\alpha$ 。求: 由 $P Q$ 与 $O P$ 连线夹角最大时开始计时 (图示状态), 直到空间站 $\mathrm{Q}$ 和同步卫星 $\mathrm{P}$的间距第一次最近时所用的时间与空间站 $\mathrm{Q}$ 的周期之比。 [图1] 你输出的所有数学公式和符号应该使用LaTeX表示！ 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER是一个不含等号的表达式，例如ANSWER=\frac{1}{2} g t^2

EX

Astronomy

ZH

multi-modal

Astronomy_395

嫦娥四号探测器成功发射, 开启了我国月球探测的新旅程, 探测器经过地月转移、近月 制动、环月飞行, 最终实现了人类首次月球背面软着陆.如图为探测器绕月球的运行轨道示意图, 其中轨道 I 为圆形轨道, 轨道 II 为制圆轨道. 下列关于探测器的说法中正确的是 [图1] A: 在轨道 I、轨道 II 上通过 $P$ 点时的动能相等 B: 在轨道 I 通过 $P$ 点的加速度比在轨道 II 通过 $P$ 点的加速度大 C: 在轨道 II 上从 $P$ 点运动到 $Q$ 点的过程中动能变大 D: 在轨道 II 上从 $P$ 点运动到 $Q$ 点的过程中的机械能守恒

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个多选题（有多个正确答案）。 问题: 嫦娥四号探测器成功发射, 开启了我国月球探测的新旅程, 探测器经过地月转移、近月 制动、环月飞行, 最终实现了人类首次月球背面软着陆.如图为探测器绕月球的运行轨道示意图, 其中轨道 I 为圆形轨道, 轨道 II 为制圆轨道. 下列关于探测器的说法中正确的是 [图1] A: 在轨道 I、轨道 II 上通过 $P$ 点时的动能相等 B: 在轨道 I 通过 $P$ 点的加速度比在轨道 II 通过 $P$ 点的加速度大 C: 在轨道 II 上从 $P$ 点运动到 $Q$ 点的过程中动能变大 D: 在轨道 II 上从 $P$ 点运动到 $Q$ 点的过程中的机械能守恒 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]

MC

Astronomy

ZH

multi-modal

Astronomy_640

如图所示, 2022 年 7 月 15 日, 由清华大学天文系祝伟教授牵头的国际团队近日宣布在宇宙中发现两个罕见的恒星系统。该系统均是由两颗互相绕行的中央恒星组成, 被气体和尘埃盘包围, 且该盘与中央恒星的轨道成一定角度, 呈现出“雾绕双星”的奇幻效 果。如图所示为该双星模型的简化图, 已知 $O_{1} O_{2}=L_{1}, \quad O_{1} O-O O_{2}=L_{2}>0$, 假设两星球的半径远小于两球球心之间的距离。则下列说法正确的是（） [图1] A: 星球 $\mathrm{P} 、 \mathrm{Q}$ 的轨道半径之比为 $\frac{L_{1}+L_{2}}{L_{1}-L_{2}}$ B: 星球 $\mathrm{P}$ 的质量大于星球 $\mathrm{Q}$ 的质量 C: 星球 $\mathrm{P} 、 \mathrm{Q}$ 的线速度之和与线速度之差的比值 $\frac{L_{1}}{L_{2}}$ D: 星球 $\mathrm{P} 、 \mathrm{Q}$ 的质量之和与 $\mathrm{P} 、 \mathrm{Q}$ 质量之差的比值 $\frac{L_{1}}{L_{2}}$

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个多选题（有多个正确答案）。 问题: 如图所示, 2022 年 7 月 15 日, 由清华大学天文系祝伟教授牵头的国际团队近日宣布在宇宙中发现两个罕见的恒星系统。该系统均是由两颗互相绕行的中央恒星组成, 被气体和尘埃盘包围, 且该盘与中央恒星的轨道成一定角度, 呈现出“雾绕双星”的奇幻效 果。如图所示为该双星模型的简化图, 已知 $O_{1} O_{2}=L_{1}, \quad O_{1} O-O O_{2}=L_{2}>0$, 假设两星球的半径远小于两球球心之间的距离。则下列说法正确的是（） [图1] A: 星球 $\mathrm{P} 、 \mathrm{Q}$ 的轨道半径之比为 $\frac{L_{1}+L_{2}}{L_{1}-L_{2}}$ B: 星球 $\mathrm{P}$ 的质量大于星球 $\mathrm{Q}$ 的质量 C: 星球 $\mathrm{P} 、 \mathrm{Q}$ 的线速度之和与线速度之差的比值 $\frac{L_{1}}{L_{2}}$ D: 星球 $\mathrm{P} 、 \mathrm{Q}$ 的质量之和与 $\mathrm{P} 、 \mathrm{Q}$ 质量之差的比值 $\frac{L_{1}}{L_{2}}$ 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]

MC

Astronomy

ZH

multi-modal

Astronomy_1138

Recent years have seen an explosion in the discovery of new exoplanets. About $85 \%$ of transiting exoplanets discovered by the NASA Kepler telescope have radii less than Neptune ( $\sim 4 R_{\oplus}$ ), meaning we are improving our understanding of what the transition between rocky Earth-size planets and gaseous Neptune-size planets looks like. Given how common these "super-Earths" and "gas dwarfs" seem to be, it was odd that we didn't have any in our own Solar System. However, Batygin \& Brown (2016) suggested that a hypothetical ninth planet (called 'Planet Nine') could explain some of the unusual properties of the orbits of objects in the Kuiper Belt. This planet is inferred to have a mass of $10 M_{\oplus}$, and so would be an example of a super-Earth. [figure1] Figure 5: A plot of planet density versus radius for 33 extrasolar planets (circles) and the planets in our solar system (diamonds). Credit: Marcy et al. (2014). Analysing exoplanets discovered by Kepler, Marcy et al. (2014) used a piecewise function to describe their planetary density data such that: $$ \begin{aligned} \text { For } R_{\mathrm{P}} \leq 1.5 R_{\oplus} & \rho & =2.32+3.18 \frac{R_{\mathrm{P}}}{R_{\oplus}}\left[\mathrm{g} \mathrm{cm}^{-3}\right] \\ \text { For } 1.5 R_{\oplus}<R_{\mathrm{P}} \leq 4.2 R_{\oplus} & \frac{M_{\mathrm{P}}}{M_{\oplus}} & =2.69\left(\frac{R_{\mathrm{P}}}{R_{\oplus}}\right)^{0.93} \end{aligned} $$ where $R_{\mathrm{P}}$ is the radius of the planet, $M_{\mathrm{P}}$ is the mass of the planet, and the model's transition between rocky super-Earth and non-rocky gas dwarf occurs at $R_{\mathrm{P}}=1.5 R_{\oplus}$. The minimum speed necessary to fully escape a planet's gravity (rather than be put into an elliptical orbit) is called the escape velocity and is calculated as $$ v_{\mathrm{esc}}=\sqrt{\frac{2 G M_{\mathrm{P}}}{R_{\mathrm{P}}}} $$ where $G$ is the universal gravitational constant. In contrast, the maximum speed a rocket can provide is determined by the ejection velocity, $v_{\mathrm{e}}$, of the gas used, as determined by the chemical energy stored in the bonds of the fuel used, and the fraction of the rocket that is fuel. Since the rocket gets lighter as it burns its fuel the final speed can be bigger than $v_{\mathrm{e}}$. The rocket equation is $$ v_{\max }=v_{\mathrm{e}} \ln \frac{m_{0}}{m_{1}} $$ where $m_{0}$ is the mass at launch and $m_{1}$ is the mass once all the fuel has been burnt. The most energetic chemical reaction we can use in a rocket is hydrogen-oxygen, which gives $v_{\mathrm{e}}=4.46 \mathrm{~km} \mathrm{~s}^{-1}$, and engineering limits us to a rocket design with a maximum of $96 \%$ of launch mass being fuel (as was used with the solid rockets that launched the space shuttle).f. In the Marcy et al. (2014) model, above $R_{P}=1.5 R_{E}$ the density of gas dwarfs rapidly decreases with radius. By looking at the piecewise function explain why this does not improve the situation for any alien civilization hoping to explore their solar system (ignoring that such planets are far less likely to be habitable). You do not need to calculate any new escape velocities.

You are participating in an international Astronomy competition and need to solve the following question. The answer to this question is an expression. Here is some context information for this question, which might assist you in solving it: Recent years have seen an explosion in the discovery of new exoplanets. About $85 \%$ of transiting exoplanets discovered by the NASA Kepler telescope have radii less than Neptune ( $\sim 4 R_{\oplus}$ ), meaning we are improving our understanding of what the transition between rocky Earth-size planets and gaseous Neptune-size planets looks like. Given how common these "super-Earths" and "gas dwarfs" seem to be, it was odd that we didn't have any in our own Solar System. However, Batygin \& Brown (2016) suggested that a hypothetical ninth planet (called 'Planet Nine') could explain some of the unusual properties of the orbits of objects in the Kuiper Belt. This planet is inferred to have a mass of $10 M_{\oplus}$, and so would be an example of a super-Earth. [figure1] Figure 5: A plot of planet density versus radius for 33 extrasolar planets (circles) and the planets in our solar system (diamonds). Credit: Marcy et al. (2014). Analysing exoplanets discovered by Kepler, Marcy et al. (2014) used a piecewise function to describe their planetary density data such that: $$ \begin{aligned} \text { For } R_{\mathrm{P}} \leq 1.5 R_{\oplus} & \rho & =2.32+3.18 \frac{R_{\mathrm{P}}}{R_{\oplus}}\left[\mathrm{g} \mathrm{cm}^{-3}\right] \\ \text { For } 1.5 R_{\oplus}<R_{\mathrm{P}} \leq 4.2 R_{\oplus} & \frac{M_{\mathrm{P}}}{M_{\oplus}} & =2.69\left(\frac{R_{\mathrm{P}}}{R_{\oplus}}\right)^{0.93} \end{aligned} $$ where $R_{\mathrm{P}}$ is the radius of the planet, $M_{\mathrm{P}}$ is the mass of the planet, and the model's transition between rocky super-Earth and non-rocky gas dwarf occurs at $R_{\mathrm{P}}=1.5 R_{\oplus}$. The minimum speed necessary to fully escape a planet's gravity (rather than be put into an elliptical orbit) is called the escape velocity and is calculated as $$ v_{\mathrm{esc}}=\sqrt{\frac{2 G M_{\mathrm{P}}}{R_{\mathrm{P}}}} $$ where $G$ is the universal gravitational constant. In contrast, the maximum speed a rocket can provide is determined by the ejection velocity, $v_{\mathrm{e}}$, of the gas used, as determined by the chemical energy stored in the bonds of the fuel used, and the fraction of the rocket that is fuel. Since the rocket gets lighter as it burns its fuel the final speed can be bigger than $v_{\mathrm{e}}$. The rocket equation is $$ v_{\max }=v_{\mathrm{e}} \ln \frac{m_{0}}{m_{1}} $$ where $m_{0}$ is the mass at launch and $m_{1}$ is the mass once all the fuel has been burnt. The most energetic chemical reaction we can use in a rocket is hydrogen-oxygen, which gives $v_{\mathrm{e}}=4.46 \mathrm{~km} \mathrm{~s}^{-1}$, and engineering limits us to a rocket design with a maximum of $96 \%$ of launch mass being fuel (as was used with the solid rockets that launched the space shuttle). problem: f. In the Marcy et al. (2014) model, above $R_{P}=1.5 R_{E}$ the density of gas dwarfs rapidly decreases with radius. By looking at the piecewise function explain why this does not improve the situation for any alien civilization hoping to explore their solar system (ignoring that such planets are far less likely to be habitable). You do not need to calculate any new escape velocities. All mathematical formulas and symbols you output should be represented with LaTeX! You can solve it step by step. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER is an expression without equals signs, e.g. ANSWER=\frac{1}{2} g t^2

EX

Astronomy

EN

multi-modal

Astronomy_1059

The surface of the Sun has a temperature of $\sim 5700 \mathrm{~K}$ yet the solar corona (a very faint region of plasma normally only visible from Earth during a solar eclipse) is considerably hotter at around $10^{6} \mathrm{~K}$. The source of coronal heating is a mystery and so understanding how this might happen is one of several key science objectives of the Solar Orbiter spacecraft. It is equipped with an array of cameras and will take photos of the Sun from distances closer than ever before (other probes will go closer, but none of those have cameras). [figure1] Figure 7: Left: The Sun's corona (coloured green) as viewed in visible light (580-640 nm) taken with the METIS coronagraph instrument onboard Solar Orbiter. The coronagraph is a disc that blocks out the light of the Sun (whose size and position is indicated with the white circle in the middle) so that the faint corona can be seen. This was taken just after first perihelion and is already at a resolution only matched by ground-based telescopes during a solar eclipse - once it gets into the main phase of the mission when it is even closer then its photos will be unrivalled. Credit: METIS Team / ESA \& NASA Right: A high-resolution image from the Extreme Ultraviolet Imager (EUI), taken with the $\mathrm{HRI}_{\mathrm{EUV}}$ telescope just before first perihelion. The circle in the lower right corner indicates the size of Earth for scale. The arrow points to one of the ubiquitous features of the solar surface, called 'campfires', that were discovered by this spacecraft and may play an important role in heating the corona. Credit: EUI Team / ESA \& NASA. Launched in February 2020 (and taken to be at aphelion at launch), it arrived at its first perihelion on $15^{\text {th }}$ June 2020 and has sent back some of the highest resolution images of the surface of the Sun (i.e. the base of the corona) we have ever seen. In them we have identified phenomena nicknamed as 'campfires' (see Fig 7) which are already being considered as a potential major contributor to the mechanism of coronal heating. Later on in its mission it will go in even closer, and so will take photos of the Sun in unprecedented detail. The highest resolution photos are taken with the Extreme Ultraviolet Imager (EUI), which consists of three separate cameras. One of them, the Extreme Ultraviolet High Resolution Imager (HRI $\mathrm{HUV}$ ), is designed to pick up an emission line from highly ionised atoms of iron in the corona. The iron being detected has lost 9 electrons (i.e. $\mathrm{Fe}^{9+}$ ) though is called $\mathrm{Fe} \mathrm{X} \mathrm{('ten')} \mathrm{by} \mathrm{astronomers} \mathrm{(as} \mathrm{Fe} \mathrm{I} \mathrm{is} \mathrm{the} \mathrm{neutral}$ atom). Its presence can be used to work out the temperature of the part of the corona being investigated by the instrument. The photons detected by $\mathrm{HRI}_{\mathrm{EUV}}$ are emitted by a rearrangement of the electrons in the $\mathrm{Fe} \mathrm{X}$ ion, corresponding to a photon energy of $71.0372 \mathrm{eV}$ (where $1 \mathrm{eV}=1.60 \times 10^{-19} \mathrm{~J}$ ). The HRI $\mathrm{HUV}_{\mathrm{EUV}}$ telescope has a $1000^{\prime \prime}$ by $1000^{\prime \prime}$ field of view (FOV, where $1^{\circ}=3600^{\prime \prime}=3600$ arcseconds), an entrance pupil diameter of $47.4 \mathrm{~mm}$, a couple of mirrors that give an effective focal length of $4187 \mathrm{~mm}$, and the image is captured by a CCD with 2048 by 2048 pixels, each of which is 10 by $10 \mu \mathrm{m}$. Although we are viewing the emissions of $\mathrm{Fe} \mathrm{X}$ ions, the vast majority of the plasma in the corona is hydrogen and helium, and the bulk motions of this determine the timescales over which visible phenomena change. In particular, the speed of sound is very important if we do not want motion blur to affect our high resolution images, as this sets the limit on exposure times.c. The Rayleigh criterion and speed of sound in a plasma are given. iv. If the mass fractions of the surface of the Sun are $X=0.7381, Y=0.2485$, and $Z=0.0134$, and treating the plasma as an ideal monatomic gas so that $\gamma=5 / 3$, determine the speed of sound of the plasma (in $\mathrm{km} \mathrm{s}^{-1}$ ) at the base of the corona. Hence, by comparison to your answer from the previous part, estimate the upper limit on the length of an exposure to avoid motion blur in the plasma. You should ignore any motion blur from the relative motion of the spacecraft or the rotation of the Sun.

You are participating in an international Astronomy competition and need to solve the following question. The answer to this question is a numerical value. Here is some context information for this question, which might assist you in solving it: The surface of the Sun has a temperature of $\sim 5700 \mathrm{~K}$ yet the solar corona (a very faint region of plasma normally only visible from Earth during a solar eclipse) is considerably hotter at around $10^{6} \mathrm{~K}$. The source of coronal heating is a mystery and so understanding how this might happen is one of several key science objectives of the Solar Orbiter spacecraft. It is equipped with an array of cameras and will take photos of the Sun from distances closer than ever before (other probes will go closer, but none of those have cameras). [figure1] Figure 7: Left: The Sun's corona (coloured green) as viewed in visible light (580-640 nm) taken with the METIS coronagraph instrument onboard Solar Orbiter. The coronagraph is a disc that blocks out the light of the Sun (whose size and position is indicated with the white circle in the middle) so that the faint corona can be seen. This was taken just after first perihelion and is already at a resolution only matched by ground-based telescopes during a solar eclipse - once it gets into the main phase of the mission when it is even closer then its photos will be unrivalled. Credit: METIS Team / ESA \& NASA Right: A high-resolution image from the Extreme Ultraviolet Imager (EUI), taken with the $\mathrm{HRI}_{\mathrm{EUV}}$ telescope just before first perihelion. The circle in the lower right corner indicates the size of Earth for scale. The arrow points to one of the ubiquitous features of the solar surface, called 'campfires', that were discovered by this spacecraft and may play an important role in heating the corona. Credit: EUI Team / ESA \& NASA. Launched in February 2020 (and taken to be at aphelion at launch), it arrived at its first perihelion on $15^{\text {th }}$ June 2020 and has sent back some of the highest resolution images of the surface of the Sun (i.e. the base of the corona) we have ever seen. In them we have identified phenomena nicknamed as 'campfires' (see Fig 7) which are already being considered as a potential major contributor to the mechanism of coronal heating. Later on in its mission it will go in even closer, and so will take photos of the Sun in unprecedented detail. The highest resolution photos are taken with the Extreme Ultraviolet Imager (EUI), which consists of three separate cameras. One of them, the Extreme Ultraviolet High Resolution Imager (HRI $\mathrm{HUV}$ ), is designed to pick up an emission line from highly ionised atoms of iron in the corona. The iron being detected has lost 9 electrons (i.e. $\mathrm{Fe}^{9+}$ ) though is called $\mathrm{Fe} \mathrm{X} \mathrm{('ten')} \mathrm{by} \mathrm{astronomers} \mathrm{(as} \mathrm{Fe} \mathrm{I} \mathrm{is} \mathrm{the} \mathrm{neutral}$ atom). Its presence can be used to work out the temperature of the part of the corona being investigated by the instrument. The photons detected by $\mathrm{HRI}_{\mathrm{EUV}}$ are emitted by a rearrangement of the electrons in the $\mathrm{Fe} \mathrm{X}$ ion, corresponding to a photon energy of $71.0372 \mathrm{eV}$ (where $1 \mathrm{eV}=1.60 \times 10^{-19} \mathrm{~J}$ ). The HRI $\mathrm{HUV}_{\mathrm{EUV}}$ telescope has a $1000^{\prime \prime}$ by $1000^{\prime \prime}$ field of view (FOV, where $1^{\circ}=3600^{\prime \prime}=3600$ arcseconds), an entrance pupil diameter of $47.4 \mathrm{~mm}$, a couple of mirrors that give an effective focal length of $4187 \mathrm{~mm}$, and the image is captured by a CCD with 2048 by 2048 pixels, each of which is 10 by $10 \mu \mathrm{m}$. Although we are viewing the emissions of $\mathrm{Fe} \mathrm{X}$ ions, the vast majority of the plasma in the corona is hydrogen and helium, and the bulk motions of this determine the timescales over which visible phenomena change. In particular, the speed of sound is very important if we do not want motion blur to affect our high resolution images, as this sets the limit on exposure times. problem: c. The Rayleigh criterion and speed of sound in a plasma are given. iv. If the mass fractions of the surface of the Sun are $X=0.7381, Y=0.2485$, and $Z=0.0134$, and treating the plasma as an ideal monatomic gas so that $\gamma=5 / 3$, determine the speed of sound of the plasma (in $\mathrm{km} \mathrm{s}^{-1}$ ) at the base of the corona. Hence, by comparison to your answer from the previous part, estimate the upper limit on the length of an exposure to avoid motion blur in the plasma. You should ignore any motion blur from the relative motion of the spacecraft or the rotation of the Sun. All mathematical formulas and symbols you output should be represented with LaTeX! You can solve it step by step. Remember, your answer should be calculated in the unit of \mathrm{~s}, but when concluding your final answer, do not include the unit. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER is the numerical value without any units.

