di-zhang-fdu/OlympiadExtended huggingface dataset:
Subfield: Electromagnetism

3. Block on an Incline Plane

Consider a block of mass, $\mathrm{M}$, and charge, $Q>0$, sliding down an incline plane at an angle $\alpha$ with the horizontal with initial position $\left(x_{0}, y_{0}\right)$. The system is exposed to an upward electric field given by $E(y)=\beta\left(y_{0}-y\right)$, with $\beta>0$.

<img_4536>

Figure 2: An incline plane submerged in a medium.

Question: (a) Find an expression for the $y_{e}$, the position where the block experiences net zero vertical force.

Solution:
The block experiences net vertical force when the upward force from the electric field equals the downward force from gravity,

$$
E\left(y_{e}\right) Q=M g
$$

Using our function for $\mathrm{E}$ and solving for $y_{e}$ we find, $y_{e}=y_{0}-\frac{g M}{\beta Q}$.

Final answer: $y_{e}=y_{0}-\frac{g M}{\beta Q}$
Subfield: Electromagnetism

3. Block on an Incline Plane

Consider a block of mass, $\mathrm{M}$, and charge, $Q>0$, sliding down an incline plane at an angle $\alpha$ with the horizontal with initial position $\left(x_{0}, y_{0}\right)$. The system is exposed to an upward electric field given by $E(y)=\beta\left(y_{0}-y\right)$, with $\beta>0$.

<img_4536>

Figure 2: An incline plane submerged in a medium.
Context question:
(a) Find an expression for the $y_{e}$, the position where the block experiences net zero vertical force.
Context answer:
\boxed{$y_{e}=y_{0}-\frac{g M}{\beta Q}$}


Question: (b) What is the force acting on the block when it is in contact with the incline plane?

Solution:
The force acting on the block while on the plane will be the sum of the electromagnetic force (which is dependent on y) and gravitational force projected onto the plane.

$$
F(y)=\left(M g-Q \beta\left(y_{0}-y\right)\right) \sin (\alpha)
$$

This force is directed parallel to the plane to the right.

Final answer: $F(y)=\left(M g-Q \beta\left(y_{0}-y\right)\right) \sin (\alpha)$
Subfield: Electromagnetism


Question: (c) While sliding down the incline plane, at what point will it lose contact with the plane?

Solution:
The block will lose contact with the surface at $y_{e}$. A qualitative answer is sufficient. This can be seen due to the fact that up until $y_{e}$ the block experiences a force parallel to the plane and thus stays in contact with the plane. Suddenly, as it passes $y_{e}$, it begins to experience and upward force due to the electric field surpassing gravity in its asserted force. At this point the block begins to peel away from the incline plane.

Final answer: 开放性回答
Subfield: Electromagnetism


Question: (d) What is the velocity vector of the block when it loses contact with the plane?

Solution:
To find the velocity vector we find the work done by the force vector on the block up until $y_{e}$. At $y_{e}$ we realize that the system has zero potential energy since the net force on the block is zero. We can then treat the total work as kinetic energy at that moment and calculate the magnitude of the velocity.

$$
W=\int_{0}^{y_{0}-y_{e}}\left(M g-Q \beta\left(y_{0}-y\right)\right) d y=M g\left(y_{0}-y_{e}\right)-\frac{Q \beta}{2}\left(y_{0}^{2}-y_{e}^{2}\right)
$$

Velocity is then found by, $W=K=\frac{1}{2} M v^{2}$,

$$
v=\sqrt{2 g\left(y_{0}-y_{e}\right)-\frac{Q \beta}{M}\left(y_{0}^{2}-y_{e}^{2}\right)}
$$

The velocity vector is then given by,

$$
\vec{v}=\sqrt{2 g\left(y_{0}-y_{e}\right)-\frac{Q \beta}{M}\left(y_{0}^{2}-y_{e}^{2}\right)} \cdot\langle\cos (\alpha),-\sin (\alpha)\rangle
$$

