Subfield: Mechanics

A Three-body Problem and LISA 

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FIGURE 1 Coplanar orbits of three bodies.
Context question:
1.1 Two gravitating masses $M$ and $m$ are moving in circular orbits of radii $R$ and $r$, respectively, about their common centre of mass. Find the angular velocity $\omega_{0}$ of the line joining $M$ and $m$ in terms of $R, r, M, m$ and the universal gravitational constant $G$.
Context answer:
\boxed{$\omega_{0}=\sqrt{\frac{G m}{R(R+r)^{2}}}$}

Question: 1.2 A third body of infinitesimal mass $\mu$ is placed in a coplanar circular orbit about the same centre of mass so that $\mu$ remains stationary relative to both $M$ and $m$ as shown in Figure 1. Assume that the infinitesimal mass is not collinear with $M$ and $m$. Find the values of the following parameters in terms of $R$ and $r$ :

1.2.1 distance from $\mu$ to $M$.

1.2.2 distance from $\mu$ to $m$.

1.2.3 distance from $\mu$ to the centre of mass.

Solution:
Since $\mu$ is infinitesimal, it has no gravitational influences on the motion of neither $M$ nor $m$. For $\mu$ to remain stationary relative to both $M$ and $m$ we must have:

$$
\frac{G M \mu}{r_{1}^{2}} \cos \theta_{1}+\frac{G m \mu}{r_{2}^{2}} \cos \theta_{2} =\mu \omega_{0}^{2} \rho=\frac{G(M+m) \mu}{(R+r)^{3}} \rho
\tag{4}
$$
$$
\frac{G M \mu}{r_{1}^{2}} \sin \theta_{1} =\frac{G m \mu}{r_{2}^{2}} \sin \theta_{2}
\tag{5}
$$

Substituting $\frac{G M}{r_{1}^{2}}$ from Eq. (5) into Eq. (4), and using the identity

$\sin \theta_{1} \cos \theta_{2}+\cos \theta_{1} \sin \theta_{2}=\sin \left(\theta_{1}+\theta_{2}\right)$, we get

$$
m \frac{\sin \left(\theta_{1}+\theta_{2}\right)}{r_{2}^{2}}=\frac{(M+m)}{(R+r)^{3}} \rho \sin \theta_{1}
\tag{6}
$$

The distances $r_{2}$ and $\rho$, the angles $\theta_{1}$ and $\theta_{2}$ are related by two Sine Rule equations

$$
\begin{aligned}
& \frac{\sin \psi_{1}}{\rho}=\frac{\sin \theta_{1}}{R} \\
& \frac{\sin \psi_{1}}{r_{2}}=\frac{\sin \left(\theta_{1}+\theta_{2}\right)}{R+r}
\end{aligned}
\tag{7}
$$

Substitute (7) into (6)

$$
\frac{1}{r_{2}^{3}}=\frac{R}{(R+r)^{4}} \frac{(M+m)}{m}
\tag{10}
$$

Since $\frac{m}{M+m}=\frac{R}{R+r}$,Eq. (10) gives

$$
r_{2}=R+r
\tag{11}
$$

By substituting $\frac{G m}{r_{2}^{2}}$ from Eq. (5) into Eq. (4), and repeat a similar procedure, we get

$$
r_{1}=R+r
\tag{12}
$$

Alternatively,

$$
\begin{aligned}
& \frac{r_{1}}{\sin \left(180^{\circ}-\phi\right)}=\frac{R}{\sin \theta_{1}} \text { and } \frac{r_{2}}{\sin \phi}=\frac{r}{\sin \theta_{2}} \\
& \frac{\sin \theta_{1}}{\sin \theta_{2}}=\frac{R}{r} \times \frac{r_{2}}{r_{1}}=\frac{m}{M} \times \frac{r_{2}}{r_{1}}
\end{aligned}
$$

Combining with Eq. (5) gives $r_{1}=r_{2}$



Hence, it is an equilateral triangle with

$$
\begin{aligned}
& \psi_{1}=60^{\circ} \\
& \psi_{2}=60^{\circ}
\end{aligned}
\tag{13}
$$

The distance $\rho$ is calculated from the Cosine Rule.