NV

Astronomy

EN

multi-modal

Astronomy_808

VOIDED A recently discovered star Recentus in a nearby galaxy has been confirmed to have a luminosity of $4 \mathrm{~L} \odot$ with a radius of $4 \mathrm{R}_{\odot}$. What is the approximate frequency of peak emission of Recentus? A: $2 \times 10^{14} \mathrm{~Hz}$ B: $4 \times 10^{14} \mathrm{~Hz}$ C: $6 \times 10^{14} \mathrm{~Hz}$ D: $8 \times 10^{14} \mathrm{~Hz}$ E: $10 \times 10^{14} \mathrm{~Hz}$

You are participating in an international Astronomy competition and need to solve the following question. This is a multiple choice question (only one correct answer). problem: VOIDED A recently discovered star Recentus in a nearby galaxy has been confirmed to have a luminosity of $4 \mathrm{~L} \odot$ with a radius of $4 \mathrm{R}_{\odot}$. What is the approximate frequency of peak emission of Recentus? A: $2 \times 10^{14} \mathrm{~Hz}$ B: $4 \times 10^{14} \mathrm{~Hz}$ C: $6 \times 10^{14} \mathrm{~Hz}$ D: $8 \times 10^{14} \mathrm{~Hz}$ E: $10 \times 10^{14} \mathrm{~Hz}$ You can solve it step by step. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER should be one of the options: [A, B, C, D, E].

SC

Astronomy

EN

text-only

Astronomy_795

At 6am on March 20th, as the Sun is rising, Leo, who is at $\left(40^{\circ} \mathrm{N}, 75^{\circ} \mathrm{W}\right)$, plants a stick vertically on the ground. At that moment, he marks out a (finite) line on the ground in the direction of the shadow of the stick at that moment, labeling it with the current time. Every hour afterwards, on the hour, he marks out a new line in the current direction of the shadow, until the sun sets at $6 \mathrm{pm}$. Three months later, Leo returns to the same spot, where the vertical stick and lines remain. Again, every hour on the hour, he marks out a line in the current direction of the shadow, until the sun sets. Let $\alpha_{12}$ and $\alpha_{6}$ be the azimuths of the lines drawn in the spring at $12 \mathrm{pm}$ and $6 \mathrm{pm}$, and $\beta_{12}$ and $\beta_{6}$ be the azimuths of the lines drawn in the summer at $12 \mathrm{pm}$ and $6 \mathrm{pm}$. Which of the following statements is true? Ignore atmospheric effects and the equation of time. A: $\alpha_{12}=\beta_{12}, \alpha_{6}=\beta_{6}$ B: $\alpha_{12}>\beta_{12}, \alpha_{6}=\beta_{6}$ C: $\alpha_{12}<\beta_{12}, \alpha_{6}=\beta_{6}$ D: $\alpha_{12}=\beta_{12}, \alpha_{6}>\beta_{6}$ E: $\alpha_{12}=\beta_{12}, \alpha_{6}<\beta_{6}$

You are participating in an international Astronomy competition and need to solve the following question. This is a multiple choice question (only one correct answer). problem: At 6am on March 20th, as the Sun is rising, Leo, who is at $\left(40^{\circ} \mathrm{N}, 75^{\circ} \mathrm{W}\right)$, plants a stick vertically on the ground. At that moment, he marks out a (finite) line on the ground in the direction of the shadow of the stick at that moment, labeling it with the current time. Every hour afterwards, on the hour, he marks out a new line in the current direction of the shadow, until the sun sets at $6 \mathrm{pm}$. Three months later, Leo returns to the same spot, where the vertical stick and lines remain. Again, every hour on the hour, he marks out a line in the current direction of the shadow, until the sun sets. Let $\alpha_{12}$ and $\alpha_{6}$ be the azimuths of the lines drawn in the spring at $12 \mathrm{pm}$ and $6 \mathrm{pm}$, and $\beta_{12}$ and $\beta_{6}$ be the azimuths of the lines drawn in the summer at $12 \mathrm{pm}$ and $6 \mathrm{pm}$. Which of the following statements is true? Ignore atmospheric effects and the equation of time. A: $\alpha_{12}=\beta_{12}, \alpha_{6}=\beta_{6}$ B: $\alpha_{12}>\beta_{12}, \alpha_{6}=\beta_{6}$ C: $\alpha_{12}<\beta_{12}, \alpha_{6}=\beta_{6}$ D: $\alpha_{12}=\beta_{12}, \alpha_{6}>\beta_{6}$ E: $\alpha_{12}=\beta_{12}, \alpha_{6}<\beta_{6}$ You can solve it step by step. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER should be one of the options: [A, B, C, D, E].

SC

Astronomy

EN

text-only

Astronomy_574

如图为高分一号北斗导航系统两颗卫星在空中某一面内运动的示意图. 导航卫星 $\mathrm{G}_{1}$和 $\mathrm{G}_{2}$ 以及高分一号均可认为统地心 $O$ 做匀速圆同运动. 卫星 $\mathrm{G}_{1}$ 和 $\mathrm{G}_{2}$ 的轨道半径为 $r$,某时刻两颗导航卫星分别位于轨道上的 $A$ 和 $B$ 两位置, 高分一号在 $C$ 位置. 若卫星均顺时针运行, $\angle A O B=60^{\circ}$, 地球表面处的重力加速度为 $g$, 地球半径为 $R$, 不计卫星间的相互作用力. 则下列说法正确的是（ ） [图1] A: 卫星 $\mathrm{G}_{1}$ 和 $\mathrm{G}_{2}$ 的加速度大小相等且为 $\frac{R}{r} g$ B: 卫星 $\mathrm{G}_{1}$ 由位置 $A$ 运动到位置 $B$ 所需的时间为 $\frac{\pi r}{3 R} \sqrt{\frac{r}{g}}$ C: 若高分一号与卫星 $\mathrm{G}_{1}$ 的周期之比为 $1: k(k>1$, 且为整数), 某时刻两者相距最近,则从此时刻起, 在卫星 $\mathrm{G}_{1}$ 运动一周的过程中二者距离最近的次数为 (k-1) D: 若高分一号与卫星 $\mathrm{G}_{1}$ 质量相等,由于高分一号的绕行速度大,则发射所需的最小能量更多

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个多选题（有多个正确答案）。 问题: 如图为高分一号北斗导航系统两颗卫星在空中某一面内运动的示意图. 导航卫星 $\mathrm{G}_{1}$和 $\mathrm{G}_{2}$ 以及高分一号均可认为统地心 $O$ 做匀速圆同运动. 卫星 $\mathrm{G}_{1}$ 和 $\mathrm{G}_{2}$ 的轨道半径为 $r$,某时刻两颗导航卫星分别位于轨道上的 $A$ 和 $B$ 两位置, 高分一号在 $C$ 位置. 若卫星均顺时针运行, $\angle A O B=60^{\circ}$, 地球表面处的重力加速度为 $g$, 地球半径为 $R$, 不计卫星间的相互作用力. 则下列说法正确的是（ ） [图1] A: 卫星 $\mathrm{G}_{1}$ 和 $\mathrm{G}_{2}$ 的加速度大小相等且为 $\frac{R}{r} g$ B: 卫星 $\mathrm{G}_{1}$ 由位置 $A$ 运动到位置 $B$ 所需的时间为 $\frac{\pi r}{3 R} \sqrt{\frac{r}{g}}$ C: 若高分一号与卫星 $\mathrm{G}_{1}$ 的周期之比为 $1: k(k>1$, 且为整数), 某时刻两者相距最近,则从此时刻起, 在卫星 $\mathrm{G}_{1}$ 运动一周的过程中二者距离最近的次数为 (k-1) D: 若高分一号与卫星 $\mathrm{G}_{1}$ 质量相等,由于高分一号的绕行速度大,则发射所需的最小能量更多 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]

MC

Astronomy

ZH

multi-modal

Astronomy_41

在太阳系外发现的某恒星 $\mathrm{a}$ 的质量为太阳系质量的 0.3 倍, 该恒星的一颗行星 $\mathrm{b}$ 的质量是地球的 4 倍, 直径是地球的 1.5 倍, 公转周期为 10 天. 设该行星与地球均为质量分布均匀的球体, 且分别绕其中心天体做匀速圆周运动, 则 ( ) A: 行星 $\mathrm{b}$ 的第一宇宙速度与地球相同 B: 行星 $\mathrm{b}$ 绕恒星 $\mathrm{a}$ 运行的角速度大于地球绕太阳运行的角速度 C: 如果将物体从地球搬到行星 $\mathrm{b}$ 上, 其重力是在地球上重力的 $\frac{16}{9}$ D: 行星 $\mathrm{b}$ 与恒星 $\mathrm{a}$ 的距离是日地距离的 $\sqrt{\frac{2}{73}}$ 倍

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个多选题（有多个正确答案）。 问题: 在太阳系外发现的某恒星 $\mathrm{a}$ 的质量为太阳系质量的 0.3 倍, 该恒星的一颗行星 $\mathrm{b}$ 的质量是地球的 4 倍, 直径是地球的 1.5 倍, 公转周期为 10 天. 设该行星与地球均为质量分布均匀的球体, 且分别绕其中心天体做匀速圆周运动, 则 ( ) A: 行星 $\mathrm{b}$ 的第一宇宙速度与地球相同 B: 行星 $\mathrm{b}$ 绕恒星 $\mathrm{a}$ 运行的角速度大于地球绕太阳运行的角速度 C: 如果将物体从地球搬到行星 $\mathrm{b}$ 上, 其重力是在地球上重力的 $\frac{16}{9}$ D: 行星 $\mathrm{b}$ 与恒星 $\mathrm{a}$ 的距离是日地距离的 $\sqrt{\frac{2}{73}}$ 倍 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]

MC

Astronomy

ZH

text-only

Astronomy_100

甲、乙两颗卫星在不同轨道上绕地球运动, 甲卫星的轨道是圆, 半径为 $R$, 乙卫星的轨道是椭圆, 其中 $P$ 点为近地点, 到地心的距离为 $a, Q$ 为远地点, 到地心的距离为 $b$ 。已知 $a<R<b$ ，则下列说法正确的是（） [图1] A: 卫星乙运动到 $P$ 点时的速度可能小于卫星甲的速度 B: 卫星乙运动到 $Q$ 点时的速度一定小于卫星甲的速度 C: 若 $a+b<2 R$, 卫星甲运行的周期一定大于卫星乙运行的周期 D: 卫星乙从 $P$ 点运动到 $Q$ 点的过程中机械能减少

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个多选题（有多个正确答案）。 问题: 甲、乙两颗卫星在不同轨道上绕地球运动, 甲卫星的轨道是圆, 半径为 $R$, 乙卫星的轨道是椭圆, 其中 $P$ 点为近地点, 到地心的距离为 $a, Q$ 为远地点, 到地心的距离为 $b$ 。已知 $a<R<b$ ，则下列说法正确的是（） [图1] A: 卫星乙运动到 $P$ 点时的速度可能小于卫星甲的速度 B: 卫星乙运动到 $Q$ 点时的速度一定小于卫星甲的速度 C: 若 $a+b<2 R$, 卫星甲运行的周期一定大于卫星乙运行的周期 D: 卫星乙从 $P$ 点运动到 $Q$ 点的过程中机械能减少 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]

MC

Astronomy

ZH

multi-modal

Astronomy_181

如图所示, $\mathrm{A} 、 \mathrm{~B} 、 \mathrm{C}$ 是在地球大气层外的圆形轨道上运行的三颗人造地球卫星,下列说法中正确的是（） [图1] A: B、C 的角速度相等, 且大于 $\mathrm{A}$ 的角速度 B: B、 $\mathrm{C}$ 的线速度大小相等，且大于 $\mathrm{A}$ 的线速度 C: $\mathrm{B} 、 \mathrm{C}$ 的向心加速度相等, 且小于 $\mathrm{A}$ 的向心加速度 D: $\mathrm{B} 、 \mathrm{C}$ 的周期相等，且大于 $\mathrm{A}$ 的周期

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个多选题（有多个正确答案）。 问题: 如图所示, $\mathrm{A} 、 \mathrm{~B} 、 \mathrm{C}$ 是在地球大气层外的圆形轨道上运行的三颗人造地球卫星,下列说法中正确的是（） [图1] A: B、C 的角速度相等, 且大于 $\mathrm{A}$ 的角速度 B: B、 $\mathrm{C}$ 的线速度大小相等，且大于 $\mathrm{A}$ 的线速度 C: $\mathrm{B} 、 \mathrm{C}$ 的向心加速度相等, 且小于 $\mathrm{A}$ 的向心加速度 D: $\mathrm{B} 、 \mathrm{C}$ 的周期相等，且大于 $\mathrm{A}$ 的周期 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]

MC

Astronomy

ZH

multi-modal

Astronomy_220

某双星系统由两颗恒星构成, 质量分别为 $m_{1}$ 和 $m_{2}$, 距中心的距离分别为 $r_{1}$ 和 $r_{2}$,且 $r_{1}>r_{2}$, 则下面的表述正确的是 ( ) [图1] A: 它们运转的周期相同 B: 它们的线速度大小相同 C: $m_{2}>m_{1}$ D: 它们的加速度大小相同

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个多选题（有多个正确答案）。 问题: 某双星系统由两颗恒星构成, 质量分别为 $m_{1}$ 和 $m_{2}$, 距中心的距离分别为 $r_{1}$ 和 $r_{2}$,且 $r_{1}>r_{2}$, 则下面的表述正确的是 ( ) [图1] A: 它们运转的周期相同 B: 它们的线速度大小相同 C: $m_{2}>m_{1}$ D: 它们的加速度大小相同 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]

MC

Astronomy

ZH

multi-modal

Astronomy_1098

The Event Horizon Telescope (EHT) is a project to use many widely-spaced radio telescopes as a Very Long Baseline Interferometer (VBLI) to create a virtual telescope as big as the Earth. This extraordinary size allows sufficient angular resolution to be able to image the space close to the event horizon of a super massive black hole (SMBH), and provide an opportunity to test the predictions of Einstein's theory of General Relativity (GR) in a very strong gravitational field. In April 2017 the EHT collaboration managed to co-ordinate time on all of the telescopes in the array so that they could observe the SMBH (called M87*) at the centre of the Virgo galaxy, M87, and they plan to also image the SMBH at the centre of our galaxy (called Sgr A*). [figure1] Figure 3: Left: The locations of all the telescopes used during the April 2017 observing run. The solid lines correspond to baselines used for observing M87, whilst the dashed lines were the baselines used for the calibration source. Credit: EHT Collaboration. Right: A simulated model of what the region near an SMBH could look like, modelled at much higher resolution than the EHT can achieve. The light comes from the accretion disc, but the paths of the photons are bent into a characteristic shape by the extreme gravity, leading to a 'shadow' in middle of the disc - this is what the EHT is trying to image. The left side of the image is brighter than the right side as light emitted from a substance moving towards an observer is brighter than that of one moving away. Credit: Hotaka Shiokawa. Some data about the locations of the eight telescopes in the array are given below in 3-D cartesian geocentric coordinates with $X$ pointing to the Greenwich meridian, $Y$ pointing $90^{\circ}$ away in the equatorial plane (eastern longitudes have positive $Y$ ), and positive $Z$ pointing in the direction of the North Pole. This is a left-handed coordinate system. | Facility | Location | $X(\mathrm{~m})$ | $Y(\mathrm{~m})$ | $Z(\mathrm{~m})$ | | :--- | :--- | :---: | :---: | :---: | | ALMA | Chile | 2225061.3 | -5440061.7 | -2481681.2 | | APEX | Chile | 2225039.5 | -5441197.6 | -2479303.4 | | JCMT | Hawaii, USA | -5464584.7 | -2493001.2 | 2150654.0 | | LMT | Mexico | -768715.6 | -5988507.1 | 2063354.9 | | PV | Spain | 5088967.8 | -301681.2 | 3825012.2 | | SMA | Hawaii, USA | -5464555.5 | -2492928.0 | 2150797.2 | | SMT | Arizona, USA | -1828796.2 | -5054406.8 | 3427865.2 | | SPT | Antarctica | 809.8 | -816.9 | -6359568.7 | The minimum angle, $\theta_{\min }$ (in radians) that can be resolved by a VLBI array is given by the equation $$ \theta_{\min }=\frac{\lambda_{\mathrm{obs}}}{d_{\max }}, $$ where $\lambda_{\text {obs }}$ is the observing wavelength and $d_{\max }$ is the longest straight line distance between two telescopes used (called the baseline), assumed perpendicular to the line of sight during the observation. An important length scale when discussing black holes is the gravitational radius, $r_{g}=\frac{G M}{c^{2}}$, where $G$ is the gravitational constant, $M$ is the mass of the black hole and $c$ is the speed of light. The familiar event horizon of a non-rotating black hole is called the Schwartzschild radius, $r_{S} \equiv 2 r_{g}$, however this is not what the EHT is able to observe - instead the closest it can see to a black hole is called the photon sphere, where photons orbit in the black hole in unstable circular orbits. On top of this the image of the black hole is gravitationally lensed by the black hole itself magnifying the apparent radius of the photon sphere to be between $(2 \sqrt{3+2 \sqrt{2}}) r_{g}$ and $(3 \sqrt{3}) r_{g}$, determined by spin and inclination; the latter corresponds to a perfectly spherical non-spinning black hole. The area within this lensed image will appear almost black and is the 'shadow' the EHT is looking for. [figure2] Figure 4: Four nights of data were taken for M87* during the observing window of the EHT, and whilst the diameter of the disk stayed relatively constant the location of bright spots moved, possibly indicating gas that is orbiting the black hole. Credit: EHT Collaboration. The EHT observed M87* on four separate occasions during the observing window (see Fig 4), and the team saw that some of the bright spots changed in that time, suggesting they may be associated with orbiting gas close to the black hole. The Innermost Stable Circular Orbit (ISCO) is the equivalent of the photon sphere but for particles with mass (and is also stable). The total conserved energy of a circular orbit close to a non-spinning black hole is given by $$ E=m c^{2}\left(\frac{1-\frac{2 r_{g}}{r}}{\sqrt{1-\frac{3 r_{g}}{r}}}\right) $$ and the radius of the ISCO, $r_{\mathrm{ISCO}}$, is the value of $r$ for which $E$ is minimised. We expect that most black holes are in fact spinning (since most stars are spinning) and the spin of a black hole is quantified with the spin parameter $a \equiv J / J_{\max }$ where $J$ is the angular momentum of the black hole and $J_{\max }=G M^{2} / c$ is the maximum possible angular momentum it can have. The value of $a$ varies from $-1 \leq a \leq 1$, where negative spins correspond to the black hole rotating in the opposite direction to its accretion disk, and positive spins in the same direction. If $a=1$ then $r_{\text {ISCO }}=r_{g}$, whilst if $a=-1$ then $r_{\text {ISCO }}=9 r_{g}$. The angular velocity of a particle in the ISCO is given by $$ \omega^{2}=\frac{G M}{\left(r_{\text {ISCO }}^{3 / 2}+a r_{g}^{3 / 2}\right)^{2}} $$ [figure3] Figure 5: Due to the curvature of spacetime, the real distance travelled by a particle moving from the ISCO to the photon sphere (indicated with the solid red arrow) is longer than you would get purely from subtracting the radial co-ordinates of those orbits (indicated with the dashed blue arrow), which would be valid for a flat spacetime. Relations between these distances are not to scale in this diagram. Credit: Modified from Bardeen et al. (1972). The spacetime near a black hole is curved, as described by the equations of GR. This means that the distance between two points can be substantially different to the distance you would expect if spacetime was flat. GR tells us that the proper distance travelled by a particle moving from radius $r_{1}$ to radius $r_{2}$ around a black hole of mass $M$ (with $r_{1}>r_{2}$ ) is given by $$ \Delta l=\int_{r_{2}}^{r_{1}}\left(1-\frac{2 r_{g}}{r}\right)^{-1 / 2} \mathrm{~d} r $$e. Taking the mass of $\mathrm{M}_{8} 7^{*}$ as $6.5 \times 10^{9} \mathrm{M}_{\odot}$ : ii. One of the bright patches in Fig 4 seemed to move a quarter of the way around the ring between April 5 and 10 (from the left hand side to the bottom). Could it be attributable to gas moving in the ISCO? If so, is the spin likely to be positive or negative?