Final answer: $\vec{v}=\sqrt{2 g\left(y_{0}-y_{e}\right)-\frac{Q \beta}{M}\left(y_{0}^{2}-y_{e}^{2}\right)} \cdot\langle\cos (\alpha),-\sin (\alpha)\rangle$
Subfield: Electromagnetism

3. Block on an Incline Plane

Consider a block of mass, $\mathrm{M}$, and charge, $Q>0$, sliding down an incline plane at an angle $\alpha$ with the horizontal with initial position $\left(x_{0}, y_{0}\right)$. The system is exposed to an upward electric field given by $E(y)=\beta\left(y_{0}-y\right)$, with $\beta>0$.

<img_4536>

Figure 2: An incline plane submerged in a medium.
Context question:
(a) Find an expression for the $y_{e}$, the position where the block experiences net zero vertical force.
Context answer:
\boxed{$y_{e}=y_{0}-\frac{g M}{\beta Q}$}


Context question:
(b) What is the force acting on the block when it is in contact with the incline plane?
Context answer:
\boxed{$F(y)=\left(M g-Q \beta\left(y_{0}-y\right)\right) \sin (\alpha)$}


Context question:
(c) While sliding down the incline plane, at what point will it lose contact with the plane?
Context answer:
开放性回答


Context question:
(d) What is the velocity vector of the block when it loses contact with the plane?
Context answer:
$\vec{v}=\sqrt{2 g\left(y_{0}-y_{e}\right)-\frac{Q \beta}{M}\left(y_{0}^{2}-y_{e}^{2}\right)} \cdot\langle\cos (\alpha),-\sin (\alpha)\rangle$


Question: (e) Once the block has lost contact with the plane it begins to oscillate in the field. What frequency does it oscillate at?

Solution:
After the block has left the incline, consider a displacement around $y_{e}, \Delta y=y_{e}-$ $y$. We can write the force experienced by the block as a function of $\Delta y, F(\Delta y)=$ $\left(M g-Q \beta\left(y_{0}-y\right)\right)=\left(M g-Q \beta\left(y_{0}-y_{e}+\Delta y\right)\right)$. Using our value for $y_{e}$, we have, $F(\Delta y)=$ $\left(M g-Q \beta\left(\frac{g M}{\beta Q}+\Delta y\right)\right)$. Thus we have $F(\Delta y)=-(\beta Q) \Delta y$. Noting that $(\beta Q)$ is constant, we see that this is an example of Hookes' Law and know that the frequency must be given by,

$$
f=\frac{1}{2 \pi} \sqrt{\frac{\beta Q}{m}}
$$

Final answer: $f=\frac{1}{2 \pi} \sqrt{\frac{\beta Q}{m}}$
Subfield: Modern Physics

4. Lorentz Boost

In Newtonian kinematics, inertial frames moving relatively to each other are related by the following transformations called Galilean boosts:

$$
\begin{aligned}
x^{\prime} & =x-v t \\
t^{\prime} & =t
\end{aligned}
$$

In relativistic kinematics, inertial frames are similarly related by the Lorentz boosts:

$$
\begin{aligned}
x^{\prime} & =\frac{1}{\sqrt{1-v^{2} / c^{2}}}(x-v t) \\
t^{\prime} & =\frac{1}{\sqrt{1-v^{2} / c^{2}}}\left(t-\frac{v}{c^{2}} x\right)
\end{aligned}
$$

In this problem you will derive the Lorentz transformations from a minimal set of postulates: the homogeneity of space and time, the isotropy of space, and the principle of relativity. You will show that these assumptions about the structure of space-time imply either (a) there is a universal "speed limit" which is frame invariant, which results in the Lorentz boost, or (b) there is no universal "speed limit," which results in the Galilean boost. For simplicity, consider a one-dimensional problem only. Let two frames $F$ and $F^{\prime}$ be such that the frame $F^{\prime}$ moves at relative velocity $v$ in the positive- $x$ direction compared to frame $F$. Denote the coordinates of $F$ as $(x, t)$ and the coordinates of $F^{\prime}$ as $\left(x^{\prime}, t^{\prime}\right)$.

