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Astronomy_550

如图所示, $a 、 b 、 c$ 均为地球卫星，且 $a$ 是地球同步卫星, $c$ 是与赤道共面的某近地卫星， $d$ 是静止在地球赤道地面上的一个物体，以下关于 $a 、 b 、 c 、 d$ 四者的线速度、角速度、周期以及向心加速度的大小关系正确的是（ ） [图1] A: $v_{c}>v_{b}>v_{a}>v_{d}$ B: $\omega_{a}>\omega_{b}>\omega_{c}>\omega_{d}$ C: $a_{a}>a_{b}>a_{c}>a_{d}$ D: $T_{c}<T_{b}<T_{a}=T_{d}$

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, $a 、 b 、 c$ 均为地球卫星，且 $a$ 是地球同步卫星, $c$ 是与赤道共面的某近地卫星， $d$ 是静止在地球赤道地面上的一个物体，以下关于 $a 、 b 、 c 、 d$ 四者的线速度、角速度、周期以及向心加速度的大小关系正确的是（ ） [图1] A: $v_{c}>v_{b}>v_{a}>v_{d}$ B: $\omega_{a}>\omega_{b}>\omega_{c}>\omega_{d}$ C: $a_{a}>a_{b}>a_{c}>a_{d}$ D: $T_{c}<T_{b}<T_{a}=T_{d}$ 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

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Astronomy_216

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 $G$ 。 若在球壳上 $A$ 点挖去半径为 $r$ 的小圆孔 ( $r \ll R$, 挖去的部分可看做质点), 求:球壳对位于球心 $O$ 处的质量为 $m$ 的物体的万有引力; [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 $G$ 。 若在球壳上 $A$ 点挖去半径为 $r$ 的小圆孔 ( $r \ll R$, 挖去的部分可看做质点), 求:球壳对位于球心 $O$ 处的质量为 $m$ 的物体的万有引力; [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

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Astronomy_700

我国航天技术走在世界的前列，探月工程“绕、落、回”三步走的最后一步即将完成,即月球探测器实现采样返回, 如图所示为该过程简化后的示意图, 探测器从圆轨道 1 上的 $\mathrm{A}$ 点减速后变轨到椭圆轨道 2 , 之后又在轨道 2 上的 $B$ 点变轨到近月圆轨道 3。已知探测器在轨道 1 上的运行周期为 $T_{1}$, 运行速度为 $v_{1}$, 在轨道 3 上运行速度为 $v_{3}, O$ 为月球球心, $C$ 为轨道 3 上的一点 $\mathrm{AC}$ 与轨道 3 相切, $A C$ 与 $A O$ 之间的最大夹角为 $\theta$, 下列说法正确的是（ ） [图1] A: 探测器在轨道 2 运行时的机械能大于在轨道 1 运行时的机械能 B: 探测器在轨道 2 运行时的线速度 $v_{2}$ 大小满足 $v_{1}<v_{2}<v_{3}$ C: 探测器在轨道 3 上运行时的周期为 $T_{1} \sqrt{\sin ^{3} \theta}$ D: 探测器在轨道 2 上运行和在圆轨道 1 上运行, 加速度大小相等的位置有两个

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: 我国航天技术走在世界的前列，探月工程“绕、落、回”三步走的最后一步即将完成,即月球探测器实现采样返回, 如图所示为该过程简化后的示意图, 探测器从圆轨道 1 上的 $\mathrm{A}$ 点减速后变轨到椭圆轨道 2 , 之后又在轨道 2 上的 $B$ 点变轨到近月圆轨道 3。已知探测器在轨道 1 上的运行周期为 $T_{1}$, 运行速度为 $v_{1}$, 在轨道 3 上运行速度为 $v_{3}, O$ 为月球球心, $C$ 为轨道 3 上的一点 $\mathrm{AC}$ 与轨道 3 相切, $A C$ 与 $A O$ 之间的最大夹角为 $\theta$, 下列说法正确的是（ ） [图1] A: 探测器在轨道 2 运行时的机械能大于在轨道 1 运行时的机械能 B: 探测器在轨道 2 运行时的线速度 $v_{2}$ 大小满足 $v_{1}<v_{2}<v_{3}$ C: 探测器在轨道 3 上运行时的周期为 $T_{1} \sqrt{\sin ^{3} \theta}$ D: 探测器在轨道 2 上运行和在圆轨道 1 上运行, 加速度大小相等的位置有两个 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

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multi-modal

Astronomy_451

如图所示, $\mathrm{A} 、 \mathrm{~B}$ 是绕地球做圆周运动的两颗卫星, $\mathrm{A} 、 \mathrm{~B}$ 两卫星与地心的连线在相等时间内扫过的面积之比为 $k: 1$, 则 $\mathrm{A} 、 \mathrm{~B}$ 两卫星的周期的比值为（） [图1] A: $k^{\frac{2}{3}}$ B: $k$ C: $k^{2}$ D: $k^{3}$

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个单选题（只有一个正确答案）。 问题: 如图所示, $\mathrm{A} 、 \mathrm{~B}$ 是绕地球做圆周运动的两颗卫星, $\mathrm{A} 、 \mathrm{~B}$ 两卫星与地心的连线在相等时间内扫过的面积之比为 $k: 1$, 则 $\mathrm{A} 、 \mathrm{~B}$ 两卫星的周期的比值为（） [图1] A: $k^{\frac{2}{3}}$ B: $k$ C: $k^{2}$ D: $k^{3}$ 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应为以下选项之一：[A, B, C, D]

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multi-modal

Astronomy_1101

On Earth the Global Positioning System (GPS) requires a minimum of 24 satellites in orbit at any one time (there are typically more than that to allow for redundancies, with the current constellation having more than 30) so that at least 4 are visible above the horizon from anywhere on Earth (necessary for an $\mathrm{x}, \mathrm{y}, \mathrm{z}$ and time co-ordinate). This is achieved by having 6 different orbital planes, separated by $60^{\circ}$, and each orbital plane has 4 satellites. [figure1] Figure 1: The current set up of the GPS system used on Earth. Credits: Left: Peter H. Dana, University of Colorado; Right: GPS Standard Positioning Service Specification, $4^{\text {th }}$ edition The orbits are essentially circular with an eccentricity $<0.02$, an inclination of $55^{\circ}$, and an orbital period of exactly half a sidereal day (called a semi-synchronous orbit). The receiving angle of each satellite's antenna needs to be about $27.8^{\circ}$, and hence about $38 \%$ of the Earth's surface is within each satellite's footprint (see Figure 1), allowing the excellent coverage required.a. Given that the Earth's sidereal day is $23 \mathrm{~h} 56$ mins, calculate the orbital radius of a GPS satellite. Express your answer in units of $R_{\oplus}$.

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 Earth the Global Positioning System (GPS) requires a minimum of 24 satellites in orbit at any one time (there are typically more than that to allow for redundancies, with the current constellation having more than 30) so that at least 4 are visible above the horizon from anywhere on Earth (necessary for an $\mathrm{x}, \mathrm{y}, \mathrm{z}$ and time co-ordinate). This is achieved by having 6 different orbital planes, separated by $60^{\circ}$, and each orbital plane has 4 satellites. [figure1] Figure 1: The current set up of the GPS system used on Earth. Credits: Left: Peter H. Dana, University of Colorado; Right: GPS Standard Positioning Service Specification, $4^{\text {th }}$ edition The orbits are essentially circular with an eccentricity $<0.02$, an inclination of $55^{\circ}$, and an orbital period of exactly half a sidereal day (called a semi-synchronous orbit). The receiving angle of each satellite's antenna needs to be about $27.8^{\circ}$, and hence about $38 \%$ of the Earth's surface is within each satellite's footprint (see Figure 1), allowing the excellent coverage required. problem: a. Given that the Earth's sidereal day is $23 \mathrm{~h} 56$ mins, calculate the orbital radius of a GPS satellite. Express your answer in units of $R_{\oplus}$. 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

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Astronomy

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Astronomy_874

Suppose a spaceship is attempting a slingshot maneuver on a gas giant with mass 100 times that of the spaceship. Because the spaceship somewhat entered the planet's atmosphere, kinetic energy was not conserved - only momentum. What is the ratio of the spaceship's change in velocity to the planet's change in velocity, $\frac{\Delta v_{s}}{\Delta v_{p}}$ ? A: 10 B: 100 C: -10 D: -100

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: Suppose a spaceship is attempting a slingshot maneuver on a gas giant with mass 100 times that of the spaceship. Because the spaceship somewhat entered the planet's atmosphere, kinetic energy was not conserved - only momentum. What is the ratio of the spaceship's change in velocity to the planet's change in velocity, $\frac{\Delta v_{s}}{\Delta v_{p}}$ ? A: 10 B: 100 C: -10 D: -100 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

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text-only

Astronomy_811

Classify the following galaxies according the Hubble galaxies classification: [figure1] Figure 1: Galaxy 1 [figure2] Figure 2: Galaxy 2 [figure3] Figure 3: Galaxy 3 [figure4] Figure 4: Galaxy 4 [figure5] Figure 5: Galaxy 5 A: $\mathrm{Sb}, \mathrm{Sc}$, Peculiar, E2, Irregular B: Sbc, E4, Irregular, Sb, Peculiar C: E3, Sbc, Sa, Peculiar, Irregular D: Sc,Sba, Sbc, E2, Peculiar E: Sa, Sbb, E3, Irregular, Peculiar

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: Classify the following galaxies according the Hubble galaxies classification: [figure1] Figure 1: Galaxy 1 [figure2] Figure 2: Galaxy 2 [figure3] Figure 3: Galaxy 3 [figure4] Figure 4: Galaxy 4 [figure5] Figure 5: Galaxy 5 A: $\mathrm{Sb}, \mathrm{Sc}$, Peculiar, E2, Irregular B: Sbc, E4, Irregular, Sb, Peculiar C: E3, Sbc, Sa, Peculiar, Irregular D: Sc,Sba, Sbc, E2, Peculiar E: Sa, Sbb, E3, Irregular, Peculiar 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

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multi-modal

Astronomy_358

光电效应和康普顿效应深入地揭示了光的粒子性的一面。前者表明光子具有能量,后者表明光子除了具有能量之外还具有动量。由狭义相对论可知, 一定的质量 $m$ 与一 定的能量 $E$ 相对应: $E=m c^{2}$, 其中 $c$ 为真空中光速。 光照射到物体表面时, 光子被物体吸收或反射时, 光都会对物体产生压强, 这就是“光压”。已知太阳半径为 $R$, 单位时间辐射的总能量为 $P_{0}$, 光速为 $c$ 。求:距离太阳中心 $r$ 处 $(r>R)$, 太阳光光子被完全吸收时产生的光压 $p$ 是多大?

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: 光电效应和康普顿效应深入地揭示了光的粒子性的一面。前者表明光子具有能量,后者表明光子除了具有能量之外还具有动量。由狭义相对论可知, 一定的质量 $m$ 与一 定的能量 $E$ 相对应: $E=m c^{2}$, 其中 $c$ 为真空中光速。 光照射到物体表面时, 光子被物体吸收或反射时, 光都会对物体产生压强, 这就是“光压”。已知太阳半径为 $R$, 单位时间辐射的总能量为 $P_{0}$, 光速为 $c$ 。求:距离太阳中心 $r$ 处 $(r>R)$, 太阳光光子被完全吸收时产生的光压 $p$ 是多大? 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

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Astronomy_38

2020 年 7 月 23 日, 我国“天问一号”火星探测器成功发射, 2021 年 2 月 10 日, 顺利进入环火星大椭圆轨道, 并变轨到近火星圆轨道运动, 将于 2021 年 5 月至 6 月择机实施火星着陆, 最终实现“绕、着、巡”三大目标。已知火星质量约为地球的 $\frac{1}{10}$, 半径约为地球的 $\frac{1}{2}$, 地球表面的重力加速度为 $g$, 火星和地球均绕太阳做逆时针方向的匀速圆周运动, 火星的公转周期是地球公转周期的两倍。质量为 $m$ 的着陆器在着陆火星前,会在火星表面附近经历一个时长为 $t$ ，速度由 $v$ 减速到零的过程. 若该减速过程可以视为一个坚直向下的匀减速直线运动, 忽略火星大气阻力, 求: 探测器分别围绕火星和地球做圆周运动一周的最短时间之比。

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: 2020 年 7 月 23 日, 我国“天问一号”火星探测器成功发射, 2021 年 2 月 10 日, 顺利进入环火星大椭圆轨道, 并变轨到近火星圆轨道运动, 将于 2021 年 5 月至 6 月择机实施火星着陆, 最终实现“绕、着、巡”三大目标。已知火星质量约为地球的 $\frac{1}{10}$, 半径约为地球的 $\frac{1}{2}$, 地球表面的重力加速度为 $g$, 火星和地球均绕太阳做逆时针方向的匀速圆周运动, 火星的公转周期是地球公转周期的两倍。质量为 $m$ 的着陆器在着陆火星前,会在火星表面附近经历一个时长为 $t$ ，速度由 $v$ 减速到零的过程. 若该减速过程可以视为一个坚直向下的匀减速直线运动, 忽略火星大气阻力, 求: 探测器分别围绕火星和地球做圆周运动一周的最短时间之比。 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。

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Astronomy

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text-only

Astronomy_490

人造地球卫星的成功发射一般经过三个阶段, 即卫星依次完成近地轨道、转移轨道、预定轨道。。其中近地轨道和预定轨道可认为是卫星做匀速圆周运动, 转移轨道是椭圆轨道。下图为同一颗人造卫星的变轨原理简图:I 为近地圆轨道, III 为预定圆轨道, II 为转移轨道即椭圆轨道, 且 I、II 轨道相切于 $P$ 点, II、III 轨道相切于 $Q$ 点。已知 I 轨道半径为 $R$ 、III 轨道半径为 $4 R$, 卫星在三个轨道上运行的周期分别是 $T_{1} 、 T_{2} 、 T_{3}$, 且已知 $T_{l}=2 \mathrm{~h}$, 卫星在各轨道经过 $P 、 Q$ 两点处的速度大小分别是 $v_{1} P 、 v_{2} P 、 v_{2} Q 、 v_{3} Q$, 则下列关于卫星正确说法是（） [图1] A: $v_{2} P>v_{1} P>v_{3} Q>v_{2} Q$ B: $T_{2} \approx 8 \mathrm{~h}, T_{3}=16 \mathrm{~h}$ C: 卫星在 II 轨道上从 $P$ 点到 $Q$ 点的运行过程中, 动能增大, 引力势能减小, 机械能大小不变 D: 卫星在三个轨道中 I 轨道的机械能最大, III 轨道的机械能最小, II 轨道的机械能居二者之间

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: 人造地球卫星的成功发射一般经过三个阶段, 即卫星依次完成近地轨道、转移轨道、预定轨道。。其中近地轨道和预定轨道可认为是卫星做匀速圆周运动, 转移轨道是椭圆轨道。下图为同一颗人造卫星的变轨原理简图:I 为近地圆轨道, III 为预定圆轨道, II 为转移轨道即椭圆轨道, 且 I、II 轨道相切于 $P$ 点, II、III 轨道相切于 $Q$ 点。已知 I 轨道半径为 $R$ 、III 轨道半径为 $4 R$, 卫星在三个轨道上运行的周期分别是 $T_{1} 、 T_{2} 、 T_{3}$, 且已知 $T_{l}=2 \mathrm{~h}$, 卫星在各轨道经过 $P 、 Q$ 两点处的速度大小分别是 $v_{1} P 、 v_{2} P 、 v_{2} Q 、 v_{3} Q$, 则下列关于卫星正确说法是（） [图1] A: $v_{2} P>v_{1} P>v_{3} Q>v_{2} Q$ B: $T_{2} \approx 8 \mathrm{~h}, T_{3}=16 \mathrm{~h}$ C: 卫星在 II 轨道上从 $P$ 点到 $Q$ 点的运行过程中, 动能增大, 引力势能减小, 机械能大小不变 D: 卫星在三个轨道中 I 轨道的机械能最大, III 轨道的机械能最小, II 轨道的机械能居二者之间 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

ZH

multi-modal

Astronomy_237

18 世纪, 数学家莫佩尔蒂和哲学家伏尔泰, 曾设想“穿透”地球 假设能够沿着地球两极连线开凿一条沿着地轴的隧道贯穿地球, 一个人可以从北极入口由静止自由落入隧道中, 忽略一切阻力, 此人可以从南极出口飞出, 则以下说法正确的是(已知此人的质量 $\mathrm{m}=50 \mathrm{~kg}$; 地球表面处重力加速度 $\mathrm{g}$ 取 $10 \mathrm{~m} / \mathrm{s}^{2}$; 地球半径 $\mathrm{R}=6.4 \times 10^{6} \mathrm{~m}$; 假设地球可视为质量分布均匀的球体, 均匀球壳对壳内任一点处的质点合引力为零)( ) A: 人与地球构成的系统, 由于重力发生变化, 故机械能不守恒 B: 人在下落过程中, 受到的万有引力与到地心的距离成正比 C: 人从北极开始下落, 直到经过地心的过程中, 万有引力对人做功 $\mathrm{W}=1.6 \times 10^{9} \mathrm{~J}$ D: 当人下落经过距地心 $\mathrm{R} / 2$ 瞬间, 人的瞬时速度大小为 $4 \times 10^{3} \mathrm{~m} / \mathrm{s}$

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: 18 世纪, 数学家莫佩尔蒂和哲学家伏尔泰, 曾设想“穿透”地球 假设能够沿着地球两极连线开凿一条沿着地轴的隧道贯穿地球, 一个人可以从北极入口由静止自由落入隧道中, 忽略一切阻力, 此人可以从南极出口飞出, 则以下说法正确的是(已知此人的质量 $\mathrm{m}=50 \mathrm{~kg}$; 地球表面处重力加速度 $\mathrm{g}$ 取 $10 \mathrm{~m} / \mathrm{s}^{2}$; 地球半径 $\mathrm{R}=6.4 \times 10^{6} \mathrm{~m}$; 假设地球可视为质量分布均匀的球体, 均匀球壳对壳内任一点处的质点合引力为零)( ) A: 人与地球构成的系统, 由于重力发生变化, 故机械能不守恒 B: 人在下落过程中, 受到的万有引力与到地心的距离成正比 C: 人从北极开始下落, 直到经过地心的过程中, 万有引力对人做功 $\mathrm{W}=1.6 \times 10^{9} \mathrm{~J}$ D: 当人下落经过距地心 $\mathrm{R} / 2$ 瞬间, 人的瞬时速度大小为 $4 \times 10^{3} \mathrm{~m} / \mathrm{s}$ 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

text-only

Astronomy_1214

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).b. Planetary formation models suggest such a gas dwarf would have a solid rocky core the size of the Earth, with similar composition and density. Calculate a simple estimate of the atmospheric pressure on the rocky surface. Compare your answer to the atmospheric pressure at sea level on the Earth. ( $p_{E}$ $=100 \mathrm{kPa}$ )

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: 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: b. Planetary formation models suggest such a gas dwarf would have a solid rocky core the size of the Earth, with similar composition and density. Calculate a simple estimate of the atmospheric pressure on the rocky surface. Compare your answer to the atmospheric pressure at sea level on the Earth. ( $p_{E}$ $=100 \mathrm{kPa}$ ) 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{~Pa}, 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_1118

What would happen to the Earth's orbit if the Sun suddenly became a black hole with the same mass? A: It would spiral inwards because of the strong gravitational forces B: It would fall on a straight line into the black hole C: It would become an open orbit and the Earth would escape from the Solar System D: Nothing

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 would happen to the Earth's orbit if the Sun suddenly became a black hole with the same mass? A: It would spiral inwards because of the strong gravitational forces B: It would fall on a straight line into the black hole C: It would become an open orbit and the Earth would escape from the Solar System D: Nothing 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_33

火星表面特征非常接近地球，适合人类居住. 近期，我国宇航员王跃正与俄罗斯宇航员一起进行“模拟登火星”实验活动. 已知火星半径是地球半径的 $\frac{1}{2}$, 质量是地球质量的 $\frac{1}{9}$. 地球表面重力加速度是 $g$, 若王跃在地面上能向上跳起的最大高度是 $h$, 在忽略地球、火星自转影响的条件下, 下述分析正确的是( ) A: 王跃在火星表面受到的万有引力是在地球表面受到的万有引力的 $\frac{2}{9}$ 倍 B: 火星表面的重力加速度是 $\frac{4 g}{9}$ C: 火星的第一宇宙速度是地球第一宇宙速度的 $\frac{2}{9}$ 倍 D: 王跃以相同的初速度在火星上起跳时, 可跳的最大高度是 $\frac{9 h}{4}$

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个多选题（有多个正确答案）。 问题: 火星表面特征非常接近地球，适合人类居住. 近期，我国宇航员王跃正与俄罗斯宇航员一起进行“模拟登火星”实验活动. 已知火星半径是地球半径的 $\frac{1}{2}$, 质量是地球质量的 $\frac{1}{9}$. 地球表面重力加速度是 $g$, 若王跃在地面上能向上跳起的最大高度是 $h$, 在忽略地球、火星自转影响的条件下, 下述分析正确的是( ) A: 王跃在火星表面受到的万有引力是在地球表面受到的万有引力的 $\frac{2}{9}$ 倍 B: 火星表面的重力加速度是 $\frac{4 g}{9}$ C: 火星的第一宇宙速度是地球第一宇宙速度的 $\frac{2}{9}$ 倍 D: 王跃以相同的初速度在火星上起跳时, 可跳的最大高度是 $\frac{9 h}{4}$ 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应为两个或更多的选项：[A, B, C, D]

MC

Astronomy

ZH

text-only

Astronomy_1144

A day on Earth can be defined in two ways: relative to the Sun (called solar or synodic time) or relative to the background stars (called sidereal time). The mean solar day is 24 hours (within a few milliseconds), whilst the mean sidereal day is shorter at 23 hours 56 minutes 4 seconds (to the nearest second). The solar day is longer as over the course of a sidereal day the Earth has moved slightly in its orbit around the Sun and so has to rotate slightly further for the Sun to be back in the same direction (see Figure 4). [figure1] Figure 4: A solar day is defined as the time between two consecutive passages of the Sun through the meridian, corresponding to local midday (which in the Northern hemisphere is in the South), whilst a sidereal day is the time for a distant star to do the same. The difference between the two is due to the Earth having moved slightly in its orbit around the Sun. Credit: Wikipedia. The length of a year on Earth is 365.25 solar days (to 2 d.p.), however some ancient civilizations used to believe that there were once exactly 360 solar days in a year, with various myths explaining where the extra days came from. In this question you will look at how to return the Earth to this time. [Note that this question is very sensitive to the precision of the fundamental constants used, so throughout please take $G=6.674 \times 10^{-11} \mathrm{~m}^{3} \mathrm{~kg}^{-1} \mathrm{~s}^{-2}, R_{\oplus}=6371 \mathrm{~km}, M_{\oplus}=5.972 \times 10^{24} \mathrm{~kg}, M_{\odot}=$ $1.989 \times 10^{30} \mathrm{~kg}$ and $1 \mathrm{au}=1.496 \times 10^{11} \mathrm{~m}$.] ## Helpful equations: The moment of inertia, $I$, of a sphere of mass $M$ and radius $R$ is $I=\frac{2}{5} M R^{2}$. The angular momentum, $L$, of a spinning object with an angular velocity of $\omega$ is $L=I \omega=r \times p$, where $p$ is the linear momentum of a point particle a distance $r$ from the axis of rotation. The speed, $v$, of an object in an elliptical orbit of semi-major axis $a$ around an object of mass $M$ when a distance $r$ away can be calculated as $$ v^{2}=G M\left(\frac{2}{r}-\frac{1}{a}\right) $$c. Given that the proton momentum is related to the average kinetic energy of a particle in the plasma by $E_{K}=p^{2} / 2 m_{p}$ calculate the value of $\lambda$ and hence calculate $T_{\text {quantum. }}$. YYou should find that it's below 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: A day on Earth can be defined in two ways: relative to the Sun (called solar or synodic time) or relative to the background stars (called sidereal time). The mean solar day is 24 hours (within a few milliseconds), whilst the mean sidereal day is shorter at 23 hours 56 minutes 4 seconds (to the nearest second). The solar day is longer as over the course of a sidereal day the Earth has moved slightly in its orbit around the Sun and so has to rotate slightly further for the Sun to be back in the same direction (see Figure 4). [figure1] Figure 4: A solar day is defined as the time between two consecutive passages of the Sun through the meridian, corresponding to local midday (which in the Northern hemisphere is in the South), whilst a sidereal day is the time for a distant star to do the same. The difference between the two is due to the Earth having moved slightly in its orbit around the Sun. Credit: Wikipedia. The length of a year on Earth is 365.25 solar days (to 2 d.p.), however some ancient civilizations used to believe that there were once exactly 360 solar days in a year, with various myths explaining where the extra days came from. In this question you will look at how to return the Earth to this time. [Note that this question is very sensitive to the precision of the fundamental constants used, so throughout please take $G=6.674 \times 10^{-11} \mathrm{~m}^{3} \mathrm{~kg}^{-1} \mathrm{~s}^{-2}, R_{\oplus}=6371 \mathrm{~km}, M_{\oplus}=5.972 \times 10^{24} \mathrm{~kg}, M_{\odot}=$ $1.989 \times 10^{30} \mathrm{~kg}$ and $1 \mathrm{au}=1.496 \times 10^{11} \mathrm{~m}$.] ## Helpful equations: The moment of inertia, $I$, of a sphere of mass $M$ and radius $R$ is $I=\frac{2}{5} M R^{2}$. The angular momentum, $L$, of a spinning object with an angular velocity of $\omega$ is $L=I \omega=r \times p$, where $p$ is the linear momentum of a point particle a distance $r$ from the axis of rotation. The speed, $v$, of an object in an elliptical orbit of semi-major axis $a$ around an object of mass $M$ when a distance $r$ away can be calculated as $$ v^{2}=G M\left(\frac{2}{r}-\frac{1}{a}\right) $$ problem: c. Given that the proton momentum is related to the average kinetic energy of a particle in the plasma by $E_{K}=p^{2} / 2 m_{p}$ calculate the value of $\lambda$ and hence calculate $T_{\text {quantum. }}$. YYou should find that it's below 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{~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_162

如图所示, 横截面积为 $A$ 、质量为 $m$ 的柱状飞行器沿半径为 $R$ 的圆形轨道在高空绕地球做无动力运行。将地球看作质量为 $M$ 的均匀球体。万有引力常量为 $G$ 。 在飞行器运行轨道附近范围内有密度为 $\rho$ (恒量) 的稀薄空气。稀薄空气可看成是由彼此没有相互作用的均匀小颗粒组成，所有小颗粒原来都静止。假设每个小颗粒与飞行器碰撞后具有与飞行器相同的速度, 且碰撞时间很短。频繁碰撞会对飞行器产生持续阻力, 飞行器的轨道高度会逐渐降低。观察发现飞行器绕地球运行很多圈之后, 其轨道高度下降了 $\Delta H$ 。由于 $\Delta H \leq R$, 可将飞行器绕地球运动的每一圈运动均视为匀速圆周运动。已知当飞行器到地球球心距离为 $r$ 时, 飞行器与地球组成的系统具有的引力势能 $E_{\mathrm{p}}=-\frac{G M m}{r}$ 。请根据上述条件推导: 飞行器在半径为 $R$ 轨道上运行时, 所受空气阻力大小 $F$ 的表达式; [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, 横截面积为 $A$ 、质量为 $m$ 的柱状飞行器沿半径为 $R$ 的圆形轨道在高空绕地球做无动力运行。将地球看作质量为 $M$ 的均匀球体。万有引力常量为 $G$ 。 在飞行器运行轨道附近范围内有密度为 $\rho$ (恒量) 的稀薄空气。稀薄空气可看成是由彼此没有相互作用的均匀小颗粒组成，所有小颗粒原来都静止。假设每个小颗粒与飞行器碰撞后具有与飞行器相同的速度, 且碰撞时间很短。频繁碰撞会对飞行器产生持续阻力, 飞行器的轨道高度会逐渐降低。观察发现飞行器绕地球运行很多圈之后, 其轨道高度下降了 $\Delta H$ 。由于 $\Delta H \leq R$, 可将飞行器绕地球运动的每一圈运动均视为匀速圆周运动。已知当飞行器到地球球心距离为 $r$ 时, 飞行器与地球组成的系统具有的引力势能 $E_{\mathrm{p}}=-\frac{G M m}{r}$ 。请根据上述条件推导: 飞行器在半径为 $R$ 轨道上运行时, 所受空气阻力大小 $F$ 的表达式; [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_191