You are participating in an international Astronomy competition and need to solve the following question. The answer to this question is a numerical value. Here is some context information for this question, which might assist you in solving it: The Event Horizon Telescope (EHT) is a project to use many widely-spaced radio telescopes as a Very Long Baseline Interferometer (VBLI) to create a virtual telescope as big as the Earth. This extraordinary size allows sufficient angular resolution to be able to image the space close to the event horizon of a super massive black hole (SMBH), and provide an opportunity to test the predictions of Einstein's theory of General Relativity (GR) in a very strong gravitational field. In April 2017 the EHT collaboration managed to co-ordinate time on all of the telescopes in the array so that they could observe the SMBH (called M87*) at the centre of the Virgo galaxy, M87, and they plan to also image the SMBH at the centre of our galaxy (called Sgr A*). [figure1] Figure 3: Left: The locations of all the telescopes used during the April 2017 observing run. The solid lines correspond to baselines used for observing M87, whilst the dashed lines were the baselines used for the calibration source. Credit: EHT Collaboration. Right: A simulated model of what the region near an SMBH could look like, modelled at much higher resolution than the EHT can achieve. The light comes from the accretion disc, but the paths of the photons are bent into a characteristic shape by the extreme gravity, leading to a 'shadow' in middle of the disc - this is what the EHT is trying to image. The left side of the image is brighter than the right side as light emitted from a substance moving towards an observer is brighter than that of one moving away. Credit: Hotaka Shiokawa. Some data about the locations of the eight telescopes in the array are given below in 3-D cartesian geocentric coordinates with $X$ pointing to the Greenwich meridian, $Y$ pointing $90^{\circ}$ away in the equatorial plane (eastern longitudes have positive $Y$ ), and positive $Z$ pointing in the direction of the North Pole. This is a left-handed coordinate system. | Facility | Location | $X(\mathrm{~m})$ | $Y(\mathrm{~m})$ | $Z(\mathrm{~m})$ | | :--- | :--- | :---: | :---: | :---: | | ALMA | Chile | 2225061.3 | -5440061.7 | -2481681.2 | | APEX | Chile | 2225039.5 | -5441197.6 | -2479303.4 | | JCMT | Hawaii, USA | -5464584.7 | -2493001.2 | 2150654.0 | | LMT | Mexico | -768715.6 | -5988507.1 | 2063354.9 | | PV | Spain | 5088967.8 | -301681.2 | 3825012.2 | | SMA | Hawaii, USA | -5464555.5 | -2492928.0 | 2150797.2 | | SMT | Arizona, USA | -1828796.2 | -5054406.8 | 3427865.2 | | SPT | Antarctica | 809.8 | -816.9 | -6359568.7 | The minimum angle, $\theta_{\min }$ (in radians) that can be resolved by a VLBI array is given by the equation $$ \theta_{\min }=\frac{\lambda_{\mathrm{obs}}}{d_{\max }}, $$ where $\lambda_{\text {obs }}$ is the observing wavelength and $d_{\max }$ is the longest straight line distance between two telescopes used (called the baseline), assumed perpendicular to the line of sight during the observation. An important length scale when discussing black holes is the gravitational radius, $r_{g}=\frac{G M}{c^{2}}$, where $G$ is the gravitational constant, $M$ is the mass of the black hole and $c$ is the speed of light. The familiar event horizon of a non-rotating black hole is called the Schwartzschild radius, $r_{S} \equiv 2 r_{g}$, however this is not what the EHT is able to observe - instead the closest it can see to a black hole is called the photon sphere, where photons orbit in the black hole in unstable circular orbits. On top of this the image of the black hole is gravitationally lensed by the black hole itself magnifying the apparent radius of the photon sphere to be between $(2 \sqrt{3+2 \sqrt{2}}) r_{g}$ and $(3 \sqrt{3}) r_{g}$, determined by spin and inclination; the latter corresponds to a perfectly spherical non-spinning black hole. The area within this lensed image will appear almost black and is the 'shadow' the EHT is looking for. [figure2] Figure 4: Four nights of data were taken for M87* during the observing window of the EHT, and whilst the diameter of the disk stayed relatively constant the location of bright spots moved, possibly indicating gas that is orbiting the black hole. Credit: EHT Collaboration. The EHT observed M87* on four separate occasions during the observing window (see Fig 4), and the team saw that some of the bright spots changed in that time, suggesting they may be associated with orbiting gas close to the black hole. The Innermost Stable Circular Orbit (ISCO) is the equivalent of the photon sphere but for particles with mass (and is also stable). The total conserved energy of a circular orbit close to a non-spinning black hole is given by $$ E=m c^{2}\left(\frac{1-\frac{2 r_{g}}{r}}{\sqrt{1-\frac{3 r_{g}}{r}}}\right) $$ and the radius of the ISCO, $r_{\mathrm{ISCO}}$, is the value of $r$ for which $E$ is minimised. We expect that most black holes are in fact spinning (since most stars are spinning) and the spin of a black hole is quantified with the spin parameter $a \equiv J / J_{\max }$ where $J$ is the angular momentum of the black hole and $J_{\max }=G M^{2} / c$ is the maximum possible angular momentum it can have. The value of $a$ varies from $-1 \leq a \leq 1$, where negative spins correspond to the black hole rotating in the opposite direction to its accretion disk, and positive spins in the same direction. If $a=1$ then $r_{\text {ISCO }}=r_{g}$, whilst if $a=-1$ then $r_{\text {ISCO }}=9 r_{g}$. The angular velocity of a particle in the ISCO is given by $$ \omega^{2}=\frac{G M}{\left(r_{\text {ISCO }}^{3 / 2}+a r_{g}^{3 / 2}\right)^{2}} $$ [figure3] Figure 5: Due to the curvature of spacetime, the real distance travelled by a particle moving from the ISCO to the photon sphere (indicated with the solid red arrow) is longer than you would get purely from subtracting the radial co-ordinates of those orbits (indicated with the dashed blue arrow), which would be valid for a flat spacetime. Relations between these distances are not to scale in this diagram. Credit: Modified from Bardeen et al. (1972). The spacetime near a black hole is curved, as described by the equations of GR. This means that the distance between two points can be substantially different to the distance you would expect if spacetime was flat. GR tells us that the proper distance travelled by a particle moving from radius $r_{1}$ to radius $r_{2}$ around a black hole of mass $M$ (with $r_{1}>r_{2}$ ) is given by $$ \Delta l=\int_{r_{2}}^{r_{1}}\left(1-\frac{2 r_{g}}{r}\right)^{-1 / 2} \mathrm{~d} r $$ problem: e. Taking the mass of $\mathrm{M}_{8} 7^{*}$ as $6.5 \times 10^{9} \mathrm{M}_{\odot}$ : ii. One of the bright patches in Fig 4 seemed to move a quarter of the way around the ring between April 5 and 10 (from the left hand side to the bottom). Could it be attributable to gas moving in the ISCO? If so, is the spin likely to be positive or negative? All mathematical formulas and symbols you output should be represented with LaTeX! You can solve it step by step. Remember, your answer should be calculated in the unit of days, but when concluding your final answer, do not include the unit. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER is the numerical value without any units.

NV

Astronomy

EN

multi-modal

Astronomy_532

《流浪地球 2 》中太空电梯非常吸引观众眼球。在影片中太空电梯高算入云, 在地表与太空间高速穿梭。太空电梯通过超级缆绳连接地球赤道上的固定基地与配重空间站, 它们随地球以同步静止状态一起旋转。随着物理学的发展, 我们越来越倾向于我们周围的环境。在本题中, 我们将跟随宇航员, 进入太空中探索。本题中, 若未特殊说明, 地球 半径为 $R$, 地球质量为 $M$, 引力常量为 $G$, 卫星与地球间的引力势能表达式为 $E_{\mathrm{p}}=-\frac{G M m}{r}$ ( $r$ 是卫星到地心的距离) 如图所示, 该系统的甲、乙两颗卫星的轨道分别是同一平面内的椭圆和圆, 甲的轨道近地点正好在地面附近, 已知远地点与地面的距离为 $3 R$, 甲轨道的远地点与乙轨道的最近距离为 $2 R$, 乙的质量为 $m_{0}$, 甲、乙的周期之比是多少? 为 [图1]

你正在参加一个国际天文竞赛，并需要解决以下问题。 这个问题的答案是一个数值。 问题: 《流浪地球 2 》中太空电梯非常吸引观众眼球。在影片中太空电梯高算入云, 在地表与太空间高速穿梭。太空电梯通过超级缆绳连接地球赤道上的固定基地与配重空间站, 它们随地球以同步静止状态一起旋转。随着物理学的发展, 我们越来越倾向于我们周围的环境。在本题中, 我们将跟随宇航员, 进入太空中探索。本题中, 若未特殊说明, 地球 半径为 $R$, 地球质量为 $M$, 引力常量为 $G$, 卫星与地球间的引力势能表达式为 $E_{\mathrm{p}}=-\frac{G M m}{r}$ ( $r$ 是卫星到地心的距离) 如图所示, 该系统的甲、乙两颗卫星的轨道分别是同一平面内的椭圆和圆, 甲的轨道近地点正好在地面附近, 已知远地点与地面的距离为 $3 R$, 甲轨道的远地点与乙轨道的最近距离为 $2 R$, 乙的质量为 $m_{0}$, 甲、乙的周期之比是多少? 为 [图1] 你输出的所有数学公式和符号应该使用LaTeX表示！ 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER是数值。

NV

Astronomy

ZH

multi-modal

Astronomy_1083

Recently a group of researchers announced that they had discovered an Earth-sized exoplanet around our nearest star, Proxima Centauri. Its closeness raises an intriguing possibility about whether or not we might be able to image it directly using telescopes. The difficulty comes from the small angular scales that need to be resolved and the extreme differences in brightness between the reflected light from the planet and the light given out by the star. [figure1] Figure 6: Artist's impression of the view from the surface of Proxima Centauri b. Credit: ESO / M. Kornmesser Data about the star and the planet are summarised below: | Proxima Centauri (star) | | Proxima Centauri b (planet) | | | :--- | :--- | :--- | :--- | | Distance | $1.295 \mathrm{pc}$ | Orbital period | 11.186 days | | Mass | $0.123 \mathrm{M}_{\odot}$ | Mass $(\mathrm{min})$ | $\approx 1.27 \mathrm{M}_{\oplus}$ | | Radius | $0.141 \mathrm{R}_{\odot}$ | Radius $(\mathrm{min})$ | $\approx 1.1 \mathrm{R}_{\oplus}$ | | Surface temperature | $3042 \mathrm{~K}$ | | | | Apparent magnitude | 11.13 | | | The following formulae may also be helpful: $$ m-\mathcal{M}=5 \log \left(\frac{d}{10}\right) \quad \mathcal{M}-\mathcal{M}_{\odot}=-2.5 \log \left(\frac{L}{\mathrm{~L}_{\odot}}\right) \quad \Delta m=2.5 \log C R $$ where $m$ is the apparent magnitude, $\mathcal{M}$ is the absolute magnitude, $d$ is the distance in parsecs, and the contrast ratio $(C R)$ is defined as the ratio of fluxes from the star and planet, $C R=\frac{f_{\text {star }}}{f_{\text {planet }}}$.d. Calculate the exposure time needed for a Keck II image of the exoplanet to have an SNR of 3. Assume that the telescope has perfect $A O$, is observed at the longest wavelength for which the planet can still be resolved from the star, all the received flux from the planet consists of photons of that longest wavelength , $\varepsilon=0.1$ and $b=10^{9}$ photons $s^{-1}$ (so $b>>$ ). Comment on your answer.

You are participating in an international Astronomy competition and need to solve the following question. The answer to this question is a numerical value. Here is some context information for this question, which might assist you in solving it: Recently a group of researchers announced that they had discovered an Earth-sized exoplanet around our nearest star, Proxima Centauri. Its closeness raises an intriguing possibility about whether or not we might be able to image it directly using telescopes. The difficulty comes from the small angular scales that need to be resolved and the extreme differences in brightness between the reflected light from the planet and the light given out by the star. [figure1] Figure 6: Artist's impression of the view from the surface of Proxima Centauri b. Credit: ESO / M. Kornmesser Data about the star and the planet are summarised below: | Proxima Centauri (star) | | Proxima Centauri b (planet) | | | :--- | :--- | :--- | :--- | | Distance | $1.295 \mathrm{pc}$ | Orbital period | 11.186 days | | Mass | $0.123 \mathrm{M}_{\odot}$ | Mass $(\mathrm{min})$ | $\approx 1.27 \mathrm{M}_{\oplus}$ | | Radius | $0.141 \mathrm{R}_{\odot}$ | Radius $(\mathrm{min})$ | $\approx 1.1 \mathrm{R}_{\oplus}$ | | Surface temperature | $3042 \mathrm{~K}$ | | | | Apparent magnitude | 11.13 | | | The following formulae may also be helpful: $$ m-\mathcal{M}=5 \log \left(\frac{d}{10}\right) \quad \mathcal{M}-\mathcal{M}_{\odot}=-2.5 \log \left(\frac{L}{\mathrm{~L}_{\odot}}\right) \quad \Delta m=2.5 \log C R $$ where $m$ is the apparent magnitude, $\mathcal{M}$ is the absolute magnitude, $d$ is the distance in parsecs, and the contrast ratio $(C R)$ is defined as the ratio of fluxes from the star and planet, $C R=\frac{f_{\text {star }}}{f_{\text {planet }}}$. problem: d. Calculate the exposure time needed for a Keck II image of the exoplanet to have an SNR of 3. Assume that the telescope has perfect $A O$, is observed at the longest wavelength for which the planet can still be resolved from the star, all the received flux from the planet consists of photons of that longest wavelength , $\varepsilon=0.1$ and $b=10^{9}$ photons $s^{-1}$ (so $b>>$ ). Comment on your answer. All mathematical formulas and symbols you output should be represented with LaTeX! You can solve it step by step. Remember, your answer should be calculated in the unit of s, but when concluding your final answer, do not include the unit. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER is the numerical value without any units.

NV

Astronomy

EN

multi-modal

Astronomy_1182

Some of the very first exoplanets to be discovered in large surveys were dubbed 'hot Jupiters' as they were similar in mass to Jupiter (i.e. a gas giant) but were much closer to their star than Mercury is to the Sun (and hence are in a very hot environment). Planetary formation models suggest that they were unlikely to have formed there, but instead formed much further out from the star and migrated inwards, due to gravitational interactions with other planets in the system. Studies of 'hot Jupiters' show that there is an overabundance of them with periods of $\sim 3-4$ days, and very few with periods shorter than that. Since large, close-in planets should be the easiest to detect in all of the main methods of finding exoplanets, this scarcity is likely to be a real effect and suggests that exoplanets which are that close to their star are in a relatively rapid (by astronomical standards) inspiral towards destruction by their star. [figure1] Figure 6: Left: The orbital radius of several 'hot Jupiters' scaled by the Roche radius of the system (where tidal forces would destroy the planet). There is an expected pile up close to radii double the Roche radius (dotted line), and very few with radii smaller than that - those that are will inevitably spiral into the star and be destroyed by the tidal forces when they get too close. Credit: Birkby et al. (2014). Right: As the planets inspiral we should see a shift in when their transits occur. This figure shows the predicted size of the shift after a period of 10 years if the tidal dissipation quality factor $Q_{\star}^{\prime}=10^{6}$, as well as the current detection limit of 5 seconds (dotted line). Therefore measuring if there is any shift in the transit times over the course of a decade of observations can put stringent limits on the value of $Q_{\star}^{\prime}$. Credit: Birkby et al. (2014). The Roche radius, where a planet will be torn apart due to the tidal forces acting on it, is defined as $$ a_{\text {Roche }} \approx 2.16 R_{P}\left(\frac{M_{\star}}{M_{P}}\right)^{1 / 3} $$ where $R_{P}$ is the radius of the planet, $M_{P}$ is the mass of the planet and $M_{\star}$ is the mass of the star. If a gas giant is knocked into a highly elliptical orbit (i.e. $e \approx 1$ ) that has a periapsis $r_{\text {peri }}<a_{\text {Roche }}$ then it will not survive. However, if the periapsis just grazes the Roche radius $\left(r_{\text {peri }} \approx a_{\text {Roche }}\right)$ then the orbit will rapidly circularise. By conserving angular momentum, it can be shown that the circular orbit will have a radius $a=2 a_{\text {Roche }}$ (see the left panel of Fig 6). Exoplanets observed to be in an orbit with a radius less than that will be unstable and angular momentum will be transferred from the planet to the star, causing the star to spin more rapidly and the planet's orbital radius to decrease. Eventually this will result in the planet's orbit crossing the Roche radius and being destroyed by the tidal forces. The duration of this inspiral will be dependent on how well the star can dissipate the orbital energy through frictional processes within the star, and can be parameterised by the tidal dissipation quality factor, $Q_{\star}^{\prime}$. By looking for changes in the orbital period of the planet, detectable by shifts in the timing of transits by the planet in front of the star, we can determine an estimate of $Q_{\star}^{\prime}$, which hence tells us about the internal structure of stars. These 'hot Jupiters' are the best laboratory we have for this, as they are the most likely to produce a measurable shift (i.e. $\sim 5 \mathrm{~s}$ ) in transit times within only $\sim 10$ years (see the right panel of Fig 6). We will try and reproduce these results in this question. The WTS-2 system consists of a star of mass $M_{\star}=0.820 M_{\odot}$, peak in its black-body spectrum at $\lambda_{\max }=580 \mathrm{~nm}$, and distance from us of $1.03 \mathrm{kpc}$, with an orbiting planet (called WTS-2b) with a period $P=1.0187$ days, mass of $1.12 M_{J}$ and radius $1.36 R_{J}$. The mass and radius of Jupiter are $M_{J}=1.90 \times 10^{27} \mathrm{~kg}$ and $R_{J}=7.15 \times 10^{7} \mathrm{~m}$ respectively. The change in the semi-major axis of the planet, $a$, due to tidal forces is given by $$ \left|\frac{\dot{a}}{a}\right|=6 k_{2} \Delta t \frac{M_{P}}{M_{\star}}\left(\frac{R_{\star}}{a}\right)^{5} n^{2} $$ where the dot notation is used to indicate the differential with respect to time (i.e. $\dot{a} \equiv \mathrm{d} a / \mathrm{d} t$ ), $k_{2}$ is a constant related to the density structure of the star, $\Delta t$ is the (assumed constant) time lag between where the planet is in its orbit and the location of the tidal bulge on the star, and $n=2 \pi / P$. By separating variables and integrating this equation, an expression can be derived for the time it takes for $a$ to decrease to zero. This is known as the inspiral time, $\tau$, and even though the planet will be destroyed when $a=a_{\text {Roche }}$ the time to go from $a=a_{\text {Roche }}$ to $a=0$ is negligible in comparison to the time to get to $a=a_{\text {Roche }}$, so $\tau$ is a good estimate of the remaining lifetime of the planet.c. Derive an equation for $\tau$, showing $\tau \propto a^{8}$. Hint: $n$ is a function of a due to Kepler's third law.