$$
\begin{aligned}
& \rho^{2}=r^{2}+(R+r)^{2}-2 r(R+r) \cos 60^{\circ} \\
& \rho=\sqrt{r^{2}+r R+R^{2}}
\end{aligned}
\tag{14}
$$
Since $\mu$ is infinitesimal, it has no gravitational influences on the motion of neither $M$ nor $m$.For $\mu$ to remain stationary relative to both $M$ and $m$ we must have:

$$
\frac{G M \mu}{r_{1}^{2}} \cos \theta_{1}+\frac{G m \mu}{r_{2}^{2}} \cos \theta_{2}=\mu \omega^{2} \rho=\frac{G(M+m) \mu}{(R+r)^{3}} \rho
\tag{4}
$$
$$
\frac{G M \mu}{r_{1}^{2}} \sin \theta_{1}=\frac{G m \mu}{r_{2}^{2}} \sin \theta_{2}
\tag{5}
$$

Note that

$$
\begin{aligned}
& \frac{r_{1}}{\sin \left(180^{\circ}-\phi\right)}=\frac{R}{\sin \theta_{1}} \\
& \frac{r_{2}}{\sin \phi}=\frac{r}{\sin \theta_{2}} \quad \text { (see figure) }
\end{aligned}
$$
$$
\frac{\sin \theta_{1}}{\sin \theta_{2}}=\frac{R}{r} \times \frac{r_{2}}{r_{1}}=\frac{m}{M} \times \frac{r_{2}}{r_{1}}
\tag{6}
$$

Equations (5) and (6):

$$
r_{1}=r_{2}
\tag{7}
$$
$$
\frac{\sin \theta_{1}}{\sin \theta_{2}}=\frac{m}{M}
\tag{8}
$$
$$
\psi_{1}=\psi_{2}
\tag{9}
$$

The equation (4) then becomes:

$$
M \cos \theta_{1}+m \cos \theta_{2}=\frac{(M+m)}{(R+r)^{3}} r_{1}^{2} \rho
\tag{10}
$$

Equations (8) and (10): 
$$
\sin \left(\theta_{1}+\theta_{2}\right)=\frac{M+m}{M} \frac{r_{1}^{2} \rho}{(R+r)^{3}} \sin \theta_{2}
\tag{11}
$$

Note that from figure, 
$$
\quad \frac{\rho}{\sin \psi_{2}}=\frac{r}{\sin \theta_{2}}
\tag{12}
$$



Equations (11) and (12): 
$$
\sin \left(\theta_{1}+\theta_{2}\right)=\frac{M+m}{M} \frac{r_{1}^{2} r}{(R+r)^{3}} \sin \psi_{2}
\tag{13}
$$

Also from figure,

$$
(R+r)^{2}=r_{2}^{2}-2 r_{1} r_{2} \cos \left(\theta_{1}+\theta_{2}\right)+r_{1}^{2}=2 r_{1}^{2}\left[1-\cos \left(\theta_{1}+\theta_{2}\right)\right]
\tag{14}
$$

Equations (13) and (14): 
$$
\sin \left(\theta_{1}+\theta_{2}\right)=\frac{\sin \psi_{2}}{2\left[1-\cos \left(\theta_{1}+\theta_{2}\right)\right]}
\tag{15}
$$

$$
\begin{aligned}
& \theta_{1}+\theta_{2}=180^{\circ}-\psi_{1}-\psi_{2}=180^{\circ}-2 \psi_{2} \quad \text { (see figure) } \\
\therefore \quad & \cos \psi_{2}=\frac{1}{2}, \psi_{2}=60^{\circ}, \psi_{1}=60^{\circ}
\end{aligned}
$$

Hence $M$ and $m$ from an equilateral triangle of sides $(R+r)$

Distance $\mu$ to $M$ is $R+r$

Distance $\mu$ to $m$ is $R+r$

Distance $\mu$ to $\mathrm{O}$ is $\rho=\sqrt{\left(\frac{R+r}{2}-R\right)^{2}+\left\{(R+r) \frac{\sqrt{3}}{2}\right\}^{2}}=\sqrt{R^{2}+R r+r^{2}}$

Final answer: $R+r$, $R+r$, $\sqrt{r^{2}+r R+R^{2}}$