Subfield: Mechanics

A Three-body Problem and LISA 

<img_4539>

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{$\omega_{0}=\sqrt{\frac{G m}{R(R+r)^{2}}}$}


Context 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.
Context answer:
distance from $\mu$ to $M$ is $R+r$
distance from $\mu$ to $m$ is $R+r$
distance from $\mu$ to the centre of mass is $\sqrt{r^{2}+r R+R^{2}}$


Question: 1.3 Consider the case $M=m$. If $\mu$ is now given a small radial perturbation (along $\mathrm{O} \mu$ ), what is the angular frequency of oscillation of $\mu$ about the unperturbed position in terms of $\omega_{0}$ ? Assume that the angular momentum of $\mu$ is conserved.

Solution:
The energy of the mass $\mu$ is given by

$$
E=-\frac{G M \mu}{r_{1}}-\frac{G m \mu}{r_{2}}+\frac{1}{2} \mu\left(\left(\frac{d \rho}{d t}\right)^{2}+\rho^{2} \omega^{2}\right)
\tag{15}
$$

Since the perturbation is in the radial direction, angular momentum is conserved $\left(r_{1}=r_{2}=\Re\right.$ and $m=M$ ),

$$
E=-\frac{2 G M \mu}{\Re}+\frac{1}{2} \mu\left(\left(\frac{d \rho}{d t}\right)^{2}+\frac{\rho_{0}{ }^{4} \omega_{0}{ }^{2}}{\rho^{2}}\right)
\tag{16}
$$

Since the energy is conserved,

$$
\frac{d E}{d t}=0
$$

$$
\frac{d E}{d t}=\frac{2 G M \mu}{\mathfrak{R}^{2}} \frac{d \Re}{d t}+\mu \frac{d \rho}{d t} \frac{d^{2} \rho}{d t^{2}}-\mu \frac{\rho_{0}{ }^{4} \omega_{0}{ }^{2}}{\rho^{3}} \frac{d \rho}{d t}=0
\tag{17}
$$

$$
\frac{d \Re}{d t}=\frac{d \Re}{d \rho} \frac{d \rho}{d t}=\frac{d \rho}{d t} \frac{\rho}{\mathfrak{R}}
\tag{18}
$$

$$
\frac{d E}{d t}=\frac{2 G M \mu}{\mathfrak{R}^{3}} \rho \frac{d \rho}{d t}+\mu \frac{d \rho}{d t} \frac{d^{2} \rho}{d t^{2}}-\mu \frac{\rho_{0}^{4} \omega_{0}{ }^{2}}{\rho^{3}} \frac{d \rho}{d t}=0
\tag{19}
$$

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Since $\frac{d \rho}{d t} \neq 0$, we have

$\frac{2 G M}{\mathfrak{R}^{3}} \rho+\frac{d^{2} \rho}{d t^{2}}-\frac{\rho_{0}{ }^{4} \omega_{0}{ }^{2}}{\rho^{3}}=0$ or

$$
\frac{d^{2} \rho}{d t^{2}}=-\frac{2 G M}{\mathfrak{R}^{3}} \rho+\frac{\rho_{0}{ }^{4} \omega_{0}{ }^{2}}{\rho^{3}}
\tag{20}
$$

The perturbation from $\mathfrak{R}_{0}$ and $\rho_{0}$ gives $\mathfrak{R}=\mathfrak{R}_{0}\left(1+\frac{\Delta \mathfrak{R}}{\mathfrak{R}_{0}}\right)$ and $\rho=\rho_{0}\left(1+\frac{\Delta \rho}{\rho_{0}}\right)$.

Then

$$
\frac{d^{2} \rho}{d t^{2}}=\frac{d^{2}}{d t^{2}}\left(\rho_{0}+\Delta \rho\right)=-\frac{2 G M}{\mathfrak{R}_{0}^{3}\left(1+\frac{\Delta \mathfrak{R}}{\mathfrak{R}_{0}}\right)^{3}} \rho_{0}\left(1+\frac{\Delta \rho}{\rho_{0}}\right)+\frac{\rho_{0}{ }^{4} \omega_{0}{ }^{2}}{\rho_{0}^{3}\left(1+\frac{\Delta \rho}{\rho_{0}}\right)^{3}}
\tag{21}
$$

Using binomial expansion $(1+\varepsilon)^{n} \approx 1+n \varepsilon$,

$$
\frac{d^{2} \Delta \rho}{d t^{2}}=-\frac{2 G M}{\mathfrak{R}_{0}^{3}} \rho_{0}\left(1+\frac{\Delta \rho}{\rho_{0}}\right)\left(1-\frac{3 \Delta \mathfrak{R}}{\mathfrak{R}_{0}}\right)+\rho_{0} \omega_{0}^{2}\left(1-\frac{3 \Delta \rho}{\rho_{0}}\right)
\tag{22}
$$