You are participating in an international Astronomy competition and need to solve the following question. The answer to this question is an expression. Here is some context information for this question, which might assist you in solving it: Some of the very first exoplanets to be discovered in large surveys were dubbed 'hot Jupiters' as they were similar in mass to Jupiter (i.e. a gas giant) but were much closer to their star than Mercury is to the Sun (and hence are in a very hot environment). Planetary formation models suggest that they were unlikely to have formed there, but instead formed much further out from the star and migrated inwards, due to gravitational interactions with other planets in the system. Studies of 'hot Jupiters' show that there is an overabundance of them with periods of $\sim 3-4$ days, and very few with periods shorter than that. Since large, close-in planets should be the easiest to detect in all of the main methods of finding exoplanets, this scarcity is likely to be a real effect and suggests that exoplanets which are that close to their star are in a relatively rapid (by astronomical standards) inspiral towards destruction by their star. [figure1] Figure 6: Left: The orbital radius of several 'hot Jupiters' scaled by the Roche radius of the system (where tidal forces would destroy the planet). There is an expected pile up close to radii double the Roche radius (dotted line), and very few with radii smaller than that - those that are will inevitably spiral into the star and be destroyed by the tidal forces when they get too close. Credit: Birkby et al. (2014). Right: As the planets inspiral we should see a shift in when their transits occur. This figure shows the predicted size of the shift after a period of 10 years if the tidal dissipation quality factor $Q_{\star}^{\prime}=10^{6}$, as well as the current detection limit of 5 seconds (dotted line). Therefore measuring if there is any shift in the transit times over the course of a decade of observations can put stringent limits on the value of $Q_{\star}^{\prime}$. Credit: Birkby et al. (2014). The Roche radius, where a planet will be torn apart due to the tidal forces acting on it, is defined as $$ a_{\text {Roche }} \approx 2.16 R_{P}\left(\frac{M_{\star}}{M_{P}}\right)^{1 / 3} $$ where $R_{P}$ is the radius of the planet, $M_{P}$ is the mass of the planet and $M_{\star}$ is the mass of the star. If a gas giant is knocked into a highly elliptical orbit (i.e. $e \approx 1$ ) that has a periapsis $r_{\text {peri }}<a_{\text {Roche }}$ then it will not survive. However, if the periapsis just grazes the Roche radius $\left(r_{\text {peri }} \approx a_{\text {Roche }}\right)$ then the orbit will rapidly circularise. By conserving angular momentum, it can be shown that the circular orbit will have a radius $a=2 a_{\text {Roche }}$ (see the left panel of Fig 6). Exoplanets observed to be in an orbit with a radius less than that will be unstable and angular momentum will be transferred from the planet to the star, causing the star to spin more rapidly and the planet's orbital radius to decrease. Eventually this will result in the planet's orbit crossing the Roche radius and being destroyed by the tidal forces. The duration of this inspiral will be dependent on how well the star can dissipate the orbital energy through frictional processes within the star, and can be parameterised by the tidal dissipation quality factor, $Q_{\star}^{\prime}$. By looking for changes in the orbital period of the planet, detectable by shifts in the timing of transits by the planet in front of the star, we can determine an estimate of $Q_{\star}^{\prime}$, which hence tells us about the internal structure of stars. These 'hot Jupiters' are the best laboratory we have for this, as they are the most likely to produce a measurable shift (i.e. $\sim 5 \mathrm{~s}$ ) in transit times within only $\sim 10$ years (see the right panel of Fig 6). We will try and reproduce these results in this question. The WTS-2 system consists of a star of mass $M_{\star}=0.820 M_{\odot}$, peak in its black-body spectrum at $\lambda_{\max }=580 \mathrm{~nm}$, and distance from us of $1.03 \mathrm{kpc}$, with an orbiting planet (called WTS-2b) with a period $P=1.0187$ days, mass of $1.12 M_{J}$ and radius $1.36 R_{J}$. The mass and radius of Jupiter are $M_{J}=1.90 \times 10^{27} \mathrm{~kg}$ and $R_{J}=7.15 \times 10^{7} \mathrm{~m}$ respectively. The change in the semi-major axis of the planet, $a$, due to tidal forces is given by $$ \left|\frac{\dot{a}}{a}\right|=6 k_{2} \Delta t \frac{M_{P}}{M_{\star}}\left(\frac{R_{\star}}{a}\right)^{5} n^{2} $$ where the dot notation is used to indicate the differential with respect to time (i.e. $\dot{a} \equiv \mathrm{d} a / \mathrm{d} t$ ), $k_{2}$ is a constant related to the density structure of the star, $\Delta t$ is the (assumed constant) time lag between where the planet is in its orbit and the location of the tidal bulge on the star, and $n=2 \pi / P$. By separating variables and integrating this equation, an expression can be derived for the time it takes for $a$ to decrease to zero. This is known as the inspiral time, $\tau$, and even though the planet will be destroyed when $a=a_{\text {Roche }}$ the time to go from $a=a_{\text {Roche }}$ to $a=0$ is negligible in comparison to the time to get to $a=a_{\text {Roche }}$, so $\tau$ is a good estimate of the remaining lifetime of the planet. problem: c. Derive an equation for $\tau$, showing $\tau \propto a^{8}$. Hint: $n$ is a function of a due to Kepler's third law. All mathematical formulas and symbols you output should be represented with LaTeX! You can solve it step by step. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER is an expression without equals signs, e.g. ANSWER=\frac{1}{2} g t^2

EX

Astronomy

EN

multi-modal

Astronomy_824

What is the shortest distance (along the surface of the Earth) between two points on the Equator separated by $30^{\circ}$ of longitude? What is the shortest distance (along the surface of the Earth) between them if both the two points lie on the $60^{\circ}$ latitude while still separated by $30^{\circ}$ of longitude? (For simplicity, assume that the Earth is a sphere) A: $3336 \mathrm{~km}, 1668 \mathrm{~km}$ B: $3336 \mathrm{~km}, 1654 \mathrm{~km}$ C: $6672 \mathrm{~km}, 3336 \mathrm{~km}$ D: $3298 \mathrm{~km}, 1649 \mathrm{~km}$ E: $3298 \mathrm{~km}, 1668 \mathrm{~km}$

You are participating in an international Astronomy competition and need to solve the following question. This is a multiple choice question (only one correct answer). problem: What is the shortest distance (along the surface of the Earth) between two points on the Equator separated by $30^{\circ}$ of longitude? What is the shortest distance (along the surface of the Earth) between them if both the two points lie on the $60^{\circ}$ latitude while still separated by $30^{\circ}$ of longitude? (For simplicity, assume that the Earth is a sphere) A: $3336 \mathrm{~km}, 1668 \mathrm{~km}$ B: $3336 \mathrm{~km}, 1654 \mathrm{~km}$ C: $6672 \mathrm{~km}, 3336 \mathrm{~km}$ D: $3298 \mathrm{~km}, 1649 \mathrm{~km}$ E: $3298 \mathrm{~km}, 1668 \mathrm{~km}$ You can solve it step by step. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER should be one of the options: [A, B, C, D, E].

SC

Astronomy

EN

text-only

Astronomy_262

2021 年 12 月 30 日 00 时 43 分, 我国在西昌卫星发射中心用长征三号乙运载火箭,成功将通信技术试验卫星九号发射升空, 卫星顺利进入预定轨道, 发射任务获得圆满成功。该卫星是一颗地球同步轨道通信技术试验卫星, 卫星不同部分运行的轨道半径不同,轨道半径不同的部分所受地球引力及向心力不同, 假设卫星能在圆轨道上正常运行, 且卫星是质量分布均匀的球体，则 A: 卫星接近地球部分受到的引力小于所需的向心力 B: 从卫星远离地球部分脱离的物体将做向心运动 C: 卫星接近地球部分对远离地球部分有指向地心的作用力 D: 卫星几何中心位置所处轨道高度略低于地球同步轨道高度

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个单选题（只有一个正确答案）。 问题: 2021 年 12 月 30 日 00 时 43 分, 我国在西昌卫星发射中心用长征三号乙运载火箭,成功将通信技术试验卫星九号发射升空, 卫星顺利进入预定轨道, 发射任务获得圆满成功。该卫星是一颗地球同步轨道通信技术试验卫星, 卫星不同部分运行的轨道半径不同,轨道半径不同的部分所受地球引力及向心力不同, 假设卫星能在圆轨道上正常运行, 且卫星是质量分布均匀的球体，则 A: 卫星接近地球部分受到的引力小于所需的向心力 B: 从卫星远离地球部分脱离的物体将做向心运动 C: 卫星接近地球部分对远离地球部分有指向地心的作用力 D: 卫星几何中心位置所处轨道高度略低于地球同步轨道高度 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]

SC

Astronomy

ZH

text-only

Astronomy_1105

In July 1969 the mission Apollo 11 was the first to successfully allow humans to walk on the Moon. This was an incredible achievement as the engineering necessary to make it a possibility was an order of magnitude more complex than anything that had come before. The Apollo 11 spacecraft was launched atop the Saturn V rocket, which still stands as the most powerful rocket ever made. [figure1] Figure 1: Left: The launch of Apollo 11 upon the Saturn V rocket. Credit: NASA. Right: Showing the three stages of the Saturn V rocket (each detached once its fuel was expended), plus the Apollo spacecraft on top (containing three astronauts) which was delivered into a translunar orbit. At the base of the rocket is a person to scale, emphasising the enormous size of the rocket. Credit: Encyclopaedia Britannica. | Stage | Initial Mass $(\mathrm{t})$ | Final mass $(\mathrm{t})$ | $I_{\mathrm{sp}}(\mathrm{s})$ | Burn duration $(\mathrm{s})$ | | :---: | :---: | :---: | :---: | :---: | | S-IC | 2283.9 | 135.6 | 263 | 168 | | S-II | 483.7 | 39.9 | 421 | 384 | | S-IV (Burn 1) | 121.0 | - | 421 | 147 | | S-IV (Burn 2) | - | 13.2 | 421 | 347 | | Apollo Spacecraft | 49.7 | - | - | - | Table 1: Data about each stage of the rocket used to launch the Apollo 11 spacecraft into a translunar orbit. Masses are given in tonnes $(1 \mathrm{t}=1000 \mathrm{~kg}$ ) and for convenience include the interstage parts of the rocket too. The specific impulse, $I_{\mathrm{sp}}$, of the stage is given at sea level atmospheric pressure for S-IC and for a vacuum for S-II and S-IVB. The Saturn V rocket consisted of three stages (see Fig 1), since this was the only practical way to get the Apollo spacecraft up to the speed necessary to make the transfer to the Moon. When fully fueled the mass of the total rocket was immense, and lots of that fuel was necessary to simply lift the fuel of the later stages into high altitude - in total about $3000 \mathrm{t}(1$ tonne, $\mathrm{t}=1000 \mathrm{~kg}$ ) of rocket on the launchpad was required to send about $50 \mathrm{t}$ on a mission to the Moon. The first stage (called S-IC) was the heaviest, the second (called S-II) was considerably lighter, and the third stage (called S-IVB) was fired twice - the first to get the spacecraft into a circular 'parking' orbit around the Earth where various safety checks were made, whilst the second burn was to get the spacecraft on its way to the Moon. Once each rocket stage was fully spent it was detached from the rest of the rocket before the next stage ignited. Data about each stage is given in Table 1. The thrust of the rocket is given as $$ F=-I_{\mathrm{sp}} g_{0} \dot{m} $$ where the specific impulse, $I_{\mathrm{sp}}$, of each stage is a constant related to the type of fuel used and the shape of the rocket nozzle, $g_{0}$ is the gravitational field strength of the Earth at sea level (i.e. $g_{0}=9.81 \mathrm{~m} \mathrm{~s}^{-2}$ ) and $\dot{m} \equiv \mathrm{d} m / \mathrm{d} t$ is the rate of change of mass of the rocket with time. The thrust generated by the first two stages (S-IC and S-II) can be taken to be constant. However, the thrust generated by the third stage (S-IVB) varied in order to give a constant acceleration (taken to be the same throughout both burns of the rocket). By the end of the second burn the Apollo spacecraft, apart from a few short burns to give mid-course corrections, coasted all the way to the far side of the Moon where the engines were then fired again to circularise the orbit. All of the early Apollo missions were on a orbit known as a 'free-return trajectory', meaning that if there was a problem then they were already on an orbit that would take them back to Earth after passing around the Moon. The real shape of such a trajectory (in a rotating frame of reference) is like a stretched figure of 8 and is shown in the top panel of Fig 2. To calculate this precisely is non-trivial and required substantial computing power in the 1960s. However, we can have two simplified models that can be used to estimate the duration of the translunar coast, and they are shown in the bottom panel of the Fig 2. The first is a Hohmann transfer orbit (dashed line), which is a single ellipse with the Earth at one focus. In this model the gravitational effect of the Moon is ignored, so the spacecraft travels from A (the perigee) to B (the apogee). The second (solid line) takes advantage of a 'patched conics' approach by having two ellipses whose apoapsides coincide at point $\mathrm{C}$ where the gravitational force on the spacecraft is equal from both the Earth and the Moon. The first ellipse has a periapsis at A and ignores the gravitational effect of the Moon, whilst the second ellipse has a periapsis at B and ignores the gravitational effect of the Earth. If the spacecraft trajectory and lunar orbit are coplanar and the Moon is in a circular orbit around the Earth then the time to travel from $\mathrm{A}$ to $\mathrm{B}$ via $\mathrm{C}$ is double the value attained if taking into account the gravitational forces of the Earth and Moon together throughout the journey, which is a much better estimate of the time of a real translunar coast. [figure2] Figure 2: Top: The real shape of a translunar free-return trajectory, with the Earth on the left and the Moon on the right (orbiting around the Earth in an anti-clockwise direction). This diagram (and the one below) is shown in a co-ordinate system co-rotating with the Earth and is not to scale. Credit: NASA. Bottom: Two simplified ways of modelling the translunar trajectory. The simplest is a Hohmann transfer orbit (dashed line, outer ellipse), which is an ellipse that has the Earth at one focus and ignores the gravitational effect of the Moon. A better model (solid line, inner ellipses) of the Apollo trajectory is the use of two ellipses that meet at point $\mathrm{C}$ where the gravitational forces of the Earth and Moon on the spacecraft are equal. For the Apollo 11 journey, the end of the second burn of the S-IVB rocket (point A) was $334 \mathrm{~km}$ above the surface of the Earth, and the end of the translunar coast (point B) was $161 \mathrm{~km}$ above the surface of the Moon. The distance between the centres of mass of the Earth and the Moon at the end of the translunar coast was $3.94 \times 10^{8} \mathrm{~m}$. Take the radius of the Earth to be $6370 \mathrm{~km}$, the radius of the Moon to be $1740 \mathrm{~km}$, and the mass of the Moon to be $7.35 \times 10^{22} \mathrm{~kg}$.a. Ignoring the effects of air resistance, the weight of the rocket, and assuming 1-D motion only iv. By using your graph or otherwise, work out the speed of the rocket when reaching the parking orbit.

You are participating in an international Astronomy competition and need to solve the following question. The answer to this question is a numerical value. Here is some context information for this question, which might assist you in solving it: In July 1969 the mission Apollo 11 was the first to successfully allow humans to walk on the Moon. This was an incredible achievement as the engineering necessary to make it a possibility was an order of magnitude more complex than anything that had come before. The Apollo 11 spacecraft was launched atop the Saturn V rocket, which still stands as the most powerful rocket ever made. [figure1] Figure 1: Left: The launch of Apollo 11 upon the Saturn V rocket. Credit: NASA. Right: Showing the three stages of the Saturn V rocket (each detached once its fuel was expended), plus the Apollo spacecraft on top (containing three astronauts) which was delivered into a translunar orbit. At the base of the rocket is a person to scale, emphasising the enormous size of the rocket. Credit: Encyclopaedia Britannica. | Stage | Initial Mass $(\mathrm{t})$ | Final mass $(\mathrm{t})$ | $I_{\mathrm{sp}}(\mathrm{s})$ | Burn duration $(\mathrm{s})$ | | :---: | :---: | :---: | :---: | :---: | | S-IC | 2283.9 | 135.6 | 263 | 168 | | S-II | 483.7 | 39.9 | 421 | 384 | | S-IV (Burn 1) | 121.0 | - | 421 | 147 | | S-IV (Burn 2) | - | 13.2 | 421 | 347 | | Apollo Spacecraft | 49.7 | - | - | - | Table 1: Data about each stage of the rocket used to launch the Apollo 11 spacecraft into a translunar orbit. Masses are given in tonnes $(1 \mathrm{t}=1000 \mathrm{~kg}$ ) and for convenience include the interstage parts of the rocket too. The specific impulse, $I_{\mathrm{sp}}$, of the stage is given at sea level atmospheric pressure for S-IC and for a vacuum for S-II and S-IVB. The Saturn V rocket consisted of three stages (see Fig 1), since this was the only practical way to get the Apollo spacecraft up to the speed necessary to make the transfer to the Moon. When fully fueled the mass of the total rocket was immense, and lots of that fuel was necessary to simply lift the fuel of the later stages into high altitude - in total about $3000 \mathrm{t}(1$ tonne, $\mathrm{t}=1000 \mathrm{~kg}$ ) of rocket on the launchpad was required to send about $50 \mathrm{t}$ on a mission to the Moon. The first stage (called S-IC) was the heaviest, the second (called S-II) was considerably lighter, and the third stage (called S-IVB) was fired twice - the first to get the spacecraft into a circular 'parking' orbit around the Earth where various safety checks were made, whilst the second burn was to get the spacecraft on its way to the Moon. Once each rocket stage was fully spent it was detached from the rest of the rocket before the next stage ignited. Data about each stage is given in Table 1. The thrust of the rocket is given as $$ F=-I_{\mathrm{sp}} g_{0} \dot{m} $$ where the specific impulse, $I_{\mathrm{sp}}$, of each stage is a constant related to the type of fuel used and the shape of the rocket nozzle, $g_{0}$ is the gravitational field strength of the Earth at sea level (i.e. $g_{0}=9.81 \mathrm{~m} \mathrm{~s}^{-2}$ ) and $\dot{m} \equiv \mathrm{d} m / \mathrm{d} t$ is the rate of change of mass of the rocket with time. The thrust generated by the first two stages (S-IC and S-II) can be taken to be constant. However, the thrust generated by the third stage (S-IVB) varied in order to give a constant acceleration (taken to be the same throughout both burns of the rocket). By the end of the second burn the Apollo spacecraft, apart from a few short burns to give mid-course corrections, coasted all the way to the far side of the Moon where the engines were then fired again to circularise the orbit. All of the early Apollo missions were on a orbit known as a 'free-return trajectory', meaning that if there was a problem then they were already on an orbit that would take them back to Earth after passing around the Moon. The real shape of such a trajectory (in a rotating frame of reference) is like a stretched figure of 8 and is shown in the top panel of Fig 2. To calculate this precisely is non-trivial and required substantial computing power in the 1960s. However, we can have two simplified models that can be used to estimate the duration of the translunar coast, and they are shown in the bottom panel of the Fig 2. The first is a Hohmann transfer orbit (dashed line), which is a single ellipse with the Earth at one focus. In this model the gravitational effect of the Moon is ignored, so the spacecraft travels from A (the perigee) to B (the apogee). The second (solid line) takes advantage of a 'patched conics' approach by having two ellipses whose apoapsides coincide at point $\mathrm{C}$ where the gravitational force on the spacecraft is equal from both the Earth and the Moon. The first ellipse has a periapsis at A and ignores the gravitational effect of the Moon, whilst the second ellipse has a periapsis at B and ignores the gravitational effect of the Earth. If the spacecraft trajectory and lunar orbit are coplanar and the Moon is in a circular orbit around the Earth then the time to travel from $\mathrm{A}$ to $\mathrm{B}$ via $\mathrm{C}$ is double the value attained if taking into account the gravitational forces of the Earth and Moon together throughout the journey, which is a much better estimate of the time of a real translunar coast. [figure2] Figure 2: Top: The real shape of a translunar free-return trajectory, with the Earth on the left and the Moon on the right (orbiting around the Earth in an anti-clockwise direction). This diagram (and the one below) is shown in a co-ordinate system co-rotating with the Earth and is not to scale. Credit: NASA. Bottom: Two simplified ways of modelling the translunar trajectory. The simplest is a Hohmann transfer orbit (dashed line, outer ellipse), which is an ellipse that has the Earth at one focus and ignores the gravitational effect of the Moon. A better model (solid line, inner ellipses) of the Apollo trajectory is the use of two ellipses that meet at point $\mathrm{C}$ where the gravitational forces of the Earth and Moon on the spacecraft are equal. For the Apollo 11 journey, the end of the second burn of the S-IVB rocket (point A) was $334 \mathrm{~km}$ above the surface of the Earth, and the end of the translunar coast (point B) was $161 \mathrm{~km}$ above the surface of the Moon. The distance between the centres of mass of the Earth and the Moon at the end of the translunar coast was $3.94 \times 10^{8} \mathrm{~m}$. Take the radius of the Earth to be $6370 \mathrm{~km}$, the radius of the Moon to be $1740 \mathrm{~km}$, and the mass of the Moon to be $7.35 \times 10^{22} \mathrm{~kg}$. problem: a. Ignoring the effects of air resistance, the weight of the rocket, and assuming 1-D motion only iv. By using your graph or otherwise, work out the speed of the rocket when reaching the parking orbit. All mathematical formulas and symbols you output should be represented with LaTeX! You can solve it step by step. Remember, your answer should be calculated in the unit of \mathrm{~m} \mathrm{~s}^{-1}, but when concluding your final answer, do not include the unit. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER is the numerical value without any units.