航天技术的发展是当今各国综合国力的直接体现，近年来，我国的航天技术取得了让世界瞩目的成绩，也引领科技爱好者思索航天技术的发展，有人就提出了一种不用火箭发射人造地球卫星的设想。其设想如下: 如图所示, 在地球上距地心 $h$ 处沿一条弦挖一光滑通道, 在通道的两个出口处 $A$ 和 $B$ 分别将质量为 $M$ 的物体和质量为 $m$ 的待发射卫星同时自由释放, $M \gg m$, 在中点 $O^{\prime}$ 弹性正撞后，质量为 $m$ 的物体，即待发射的卫星就会从通道口 $B$ 冲出通道, 设置一个装置, 卫星从 $B$ 冲出就把速度变为沿地球切线方向, 但不改变速度大小, 这样就有可能成功发射卫星。已知地球可视为质量分布均匀 的球体, 且质量分布均匀的球壳对壳内物体的引力为零, 地球半径为 $R_{0}$, 表面的重力加速度为 $g$ 。 求在地球内部距球心 $h$ 处的重力加速度大小;[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: 航天技术的发展是当今各国综合国力的直接体现，近年来，我国的航天技术取得了让世界瞩目的成绩，也引领科技爱好者思索航天技术的发展，有人就提出了一种不用火箭发射人造地球卫星的设想。其设想如下: 如图所示, 在地球上距地心 $h$ 处沿一条弦挖一光滑通道, 在通道的两个出口处 $A$ 和 $B$ 分别将质量为 $M$ 的物体和质量为 $m$ 的待发射卫星同时自由释放, $M \gg m$, 在中点 $O^{\prime}$ 弹性正撞后，质量为 $m$ 的物体，即待发射的卫星就会从通道口 $B$ 冲出通道, 设置一个装置, 卫星从 $B$ 冲出就把速度变为沿地球切线方向, 但不改变速度大小, 这样就有可能成功发射卫星。已知地球可视为质量分布均匀 的球体, 且质量分布均匀的球壳对壳内物体的引力为零, 地球半径为 $R_{0}$, 表面的重力加速度为 $g$ 。 求在地球内部距球心 $h$ 处的重力加速度大小;[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_553

已知地球半径为 $R$, 地球表面的重力加速度为 $g$ 。质量为 $m$ 的宇宙飞船在半径为 $2 R$的轨道 1 上绕地球中心 $O$ 做圆两运动。现飞船在轨道 1 的 $A$ 点加速到陏圆轨道 2 上,再在远地点 $B$ 点加速, 从而使飞船转移到半径为 $4 R$ 的轨道 3 上, 如图所示。若相距 $r$的两物体间引力势能为 $E_{\mathrm{p}}=-G \frac{M m}{r}$, 求: 若要实现由 $A$ 点从轨道 1 转移到轨道 2 , 需要在轨道 1 上对飞船做多少功? [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: 已知地球半径为 $R$, 地球表面的重力加速度为 $g$ 。质量为 $m$ 的宇宙飞船在半径为 $2 R$的轨道 1 上绕地球中心 $O$ 做圆两运动。现飞船在轨道 1 的 $A$ 点加速到陏圆轨道 2 上,再在远地点 $B$ 点加速, 从而使飞船转移到半径为 $4 R$ 的轨道 3 上, 如图所示。若相距 $r$的两物体间引力势能为 $E_{\mathrm{p}}=-G \frac{M m}{r}$, 求: 若要实现由 $A$ 点从轨道 1 转移到轨道 2 , 需要在轨道 1 上对飞船做多少功? [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_1049

The Sun is our closest star so it is arguably the most studied. Results from detailed observations of how sound waves propagate through the plasma of the Sun allow us to get a sense of the general structure of the Sun, with an upper layer of convecting plasma (leading to the 'bubbling' we see with granulation on the surface) and radiative heat transfer below, including a central core region where the pressure and temperature are large enough for nuclear fusion to occur (see Figure 3). [figure1] Figure 3: Left: The general structure of the Sun, with a core in which all the nuclear reactions take place, and convective cells in an outer layer. Credit: Dmitri Pogosyan / University of Alberta. Right: The fraction of the Sun's mass and the contribution to the Sun's luminosity as a function of solar radius as determined from detailed computer simulations. Essentially all of the nuclear reactions creating the photons for the Sun's luminosity take place in a core with a radius of $0.20 R_{\odot}$ and a mass of $0.35 M_{\odot}$. Credit: Kevin France / University of Colorado. Estimating the conditions in the cores of stars is an important aspect of constructing stellar models. This question explores some of the equations governing stellar structure and estimates the central temperature and pressure of the Sun. The primary nuclear fusion pathway responsible for much of the Sun's luminous output is called the proton-proton chain (p-p chain). All of the most common steps are shown in Figure 4. [figure2] Figure 4: An overview of the all the steps in the most common form of the p-p chain, turning four protons into one helium-4 nucleus plus some light particles and photons. Credit: Wikipedia. The net reaction of the p-p chain is $$ 4_{1}^{1} \mathrm{H} \rightarrow{ }_{2}^{4} \mathrm{He}+2 e^{+}+2 \nu_{e}+2 \gamma . $$ The rate-limiting step is the first one $\left({ }_{1}^{1} \mathrm{H}+{ }_{1}^{1} \mathrm{H}\right)$ as the probability of interaction is so low, since it requires the decay of a proton into a neutron and so involves the weak nuclear force. Nuclear physics allow us to calculate the reaction rate coefficient, $R$, energy produced per reaction, $Q$, and hence the energy generation rate, $q$. The reaction rate is related to the number of particles that have enough energy to undergo quantum tunnelling, and the distribution as a function of energy is known as the Gamow peak, with the top of the curve at energy $E_{0}$. For two nuclei, ${ }_{Z_{i}}^{A_{i}} C_{i}$ and ${ }_{Z_{j}}^{A_{j}} C_{j}$, with mass fractions $X_{i}$ and $X_{j}$, then to first order and ignoring electron screening, $$ R=\frac{4}{3^{2.5} \pi^{2}} \frac{h}{\mu_{r} m_{p}} \frac{4 \pi \varepsilon_{0}}{Z_{i} Z_{j} e^{2}} S\left(E_{0}\right) \tau^{2} e^{-\tau}, \quad \text { where } \quad \mu_{r}=\frac{A_{i} A_{j}}{A_{i}+A_{j}} $$ and $S\left(E_{0}\right)$ measures the probability of interaction at the maximum of the Gamow peak whilst $\tau$ is a characteristic width of the Gamow peak, $$ \tau=\frac{3 E_{0}}{k_{B} T} \quad \text { where } \quad E_{0}=\left(\frac{b k_{B} T}{2}\right)^{2 / 3} \quad \text { given } \quad b=\sqrt{\frac{\mu_{r} m_{p}}{2}} \frac{\pi Z_{i} Z_{j} e^{2}}{h \varepsilon_{0}} $$ Here $h$ is Planck's constant, $\varepsilon_{0}$ is the permittivity of free space and $e$ is the elementary charge (the charge on a proton). Finally, the energy generation rate per unit mass is $$ q=\frac{\rho}{m_{p}^{2}}\left(\frac{1}{1+\delta_{i j}}\right) \frac{X_{i} X_{j}}{A_{i} A_{j}} R Q $$ where $\delta_{i j}$ is the Kronecker delta, so equals 1 when $i=j$ and 0 otherwise. Evaluating the fundamental constants and defining $T_{6} \equiv \frac{T}{10^{6} \mathrm{~K}}$ gives $$ \tau=42.59\left[Z_{i}^{2} Z_{j}^{2} \mu_{r} T_{6}^{-1}\right]^{1 / 3}, $$ whilst for the proton-proton interaction $Q=13.366 \mathrm{MeV}$ (half the overall energy of the p-p chain), $Z_{i}=Z_{j}=A_{i}=A_{j}=\delta_{i j}=1$, and $S\left(E_{0}\right)$ is $4.01 \times 10^{-50} \mathrm{keV} \mathrm{m}^{2}$ (Adelberger et al. 2011), so $R=6.55 \times 10^{-43} T_{6}^{-2 / 3} e^{-33.80 T_{6}^{-1 / 3}} \mathrm{~m}^{3} \mathrm{~s}^{-1}$ and $q=0.251 \rho X^{2} T_{6}^{-2 / 3} e^{-33.80 T_{6}^{-1 / 3}} \mathrm{~W} \mathrm{~kg}^{-1}$.c. Considering the evaluated equations for $\tau, R$, and $q$ we can use this with the measured luminosity of the Sun to get a new estimate for the central temperature. iv. The carbon-nitrogen-oxygen (CNO) cycle is an alternative pathway that becomes important at higher temperatures, where heavier elements catalyse the process of turning hydrogen into helium. There the rate limiting step is between nitrogen-14 and hydrogen-1. Compare the temperature dependence of the CNO cycle to the p-p chain at the Sun's central temperature.

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 Sun is our closest star so it is arguably the most studied. Results from detailed observations of how sound waves propagate through the plasma of the Sun allow us to get a sense of the general structure of the Sun, with an upper layer of convecting plasma (leading to the 'bubbling' we see with granulation on the surface) and radiative heat transfer below, including a central core region where the pressure and temperature are large enough for nuclear fusion to occur (see Figure 3). [figure1] Figure 3: Left: The general structure of the Sun, with a core in which all the nuclear reactions take place, and convective cells in an outer layer. Credit: Dmitri Pogosyan / University of Alberta. Right: The fraction of the Sun's mass and the contribution to the Sun's luminosity as a function of solar radius as determined from detailed computer simulations. Essentially all of the nuclear reactions creating the photons for the Sun's luminosity take place in a core with a radius of $0.20 R_{\odot}$ and a mass of $0.35 M_{\odot}$. Credit: Kevin France / University of Colorado. Estimating the conditions in the cores of stars is an important aspect of constructing stellar models. This question explores some of the equations governing stellar structure and estimates the central temperature and pressure of the Sun. The primary nuclear fusion pathway responsible for much of the Sun's luminous output is called the proton-proton chain (p-p chain). All of the most common steps are shown in Figure 4. [figure2] Figure 4: An overview of the all the steps in the most common form of the p-p chain, turning four protons into one helium-4 nucleus plus some light particles and photons. Credit: Wikipedia. The net reaction of the p-p chain is $$ 4_{1}^{1} \mathrm{H} \rightarrow{ }_{2}^{4} \mathrm{He}+2 e^{+}+2 \nu_{e}+2 \gamma . $$ The rate-limiting step is the first one $\left({ }_{1}^{1} \mathrm{H}+{ }_{1}^{1} \mathrm{H}\right)$ as the probability of interaction is so low, since it requires the decay of a proton into a neutron and so involves the weak nuclear force. Nuclear physics allow us to calculate the reaction rate coefficient, $R$, energy produced per reaction, $Q$, and hence the energy generation rate, $q$. The reaction rate is related to the number of particles that have enough energy to undergo quantum tunnelling, and the distribution as a function of energy is known as the Gamow peak, with the top of the curve at energy $E_{0}$. For two nuclei, ${ }_{Z_{i}}^{A_{i}} C_{i}$ and ${ }_{Z_{j}}^{A_{j}} C_{j}$, with mass fractions $X_{i}$ and $X_{j}$, then to first order and ignoring electron screening, $$ R=\frac{4}{3^{2.5} \pi^{2}} \frac{h}{\mu_{r} m_{p}} \frac{4 \pi \varepsilon_{0}}{Z_{i} Z_{j} e^{2}} S\left(E_{0}\right) \tau^{2} e^{-\tau}, \quad \text { where } \quad \mu_{r}=\frac{A_{i} A_{j}}{A_{i}+A_{j}} $$ and $S\left(E_{0}\right)$ measures the probability of interaction at the maximum of the Gamow peak whilst $\tau$ is a characteristic width of the Gamow peak, $$ \tau=\frac{3 E_{0}}{k_{B} T} \quad \text { where } \quad E_{0}=\left(\frac{b k_{B} T}{2}\right)^{2 / 3} \quad \text { given } \quad b=\sqrt{\frac{\mu_{r} m_{p}}{2}} \frac{\pi Z_{i} Z_{j} e^{2}}{h \varepsilon_{0}} $$ Here $h$ is Planck's constant, $\varepsilon_{0}$ is the permittivity of free space and $e$ is the elementary charge (the charge on a proton). Finally, the energy generation rate per unit mass is $$ q=\frac{\rho}{m_{p}^{2}}\left(\frac{1}{1+\delta_{i j}}\right) \frac{X_{i} X_{j}}{A_{i} A_{j}} R Q $$ where $\delta_{i j}$ is the Kronecker delta, so equals 1 when $i=j$ and 0 otherwise. Evaluating the fundamental constants and defining $T_{6} \equiv \frac{T}{10^{6} \mathrm{~K}}$ gives $$ \tau=42.59\left[Z_{i}^{2} Z_{j}^{2} \mu_{r} T_{6}^{-1}\right]^{1 / 3}, $$ whilst for the proton-proton interaction $Q=13.366 \mathrm{MeV}$ (half the overall energy of the p-p chain), $Z_{i}=Z_{j}=A_{i}=A_{j}=\delta_{i j}=1$, and $S\left(E_{0}\right)$ is $4.01 \times 10^{-50} \mathrm{keV} \mathrm{m}^{2}$ (Adelberger et al. 2011), so $R=6.55 \times 10^{-43} T_{6}^{-2 / 3} e^{-33.80 T_{6}^{-1 / 3}} \mathrm{~m}^{3} \mathrm{~s}^{-1}$ and $q=0.251 \rho X^{2} T_{6}^{-2 / 3} e^{-33.80 T_{6}^{-1 / 3}} \mathrm{~W} \mathrm{~kg}^{-1}$. problem: c. Considering the evaluated equations for $\tau, R$, and $q$ we can use this with the measured luminosity of the Sun to get a new estimate for the central temperature. iv. The carbon-nitrogen-oxygen (CNO) cycle is an alternative pathway that becomes important at higher temperatures, where heavier elements catalyse the process of turning hydrogen into helium. There the rate limiting step is between nitrogen-14 and hydrogen-1. Compare the temperature dependence of the CNO cycle to the p-p chain at the Sun's central temperature. 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 the numerical value.

NV

Astronomy

EN

multi-modal

Astronomy_10

如图所示, 人造地球卫星 $M 、 N$ 在同一平面内同方向绕地心 $O$ 做匀速圆周运动, $M 、 N$ 连线与 $M 、 O$ 连线间的夹角用 $\theta$ 表示. 已知 $\theta$ 从 $0^{\circ}$ 变到最大值 $30^{\circ}$ 经历的最短时间为 $t$, 万有引力常致为 $G$, 由已知的数据, 以下说法正确的是 ( ) [图1] A: 可以求出地球的质量 B: $M 、 N$ 的线速度之比为 $1: 2$ C: $M 、 N$ 的周期之比为 3: 1 D: 卫星 $N$ 的角速度为 $\frac{2 \pi(4+\sqrt{2})}{21 t}$

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: 如图所示, 人造地球卫星 $M 、 N$ 在同一平面内同方向绕地心 $O$ 做匀速圆周运动, $M 、 N$ 连线与 $M 、 O$ 连线间的夹角用 $\theta$ 表示. 已知 $\theta$ 从 $0^{\circ}$ 变到最大值 $30^{\circ}$ 经历的最短时间为 $t$, 万有引力常致为 $G$, 由已知的数据, 以下说法正确的是 ( ) [图1] A: 可以求出地球的质量 B: $M 、 N$ 的线速度之比为 $1: 2$ C: $M 、 N$ 的周期之比为 3: 1 D: 卫星 $N$ 的角速度为 $\frac{2 \pi(4+\sqrt{2})}{21 t}$ 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_1084

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 $$ \Delta l=\int_{r_{2}}^{r_{1}}\left(1-\frac{2 r_{g}}{r}\right)^{-1 / 2} \mathrm{~d} r $$c. The angular diameter of M87* as determined from the images gained by the EHT (shown in Fig 4) is 42 microarcseconds, and the galaxy is $16.8 \mathrm{Mpc}$ away from us. Determine the minimum and maximum possible masses of the $\mathrm{SMBH}$ in units of $\mathrm{M}_{\odot}$.

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 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: c. The angular diameter of M87* as determined from the images gained by the EHT (shown in Fig 4) is 42 microarcseconds, and the galaxy is $16.8 \mathrm{Mpc}$ away from us. Determine the minimum and maximum possible masses of the $\mathrm{SMBH}$ in units of $\mathrm{M}_{\odot}$. 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_686

如图所示, “火星”探测飞行器 $P$ 绕火星做匀速圆周运动, 若“火星”探测飞行器某时刻的轨道半径为 $r$, 探测飞行器 $\mathrm{P}$ 观测火星的最大张角为 $\beta$, 下列说法正确的是 ( ) [图1] A: 探测飞行器 $\mathrm{P}$ 的轨道半径 $r$ 越大, 其周期越小 B: 探测飞行器 $\mathrm{P}$ 的轨道半径 $r$ 越大, 其速度越大 C: 若测得周期和张角, 可得到火星的平均密度 D: 若测得周期和轨道半径, 可得到探测器 $P$ 的质量

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: 如图所示, “火星”探测飞行器 $P$ 绕火星做匀速圆周运动, 若“火星”探测飞行器某时刻的轨道半径为 $r$, 探测飞行器 $\mathrm{P}$ 观测火星的最大张角为 $\beta$, 下列说法正确的是 ( ) [图1] A: 探测飞行器 $\mathrm{P}$ 的轨道半径 $r$ 越大, 其周期越小 B: 探测飞行器 $\mathrm{P}$ 的轨道半径 $r$ 越大, 其速度越大 C: 若测得周期和张角, 可得到火星的平均密度 D: 若测得周期和轨道半径, 可得到探测器 $P$ 的质量 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_11

2019 年 1 月 3 日, 我国“嫦娥四号”探测器在月球背面成功着陆并发回大量月背影 像. 如图所示为位于月球背面的“嫦娥四号”探测器 $A$ 通过“鹊桥”中继站 $B$ 向地球传输电磁波信息的示意图. 拉格朗日 $L_{2}$ 点位于地月连线延长线上, “鹊桥”的运动可看成如下两种运动的合运动: 一是在地球和月球引力共同作用下, “鹊桥”在 $L_{2}$ 点附近与月球以相同的周期 $T_{0}$ 一起绕地球做匀速圆周运动; 二是在与地月连线垂直的平面内绕 $L_{2}$ 点做匀速圆周运动. 已知地球的质量为月球质量的 $n$ 倍, 地球到 $L_{2}$ 点的距离为月球到 $L_{2}$ 点的距离的 $k$ 倍, 地球半径、月球半径以及“鹊桥”绕 $L_{2}$ 点做匀速圆周运动的半径均远小于月球到 $L_{2}$ 点的距离 (提示: “鹊桥”绕 $L_{2}$ 点做匀速圆周运动的向心力由地球和月球对其引力在过 $L_{2}$ 点与地月连线垂直的平面内的分量提供).[图1] 若月球到 $L_{2}$ 点的距离 $r=6.5 \times 10^{7} \mathrm{~m}, k=7$, “鹊桥”接收到“嫦娥四号”传来的信息后需经 $t_{0}=60.0 \mathrm{~s}$ 处理才能发出, 试估算“嫦娥四号”从发出信息到传回地球的最短时间（保留三位有效数字);

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: 2019 年 1 月 3 日, 我国“嫦娥四号”探测器在月球背面成功着陆并发回大量月背影 像. 如图所示为位于月球背面的“嫦娥四号”探测器 $A$ 通过“鹊桥”中继站 $B$ 向地球传输电磁波信息的示意图. 拉格朗日 $L_{2}$ 点位于地月连线延长线上, “鹊桥”的运动可看成如下两种运动的合运动: 一是在地球和月球引力共同作用下, “鹊桥”在 $L_{2}$ 点附近与月球以相同的周期 $T_{0}$ 一起绕地球做匀速圆周运动; 二是在与地月连线垂直的平面内绕 $L_{2}$ 点做匀速圆周运动. 已知地球的质量为月球质量的 $n$ 倍, 地球到 $L_{2}$ 点的距离为月球到 $L_{2}$ 点的距离的 $k$ 倍, 地球半径、月球半径以及“鹊桥”绕 $L_{2}$ 点做匀速圆周运动的半径均远小于月球到 $L_{2}$ 点的距离 (提示: “鹊桥”绕 $L_{2}$ 点做匀速圆周运动的向心力由地球和月球对其引力在过 $L_{2}$ 点与地月连线垂直的平面内的分量提供).[图1] 若月球到 $L_{2}$ 点的距离 $r=6.5 \times 10^{7} \mathrm{~m}, k=7$, “鹊桥”接收到“嫦娥四号”传来的信息后需经 $t_{0}=60.0 \mathrm{~s}$ 处理才能发出, 试估算“嫦娥四号”从发出信息到传回地球的最短时间（保留三位有效数字); 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。

NV

Astronomy

ZH

multi-modal

Astronomy_671

如图所示, 北斗卫星导航系统中的一颗卫星 $a$ 位于赤道上空, 其对地张角为 $60^{\circ}$ 。已知地球的半径为 $R$, 自转周期为 $T_{0}$, 表面的重力加速度为 $g$, 万有引力常量为 $G$ 。根据题中条件，可求出（） [图1] A: 地球的平均密度为 $\frac{3 \pi}{G T_{0}^{2}}$ B: 静止卫星的轨道半径为 $\sqrt[3]{\frac{g T_{0}^{2} R^{2}}{4 \pi^{2}}}$ C: 卫星 $a$ 的周期为 $2 \sqrt{2} T_{0}$ D: $a$ 与近地卫星运行方向相反时, 二者不能直接通讯的连续时间为 $\frac{8 \pi \sqrt{2 g R}}{3(2 \sqrt{2}+1) g}$

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: 如图所示, 北斗卫星导航系统中的一颗卫星 $a$ 位于赤道上空, 其对地张角为 $60^{\circ}$ 。已知地球的半径为 $R$, 自转周期为 $T_{0}$, 表面的重力加速度为 $g$, 万有引力常量为 $G$ 。根据题中条件，可求出（） [图1] A: 地球的平均密度为 $\frac{3 \pi}{G T_{0}^{2}}$ B: 静止卫星的轨道半径为 $\sqrt[3]{\frac{g T_{0}^{2} R^{2}}{4 \pi^{2}}}$ C: 卫星 $a$ 的周期为 $2 \sqrt{2} T_{0}$ D: $a$ 与近地卫星运行方向相反时, 二者不能直接通讯的连续时间为 $\frac{8 \pi \sqrt{2 g R}}{3(2 \sqrt{2}+1) g}$ 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

multi-modal

Astronomy_284

如图为某双星系统 $\mathrm{A} 、 \mathrm{~B}$ 绕其连线上的 $O$ 点做匀速圆周运动的示意图, 若 $\mathrm{A}$ 星的轨道半径大于 $\mathrm{B}$ 星的轨道半径, 双星的总质量 $M$, 双星间的距离为 $L$, 其运动周期为 $T$, 则 ( ) [图1] A: $\mathrm{A}$ 的质量一定大于 $\mathrm{B}$ 的质量 B: $\mathrm{A}$ 的加速度一定大于 $\mathrm{B}$ 的加速度 C: $L$ 一定时, $M$ 越小, $T$ 越大 D: $L$ 一定时, $\mathrm{A}$ 的质量减小 $\Delta m$ 而 $\mathrm{B}$ 的质量增加 $\Delta m$, 它们的向心力减小

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个多选题（有多个正确答案）。 问题: 如图为某双星系统 $\mathrm{A} 、 \mathrm{~B}$ 绕其连线上的 $O$ 点做匀速圆周运动的示意图, 若 $\mathrm{A}$ 星的轨道半径大于 $\mathrm{B}$ 星的轨道半径, 双星的总质量 $M$, 双星间的距离为 $L$, 其运动周期为 $T$, 则 ( ) [图1] A: $\mathrm{A}$ 的质量一定大于 $\mathrm{B}$ 的质量 B: $\mathrm{A}$ 的加速度一定大于 $\mathrm{B}$ 的加速度 C: $L$ 一定时, $M$ 越小, $T$ 越大 D: $L$ 一定时, $\mathrm{A}$ 的质量减小 $\Delta m$ 而 $\mathrm{B}$ 的质量增加 $\Delta m$, 它们的向心力减小 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应为两个或更多的选项：[A, B, C, D]

MC

Astronomy

ZH

multi-modal

Astronomy_743

What is the name of the JWST component highlighted below? [figure1] A: Stabilization flap B: Spacecraft bus C: Antenna D: Star tracker

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 name of the JWST component highlighted below? [figure1] A: Stabilization flap B: Spacecraft bus C: Antenna D: Star tracker 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_495

当某一地外行星 (火星、木星、土星、天王星、海王星) 于绕日公转过程中运行到试卷第 48 页，共 150 页 与地球、太阳成一直线的状态, 且地球恰好位于太阳和外行星之间的这种天文现象叫“冲日”, 冲日前后是观测地外行星的好时机。如图所示是土星冲日示意图, 已知地球质量为 $M$, 半径为 $R$, 公转周期是 1 年, 公转半径为 $r$, 土星质量是地球的 95 倍, 土星半径是地球的 9.5 倍, 土星的公转半径是地球的 9.5 倍。求: $\left(\sqrt{9.5^{3}} \approx 29\right)$ 土星的第一宇宙速度是地球的几倍? [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: 当某一地外行星 (火星、木星、土星、天王星、海王星) 于绕日公转过程中运行到试卷第 48 页，共 150 页 与地球、太阳成一直线的状态, 且地球恰好位于太阳和外行星之间的这种天文现象叫“冲日”, 冲日前后是观测地外行星的好时机。如图所示是土星冲日示意图, 已知地球质量为 $M$, 半径为 $R$, 公转周期是 1 年, 公转半径为 $r$, 土星质量是地球的 95 倍, 土星半径是地球的 9.5 倍, 土星的公转半径是地球的 9.5 倍。求: $\left(\sqrt{9.5^{3}} \approx 29\right)$ 土星的第一宇宙速度是地球的几倍? [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_129

地球同步卫星的发射方法是变轨发射, 如图所示, 先把卫星发射到近地圆形轨道I 上, 当卫星到达 $P$ 点时, 发动机点火。使卫星进入粗圆轨道II, 其远地点恰好在地球赤道上空约 $36000 \mathrm{~km}$ 处, 当卫星到达远地点 $Q$ 时, 发动机再次点火。使之进入同步轨道 III, 已知地球赤道上的重力加速度为 $g$, 物体在赤道表面上随地球自转的向心加速度大小为 $a$, 下列说法正确的是如果地球自转的（） [图1] A: 角速度突然变为原来的 $\frac{g+a}{a}$ 倍, 那么赤道上的物体将会飘起来 B: 卫星与地心连线在轨道II上单位时间内扫过的面积小于在轨道III上单位时间内扫过的面积 C: 卫星在轨道III上运行时的机械能小于在轨道I上运行时的机械能 D: 卫星在远地点 $Q$ 时的速度可能大于第一宇宙速度

