Circular Restricted Three-Body Problem

Consider an isolated dynamical system consisting of three gravitationally interacting point masses, $m_1$, $m_2$, and $m_3$. Suppose, however, that the third mass, $m_3$, is so much smaller than the other two that it has a negligible effect on their motion. Suppose, further, that the first two masses, $m_1$ and $m_2$, execute circular orbits about their common center of mass. In the following, we shall investigate this simplified problem, which is generally known as the circular restricted three-body problem.

Figure 47: The circular restricted three-body problem.

Let us define a Cartesian coordinate system $(\xi,\,\eta,\,\zeta)$ in an inertial reference frame whose origin coincides with the center of mass, $C$, of the two orbiting masses. Furthermore, let the orbital plane of these masses coincide with the $\xi$-$\eta$ plane, and let them both lie on the $\xi$-axis at time $t=0$--see Figure 47. Suppose that $R$ is the constant distance between the two orbiting masses, $r_1$ the constant distance between mass $m_1$ and the origin, and $r_2$ the constant distance between mass $m_2$ and the origin. Moreover, let $\omega$ be the constant orbital angular velocity. It follows, from Section 6.3, that

$$\omega^2 = \frac{G\,M}{R^3},$$ (1030)

$$\frac{r_1}{r_2} = \frac{m_2}{m_1},$$ (1031)

where $M=m_1+m_2$.

It is convenient to choose our unit of length such that $R=1$, and our unit of mass such that $G\,M=1$. It follows, from Equation (1030), that $\omega =1$. However, we shall continue to retain $\omega$ in our equations, for the sake of clarity. Let $\mu_1=G\,m_1$, and $\mu_2=G\,m_2=1-\mu_1$. It is easily demonstrated that $r_1 = \mu_2$, and $r_2=1-r_1=\mu_1$. Hence, the two orbiting masses, $m_1$ and $m_2$, have position vectors ${\bf r}_1=(\xi_1,\,\eta_1,\,0)$ and ${\bf r}_2=(\xi_2,\,\eta_2,\,0)$, respectively, where (see Figure 47)

$${\bf r}_1 = \mu_2\,(-\cos \omega t,\,-\sin\omega t,\,0),$$ (1032)

$${\bf r}_2 = \mu_1\,(\cos \omega t,\,\sin\omega t,\,0).$$ (1033)

Let the third mass have position vector ${\bf r} = (\xi,\,\eta,\,\zeta)$. The Cartesian components of the equation of motion of this mass are thus

$$\ddot{\xi} = -\mu_1\,\frac{(\xi-\xi_1)}{\rho_1^{\,3}} - \mu_2\,\frac{(\xi-\xi_2)}{\rho_2^{\,3}},$$ (1034)

$$\ddot{\eta} = -\mu_1\,\frac{(\eta-\eta_1)}{\rho_1^{\,3}} - \mu_2\,\frac{(\eta-\eta_2)}{\rho_2^{\,3}},$$ (1035)

$$\ddot{\zeta} = -\mu_1\,\frac{\zeta}{\rho_1^{\,3}} - \mu_2\,\frac{\zeta}{\rho_2^{\,3}},$$ (1036)

where

$$\rho_1^{\,2} = (\xi-\xi_1)^2+(\eta-\eta_1)^2 + \zeta^2,$$ (1037)

$$\rho_2^{\,2} = (\xi-\xi_2)^2+(\eta-\eta_2)^2 + \zeta^2.$$ (1038)