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 $M_{2}$.
Context question:
(a) Express $\alpha$ in terms of $M_{1}$ and $M_{2}$.
Context answer:
\boxed{$\alpha=\frac{M_{2}}{M_{1}+M_{2}}$}


Context question:
(b) Let $\rho_{1}(t)$ and $\rho_{2}(t)$ be the distances from $m$ to $M_{1}$ and $M_{2}$ respectively. Express $\rho_{1}(t)$ and $\rho_{2}(t)$ in terms of the coordinates and constants given.
Context answer:
\boxed{$\rho_{1}(t)=\sqrt{(x(t)+\alpha R)^{2}+(y(t))^{2}}$ , $\rho_{2}(t)=\sqrt{(x(t)-(1-\alpha) R)^{2}+(y(t))^{2}}$}


Context question:
(c) By considering the centrifugal acceleration $\omega^{2} \vec{r}$ and Coriolis acceleration $-2 \omega \times$ $\vec{v}$, find the acceleration $\frac{d^{2}}{d t^{2}} \vec{r}$ of the third mass in terms of the coordinates and constants given, including $\rho_{1}$ and $\rho_{2}$.
Context answer:
\boxed{$\frac{d^{2}}{d t^{2}} \vec{r}=-G M_{1} \frac{\vec{r}-\vec{r_{1}}}{\rho_{1}^{3}}-G M_{2} \frac{\vec{r}-\vec{r_{2}}}{\rho_{2}^{3}}+\omega^{2} \vec{r}-2 \omega \times \vec{r}$}


Context question:
(d) Express $\frac{d^{2} x}{d t^{2}}$ and $\frac{d^{2} y}{d t^{2}}$ in terms of $U$, where $U=-\frac{G M_{1}}{\rho_{1}}-\frac{G M_{2}}{\rho_{2}}-\frac{\omega^{2}}{2}\left(x^{2}+y^{2}\right)$.
Context answer:
\boxed{$\ddot{x} =2 \omega \dot{y}-\frac{\partial U}{\partial x}$ , $\ddot{y} =-2 \omega \dot{x}-\frac{\partial U}{\partial y}$}


Question: (e) Hence, write down an expression of the motion of $m$ which is a constant.

Solution:
From d),

$$
\begin{gathered}
\ddot{x} \dot{x}=2 \omega \dot{x} \dot{y}-\dot{x} \frac{\partial U}{\partial x} \\
\ddot{y} \dot{y}=-2 \omega \dot{x} \dot{y}-\dot{y} \frac{\partial U}{\partial y} \\
\ddot{x} \dot{x}+\ddot{y} \dot{y}=-\dot{x} \frac{\partial U}{\partial x}-\dot{y} \frac{\partial U}{\partial y} \\
\frac{d}{d t}\left[\frac{1}{2}\left(\dot{x}^{2}+\dot{y}^{2}\right)\right]=-\left(\dot{x} \frac{\partial U}{\partial x}+\dot{y} \frac{\partial U}{\partial y}\right)=-\frac{d U}{d t} \\
\frac{d}{d t}\left[\frac{1}{2}\left(\dot{x}^{2}+\dot{y}^{2}\right)+U\right]=0 \\
\text { Constant }=-2 U-v^{2}
\end{gathered}
$$

Final answer: $-2 U-v^{2}$