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个单选题（只有一个正确答案）。 问题: 地球同步卫星的发射方法是变轨发射, 如图所示, 先把卫星发射到近地圆形轨道I 上, 当卫星到达 $P$ 点时, 发动机点火。使卫星进入粗圆轨道II, 其远地点恰好在地球赤道上空约 $36000 \mathrm{~km}$ 处, 当卫星到达远地点 $Q$ 时, 发动机再次点火。使之进入同步轨道 III, 已知地球赤道上的重力加速度为 $g$, 物体在赤道表面上随地球自转的向心加速度大小为 $a$, 下列说法正确的是如果地球自转的（） [图1] A: 角速度突然变为原来的 $\frac{g+a}{a}$ 倍, 那么赤道上的物体将会飘起来 B: 卫星与地心连线在轨道II上单位时间内扫过的面积小于在轨道III上单位时间内扫过的面积 C: 卫星在轨道III上运行时的机械能小于在轨道I上运行时的机械能 D: 卫星在远地点 $Q$ 时的速度可能大于第一宇宙速度 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

multi-modal

Astronomy_892

One possible theory for why the gas giants have ring systems is that a small moon got too close to the parent planet. When the gravitational tidal forces (due to the difference between the strength of the planet's pull on the near and far sides of the moon) became greater than the gravitational forces holding the moon together, it was ripped apart. This minimum distance is called the "Roche limit", named after the French astronomer Edouard Roche who first calculated it. It is defined as when the gravitational force generated by the moon at its surface is equal to the tidal forces it experiences at that distance. [figure1] Consider a spherical planet with mass $M$ and radius $R$, and a perfectly rigid spherical moon with mass $m$ and radius $r$, orbiting the planet in a circular orbit of radius $d$. For a small particle of mass $u$ on the surface of the moon, the gravitational and tidal forces it experiences will be $$ F_{\text {grav }}=\frac{G m u}{r^{2}} \quad F_{\text {tidal }}=\frac{2 G M u r}{d^{3}} $$ By making these two expressions equal, derive an expression for the Roche limit, $d_{R L}$, purely in terms of $R$ and the uniform densities of the planet and the moon ( $\rho_{P}$ and $\rho_{m}$ respectively)

You are participating in an international Astronomy competition and need to solve the following question. The answer to this question is an expression. problem: One possible theory for why the gas giants have ring systems is that a small moon got too close to the parent planet. When the gravitational tidal forces (due to the difference between the strength of the planet's pull on the near and far sides of the moon) became greater than the gravitational forces holding the moon together, it was ripped apart. This minimum distance is called the "Roche limit", named after the French astronomer Edouard Roche who first calculated it. It is defined as when the gravitational force generated by the moon at its surface is equal to the tidal forces it experiences at that distance. [figure1] Consider a spherical planet with mass $M$ and radius $R$, and a perfectly rigid spherical moon with mass $m$ and radius $r$, orbiting the planet in a circular orbit of radius $d$. For a small particle of mass $u$ on the surface of the moon, the gravitational and tidal forces it experiences will be $$ F_{\text {grav }}=\frac{G m u}{r^{2}} \quad F_{\text {tidal }}=\frac{2 G M u r}{d^{3}} $$ By making these two expressions equal, derive an expression for the Roche limit, $d_{R L}$, purely in terms of $R$ and the uniform densities of the planet and the moon ( $\rho_{P}$ and $\rho_{m}$ respectively) 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_975

Special Relativity (SR) tells us that two observers will disagree about the duration of a time interval measured by each one's clock if one is moving at speed $v$ relative to the other, a phenomenon called time dilation. General Relativity (GR) tells us that gravitational fields dilate time too. This has an impact on satellites, since they travel at high orbital speeds (slowing down their clocks relative to the surface) but due to their altitude they are in a weaker gravitational field (speeding up their clocks relative to the surface). Which effect is dominant varies with orbital radius. Global Positioning System (GPS) satellites must compensate for this effect, since the satellites rely on accurate measurements of the time between sending and receiving a radio signal. [figure1] Figure 4: A scale diagram of the positions of the orbits for the International Space Station (ISS), GPS satellites and geostationary satellites, along with their orbital periods In $\mathrm{SR}$, time dilation can be calculated with $$ t^{\prime}=\gamma t_{0} \quad \text { where } \quad \gamma=\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}} \quad \text { so } \quad \Delta t_{\mathrm{SR}}=t_{0}-t^{\prime}=(1-\gamma) t_{0} $$ where $t_{0}$ is the time measured by the moving clock, $t^{\prime}$ is the time measured by the observer, $c$ is the speed of light and $v$ is the speed of the object. A negative $\Delta t$ indicates that the clocks are passing time slower relative to the observer, whilst a positive indicates they are passing quicker. GPS satellites have a period of exactly half a day. Use this to determine their orbital speed and hence show that for them $\Delta t_{\mathrm{SR}} \approx-7 \mu$ when $t_{0}=1$ day. In GR, the overall effect (taking into account both the orbital motion and changing gravitational field strength) can be calculated by considering the measurements of time passing on the surface of the Earth and on the satellite as taken by an observer infinitely far away from Earth (so outside the gravitational field). It can be shown that $$ \Delta t_{\text {overall }}=\left(\Gamma_{\mathrm{GPS}}-\Gamma_{\mathrm{E}}\right) t_{0} \quad \text { where } \quad \Gamma_{\mathrm{GPS}}=\sqrt{1-\frac{3 G M_{E}}{a_{\mathrm{GPS}} c^{2}}} \quad \text { and } \quad \Gamma_{\mathrm{E}}=\sqrt{1-\frac{2 G M_{E}}{R_{E} c^{2}}} $$

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: Special Relativity (SR) tells us that two observers will disagree about the duration of a time interval measured by each one's clock if one is moving at speed $v$ relative to the other, a phenomenon called time dilation. General Relativity (GR) tells us that gravitational fields dilate time too. This has an impact on satellites, since they travel at high orbital speeds (slowing down their clocks relative to the surface) but due to their altitude they are in a weaker gravitational field (speeding up their clocks relative to the surface). Which effect is dominant varies with orbital radius. Global Positioning System (GPS) satellites must compensate for this effect, since the satellites rely on accurate measurements of the time between sending and receiving a radio signal. [figure1] Figure 4: A scale diagram of the positions of the orbits for the International Space Station (ISS), GPS satellites and geostationary satellites, along with their orbital periods In $\mathrm{SR}$, time dilation can be calculated with $$ t^{\prime}=\gamma t_{0} \quad \text { where } \quad \gamma=\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}} \quad \text { so } \quad \Delta t_{\mathrm{SR}}=t_{0}-t^{\prime}=(1-\gamma) t_{0} $$ where $t_{0}$ is the time measured by the moving clock, $t^{\prime}$ is the time measured by the observer, $c$ is the speed of light and $v$ is the speed of the object. A negative $\Delta t$ indicates that the clocks are passing time slower relative to the observer, whilst a positive indicates they are passing quicker. GPS satellites have a period of exactly half a day. Use this to determine their orbital speed and hence show that for them $\Delta t_{\mathrm{SR}} \approx-7 \mu$ when $t_{0}=1$ day. In GR, the overall effect (taking into account both the orbital motion and changing gravitational field strength) can be calculated by considering the measurements of time passing on the surface of the Earth and on the satellite as taken by an observer infinitely far away from Earth (so outside the gravitational field). It can be shown that $$ \Delta t_{\text {overall }}=\left(\Gamma_{\mathrm{GPS}}-\Gamma_{\mathrm{E}}\right) t_{0} \quad \text { where } \quad \Gamma_{\mathrm{GPS}}=\sqrt{1-\frac{3 G M_{E}}{a_{\mathrm{GPS}} c^{2}}} \quad \text { and } \quad \Gamma_{\mathrm{E}}=\sqrt{1-\frac{2 G M_{E}}{R_{E} c^{2}}} $$ 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_64

由多颗星体构成的系统, 叫做多星系统. 有这样一种简单的四星系统: 质量刚好都相同的四个星体 $\mathrm{A} 、 \mathrm{~B} 、 \mathrm{C} 、 \mathrm{D}, \mathrm{A} 、 \mathrm{~B} 、 \mathrm{C}$ 分别位于等边三角形的三个顶点上, $\mathrm{D}$ 位于等边三角形的中心. 在四者相互之间的万有引力作用下, $\mathrm{D}$ 静止不动, $\mathrm{A} 、 \mathrm{~B} 、 \mathrm{C}$ 绕共同的圆心 $\mathrm{D}$ 在等边三角形所在的平面内做相同周期的圆周运动. 若四个星体的质量均为 $m$, 三角形的边长为 $a$, 引力常量为 $G$, 则下列说法正确的是 [图1] A: $\mathrm{A} 、 \mathrm{~B} 、 \mathrm{C}$ 三个星体做圆周运动的半径均为 $\frac{\sqrt{3}}{2} a$ B: A、B 两个星体之间的万有引力大小为 $\frac{G m^{2}}{a^{2}}$ C: $\mathrm{A} 、 \mathrm{~B} 、 \mathrm{C}$ 三个星体做圆周运动的向心加速度大小均为 $\frac{(\sqrt{3}+3) G m}{a^{2}}$ D: A、B、C 三个星体做圆周运动的周期均为 $2 \pi a \sqrt{\frac{a}{(3+\sqrt{3}) G m}}$

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个多选题（有多个正确答案）。 问题: 由多颗星体构成的系统, 叫做多星系统. 有这样一种简单的四星系统: 质量刚好都相同的四个星体 $\mathrm{A} 、 \mathrm{~B} 、 \mathrm{C} 、 \mathrm{D}, \mathrm{A} 、 \mathrm{~B} 、 \mathrm{C}$ 分别位于等边三角形的三个顶点上, $\mathrm{D}$ 位于等边三角形的中心. 在四者相互之间的万有引力作用下, $\mathrm{D}$ 静止不动, $\mathrm{A} 、 \mathrm{~B} 、 \mathrm{C}$ 绕共同的圆心 $\mathrm{D}$ 在等边三角形所在的平面内做相同周期的圆周运动. 若四个星体的质量均为 $m$, 三角形的边长为 $a$, 引力常量为 $G$, 则下列说法正确的是 [图1] A: $\mathrm{A} 、 \mathrm{~B} 、 \mathrm{C}$ 三个星体做圆周运动的半径均为 $\frac{\sqrt{3}}{2} a$ B: A、B 两个星体之间的万有引力大小为 $\frac{G m^{2}}{a^{2}}$ C: $\mathrm{A} 、 \mathrm{~B} 、 \mathrm{C}$ 三个星体做圆周运动的向心加速度大小均为 $\frac{(\sqrt{3}+3) G m}{a^{2}}$ D: A、B、C 三个星体做圆周运动的周期均为 $2 \pi a \sqrt{\frac{a}{(3+\sqrt{3}) G m}}$ 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应为两个或更多的选项：[A, B, C, D]

MC

Astronomy

ZH

multi-modal

Astronomy_587

天问一号探测器在 2 月 10 日成功被火星捕获, 进入环火星轨道。如图所示, 假设探测器绕火星先后在椭圆轨道 1 、近火圆轨道 2 上运行, $\mathrm{A}$ 是两轨道的切点, $B$ 是粗圆轨道的远地点, 已知火星的质量与半径分别为 $M 、 R$, 火星的球心与 $B$ 点的距离为 $3 R$,探测器的质量为 $m$ 。若规定探测器距火星无限远时探测器的引力势能为 0 , 则探测器的引力势能的表达式为 $E_{\mathrm{p}}=-\frac{G M m}{r}$, 其中 $r$ 是探测器与火星的球心之间的距离, $G$ 为引力常量。下列说法正确的是 ( ) [图1] A: 探测器在轨道 $1 、 2$ 上运行周期的比值为 $3 \sqrt{3}$ B: 探测器在轨道 2 上运行时, 动量大小为 $\frac{m}{2} \sqrt{\frac{G M}{R}}$ C: 探测器在轨道 1 上运行, 经过 $\mathrm{A}$ 点时, 动能为 $\frac{G M m}{2 R}$ D: 探测器在轨道 1 上运行, 经过 $\mathrm{A}$ 点时, 动能大于 $\frac{G M m}{2 R}$

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: 天问一号探测器在 2 月 10 日成功被火星捕获, 进入环火星轨道。如图所示, 假设探测器绕火星先后在椭圆轨道 1 、近火圆轨道 2 上运行, $\mathrm{A}$ 是两轨道的切点, $B$ 是粗圆轨道的远地点, 已知火星的质量与半径分别为 $M 、 R$, 火星的球心与 $B$ 点的距离为 $3 R$,探测器的质量为 $m$ 。若规定探测器距火星无限远时探测器的引力势能为 0 , 则探测器的引力势能的表达式为 $E_{\mathrm{p}}=-\frac{G M m}{r}$, 其中 $r$ 是探测器与火星的球心之间的距离, $G$ 为引力常量。下列说法正确的是 ( ) [图1] A: 探测器在轨道 $1 、 2$ 上运行周期的比值为 $3 \sqrt{3}$ B: 探测器在轨道 2 上运行时, 动量大小为 $\frac{m}{2} \sqrt{\frac{G M}{R}}$ C: 探测器在轨道 1 上运行, 经过 $\mathrm{A}$ 点时, 动能为 $\frac{G M m}{2 R}$ D: 探测器在轨道 1 上运行, 经过 $\mathrm{A}$ 点时, 动能大于 $\frac{G M m}{2 R}$ 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

multi-modal

Astronomy_543

如图, 地球和某行星在同一轨道平面内同向绕太阳做匀速圆周运动。地球的运转周期为 $T$ 。地球和太阳的连线与地球和行星的连线所夹的角叫地球对该行星的观察视角 (简称视角)。已知该行星的最大视角为 $\theta$, 当行星处于最大视角处时, 是地球上天文爱好者观察该行星的最佳时期。 若某时刻该行星正处于最佳观察期，则该行星下一次处于最佳观察期至少需经历 多长时间？ [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, 地球和某行星在同一轨道平面内同向绕太阳做匀速圆周运动。地球的运转周期为 $T$ 。地球和太阳的连线与地球和行星的连线所夹的角叫地球对该行星的观察视角 (简称视角)。已知该行星的最大视角为 $\theta$, 当行星处于最大视角处时, 是地球上天文爱好者观察该行星的最佳时期。 若某时刻该行星正处于最佳观察期，则该行星下一次处于最佳观察期至少需经历 多长时间？ [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_732

如图所示, 发射地球同步卫星时, 先将卫星发射至近地圆轨道 1 , 然后经点火, 使其沿椭圆轨道 2 运行, 最后再次点火, 将卫星送入同步圆轨道 3. 轨道 $1 、 2$ 相切于 $Q$ 点,轨道 2、3 相切于 $P$ 点, 假设在整个过程中卫星的质量保持不变, 则当卫星分别在 $1 、 2$ 、 3 轨道上正常运行时, 下列说法正确的是（） [图1] A: 卫星在轨道 3 上的动能小于在轨道 1 上的动能 B: 卫星在轨道 3 上的机械能小于在轨道 1 上的机械能 C: 卫星在轨道 1 上经过 $Q$ 点时的动能等于它在轨道 2 上经过 $Q$ 点时的动能 D: 卫星在轨道 2 上经过 $P$ 点时的机械能小于它在轨道 3 上经过 $P$ 点时的机械能

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: 如图所示, 发射地球同步卫星时, 先将卫星发射至近地圆轨道 1 , 然后经点火, 使其沿椭圆轨道 2 运行, 最后再次点火, 将卫星送入同步圆轨道 3. 轨道 $1 、 2$ 相切于 $Q$ 点,轨道 2、3 相切于 $P$ 点, 假设在整个过程中卫星的质量保持不变, 则当卫星分别在 $1 、 2$ 、 3 轨道上正常运行时, 下列说法正确的是（） [图1] A: 卫星在轨道 3 上的动能小于在轨道 1 上的动能 B: 卫星在轨道 3 上的机械能小于在轨道 1 上的机械能 C: 卫星在轨道 1 上经过 $Q$ 点时的动能等于它在轨道 2 上经过 $Q$ 点时的动能 D: 卫星在轨道 2 上经过 $P$ 点时的机械能小于它在轨道 3 上经过 $P$ 点时的机械能 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_1208

The James Webb Space Telescope (JWST) is an incredibly exciting next generation telescope that was successfully launched on $25^{\text {th }}$ December 2021 . Its mirror is approximately $6.5 \mathrm{~m}$ in diameter, much larger than the $2.4 \mathrm{~m}$ mirror of the Hubble Space Telescope (HST), and so it has far greater resolution and sensitivity. Whilst HST largely imaged in the visible, JWST will do most of its work in the nearand mid-infrared (NIR and MIR respectively). This will allow it to pick up heavily redshifted light, such as that from the first generation of stars in the very first galaxies. [figure1] Figure 5: Left: A full-scale model of JWST next to some of the scientists and engineers involved in its development at the Goddard Space Flight Center. Credit: NASA / Goddard Space Flight Center / Pat Izzo. Right: The position of the second Lagrangian point, $L_{2}$, relative to the Earth. Credit: ESA. The resolution limit of a telescope is set by the amount of diffraction light rays experience as they enter the system, and is related to the diameter of a telescope, $D$, and the wavelength being observed, $\lambda$. The resolution limit of a CCD is set by the size of the pixels. Three of the imaging cameras on JWST are tabulated with some properties below: | Instrument | Wavelength range $(\mu \mathrm{m})$ | CCD plate scale (arcseconds / pixel) | | :---: | :---: | :---: | | NIRCam (short wave) | $0.6-2.3$ | 0.031 | | NIRCam (long wave) | $2.4-5.0$ | 0.065 | | MIRI | $5.6-25.5$ | 0.11 | An arcsecond is a measure of angle where $1^{\circ}=3600$ arcseconds. The familiar variation in intensity on a screen, $I_{\text {slit }}$, due to diffraction through an infinitely tall single slit is given as $$ I_{\text {slit }}=I_{0}\left(\frac{\sin (x)}{x}\right)^{2}, \text { where } \quad x=\frac{\pi D \theta}{\lambda} $$ and $I_{0}$ is the initial intensity. For a circular aperture, the formula is slightly different and is given as $$ I_{\mathrm{circ}}=I_{0}\left(\frac{2 J_{1}(x)}{x}\right)^{2} . $$ Here $J_{1}(x)$ is the Bessel function of the first kind and is calculated as $$ J_{n}(x)=\sum_{r=0}^{\infty} \frac{(-1)^{r}}{r !(n+r) !}\left(\frac{x}{2}\right)^{n+2 r} \quad \text { so } \quad J_{1}(x)=\frac{x}{2}\left(1-\frac{x^{2}}{8}+\frac{x^{4}}{192}-\ldots\right) . $$ The $x$-axis intercepts and shape of the maxima are quite different, as shown in Figure 6. The position of the first minimum of $I_{\text {slit }}$ is at $x_{\min }=\pi$ meaning that $\theta_{\min , \text { slit }}=\lambda / D$, whilst for $I_{\text {circ }}$ it is at $x_{\min }=3.8317 \ldots$ so $\theta_{\min , \mathrm{circ}} \approx 1.22 \lambda / D$. This is one way of defining the minimum angular resolution, although since the flux drops off so steeply away from the central maximum a more convenient one for use with CCDs is the angle corresponding to the full width half maximum (FWHM). [figure2] Figure 6: Left: The $I_{\text {slit }}$ (purple) and $I_{\text {circ }}$ (blue - the wider central maximum) functions, normalised so that $I_{0}=1$. You can see the shapes and $x$-intercepts are different. Right: How $x_{\min }$ and the full width half maximum (FWHM) are defined. Here it is shown for $I_{\text {circ }}$. As well as having the largest mirror of any space telescope ever launched, it is also one of the most sensitive, with its greatest sensitivity in the NIRCam F200W filter (centred on a wavelength of $1.989 \mu \mathrm{m})$ where after $10^{4}$ seconds it can detect a flux of $9.1 \mathrm{nJy}\left(1 \mathrm{Jy}=10^{-26} \mathrm{~W} \mathrm{~m}^{-2} \mathrm{~Hz}^{-1}\right.$ ) with a signal-to-noise ratio (S/N) of 10 , corresponding to an apparent magnitude of $m=29.0$. This extraordinary sensitivity can be used to pick up light from the earliest galaxies in the Universe. The scale factor, $a$, parameterises the expansion of the Universe since the Big Bang, and is related to the redshift, $z$, as $$ a=(1+z)^{-1} \quad \text { where } \quad z \equiv \frac{\lambda_{\text {obs }}-\lambda_{\mathrm{emit}}}{\lambda_{\mathrm{emit}}} $$ with $\lambda_{\text {obs }}$ the observed wavelength and $\lambda_{\text {emit }}$ the rest frame wavelength. The current rate of expansion of the Universe is given by the Hubble constant, $H_{0}$, and this is related to the current Hubble time, $t_{\mathrm{H}_{0}}$, and current Hubble distance, $D_{\mathrm{H}_{0}}$, as $$ t_{\mathrm{H}_{0}} \equiv H_{0}^{-1} \quad \text { and } \quad D_{\mathrm{H}_{0}} \equiv c t_{\mathrm{H}_{0}} \text {. } $$ Here the subscript 0 indicates the values are measured today. The Hubble constant is more appropriately known as the Hubble parameter as it is a function of time, and the evolution of $H$ as a function of $z$ is $$ E(z)=\frac{H}{H_{0}} \equiv\left[\Omega_{0, m}(1+z)^{3}+\Omega_{0, \Lambda}+\Omega_{0, r}(1+z)^{4}\right]^{1 / 2}, $$ where $\Omega$ is the normalised density parameter, and the subscript $m, r$, and $\Lambda$ indicate the contribution to $\Omega$ from matter, radiation, and dark energy, respectively. The proper age of the Universe $t(z)$ at redshift $z$ is best evaluated in terms of $a$ as $$ t=t_{\mathrm{H}_{0}} \int_{0}^{(1+z)^{-1}} \frac{a}{\left(\Omega_{0, m} a+\Omega_{0, \Lambda} a^{4}+\Omega_{0, r}\right)^{1 / 2}} \mathrm{~d} a $$ If $\Omega_{0, r}=0$ and $\Omega_{0, m}+\Omega_{0, \Lambda}=1$ (corresponding to what it known as a flat Universe), then via the standard integral $\int\left(b^{2}+x^{2}\right)^{-1 / 2} \mathrm{~d} x=\ln \left(x+\sqrt{b^{2}+x^{2}}\right)+C$ this integral can be evaluated analytically to give $$ t=t_{\mathrm{H}_{0}} \frac{2}{3 \Omega_{0, \Lambda}^{1 / 2}} \ln \left[\left(\frac{\Omega_{0, \Lambda}}{\Omega_{0, m}}\right)^{1 / 2}(1+z)^{-3 / 2}+\left(\frac{\Omega_{0, \Lambda}}{\Omega_{0, m}(1+z)^{3}}+1\right)^{1 / 2}\right] $$ Finally, the luminosity distance, $D_{L}(z)$, corresponding to the distance away that an object appears to be due to its measured flux given its intrinsic luminosity (i.e. $f \equiv L / 4 \pi D_{L}^{2}$ ) is given as $$ D_{L}=\left(1+z_{i}\right) D_{\mathrm{H}_{0}} \int_{0}^{z_{i}} \frac{1}{E(z)} \mathrm{d} z=\left(1+z_{i}\right) D_{\mathrm{H}_{0}} \int_{a_{i}}^{1} \frac{1}{\left(\Omega_{0, m} a+\Omega_{0, \Lambda} a^{4}+\Omega_{0, r}\right)^{1 / 2}} \mathrm{~d} a $$ where $z_{i}$ is the redshift of interest and $a_{i}$ is the equivalent scale factor. Even for the flat Universe case with $\Omega_{0, r}=0$ this integral cannot be be done analytically so must be evaluated numerically.b. To achieve suitable sampling, an image will be considered diffraction limited when it has $\geq 2$ pixels per $\theta_{\text {FWHM. }}$. The diameter of the JWST primary mirror is $6.5 \mathrm{~m}$, however since it is composed of hexagons and hexagonal in shape, it is not straightforward to work out the equivalent circular mirror diameter. To a good approximation it can be taken to be $6.0 \mathrm{~m}$. ii. Hence, determine which of the three imaging instruments is diffraction limited for the greatest fraction of its wavelength range.