The most general coordinate transformations between $F$ and $F^{\prime}$ are given by functions $X, T$,

$$
\begin{aligned}
x^{\prime} & =X(x, t, v) \\
t^{\prime} & =T(x, t, v)
\end{aligned}
$$

which we will refer to as the generalized boost.

Question: (a) The homogeneity of space and time imply that the laws of physics are the same no matter where in space and time you are. In other words, they do not depend on a choice of origin for coordinates $x$ and $t$. Use this fact to show that $\frac{\partial X}{\partial x}$ is independent of the position $x$ and $\frac{\partial T}{\partial t}$ is independent of the time $t$. (Hint: Recall the definition of the partial derivative.)

Solution:
The homogeneity of space implies the following:

$$
X\left(x_{2}+h, t, v\right)-X\left(x_{2}, t, v\right)=X\left(x_{1}+h, t, v\right)-X\left(x_{1}, t, v\right)
$$



<img_4548>

Figure 3: The shaded area is the ruler's trajectory in spacetime. Any measurement of its length must intersect the shaded area.



now dividing by $h$ and sending $h \rightarrow 0$ is the partial derivative, therefore

$$
\left.\frac{\partial X}{\partial x}\right|_{x_{2}}=\left.\frac{\partial X}{\partial x}\right|_{x_{1}}
$$

The same method is repeated for the other variables.
Subfield: Electromagnetism

Computing electric fields

Electrostatics relies on multiple methods for computing electric fields and potentials. In this problem, we will explore two of them, Gauss's Law and Legendre polynomials.

Uniform charge distributions

Let us consider a hollow conducting sphere of radius $R$ charged with the electric charge $Q$, uniformly distributed on its surface. In order to calculate its potential, we can use Gauss's Law, which states that the flux of the electric field $d F=\mathbf{E} \cdot \mathbf{d} \mathbf{A}$ across a closed surface is proportional to the charge enclosed by that surface: $F=Q / \varepsilon_{0}$. We have denoted $\mathbf{d A}=d A \mathbf{n}$ the elementary oriented (towards the exterior) surface element.

Question: (a) Compute the electric potential inside and outside the sphere.

Solution:
From Gauss's theorem:

$$
\frac{Q}{\varepsilon_{0}}=4 \pi R^{2} E
$$

So:

$$
E=\frac{Q}{4 \pi \varepsilon_{0}}
$$

But:

$$
\mathbf{E}=-\nabla \Phi
$$

In this case $E=-\frac{\partial \Phi}{\partial r}$. To obtain a zero potential at infinity we get:

$$
\Phi_{+}=\frac{Q}{4 \pi \varepsilon_{0} r}
$$

The electric field inside the sphere is 0 (no charges), so the potential is constant. Because it is continuous on the surface of the sphere (no jumps):

$$
\Phi_{-}=\frac{Q}{4 \pi \varepsilon_{0} R}
$$

Final answer: $\Phi_{-}=\frac{Q}{4 \pi \varepsilon_{0} R}$,$\Phi_{+}=\frac{Q}{4 \pi \varepsilon_{0} r}$
Subfield: Electromagnetism

2. Johnson-Nyquist noise

In this problem we study thermal noise in electrical circuits. The goal is to derive the JohnsonNyquist spectral (per-frequency, $f$ ) density of noise produced by a resistor, $R$ :

$$
\frac{d\left\langle V^{2}\right\rangle}{d f}=4 k T R
\tag{2}
$$

Here, \langle\rangle denotes an average over time, so $\left\langle V^{2}\right\rangle$ is the mean-square value of the voltage fluctuations due to thermal noise. $f$ is the angular frequency, $k$ is Boltzmann's constant, and $T$ is temperature. It says that every frequency range $[f, f+d f]$ contributes a roughly equal amount of noise to the total noise in the resistor; this is called white noise.