NV

Astronomy

EN

multi-modal

Astronomy_1209

Plotting the position of the Sun in the sky at the same time every day, you get an interesting figure-ofeight shape known as an analemma (see Figure 1). For observers in the Northern hemisphere, you might expect to always see the Sun due South at midday, however on some days the Sun has already passed through that bearing and on others it needs a few more minutes before it gets there. This is due to two effects: the axial tilt of the Earth, and the fact the Earth's orbit is not perfectly circular [figure1] Figure 1: The analemma above was composed from images taken every few days at noon near the village of Callanish in the Outer Hebrides in Scotland. In the foreground are the Callanish Stones and the main photo was taken on the winter solstice (when the maximum angle the Sun reaches above the horizon is the lowest of the year, so is at the bottom of the analemma). Credit: Giuseppe Petricca. The vertical co-ordinate of a point in the analemma is entirely determined by the Earth's axial tilt. This is known as the solar declination, $\delta$, and varies sinusoidally throughout the year. The horizontal coordinate of a point in the analemma is determined by a combination of the Earth's axial tilt and the eccentricity of the Earth's orbit. Both of these individually vary sinusoidally, but the superposition of the two is no longer sinusoidal. We will define $\alpha$ as the angle between due South and the Sun at local midday as seen from Oxford, where a positive value means the Sun has already passed through due South (so is on the right of the figure above) whilst a negative value means the Sun has yet to pass through due South. If $\alpha_{\text {tilt }}$ is the contribution due to the axial tilt and $\alpha_{\text {ecc }}$ is the contribution due to the Earth's orbital eccentricity, then $$ \alpha=\alpha_{\text {tilt }}+\alpha_{\text {ecc }} $$ If the angle of the axial tilt is $\varepsilon$ and the eccentricity of the Earth's orbit is $e$, and we assume that both are small enough that the sinusoidal approximation of $\delta, \alpha_{\text {tilt }}$, and $\alpha_{\text {ecc }}$ apply, then we find the following boundary conditions: - $\delta$ has a period of 1 year, an amplitude of $\varepsilon$, is maximum at the summer solstice (21 $21^{\text {st }}$ June) and minimum at the winter solstice $\left(21^{\text {st }}\right.$ December $)$ - $\alpha_{\text {tilt }}$ has a period of 0.5 years, an amplitude (in radians) of $\tan ^{2}(\varepsilon / 2)$, is zero at the solstices and the equinoxes (vernal equinox $=21^{\text {st }}$ March, autumnal equinox $=21^{\text {st }}$ September), and (using our sign convention) positive just after the vernal equinox - $\alpha_{\text {ecc }}$ has a period of 1 year, an amplitude (in radians) of $2 e$, is zero at the perihelion (4 $4^{\text {th }}$ January) and the aphelion ( $6^{\text {th }}$ July), and (using our sign convention) negative just after the perihelion Given the $n^{\text {th }}$ day of the year, a value can be calculated for $\delta$ and $\alpha$, and these are the co-ordinates for the analemma (it is drawn by these parametric equations). For the Earth, $\varepsilon=23.44^{\circ}$ and $e=0.0167$. Consider an alternative version of Earth, known as Earth 2.0. On this planet, the year is unchanged and the perihelion and aphelion are at the same time, but it has a different axial tilt, a different orbital eccentricity, and a different month for the vernal equinox (although it is still on the $21^{\text {st }}$ day of that month). The analemma as viewed from Earth 2.0 is show in Figure 2 below. [figure2] Figure 2: The analemma of the Sun at midday as seen by an observer on Earth 2.0. In this situation, $\alpha$ ranges from -26 mins 47 secs to 18 mins 56 secs. The circled letters correspond to the same (unknown) day of each month (for example $5^{\text {th }}$ Jan, $5^{\text {th }}$ Feb, $5^{\text {th }}$ March etc.). Credit: Bob Urschel.c. Using your graph to guide you to the relevant point of the year, but using your precise algebraic expressions (rather than reading off the graph): ii. What are the dates when the Sun is due South at midday?

You are participating in an international Astronomy competition and need to solve the following question. The answer to this question is an expression. Here is some context information for this question, which might assist you in solving it: Plotting the position of the Sun in the sky at the same time every day, you get an interesting figure-ofeight shape known as an analemma (see Figure 1). For observers in the Northern hemisphere, you might expect to always see the Sun due South at midday, however on some days the Sun has already passed through that bearing and on others it needs a few more minutes before it gets there. This is due to two effects: the axial tilt of the Earth, and the fact the Earth's orbit is not perfectly circular [figure1] Figure 1: The analemma above was composed from images taken every few days at noon near the village of Callanish in the Outer Hebrides in Scotland. In the foreground are the Callanish Stones and the main photo was taken on the winter solstice (when the maximum angle the Sun reaches above the horizon is the lowest of the year, so is at the bottom of the analemma). Credit: Giuseppe Petricca. The vertical co-ordinate of a point in the analemma is entirely determined by the Earth's axial tilt. This is known as the solar declination, $\delta$, and varies sinusoidally throughout the year. The horizontal coordinate of a point in the analemma is determined by a combination of the Earth's axial tilt and the eccentricity of the Earth's orbit. Both of these individually vary sinusoidally, but the superposition of the two is no longer sinusoidal. We will define $\alpha$ as the angle between due South and the Sun at local midday as seen from Oxford, where a positive value means the Sun has already passed through due South (so is on the right of the figure above) whilst a negative value means the Sun has yet to pass through due South. If $\alpha_{\text {tilt }}$ is the contribution due to the axial tilt and $\alpha_{\text {ecc }}$ is the contribution due to the Earth's orbital eccentricity, then $$ \alpha=\alpha_{\text {tilt }}+\alpha_{\text {ecc }} $$ If the angle of the axial tilt is $\varepsilon$ and the eccentricity of the Earth's orbit is $e$, and we assume that both are small enough that the sinusoidal approximation of $\delta, \alpha_{\text {tilt }}$, and $\alpha_{\text {ecc }}$ apply, then we find the following boundary conditions: - $\delta$ has a period of 1 year, an amplitude of $\varepsilon$, is maximum at the summer solstice (21 $21^{\text {st }}$ June) and minimum at the winter solstice $\left(21^{\text {st }}\right.$ December $)$ - $\alpha_{\text {tilt }}$ has a period of 0.5 years, an amplitude (in radians) of $\tan ^{2}(\varepsilon / 2)$, is zero at the solstices and the equinoxes (vernal equinox $=21^{\text {st }}$ March, autumnal equinox $=21^{\text {st }}$ September), and (using our sign convention) positive just after the vernal equinox - $\alpha_{\text {ecc }}$ has a period of 1 year, an amplitude (in radians) of $2 e$, is zero at the perihelion (4 $4^{\text {th }}$ January) and the aphelion ( $6^{\text {th }}$ July), and (using our sign convention) negative just after the perihelion Given the $n^{\text {th }}$ day of the year, a value can be calculated for $\delta$ and $\alpha$, and these are the co-ordinates for the analemma (it is drawn by these parametric equations). For the Earth, $\varepsilon=23.44^{\circ}$ and $e=0.0167$. Consider an alternative version of Earth, known as Earth 2.0. On this planet, the year is unchanged and the perihelion and aphelion are at the same time, but it has a different axial tilt, a different orbital eccentricity, and a different month for the vernal equinox (although it is still on the $21^{\text {st }}$ day of that month). The analemma as viewed from Earth 2.0 is show in Figure 2 below. [figure2] Figure 2: The analemma of the Sun at midday as seen by an observer on Earth 2.0. In this situation, $\alpha$ ranges from -26 mins 47 secs to 18 mins 56 secs. The circled letters correspond to the same (unknown) day of each month (for example $5^{\text {th }}$ Jan, $5^{\text {th }}$ Feb, $5^{\text {th }}$ March etc.). Credit: Bob Urschel. problem: c. Using your graph to guide you to the relevant point of the year, but using your precise algebraic expressions (rather than reading off the graph): ii. What are the dates when the Sun is due South at midday? All mathematical formulas and symbols you output should be represented with LaTeX! You can solve it step by step. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER is an expression without equals signs, e.g. ANSWER=\frac{1}{2} g t^2

EX

Astronomy

EN

multi-modal

Astronomy_803

How far away is the horizon from you, when your height is about $5 \mathrm{ft} 11 \mathrm{in}(1.8 \mathrm{~m})$ ? You are standing on a plain that has no mountains near you and the elevation is Oft. Ignore the effect of atmospheric refraction and the oblateness of the Earth. A: $2.4 \mathrm{~km}$ B: $4.8 \mathrm{~km}$ C: $24 \mathrm{~km}$ D: $48 \mathrm{~km}$ E: $240 \mathrm{~km}$

You are participating in an international Astronomy competition and need to solve the following question. This is a multiple choice question (only one correct answer). problem: How far away is the horizon from you, when your height is about $5 \mathrm{ft} 11 \mathrm{in}(1.8 \mathrm{~m})$ ? You are standing on a plain that has no mountains near you and the elevation is Oft. Ignore the effect of atmospheric refraction and the oblateness of the Earth. A: $2.4 \mathrm{~km}$ B: $4.8 \mathrm{~km}$ C: $24 \mathrm{~km}$ D: $48 \mathrm{~km}$ E: $240 \mathrm{~km}$ You can solve it step by step. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER should be one of the options: [A, B, C, D, E].

SC

Astronomy

EN

text-only

Astronomy_271

宇宙间存在一些离其他恒星较远的三星系统, 其中有一种三星系统如图所示, 三颗质量为 $m$ 的星球位于等边三角形的三个顶点上, 任意两颗星球的距离均为 $L$, 并绕其中心 $O$ 做匀速圆周运动. 忽略其他星球对它们的引力作用, 引力常量为 $G$, 以下对该三星系统的说法正确的是 ( ) [图1] A: 每颗星球做圆周运动的半径都等于 $L$ B: 每颗星球做圆周运动的加速度与星球的质量无关 C: 每颗星球做圆周运动的线速度 $v=\sqrt{\frac{G m}{L}}$ D: 每颗星球做圆周运动的周期为 $T=2 \pi L \sqrt{\frac{L}{G m}}$

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个单选题（只有一个正确答案）。 问题: 宇宙间存在一些离其他恒星较远的三星系统, 其中有一种三星系统如图所示, 三颗质量为 $m$ 的星球位于等边三角形的三个顶点上, 任意两颗星球的距离均为 $L$, 并绕其中心 $O$ 做匀速圆周运动. 忽略其他星球对它们的引力作用, 引力常量为 $G$, 以下对该三星系统的说法正确的是 ( ) [图1] A: 每颗星球做圆周运动的半径都等于 $L$ B: 每颗星球做圆周运动的加速度与星球的质量无关 C: 每颗星球做圆周运动的线速度 $v=\sqrt{\frac{G m}{L}}$ D: 每颗星球做圆周运动的周期为 $T=2 \pi L \sqrt{\frac{L}{G m}}$ 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]

SC

Astronomy

ZH

multi-modal

Astronomy_370

如图所示, 半径为 $R$ 的薄球壳质量均匀分布, 其单位面积的质量为 $\rho_{0}$, 已知引力常量为 $G$ 。 设球形空腔体积为 $V$, 球心深度为 $d$ (远小于地球半径), $\overline{P Q}=x$, 求空腔所引起的 $Q$点处的重力加速度反常 $\Delta g^{\prime}$ 的大小。 [图1]

你正在参加一个国际天文竞赛，并需要解决以下问题。 这个问题的答案是一个表达式。 问题: 如图所示, 半径为 $R$ 的薄球壳质量均匀分布, 其单位面积的质量为 $\rho_{0}$, 已知引力常量为 $G$ 。 设球形空腔体积为 $V$, 球心深度为 $d$ (远小于地球半径), $\overline{P Q}=x$, 求空腔所引起的 $Q$点处的重力加速度反常 $\Delta g^{\prime}$ 的大小。 [图1] 你输出的所有数学公式和符号应该使用LaTeX表示！ 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER是一个不含等号的表达式，例如ANSWER=\frac{1}{2} g t^2

EX

Astronomy

ZH

multi-modal

Astronomy_371

已知某卫星在赤道上空轨道半径为 $r_{1}$ 的圆形轨道上绕地运行的周期为 $T$, 卫星运动方向与地球自转方向相同, 赤道上某城市的人每三天恰好五次看到卫星掠过其正上 方. 假设某时刻, 该卫星如图在 $A$ 点变轨进入椭圆轨道, 近地点 $B$ 到地心距离为 $r_{2}$. 设卫星由 $A$ 到 $B$ 运动的时间为 $t$, 地球自转周期为 $T_{0}$, 不计空气阻力. 则 [图1] A: $T=\frac{3}{5} T_{0}$ B: $t=\frac{\left(r_{1}+r_{2}\right) T}{4 r_{1}} \sqrt{\frac{r_{1}+r_{2}}{2 r_{1}}}$ C: 卫星在图中椭圆轨道由 $A$ 到 $B$ 时, 机械能不变 D: 卫星由图中圆轨道进入粗圆轨道过程中，机械能不变

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个多选题（有多个正确答案）。 问题: 已知某卫星在赤道上空轨道半径为 $r_{1}$ 的圆形轨道上绕地运行的周期为 $T$, 卫星运动方向与地球自转方向相同, 赤道上某城市的人每三天恰好五次看到卫星掠过其正上 方. 假设某时刻, 该卫星如图在 $A$ 点变轨进入椭圆轨道, 近地点 $B$ 到地心距离为 $r_{2}$. 设卫星由 $A$ 到 $B$ 运动的时间为 $t$, 地球自转周期为 $T_{0}$, 不计空气阻力. 则 [图1] A: $T=\frac{3}{5} T_{0}$ B: $t=\frac{\left(r_{1}+r_{2}\right) T}{4 r_{1}} \sqrt{\frac{r_{1}+r_{2}}{2 r_{1}}}$ C: 卫星在图中椭圆轨道由 $A$ 到 $B$ 时, 机械能不变 D: 卫星由图中圆轨道进入粗圆轨道过程中，机械能不变 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]

MC

Astronomy

ZH

multi-modal

Astronomy_880

A comet called "SMukherjee2017" which has physical values of $e=0.12, a=4 \mathrm{AU}$ was visible from the Earth in 2017. In which year is this comet visible again? A: 2021 B: 2022 C: 2023 D: 2024 E: 2025

You are participating in an international Astronomy competition and need to solve the following question. This is a multiple choice question (only one correct answer). problem: A comet called "SMukherjee2017" which has physical values of $e=0.12, a=4 \mathrm{AU}$ was visible from the Earth in 2017. In which year is this comet visible again? A: 2021 B: 2022 C: 2023 D: 2024 E: 2025 You can solve it step by step. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER should be one of the options: [A, B, C, D, E].

SC

Astronomy

EN

text-only

Astronomy_1057

Wolf-Rayet (WR) stars are some of the hottest stars known, with very strong stellar winds causing considerable mass to the be lost to the interstellar medium (ISM). In binary systems between a WR star and a very large $\mathrm{O}$ or $\mathrm{B}$ spectral class star, where their strong stellar winds collide can create the conditions for the formation of dust which goes on to enrich the ISM. [figure1] Figure 3: Left: A view of the WR140 binary system taken with the James Webb Space Telescope (JWST) in July 2022, showing clearly at least 17 nested dust shells. Credit: NASA/ESA/CSA/STScI/JPL-Caltech. Right: A radial plot along the image in the three mid infrared JWST filters used corresponding to 7.7, 15 and $21 \mu \mathrm{m}$, as well as the model of the dust production. The peaks correspond to each shell. Shells 2 and 17 are indicated on the model and the median shell separation is shown by the grey vertical lines. The projected distance is given in arcseconds. Credit: Lau et al. (2022). The WR140 system consists of a WR and an O star which produce dust very regularly when the two stars are close together, around periastron. They are in a highly elliptical orbit $(e=0.8993)$ with a period of 2895 days. Once far from the stars, these dust shells move through space at a remarkably constant speed as indicated by the regularity of the shells in the recent image taken with the James Webb Space Telescope (JWST), shown above in Figure 3. An artist's impression of the two stars in the system and the orbit (in the reference frame of the WR star) is shown in Figure 4 below. [figure2] Figure 4: Left: The relative size of the Sun, upper left, compared to the two stars in the system WR140. The O-type star is $\sim 30 M_{\odot}$, while its companion is $\sim 10 M_{\odot}$. Credit: NASA/JPL-Caltech. Right: The projected orbital configuration of WR 140 in the reference frame of the WR star. The red solid region around the periastron passage is where the O star is when dust is being formed. Credit: Lau et al. (2022).b. In the Han et al. (2022) model they use $r_{\text {inner }}=50$ au, $r_{\text {outer }}=220$ au, and assume spherical dust grains with a radius of $40 \mathrm{~nm}$ and density $1.6 \mathrm{~g} \mathrm{~cm}^{-3}$, as well as a dust-to-gas ratio of 0.019 . Take the apparent magnitude of the system (after correcting for substantial extinction by the dust shells and the general ISM) as $m=0.426$, and you are given the absolute magnitude of the Sun is $\mathcal{M}_{\odot}=4.74$. ii. Hence, sketch a(r) from $0 \leq r \leq 2500$ au.