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 James Webb Space Telescope (JWST) is an incredibly exciting next generation telescope that was successfully launched on $25^{\text {th }}$ December 2021 . Its mirror is approximately $6.5 \mathrm{~m}$ in diameter, much larger than the $2.4 \mathrm{~m}$ mirror of the Hubble Space Telescope (HST), and so it has far greater resolution and sensitivity. Whilst HST largely imaged in the visible, JWST will do most of its work in the nearand mid-infrared (NIR and MIR respectively). This will allow it to pick up heavily redshifted light, such as that from the first generation of stars in the very first galaxies. [figure1] Figure 5: Left: A full-scale model of JWST next to some of the scientists and engineers involved in its development at the Goddard Space Flight Center. Credit: NASA / Goddard Space Flight Center / Pat Izzo. Right: The position of the second Lagrangian point, $L_{2}$, relative to the Earth. Credit: ESA. The resolution limit of a telescope is set by the amount of diffraction light rays experience as they enter the system, and is related to the diameter of a telescope, $D$, and the wavelength being observed, $\lambda$. The resolution limit of a CCD is set by the size of the pixels. Three of the imaging cameras on JWST are tabulated with some properties below: | Instrument | Wavelength range $(\mu \mathrm{m})$ | CCD plate scale (arcseconds / pixel) | | :---: | :---: | :---: | | NIRCam (short wave) | $0.6-2.3$ | 0.031 | | NIRCam (long wave) | $2.4-5.0$ | 0.065 | | MIRI | $5.6-25.5$ | 0.11 | An arcsecond is a measure of angle where $1^{\circ}=3600$ arcseconds. The familiar variation in intensity on a screen, $I_{\text {slit }}$, due to diffraction through an infinitely tall single slit is given as $$ I_{\text {slit }}=I_{0}\left(\frac{\sin (x)}{x}\right)^{2}, \text { where } \quad x=\frac{\pi D \theta}{\lambda} $$ and $I_{0}$ is the initial intensity. For a circular aperture, the formula is slightly different and is given as $$ I_{\mathrm{circ}}=I_{0}\left(\frac{2 J_{1}(x)}{x}\right)^{2} . $$ Here $J_{1}(x)$ is the Bessel function of the first kind and is calculated as $$ J_{n}(x)=\sum_{r=0}^{\infty} \frac{(-1)^{r}}{r !(n+r) !}\left(\frac{x}{2}\right)^{n+2 r} \quad \text { so } \quad J_{1}(x)=\frac{x}{2}\left(1-\frac{x^{2}}{8}+\frac{x^{4}}{192}-\ldots\right) . $$ The $x$-axis intercepts and shape of the maxima are quite different, as shown in Figure 6. The position of the first minimum of $I_{\text {slit }}$ is at $x_{\min }=\pi$ meaning that $\theta_{\min , \text { slit }}=\lambda / D$, whilst for $I_{\text {circ }}$ it is at $x_{\min }=3.8317 \ldots$ so $\theta_{\min , \mathrm{circ}} \approx 1.22 \lambda / D$. This is one way of defining the minimum angular resolution, although since the flux drops off so steeply away from the central maximum a more convenient one for use with CCDs is the angle corresponding to the full width half maximum (FWHM). [figure2] Figure 6: Left: The $I_{\text {slit }}$ (purple) and $I_{\text {circ }}$ (blue - the wider central maximum) functions, normalised so that $I_{0}=1$. You can see the shapes and $x$-intercepts are different. Right: How $x_{\min }$ and the full width half maximum (FWHM) are defined. Here it is shown for $I_{\text {circ }}$. As well as having the largest mirror of any space telescope ever launched, it is also one of the most sensitive, with its greatest sensitivity in the NIRCam F200W filter (centred on a wavelength of $1.989 \mu \mathrm{m})$ where after $10^{4}$ seconds it can detect a flux of $9.1 \mathrm{nJy}\left(1 \mathrm{Jy}=10^{-26} \mathrm{~W} \mathrm{~m}^{-2} \mathrm{~Hz}^{-1}\right.$ ) with a signal-to-noise ratio (S/N) of 10 , corresponding to an apparent magnitude of $m=29.0$. This extraordinary sensitivity can be used to pick up light from the earliest galaxies in the Universe. The scale factor, $a$, parameterises the expansion of the Universe since the Big Bang, and is related to the redshift, $z$, as $$ a=(1+z)^{-1} \quad \text { where } \quad z \equiv \frac{\lambda_{\text {obs }}-\lambda_{\mathrm{emit}}}{\lambda_{\mathrm{emit}}} $$ with $\lambda_{\text {obs }}$ the observed wavelength and $\lambda_{\text {emit }}$ the rest frame wavelength. The current rate of expansion of the Universe is given by the Hubble constant, $H_{0}$, and this is related to the current Hubble time, $t_{\mathrm{H}_{0}}$, and current Hubble distance, $D_{\mathrm{H}_{0}}$, as $$ t_{\mathrm{H}_{0}} \equiv H_{0}^{-1} \quad \text { and } \quad D_{\mathrm{H}_{0}} \equiv c t_{\mathrm{H}_{0}} \text {. } $$ Here the subscript 0 indicates the values are measured today. The Hubble constant is more appropriately known as the Hubble parameter as it is a function of time, and the evolution of $H$ as a function of $z$ is $$ E(z)=\frac{H}{H_{0}} \equiv\left[\Omega_{0, m}(1+z)^{3}+\Omega_{0, \Lambda}+\Omega_{0, r}(1+z)^{4}\right]^{1 / 2}, $$ where $\Omega$ is the normalised density parameter, and the subscript $m, r$, and $\Lambda$ indicate the contribution to $\Omega$ from matter, radiation, and dark energy, respectively. The proper age of the Universe $t(z)$ at redshift $z$ is best evaluated in terms of $a$ as $$ t=t_{\mathrm{H}_{0}} \int_{0}^{(1+z)^{-1}} \frac{a}{\left(\Omega_{0, m} a+\Omega_{0, \Lambda} a^{4}+\Omega_{0, r}\right)^{1 / 2}} \mathrm{~d} a $$ If $\Omega_{0, r}=0$ and $\Omega_{0, m}+\Omega_{0, \Lambda}=1$ (corresponding to what it known as a flat Universe), then via the standard integral $\int\left(b^{2}+x^{2}\right)^{-1 / 2} \mathrm{~d} x=\ln \left(x+\sqrt{b^{2}+x^{2}}\right)+C$ this integral can be evaluated analytically to give $$ t=t_{\mathrm{H}_{0}} \frac{2}{3 \Omega_{0, \Lambda}^{1 / 2}} \ln \left[\left(\frac{\Omega_{0, \Lambda}}{\Omega_{0, m}}\right)^{1 / 2}(1+z)^{-3 / 2}+\left(\frac{\Omega_{0, \Lambda}}{\Omega_{0, m}(1+z)^{3}}+1\right)^{1 / 2}\right] $$ Finally, the luminosity distance, $D_{L}(z)$, corresponding to the distance away that an object appears to be due to its measured flux given its intrinsic luminosity (i.e. $f \equiv L / 4 \pi D_{L}^{2}$ ) is given as $$ D_{L}=\left(1+z_{i}\right) D_{\mathrm{H}_{0}} \int_{0}^{z_{i}} \frac{1}{E(z)} \mathrm{d} z=\left(1+z_{i}\right) D_{\mathrm{H}_{0}} \int_{a_{i}}^{1} \frac{1}{\left(\Omega_{0, m} a+\Omega_{0, \Lambda} a^{4}+\Omega_{0, r}\right)^{1 / 2}} \mathrm{~d} a $$ where $z_{i}$ is the redshift of interest and $a_{i}$ is the equivalent scale factor. Even for the flat Universe case with $\Omega_{0, r}=0$ this integral cannot be be done analytically so must be evaluated numerically. problem: b. To achieve suitable sampling, an image will be considered diffraction limited when it has $\geq 2$ pixels per $\theta_{\text {FWHM. }}$. The diameter of the JWST primary mirror is $6.5 \mathrm{~m}$, however since it is composed of hexagons and hexagonal in shape, it is not straightforward to work out the equivalent circular mirror diameter. To a good approximation it can be taken to be $6.0 \mathrm{~m}$. ii. Hence, determine which of the three imaging instruments is diffraction limited for the greatest fraction of its wavelength range. 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 \mu \mathrm{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.

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Astronomy

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multi-modal

Astronomy_524

地球的公转轨道接近圆, 但彗星的运动轨道则是一个非常扁的椭圆 (如图)。天文学家哈雷成功预言哈雷彗星的回归, 哈雷彗星最近出现的时间是 1986 年, 预测下次飞近地球将在 2061 年。设哈雷彗星在近日点与太阳中心的距离为 $r_{1}$, 在远日点与太阳中心的距离为 $r_{2}$ 。地球公转半径为 $R$ 。则 ( ) [图1] A: $r_{l} \approx 18 R$ B: $r_{1}+r_{2} \approx 36 R$ C: 哈雷彗星在近日点和远日点的加速度大小之比为 $\frac{r_{2}^{2}}{r_{1}^{2}}$ D: 哈雷彗星在近日点和远日点的速度大小之比为 $\sqrt{\frac{r_{2}}{r_{1}}}$

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个多选题（有多个正确答案）。 问题: 地球的公转轨道接近圆, 但彗星的运动轨道则是一个非常扁的椭圆 (如图)。天文学家哈雷成功预言哈雷彗星的回归, 哈雷彗星最近出现的时间是 1986 年, 预测下次飞近地球将在 2061 年。设哈雷彗星在近日点与太阳中心的距离为 $r_{1}$, 在远日点与太阳中心的距离为 $r_{2}$ 。地球公转半径为 $R$ 。则 ( ) [图1] A: $r_{l} \approx 18 R$ B: $r_{1}+r_{2} \approx 36 R$ C: 哈雷彗星在近日点和远日点的加速度大小之比为 $\frac{r_{2}^{2}}{r_{1}^{2}}$ D: 哈雷彗星在近日点和远日点的速度大小之比为 $\sqrt{\frac{r_{2}}{r_{1}}}$ 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应为两个或更多的选项：[A, B, C, D]

MC

Astronomy

ZH

multi-modal

Astronomy_856

We observe that a quasar's brightness varies within less than a day. What is the best upper bound on the quasar's size that you can derive from this information? A: $8 \mathrm{kpc}$ B: $170 \mathrm{AU}$ C: $3 \mathrm{AU}$ D: 3 Sun Radii E: $1 \mathrm{pc}$

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: We observe that a quasar's brightness varies within less than a day. What is the best upper bound on the quasar's size that you can derive from this information? A: $8 \mathrm{kpc}$ B: $170 \mathrm{AU}$ C: $3 \mathrm{AU}$ D: 3 Sun Radii E: $1 \mathrm{pc}$ 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

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text-only

Astronomy_893

Figure 4 below is a composite image which depicts a transit of the International Space Station (ISS) across the disc of the Sun. The image comprises 26 individual photographs which were taken at regular time intervals during the transit. The total duration of the transit was less than one second. In this question we will ignore any effects caused by the rotation of the Earth. [figure1] Figure 4: A composite of a selection of the frames taken with a high-speed camera of a transit of the ISS in front of the Sun, taken from Northamptonshire at 10:22 BST on $17^{\text {th }}$ June 2022. Credit: Jamie Cooper Photography During the transit, the angular diameter of the Sun as viewed from the position of the camera was $31^{\prime} 29^{\prime \prime}$. Use the image to calculate the angle $\theta_{1}$ subtended by the ISS between the first and last photographs, as viewed from the position of the camera. Note that the centre of the solar disc is NOT in the field of view of the photograph. [You are given that $1^{\circ}=60^{\prime}=$ $3600^{\prime \prime}$.]

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: Figure 4 below is a composite image which depicts a transit of the International Space Station (ISS) across the disc of the Sun. The image comprises 26 individual photographs which were taken at regular time intervals during the transit. The total duration of the transit was less than one second. In this question we will ignore any effects caused by the rotation of the Earth. [figure1] Figure 4: A composite of a selection of the frames taken with a high-speed camera of a transit of the ISS in front of the Sun, taken from Northamptonshire at 10:22 BST on $17^{\text {th }}$ June 2022. Credit: Jamie Cooper Photography During the transit, the angular diameter of the Sun as viewed from the position of the camera was $31^{\prime} 29^{\prime \prime}$. Use the image to calculate the angle $\theta_{1}$ subtended by the ISS between the first and last photographs, as viewed from the position of the camera. Note that the centre of the solar disc is NOT in the field of view of the photograph. [You are given that $1^{\circ}=60^{\prime}=$ $3600^{\prime \prime}$.] 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 the numerical value.

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Astronomy

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multi-modal

Astronomy_22

$\mathrm{A} 、 \mathrm{~B}$ 两颗卫星在同一平面内沿同一方向绕地球做匀速圆周运动, 它们之间的距离 $\Delta r$随时间变化的关系如图所示, 不考虑 $\mathrm{A} 、 \mathrm{~B}$ 之间的万有引力, 已知地球的半径为 $0.8 r$,万有引力常量为 $G$, 卫星 $\mathrm{A}$ 的线速度大于卫星 $\mathrm{B}$ 的线速度, 则以下说法正确的是 $(\quad)$ [图1] A: 卫星 A 的发射速度可能大于第二宇宙速度 B: 地球的第一宇宙速度为 $\frac{8 \sqrt{5} \pi r}{7 T}$ C: 地球的密度为 $\frac{192 \pi}{49 G T^{2}}$ D: 卫星 A 的加速度大小为 $\frac{256 \pi^{2} r}{49 T^{2}}$

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: $\mathrm{A} 、 \mathrm{~B}$ 两颗卫星在同一平面内沿同一方向绕地球做匀速圆周运动, 它们之间的距离 $\Delta r$随时间变化的关系如图所示, 不考虑 $\mathrm{A} 、 \mathrm{~B}$ 之间的万有引力, 已知地球的半径为 $0.8 r$,万有引力常量为 $G$, 卫星 $\mathrm{A}$ 的线速度大于卫星 $\mathrm{B}$ 的线速度, 则以下说法正确的是 $(\quad)$ [图1] A: 卫星 A 的发射速度可能大于第二宇宙速度 B: 地球的第一宇宙速度为 $\frac{8 \sqrt{5} \pi r}{7 T}$ C: 地球的密度为 $\frac{192 \pi}{49 G T^{2}}$ D: 卫星 A 的加速度大小为 $\frac{256 \pi^{2} r}{49 T^{2}}$ 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]

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Astronomy

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multi-modal

Astronomy_965

On $21^{\text {st }}$ March the Sun is in the constellation of Pisces. In which of these constellations would it be possible to find Venus? You are given that Venus has a circular orbit of radius 0.723 au. A: Aries B: Leo C: Libra D: Gemini

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: On $21^{\text {st }}$ March the Sun is in the constellation of Pisces. In which of these constellations would it be possible to find Venus? You are given that Venus has a circular orbit of radius 0.723 au. A: Aries B: Leo C: Libra D: Gemini 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

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text-only

Astronomy_555

2023 年 10 月 26 日, 神舟十七号载人飞船与天和核心舱进行了对接, “太空之家”迎来汤洪波、唐胜杰、江新林 3 名中国航天史上最年轻的乘组入驻。如图为神舟十七号的发射与交会对接过程示意图, 图中(1)为飞船的近地圆轨道, 其轨道半径为 $R_{1}$, (2)为椭圆变轨轨道, (3)为天和核心舱所在的圆轨道, 其轨道半径为 $R_{2}, P 、 Q$ 分别为(2)轨道与(1)、(3)轨道的交会点。关于神舟十七号载人飞船与天和核心舱交会对接过程，下列说法正确的是 ( ) [图1] A: 飞船在轨道 3 上运行的速度大于第一宇宙速度 B: 飞船从(2)轨道到变轨到(3)轨道需要在 $Q$ 点点火加速 C: 飞船在(1)轨道的动能一定大于天和核心舱在(3)轨道的动能 D: 若核心舱在(3)轨道运行周期为 $T$, 则飞船在(2)轨道从 $P$ 到 $Q$ 的时间为 $$ \frac{1}{2} \sqrt{\left(\frac{R_{1}+R_{2}}{2 R_{2}}\right)^{3} T} $$

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: 2023 年 10 月 26 日, 神舟十七号载人飞船与天和核心舱进行了对接, “太空之家”迎来汤洪波、唐胜杰、江新林 3 名中国航天史上最年轻的乘组入驻。如图为神舟十七号的发射与交会对接过程示意图, 图中(1)为飞船的近地圆轨道, 其轨道半径为 $R_{1}$, (2)为椭圆变轨轨道, (3)为天和核心舱所在的圆轨道, 其轨道半径为 $R_{2}, P 、 Q$ 分别为(2)轨道与(1)、(3)轨道的交会点。关于神舟十七号载人飞船与天和核心舱交会对接过程，下列说法正确的是 ( ) [图1] A: 飞船在轨道 3 上运行的速度大于第一宇宙速度 B: 飞船从(2)轨道到变轨到(3)轨道需要在 $Q$ 点点火加速 C: 飞船在(1)轨道的动能一定大于天和核心舱在(3)轨道的动能 D: 若核心舱在(3)轨道运行周期为 $T$, 则飞船在(2)轨道从 $P$ 到 $Q$ 的时间为 $$ \frac{1}{2} \sqrt{\left(\frac{R_{1}+R_{2}}{2 R_{2}}\right)^{3} T} $$ 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_297

2020 年 5 月 5 日 18 时, 长征五号 $B$ 运载火箭成功将新一代载人飞船试验船送入预定轨道, 拉开了我国载人航天工程“第三步”任务序幕, 下图是试验船返回地球的示意图。假设地球半径为 $R$, 地球表面的重力加速度为 $g_{0}$, 试验船初始时在距地球表面高度为 $3 R$的圆形轨道 $\mathrm{I}$ 上运动, 然后在 $A$ 点点火变轨进入粗圆轨道 II, 到达近地点 $B$ (忽略距离地表的高度)再次点火进入近地轨道 III 绕地球位圆周运动。下列说法中正确的是 ( ) [图1] A: 在 $A$ 点点火的目的是增大试验船的速度 B: 在 $B$ 点点火的目的是增大试验船的速度 C: 试验船在轨道 I 上运行时的加速度大于在轨道 II 上运行时的加速度 D: 试验船在轨道 III 上绕地球运行一周所需的时间为 $2 \pi \sqrt{\frac{R}{g_{0}}}$

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: 2020 年 5 月 5 日 18 时, 长征五号 $B$ 运载火箭成功将新一代载人飞船试验船送入预定轨道, 拉开了我国载人航天工程“第三步”任务序幕, 下图是试验船返回地球的示意图。假设地球半径为 $R$, 地球表面的重力加速度为 $g_{0}$, 试验船初始时在距地球表面高度为 $3 R$的圆形轨道 $\mathrm{I}$ 上运动, 然后在 $A$ 点点火变轨进入粗圆轨道 II, 到达近地点 $B$ (忽略距离地表的高度)再次点火进入近地轨道 III 绕地球位圆周运动。下列说法中正确的是 ( ) [图1] A: 在 $A$ 点点火的目的是增大试验船的速度 B: 在 $B$ 点点火的目的是增大试验船的速度 C: 试验船在轨道 I 上运行时的加速度大于在轨道 II 上运行时的加速度 D: 试验船在轨道 III 上绕地球运行一周所需的时间为 $2 \pi \sqrt{\frac{R}{g_{0}}}$ 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]

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Astronomy

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multi-modal

Astronomy_340

在宇宙中有两颗星组成的孤立“双星系统”, “双星系统”离其他恒星较远，通常可忽略其他星体对“双星系统”的引力作用。星 $\mathrm{A}$ 和星 $\mathrm{B}$ 的质量分别为 $M_{1}$ 和 $M_{2}$, 它们都绕二者连线上的某点做周期为 $T$ 的匀速圆周运动。已知引力常量为 $G$, 求星 $\mathrm{A}$ 和星 $\mathrm{B}$间的距离 $L$ 。

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: 在宇宙中有两颗星组成的孤立“双星系统”, “双星系统”离其他恒星较远，通常可忽略其他星体对“双星系统”的引力作用。星 $\mathrm{A}$ 和星 $\mathrm{B}$ 的质量分别为 $M_{1}$ 和 $M_{2}$, 它们都绕二者连线上的某点做周期为 $T$ 的匀速圆周运动。已知引力常量为 $G$, 求星 $\mathrm{A}$ 和星 $\mathrm{B}$间的距离 $L$ 。 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

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text-only

Astronomy_689

2021 年 5 月, “天问一号”探测器成功在火星软着陆, 我国成为第一个首次探测火星就实现“绕、落、巡”任务的国家。为了简化问题，可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 $T_{1}$, 公转轨道半径为 $r_{1}$, 火星的公转周期为 $T_{2}$, 火星质量为 $M$ 。如图 2 所示, 以火星为参考系, 质量为 $m_{1}$ 的探测器沿 1 号轨道到达 $B$ 点时速度为 $v_{1}, B$ 点到火星球心的距离为 $r_{3}$, 此时启动发动机, 在极短时间内一次性喷出部分气体, 喷气后探测器质量变为 $m_{2}$ 、速度变为与 $v_{1}$ 垂直的 $v_{2}$,然后进入以 $B$ 点为远火点的椭圆轨道 2 。已知万有引力势能公式 $E_{\mathrm{p}}=-\frac{G M m}{r}$, 其中 $M$为中心天体的质量, $m$ 为卫星的质量, $G$ 为引力常量, $r$ 为卫星到中心天体球心的距离。求探测器沿 2 号轨道运动至近火点的速度 $v_{3}$ 的大小。[图1] 图1 [图2] 图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, 我国成为第一个首次探测火星就实现“绕、落、巡”任务的国家。为了简化问题，可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 $T_{1}$, 公转轨道半径为 $r_{1}$, 火星的公转周期为 $T_{2}$, 火星质量为 $M$ 。如图 2 所示, 以火星为参考系, 质量为 $m_{1}$ 的探测器沿 1 号轨道到达 $B$ 点时速度为 $v_{1}, B$ 点到火星球心的距离为 $r_{3}$, 此时启动发动机, 在极短时间内一次性喷出部分气体, 喷气后探测器质量变为 $m_{2}$ 、速度变为与 $v_{1}$ 垂直的 $v_{2}$,然后进入以 $B$ 点为远火点的椭圆轨道 2 。已知万有引力势能公式 $E_{\mathrm{p}}=-\frac{G M m}{r}$, 其中 $M$为中心天体的质量, $m$ 为卫星的质量, $G$ 为引力常量, $r$ 为卫星到中心天体球心的距离。求探测器沿 2 号轨道运动至近火点的速度 $v_{3}$ 的大小。[图1] 图1 [图2] 图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

ZH

multi-modal

Astronomy_406

“双星系统”由相距较近的恒星组成, 每个恒星的半径远小于两个恒星之间的距离,而且双星系统一般远离其他天体, 它们在相互间的万有引力作用下, 绕某一点做匀速圆周运动, 如图所示为某一双星系统, $\mathrm{A}$ 星球的质量为 $m_{1}, \mathrm{~B}$ 星球的质量为 $m_{2}$, 它们中心之间的距离为 $L$, 引力常量为 $G$, 则下列说法正确的是 ( ) [图1] A: A 星球的轨道半径为 $R=\frac{m_{1}}{m_{1}+m_{2}} L$ B: 双星运行的周期为 $T=2 \pi L \sqrt{\frac{L}{G\left(m_{1}+m_{2}\right)}}$ C: B 星球的轨道半径为 $r=\frac{m_{2}}{m_{1}} L$ D: 若近似认为 $\mathrm{B}$ 星球绕 $\mathrm{A}$ 星球中心做圆周运动, 则 $\mathrm{B}$ 星球的运行周期为 $$ T=2 \pi L \sqrt{\frac{L}{G m_{2}}} $$

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: “双星系统”由相距较近的恒星组成, 每个恒星的半径远小于两个恒星之间的距离,而且双星系统一般远离其他天体, 它们在相互间的万有引力作用下, 绕某一点做匀速圆周运动, 如图所示为某一双星系统, $\mathrm{A}$ 星球的质量为 $m_{1}, \mathrm{~B}$ 星球的质量为 $m_{2}$, 它们中心之间的距离为 $L$, 引力常量为 $G$, 则下列说法正确的是 ( ) [图1] A: A 星球的轨道半径为 $R=\frac{m_{1}}{m_{1}+m_{2}} L$ B: 双星运行的周期为 $T=2 \pi L \sqrt{\frac{L}{G\left(m_{1}+m_{2}\right)}}$ C: B 星球的轨道半径为 $r=\frac{m_{2}}{m_{1}} L$ D: 若近似认为 $\mathrm{B}$ 星球绕 $\mathrm{A}$ 星球中心做圆周运动, 则 $\mathrm{B}$ 星球的运行周期为 $$ T=2 \pi L \sqrt{\frac{L}{G m_{2}}} $$ 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_419

地球表面上两极的重力加速度约为 $9.83 \mathrm{~m} / \mathrm{s}^{2}$, 而赤道上的重力加速度约为 $9.78 \mathrm{~m} / \mathrm{s}^{2}$,即赤道上的重力加速度比两极的重力加速度小约 $\frac{1}{200}$, 赤道上有一观察者, 日落后, 他用天文望远镜观察被太阳光照射的地球同步卫星, 他在一天的时间内看不到此卫星的时间为 $t$, 若将地球看成球体, 且地球的质量分布均匀, 半径约为 $6.4 \times 10^{3} \mathrm{~km}$, 取 $\sqrt[3]{200} \approx 6, \sin 10^{\circ}=\frac{1}{6}$, 则通过以上数据估算可得（） A: 同步卫星的高度约为 $3.2 \times 10^{3} \mathrm{~km}$ B: 同步卫星的高度约为 $3.2 \times 10^{5} \mathrm{~km}$ C: 看不到同步卫星的时间与看得到卫星的时间之比约为 $1: 17$ D: 看不到同步卫星的时间与看得到卫星的时间之比约为 $2: 17$

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: 地球表面上两极的重力加速度约为 $9.83 \mathrm{~m} / \mathrm{s}^{2}$, 而赤道上的重力加速度约为 $9.78 \mathrm{~m} / \mathrm{s}^{2}$,即赤道上的重力加速度比两极的重力加速度小约 $\frac{1}{200}$, 赤道上有一观察者, 日落后, 他用天文望远镜观察被太阳光照射的地球同步卫星, 他在一天的时间内看不到此卫星的时间为 $t$, 若将地球看成球体, 且地球的质量分布均匀, 半径约为 $6.4 \times 10^{3} \mathrm{~km}$, 取 $\sqrt[3]{200} \approx 6, \sin 10^{\circ}=\frac{1}{6}$, 则通过以上数据估算可得（） A: 同步卫星的高度约为 $3.2 \times 10^{3} \mathrm{~km}$ B: 同步卫星的高度约为 $3.2 \times 10^{5} \mathrm{~km}$ C: 看不到同步卫星的时间与看得到卫星的时间之比约为 $1: 17$ D: 看不到同步卫星的时间与看得到卫星的时间之比约为 $2: 17$ 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_931

In the UK we use the Gregorian calendar; it is a solar calendar so that a year corresponds to the time to orbit the Sun once, where 1 solar year is $\approx 365.25$ days. Several cultures use a lunar calendar, where each month is determined by the time it takes to go from New Moon to New Moon, and have a lunar year that is exactly 12 lunar months. An example of this is the Islamic calendar. Since the length of a lunar month (29.53 days) is a little shorter than the average month length in our solar calendar (see Figure 3), it means the start date of each month in the Islamic calendar is not tied to the seasons and gradually moves earlier in the solar year. [figure1] Figure 3: All the moon phases in May 2022, showing that the time measured from New Moon to New Moon (a lunar month) is shorter than a month in the Gregorian calendar. Credit: MoonConnection.com Calculate (to the nearest day) how much earlier in the Gregorian calendar (on average) a given month in the Islamic calendar will start compared to the previous year.

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: In the UK we use the Gregorian calendar; it is a solar calendar so that a year corresponds to the time to orbit the Sun once, where 1 solar year is $\approx 365.25$ days. Several cultures use a lunar calendar, where each month is determined by the time it takes to go from New Moon to New Moon, and have a lunar year that is exactly 12 lunar months. An example of this is the Islamic calendar. Since the length of a lunar month (29.53 days) is a little shorter than the average month length in our solar calendar (see Figure 3), it means the start date of each month in the Islamic calendar is not tied to the seasons and gradually moves earlier in the solar year. [figure1] Figure 3: All the moon phases in May 2022, showing that the time measured from New Moon to New Moon (a lunar month) is shorter than a month in the Gregorian calendar. Credit: MoonConnection.com Calculate (to the nearest day) how much earlier in the Gregorian calendar (on average) a given month in the Islamic calendar will start compared to the previous year. 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 the numerical value.

NV

Astronomy

EN

multi-modal

Astronomy_82

已知地球的质量为 $M$, 半径为 $R$, 引力常量为 $G$ 。赤道上地球表面附近的重力加速度用 $g_{e}$ 表示, 北极处地球表面附近的重力加速度用 $g_{N}$ 表示, 将地球视为均匀球体。 用 $g_{e} 、 g_{N}$ 和半径 $R$ 表示地球自转周期; [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: 已知地球的质量为 $M$, 半径为 $R$, 引力常量为 $G$ 。赤道上地球表面附近的重力加速度用 $g_{e}$ 表示, 北极处地球表面附近的重力加速度用 $g_{N}$ 表示, 将地球视为均匀球体。 用 $g_{e} 、 g_{N}$ 和半径 $R$ 表示地球自转周期; [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_979

From the upper parts of the Sun's corona comes a stream of charged particles called the solar wind, meaning the Sun is slowly losing some of its mass (although the effect is negligible compared to the mass loss in nuclear fusion). The particles travel at supersonic speeds until the pressure from interstellar space causes their speed to drop to subsonic speeds instead - this transition is called the termination shock, and has been explored by the two Voyager probes as they leave the solar system. [figure1] Figure 3: Left: A demonstration of a termination shock formed with water flowing from a tap into a sink. Right: The Voyager spacecraft crossing the termination shock of the Solar System. At the orbit of the Earth, the solar wind is measured to have a density of 7 protons $\mathrm{cm}^{-3}$ and to be travelling at $500 \mathrm{~km} \mathrm{~s}^{-1}$. Calculate $\Delta M / \Delta t$, giving your answer in units of $\mathrm{M}_{\odot}$ year $^{-1}$.