Electromagnetic modes in a resistor

We first establish the properties of thermally excited electromagnetic modes

$$
V_{n}(x)=V_{0} \cos \left(k_{n} x-\omega_{n} t\right)
\tag{3}
$$

in a resistor of length $L$. The speed of light $c^{\prime}=\omega_{n} / k_{n}$ in the resistor is independent of $n$.

Question: (a) The electromagnetic modes travel through the ends, $x=0$ and $x=L$, of the resistor. Show that the wavevectors corresponding to periodic waves on the interval $[0, L]$ are $k_{n}=\frac{2 \pi n}{L}$.

Then, show that the number of states per angular frequency is $\frac{d n}{d \omega_{n}}=\frac{L}{2 \pi c^{\prime}}$.

Solution:
A periodic wave on interval $[0, L]$ must fit an integer number of wavelengths $\lambda$ into the length $L$,

$$
n \lambda=L
\tag{1}
$$

Therefore $k_{n}=\frac{2 \pi}{\lambda_{n}}=\frac{2 \pi n}{L}$.

For the next part, use $c^{\prime}=\frac{\omega}{k}$. Therefore,

$$
\omega_{n}=c^{\prime} k_{n}=\frac{2 \pi c^{\prime} n}{L} \Rightarrow d \omega_{n}=\frac{2 \pi c^{\prime}}{L} d n \Rightarrow \frac{d n}{d \omega_{n}}=\frac{L}{2 \pi c^{\prime}}
\tag{2}
$$
Subfield: Electromagnetism

2. Johnson-Nyquist noise

In this problem we study thermal noise in electrical circuits. The goal is to derive the JohnsonNyquist spectral (per-frequency, $f$ ) density of noise produced by a resistor, $R$ :

$$
\frac{d\left\langle V^{2}\right\rangle}{d f}=4 k T R
\tag{2}
$$

Here, \langle\rangle denotes an average over time, so $\left\langle V^{2}\right\rangle$ is the mean-square value of the voltage fluctuations due to thermal noise. $f$ is the angular frequency, $k$ is Boltzmann's constant, and $T$ is temperature. It says that every frequency range $[f, f+d f]$ contributes a roughly equal amount of noise to the total noise in the resistor; this is called white noise.

Electromagnetic modes in a resistor

We first establish the properties of thermally excited electromagnetic modes

$$
V_{n}(x)=V_{0} \cos \left(k_{n} x-\omega_{n} t\right)
\tag{3}
$$

in a resistor of length $L$. The speed of light $c^{\prime}=\omega_{n} / k_{n}$ in the resistor is independent of $n$.
Context question:
(a) The electromagnetic modes travel through the ends, $x=0$ and $x=L$, of the resistor. Show that the wavevectors corresponding to periodic waves on the interval $[0, L]$ are $k_{n}=\frac{2 \pi n}{L}$.

Then, show that the number of states per angular frequency is $\frac{d n}{d \omega_{n}}=\frac{L}{2 \pi c^{\prime}}$.
Context answer:
\boxed{证明题}


Question: (b) Each mode $n$ in the resistor can be thought of as a species of particle, called a bosonic collective mode. This particle obeys Bose-Einstein statistics: the average number of particles $\left\langle N_{n}\right\rangle$ in the mode $n$ is

$$
\left\langle N_{n}\right\rangle=\frac{1}{\exp \frac{\hbar \omega_{n}}{k T}-1}
\tag{4}
$$

In the low-energy limit $\hbar \omega_{n} \ll k T$, show that

$$
\left\langle N_{n}\right\rangle \approx \frac{k T}{\hbar \omega_{n}}
\tag{5}
$$

You can use the Taylor expansion $e^{x} \approx 1+x$ for small $x$.