You are participating in an international Astronomy competition and need to solve the following question. The answer to this question is a range interval. Here is some context information for this question, which might assist you in solving it: Wolf-Rayet (WR) stars are some of the hottest stars known, with very strong stellar winds causing considerable mass to the be lost to the interstellar medium (ISM). In binary systems between a WR star and a very large $\mathrm{O}$ or $\mathrm{B}$ spectral class star, where their strong stellar winds collide can create the conditions for the formation of dust which goes on to enrich the ISM. [figure1] Figure 3: Left: A view of the WR140 binary system taken with the James Webb Space Telescope (JWST) in July 2022, showing clearly at least 17 nested dust shells. Credit: NASA/ESA/CSA/STScI/JPL-Caltech. Right: A radial plot along the image in the three mid infrared JWST filters used corresponding to 7.7, 15 and $21 \mu \mathrm{m}$, as well as the model of the dust production. The peaks correspond to each shell. Shells 2 and 17 are indicated on the model and the median shell separation is shown by the grey vertical lines. The projected distance is given in arcseconds. Credit: Lau et al. (2022). The WR140 system consists of a WR and an O star which produce dust very regularly when the two stars are close together, around periastron. They are in a highly elliptical orbit $(e=0.8993)$ with a period of 2895 days. Once far from the stars, these dust shells move through space at a remarkably constant speed as indicated by the regularity of the shells in the recent image taken with the James Webb Space Telescope (JWST), shown above in Figure 3. An artist's impression of the two stars in the system and the orbit (in the reference frame of the WR star) is shown in Figure 4 below. [figure2] Figure 4: Left: The relative size of the Sun, upper left, compared to the two stars in the system WR140. The O-type star is $\sim 30 M_{\odot}$, while its companion is $\sim 10 M_{\odot}$. Credit: NASA/JPL-Caltech. Right: The projected orbital configuration of WR 140 in the reference frame of the WR star. The red solid region around the periastron passage is where the O star is when dust is being formed. Credit: Lau et al. (2022). problem: b. In the Han et al. (2022) model they use $r_{\text {inner }}=50$ au, $r_{\text {outer }}=220$ au, and assume spherical dust grains with a radius of $40 \mathrm{~nm}$ and density $1.6 \mathrm{~g} \mathrm{~cm}^{-3}$, as well as a dust-to-gas ratio of 0.019 . Take the apparent magnitude of the system (after correcting for substantial extinction by the dust shells and the general ISM) as $m=0.426$, and you are given the absolute magnitude of the Sun is $\mathcal{M}_{\odot}=4.74$. ii. Hence, sketch a(r) from $0 \leq r \leq 2500$ au. All mathematical formulas and symbols you output should be represented with LaTeX! You can solve it step by step. Remember, your answer should be calculated in the unit of au, but when concluding your final answer, do not include the unit. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER is an interval without any units, e.g. ANSWER=(1,2] \cup[7,+\infty)

IN

Astronomy

EN

multi-modal

Astronomy_12

我国北斗中圆地球轨道卫星, 轨道离地高度 $21500 \mathrm{~km}$, 美国 GPS 导航卫星在轨的运行周期约为 12 小时。已知地球同步卫星离地高度约 $36000 \mathrm{~km}$, 地球的半径为 $6400 \mathrm{~km}$, 若北斗中圆地球轨道卫星和美国 GPS 导航卫星的质量相同, 北斗中圆地球轨道卫星在轨运行时的速度大小为 $v_{1}$, 美国 GPS 导航卫星在轨运行时的速度大小为 $v_{2}$, （已知 $\sqrt[3]{4} \approx 1.6$ ), 则 $\frac{v_{1}}{v_{2}}$ 约为 ( ) A: 0.81 B: 0.98 C: 1.11 D: 2.03

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个单选题（只有一个正确答案）。 问题: 我国北斗中圆地球轨道卫星, 轨道离地高度 $21500 \mathrm{~km}$, 美国 GPS 导航卫星在轨的运行周期约为 12 小时。已知地球同步卫星离地高度约 $36000 \mathrm{~km}$, 地球的半径为 $6400 \mathrm{~km}$, 若北斗中圆地球轨道卫星和美国 GPS 导航卫星的质量相同, 北斗中圆地球轨道卫星在轨运行时的速度大小为 $v_{1}$, 美国 GPS 导航卫星在轨运行时的速度大小为 $v_{2}$, （已知 $\sqrt[3]{4} \approx 1.6$ ), 则 $\frac{v_{1}}{v_{2}}$ 约为 ( ) A: 0.81 B: 0.98 C: 1.11 D: 2.03 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]

SC

Astronomy

ZH

text-only

Astronomy_641

地球半径约为 $6400 \mathrm{~km}$, 地球表面的大气随海拔高度增加而变薄, 大气压强也随之减小到零, 海拔 $100 \mathrm{~km}$ 的高度被定义为卡门线, 为大气层与太空的分界线。有人设想给太空飞船安装“太阳帆”, 用太阳光的“光子流”为飞船提供动力来实现星际旅行。已知在卡门线附近, 一个正对太阳光、面积为 $1.0 \times 10^{6} \mathrm{~m}^{2}$ 的平整光亮表面, 受到光的压力约为 $9 \mathrm{~N}$; 力虽小, 但假设以同样材料做成面积为 $1.0 \times 10^{4} \mathrm{~m}^{2}$ 的 “帆”安装在飞船上, 若只在光压作用下，从卡门线附近出发，一个月后飞船的速度可达到 2 倍声速。设想实际中有一艘安装了“帆” (面积为 $1.0 \times 10^{4} \mathrm{~m}^{2}$ ) 的飞船, 在卡门线上正对太阳光, 下列说法正确的是 ( ) A: 飞船无需其他动力, 即可不断远离太阳 B: 一年后, 飞船的速度将达到 24 倍声速 C: 与太阳中心的距离为日地间距离 2 倍时, “帆”上的压力约为 $2.25 \times 10^{-2} \mathrm{~N}$ D: 与太阳中心的距离为日地间距离 2 倍时, 飞船的加速度为出发时的 $\frac{1}{4}$

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个单选题（只有一个正确答案）。 问题: 地球半径约为 $6400 \mathrm{~km}$, 地球表面的大气随海拔高度增加而变薄, 大气压强也随之减小到零, 海拔 $100 \mathrm{~km}$ 的高度被定义为卡门线, 为大气层与太空的分界线。有人设想给太空飞船安装“太阳帆”, 用太阳光的“光子流”为飞船提供动力来实现星际旅行。已知在卡门线附近, 一个正对太阳光、面积为 $1.0 \times 10^{6} \mathrm{~m}^{2}$ 的平整光亮表面, 受到光的压力约为 $9 \mathrm{~N}$; 力虽小, 但假设以同样材料做成面积为 $1.0 \times 10^{4} \mathrm{~m}^{2}$ 的 “帆”安装在飞船上, 若只在光压作用下，从卡门线附近出发，一个月后飞船的速度可达到 2 倍声速。设想实际中有一艘安装了“帆” (面积为 $1.0 \times 10^{4} \mathrm{~m}^{2}$ ) 的飞船, 在卡门线上正对太阳光, 下列说法正确的是 ( ) A: 飞船无需其他动力, 即可不断远离太阳 B: 一年后, 飞船的速度将达到 24 倍声速 C: 与太阳中心的距离为日地间距离 2 倍时, “帆”上的压力约为 $2.25 \times 10^{-2} \mathrm{~N}$ D: 与太阳中心的距离为日地间距离 2 倍时, 飞船的加速度为出发时的 $\frac{1}{4}$ 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]

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Astronomy_447

“天问一号”的发射为人类探索火星的奥秘打下了坚实的基础, 已知地球与火星的半 径之比为 $a$, 地球与火星的质量之比为 $b$, 地球表面的重力加速度为 $g$ 。现将一汽缸开口向上放在火星表面的水平面上, 质量均为 $m$ 的活塞 $P 、 Q$ 将一定质量的气体密封且分成甲、乙两部分, 活塞 $P 、 Q$ 导热性能良好。已知火星表面的大气压为 $p_{0}$, 温度为 $T_{0}$,活塞的截面积为 $S$, 当系统平衡时, 甲、乙两部分气体的长度均为 $l_{0}, \frac{m a^{2} g}{b}=p_{0} S$, 外界环境温度保持不变。现在活塞 $P$ 上缓慢地添加质量为 $2 \mathrm{~m}$ 的砝码, 经过一段时间系统再次达到平衡。求: 再次平衡时, 活塞 $Q$ 向下移动的距离为多少? [图1]

你正在参加一个国际天文竞赛，并需要解决以下问题。 这个问题的答案是一个表达式。 问题: “天问一号”的发射为人类探索火星的奥秘打下了坚实的基础, 已知地球与火星的半 径之比为 $a$, 地球与火星的质量之比为 $b$, 地球表面的重力加速度为 $g$ 。现将一汽缸开口向上放在火星表面的水平面上, 质量均为 $m$ 的活塞 $P 、 Q$ 将一定质量的气体密封且分成甲、乙两部分, 活塞 $P 、 Q$ 导热性能良好。已知火星表面的大气压为 $p_{0}$, 温度为 $T_{0}$,活塞的截面积为 $S$, 当系统平衡时, 甲、乙两部分气体的长度均为 $l_{0}, \frac{m a^{2} g}{b}=p_{0} S$, 外界环境温度保持不变。现在活塞 $P$ 上缓慢地添加质量为 $2 \mathrm{~m}$ 的砝码, 经过一段时间系统再次达到平衡。求: 再次平衡时, 活塞 $Q$ 向下移动的距离为多少? [图1] 你输出的所有数学公式和符号应该使用LaTeX表示！ 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER是一个不含等号的表达式，例如ANSWER=\frac{1}{2} g t^2

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Astronomy_236

2015 年 9 月 14 日, 美国的 LIGO 探测设施接收到一个来自 GW150914 的引力波信号, 此信号是由两个黑洞的合并过程产生的。如果将某个双黑洞系统简化为如图所示的圆周运动模型, 两黑洞绕 $O$ 点做匀速圆周运动。在相互强大的引力作用下, 两黑洞间的距离逐渐减小, 在此过程中, 两黑洞做圆周运动的（） [图1] A: 向心力均逐渐减小 B: 线速度均逐渐减小 C: 周期均不变 D: 角速度均逐渐增大

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个单选题（只有一个正确答案）。 问题: 2015 年 9 月 14 日, 美国的 LIGO 探测设施接收到一个来自 GW150914 的引力波信号, 此信号是由两个黑洞的合并过程产生的。如果将某个双黑洞系统简化为如图所示的圆周运动模型, 两黑洞绕 $O$ 点做匀速圆周运动。在相互强大的引力作用下, 两黑洞间的距离逐渐减小, 在此过程中, 两黑洞做圆周运动的（） [图1] A: 向心力均逐渐减小 B: 线速度均逐渐减小 C: 周期均不变 D: 角速度均逐渐增大 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]

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Astronomy_104

半径为 $R$ 、密度为 $\rho$ 的球内部有半径为 $r(r<\mathrm{R})$ 的球形空腔, 空腔中心位于离球心 $s$ 处 (如图)。质量为 $m$ 的质点离球心距离为 $l$, 如果三角形 $A B C$ 是直角三角形:$\angle A C B$ 为直角时, 求该质点被多大力吸向球; [图1]

你正在参加一个国际天文竞赛，并需要解决以下问题。 这个问题的答案是一个表达式。 问题: 半径为 $R$ 、密度为 $\rho$ 的球内部有半径为 $r(r<\mathrm{R})$ 的球形空腔, 空腔中心位于离球心 $s$ 处 (如图)。质量为 $m$ 的质点离球心距离为 $l$, 如果三角形 $A B C$ 是直角三角形:$\angle A C B$ 为直角时, 求该质点被多大力吸向球; [图1] 你输出的所有数学公式和符号应该使用LaTeX表示！ 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER是一个不含等号的表达式，例如ANSWER=\frac{1}{2} g t^2

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Astronomy_244

2021 年 5 月, “天问一号”探测器成功在火星软着陆, 我国成为世界上第一个首次探测火星就实现“绕、落、巡”三项任务的国家。 (1) 为了简化问题, 可以认为地球和火星在同一平面上绕太阳做匀速圆周运动, 如图 1 所示。已知地球的公转周期为 $T_{1}$, 火星的公转周期为 $T_{2}$ 已知地球公转轨道半径为 $r_{1}$, 求火星公转轨道半径 $r_{2}$ 。

你正在参加一个国际天文竞赛，并需要解决以下问题。 这个问题的答案是一个表达式。 问题: 2021 年 5 月, “天问一号”探测器成功在火星软着陆, 我国成为世界上第一个首次探测火星就实现“绕、落、巡”三项任务的国家。 (1) 为了简化问题, 可以认为地球和火星在同一平面上绕太阳做匀速圆周运动, 如图 1 所示。已知地球的公转周期为 $T_{1}$, 火星的公转周期为 $T_{2}$ 已知地球公转轨道半径为 $r_{1}$, 求火星公转轨道半径 $r_{2}$ 。 你输出的所有数学公式和符号应该使用LaTeX表示！ 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER是一个不含等号的表达式，例如ANSWER=\frac{1}{2} g t^2

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Astronomy_76

两颗靠得很近的天体称为双星, 它们以两者连线上的某点为圆心做匀速圆周运动,则下列说法正确的是 A: 它们做匀速圆周运动的半径与其质量成正比 B: 它们做匀速圆周运动的向心加速度大小相等 C: 它们做匀速圆周运动的向心力大小相等 D: 它们做匀速圆周运动的周期相等

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个多选题（有多个正确答案）。 问题: 两颗靠得很近的天体称为双星, 它们以两者连线上的某点为圆心做匀速圆周运动,则下列说法正确的是 A: 它们做匀速圆周运动的半径与其质量成正比 B: 它们做匀速圆周运动的向心加速度大小相等 C: 它们做匀速圆周运动的向心力大小相等 D: 它们做匀速圆周运动的周期相等 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]

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Astronomy_746

Keplers 2nd law states that ... A: the orbit of a planet is an ellipse. B: $d A / d t$ is constant. C: $a^{3} / T^{2}$ is constant. D: $a^{2} / T^{3}$ is constant.

You are participating in an international Astronomy competition and need to solve the following question. This is a multiple choice question (only one correct answer). problem: Keplers 2nd law states that ... A: the orbit of a planet is an ellipse. B: $d A / d t$ is constant. C: $a^{3} / T^{2}$ is constant. D: $a^{2} / T^{3}$ is constant. You can solve it step by step. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER should be one of the options: [A, B, C, D].

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Astronomy_316

中国行星探测任务名称为“天问系列”, 首次火星探测任务被命名为“天问一号”。若已知“天问一号”探测器在距离火星中心为 $r_{1}$ 的轨道上做匀速圆周运动, 其周期为 $T_{1}$ 。火星半径为 $R_{0}$, 自转周期为 $T_{0}$ 。引力常量为 $G$, 下列说法正确的是 ( ) A: 火星的质量为 $\frac{4 \pi^{2} R_{0}^{2}}{G T_{0}^{2}}$ B: 火星表面两极的重力加速度为 $\frac{4 \pi^{2} r_{1}^{3}}{T_{1}^{2} R_{0}^{2}}$ C: 火星的第一宇宙速度为 $\frac{2 \pi r_{1}}{T_{1}} \sqrt{\frac{r_{1}}{R_{0}}}$ D: 火星的同步卫星距离星球表面高度为 $r_{1} \sqrt{\frac{T_{0}^{2}}{T_{1}^{2}}}-R_{0}$

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个多选题（有多个正确答案）。 问题: 中国行星探测任务名称为“天问系列”, 首次火星探测任务被命名为“天问一号”。若已知“天问一号”探测器在距离火星中心为 $r_{1}$ 的轨道上做匀速圆周运动, 其周期为 $T_{1}$ 。火星半径为 $R_{0}$, 自转周期为 $T_{0}$ 。引力常量为 $G$, 下列说法正确的是 ( ) A: 火星的质量为 $\frac{4 \pi^{2} R_{0}^{2}}{G T_{0}^{2}}$ B: 火星表面两极的重力加速度为 $\frac{4 \pi^{2} r_{1}^{3}}{T_{1}^{2} R_{0}^{2}}$ C: 火星的第一宇宙速度为 $\frac{2 \pi r_{1}}{T_{1}} \sqrt{\frac{r_{1}}{R_{0}}}$ D: 火星的同步卫星距离星球表面高度为 $r_{1} \sqrt{\frac{T_{0}^{2}}{T_{1}^{2}}}-R_{0}$ 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]

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Astronomy_782

The picture below shows a very famous nebula: What is the name of this nebula? [figure1] Figure 1: Nebula. Credit: NASA A: Crab Nebula B: Orion Nebula C: Ring Nebula D: Carina Nebula

You are participating in an international Astronomy competition and need to solve the following question. This is a multiple choice question (only one correct answer). problem: The picture below shows a very famous nebula: What is the name of this nebula? [figure1] Figure 1: Nebula. Credit: NASA A: Crab Nebula B: Orion Nebula C: Ring Nebula D: Carina Nebula You can solve it step by step. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER should be one of the options: [A, B, C, D].

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Astronomy_691

质量为 $100 \mathrm{~kg}$ 的“勇气”号火星车于 2004 年成功登陆在火星表面。若“勇气”号在离火星表面 $12 \mathrm{~m}$ 时与降落伞自动脱离, 被气囊包裹的“勇气”号下落到地面后又弹跳到 $18 \mathrm{~m}$ 高处, 这样上下碰撞了若干次后, 才静止在火星表面上。已知火星的半径为地球半径的 0.5 倍, 质量为地球质量的 0.1 倍。若“勇气”号第一次碰撞火星地面时, 气囊和地面的接触时间为 $0.7 \mathrm{~s}$, 其损失的机械能为它与降落伞自动脱离处 (即离火星地面 $12 \mathrm{~m}$ 时) 动能的 70\%, (地球表面的重力加速度 $g=10 \mathrm{~m} / \mathrm{s}^{2}$, 不考虑火星表面空气阻力) 求: 火星表面的重力加速度;

你正在参加一个国际天文竞赛，并需要解决以下问题。 这个问题的答案是一个数值。 问题: 质量为 $100 \mathrm{~kg}$ 的“勇气”号火星车于 2004 年成功登陆在火星表面。若“勇气”号在离火星表面 $12 \mathrm{~m}$ 时与降落伞自动脱离, 被气囊包裹的“勇气”号下落到地面后又弹跳到 $18 \mathrm{~m}$ 高处, 这样上下碰撞了若干次后, 才静止在火星表面上。已知火星的半径为地球半径的 0.5 倍, 质量为地球质量的 0.1 倍。若“勇气”号第一次碰撞火星地面时, 气囊和地面的接触时间为 $0.7 \mathrm{~s}$, 其损失的机械能为它与降落伞自动脱离处 (即离火星地面 $12 \mathrm{~m}$ 时) 动能的 70\%, (地球表面的重力加速度 $g=10 \mathrm{~m} / \mathrm{s}^{2}$, 不考虑火星表面空气阻力) 求: 火星表面的重力加速度; 你输出的所有数学公式和符号应该使用LaTeX表示！ 你可以一步一步来解决这个问题，并输出详细的解答过程。 请记住，你的答案应以m/s为单位计算，但在给出最终答案时，请不要包含单位。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER是不包含任何单位的数值。

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Astronomy_778

Sub-Neptune planets are ... A: smaller than Neptune. B: bigger than Neptune. C: farther away than Neptune. D: closer than Neptune.

You are participating in an international Astronomy competition and need to solve the following question. This is a multiple choice question (only one correct answer). problem: Sub-Neptune planets are ... A: smaller than Neptune. B: bigger than Neptune. C: farther away than Neptune. D: closer than Neptune. You can solve it step by step. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER should be one of the options: [A, B, C, D].