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: From the upper parts of the Sun's corona comes a stream of charged particles called the solar wind, meaning the Sun is slowly losing some of its mass (although the effect is negligible compared to the mass loss in nuclear fusion). The particles travel at supersonic speeds until the pressure from interstellar space causes their speed to drop to subsonic speeds instead - this transition is called the termination shock, and has been explored by the two Voyager probes as they leave the solar system. [figure1] Figure 3: Left: A demonstration of a termination shock formed with water flowing from a tap into a sink. Right: The Voyager spacecraft crossing the termination shock of the Solar System. At the orbit of the Earth, the solar wind is measured to have a density of 7 protons $\mathrm{cm}^{-3}$ and to be travelling at $500 \mathrm{~km} \mathrm{~s}^{-1}$. Calculate $\Delta M / \Delta t$, giving your answer in units of $\mathrm{M}_{\odot}$ year $^{-1}$. 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 the numerical value.

NV

Astronomy

EN

multi-modal

Astronomy_19

将太阳系八大行星绕太阳的运转视作匀速圆周运动。设行星的轨道半径为 $R$, 环绕周期为 $T$, 角速度为 $\omega$, 环绕速度为 $v$, 下列描述它们之间的关系图像中正确的是 ( ) A: [图1] B: [图2] C: [图3] D: [图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: 将太阳系八大行星绕太阳的运转视作匀速圆周运动。设行星的轨道半径为 $R$, 环绕周期为 $T$, 角速度为 $\omega$, 环绕速度为 $v$, 下列描述它们之间的关系图像中正确的是 ( ) A: [图1] B: [图2] C: [图3] D: [图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]

SC

Astronomy

ZH

multi-modal

Astronomy_452

我国航天技术水平在世界处于领先地位，对于人造卫星的发射，有人提出了利用“地球隧道”发射人造卫星的构想：沿地球的一条弦挖一通道，在通道的两个出口处分别将等质量的待发射卫星部件同时释放，部件将在通道中间位置“碰撞组装”成卫星并静止下来; 另在通道的出口处由静止释放一个大质量物体，大质量物体会在通道与待发射的卫星碰撞, 只要物体质量相比卫星质量足够大, 卫星获得足够速度就会从对向通道口射出。 (以下计算中, 已知地球的质量为 $M_{0}$, 地球半径为 $R_{0}$, 引力常量为 $G$, 可忽略通道 $A B$的内径大小和地球自转影响。) 如图丙所示, 如果质量为 $m$ 的待发射卫星已静止在通道中心 $O^{\prime}$ 处, 由 $A$ 处静止释放另一质量为 $M$ 的物体, 物体到达 $O^{\prime}$ 处与卫星发生弹性正碰, 设 $M$ 远大于 $m$, 计算时可取 $\frac{m}{M} \approx 0$ 。卫星从图丙示通道右侧 $B$ 处飞出, 为使飞出速度达到地球第一宇宙速度, $h$ 应为多大? [图1] 丙

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: 我国航天技术水平在世界处于领先地位，对于人造卫星的发射，有人提出了利用“地球隧道”发射人造卫星的构想：沿地球的一条弦挖一通道，在通道的两个出口处分别将等质量的待发射卫星部件同时释放，部件将在通道中间位置“碰撞组装”成卫星并静止下来; 另在通道的出口处由静止释放一个大质量物体，大质量物体会在通道与待发射的卫星碰撞, 只要物体质量相比卫星质量足够大, 卫星获得足够速度就会从对向通道口射出。 (以下计算中, 已知地球的质量为 $M_{0}$, 地球半径为 $R_{0}$, 引力常量为 $G$, 可忽略通道 $A B$的内径大小和地球自转影响。) 如图丙所示, 如果质量为 $m$ 的待发射卫星已静止在通道中心 $O^{\prime}$ 处, 由 $A$ 处静止释放另一质量为 $M$ 的物体, 物体到达 $O^{\prime}$ 处与卫星发生弹性正碰, 设 $M$ 远大于 $m$, 计算时可取 $\frac{m}{M} \approx 0$ 。卫星从图丙示通道右侧 $B$ 处飞出, 为使飞出速度达到地球第一宇宙速度, $h$ 应为多大? [图1] 丙 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_1010

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] Using a simple approximation, the equilibrium temperature of a planet can be calculated as $$ T_{\text {Planet }}=T_{\text {Star }} \sqrt{\frac{R_{\text {Star }}}{2 d}} $$ where $d$ is the distance between the star and the planet. Given that the astronomers discovered that the orbit of the planet is in fact an ellipse with an eccentricity of 0.35 , and that the star has a surface temperature of $3000 \mathrm{~K}$ and a radius of $0.14 R_{\odot^{\prime}}$ what are the minimum and maximum equilibrium temperatures of Proxima Centauri B? Comment on whether or not the planet is in the habitable zone of Proxima Centauri. [The habitable zone is the band around a star where a planet can have water on its surface in liquid form, at normal pressure.]

You are participating in an international Astronomy competition and need to solve the following question. The answer to this question is a range interval. 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] Using a simple approximation, the equilibrium temperature of a planet can be calculated as $$ T_{\text {Planet }}=T_{\text {Star }} \sqrt{\frac{R_{\text {Star }}}{2 d}} $$ where $d$ is the distance between the star and the planet. Given that the astronomers discovered that the orbit of the planet is in fact an ellipse with an eccentricity of 0.35 , and that the star has a surface temperature of $3000 \mathrm{~K}$ and a radius of $0.14 R_{\odot^{\prime}}$ what are the minimum and maximum equilibrium temperatures of Proxima Centauri B? Comment on whether or not the planet is in the habitable zone of Proxima Centauri. [The habitable zone is the band around a star where a planet can have water on its surface in liquid form, at normal pressure.] 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 interval, e.g. ANSWER=(1,2] \cup[7,+\infty)

IN

Astronomy

EN

multi-modal

Astronomy_478

电影中的太空电梯非常吸引人。现假设已经建成了如图所示的太空电梯, 其通过超级缆绳将地球赤道上的固定基地、同步空间站和配重空间站连接在一起，它们随地球同步旋转。图中配重空间站比同步空间站更高, $P$ 是缆绳上的一个平台。则下列说法正确的是 ( ) [图1] A: 太空电梯上各点线速度的平方与该点离地球球心的距离成反比 B: 宇航员在配重空间站时处于完全失重状态 C: 若从 $P$ 平台向外自由释放一个小物块, 则小物块会一边朝 $P$ 点转动的方向向前运动一边落向地球 D: 若两空间站之间缆绳断裂, 配重空间站将绕地球做椭圆运动, 且断裂处为椭圆的远地点

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: 电影中的太空电梯非常吸引人。现假设已经建成了如图所示的太空电梯, 其通过超级缆绳将地球赤道上的固定基地、同步空间站和配重空间站连接在一起，它们随地球同步旋转。图中配重空间站比同步空间站更高, $P$ 是缆绳上的一个平台。则下列说法正确的是 ( ) [图1] A: 太空电梯上各点线速度的平方与该点离地球球心的距离成反比 B: 宇航员在配重空间站时处于完全失重状态 C: 若从 $P$ 平台向外自由释放一个小物块, 则小物块会一边朝 $P$ 点转动的方向向前运动一边落向地球 D: 若两空间站之间缆绳断裂, 配重空间站将绕地球做椭圆运动, 且断裂处为椭圆的远地点 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_150

在寻找假想的第九大行星的过程中，天文学家发现了 2018VG18“外海王星天体”,外海王星天体的实际运行轨道与计算中的轨道数据有偏差, 而这通常被认为是受到了假想中第九行星（轨道处在太阳与外海王星天体之间）的引力扰动所致。如图所示, 在运行轨道不变的情况下, 当存在假想中的第九行星时, 跟没有假想中的第九行星相比, 2018VG18 ( ) [图1] A: 公转周期更大 B: 平均速率更小 C: 自转周期更大 D: 公转向心加速度更大

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: 在寻找假想的第九大行星的过程中，天文学家发现了 2018VG18“外海王星天体”,外海王星天体的实际运行轨道与计算中的轨道数据有偏差, 而这通常被认为是受到了假想中第九行星（轨道处在太阳与外海王星天体之间）的引力扰动所致。如图所示, 在运行轨道不变的情况下, 当存在假想中的第九行星时, 跟没有假想中的第九行星相比, 2018VG18 ( ) [图1] A: 公转周期更大 B: 平均速率更小 C: 自转周期更大 D: 公转向心加速度更大 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]

SC

Astronomy

ZH

multi-modal

Astronomy_860

The fictional towns of Baia and Caia are located at $\left(66.56^{\circ} \mathrm{N}, 67.55^{\circ} \mathrm{E}\right)$ and $\left(\delta, 18.95^{\circ} \mathrm{E}\right)$, respectively. It is known that the spherical triangle with vertices at Baia, Caia, and the North Pole covers $6.75 \%$ of Earth's surface. Compute $\delta$. A: $66.56^{\circ} \mathrm{N}$ B: $55.25^{\circ} \mathrm{N}$ C: $23.44^{\circ} \mathrm{N}$ D: $55.25^{\circ} \mathrm{S}$ E: $66.56^{\circ} \mathrm{S}$

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 fictional towns of Baia and Caia are located at $\left(66.56^{\circ} \mathrm{N}, 67.55^{\circ} \mathrm{E}\right)$ and $\left(\delta, 18.95^{\circ} \mathrm{E}\right)$, respectively. It is known that the spherical triangle with vertices at Baia, Caia, and the North Pole covers $6.75 \%$ of Earth's surface. Compute $\delta$. A: $66.56^{\circ} \mathrm{N}$ B: $55.25^{\circ} \mathrm{N}$ C: $23.44^{\circ} \mathrm{N}$ D: $55.25^{\circ} \mathrm{S}$ E: $66.56^{\circ} \mathrm{S}$ 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_101

如图所示设地球的质量为 $M$ 且绕太阳做匀速圆周运动, 当地球运动到 $D$ 点时, 有一质量为 $m$ 的飞船由静止开始从 $D$ 点只在恒力 $F$ 的作用下沿 $D C$ 方向做匀加速直线运动, 再过两个月, 飞船在 $C$ 处再次掠过地球上空, 假设太阳与地球的万有引力作用不改变飞船所受恒力 $F$ 的大小和方向, 飞船到地球表面的距离远小于地球与太阳间的距离,则地球与太阳间的万有引力大小（） [图1] A: $\frac{M F \pi^{2}}{3 m}$ B: $\frac{M F \pi^{2}}{6 m}$ C: $\frac{M F \pi^{2}}{9 m}$ D: $\frac{M F \pi^{2}}{18 m}$

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: 如图所示设地球的质量为 $M$ 且绕太阳做匀速圆周运动, 当地球运动到 $D$ 点时, 有一质量为 $m$ 的飞船由静止开始从 $D$ 点只在恒力 $F$ 的作用下沿 $D C$ 方向做匀加速直线运动, 再过两个月, 飞船在 $C$ 处再次掠过地球上空, 假设太阳与地球的万有引力作用不改变飞船所受恒力 $F$ 的大小和方向, 飞船到地球表面的距离远小于地球与太阳间的距离,则地球与太阳间的万有引力大小（） [图1] A: $\frac{M F \pi^{2}}{3 m}$ B: $\frac{M F \pi^{2}}{6 m}$ C: $\frac{M F \pi^{2}}{9 m}$ D: $\frac{M F \pi^{2}}{18 m}$ 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_548

将物体以一定初速度 $v_{0}=4 \mathrm{~m} / \mathrm{s}$ 在某行星表面坚直上抛, 从抛出开始计时, 落回抛出点前, 物体第 $1 \mathrm{~s}$ 内和第 $4 \mathrm{~s}$ 内通过的位移大小相等。已知该行星半径和地球半径相同的,地球表面的重力加速度为 $g=10 \mathrm{~m} / \mathrm{s}^{2}$, 不计空气阻力, 则下列说法正确的是 ( ) A: 地球的平均密度和该行星的平均密度之比为 1:5 B: 该行星和地球的第一宇宙速度之比为 $1: \sqrt{5}$ C: 将物体在地球表面以相同的初速度 $v_{0}$ 坚直抛出后上升的最大高度 $1.6 \mathrm{~m}$ D: 该行星表面的自由落体加速度大小为 $1 \mathrm{~m} / \mathrm{s}^{2}$

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: 将物体以一定初速度 $v_{0}=4 \mathrm{~m} / \mathrm{s}$ 在某行星表面坚直上抛, 从抛出开始计时, 落回抛出点前, 物体第 $1 \mathrm{~s}$ 内和第 $4 \mathrm{~s}$ 内通过的位移大小相等。已知该行星半径和地球半径相同的,地球表面的重力加速度为 $g=10 \mathrm{~m} / \mathrm{s}^{2}$, 不计空气阻力, 则下列说法正确的是 ( ) A: 地球的平均密度和该行星的平均密度之比为 1:5 B: 该行星和地球的第一宇宙速度之比为 $1: \sqrt{5}$ C: 将物体在地球表面以相同的初速度 $v_{0}$ 坚直抛出后上升的最大高度 $1.6 \mathrm{~m}$ D: 该行星表面的自由落体加速度大小为 $1 \mathrm{~m} / \mathrm{s}^{2}$ 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_91

2020 年 7 月 31 日, 北斗闪耀, 泽沐八方。北斗三号全球卫星导航系统 (如图甲所示）建成暨开通仪式在北京举行。如图乙所示为 55 颗卫星绕地球在不同轨道上运动的 $\lg T-\lg r$ 图像, 其中 $\mathrm{T}$ 为卫星的周期, $r$ 为卫星的轨道半径, 1 和 2 为其中的两颗卫星。 已知引力常量为 $G$, 下列说法正确的是 ( ) [图1] 图甲 [图2] 图乙 A: 地球的半径为 $x_{0}$ B: 地球质量为 $\frac{4 \pi^{2} 10^{b}}{G}$ C: 卫星 1 和 2 运动的线速度大小之比为 $x_{1}: x_{2}$ D: 卫星 1 和 2 向心加速度大小之比为 $10^{2 x_{2}}: 10^{2 x_{1}}$

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: 2020 年 7 月 31 日, 北斗闪耀, 泽沐八方。北斗三号全球卫星导航系统 (如图甲所示）建成暨开通仪式在北京举行。如图乙所示为 55 颗卫星绕地球在不同轨道上运动的 $\lg T-\lg r$ 图像, 其中 $\mathrm{T}$ 为卫星的周期, $r$ 为卫星的轨道半径, 1 和 2 为其中的两颗卫星。 已知引力常量为 $G$, 下列说法正确的是 ( ) [图1] 图甲 [图2] 图乙 A: 地球的半径为 $x_{0}$ B: 地球质量为 $\frac{4 \pi^{2} 10^{b}}{G}$ C: 卫星 1 和 2 运动的线速度大小之比为 $x_{1}: x_{2}$ D: 卫星 1 和 2 向心加速度大小之比为 $10^{2 x_{2}}: 10^{2 x_{1}}$ 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]

SC

Astronomy

ZH

multi-modal

Astronomy_1062

On $21^{\text {st }}$ August 2017 the continental United States experienced a total solar eclipse. Dubbed the 'Great American Eclipse', it was estimated to be one of the most watched eclipses in history. ## Total Solar Eclipse of 2017 Aug 21 [figure1] Figure 3: The path of totality for the Great American Eclipse. The narrow dimension is its width. Credit: Fred Espenak, NASA's GSFC. The path of totality (where the Moon completely obscures the Sun) is shown in Figure 3, and the point of greatest eclipse ("GE"; where the path was widest since the axis of the cone of the Moon's shadow passed closest to the centre of the Earth) was near the village of Cerulean, Kentucky. The following data can be used for this question: - The angular radii of the Sun and the Moon (if observed from the centre of the Earth) at the moment of GE are $15^{\prime} 48.7^{\prime \prime}$ and $16^{\prime} 03.4^{\prime \prime}$, respectively, where the notation $x x^{\prime} y y . y^{\prime \prime}$ corresponds to $x x$ arcminutes and $y y . y$ arcseconds ( 60 arcminutes $=1$ degree, and 60 arcseconds $=1$ arcminute $)$ - The latitude and longitude of the location of GE are $36^{\circ} 58.0^{\prime} \mathrm{N}$ and $87^{\circ} 40.3^{\prime} \mathrm{W}$, respectively - Take the mean radii of the Sun, Earth and Moon to be respectively $R_{\odot}=695700 \mathrm{~km}, R_{\oplus}=$ $6371 \mathrm{~km}$ and $R_{\text {Moon }}=1737 \mathrm{~km}$, and a day to be 24 hours - Take the semi-major axes of the Sun-Earth and Earth-Moon systems to be $149600000 \mathrm{~km}$ and $384400 \mathrm{~km}$, respectively - As viewed from a location far above the North Pole, the Moon orbits in an anticlockwise direction around the Earth, and the Earth spins in an anticlockwise direction For an ellipse with semi-major axis $a$ it can be shown that the velocity $v$, at a distance $r$ from mass $M$, can be written as: $$ v^{2}=G M\left(\frac{2}{r}-\frac{1}{a}\right) $$ The point of greatest eclipse and greatest duration do not generally coincide, as a more elliptical shadow with a major axis aligned with the path of maximum totality (and thinner path width, equal to the minor axis) can compensate for the shadow moving faster at higher latitudes. For this eclipse the point of greatest duration ("GD") was at co-ordinates of $37^{\circ} 35^{\prime} \mathrm{N}$ latitude and $89^{\circ} 07^{\prime} \mathrm{W}$ longitude, reached about 4 minutes before GE, and where totality lasted $0.1 \mathrm{~s}$ longer than the value calculated in part $\mathrm{c}$. [figure2] Figure 4: The route of the Moon's shadow in the vicinity of the points of greatest duration (GD, near Carbondale) and greatest eclipse (GE, near Hopkinsville). Any places between the two limits on the path of totality will experience at least a very short period of totality - outside that region will only be a partial eclipse and the perpendicular distance between them is the path width. The longest duration of totality at that point of the shadow's journey is indicated as the path of maximum totality, which both GE and GD sit on. The closest part of that path to Carbondale is indicated as CP. Credit: Fred Espenak \& Google Maps.d. The town of Carbondale, Illinois, is the closest big town to the point of GD, with co-ordinates $37^{\circ} 44^{\prime} \mathrm{N}$ latitude and $89^{\circ} 13^{\prime} \mathrm{W}$ longitude. Assuming the path of maximum totality can be treated as linear as it passes through the region around GD and GE: iii. Hence calculate (to the nearest $0.1 \mathrm{~s}$ ) how much shorter totality was for residents in Carbondale compared with CP. Take the duration at CP to be the same as GD, the width of the path to be $0.5 \mathrm{~km}$ less than for GE (so the Moon's shadow is elliptical), and the speed of the Moon's shadow to have only been affected by the change in latitude.

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: On $21^{\text {st }}$ August 2017 the continental United States experienced a total solar eclipse. Dubbed the 'Great American Eclipse', it was estimated to be one of the most watched eclipses in history. ## Total Solar Eclipse of 2017 Aug 21 [figure1] Figure 3: The path of totality for the Great American Eclipse. The narrow dimension is its width. Credit: Fred Espenak, NASA's GSFC. The path of totality (where the Moon completely obscures the Sun) is shown in Figure 3, and the point of greatest eclipse ("GE"; where the path was widest since the axis of the cone of the Moon's shadow passed closest to the centre of the Earth) was near the village of Cerulean, Kentucky. The following data can be used for this question: - The angular radii of the Sun and the Moon (if observed from the centre of the Earth) at the moment of GE are $15^{\prime} 48.7^{\prime \prime}$ and $16^{\prime} 03.4^{\prime \prime}$, respectively, where the notation $x x^{\prime} y y . y^{\prime \prime}$ corresponds to $x x$ arcminutes and $y y . y$ arcseconds ( 60 arcminutes $=1$ degree, and 60 arcseconds $=1$ arcminute $)$ - The latitude and longitude of the location of GE are $36^{\circ} 58.0^{\prime} \mathrm{N}$ and $87^{\circ} 40.3^{\prime} \mathrm{W}$, respectively - Take the mean radii of the Sun, Earth and Moon to be respectively $R_{\odot}=695700 \mathrm{~km}, R_{\oplus}=$ $6371 \mathrm{~km}$ and $R_{\text {Moon }}=1737 \mathrm{~km}$, and a day to be 24 hours - Take the semi-major axes of the Sun-Earth and Earth-Moon systems to be $149600000 \mathrm{~km}$ and $384400 \mathrm{~km}$, respectively - As viewed from a location far above the North Pole, the Moon orbits in an anticlockwise direction around the Earth, and the Earth spins in an anticlockwise direction For an ellipse with semi-major axis $a$ it can be shown that the velocity $v$, at a distance $r$ from mass $M$, can be written as: $$ v^{2}=G M\left(\frac{2}{r}-\frac{1}{a}\right) $$ The point of greatest eclipse and greatest duration do not generally coincide, as a more elliptical shadow with a major axis aligned with the path of maximum totality (and thinner path width, equal to the minor axis) can compensate for the shadow moving faster at higher latitudes. For this eclipse the point of greatest duration ("GD") was at co-ordinates of $37^{\circ} 35^{\prime} \mathrm{N}$ latitude and $89^{\circ} 07^{\prime} \mathrm{W}$ longitude, reached about 4 minutes before GE, and where totality lasted $0.1 \mathrm{~s}$ longer than the value calculated in part $\mathrm{c}$. [figure2] Figure 4: The route of the Moon's shadow in the vicinity of the points of greatest duration (GD, near Carbondale) and greatest eclipse (GE, near Hopkinsville). Any places between the two limits on the path of totality will experience at least a very short period of totality - outside that region will only be a partial eclipse and the perpendicular distance between them is the path width. The longest duration of totality at that point of the shadow's journey is indicated as the path of maximum totality, which both GE and GD sit on. The closest part of that path to Carbondale is indicated as CP. Credit: Fred Espenak \& Google Maps. problem: d. The town of Carbondale, Illinois, is the closest big town to the point of GD, with co-ordinates $37^{\circ} 44^{\prime} \mathrm{N}$ latitude and $89^{\circ} 13^{\prime} \mathrm{W}$ longitude. Assuming the path of maximum totality can be treated as linear as it passes through the region around GD and GE: iii. Hence calculate (to the nearest $0.1 \mathrm{~s}$ ) how much shorter totality was for residents in Carbondale compared with CP. Take the duration at CP to be the same as GD, the width of the path to be $0.5 \mathrm{~km}$ less than for GE (so the Moon's shadow is elliptical), and the speed of the Moon's shadow to have only been affected by the change in latitude. 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_1177

On $21^{\text {st }}$ August 2017 the continental United States experienced a total solar eclipse. Dubbed the 'Great American Eclipse', it was estimated to be one of the most watched eclipses in history. ## Total Solar Eclipse of 2017 Aug 21 [figure1] Figure 3: The path of totality for the Great American Eclipse. The narrow dimension is its width. Credit: Fred Espenak, NASA's GSFC. The path of totality (where the Moon completely obscures the Sun) is shown in Figure 3, and the point of greatest eclipse ("GE"; where the path was widest since the axis of the cone of the Moon's shadow passed closest to the centre of the Earth) was near the village of Cerulean, Kentucky. The following data can be used for this question: - The angular radii of the Sun and the Moon (if observed from the centre of the Earth) at the moment of GE are $15^{\prime} 48.7^{\prime \prime}$ and $16^{\prime} 03.4^{\prime \prime}$, respectively, where the notation $x x^{\prime} y y . y^{\prime \prime}$ corresponds to $x x$ arcminutes and $y y . y$ arcseconds ( 60 arcminutes $=1$ degree, and 60 arcseconds $=1$ arcminute $)$ - The latitude and longitude of the location of GE are $36^{\circ} 58.0^{\prime} \mathrm{N}$ and $87^{\circ} 40.3^{\prime} \mathrm{W}$, respectively - Take the mean radii of the Sun, Earth and Moon to be respectively $R_{\odot}=695700 \mathrm{~km}, R_{\oplus}=$ $6371 \mathrm{~km}$ and $R_{\text {Moon }}=1737 \mathrm{~km}$, and a day to be 24 hours - Take the semi-major axes of the Sun-Earth and Earth-Moon systems to be $149600000 \mathrm{~km}$ and $384400 \mathrm{~km}$, respectively - As viewed from a location far above the North Pole, the Moon orbits in an anticlockwise direction around the Earth, and the Earth spins in an anticlockwise direction For an ellipse with semi-major axis $a$ it can be shown that the velocity $v$, at a distance $r$ from mass $M$, can be written as: $$ v^{2}=G M\left(\frac{2}{r}-\frac{1}{a}\right) $$ The point of greatest eclipse and greatest duration do not generally coincide, as a more elliptical shadow with a major axis aligned with the path of maximum totality (and thinner path width, equal to the minor axis) can compensate for the shadow moving faster at higher latitudes. For this eclipse the point of greatest duration ("GD") was at co-ordinates of $37^{\circ} 35^{\prime} \mathrm{N}$ latitude and $89^{\circ} 07^{\prime} \mathrm{W}$ longitude, reached about 4 minutes before GE, and where totality lasted $0.1 \mathrm{~s}$ longer than the value calculated in part $\mathrm{c}$. [figure2] Figure 4: The route of the Moon's shadow in the vicinity of the points of greatest duration (GD, near Carbondale) and greatest eclipse (GE, near Hopkinsville). Any places between the two limits on the path of totality will experience at least a very short period of totality - outside that region will only be a partial eclipse and the perpendicular distance between them is the path width. The longest duration of totality at that point of the shadow's journey is indicated as the path of maximum totality, which both GE and GD sit on. The closest part of that path to Carbondale is indicated as CP. Credit: Fred Espenak \& Google Maps.d. The town of Carbondale, Illinois, is the closest big town to the point of GD, with co-ordinates $37^{\circ} 44^{\prime} \mathrm{N}$ latitude and $89^{\circ} 13^{\prime} \mathrm{W}$ longitude. Assuming the path of maximum totality can be treated as linear as it passes through the region around GD and GE: i. Calculate the co-ordinates of the closest point (" $\mathrm{CP}^{\prime \prime}$ ) to Carbondale on the path of maximum totality.