Solution:
Use the given Taylor expansion,

$$
\left\langle N_{n}\right\rangle=\frac{1}{\exp \frac{\hbar \omega_{n}}{k T}-1} \approx \frac{1}{\left(1+\frac{\hbar \omega_{n}}{k T}\right)-1}=\frac{k T}{\hbar \omega_{n}}
\tag{3}
$$
Subfield: Electromagnetism

2. Johnson-Nyquist noise

In this problem we study thermal noise in electrical circuits. The goal is to derive the JohnsonNyquist spectral (per-frequency, $f$ ) density of noise produced by a resistor, $R$ :

$$
\frac{d\left\langle V^{2}\right\rangle}{d f}=4 k T R
\tag{2}
$$

Here, \langle\rangle denotes an average over time, so $\left\langle V^{2}\right\rangle$ is the mean-square value of the voltage fluctuations due to thermal noise. $f$ is the angular frequency, $k$ is Boltzmann's constant, and $T$ is temperature. It says that every frequency range $[f, f+d f]$ contributes a roughly equal amount of noise to the total noise in the resistor; this is called white noise.

Electromagnetic modes in a resistor

We first establish the properties of thermally excited electromagnetic modes

$$
V_{n}(x)=V_{0} \cos \left(k_{n} x-\omega_{n} t\right)
\tag{3}
$$

in a resistor of length $L$. The speed of light $c^{\prime}=\omega_{n} / k_{n}$ in the resistor is independent of $n$.
Context question:
(a) The electromagnetic modes travel through the ends, $x=0$ and $x=L$, of the resistor. Show that the wavevectors corresponding to periodic waves on the interval $[0, L]$ are $k_{n}=\frac{2 \pi n}{L}$.

Then, show that the number of states per angular frequency is $\frac{d n}{d \omega_{n}}=\frac{L}{2 \pi c^{\prime}}$.
Context answer:
\boxed{证明题}


Context question:
(b) Each mode $n$ in the resistor can be thought of as a species of particle, called a bosonic collective mode. This particle obeys Bose-Einstein statistics: the average number of particles $\left\langle N_{n}\right\rangle$ in the mode $n$ is

$$
\left\langle N_{n}\right\rangle=\frac{1}{\exp \frac{\hbar \omega_{n}}{k T}-1}
\tag{4}
$$

In the low-energy limit $\hbar \omega_{n} \ll k T$, show that

$$
\left\langle N_{n}\right\rangle \approx \frac{k T}{\hbar \omega_{n}}
\tag{5}
$$

You can use the Taylor expansion $e^{x} \approx 1+x$ for small $x$.
Context answer:
\boxed{证明题}


Question: (c) By analogy to the photon, explain why the energy of each particle in the mode $n$ is $\hbar \omega_{n}$.

Solution:
The electromagnetic modes are excitations of the electromagnetic field like photons, so they are also massless. The energy of a photon is $E=h f=\hbar \omega$.

Alternatively, you can start from $E^{2}=\left(m c^{2}\right)^{2}+(p c)^{2}$ and use $m=0, p=h / \lambda$.
Subfield: Mechanics


Question: (a) Express the equations of motion of $M_{1}$ and $M_{2}$ in terms of the gravitational constant $G$ and the position vectors $\overrightarrow{r_{1}}$ and $\overrightarrow{r_{2}}$ which point from the origin to $M_{1}$ and $M_{2}$, respectively.

Solution:
$$
\begin{aligned}
& M_{1} \frac{d^{2}}{d t^{2}} \overrightarrow{r_{1}}=\frac{G M_{1} M_{2}}{\left|\overrightarrow{r_{2}}-\overrightarrow{r_{1}}\right|^{3}}\left(\overrightarrow{r_{2}}-\overrightarrow{r_{1}}\right) \\
& M_{2} \frac{d^{2}}{d t^{2}} \overrightarrow{r_{2}}=\frac{G M_{1} M_{2}}{\left|\overrightarrow{r_{1}}-\overrightarrow{r_{2}}\right|^{3}}\left(\overrightarrow{r_{1}}-\overrightarrow{r_{2}}\right)
\end{aligned}
$$