Using $\Delta \rho=\frac{\Re}{\rho} \Delta \mathfrak{R}$,

$$
\frac{d^{2} \Delta \rho}{d t^{2}}=-\frac{2 G M}{\mathfrak{R}_{0}^{3}} \rho_{0}\left(1+\frac{\Delta \rho}{\rho_{0}}-\frac{3 \rho_{0} \Delta \rho}{\mathfrak{R}_{0}^{2}}\right)+\rho_{0} \omega_{0}^{2}\left(1-\frac{3 \Delta \rho}{\rho_{0}}\right)
\tag{23}
$$

Since $\omega_{0}^{2}=\frac{2 G M}{\Re_{0}^{3}}$,

$$
\frac{d^{2} \Delta \rho}{d t^{2}}=-\omega_{0}^{2} \rho_{0}\left(1+\frac{\Delta \rho}{\rho_{0}}-\frac{3 \rho_{0} \Delta \rho}{\mathfrak{R}_{0}^{2}}\right)+\omega_{0}^{2} \rho_{0}\left(1-\frac{3 \Delta \rho}{\rho_{0}}\right)
\tag{24}
$$

$$
\frac{d^{2} \Delta \rho}{d t^{2}}=-\omega_{0}^{2} \rho_{0}\left(\frac{4 \Delta \rho}{\rho_{0}}-\frac{3 \rho_{0} \Delta \rho}{\mathfrak{R}_{0}^{2}}\right)
\tag{25}
$$

$$
\frac{d^{2} \Delta \rho}{d t^{2}}=-\omega_{0}^{2} \Delta \rho\left(4-\frac{3 \rho_{0}^{2}}{\mathfrak{R}_{0}^{2}}\right)
\tag{26}
$$

From the figure, $\rho_{0}=\mathfrak{R}_{0} \cos 30^{\circ}$ or $\frac{\rho_{0}{ }^{2}}{\mathfrak{R}_{0}{ }^{2}}=\frac{3}{4}$,

$$
\frac{d^{2} \Delta \rho}{d t^{2}}=-\omega_{0}^{2} \Delta \rho\left(4-\frac{9}{4}\right)=-\frac{7}{4} \omega_{0}^{2} \Delta \rho
\tag{27}
$$



Angular frequency of oscillation is $\frac{\sqrt{7}}{2} \omega_{0}$.
$M=m$ gives $R=r$ and $\omega_{0}{ }^{2}=\frac{G(M+M)}{(R+R)^{3}}=\frac{G M}{4 R^{3}}$. The unperturbed radial distance of $\mu$ is $\sqrt{3} R$, so the perturbed radial distance can be represented by $\sqrt{3} R+\zeta$ where $\zeta<<\sqrt{3} R$ as shown in the following figure.

Using Newton's $2^{\text {nd }}$ law, $-\frac{2 G M \mu}{\left\{R^{2}+(\sqrt{3} R+\zeta)^{2}\right\}^{3 / 2}}(\sqrt{3} R+\zeta)=\mu \frac{d^{2}}{d t^{2}}(\sqrt{3} R+\zeta)-\mu \omega^{2}(\sqrt{3} R+\zeta)$.

(1)

The conservation of angular momentum gives $\mu \omega_{0}(\sqrt{3} R)^{2}=\mu \omega(\sqrt{3} R+\zeta)^{2}$.

(2)

Manipulate (1) and (2) algebraically, applying $\zeta^{2} \approx 0$ and binomial approximation.

$-\frac{2 G M}{\left\{R^{2}+(\sqrt{3} R+\zeta)^{2}\right\}^{3 / 2}}(\sqrt{3} R+\zeta)=\frac{d^{2} \zeta}{d t^{2}}-\frac{\omega_{0}^{2} \sqrt{3} R}{(1+\zeta / \sqrt{3} R)^{3}}$

$-\frac{2 G M}{\left\{4 R^{2}+2 \sqrt{3} \zeta R\right\}^{3 / 2}}(\sqrt{3} R+\zeta) \approx \frac{d^{2} \zeta}{d t^{2}}-\frac{\omega_{0}^{2} \sqrt{3} R}{(1+\zeta / \sqrt{3} R)^{3}}$

$-\frac{G M}{4 R^{3}} \sqrt{3} R \frac{(1+\zeta / \sqrt{3} R)}{(1+\sqrt{3} \zeta / 2 R)^{3 / 2}}=\frac{d^{2} \zeta}{d t^{2}}-\frac{\omega_{0}^{2} \sqrt{3} R}{(1+\zeta / \sqrt{3} R)^{3}}$

$-\omega_{0}^{2} \sqrt{3} R\left(1-\frac{3 \sqrt{3} \zeta}{4 R}\right)\left(1+\frac{\zeta}{\sqrt{3} R}\right) \approx \frac{d^{2} \zeta}{d t^{2}}-\omega_{0}^{2} \sqrt{3} R\left(1-\frac{3 \zeta}{\sqrt{3} R}\right)$

$\frac{d^{2}}{d t^{2}} \zeta=-\left(\frac{7}{4} \omega_{0}^{2}\right) \zeta$

Final answer: $\frac{\sqrt{7}}{2} \omega_{0}$