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: On $21^{\text {st }}$ August 2017 the continental United States experienced a total solar eclipse. Dubbed the 'Great American Eclipse', it was estimated to be one of the most watched eclipses in history. ## Total Solar Eclipse of 2017 Aug 21 [figure1] Figure 3: The path of totality for the Great American Eclipse. The narrow dimension is its width. Credit: Fred Espenak, NASA's GSFC. The path of totality (where the Moon completely obscures the Sun) is shown in Figure 3, and the point of greatest eclipse ("GE"; where the path was widest since the axis of the cone of the Moon's shadow passed closest to the centre of the Earth) was near the village of Cerulean, Kentucky. The following data can be used for this question: - The angular radii of the Sun and the Moon (if observed from the centre of the Earth) at the moment of GE are $15^{\prime} 48.7^{\prime \prime}$ and $16^{\prime} 03.4^{\prime \prime}$, respectively, where the notation $x x^{\prime} y y . y^{\prime \prime}$ corresponds to $x x$ arcminutes and $y y . y$ arcseconds ( 60 arcminutes $=1$ degree, and 60 arcseconds $=1$ arcminute $)$ - The latitude and longitude of the location of GE are $36^{\circ} 58.0^{\prime} \mathrm{N}$ and $87^{\circ} 40.3^{\prime} \mathrm{W}$, respectively - Take the mean radii of the Sun, Earth and Moon to be respectively $R_{\odot}=695700 \mathrm{~km}, R_{\oplus}=$ $6371 \mathrm{~km}$ and $R_{\text {Moon }}=1737 \mathrm{~km}$, and a day to be 24 hours - Take the semi-major axes of the Sun-Earth and Earth-Moon systems to be $149600000 \mathrm{~km}$ and $384400 \mathrm{~km}$, respectively - As viewed from a location far above the North Pole, the Moon orbits in an anticlockwise direction around the Earth, and the Earth spins in an anticlockwise direction For an ellipse with semi-major axis $a$ it can be shown that the velocity $v$, at a distance $r$ from mass $M$, can be written as: $$ v^{2}=G M\left(\frac{2}{r}-\frac{1}{a}\right) $$ The point of greatest eclipse and greatest duration do not generally coincide, as a more elliptical shadow with a major axis aligned with the path of maximum totality (and thinner path width, equal to the minor axis) can compensate for the shadow moving faster at higher latitudes. For this eclipse the point of greatest duration ("GD") was at co-ordinates of $37^{\circ} 35^{\prime} \mathrm{N}$ latitude and $89^{\circ} 07^{\prime} \mathrm{W}$ longitude, reached about 4 minutes before GE, and where totality lasted $0.1 \mathrm{~s}$ longer than the value calculated in part $\mathrm{c}$. [figure2] Figure 4: The route of the Moon's shadow in the vicinity of the points of greatest duration (GD, near Carbondale) and greatest eclipse (GE, near Hopkinsville). Any places between the two limits on the path of totality will experience at least a very short period of totality - outside that region will only be a partial eclipse and the perpendicular distance between them is the path width. The longest duration of totality at that point of the shadow's journey is indicated as the path of maximum totality, which both GE and GD sit on. The closest part of that path to Carbondale is indicated as CP. Credit: Fred Espenak \& Google Maps. problem: d. The town of Carbondale, Illinois, is the closest big town to the point of GD, with co-ordinates $37^{\circ} 44^{\prime} \mathrm{N}$ latitude and $89^{\circ} 13^{\prime} \mathrm{W}$ longitude. Assuming the path of maximum totality can be treated as linear as it passes through the region around GD and GE: i. Calculate the co-ordinates of the closest point (" $\mathrm{CP}^{\prime \prime}$ ) to Carbondale on the path of maximum totality. 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 \circ, 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_229

2021 年 6 月 17 日, 搭载神舟十二号载人飞船的长征二号 $\mathrm{F}$ 遥十二运载火箭在酒泉卫星发射中心点成功发射升空, 神舟十二号载人飞船与天和核心舱及天舟二号组合体成功对接，将中国三名航天员送入“太空家园”，核心舱绕地球飞行的轨道可视为圆轨道，轨道离地面的高度约为地球半径的 $\frac{1}{16}$, 运行周期约为 $90 \mathrm{~min}$, 引力常量 $G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2}$ 。下列说法中正确的是（） [图1] A: 核心舱在轨道上飞行的速度约为 $7.9 \mathrm{~km} / \mathrm{s}$ B: 仅根据题中数据即可估算出地球密度 C: “太空家园”中静止状态的宇航员的加速度为 0 D: 理论上火箭从酒泉发射比从文昌发射更节省燃料

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, 搭载神舟十二号载人飞船的长征二号 $\mathrm{F}$ 遥十二运载火箭在酒泉卫星发射中心点成功发射升空, 神舟十二号载人飞船与天和核心舱及天舟二号组合体成功对接，将中国三名航天员送入“太空家园”，核心舱绕地球飞行的轨道可视为圆轨道，轨道离地面的高度约为地球半径的 $\frac{1}{16}$, 运行周期约为 $90 \mathrm{~min}$, 引力常量 $G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2}$ 。下列说法中正确的是（） [图1] A: 核心舱在轨道上飞行的速度约为 $7.9 \mathrm{~km} / \mathrm{s}$ B: 仅根据题中数据即可估算出地球密度 C: “太空家园”中静止状态的宇航员的加速度为 0 D: 理论上火箭从酒泉发射比从文昌发射更节省燃料 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_384

拉格朗日点指在两个大天体引力作用下, 能使小物体稳定的点 (小物体质量相对两 大天体可忽略不计)。这些点的存在由法国数学家拉格朗日于 1772 年推导证明的, 1906 年首次发现运动于木星轨道上的小行星 (见脱罗央群小行星) 在木星和太阳的作用下处于拉格朗日点上。在每个由两大天体构成的系统中, 按推论有 5 个拉格朗日点, 其中连线上有三个拉格朗日点, 分别是 $L_{1} 、 L_{2} 、 L_{3}$, 如图所示。我国发射的“鹊桥”卫星就在地月系统平衡点 $L_{2}$ 点做周期运动, 通过定期轨控保持轨道的稳定性, 可实现对着陆器和巡视器的中继通信覆盖, 首次实现地月 $L_{2}$ 点周期轨道的长期稳定运行。设某两个天体系统的中心天体质量为 $M$, 环绕天体质量为 $m$, 两天体间距离为 $L$, 万有引力常量为 $G, L_{1}$ 点到中心天体的距离为 $R_{1}, L_{2}$ 点到中心天体的距离为 $R_{2}$ 。求: 为了进一步的通信覆盖, 发射两颗质量均为 $m_{0}$ 的卫星, 分别处于 $L_{1} 、 L_{2}$ 点, $L_{1}$ 、 $L_{2}$ 到环绕天体的距离近似相等 (远小于 $L$ ), 两卫星与环绕天体同步绕中心天体做圆周运动, 忽略卫星间的引力, 求中心天体对环绕天体的引力与它对两卫星的引力之和的比值? [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: 拉格朗日点指在两个大天体引力作用下, 能使小物体稳定的点 (小物体质量相对两 大天体可忽略不计)。这些点的存在由法国数学家拉格朗日于 1772 年推导证明的, 1906 年首次发现运动于木星轨道上的小行星 (见脱罗央群小行星) 在木星和太阳的作用下处于拉格朗日点上。在每个由两大天体构成的系统中, 按推论有 5 个拉格朗日点, 其中连线上有三个拉格朗日点, 分别是 $L_{1} 、 L_{2} 、 L_{3}$, 如图所示。我国发射的“鹊桥”卫星就在地月系统平衡点 $L_{2}$ 点做周期运动, 通过定期轨控保持轨道的稳定性, 可实现对着陆器和巡视器的中继通信覆盖, 首次实现地月 $L_{2}$ 点周期轨道的长期稳定运行。设某两个天体系统的中心天体质量为 $M$, 环绕天体质量为 $m$, 两天体间距离为 $L$, 万有引力常量为 $G, L_{1}$ 点到中心天体的距离为 $R_{1}, L_{2}$ 点到中心天体的距离为 $R_{2}$ 。求: 为了进一步的通信覆盖, 发射两颗质量均为 $m_{0}$ 的卫星, 分别处于 $L_{1} 、 L_{2}$ 点, $L_{1}$ 、 $L_{2}$ 到环绕天体的距离近似相等 (远小于 $L$ ), 两卫星与环绕天体同步绕中心天体做圆周运动, 忽略卫星间的引力, 求中心天体对环绕天体的引力与它对两卫星的引力之和的比值? [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_1074

The James Webb Space Telescope (JWST) is an incredibly exciting next generation telescope that was successfully launched on $25^{\text {th }}$ December 2021 . Its mirror is approximately $6.5 \mathrm{~m}$ in diameter, much larger than the $2.4 \mathrm{~m}$ mirror of the Hubble Space Telescope (HST), and so it has far greater resolution and sensitivity. Whilst HST largely imaged in the visible, JWST will do most of its work in the nearand mid-infrared (NIR and MIR respectively). This will allow it to pick up heavily redshifted light, such as that from the first generation of stars in the very first galaxies. [figure1] Figure 5: Left: A full-scale model of JWST next to some of the scientists and engineers involved in its development at the Goddard Space Flight Center. Credit: NASA / Goddard Space Flight Center / Pat Izzo. Right: The position of the second Lagrangian point, $L_{2}$, relative to the Earth. Credit: ESA. The resolution limit of a telescope is set by the amount of diffraction light rays experience as they enter the system, and is related to the diameter of a telescope, $D$, and the wavelength being observed, $\lambda$. The resolution limit of a CCD is set by the size of the pixels. Three of the imaging cameras on JWST are tabulated with some properties below: | Instrument | Wavelength range $(\mu \mathrm{m})$ | CCD plate scale (arcseconds / pixel) | | :---: | :---: | :---: | | NIRCam (short wave) | $0.6-2.3$ | 0.031 | | NIRCam (long wave) | $2.4-5.0$ | 0.065 | | MIRI | $5.6-25.5$ | 0.11 | An arcsecond is a measure of angle where $1^{\circ}=3600$ arcseconds. The familiar variation in intensity on a screen, $I_{\text {slit }}$, due to diffraction through an infinitely tall single slit is given as $$ I_{\text {slit }}=I_{0}\left(\frac{\sin (x)}{x}\right)^{2}, \text { where } \quad x=\frac{\pi D \theta}{\lambda} $$ and $I_{0}$ is the initial intensity. For a circular aperture, the formula is slightly different and is given as $$ I_{\mathrm{circ}}=I_{0}\left(\frac{2 J_{1}(x)}{x}\right)^{2} . $$ Here $J_{1}(x)$ is the Bessel function of the first kind and is calculated as $$ J_{n}(x)=\sum_{r=0}^{\infty} \frac{(-1)^{r}}{r !(n+r) !}\left(\frac{x}{2}\right)^{n+2 r} \quad \text { so } \quad J_{1}(x)=\frac{x}{2}\left(1-\frac{x^{2}}{8}+\frac{x^{4}}{192}-\ldots\right) . $$ The $x$-axis intercepts and shape of the maxima are quite different, as shown in Figure 6. The position of the first minimum of $I_{\text {slit }}$ is at $x_{\min }=\pi$ meaning that $\theta_{\min , \text { slit }}=\lambda / D$, whilst for $I_{\text {circ }}$ it is at $x_{\min }=3.8317 \ldots$ so $\theta_{\min , \mathrm{circ}} \approx 1.22 \lambda / D$. This is one way of defining the minimum angular resolution, although since the flux drops off so steeply away from the central maximum a more convenient one for use with CCDs is the angle corresponding to the full width half maximum (FWHM). [figure2] Figure 6: Left: The $I_{\text {slit }}$ (purple) and $I_{\text {circ }}$ (blue - the wider central maximum) functions, normalised so that $I_{0}=1$. You can see the shapes and $x$-intercepts are different. Right: How $x_{\min }$ and the full width half maximum (FWHM) are defined. Here it is shown for $I_{\text {circ }}$. As well as having the largest mirror of any space telescope ever launched, it is also one of the most sensitive, with its greatest sensitivity in the NIRCam F200W filter (centred on a wavelength of $1.989 \mu \mathrm{m})$ where after $10^{4}$ seconds it can detect a flux of $9.1 \mathrm{nJy}\left(1 \mathrm{Jy}=10^{-26} \mathrm{~W} \mathrm{~m}^{-2} \mathrm{~Hz}^{-1}\right.$ ) with a signal-to-noise ratio (S/N) of 10 , corresponding to an apparent magnitude of $m=29.0$. This extraordinary sensitivity can be used to pick up light from the earliest galaxies in the Universe. The scale factor, $a$, parameterises the expansion of the Universe since the Big Bang, and is related to the redshift, $z$, as $$ a=(1+z)^{-1} \quad \text { where } \quad z \equiv \frac{\lambda_{\text {obs }}-\lambda_{\mathrm{emit}}}{\lambda_{\mathrm{emit}}} $$ with $\lambda_{\text {obs }}$ the observed wavelength and $\lambda_{\text {emit }}$ the rest frame wavelength. The current rate of expansion of the Universe is given by the Hubble constant, $H_{0}$, and this is related to the current Hubble time, $t_{\mathrm{H}_{0}}$, and current Hubble distance, $D_{\mathrm{H}_{0}}$, as $$ t_{\mathrm{H}_{0}} \equiv H_{0}^{-1} \quad \text { and } \quad D_{\mathrm{H}_{0}} \equiv c t_{\mathrm{H}_{0}} \text {. } $$ Here the subscript 0 indicates the values are measured today. The Hubble constant is more appropriately known as the Hubble parameter as it is a function of time, and the evolution of $H$ as a function of $z$ is $$ E(z)=\frac{H}{H_{0}} \equiv\left[\Omega_{0, m}(1+z)^{3}+\Omega_{0, \Lambda}+\Omega_{0, r}(1+z)^{4}\right]^{1 / 2}, $$ where $\Omega$ is the normalised density parameter, and the subscript $m, r$, and $\Lambda$ indicate the contribution to $\Omega$ from matter, radiation, and dark energy, respectively. The proper age of the Universe $t(z)$ at redshift $z$ is best evaluated in terms of $a$ as $$ t=t_{\mathrm{H}_{0}} \int_{0}^{(1+z)^{-1}} \frac{a}{\left(\Omega_{0, m} a+\Omega_{0, \Lambda} a^{4}+\Omega_{0, r}\right)^{1 / 2}} \mathrm{~d} a $$ If $\Omega_{0, r}=0$ and $\Omega_{0, m}+\Omega_{0, \Lambda}=1$ (corresponding to what it known as a flat Universe), then via the standard integral $\int\left(b^{2}+x^{2}\right)^{-1 / 2} \mathrm{~d} x=\ln \left(x+\sqrt{b^{2}+x^{2}}\right)+C$ this integral can be evaluated analytically to give $$ t=t_{\mathrm{H}_{0}} \frac{2}{3 \Omega_{0, \Lambda}^{1 / 2}} \ln \left[\left(\frac{\Omega_{0, \Lambda}}{\Omega_{0, m}}\right)^{1 / 2}(1+z)^{-3 / 2}+\left(\frac{\Omega_{0, \Lambda}}{\Omega_{0, m}(1+z)^{3}}+1\right)^{1 / 2}\right] $$ Finally, the luminosity distance, $D_{L}(z)$, corresponding to the distance away that an object appears to be due to its measured flux given its intrinsic luminosity (i.e. $f \equiv L / 4 \pi D_{L}^{2}$ ) is given as $$ D_{L}=\left(1+z_{i}\right) D_{\mathrm{H}_{0}} \int_{0}^{z_{i}} \frac{1}{E(z)} \mathrm{d} z=\left(1+z_{i}\right) D_{\mathrm{H}_{0}} \int_{a_{i}}^{1} \frac{1}{\left(\Omega_{0, m} a+\Omega_{0, \Lambda} a^{4}+\Omega_{0, r}\right)^{1 / 2}} \mathrm{~d} a $$ where $z_{i}$ is the redshift of interest and $a_{i}$ is the equivalent scale factor. Even for the flat Universe case with $\Omega_{0, r}=0$ this integral cannot be be done analytically so must be evaluated numerically.c. Computer models suggest the first galaxies formed around $z \sim 10-20$. One of the best ways to look for high-redshift galaxies is to try and detect the emission from the Lyman alpha (Lya) emission line at $\lambda_{\text {emit }}=121.6 \mathrm{~nm}$ as it is a relatively bright line. Some of the brightest galaxies in that initial era of galaxy formation would have an absolute magnitude of $\mathcal{M} \sim 20$. In this question, you are given that $\Omega_{0, \mathrm{~m}}=0.3, \Omega_{0, \Lambda}=0.7, \Omega_{0, \mathrm{r}}=0$ and $\mathrm{H}_{0}=70 \mathrm{~km} \mathrm{~s}^{-1} \mathrm{Mpc}^{-1}$. i. Calculate the redshift at which the Lya line is detected in the centre of the F200W filter.

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 James Webb Space Telescope (JWST) is an incredibly exciting next generation telescope that was successfully launched on $25^{\text {th }}$ December 2021 . Its mirror is approximately $6.5 \mathrm{~m}$ in diameter, much larger than the $2.4 \mathrm{~m}$ mirror of the Hubble Space Telescope (HST), and so it has far greater resolution and sensitivity. Whilst HST largely imaged in the visible, JWST will do most of its work in the nearand mid-infrared (NIR and MIR respectively). This will allow it to pick up heavily redshifted light, such as that from the first generation of stars in the very first galaxies. [figure1] Figure 5: Left: A full-scale model of JWST next to some of the scientists and engineers involved in its development at the Goddard Space Flight Center. Credit: NASA / Goddard Space Flight Center / Pat Izzo. Right: The position of the second Lagrangian point, $L_{2}$, relative to the Earth. Credit: ESA. The resolution limit of a telescope is set by the amount of diffraction light rays experience as they enter the system, and is related to the diameter of a telescope, $D$, and the wavelength being observed, $\lambda$. The resolution limit of a CCD is set by the size of the pixels. Three of the imaging cameras on JWST are tabulated with some properties below: | Instrument | Wavelength range $(\mu \mathrm{m})$ | CCD plate scale (arcseconds / pixel) | | :---: | :---: | :---: | | NIRCam (short wave) | $0.6-2.3$ | 0.031 | | NIRCam (long wave) | $2.4-5.0$ | 0.065 | | MIRI | $5.6-25.5$ | 0.11 | An arcsecond is a measure of angle where $1^{\circ}=3600$ arcseconds. The familiar variation in intensity on a screen, $I_{\text {slit }}$, due to diffraction through an infinitely tall single slit is given as $$ I_{\text {slit }}=I_{0}\left(\frac{\sin (x)}{x}\right)^{2}, \text { where } \quad x=\frac{\pi D \theta}{\lambda} $$ and $I_{0}$ is the initial intensity. For a circular aperture, the formula is slightly different and is given as $$ I_{\mathrm{circ}}=I_{0}\left(\frac{2 J_{1}(x)}{x}\right)^{2} . $$ Here $J_{1}(x)$ is the Bessel function of the first kind and is calculated as $$ J_{n}(x)=\sum_{r=0}^{\infty} \frac{(-1)^{r}}{r !(n+r) !}\left(\frac{x}{2}\right)^{n+2 r} \quad \text { so } \quad J_{1}(x)=\frac{x}{2}\left(1-\frac{x^{2}}{8}+\frac{x^{4}}{192}-\ldots\right) . $$ The $x$-axis intercepts and shape of the maxima are quite different, as shown in Figure 6. The position of the first minimum of $I_{\text {slit }}$ is at $x_{\min }=\pi$ meaning that $\theta_{\min , \text { slit }}=\lambda / D$, whilst for $I_{\text {circ }}$ it is at $x_{\min }=3.8317 \ldots$ so $\theta_{\min , \mathrm{circ}} \approx 1.22 \lambda / D$. This is one way of defining the minimum angular resolution, although since the flux drops off so steeply away from the central maximum a more convenient one for use with CCDs is the angle corresponding to the full width half maximum (FWHM). [figure2] Figure 6: Left: The $I_{\text {slit }}$ (purple) and $I_{\text {circ }}$ (blue - the wider central maximum) functions, normalised so that $I_{0}=1$. You can see the shapes and $x$-intercepts are different. Right: How $x_{\min }$ and the full width half maximum (FWHM) are defined. Here it is shown for $I_{\text {circ }}$. As well as having the largest mirror of any space telescope ever launched, it is also one of the most sensitive, with its greatest sensitivity in the NIRCam F200W filter (centred on a wavelength of $1.989 \mu \mathrm{m})$ where after $10^{4}$ seconds it can detect a flux of $9.1 \mathrm{nJy}\left(1 \mathrm{Jy}=10^{-26} \mathrm{~W} \mathrm{~m}^{-2} \mathrm{~Hz}^{-1}\right.$ ) with a signal-to-noise ratio (S/N) of 10 , corresponding to an apparent magnitude of $m=29.0$. This extraordinary sensitivity can be used to pick up light from the earliest galaxies in the Universe. The scale factor, $a$, parameterises the expansion of the Universe since the Big Bang, and is related to the redshift, $z$, as $$ a=(1+z)^{-1} \quad \text { where } \quad z \equiv \frac{\lambda_{\text {obs }}-\lambda_{\mathrm{emit}}}{\lambda_{\mathrm{emit}}} $$ with $\lambda_{\text {obs }}$ the observed wavelength and $\lambda_{\text {emit }}$ the rest frame wavelength. The current rate of expansion of the Universe is given by the Hubble constant, $H_{0}$, and this is related to the current Hubble time, $t_{\mathrm{H}_{0}}$, and current Hubble distance, $D_{\mathrm{H}_{0}}$, as $$ t_{\mathrm{H}_{0}} \equiv H_{0}^{-1} \quad \text { and } \quad D_{\mathrm{H}_{0}} \equiv c t_{\mathrm{H}_{0}} \text {. } $$ Here the subscript 0 indicates the values are measured today. The Hubble constant is more appropriately known as the Hubble parameter as it is a function of time, and the evolution of $H$ as a function of $z$ is $$ E(z)=\frac{H}{H_{0}} \equiv\left[\Omega_{0, m}(1+z)^{3}+\Omega_{0, \Lambda}+\Omega_{0, r}(1+z)^{4}\right]^{1 / 2}, $$ where $\Omega$ is the normalised density parameter, and the subscript $m, r$, and $\Lambda$ indicate the contribution to $\Omega$ from matter, radiation, and dark energy, respectively. The proper age of the Universe $t(z)$ at redshift $z$ is best evaluated in terms of $a$ as $$ t=t_{\mathrm{H}_{0}} \int_{0}^{(1+z)^{-1}} \frac{a}{\left(\Omega_{0, m} a+\Omega_{0, \Lambda} a^{4}+\Omega_{0, r}\right)^{1 / 2}} \mathrm{~d} a $$ If $\Omega_{0, r}=0$ and $\Omega_{0, m}+\Omega_{0, \Lambda}=1$ (corresponding to what it known as a flat Universe), then via the standard integral $\int\left(b^{2}+x^{2}\right)^{-1 / 2} \mathrm{~d} x=\ln \left(x+\sqrt{b^{2}+x^{2}}\right)+C$ this integral can be evaluated analytically to give $$ t=t_{\mathrm{H}_{0}} \frac{2}{3 \Omega_{0, \Lambda}^{1 / 2}} \ln \left[\left(\frac{\Omega_{0, \Lambda}}{\Omega_{0, m}}\right)^{1 / 2}(1+z)^{-3 / 2}+\left(\frac{\Omega_{0, \Lambda}}{\Omega_{0, m}(1+z)^{3}}+1\right)^{1 / 2}\right] $$ Finally, the luminosity distance, $D_{L}(z)$, corresponding to the distance away that an object appears to be due to its measured flux given its intrinsic luminosity (i.e. $f \equiv L / 4 \pi D_{L}^{2}$ ) is given as $$ D_{L}=\left(1+z_{i}\right) D_{\mathrm{H}_{0}} \int_{0}^{z_{i}} \frac{1}{E(z)} \mathrm{d} z=\left(1+z_{i}\right) D_{\mathrm{H}_{0}} \int_{a_{i}}^{1} \frac{1}{\left(\Omega_{0, m} a+\Omega_{0, \Lambda} a^{4}+\Omega_{0, r}\right)^{1 / 2}} \mathrm{~d} a $$ where $z_{i}$ is the redshift of interest and $a_{i}$ is the equivalent scale factor. Even for the flat Universe case with $\Omega_{0, r}=0$ this integral cannot be be done analytically so must be evaluated numerically. problem: c. Computer models suggest the first galaxies formed around $z \sim 10-20$. One of the best ways to look for high-redshift galaxies is to try and detect the emission from the Lyman alpha (Lya) emission line at $\lambda_{\text {emit }}=121.6 \mathrm{~nm}$ as it is a relatively bright line. Some of the brightest galaxies in that initial era of galaxy formation would have an absolute magnitude of $\mathcal{M} \sim 20$. In this question, you are given that $\Omega_{0, \mathrm{~m}}=0.3, \Omega_{0, \Lambda}=0.7, \Omega_{0, \mathrm{r}}=0$ and $\mathrm{H}_{0}=70 \mathrm{~km} \mathrm{~s}^{-1} \mathrm{Mpc}^{-1}$. i. Calculate the redshift at which the Lya line is detected in the centre of the F200W filter. 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 the numerical value.

NV

Astronomy

EN

multi-modal

Astronomy_336

设想地球没有自转, 坚直向下通过地心把地球钻通. 如果在这个通过地心的笔直的管道的一端无初速度地放下一物体，下列说法正确的是（） A: 物体在地心时, 它与地心间距离为零, 地球对物体的万有引力无穷大 B: 物体在地心时, 地球对它的万有引力为零 C: 物体在管道中将往返运动, 通过地心时加速度为零, 速率最大 D: 物体运动到地心时由于万有引力为零, 它将静止在地心不动

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: 设想地球没有自转, 坚直向下通过地心把地球钻通. 如果在这个通过地心的笔直的管道的一端无初速度地放下一物体，下列说法正确的是（） A: 物体在地心时, 它与地心间距离为零, 地球对物体的万有引力无穷大 B: 物体在地心时, 地球对它的万有引力为零 C: 物体在管道中将往返运动, 通过地心时加速度为零, 速率最大 D: 物体运动到地心时由于万有引力为零, 它将静止在地心不动 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_74

有人提出了一种不用火箭发射人造地球卫星的设想。其设想如下: 沿地球的一条弦挖一通道, 如图乙所示. 在通道的两个出口处 $A$ 和 $B$, 分别将质量为 $M$ 的物体和质量为 $m$ 的待发射卫星同时自由释放, 只要 $M$ 比 $m$ 足够大, 碰撞后, 质量为 $m$ 的物体, 即待发射的卫星就会从通道口 $B$ 冲出通道。(已知地球表面的重力加速度为 $g$, 地球半径为 $\left.R_{0}\right)$ [图1] 甲 [图2] 乙 如图乙所示, 是在地球上距地心 $h$ 处沿一条弦挖了一条光滑的通道 $A B$, 从 $A$ 点处静止释放一个质量为 $m$ 的物体，物体下落到通道中点 $O^{\prime}$ 处的速度多大?