Hence,

$$
\begin{aligned}
& \frac{d^{2}}{d t^{2}} \overrightarrow{r_{1}}=\frac{G M_{2}}{\left|\overrightarrow{r_{2}}-\overrightarrow{r_{1}}\right|^{3}}\left(\overrightarrow{r_{2}}-\overrightarrow{r_{1}}\right) \\
& \frac{d^{2}}{d t^{2}} \overrightarrow{r_{2}}=\frac{G M_{1}}{\left|\overrightarrow{r_{1}}-\overrightarrow{r_{2}}\right|^{3}}\left(\overrightarrow{r_{1}}-\overrightarrow{r_{2}}\right)
\end{aligned}
$$

Final answer: $$
\begin{aligned}
& \frac{d^{2}}{d t^{2}} \overrightarrow{r_{1}}=\frac{G M_{2}}{\left|\overrightarrow{r_{2}}-\overrightarrow{r_{1}}\right|^{3}}\left(\overrightarrow{r_{2}}-\overrightarrow{r_{1}}\right) \\
& \frac{d^{2}}{d t^{2}} \overrightarrow{r_{2}}=\frac{G M_{1}}{\left|\overrightarrow{r_{1}}-\overrightarrow{r_{2}}\right|^{3}}\left(\overrightarrow{r_{1}}-\overrightarrow{r_{2}}\right)
\end{aligned}
$$
Subfield: Mechanics

3. The circular restricted three-body problem

In general, there is no exact solution of the three-body problem, in which three masses move under their mutual gravitational attraction. However, it is possible to make some progress by adding some constraints to the motion.

Two-body problem

Let's start with the motion of two masses, $M_{1}$ and $M_{2}$. Assume both masses move in circular orbits about their center of mass. Consider the inertial frame whose origin coincides with the center of mass of the system.
Context question:
(a) Express the equations of motion of $M_{1}$ and $M_{2}$ in terms of the gravitational constant $G$ and the position vectors $\overrightarrow{r_{1}}$ and $\overrightarrow{r_{2}}$ which point from the origin to $M_{1}$ and $M_{2}$, respectively.
Context answer:
$$
\begin{aligned}
& \frac{d^{2}}{d t^{2}} \overrightarrow{r_{1}}=\frac{G M_{2}}{\left|\overrightarrow{r_{2}}-\overrightarrow{r_{1}}\right|^{3}}\left(\overrightarrow{r_{2}}-\overrightarrow{r_{1}}\right) \\
& \frac{d^{2}}{d t^{2}} \overrightarrow{r_{2}}=\frac{G M_{1}}{\left|\overrightarrow{r_{1}}-\overrightarrow{r_{2}}\right|^{3}}\left(\overrightarrow{r_{1}}-\overrightarrow{r_{2}}\right)
\end{aligned}
$$


Question: (b) Find the period $T$ and angular frequency $\omega$ of the orbital motion.

Solution:
$$
\frac{d^{2}}{d t^{2}}\left(\overrightarrow{r_{1}}-\overrightarrow{r_{2}}\right)=-\frac{G\left(M_{1}+M_{2}\right)}{\left|\overrightarrow{r_{1}}-\overrightarrow{r_{2}}\right|^{3}}\left(\overrightarrow{r_{1}}-\overrightarrow{r_{2}}\right)
$$

Using polar coordinate, the time derivative of a position vector $\vec{r}$ with constant magnitude,

$$
\begin{aligned}
& \frac{d}{d t} \vec{r}=\frac{d}{d t} r \hat{r}=r \frac{d}{d t} \hat{r}=\vec{\omega} \times \vec{r} \\
& \frac{d 2}{d t^{2}} \vec{r}=\vec{\omega} \times(\vec{\omega} \times \vec{r})=-\omega^{2} \vec{r}
\end{aligned}
$$