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Astronomy_645

华为 Mate 60 Pro 通过中国自主研制的天通一号卫星通信系统实现了卫星电话功能,天通一号卫星是地球同步卫星。天通一号卫星发射首先利用火箭将卫星运载至地球附近圆形轨道 1, 通过多次变轨最终进入地球同步轨道 3。其变轨简化示意图如图所示, 轨道 1 离地面高度为 $h$ 。已知天通一号卫星质量为 $m$, 地球自转周期为 $T_{0}$, 地球半径为 $R$,地球表面的重力加速度为 $g, O$ 为地球中心，引力常量为 $G$ 。如果规定距地球无限远处为地球引力零势能点, 地球附近物体的引力势能可表示为 $E_{\mathrm{p}}=-\frac{G M m}{r}$, 其中 $M$ (未知)为地球质量, $m$ 为物体质量, $r$ 为物体到地心距离。求: “天通一号”卫星在轨道 1 的运行周期 $T$; [图1]

你正在参加一个国际天文竞赛，并需要解决以下问题。 这个问题的答案是一个表达式。 问题: 华为 Mate 60 Pro 通过中国自主研制的天通一号卫星通信系统实现了卫星电话功能,天通一号卫星是地球同步卫星。天通一号卫星发射首先利用火箭将卫星运载至地球附近圆形轨道 1, 通过多次变轨最终进入地球同步轨道 3。其变轨简化示意图如图所示, 轨道 1 离地面高度为 $h$ 。已知天通一号卫星质量为 $m$, 地球自转周期为 $T_{0}$, 地球半径为 $R$,地球表面的重力加速度为 $g, O$ 为地球中心，引力常量为 $G$ 。如果规定距地球无限远处为地球引力零势能点, 地球附近物体的引力势能可表示为 $E_{\mathrm{p}}=-\frac{G M m}{r}$, 其中 $M$ (未知)为地球质量, $m$ 为物体质量, $r$ 为物体到地心距离。求: “天通一号”卫星在轨道 1 的运行周期 $T$; [图1] 你输出的所有数学公式和符号应该使用LaTeX表示！ 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER是一个不含等号的表达式，例如ANSWER=\frac{1}{2} g t^2

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Astronomy

ZH

multi-modal

Astronomy_894

On $24^{\text {th }}$ August 2016, astronomers discovered a planet orbiting the closest star to the Sun, Proxima Centauri, situated 4.22 light years away, which fulfils a long-standing dream of science-fiction writers: a world that is close enough for humans to send their first interstellar spacecraft. Astronomers have noted how the motion of Proxima Centauri changed in the first months of 2016, with the star moving towards and away from the Earth, as seen in the figure below. Sometimes Proxima Centauri is approaching Earth at $5 \mathrm{~km} \mathrm{hour}^{-1}-$ normal human walking pace - and at times receding at the same speed. This regular pattern of changing radial velocities caused by an unseen planet, which they named Proxima Centauri B, repeats and results in tiny Doppler shifts in the star's light, making the light appear slightly redder, then bluer. [figure1] Proxima Centauri is a red dwarf star, unlike our Sun, with a mass of only $0.12 \mathrm{M}_{\odot}$. What is the semi-major axis of the planet's orbit in $\mathrm{AU}$ ?

You are participating in an international Astronomy competition and need to solve the following question. The answer to this question is a numerical value. problem: On $24^{\text {th }}$ August 2016, astronomers discovered a planet orbiting the closest star to the Sun, Proxima Centauri, situated 4.22 light years away, which fulfils a long-standing dream of science-fiction writers: a world that is close enough for humans to send their first interstellar spacecraft. Astronomers have noted how the motion of Proxima Centauri changed in the first months of 2016, with the star moving towards and away from the Earth, as seen in the figure below. Sometimes Proxima Centauri is approaching Earth at $5 \mathrm{~km} \mathrm{hour}^{-1}-$ normal human walking pace - and at times receding at the same speed. This regular pattern of changing radial velocities caused by an unseen planet, which they named Proxima Centauri B, repeats and results in tiny Doppler shifts in the star's light, making the light appear slightly redder, then bluer. [figure1] Proxima Centauri is a red dwarf star, unlike our Sun, with a mass of only $0.12 \mathrm{M}_{\odot}$. What is the semi-major axis of the planet's orbit in $\mathrm{AU}$ ? All mathematical formulas and symbols you output should be represented with LaTeX! You can solve it step by step. Remember, your answer should be calculated in the unit of AU, but when concluding your final answer, do not include the unit. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER is the numerical value without any units.

NV

Astronomy

EN

multi-modal

Astronomy_217

如图所示, 质量分别为 $m$ 和 $M$ 的两个星球 $A$ 和 $B$ 在引力作用下都绕 $O$ 点做匀速圆周运动, 星球 $A$ 和 $B$ 两者中心之间距离为 $L$ 。已知 $A 、 B$ 的中心和 $O$ 三点始终共线, $A$和 $B$ 分别在 $O$ 的两侧, 引力常量为 $G$ 。求: $A$ 星球做圆周运动的半径 $R$ 和 $B$ 星球做圆周运动的半径 $r$; [图1]

你正在参加一个国际天文竞赛，并需要解决以下问题。 这个问题包含多个待求解的量。 问题: 如图所示, 质量分别为 $m$ 和 $M$ 的两个星球 $A$ 和 $B$ 在引力作用下都绕 $O$ 点做匀速圆周运动, 星球 $A$ 和 $B$ 两者中心之间距离为 $L$ 。已知 $A 、 B$ 的中心和 $O$ 三点始终共线, $A$和 $B$ 分别在 $O$ 的两侧, 引力常量为 $G$ 。求: $A$ 星球做圆周运动的半径 $R$ 和 $B$ 星球做圆周运动的半径 $r$; [图1] 你输出的所有数学公式和符号应该使用LaTeX表示！ 你可以一步一步来解决这个问题，并输出详细的解答过程。 你的最终解答的量应该按以下顺序输出：[r, R] 它们的答案类型依次是[表达式, 表达式] 你需要在输出的最后用以下格式总结答案：“最终答案是\boxed{ANSWER}”，其中ANSWER应为你的最终答案序列，用英文逗号分隔，例如：5, 7, 2.5

MPV

Astronomy

ZH

multi-modal

Astronomy_915

The tail of a comet A: is gas and dust pulled off the comet by the Sun's gravity B: always points towards the Sun C: is a trail behind the comet, pointing away from the Sun as the comet approaches it, and toward the Sun as the comet moves out of the inner Solar System D: is gas and dust expelled from the comet's nucleus and blown outward by radiation pressure and solar wind

You are participating in an international Astronomy competition and need to solve the following question. This is a multiple choice question (only one correct answer). problem: The tail of a comet A: is gas and dust pulled off the comet by the Sun's gravity B: always points towards the Sun C: is a trail behind the comet, pointing away from the Sun as the comet approaches it, and toward the Sun as the comet moves out of the inner Solar System D: is gas and dust expelled from the comet's nucleus and blown outward by radiation pressure and solar wind You can solve it step by step. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER should be one of the options: [A, B, C, D].

SC

Astronomy

EN

text-only

Astronomy_365

科学家在太空发现了一颗未知天体, 并发射探测器对该天体进行观测。探测器在圆形轨道 $\mathrm{I}$ 上绕行, 探测器的周期为 $T$, 距该星体表面的高度为 $h$, 飞行器对星球观测张角为 $\theta$, 如图所示。观测到该星体有一颗低轨道卫星在圆形轨道II上运动（绕行方向与探测器方向相同), 探测器和卫星两次靠得最近的时间间隔为 $t$ 。已知万有引力常量为 $G$,不计探测器和卫星之间的引力及其他天体引力影响, 根据题目所给的信息, 可得出的是 [图1] A: 可以求出探测器在轨道I上运行时探测器与该未知天体之间的万有引力 B: 可以求出该未知天体的质量 $M$ C: 可以求出轨道II离该天体表面的高度 D: 探测器在轨道II上运行的速度一定小于 $7.9 \mathrm{~km} / \mathrm{s}$

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个多选题（有多个正确答案）。 问题: 科学家在太空发现了一颗未知天体, 并发射探测器对该天体进行观测。探测器在圆形轨道 $\mathrm{I}$ 上绕行, 探测器的周期为 $T$, 距该星体表面的高度为 $h$, 飞行器对星球观测张角为 $\theta$, 如图所示。观测到该星体有一颗低轨道卫星在圆形轨道II上运动（绕行方向与探测器方向相同), 探测器和卫星两次靠得最近的时间间隔为 $t$ 。已知万有引力常量为 $G$,不计探测器和卫星之间的引力及其他天体引力影响, 根据题目所给的信息, 可得出的是 [图1] A: 可以求出探测器在轨道I上运行时探测器与该未知天体之间的万有引力 B: 可以求出该未知天体的质量 $M$ C: 可以求出轨道II离该天体表面的高度 D: 探测器在轨道II上运行的速度一定小于 $7.9 \mathrm{~km} / \mathrm{s}$ 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]

MC

Astronomy

ZH

multi-modal

Astronomy_261

如图所示, 宇宙中三颗质量分别为 $4 m 、 m 、 m$ 的恒星 $a 、 b 、 c$ 的球心位于等边三角形的三个顶点, 它们在相互之间的万有引力作用下共同绕三角形内某一点做匀速圆周运动, 运行周期相同, 等边三角形边长为 $L$ 。已知恒星 $a$ 表面重力加速度为 $g$, 引力恒量为 $G$, 将恒星视为均匀球体, 忽略星球自转, 求: 恒星的运行周期 $T$ 。 [图1]

你正在参加一个国际天文竞赛，并需要解决以下问题。 这个问题的答案是一个表达式。 问题: 如图所示, 宇宙中三颗质量分别为 $4 m 、 m 、 m$ 的恒星 $a 、 b 、 c$ 的球心位于等边三角形的三个顶点, 它们在相互之间的万有引力作用下共同绕三角形内某一点做匀速圆周运动, 运行周期相同, 等边三角形边长为 $L$ 。已知恒星 $a$ 表面重力加速度为 $g$, 引力恒量为 $G$, 将恒星视为均匀球体, 忽略星球自转, 求: 恒星的运行周期 $T$ 。 [图1] 你输出的所有数学公式和符号应该使用LaTeX表示！ 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER是一个不含等号的表达式，例如ANSWER=\frac{1}{2} g t^2

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Astronomy

ZH

multi-modal

Astronomy_213

北京时间 2021 年 2 月 19 日 4 时 55 分, 美国“毅力号”火星车成功登陆火星。“空中起重机”和“毅力号”火星车组合体到达着陆点上空 $20 \mathrm{~m}$ 处后, “空中起重机”保持大小为 $0.75 \mathrm{~m} / \mathrm{s}$ 的速度坚直下降，同时，在着陆点上空 $20 \mathrm{~m}$ 处时，以相对“空中起重机”大小也为 $0.75 \mathrm{~m} / \mathrm{s}$ 的速度立即（时间很短, 可忽略）坚直向下释放火星车; 当全长为 $7.6 \mathrm{~m}$ 的吊索完全释放后，组合体又立即（时间很短，可忽略）共同以 $0.75 \mathrm{~m} / \mathrm{s}$ 的速度下降，直到火星车着陆, 然后断开吊索, “空中起重机”飘离。设火星质量是地球质量的 $p$ 倍, 火星半径是地球半径的 $q$ 倍, 地球表面重力加速度为 $g$, 引力常量为 $G$ 。假设工作中组合体 (含燃料) 的总质量 $M$ 保持不变, 不考虑下降过程中重力的变化, 工作时喷出的气体密度为 $\rho$, “空中起重机”共四台发动机, 每台发动机喷口截面为 $S$; 下列说法正确的是 ( ) [图1] A: 火星表面的重力加速度大小为 $g_{\text {火 }}=\frac{p}{q^{2}} g$ B: 匀速坚直下降的过程中, 发动机喷出气体相对火星表面的速度大小为 $\frac{1}{2 q} \sqrt{\frac{M p g}{\rho S}}$ C: 从火星车刚被释放直到火星车着陆的整个过程中, 空中起重机下降的时间约为 $16.5 \mathrm{~s}$ D: 从火星车刚被释放直到火星车着陆的整个过程中, 吊索的拉力始终保持不变

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个多选题（有多个正确答案）。 问题: 北京时间 2021 年 2 月 19 日 4 时 55 分, 美国“毅力号”火星车成功登陆火星。“空中起重机”和“毅力号”火星车组合体到达着陆点上空 $20 \mathrm{~m}$ 处后, “空中起重机”保持大小为 $0.75 \mathrm{~m} / \mathrm{s}$ 的速度坚直下降，同时，在着陆点上空 $20 \mathrm{~m}$ 处时，以相对“空中起重机”大小也为 $0.75 \mathrm{~m} / \mathrm{s}$ 的速度立即（时间很短, 可忽略）坚直向下释放火星车; 当全长为 $7.6 \mathrm{~m}$ 的吊索完全释放后，组合体又立即（时间很短，可忽略）共同以 $0.75 \mathrm{~m} / \mathrm{s}$ 的速度下降，直到火星车着陆, 然后断开吊索, “空中起重机”飘离。设火星质量是地球质量的 $p$ 倍, 火星半径是地球半径的 $q$ 倍, 地球表面重力加速度为 $g$, 引力常量为 $G$ 。假设工作中组合体 (含燃料) 的总质量 $M$ 保持不变, 不考虑下降过程中重力的变化, 工作时喷出的气体密度为 $\rho$, “空中起重机”共四台发动机, 每台发动机喷口截面为 $S$; 下列说法正确的是 ( ) [图1] A: 火星表面的重力加速度大小为 $g_{\text {火 }}=\frac{p}{q^{2}} g$ B: 匀速坚直下降的过程中, 发动机喷出气体相对火星表面的速度大小为 $\frac{1}{2 q} \sqrt{\frac{M p g}{\rho S}}$ C: 从火星车刚被释放直到火星车着陆的整个过程中, 空中起重机下降的时间约为 $16.5 \mathrm{~s}$ D: 从火星车刚被释放直到火星车着陆的整个过程中, 吊索的拉力始终保持不变 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]

MC

Astronomy

ZH

multi-modal

Astronomy_570

一颗人造地球卫星在地面附近绕地球做匀速圆周运动的速率为 $v$, 角速度为 $\omega$, 加速度为 $g$, 周期为 $T$. 另一颗人造地球卫星在离地面高度为地球半径的轨道上做匀速圆周运动，则（） A: 它的速率为 $\frac{\sqrt{2}}{2} v$ B: 它的加速度为 $\frac{g}{4}$ C: 它的运动周期为 $\sqrt{2} T$ D: 它的角速度也为 $\omega$

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个多选题（有多个正确答案）。 问题: 一颗人造地球卫星在地面附近绕地球做匀速圆周运动的速率为 $v$, 角速度为 $\omega$, 加速度为 $g$, 周期为 $T$. 另一颗人造地球卫星在离地面高度为地球半径的轨道上做匀速圆周运动，则（） A: 它的速率为 $\frac{\sqrt{2}}{2} v$ B: 它的加速度为 $\frac{g}{4}$ C: 它的运动周期为 $\sqrt{2} T$ D: 它的角速度也为 $\omega$ 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]

MC

Astronomy

ZH

text-only

Astronomy_832

Figure 6 shows a 6-hr root-mean-square (rms) Combined Differential Photometric Precision (CDPP) curve for 150,000 stars observed by the Kepler space telescope. CDPP is a measure of the white noise contained in a light curve, so for a target with 6 -hour CDPP of 100 parts per million (ppm), a 6-hour transit with depth $100 \mathrm{ppm}$ would be considered a 1- $\sigma$ detection. [figure1] Figure 6: From Christiansen et al. (https://arxiv.org/abs/1208.0595). Original caption: The distribution of the 6-hour rms CDPP values with Kp magnitude for all Quarter 3 planetary targets. Consider a $1 R_{\odot}$ target with Kepler magnitude $K_{p}=13.5$ that's among the best targets for its magnitude with respect to noise in Figure 6. Also consider three independent exoplanet scenarios for exoplanets with radii: - I. $0.5 R_{\oplus}$ - II. $1 R_{\oplus}$ - III. $10 R_{\oplus}$ Using a $1-\sigma$ detection threshold (and assuming 6-hour transit durations), which planet(s) transits would we likely fail to observe due to noise? A: I B: III C: I and II D: II and III E: I, II, and III

You are participating in an international Astronomy competition and need to solve the following question. This is a multiple choice question (only one correct answer). problem: Figure 6 shows a 6-hr root-mean-square (rms) Combined Differential Photometric Precision (CDPP) curve for 150,000 stars observed by the Kepler space telescope. CDPP is a measure of the white noise contained in a light curve, so for a target with 6 -hour CDPP of 100 parts per million (ppm), a 6-hour transit with depth $100 \mathrm{ppm}$ would be considered a 1- $\sigma$ detection. [figure1] Figure 6: From Christiansen et al. (https://arxiv.org/abs/1208.0595). Original caption: The distribution of the 6-hour rms CDPP values with Kp magnitude for all Quarter 3 planetary targets. Consider a $1 R_{\odot}$ target with Kepler magnitude $K_{p}=13.5$ that's among the best targets for its magnitude with respect to noise in Figure 6. Also consider three independent exoplanet scenarios for exoplanets with radii: - I. $0.5 R_{\oplus}$ - II. $1 R_{\oplus}$ - III. $10 R_{\oplus}$ Using a $1-\sigma$ detection threshold (and assuming 6-hour transit durations), which planet(s) transits would we likely fail to observe due to noise? A: I B: III C: I and II D: II and III E: I, II, and III You can solve it step by step. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER should be one of the options: [A, B, C, D, E].

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Astronomy

EN

multi-modal

Astronomy_57

两颗卫星在同一轨道平面绕地球做匀速圆周运动, 地球半径为 $R, a$ 卫星离地面的高度等于 $R, b$ 卫星离地面高度为 $3 R$, 则: $a 、 b$ 两卫星周期之比 $T a ： T b$ 是多少？

你正在参加一个国际天文竞赛，并需要解决以下问题。 这个问题的答案是一个数值。 问题: 两颗卫星在同一轨道平面绕地球做匀速圆周运动, 地球半径为 $R, a$ 卫星离地面的高度等于 $R, b$ 卫星离地面高度为 $3 R$, 则: $a 、 b$ 两卫星周期之比 $T a ： T b$ 是多少？ 你输出的所有数学公式和符号应该使用LaTeX表示！ 你可以一步一步来解决这个问题，并输出详细的解答过程。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER是数值。

NV

Astronomy

ZH

text-only

Astronomy_1041

In the heart of every star, nuclear fusion is taking place. For most stars that involves hydrogen being turned into helium, a process that starts by bringing two protons close enough that the strong nuclear force can act upon them. The smallest stars are the ones that have a core that is only just hot enough for fusion to occur, whilst in the biggest ones the radiation pressure of the photons given out by the fusion reaction pushing on the stellar material can overcome the gravitational forces holding it together. [figure1] Figure 5: Left: The lowest mass star we know of, EBLM J0555-57Ab, was found by von Boetticher et al. (2017) and is about the size of Saturn with a mass of $0.081 M_{\odot}$. Credit: Amanda Smith, University of Cambridge. Right: The highest mass star we know of, R136a1, is in the centre of the clump of stars on the right of this HST image of the Tarantula Nebula. Schneider et al. (2014) suggest it has a mass of $315 M_{\odot}$, which is above what stellar evolution models allow. Despite its large mass, other stars have far bigger radii. Credit: NASA \& ESA. For a spherical main sequence star made of a plasma (a fully ionized gas of electrons and nuclei) that is acting like an ideal gas, the temperature at the core can be approximately calculated as $$ T_{\mathrm{int}} \simeq \frac{G M \bar{\mu}}{k_{\mathrm{B}} R} \quad \text { where } \quad \bar{\mu}=\frac{m_{\mathrm{p}}}{2 X+3 Y / 4+Z / 2} . $$ In this equation, $M$ is the mass of the star, $R$ is its radius, $k_{\mathrm{B}}$ is the Boltzmann constant, and $\bar{\mu}$ is the mean mass of the plasma particles (i.e nuclei and electrons) with $m_{\mathrm{p}}$ the mass of a proton. Classically, the core of the Sun is not hot enough for fusion, and yet fusion is clearly happening. The key is that it is a fundamentally quantum process, and so protons are able to 'quantum tunnel' through the Coloumb barrier (see Figure 6), allowing fusion to occur at lower temperatures. In quantum mechanics, fusion will happen when $b=\lambda$ where $\lambda$ is the de Broglie wavelength of the proton, related to the momentum of the proton by $\lambda=h / p$. [figure2] Figure 6: A diagram showing the way a particle can pass through a classically impenetrable potential barrier due to its wave-like properties on the quantum scale. Credit: Brooks/Cole - Thomson Learning. In the smallest stars, electron degeneracy prevents them from compressing in radius and thus stops the core reaching $T_{\text {int }} \gtrsim T_{\text {quantum }}$. At the limit of electron degeneracy, the number density of electrons $n_{\mathrm{e}}=1 / \lambda_{\mathrm{e}}^{3}$ where $\lambda_{\mathrm{e}}$ is the de Broglie wavelength of the electrons. In the largest stars, radiation pressure pushes on the outer layers of the star stronger than gravity pulls them in. The brightest luminosity for a star is known as the Eddington luminosity, $L_{\text {Edd }}$. The acceleration due to radiation pressure can be calculated as $$ g_{\mathrm{rad}}=\frac{\kappa_{\mathrm{e}} I}{c} \quad \text { where } \quad \kappa_{\mathrm{e}}=\frac{\sigma_{\mathrm{T}}}{2 m_{\mathrm{p}}}(1+X) $$ and $\kappa_{\mathrm{e}}$ is the electron opacity of the stellar material, $\sigma_{\mathrm{T}}$ is the Thomson scattering cross-section for electrons $\left(=66.5 \mathrm{fm}^{2}\right.$ ), $X$ is the hydrogen fraction, and $I$ is the intensity of radiation (in $\mathrm{W} \mathrm{m}^{-2}$ ). Assuming main-sequence stars follow a mass-luminosity relation of $L \propto M^{3}$, the maximum mass of a star can be found by considering one that is radiating at $L_{\text {Edd }}$.b. Classically, two protons need to have enough energy to overcome their electrostatic repulsion in order to fuse. Calculate the value of $T_{\text {classical }}$ necessary to allow fusion to occur, given that at that point $b=1 \mathrm{fm}\left(=10^{-15} \mathrm{~m}\right)$. [You should find that it's much larger than your answer to part a.]