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: 有人提出了一种不用火箭发射人造地球卫星的设想。其设想如下: 沿地球的一条弦挖一通道, 如图乙所示. 在通道的两个出口处 $A$ 和 $B$, 分别将质量为 $M$ 的物体和质量为 $m$ 的待发射卫星同时自由释放, 只要 $M$ 比 $m$ 足够大, 碰撞后, 质量为 $m$ 的物体, 即待发射的卫星就会从通道口 $B$ 冲出通道。(已知地球表面的重力加速度为 $g$, 地球半径为 $\left.R_{0}\right)$ [图1] 甲 [图2] 乙 如图乙所示, 是在地球上距地心 $h$ 处沿一条弦挖了一条光滑的通道 $A B$, 从 $A$ 点处静止释放一个质量为 $m$ 的物体，物体下落到通道中点 $O^{\prime}$ 处的速度多大? 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_263

在宇宙中也存在由质量相等的四颗星组成的“四星系统”, “四星系统”离其他恒星较远，通常可忽略其他星体对“四星系统”的引力作用。已观测到稳定的“四星系统”存在两种基本的构成形式: 一种是四颗星稳定地分布在边长为 $a$ 的正方形的四个顶点上, 均围绕正方形对角线的交点做匀速圆周运动, 如下图 (1) 所示。另一种形式是有三颗星位于等边三角形的三个顶点上, 第四颗星刚好位于三角形的中心不动, 三颗星沿外接于等边三角形的半径为 $a$ 的圆形轨道运行, 如下图 (2) 所示。假设两种形式的“四星系统”中每个星的质量均为 $m$, 已知引力常量为 $G$, 求这两种形式下的周期 $T_{1}$ 和 $T_{2}$ 。 [图1] (1) [图2]

你正在参加一个国际天文竞赛，并需要解决以下问题。 这个问题包含多个待求解的量。 问题: 在宇宙中也存在由质量相等的四颗星组成的“四星系统”, “四星系统”离其他恒星较远，通常可忽略其他星体对“四星系统”的引力作用。已观测到稳定的“四星系统”存在两种基本的构成形式: 一种是四颗星稳定地分布在边长为 $a$ 的正方形的四个顶点上, 均围绕正方形对角线的交点做匀速圆周运动, 如下图 (1) 所示。另一种形式是有三颗星位于等边三角形的三个顶点上, 第四颗星刚好位于三角形的中心不动, 三颗星沿外接于等边三角形的半径为 $a$ 的圆形轨道运行, 如下图 (2) 所示。假设两种形式的“四星系统”中每个星的质量均为 $m$, 已知引力常量为 $G$, 求这两种形式下的周期 $T_{1}$ 和 $T_{2}$ 。 [图1] (1) [图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。 你的最终解答的量应该按以下顺序输出：[ $T_{1}$, $T_{2}$] 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_357

如图所示, 发射同步卫星的一种程序是: 先让卫星进入一个近地的圆轨道, 然后在 $P$ 点点火加速, 进入粗圆形转移轨道 (该陏圆轨道的近地点为近地轨道上的 $P$, 远地点为同步轨道上的 $Q)$, 到达远地点时再次自动点火加速, 进入同步轨道. 设卫星在近地圆轨道上运行的速率为 $v_{1}$, 在 $P$ 点短时间加速后的速率为 $v_{2}$, 沿转移轨道刚到达远地点 $Q$ 时的速率为 $v_{3}$, 在 $Q$ 点短时间加速后进入同步轨道后的速率为 $v_{4}$, 则四个速率的大小排列正确的是（ [图1] A: $v_{1}>v_{2}>v_{3}>v_{4}$ B: $v_{2}>v_{1}>v_{3}>v_{4}$ C: $v_{1}>v_{2}>v_{4}>v_{3}$ D: $v_{2}>v_{1}>v_{4}>v_{3}$

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个单选题（只有一个正确答案）。 问题: 如图所示, 发射同步卫星的一种程序是: 先让卫星进入一个近地的圆轨道, 然后在 $P$ 点点火加速, 进入粗圆形转移轨道 (该陏圆轨道的近地点为近地轨道上的 $P$, 远地点为同步轨道上的 $Q)$, 到达远地点时再次自动点火加速, 进入同步轨道. 设卫星在近地圆轨道上运行的速率为 $v_{1}$, 在 $P$ 点短时间加速后的速率为 $v_{2}$, 沿转移轨道刚到达远地点 $Q$ 时的速率为 $v_{3}$, 在 $Q$ 点短时间加速后进入同步轨道后的速率为 $v_{4}$, 则四个速率的大小排列正确的是（ [图1] A: $v_{1}>v_{2}>v_{3}>v_{4}$ B: $v_{2}>v_{1}>v_{3}>v_{4}$ C: $v_{1}>v_{2}>v_{4}>v_{3}$ D: $v_{2}>v_{1}>v_{4}>v_{3}$ 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应为以下选项之一：[A, B, C, D]

SC

Astronomy

ZH

multi-modal

Astronomy_265

如图所示为科学家模拟水星探测器进入水星表面绕行轨道的过程示意图, 假设水星的半径为 $R$, 探测器在距离水星表面高度为 $3 R$ 的圆形轨道 $\mathrm{I}$ 上做匀速圆周运动, 运行的周期为 $T$, 在到达轨道的 $P$ 点时变轨进入椭圆轨道 II, 到达轨道 II 的“近水星点” $Q$ 时,再次变轨进入近水星轨道III绕水星做匀速圆周运动, 从而实施对水星探测的任务, 则下列说法正确的是（） [图1] A: 水星探测器在 $P 、 Q$ 两点变轨的过程中速度均减小 B: 水星探测器在轨道 II 上运行的周期小于 $T$ C: 水星探测器在轨道 I 和轨道 II 上稳定运行经过 $P$ 时加速度大小不相等 D: 若水星探测器在轨道 II 上经过 $P$ 点时的速度大小为 $\mathrm{v}_{\mathrm{P}}$, 在轨道III上做圆周运动的速度大小为 $v_{3}$, 则有 $v_{3}>\mathrm{v}_{\mathrm{P}}$

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: 如图所示为科学家模拟水星探测器进入水星表面绕行轨道的过程示意图, 假设水星的半径为 $R$, 探测器在距离水星表面高度为 $3 R$ 的圆形轨道 $\mathrm{I}$ 上做匀速圆周运动, 运行的周期为 $T$, 在到达轨道的 $P$ 点时变轨进入椭圆轨道 II, 到达轨道 II 的“近水星点” $Q$ 时,再次变轨进入近水星轨道III绕水星做匀速圆周运动, 从而实施对水星探测的任务, 则下列说法正确的是（） [图1] A: 水星探测器在 $P 、 Q$ 两点变轨的过程中速度均减小 B: 水星探测器在轨道 II 上运行的周期小于 $T$ C: 水星探测器在轨道 I 和轨道 II 上稳定运行经过 $P$ 时加速度大小不相等 D: 若水星探测器在轨道 II 上经过 $P$ 点时的速度大小为 $\mathrm{v}_{\mathrm{P}}$, 在轨道III上做圆周运动的速度大小为 $v_{3}$, 则有 $v_{3}>\mathrm{v}_{\mathrm{P}}$ 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_682

如图所示, 质量相等、周期均为 $T$ 的两颗人造地球卫星, 1 轨卫星道为圆、 2 轨道为椭圆。 $A 、 B$ 两点是椭圆长轴两端, $A$ 距离地心为 $r \circ C$ 为椭圆短轴端点且是两轨道的交点, 到地心距离为 $2 r$, 卫星 1 的速率为 $v$, 下列说法正确的是 ( ) [图1] A: $C$ 点到椭圆中心的距离为 $r$ B: 卫星 2 在 $C$ 点的速率等于 $v$ C: 卫星 2 在 $C$ 点的向心加速度等于 $\frac{v^{2}}{2 r}$ D: 卫星 2 由 $A$ 到 $C$ 的时间等于 $\frac{T}{4}$

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: 如图所示, 质量相等、周期均为 $T$ 的两颗人造地球卫星, 1 轨卫星道为圆、 2 轨道为椭圆。 $A 、 B$ 两点是椭圆长轴两端, $A$ 距离地心为 $r \circ C$ 为椭圆短轴端点且是两轨道的交点, 到地心距离为 $2 r$, 卫星 1 的速率为 $v$, 下列说法正确的是 ( ) [图1] A: $C$ 点到椭圆中心的距离为 $r$ B: 卫星 2 在 $C$ 点的速率等于 $v$ C: 卫星 2 在 $C$ 点的向心加速度等于 $\frac{v^{2}}{2 r}$ D: 卫星 2 由 $A$ 到 $C$ 的时间等于 $\frac{T}{4}$ 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]

SC

Astronomy

ZH

multi-modal

Astronomy_998

Two light sources, $\mathrm{A}$ and $\mathrm{B}$, emitting their light isotropically (i.e. equally in all directions) are placed at distance $r$ and $2 r$ respectively from a detector, which shows they have the same apparent brightness (i.e. $b_{A} / b_{B}=1$ ). If $\mathrm{A}$ is moved to $2 r$ and $\mathrm{B}$ is moved to $3 r$, what is the new ratio of apparent brightness $b_{A}^{\prime} / b_{B}^{\prime}$ ? A: $2 / 3$ B: $4 / 9$ C: $3 / 4$ D: $9 / 16$

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 light sources, $\mathrm{A}$ and $\mathrm{B}$, emitting their light isotropically (i.e. equally in all directions) are placed at distance $r$ and $2 r$ respectively from a detector, which shows they have the same apparent brightness (i.e. $b_{A} / b_{B}=1$ ). If $\mathrm{A}$ is moved to $2 r$ and $\mathrm{B}$ is moved to $3 r$, what is the new ratio of apparent brightness $b_{A}^{\prime} / b_{B}^{\prime}$ ? A: $2 / 3$ B: $4 / 9$ C: $3 / 4$ D: $9 / 16$ 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_296

北斗卫星导航系统空间段计划由 35 颗卫星组成,包括 5 颗静止同步轨道卫星和 3 颗倾斜同步轨道卫星, 以及 27 颗相同高度的中轨道卫星. 中轨道卫星运行在 3 个轨道面上, 轨道面之间相隔 $120^{\circ}$ 均匀分布, 如图所示. 已知同步轨道、中轨道、倾斜同步轨道卫星距地面的高度分别约为 $6 R 、 4 R 、 6 R$ ( $R$ 为地球半径), 则 [图1] A: 静止同步轨道卫星和倾斜同步轨道卫星的周期不同 B: 3 个轨道面上的中轨道卫星角速度的值均相同 C: 同步轨道卫星与中轨道卫星周期的比值约为 $\frac{7}{5} \sqrt{\frac{7}{5}}$ D: 倾斜同步轨道卫星与中轨道卫星角速度的比值约为 $\frac{5}{7}$

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: 北斗卫星导航系统空间段计划由 35 颗卫星组成,包括 5 颗静止同步轨道卫星和 3 颗倾斜同步轨道卫星, 以及 27 颗相同高度的中轨道卫星. 中轨道卫星运行在 3 个轨道面上, 轨道面之间相隔 $120^{\circ}$ 均匀分布, 如图所示. 已知同步轨道、中轨道、倾斜同步轨道卫星距地面的高度分别约为 $6 R 、 4 R 、 6 R$ ( $R$ 为地球半径), 则 [图1] A: 静止同步轨道卫星和倾斜同步轨道卫星的周期不同 B: 3 个轨道面上的中轨道卫星角速度的值均相同 C: 同步轨道卫星与中轨道卫星周期的比值约为 $\frac{7}{5} \sqrt{\frac{7}{5}}$ D: 倾斜同步轨道卫星与中轨道卫星角速度的比值约为 $\frac{5}{7}$ 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_926

Jupiter's apparent magnitude at opposition in 2022 of $m=-2.94$ is the brightest for $\sim 70$ years due to it happening close to Jupiter's perihelion. What is the difference in apparent magnitude between Jupiter's brightest and faintest possible oppositions? Jupiter has a semi-major axis of 5.20 au and an eccentricity of 0.0489 . Assume the Earth's orbit is circular. A: 0.38 B: 0.43 C: 0.48 D: 0.53

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 apparent magnitude at opposition in 2022 of $m=-2.94$ is the brightest for $\sim 70$ years due to it happening close to Jupiter's perihelion. What is the difference in apparent magnitude between Jupiter's brightest and faintest possible oppositions? Jupiter has a semi-major axis of 5.20 au and an eccentricity of 0.0489 . Assume the Earth's orbit is circular. A: 0.38 B: 0.43 C: 0.48 D: 0.53 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_1065

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 $L_{\text {Edd }}$.d. Assuming the star to be of uniform density at this limit with $\rho=m_{p} n_{e}$ and the electrons to be in thermal equilibrium with the plasma, show that the minimum mass of a star for which $T_{\text {int }}=T_{\text {quantum }}$ is $\approx 0.1 \mathrm{M}_{\odot}$.

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: 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: d. Assuming the star to be of uniform density at this limit with $\rho=m_{p} n_{e}$ and the electrons to be in thermal equilibrium with the plasma, show that the minimum mass of a star for which $T_{\text {int }}=T_{\text {quantum }}$ is $\approx 0.1 \mathrm{M}_{\odot}$. 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_977

In the UK we use the Gregorian calendar; it is a solar calendar so that a year corresponds to the time to orbit the Sun once, where 1 solar year is $\approx 365.25$ days. Several cultures use a lunar calendar, where each month is determined by the time it takes to go from New Moon to New Moon, and have a lunar year that is exactly 12 lunar months. An example of this is the Islamic calendar. Since the length of a lunar month (29.53 days) is a little shorter than the average month length in our solar calendar (see Figure 3), it means the start date of each month in the Islamic calendar is not tied to the seasons and gradually moves earlier in the solar year. [figure1] Figure 3: All the moon phases in May 2022, showing that the time measured from New Moon to New Moon (a lunar month) is shorter than a month in the Gregorian calendar. Credit: MoonConnection.com In other Islamic countries, the start of a new month is determined through direct telescopic observations of the Moon, looking for a very thin crescent. This is a very hard measurement to make and the human eye struggles to recognise the presence of a crescent until about $0.6 \%$ of the lunar disc is illuminated. Calculate how many hours this is after the end of the astronomical New Moon. Assume the lunar orbit is circular.

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: In the UK we use the Gregorian calendar; it is a solar calendar so that a year corresponds to the time to orbit the Sun once, where 1 solar year is $\approx 365.25$ days. Several cultures use a lunar calendar, where each month is determined by the time it takes to go from New Moon to New Moon, and have a lunar year that is exactly 12 lunar months. An example of this is the Islamic calendar. Since the length of a lunar month (29.53 days) is a little shorter than the average month length in our solar calendar (see Figure 3), it means the start date of each month in the Islamic calendar is not tied to the seasons and gradually moves earlier in the solar year. [figure1] Figure 3: All the moon phases in May 2022, showing that the time measured from New Moon to New Moon (a lunar month) is shorter than a month in the Gregorian calendar. Credit: MoonConnection.com In other Islamic countries, the start of a new month is determined through direct telescopic observations of the Moon, looking for a very thin crescent. This is a very hard measurement to make and the human eye struggles to recognise the presence of a crescent until about $0.6 \%$ of the lunar disc is illuminated. Calculate how many hours this is after the end of the astronomical New Moon. Assume the lunar orbit is circular. 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 hours, 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_752

The moment of inertia of a solid cylinder with radius $R$ and mass $M$ is given by .. A: $M R^{2} / 2$ B: $2 M R^{2} / 5$ C: $3 M R^{2} / 10$ D: $M R^{2} / 3$

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 moment of inertia of a solid cylinder with radius $R$ and mass $M$ is given by .. A: $M R^{2} / 2$ B: $2 M R^{2} / 5$ C: $3 M R^{2} / 10$ D: $M R^{2} / 3$ 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_227

据报道, 天文学家近日发现了一颗距地球 40 光年的“超级地球”, 命名为 “55Cancrie", 该行星绕母星 (中心天体) 运行的周期约为地球绕太阳运行周期的 $\frac{1}{480}$,母星的体积约为太阳的 60 倍. 假设母星与太阳密度相同, “ 55 Cancrie”与地球均做匀速圆周运动，则“55Cancrie”与地球的（） A: 轨道半径之比约为 $\sqrt[3]{\frac{60}{480}}$ B: 轨道半径之比约为 $\sqrt[3]{\frac{60}{480^{2}}}$ C: 向心加速度之比约为 $\sqrt[3]{60 \times 480^{2}}$ D: 向心加速度之比约为 $\sqrt[3]{60^{2} \times 480}$

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: 据报道, 天文学家近日发现了一颗距地球 40 光年的“超级地球”, 命名为 “55Cancrie", 该行星绕母星 (中心天体) 运行的周期约为地球绕太阳运行周期的 $\frac{1}{480}$,母星的体积约为太阳的 60 倍. 假设母星与太阳密度相同, “ 55 Cancrie”与地球均做匀速圆周运动，则“55Cancrie”与地球的（） A: 轨道半径之比约为 $\sqrt[3]{\frac{60}{480}}$ B: 轨道半径之比约为 $\sqrt[3]{\frac{60}{480^{2}}}$ C: 向心加速度之比约为 $\sqrt[3]{60 \times 480^{2}}$ D: 向心加速度之比约为 $\sqrt[3]{60^{2} \times 480}$ 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_69

地球赤道上有一物体随地球的自转而做圆周运动, 所受的向心力为 $F_{1}$, 向心加速度为 $a_{1}$, 线速度为 $v_{1}$, 角速度为 $\omega_{1}$, 绕地球表面附近做圆周运动的人造卫星（高度忽略）所受的向心力为 $F_{2}$, 向心加速度为 $a_{2}$, 线速度为 $v_{2}$, 角速度为 $\omega_{2}$; 地球同步卫星所受的向心力为 $F_{3}$, 向心加速度为 $a_{3}$, 线速度为 $v_{3}$, 角速度为 $\omega_{3}$; 地球表面重力加速度为 $g$,第一宇宙速度为 $v$, 假设三者质量相等, 则 ( ) A: $F_{2}>F_{1}>F_{3}$ B: $\omega_{1}=\omega_{3}<\omega_{2}$ C: $v_{1}=v_{2}=v>v_{3}$ D: $a_{1}>a_{2}=g>a_{3}$

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个单选题（只有一个正确答案）。 问题: 地球赤道上有一物体随地球的自转而做圆周运动, 所受的向心力为 $F_{1}$, 向心加速度为 $a_{1}$, 线速度为 $v_{1}$, 角速度为 $\omega_{1}$, 绕地球表面附近做圆周运动的人造卫星（高度忽略）所受的向心力为 $F_{2}$, 向心加速度为 $a_{2}$, 线速度为 $v_{2}$, 角速度为 $\omega_{2}$; 地球同步卫星所受的向心力为 $F_{3}$, 向心加速度为 $a_{3}$, 线速度为 $v_{3}$, 角速度为 $\omega_{3}$; 地球表面重力加速度为 $g$,第一宇宙速度为 $v$, 假设三者质量相等, 则 ( ) A: $F_{2}>F_{1}>F_{3}$ B: $\omega_{1}=\omega_{3}<\omega_{2}$ C: $v_{1}=v_{2}=v>v_{3}$ D: $a_{1}>a_{2}=g>a_{3}$ 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应为以下选项之一：[A, B, C, D]

SC

Astronomy

ZH

text-only

Astronomy_783

As a star collapses at the end of its life, the triple-alpha reaction takes place. Which one of these equations describes this reaction correctly? A: ${ }_{1}^{2} \mathrm{H}+{ }_{1}^{2} \mathrm{H}+{ }_{1}^{2} \mathrm{H} \rightarrow{ }_{3}^{6} \mathrm{Li}+\gamma$ B: ${ }_{2}^{4} \mathrm{He}+{ }_{2}^{4} \mathrm{He}+{ }_{2}^{4} \mathrm{He} \rightarrow{ }_{6}^{12} \mathrm{C}+\gamma$ C: ${ }_{1}^{2} \mathrm{H}+{ }_{1}^{2} \mathrm{H}+{ }_{2}^{4} \mathrm{He}+{ }_{2}^{4} \mathrm{He} \rightarrow{ }_{6}^{12} \mathrm{C}+\gamma$ D: ${ }_{1}^{2} \mathrm{H}+{ }_{2}^{4} \mathrm{He} \rightarrow{ }_{3}^{6} \mathrm{Li}+\gamma$

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: As a star collapses at the end of its life, the triple-alpha reaction takes place. Which one of these equations describes this reaction correctly? A: ${ }_{1}^{2} \mathrm{H}+{ }_{1}^{2} \mathrm{H}+{ }_{1}^{2} \mathrm{H} \rightarrow{ }_{3}^{6} \mathrm{Li}+\gamma$ B: ${ }_{2}^{4} \mathrm{He}+{ }_{2}^{4} \mathrm{He}+{ }_{2}^{4} \mathrm{He} \rightarrow{ }_{6}^{12} \mathrm{C}+\gamma$ C: ${ }_{1}^{2} \mathrm{H}+{ }_{1}^{2} \mathrm{H}+{ }_{2}^{4} \mathrm{He}+{ }_{2}^{4} \mathrm{He} \rightarrow{ }_{6}^{12} \mathrm{C}+\gamma$ D: ${ }_{1}^{2} \mathrm{H}+{ }_{2}^{4} \mathrm{He} \rightarrow{ }_{3}^{6} \mathrm{Li}+\gamma$ 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_572

人类对来知事物的好奇和科学家们的不懈努力, 使人类对宇宙的认识越来越丰富。 开普勒坚信哥白尼的“日心说”, 在研究了导师第谷在 20 余年中坚持对天体进行系统观测得到的大量精确资料后, 得出了开普勒三定律。为人们解决行星运动问题提供了依据, 也为牛顿发现万有引力定律提供了基础。开普勒认为, 所有行星围绕太阳运动的轨道都是椭圆, 太阳处在所有粗圆的一个焦点上、行星轨道半长轴的三次方与其公转周期的二次方的比值是一个常量。实际上行星的轨道与圆十分接近, 在中学阶段的研究中我们按圆轨道处理, 请你以地球绕太阳公转为例, 若太阳的质量为 $M$, 引力常量为 $G$ 。根据万有引力定律和牛牛顿运动定律推导出此常量的表达式;

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: 人类对来知事物的好奇和科学家们的不懈努力, 使人类对宇宙的认识越来越丰富。 开普勒坚信哥白尼的“日心说”, 在研究了导师第谷在 20 余年中坚持对天体进行系统观测得到的大量精确资料后, 得出了开普勒三定律。为人们解决行星运动问题提供了依据, 也为牛顿发现万有引力定律提供了基础。开普勒认为, 所有行星围绕太阳运动的轨道都是椭圆, 太阳处在所有粗圆的一个焦点上、行星轨道半长轴的三次方与其公转周期的二次方的比值是一个常量。实际上行星的轨道与圆十分接近, 在中学阶段的研究中我们按圆轨道处理, 请你以地球绕太阳公转为例, 若太阳的质量为 $M$, 引力常量为 $G$ 。根据万有引力定律和牛牛顿运动定律推导出此常量的表达式; 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

text-only

Astronomy_311

预计于 2022 年建成的中国空间站将成为中国空间和新技术研究的重要基地。假设空间站中的宇航员利用电热器对食品加热, 电热器的加热线圈可以简化成如图甲所示的圆形闭合线圈, 其匝数为 $n$, 半径为 $r_{1}$, 总电阻为 $R_{0}$ 。将此线圈垂直放在圆形磁场中,且保证两圆心重合, 圆形磁场的半径为 $r_{2}\left(r_{2}>r_{1}\right)$, 磁感应强度大小 $B$ 随时间 $t$ 的变化关系如图乙所示。求: 在 $t=1.5 t_{0}$ 时线圈中产生感应电流的大小; [图1] 图甲 [图2] 图乙

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: 预计于 2022 年建成的中国空间站将成为中国空间和新技术研究的重要基地。假设空间站中的宇航员利用电热器对食品加热, 电热器的加热线圈可以简化成如图甲所示的圆形闭合线圈, 其匝数为 $n$, 半径为 $r_{1}$, 总电阻为 $R_{0}$ 。将此线圈垂直放在圆形磁场中,且保证两圆心重合, 圆形磁场的半径为 $r_{2}\left(r_{2}>r_{1}\right)$, 磁感应强度大小 $B$ 随时间 $t$ 的变化关系如图乙所示。求: 在 $t=1.5 t_{0}$ 时线圈中产生感应电流的大小; [图1] 图甲 [图2] 图乙 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_1056

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.b. Given the apparent magnitude of WTS-2 in the visible is $m=16.14$ and the absolute magnitude of the Sun in the same part of the EM spectrum is $\mathcal{M}_{\odot}=4.83$ : i. Calculate the luminosity of the star, $L_{*}$. Give your answer in units of $L_{\odot}$.

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: b. Given the apparent magnitude of WTS-2 in the visible is $m=16.14$ and the absolute magnitude of the Sun in the same part of the EM spectrum is $\mathcal{M}_{\odot}=4.83$ : i. Calculate the luminosity of the star, $L_{*}$. Give your answer in units of $L_{\odot}$. 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_853

Deneb is a very important star in the Northern hemisphere as it is one of the three stars in the Summer Triangle. Deneb ( $\alpha \mathrm{Cyg})$ is also the brightest star in the Cygnus constellation. Knowing the following information calculate the distance between Deneb and Albireo ( $\beta \mathrm{Cyg}$ ). | | Deneb | Albireo | | :---: | :---: | :---: | | Parallax $(\pi)$ | 2.29 mas | 7.51 mas | | Declination $(\delta)$ | $45^{\circ} 17^{\prime}$ | $27^{\circ} 57^{\prime}$ | | Right ascension $(\alpha)$ | $20 \mathrm{~h} 41 \mathrm{~min}$ | $19 \mathrm{~h} 31 \mathrm{~min}$ | A: $569 \mathrm{pc}$ B: $102 \mathrm{pc}$ C: $432 \mathrm{pc}$ D: $317 \mathrm{pc}$ E: $459 \mathrm{pc}$

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: Deneb is a very important star in the Northern hemisphere as it is one of the three stars in the Summer Triangle. Deneb ( $\alpha \mathrm{Cyg})$ is also the brightest star in the Cygnus constellation. Knowing the following information calculate the distance between Deneb and Albireo ( $\beta \mathrm{Cyg}$ ). | | Deneb | Albireo | | :---: | :---: | :---: | | Parallax $(\pi)$ | 2.29 mas | 7.51 mas | | Declination $(\delta)$ | $45^{\circ} 17^{\prime}$ | $27^{\circ} 57^{\prime}$ | | Right ascension $(\alpha)$ | $20 \mathrm{~h} 41 \mathrm{~min}$ | $19 \mathrm{~h} 31 \mathrm{~min}$ | A: $569 \mathrm{pc}$ B: $102 \mathrm{pc}$ C: $432 \mathrm{pc}$ D: $317 \mathrm{pc}$ E: $459 \mathrm{pc}$ 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_552

如图所示, 两个星球 $\mathrm{A}$ 和 $\mathrm{B}$ 在引力作用下都绕 $O$ 点做匀速圆周运动, 星球 $\mathrm{A}$ 和 $\mathrm{B}$的距离为 $L$ 。已知 $\mathrm{A} 、 \mathrm{~B}$ 和 $O$ 点三点始终共线, $\mathrm{A}$ 和 $\mathrm{B}$ 分别在 $O$ 点的两侧, 引力常量为 $G$, 星球 $\mathrm{A}$ 的质量为 $m_{A}$, 星球 $\mathrm{A}$ 的轨道半径为 $r_{A}$, 两星球均可视为质点。则 ( ) [图1] A: 星球 $\mathrm{B}$ 的质量为 $\frac{m_{A} r_{A}}{L}$ B: 星球 B 的质量为 $\frac{m_{A} r_{A}}{L-r_{A}}$ C: 两星球做圆周运动的周期为 $2 \pi L \sqrt{\frac{r_{A}}{G m_{A}}}$ D: 两星球做圆周运动的周期为 $2 \pi L \sqrt{\frac{\left(L-r_{A}\right)}{G m_{A}}}$

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: 如图所示, 两个星球 $\mathrm{A}$ 和 $\mathrm{B}$ 在引力作用下都绕 $O$ 点做匀速圆周运动, 星球 $\mathrm{A}$ 和 $\mathrm{B}$的距离为 $L$ 。已知 $\mathrm{A} 、 \mathrm{~B}$ 和 $O$ 点三点始终共线, $\mathrm{A}$ 和 $\mathrm{B}$ 分别在 $O$ 点的两侧, 引力常量为 $G$, 星球 $\mathrm{A}$ 的质量为 $m_{A}$, 星球 $\mathrm{A}$ 的轨道半径为 $r_{A}$, 两星球均可视为质点。则 ( ) [图1] A: 星球 $\mathrm{B}$ 的质量为 $\frac{m_{A} r_{A}}{L}$ B: 星球 B 的质量为 $\frac{m_{A} r_{A}}{L-r_{A}}$ C: 两星球做圆周运动的周期为 $2 \pi L \sqrt{\frac{r_{A}}{G m_{A}}}$ D: 两星球做圆周运动的周期为 $2 \pi L \sqrt{\frac{\left(L-r_{A}\right)}{G m_{A}}}$ 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]

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multi-modal

Astronomy_165

国际科研团队发现了两颗距离地球仅 100 光年的新行星, 其中一颗可能适合生命生存。这两颗行星分别是 LP890-9b（以下简称行星 A) 和 LP890-9c (以下简称行星 B)。行星 A 的半径约为 8370 公里, 仅需 2.7 天就能绕恒星 C一圈; 行星 B 半径约为 8690 公里, 8.5 天能绕恒星 $\mathrm{C}$ 一圈, 行星 $\mathrm{B}$ 到恒星 $\mathrm{C}$ 的距离约为水星与太阳间距离的 0.1 倍,水星的公转周期约为 88 天。假设行星 $\mathrm{A} 、 \mathrm{~B}$ 绕恒星 $\mathrm{C}$ 做匀速圆周运动。则 ( ) A: 行星 A 表面的重力加速度大于行星 B 表面的重力加速度 B: 行星 A 的公转轨道半径大于行星 B 的公转轨道半径 C: 太阳的质量大于恒星 $\mathrm{C}$ 的质量 D: 水星的公转速度大于行星 B 的公转速度