Since $\left|\overrightarrow{r_{1}}-\overrightarrow{r_{2}}\right|$ is constant,

$$
\begin{gathered}
\omega=\sqrt{\frac{G\left(M_{1}+M_{2}\right)}{\left|\overrightarrow{r_{2}}-\overrightarrow{r_{1}}\right|^{3}}} \\
T=2 \pi / \omega=2 \pi \sqrt{\frac{\left|\overrightarrow{r_{2}}-\overrightarrow{r_{1}}\right|^{3}}{G\left(M_{1}+M_{2}\right)}}
\end{gathered}
$$

Final answer: $\omega=\sqrt{\frac{G\left(M_{1}+M_{2}\right)}{\left|\overrightarrow{r_{2}}-\overrightarrow{r_{1}}\right|^{3}}}$ , $T=2 \pi \sqrt{\frac{\left|\overrightarrow{r_{2}}-\overrightarrow{r_{1}}\right|^{3}}{G\left(M_{1}+M_{2}\right)}}$
Subfield: Mechanics

3. The circular restricted three-body problem

In general, there is no exact solution of the three-body problem, in which three masses move under their mutual gravitational attraction. However, it is possible to make some progress by adding some constraints to the motion.

Two-body problem

Let's start with the motion of two masses, $M_{1}$ and $M_{2}$. Assume both masses move in circular orbits about their center of mass. Consider the inertial frame whose origin coincides with the center of mass of the system.
Context question:
(a) Express the equations of motion of $M_{1}$ and $M_{2}$ in terms of the gravitational constant $G$ and the position vectors $\overrightarrow{r_{1}}$ and $\overrightarrow{r_{2}}$ which point from the origin to $M_{1}$ and $M_{2}$, respectively.
Context answer:
$$
\begin{aligned}
& \frac{d^{2}}{d t^{2}} \overrightarrow{r_{1}}=\frac{G M_{2}}{\left|\overrightarrow{r_{2}}-\overrightarrow{r_{1}}\right|^{3}}\left(\overrightarrow{r_{2}}-\overrightarrow{r_{1}}\right) \\
& \frac{d^{2}}{d t^{2}} \overrightarrow{r_{2}}=\frac{G M_{1}}{\left|\overrightarrow{r_{1}}-\overrightarrow{r_{2}}\right|^{3}}\left(\overrightarrow{r_{1}}-\overrightarrow{r_{2}}\right)
\end{aligned}
$$


Context question:
(b) Find the period $T$ and angular frequency $\omega$ of the orbital motion.
Context answer:
\boxed{$\omega=\sqrt{\frac{G\left(M_{1}+M_{2}\right)}{\left|\overrightarrow{r_{2}}-\overrightarrow{r_{1}}\right|^{3}}}$ , $T=2 \pi \sqrt{\frac{\left|\overrightarrow{r_{2}}-\overrightarrow{r_{1}}\right|^{3}}{G\left(M_{1}+M_{2}\right)}}$}


Extra Supplementary Reading Materials:

Circular restricted three-body problem

Let us transform to a non-inertial frame rotating with angular velocity $\vec{\omega}=(0,0, \omega)$ about an axis normal to the orbital plane of masses $M_{1}$ and $M_{2}$, with the origin at their center of mass. In this frame, $M_{1}$ and $M_{2}$ are stationary at the Cartesian coordinates $(-\alpha R, 0,0)$ and $((1-\alpha) R, 0,0)$ respectively. The third mass $m$ is not stationary in this frame; in this non-inertial frame its position is $\vec{r}(t)=(x(t), y(t), 0)$.

The masses satisfy $M_{1}, M_{2} \gg m$. Consider $m$ to be so small that it does not affect the motion of $M_{1}$ or $M_{2}$.

Question: (a) Express $\alpha$ in terms of $M_{1}$ and $M_{2}$.

Solution:
$$
\begin{gathered}
M_{1}(-\alpha R)+M_{2}((1-\alpha) R)=0 \\
\alpha=\frac{M_{2}}{M_{1}+M_{2}}
\end{gathered}
$$

Final answer: $\alpha=\frac{M_{2}}{M_{1}+M_{2}}$