You are participating in an international Astronomy competition and need to solve the following question. The answer to this question is a numerical value. Here is some context information for this question, which might assist you in solving it: In the heart of every star, nuclear fusion is taking place. For most stars that involves hydrogen being turned into helium, a process that starts by bringing two protons close enough that the strong nuclear force can act upon them. The smallest stars are the ones that have a core that is only just hot enough for fusion to occur, whilst in the biggest ones the radiation pressure of the photons given out by the fusion reaction pushing on the stellar material can overcome the gravitational forces holding it together. [figure1] Figure 5: Left: The lowest mass star we know of, EBLM J0555-57Ab, was found by von Boetticher et al. (2017) and is about the size of Saturn with a mass of $0.081 M_{\odot}$. Credit: Amanda Smith, University of Cambridge. Right: The highest mass star we know of, R136a1, is in the centre of the clump of stars on the right of this HST image of the Tarantula Nebula. Schneider et al. (2014) suggest it has a mass of $315 M_{\odot}$, which is above what stellar evolution models allow. Despite its large mass, other stars have far bigger radii. Credit: NASA \& ESA. For a spherical main sequence star made of a plasma (a fully ionized gas of electrons and nuclei) that is acting like an ideal gas, the temperature at the core can be approximately calculated as $$ T_{\mathrm{int}} \simeq \frac{G M \bar{\mu}}{k_{\mathrm{B}} R} \quad \text { where } \quad \bar{\mu}=\frac{m_{\mathrm{p}}}{2 X+3 Y / 4+Z / 2} . $$ In this equation, $M$ is the mass of the star, $R$ is its radius, $k_{\mathrm{B}}$ is the Boltzmann constant, and $\bar{\mu}$ is the mean mass of the plasma particles (i.e nuclei and electrons) with $m_{\mathrm{p}}$ the mass of a proton. Classically, the core of the Sun is not hot enough for fusion, and yet fusion is clearly happening. The key is that it is a fundamentally quantum process, and so protons are able to 'quantum tunnel' through the Coloumb barrier (see Figure 6), allowing fusion to occur at lower temperatures. In quantum mechanics, fusion will happen when $b=\lambda$ where $\lambda$ is the de Broglie wavelength of the proton, related to the momentum of the proton by $\lambda=h / p$. [figure2] Figure 6: A diagram showing the way a particle can pass through a classically impenetrable potential barrier due to its wave-like properties on the quantum scale. Credit: Brooks/Cole - Thomson Learning. In the smallest stars, electron degeneracy prevents them from compressing in radius and thus stops the core reaching $T_{\text {int }} \gtrsim T_{\text {quantum }}$. At the limit of electron degeneracy, the number density of electrons $n_{\mathrm{e}}=1 / \lambda_{\mathrm{e}}^{3}$ where $\lambda_{\mathrm{e}}$ is the de Broglie wavelength of the electrons. In the largest stars, radiation pressure pushes on the outer layers of the star stronger than gravity pulls them in. The brightest luminosity for a star is known as the Eddington luminosity, $L_{\text {Edd }}$. The acceleration due to radiation pressure can be calculated as $$ g_{\mathrm{rad}}=\frac{\kappa_{\mathrm{e}} I}{c} \quad \text { where } \quad \kappa_{\mathrm{e}}=\frac{\sigma_{\mathrm{T}}}{2 m_{\mathrm{p}}}(1+X) $$ and $\kappa_{\mathrm{e}}$ is the electron opacity of the stellar material, $\sigma_{\mathrm{T}}$ is the Thomson scattering cross-section for electrons $\left(=66.5 \mathrm{fm}^{2}\right.$ ), $X$ is the hydrogen fraction, and $I$ is the intensity of radiation (in $\mathrm{W} \mathrm{m}^{-2}$ ). Assuming main-sequence stars follow a mass-luminosity relation of $L \propto M^{3}$, the maximum mass of a star can be found by considering one that is radiating at $L_{\text {Edd }}$. problem: b. Classically, two protons need to have enough energy to overcome their electrostatic repulsion in order to fuse. Calculate the value of $T_{\text {classical }}$ necessary to allow fusion to occur, given that at that point $b=1 \mathrm{fm}\left(=10^{-15} \mathrm{~m}\right)$. [You should find that it's much larger than your answer to part a.] All mathematical formulas and symbols you output should be represented with LaTeX! You can solve it step by step. Remember, your answer should be calculated in the unit of \mathrm{~K}, but when concluding your final answer, do not include the unit. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER is the numerical value without any units.

NV

Astronomy

EN

multi-modal

Astronomy_1216

Hanny's Voorwerp (Dutch for 'object') is a rare type of astronomical object discovered in 2007 by the school teacher Hanny van Arkel whilst participating as a volunteer in the Galaxy Zoo project. When inspecting the image of the galaxy IC 2497 in the constellation Leo Minor, she observed a bright green blob close to the galaxy. [figure1] Figure 5: HST image of galaxy IC 2497 and the glowing Voorwerp below it. Credit: Keel et al. (2012) \& Galaxy Zoo. Subsequent observations have shown that the galaxy IC 2497 is at a redshift of $z=0.05$, with the Voorwerp at a similar distance and with a projected angular separation of 20 arcseconds from the centre of the galaxy $\left(3600\right.$ arcseconds $\left.=1^{\circ}\right)$. Radio observations suggest that the Voorwerp is a massive cloud of gas, made of ionized hydrogen, with a size of $10 \mathrm{kpc}$ and a mass of $10^{11} \mathrm{M}_{\odot}$. It is probably a cloud of gas that was stripped from the galaxy during a merger with another nearby galaxy. In this question you will explore the cause of the 'glow' of the Voorwerp and will learn about a new type of an astronomical object; a quasar.a. Given that Hubble's constant is measured as $H_{0}=70 \mathrm{~km} \mathrm{~s}^{-1} \mathrm{Mpc}^{-1}$, calculate the distance to the galaxy (in $\mathrm{Mpc}$ ).

You are participating in an international Astronomy competition and need to solve the following question. The answer to this question is a numerical value. Here is some context information for this question, which might assist you in solving it: Hanny's Voorwerp (Dutch for 'object') is a rare type of astronomical object discovered in 2007 by the school teacher Hanny van Arkel whilst participating as a volunteer in the Galaxy Zoo project. When inspecting the image of the galaxy IC 2497 in the constellation Leo Minor, she observed a bright green blob close to the galaxy. [figure1] Figure 5: HST image of galaxy IC 2497 and the glowing Voorwerp below it. Credit: Keel et al. (2012) \& Galaxy Zoo. Subsequent observations have shown that the galaxy IC 2497 is at a redshift of $z=0.05$, with the Voorwerp at a similar distance and with a projected angular separation of 20 arcseconds from the centre of the galaxy $\left(3600\right.$ arcseconds $\left.=1^{\circ}\right)$. Radio observations suggest that the Voorwerp is a massive cloud of gas, made of ionized hydrogen, with a size of $10 \mathrm{kpc}$ and a mass of $10^{11} \mathrm{M}_{\odot}$. It is probably a cloud of gas that was stripped from the galaxy during a merger with another nearby galaxy. In this question you will explore the cause of the 'glow' of the Voorwerp and will learn about a new type of an astronomical object; a quasar. problem: a. Given that Hubble's constant is measured as $H_{0}=70 \mathrm{~km} \mathrm{~s}^{-1} \mathrm{Mpc}^{-1}$, calculate the distance to the galaxy (in $\mathrm{Mpc}$ ). All mathematical formulas and symbols you output should be represented with LaTeX! You can solve it step by step. Remember, your answer should be calculated in the unit of MPc, but when concluding your final answer, do not include the unit. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER is the numerical value without any units.

NV

Astronomy

EN

multi-modal

Astronomy_1174

The speed of light is considered to be the speed limit of the Universe, however knots of plasma in the jets from active galactic nuclei (AGN) have been observed to be moving with apparent transverse speeds in excess of this, called superluminal speeds. Some of the more extreme examples can be appearing to move at up to 6 times the speed of light (see Figure 7). [figure1] Figure 7: Left: The jet coming from the elliptical galaxy M87 as viewed by the Hubble Space Telescope (HST). Right: Sequence of HST images showing motion at six times the speed of light. The slanting lines track the moving features, and the speeds are given in units of the velocity of light, $c$. Credit: NASA / Space Telescope Science Institute / John Biretta. This can be explained by understanding that the jet is offset by an angle $\theta$ from the sightline to Earth, and that the real speed of the plasma knot, $v$, is less than $c$, and from it we can define the scaled speed $\beta \equiv v / c$. Superluminal jets are not limited just to AGN, as they have also been observed from systems within our own galaxy. A particularly famous one is the 'microquasar' GRS $1915+105$, which is a low mass X-ray binary consisting of a small star orbiting a black hole. A symmetrical jet with components approaching and receding from us is observed (as expected for jets coming from the poles of the black hole), and the apparent transverse motion of material in those jets has been measured using very high resolution radio imaging. Fender et. al (1999) measure these motions to be $\mu_{a}=23.6$ mas day $^{-1}$ and $\mu_{r}=$ 10.0 mas day $^{-1}$ for the approaching and receding jet respectively ( 1 mas $=1$ milliarcsecond, a unit of angle, and there are 3600 arcseconds in a degree) and the distance to the system as $11 \mathrm{kpc}$. In practice, for a given $\beta_{\text {app }}$ the values of $\beta$ and $\theta$ are degenerate and it is unlikely that the orientation of the jet is such that $\beta_{\text {app }}$ has been maximised, so the value in part $\mathrm{c}$. is just a lower limit. However, since there are two jets then if we assume that they are from the same event (and so equal in speed but opposite in direction) we can break this degeneracy. Since it is a binary system, we can gain information about the masses of the objects by looking at their period and radial velocity. Formally, the relationship is $$ \frac{\left(M_{\mathrm{BH}} \sin i\right)^{3}}{\left(M_{\mathrm{BH}}+M_{\star}\right)^{2}}=\frac{P_{\mathrm{orb}} K_{d}^{3}}{2 \pi G} $$ where $M_{\mathrm{BH}}$ is the mass of the black hole, $M_{\star}$ is the mass of the orbiting star, $i$ is the inclination of the orbit, $P_{\text {orb }}$ is the orbital period, and $K_{d}$ is the amplitude of the radial velocity curve. Normally the inclination can't be measured, however if we assume that the orbit is perpendicular to the jets then $i=\theta$ and we can measure the mass of the black hole.a. Show, with use of an appropriate diagram, that the apparent value of the scaled transverse speed (for a jet coming towards us) is $\beta_{\text {app }}=\frac{\beta \sin \theta}{1-\beta \cos \theta}$.

You are participating in an international Astronomy competition and need to solve the following question. The answer to this question is an expression. Here is some context information for this question, which might assist you in solving it: The speed of light is considered to be the speed limit of the Universe, however knots of plasma in the jets from active galactic nuclei (AGN) have been observed to be moving with apparent transverse speeds in excess of this, called superluminal speeds. Some of the more extreme examples can be appearing to move at up to 6 times the speed of light (see Figure 7). [figure1] Figure 7: Left: The jet coming from the elliptical galaxy M87 as viewed by the Hubble Space Telescope (HST). Right: Sequence of HST images showing motion at six times the speed of light. The slanting lines track the moving features, and the speeds are given in units of the velocity of light, $c$. Credit: NASA / Space Telescope Science Institute / John Biretta. This can be explained by understanding that the jet is offset by an angle $\theta$ from the sightline to Earth, and that the real speed of the plasma knot, $v$, is less than $c$, and from it we can define the scaled speed $\beta \equiv v / c$. Superluminal jets are not limited just to AGN, as they have also been observed from systems within our own galaxy. A particularly famous one is the 'microquasar' GRS $1915+105$, which is a low mass X-ray binary consisting of a small star orbiting a black hole. A symmetrical jet with components approaching and receding from us is observed (as expected for jets coming from the poles of the black hole), and the apparent transverse motion of material in those jets has been measured using very high resolution radio imaging. Fender et. al (1999) measure these motions to be $\mu_{a}=23.6$ mas day $^{-1}$ and $\mu_{r}=$ 10.0 mas day $^{-1}$ for the approaching and receding jet respectively ( 1 mas $=1$ milliarcsecond, a unit of angle, and there are 3600 arcseconds in a degree) and the distance to the system as $11 \mathrm{kpc}$. In practice, for a given $\beta_{\text {app }}$ the values of $\beta$ and $\theta$ are degenerate and it is unlikely that the orientation of the jet is such that $\beta_{\text {app }}$ has been maximised, so the value in part $\mathrm{c}$. is just a lower limit. However, since there are two jets then if we assume that they are from the same event (and so equal in speed but opposite in direction) we can break this degeneracy. Since it is a binary system, we can gain information about the masses of the objects by looking at their period and radial velocity. Formally, the relationship is $$ \frac{\left(M_{\mathrm{BH}} \sin i\right)^{3}}{\left(M_{\mathrm{BH}}+M_{\star}\right)^{2}}=\frac{P_{\mathrm{orb}} K_{d}^{3}}{2 \pi G} $$ where $M_{\mathrm{BH}}$ is the mass of the black hole, $M_{\star}$ is the mass of the orbiting star, $i$ is the inclination of the orbit, $P_{\text {orb }}$ is the orbital period, and $K_{d}$ is the amplitude of the radial velocity curve. Normally the inclination can't be measured, however if we assume that the orbit is perpendicular to the jets then $i=\theta$ and we can measure the mass of the black hole. problem: a. Show, with use of an appropriate diagram, that the apparent value of the scaled transverse speed (for a jet coming towards us) is $\beta_{\text {app }}=\frac{\beta \sin \theta}{1-\beta \cos \theta}$. All mathematical formulas and symbols you output should be represented with LaTeX! You can solve it step by step. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER is an expression without equals signs, e.g. ANSWER=\frac{1}{2} g t^2

EX

Astronomy

EN

multi-modal

Astronomy_813

Consider the following elliptical orbit of a comet around a star: [figure1] Which of the following expressions corresponds to the time that the comet takes to go from point A to point B as a function of the period of the comet $(T)$ and the eccentricity of the orbit $(e) ?$ Assume that the direction of the orbit is counterclockwise. A: $\frac{T}{2}$ B: $\left(\frac{e}{\pi}+\frac{1}{2}\right) * T$ C: $\left(\frac{1}{2}-\frac{e}{\pi}\right) * T$ D: $(1+e) * \frac{T}{2}$ E: $\frac{T * e}{2}$

You are participating in an international Astronomy competition and need to solve the following question. This is a multiple choice question (only one correct answer). problem: Consider the following elliptical orbit of a comet around a star: [figure1] Which of the following expressions corresponds to the time that the comet takes to go from point A to point B as a function of the period of the comet $(T)$ and the eccentricity of the orbit $(e) ?$ Assume that the direction of the orbit is counterclockwise. A: $\frac{T}{2}$ B: $\left(\frac{e}{\pi}+\frac{1}{2}\right) * T$ C: $\left(\frac{1}{2}-\frac{e}{\pi}\right) * T$ D: $(1+e) * \frac{T}{2}$ E: $\frac{T * e}{2}$ You can solve it step by step. Please end your response with: "The final answer is $\boxed{ANSWER}$", where ANSWER should be one of the options: [A, B, C, D, E].

SC

Astronomy

EN

multi-modal

Astronomy_409

太阳主要是由 e、 ${ }_{1}^{1} \mathrm{H}$ 和 ${ }_{2}^{4} \mathrm{He}$ 等粒子组成的。维持太阳辐射的是其内部的核聚变反应, 核反应方程是 $2 \mathrm{e}+4{ }_{1}^{1} \mathrm{H} \rightarrow{ }_{2}^{4} \mathrm{He}$, 该核反应产生的核能最后转化为辐射能。根据目前关于 恒星演化的理论, 若由于聚变反应而使太阳中的 ${ }_{1}^{1} \mathrm{H}$ 核数目从现有数减少 $10 \%$, 太阳将离开主序星阶段而转入红巨星的演化阶段。为了简化模型, 假定目前太阳全部由 $\mathrm{e}$ 和 ${ }_{1}{ }^{1} \mathrm{H}$核组成, 并据此回答下列问题。 已知地球半径 $R=6.4 \times 10^{6} \mathrm{~m}$, 地球质量 $m=6.0 \times 10^{24} \mathrm{~kg}$, 日地中心的距离 $r=1.5 \times 10^{11} \mathrm{~m}$, 地球表面处的重力加速度 $g=10 \mathrm{~m} / \mathrm{s}^{2}, 1$ 年约为 $3.2 \times 10^{7} \mathrm{~s}$, 试估算目前太阳的质量 $M$ 。

你正在参加一个国际天文竞赛，并需要解决以下问题。 这个问题的答案是一个数值。 问题: 太阳主要是由 e、 ${ }_{1}^{1} \mathrm{H}$ 和 ${ }_{2}^{4} \mathrm{He}$ 等粒子组成的。维持太阳辐射的是其内部的核聚变反应, 核反应方程是 $2 \mathrm{e}+4{ }_{1}^{1} \mathrm{H} \rightarrow{ }_{2}^{4} \mathrm{He}$, 该核反应产生的核能最后转化为辐射能。根据目前关于 恒星演化的理论, 若由于聚变反应而使太阳中的 ${ }_{1}^{1} \mathrm{H}$ 核数目从现有数减少 $10 \%$, 太阳将离开主序星阶段而转入红巨星的演化阶段。为了简化模型, 假定目前太阳全部由 $\mathrm{e}$ 和 ${ }_{1}{ }^{1} \mathrm{H}$核组成, 并据此回答下列问题。 已知地球半径 $R=6.4 \times 10^{6} \mathrm{~m}$, 地球质量 $m=6.0 \times 10^{24} \mathrm{~kg}$, 日地中心的距离 $r=1.5 \times 10^{11} \mathrm{~m}$, 地球表面处的重力加速度 $g=10 \mathrm{~m} / \mathrm{s}^{2}, 1$ 年约为 $3.2 \times 10^{7} \mathrm{~s}$, 试估算目前太阳的质量 $M$ 。 你输出的所有数学公式和符号应该使用LaTeX表示！ 你可以一步一步来解决这个问题，并输出详细的解答过程。 请记住，你的答案应以kg为单位计算，但在给出最终答案时，请不要包含单位。 你需要在输出的最后用以下格式总结答案：“最终答案是$\boxed{ANSWER}$”，其中ANSWER是不包含任何单位的数值。

NV

Astronomy

ZH

text-only