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个单选题（只有一个正确答案）。 问题: 国际科研团队发现了两颗距离地球仅 100 光年的新行星, 其中一颗可能适合生命生存。这两颗行星分别是 LP890-9b（以下简称行星 A) 和 LP890-9c (以下简称行星 B)。行星 A 的半径约为 8370 公里, 仅需 2.7 天就能绕恒星 C一圈; 行星 B 半径约为 8690 公里, 8.5 天能绕恒星 $\mathrm{C}$ 一圈, 行星 $\mathrm{B}$ 到恒星 $\mathrm{C}$ 的距离约为水星与太阳间距离的 0.1 倍,水星的公转周期约为 88 天。假设行星 $\mathrm{A} 、 \mathrm{~B}$ 绕恒星 $\mathrm{C}$ 做匀速圆周运动。则 ( ) A: 行星 A 表面的重力加速度大于行星 B 表面的重力加速度 B: 行星 A 的公转轨道半径大于行星 B 的公转轨道半径 C: 太阳的质量大于恒星 $\mathrm{C}$ 的质量 D: 水星的公转速度大于行星 B 的公转速度 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]

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Astronomy_310

如图所示是“天宫二号”空间实验室轨道控制时在近地点高度 $(\mathrm{Q}$ 点)200 千米、远地点高度(P 点) 394 千米的椭圆轨道运行, 已知地球半径取 $6400 \mathrm{~km}, \mathrm{M} 、 \mathrm{~N}$ 为短轴与粗圆轨道的交点, 对于“天宫二号” 空间实验室在椭圆轨道上的运行, 下列说法正确的是 [图1] A: “天宫二号”空间实验室在 $P$ 点时的加速度一定比 $\mathrm{Q}$ 点小, 速度可能比 $\mathrm{Q}$ 点大 B: “天宫二号”空间实验室从 $N$ 点经 $P$ 点运动到 $M$ 点的时间可能小于“天宫二号”空间实验室从 $M$ 点经 $\mathrm{Q}$ 点运动到 $\mathrm{N}$ 点的时间 C: “天宫二号”空间实验室在远地点 (P 点) 所受地球的万有引力大约是在近地点 (Q 点) 的 $\frac{1}{4}$ D: “天宫二号”空间实验室从 $\mathrm{P}$ 点经 $\mathrm{M}$ 点运动到 $\mathrm{Q}$ 点的过程中万有引力做正功,从 $\mathrm{Q}$ 点经 $\mathrm{N}$ 点运动到 $\mathrm{P}$ 点的过程中要克服万有引力做功

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: 如图所示是“天宫二号”空间实验室轨道控制时在近地点高度 $(\mathrm{Q}$ 点)200 千米、远地点高度(P 点) 394 千米的椭圆轨道运行, 已知地球半径取 $6400 \mathrm{~km}, \mathrm{M} 、 \mathrm{~N}$ 为短轴与粗圆轨道的交点, 对于“天宫二号” 空间实验室在椭圆轨道上的运行, 下列说法正确的是 [图1] A: “天宫二号”空间实验室在 $P$ 点时的加速度一定比 $\mathrm{Q}$ 点小, 速度可能比 $\mathrm{Q}$ 点大 B: “天宫二号”空间实验室从 $N$ 点经 $P$ 点运动到 $M$ 点的时间可能小于“天宫二号”空间实验室从 $M$ 点经 $\mathrm{Q}$ 点运动到 $\mathrm{N}$ 点的时间 C: “天宫二号”空间实验室在远地点 (P 点) 所受地球的万有引力大约是在近地点 (Q 点) 的 $\frac{1}{4}$ D: “天宫二号”空间实验室从 $\mathrm{P}$ 点经 $\mathrm{M}$ 点运动到 $\mathrm{Q}$ 点的过程中万有引力做正功,从 $\mathrm{Q}$ 点经 $\mathrm{N}$ 点运动到 $\mathrm{P}$ 点的过程中要克服万有引力做功 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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multi-modal

Astronomy_55

人类太空探测计划旨在探测恒星亮度以寻找适合人类居住的宜居行星。在某次探测中发现距地球数光年处有一颗相对太阳静止的质量为 $M$ 的恒星 $\mathrm{A}$, 将恒星 $\mathrm{A}$ 视为黑体,根据斯特藩-玻尔兹曼定律: 一个黑体表面单位面积辐射出的功率与黑体本身的热力学温度 $T$ 的四次方成正比, 即黑体表面单位面积辐射出的功率为 $\sigma T^{4}$ (其中 $\sigma$ 为常数), A 的表面温度为 $T_{0}$, 地球上正对 $\mathrm{A}$ 的单位面积接收到 $\mathrm{A}$ 辐射出的功率为 $I$ 。已知 $\mathrm{A}$ 在地球轨道平面上，地球公转半径为 $R_{0}$, 一年内地球上的观测者测得地球与 $\mathrm{A}$ 的连线之间的最大夹角为 $\theta$ (角 $\theta$ 很小, 可认为 $\sin \theta \approx \tan \theta \approx \theta$ )。恒星 $\mathrm{A}$ 有一颗绕它做匀速圆周运动的行星 $\mathrm{B}$, 该行星也可视为黑体, 其表面的温度保持为 $T_{1}$, 恒星 $\mathrm{A}$ 射向行星 $\mathrm{B}$ 的光可看作平行光。已知引力常量为 $G$, 求: 恒星 $\mathrm{A}$ 的半径 $R_{\mathrm{A}}$;

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: 人类太空探测计划旨在探测恒星亮度以寻找适合人类居住的宜居行星。在某次探测中发现距地球数光年处有一颗相对太阳静止的质量为 $M$ 的恒星 $\mathrm{A}$, 将恒星 $\mathrm{A}$ 视为黑体,根据斯特藩-玻尔兹曼定律: 一个黑体表面单位面积辐射出的功率与黑体本身的热力学温度 $T$ 的四次方成正比, 即黑体表面单位面积辐射出的功率为 $\sigma T^{4}$ (其中 $\sigma$ 为常数), A 的表面温度为 $T_{0}$, 地球上正对 $\mathrm{A}$ 的单位面积接收到 $\mathrm{A}$ 辐射出的功率为 $I$ 。已知 $\mathrm{A}$ 在地球轨道平面上，地球公转半径为 $R_{0}$, 一年内地球上的观测者测得地球与 $\mathrm{A}$ 的连线之间的最大夹角为 $\theta$ (角 $\theta$ 很小, 可认为 $\sin \theta \approx \tan \theta \approx \theta$ )。恒星 $\mathrm{A}$ 有一颗绕它做匀速圆周运动的行星 $\mathrm{B}$, 该行星也可视为黑体, 其表面的温度保持为 $T_{1}$, 恒星 $\mathrm{A}$ 射向行星 $\mathrm{B}$ 的光可看作平行光。已知引力常量为 $G$, 求: 恒星 $\mathrm{A}$ 的半径 $R_{\mathrm{A}}$; 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_1156

A comet follows an elliptical orbit that is $31.5 \mathrm{AU}$ at aphelion and $0.5 \mathrm{AU}$ at perihelion. What is the period of the comet? A: 181 years B: 16 years C: 64 years D: 6.3 years

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 follows an elliptical orbit that is $31.5 \mathrm{AU}$ at aphelion and $0.5 \mathrm{AU}$ at perihelion. What is the period of the comet? A: 181 years B: 16 years C: 64 years D: 6.3 years 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_407

55. 已知地球半径为 $6400 \mathrm{~km}$, 可能用到数学近似算法 $(1+x) k \approx 1+k x,(x<<1)$, 以下说法正确的是（ ） A: 若地球质量减少 $2 \%$, 则地球公转要加快约 $1 \%$ B: 已知太阳半径和地球绕太阳公转的周期, 可估算出太阳质量 C: 离地高 $3.2 \mathrm{~km}$ 处的重力加速度比地面处重力加速度少约 $0.1 \%$ D: 月球上的重力加速度是地球上的 $\frac{1}{6}$, 可知月球质量是地球质量的 $\frac{1}{6}$

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个单选题（只有一个正确答案）。 问题: 55. 已知地球半径为 $6400 \mathrm{~km}$, 可能用到数学近似算法 $(1+x) k \approx 1+k x,(x<<1)$, 以下说法正确的是（ ） A: 若地球质量减少 $2 \%$, 则地球公转要加快约 $1 \%$ B: 已知太阳半径和地球绕太阳公转的周期, 可估算出太阳质量 C: 离地高 $3.2 \mathrm{~km}$ 处的重力加速度比地面处重力加速度少约 $0.1 \%$ D: 月球上的重力加速度是地球上的 $\frac{1}{6}$, 可知月球质量是地球质量的 $\frac{1}{6}$ 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应为以下选项之一：[A, B, C, D]

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Astronomy_726

如图所示。甲、乙为地球赤道面内围绕地球运转的通讯卫星。已知甲是与地面相对静止的同步卫星; 乙的运转方向与地球自转方向相反, 轨道半径为地球半径的 2 倍, 周期为 $T$, 在地球赤道上的 $P$ 点有一位观测者，观测者始终相对于地面静止。若地球半径为 $R$, 地球的自转周期为 $T_{0}$ 。求: 若某一时刻卫星乙刚好在 $P$ 点的观察者的正上方, 则至少再经多长时间卫星乙会再次经过该观察者的正上方? [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。甲、乙为地球赤道面内围绕地球运转的通讯卫星。已知甲是与地面相对静止的同步卫星; 乙的运转方向与地球自转方向相反, 轨道半径为地球半径的 2 倍, 周期为 $T$, 在地球赤道上的 $P$ 点有一位观测者，观测者始终相对于地面静止。若地球半径为 $R$, 地球的自转周期为 $T_{0}$ 。求: 若某一时刻卫星乙刚好在 $P$ 点的观察者的正上方, 则至少再经多长时间卫星乙会再次经过该观察者的正上方? [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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multi-modal

Astronomy_156

地球可视为质量均匀分布的球体。用弹簧科测某物体重力大小, 在北极点称时读数为 $F_{1}$, 在赤道上称时读数为 $F_{2}$; 地球自转周期为 $T$, 万有引力常量为 $G$ 。则地球密度的表达式为（ ） A: $\frac{3 \pi F_{1}}{G T^{2}\left(F_{1}-F_{2}\right)}$ B: $\frac{3 \pi\left(F_{1}-F_{2}\right)}{G T^{2} F_{1}}$ C: $\frac{3 \pi}{G T^{2}}$ D: $\frac{3 \pi F_{1}}{G T^{2} F_{2}}$

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: 地球可视为质量均匀分布的球体。用弹簧科测某物体重力大小, 在北极点称时读数为 $F_{1}$, 在赤道上称时读数为 $F_{2}$; 地球自转周期为 $T$, 万有引力常量为 $G$ 。则地球密度的表达式为（ ） A: $\frac{3 \pi F_{1}}{G T^{2}\left(F_{1}-F_{2}\right)}$ B: $\frac{3 \pi\left(F_{1}-F_{2}\right)}{G T^{2} F_{1}}$ C: $\frac{3 \pi}{G T^{2}}$ D: $\frac{3 \pi F_{1}}{G T^{2} F_{2}}$ 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]

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Astronomy_684

质量为 $m$ 的宇宙飞船, 在离月球地面高度 $h$ 处沿圆形轨道绕月球运行。为使飞船到达月球表面 $B$ 点, 喷气发动机在 $A$ 点做一次极短时间的喷气。从喷口射出的气流方向与圆周轨道相切且相对飞船的速度为 $u$, 月球半径为 $R, h=\frac{R}{16}, A 、 B$ 两点与球心在一直线上, 其速度与飞船到球心的距离成反比。月球表面重力加速度为 $g$ 。若以无穷远处为飞船引力势能的零势能点, 飞船和球心距离为 $r$ 时, 引力势能的表达式为 $E_{\mathrm{P}}=-\frac{G M m}{r}$ 。 ( $M$ 是月球质量, 本题中未知)。求: 需要喷出气流的质量 $\Delta m$ 。 [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: 质量为 $m$ 的宇宙飞船, 在离月球地面高度 $h$ 处沿圆形轨道绕月球运行。为使飞船到达月球表面 $B$ 点, 喷气发动机在 $A$ 点做一次极短时间的喷气。从喷口射出的气流方向与圆周轨道相切且相对飞船的速度为 $u$, 月球半径为 $R, h=\frac{R}{16}, A 、 B$ 两点与球心在一直线上, 其速度与飞船到球心的距离成反比。月球表面重力加速度为 $g$ 。若以无穷远处为飞船引力势能的零势能点, 飞船和球心距离为 $r$ 时, 引力势能的表达式为 $E_{\mathrm{P}}=-\frac{G M m}{r}$ 。 ( $M$ 是月球质量, 本题中未知)。求: 需要喷出气流的质量 $\Delta m$ 。 [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

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multi-modal

Astronomy_980

A consequence of the expansion of the Universe is that galaxies appear to be moving away from us, and the further they are away the quicker they seem to be moving. Edwin Hubble showed that if the expansion was uniform in all directions then the relationship can be expressed as $v=H_{0} d$ where $v$ is the recessional velocity, $d$ is the distance to the galaxy, and $H_{0}$ is called the Hubble constant. His original compilation of this data from his 1929 paper is shown in Figure 2. [figure1] Figure 2: The original plot of recessional velocity (based upon redshifts) given in $\mathrm{km} \mathrm{s}^{-1}$ (despite the incorrect axes labels) and distances given in $\mathrm{Mpc}\left(=10^{6} \mathrm{pc}\right.$ ) to 32 nearby galaxies. The solid line shows the best fit to the individual data points. Credit: Hubble (1929). Whilst the measured redshifts used to derive the values of $v$ are largely consistent with the modern values, the distances are considerably different from the ones we accept today. Even so, we can repeat the analysis he did to get a rough value for the age of the Universe, despite it being rather different from our current estimates. The age of the Universe, $t$, is related to $H_{0}$ by $t=k H_{0}^{n}$ where $k$ is a unitless numerical factor. By considering the dimensions of $H_{0}$ in SI units, find a value for $n$ and, given $k=1$, calculate $t$ from your value of $H_{0}$. Give your answer in $\operatorname{Gyr}\left(=10^{9}\right.$ years).

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: A consequence of the expansion of the Universe is that galaxies appear to be moving away from us, and the further they are away the quicker they seem to be moving. Edwin Hubble showed that if the expansion was uniform in all directions then the relationship can be expressed as $v=H_{0} d$ where $v$ is the recessional velocity, $d$ is the distance to the galaxy, and $H_{0}$ is called the Hubble constant. His original compilation of this data from his 1929 paper is shown in Figure 2. [figure1] Figure 2: The original plot of recessional velocity (based upon redshifts) given in $\mathrm{km} \mathrm{s}^{-1}$ (despite the incorrect axes labels) and distances given in $\mathrm{Mpc}\left(=10^{6} \mathrm{pc}\right.$ ) to 32 nearby galaxies. The solid line shows the best fit to the individual data points. Credit: Hubble (1929). Whilst the measured redshifts used to derive the values of $v$ are largely consistent with the modern values, the distances are considerably different from the ones we accept today. Even so, we can repeat the analysis he did to get a rough value for the age of the Universe, despite it being rather different from our current estimates. The age of the Universe, $t$, is related to $H_{0}$ by $t=k H_{0}^{n}$ where $k$ is a unitless numerical factor. By considering the dimensions of $H_{0}$ in SI units, find a value for $n$ and, given $k=1$, calculate $t$ from your value of $H_{0}$. Give your answer in $\operatorname{Gyr}\left(=10^{9}\right.$ years). 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 Gyr, 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_491

地球的两个卫星 $P 、 Q$ 绕地球做匀速圆周运动, $P$ 的运行周期大于 $Q$ 的运行周 期. 设卫星与地球中心的连线在单位时间内扫过的面积为 $S$, 下列图像中能大致描述 $S$ 与两卫星的线速度 $v$ 之间关系的是 ( ) A: [图1] B: [图2] C: [图3] D: [图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: 地球的两个卫星 $P 、 Q$ 绕地球做匀速圆周运动, $P$ 的运行周期大于 $Q$ 的运行周 期. 设卫星与地球中心的连线在单位时间内扫过的面积为 $S$, 下列图像中能大致描述 $S$ 与两卫星的线速度 $v$ 之间关系的是 ( ) A: [图1] B: [图2] C: [图3] D: [图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]

SC

Astronomy

ZH

multi-modal

Astronomy_598

中国火星探测器 “天问一号”已于 2021 年春节期间抵达火星轨道, 随后将择机着陆火星对火星进行科学探测。现将探测器抵达火星轨道的过程, 简化成如图所示的三个阶段, 沿轨道 I 的地火转移轨道, 在轨道II上运行的火星停泊轨道及沿圆轨道III运行的科学探测轨道。已知三条轨道相切于 $A$ 点, 且 $A 、 B$ 两点分别为轨道II的近火点和远火点,其距离火星地面的高度分别为 $h_{1}$ 和 $h_{2}$, 火星探测器在轨道III上运行的周期为 $T$, 火星的半径为 $R$, 引力常量为 $G$, 则下列判断正确的是 ( ) [图1] A: 探测器在轨道 I、II、III上经过 $A$ 点时的速度 $v_{1} 、 v_{2} 、 v_{3}$ 的大小关系为 $v_{1}<v_{2}<v_{3}$ B: 探测器在轨道II上运行的周期为 $\sqrt{\left[\frac{h_{1}+h_{2}+2 R}{2\left(h_{1}+R\right)}\right]^{3}} T$ C: 火星表面的重力加速度为 $\frac{4 \pi^{2}\left(R+h_{1}\right)^{3}}{T^{2} R^{2}}$ D: 火星的平均密度为 $\frac{3 \pi}{G T^{2}}$

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: 中国火星探测器 “天问一号”已于 2021 年春节期间抵达火星轨道, 随后将择机着陆火星对火星进行科学探测。现将探测器抵达火星轨道的过程, 简化成如图所示的三个阶段, 沿轨道 I 的地火转移轨道, 在轨道II上运行的火星停泊轨道及沿圆轨道III运行的科学探测轨道。已知三条轨道相切于 $A$ 点, 且 $A 、 B$ 两点分别为轨道II的近火点和远火点,其距离火星地面的高度分别为 $h_{1}$ 和 $h_{2}$, 火星探测器在轨道III上运行的周期为 $T$, 火星的半径为 $R$, 引力常量为 $G$, 则下列判断正确的是 ( ) [图1] A: 探测器在轨道 I、II、III上经过 $A$ 点时的速度 $v_{1} 、 v_{2} 、 v_{3}$ 的大小关系为 $v_{1}<v_{2}<v_{3}$ B: 探测器在轨道II上运行的周期为 $\sqrt{\left[\frac{h_{1}+h_{2}+2 R}{2\left(h_{1}+R\right)}\right]^{3}} T$ C: 火星表面的重力加速度为 $\frac{4 \pi^{2}\left(R+h_{1}\right)^{3}}{T^{2} R^{2}}$ D: 火星的平均密度为 $\frac{3 \pi}{G T^{2}}$ 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_320

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 $a$ 的星球半径 $R$; [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 $a$ 的星球半径 $R$; [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_239

人造卫星在绕地球运行时, 会遇到稀薄大气的阻力。如果不进行必要的轨道维持,稀薄大气对卫星的这种微小阻力会导致卫星轨道半径逐渐减小, 以至最终落回地球。这个过程是非常漫长的, 因此卫星每一圈的运动仍可以认为是匀速圆周运动。规定两质点相距无穷远时的引力势能为零, 理论上可以得出质量分别 $m_{1} 、 m_{2}$ 的两个物体相距 $r$ 时,系统的引力势能为 $E_{p}=\frac{G m_{1} m_{2}}{r}$ 。已知人造卫星的质量为 $m$, 某时刻绕地球做匀速圆周运动的轨道半径为 $r$, 地球半径为 $R$, 地球表面附近的重力加速度为 $g$ 。 已知地球半径为 $6400 \mathrm{~km}$ 。当卫星轨道离地面的高度为 $200 \mathrm{~km}$ 时, 由于大气阻力的影响, 测得卫星每绕地球一周, 轨道高度降低 $20 \mathrm{~m}$ 。试估算在此高度大气对卫星的阻力大小 $f$ 与卫星所受地球引力大小 $F$ 的比值 $k$ (答案保留 1 位有效数字)。

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: 人造卫星在绕地球运行时, 会遇到稀薄大气的阻力。如果不进行必要的轨道维持,稀薄大气对卫星的这种微小阻力会导致卫星轨道半径逐渐减小, 以至最终落回地球。这个过程是非常漫长的, 因此卫星每一圈的运动仍可以认为是匀速圆周运动。规定两质点相距无穷远时的引力势能为零, 理论上可以得出质量分别 $m_{1} 、 m_{2}$ 的两个物体相距 $r$ 时,系统的引力势能为 $E_{p}=\frac{G m_{1} m_{2}}{r}$ 。已知人造卫星的质量为 $m$, 某时刻绕地球做匀速圆周运动的轨道半径为 $r$, 地球半径为 $R$, 地球表面附近的重力加速度为 $g$ 。 已知地球半径为 $6400 \mathrm{~km}$ 。当卫星轨道离地面的高度为 $200 \mathrm{~km}$ 时, 由于大气阻力的影响, 测得卫星每绕地球一周, 轨道高度降低 $20 \mathrm{~m}$ 。试估算在此高度大气对卫星的阻力大小 $f$ 与卫星所受地球引力大小 $F$ 的比值 $k$ (答案保留 1 位有效数字)。 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_325

宇宙飞船以周期 $T$ 绕地球做圆周运动时，由于地球遮挡阳光，会经历“日全食”过程 (宇航员看不见太阳), 如图所示. 已知地球的半径为 $R$, 地球质量为 $M$, 引力常量为 $G$,地球自转周期为 $T_{0}$, 太阳光可看作平行光, 飞船上的宇航员在 $A$ 点测出对地球的张角为 $\alpha$ ，则以下判断正确的是( ) [图1] A: 飞船绕地球运动的线速度为 $\frac{2 \pi R}{T \sin \left(\frac{\alpha}{2}\right)}$ B: 一天内飞船经历“日全食”的次数为 $\frac{T_{0}}{T}$ C: 飞船每次“日全食”过程的时间为 $\frac{\alpha T_{0}}{2 \pi} D: 飞船周期为 $T=\frac{2 \pi R}{\sin \left(\frac{\alpha}{2}\right)} \sqrt{\frac{R}{G M \sin \left(\frac{\alpha}{2}\right)}}$

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: 宇宙飞船以周期 $T$ 绕地球做圆周运动时，由于地球遮挡阳光，会经历“日全食”过程 (宇航员看不见太阳), 如图所示. 已知地球的半径为 $R$, 地球质量为 $M$, 引力常量为 $G$,地球自转周期为 $T_{0}$, 太阳光可看作平行光, 飞船上的宇航员在 $A$ 点测出对地球的张角为 $\alpha$ ，则以下判断正确的是( ) [图1] A: 飞船绕地球运动的线速度为 $\frac{2 \pi R}{T \sin \left(\frac{\alpha}{2}\right)}$ B: 一天内飞船经历“日全食”的次数为 $\frac{T_{0}}{T}$ C: 飞船每次“日全食”过程的时间为 $\frac{\alpha T_{0}}{2 \pi} D: 飞船周期为 $T=\frac{2 \pi R}{\sin \left(\frac{\alpha}{2}\right)} \sqrt{\frac{R}{G M \sin \left(\frac{\alpha}{2}\right)}}$ 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_542

如图所示是卫星绕不同行星在不同轨道上运动的 $\lg T-\lg r$ 图像, 其中 $T$ 为卫星的周期, $r$ 为卫星的轨道半径。卫星 $\mathrm{M}$ 绕行星 $P$ 运动的图线是 $a$, 卫星 $\mathrm{N}$ 绕行星 $\mathrm{Q}$ 运动的图线是 $b$, 若卫星绕行星的运动可以看成匀速圆周运动, 则 ( ) [图1] A: 直线 $a$ 的斜率与行星 $\mathrm{P}$ 质量无关 B: 行星 $\mathrm{P}$ 的质量大于行星 $\mathrm{Q}$ 的质量 C: 卫星 $\mathrm{M}$ 在 1 处的向心加速度小于在 2 处的向心加速度 D: 卫星 $\mathrm{M}$ 在 2 处的向心加速度小于卫星 $\mathrm{N}$ 在 3 处的向心加速度

你正在参加一个国际天文竞赛，并需要解决以下问题。 这是一个多选题（有多个正确答案）。 问题: 如图所示是卫星绕不同行星在不同轨道上运动的 $\lg T-\lg r$ 图像, 其中 $T$ 为卫星的周期, $r$ 为卫星的轨道半径。卫星 $\mathrm{M}$ 绕行星 $P$ 运动的图线是 $a$, 卫星 $\mathrm{N}$ 绕行星 $\mathrm{Q}$ 运动的图线是 $b$, 若卫星绕行星的运动可以看成匀速圆周运动, 则 ( ) [图1] A: 直线 $a$ 的斜率与行星 $\mathrm{P}$ 质量无关 B: 行星 $\mathrm{P}$ 的质量大于行星 $\mathrm{Q}$ 的质量 C: 卫星 $\mathrm{M}$ 在 1 处的向心加速度小于在 2 处的向心加速度 D: 卫星 $\mathrm{M}$ 在 2 处的向心加速度小于卫星 $\mathrm{N}$ 在 3 处的向心加速度 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应为两个或更多的选项：[A, B, C, D]

MC

Astronomy

ZH

multi-modal

Astronomy_901

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] Assuming that the orbit of Proxima Centauri B is circular, what is the planet's orbital velocity?

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] Assuming that the orbit of Proxima Centauri B is circular, what is the planet's orbital velocity? 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 km/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_386

如图为某双星系统 $\mathrm{A} 、 \mathrm{~B}$ 绕其连线上的 $O$ 点做匀速圆周运动的示意图, 若 $\mathrm{A}$ 星的轨道半径大于 $\mathrm{B}$ 星的轨道半径, 双星的总质量 $M$, 双星间的距离为 $L$, 其运动周期为 $T$, 则不正确的是 ( ) [图1] A: A 的加速度一定大于 $\mathrm{B}$ 的加速度 B: $L$ 一定时, $m$ 越小, $r$ 越大 C: $L$ 一定时, $\mathrm{A}$ 的质量减小 $\Delta m$ 而 $\mathrm{B}$ 的质量增加 $\Delta m$, 它们的向心力减小 D: $\mathrm{A}$ 的质量一定大于 $\mathrm{B}$ 的质量

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: 如图为某双星系统 $\mathrm{A} 、 \mathrm{~B}$ 绕其连线上的 $O$ 点做匀速圆周运动的示意图, 若 $\mathrm{A}$ 星的轨道半径大于 $\mathrm{B}$ 星的轨道半径, 双星的总质量 $M$, 双星间的距离为 $L$, 其运动周期为 $T$, 则不正确的是 ( ) [图1] A: A 的加速度一定大于 $\mathrm{B}$ 的加速度 B: $L$ 一定时, $m$ 越小, $r$ 越大 C: $L$ 一定时, $\mathrm{A}$ 的质量减小 $\Delta m$ 而 $\mathrm{B}$ 的质量增加 $\Delta m$, 它们的向心力减小 D: $\mathrm{A}$ 的质量一定大于 $\mathrm{B}$ 的质量 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