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 examine this simplified problem, which is usually referred to as the circular restricted three-body problem. The problem under investigation has obvious applications to the solar system. For instance, the first two masses might represent the Sun and a planet (recall that a given planet and the Sun do indeed execute almost circular orbits about their common center of mass), whereas the third mass might represent an asteroid or a comet (asteroids and comets do indeed have much smaller masses than the Sun or any of the planets).

Figure 9.1: 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, $m_1$ and $m_2$. 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 9.1. Suppose that $a$ 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 4.16, that

$\displaystyle \omega^{\,2}$ $\displaystyle = \frac{G\,M}{a^{\,3}},$ (9.1)
$\displaystyle \frac{r_1}{r_2}$ $\displaystyle = \frac{m_2}{m_1},$ (9.2)

where $M=m_1+m_2$.

It is convenient to choose our unit of length such that $a=1$, and our unit of mass such that $G\,M=1$. It follows, from Equation (9.1), 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

$\displaystyle {\bf r}_1$ $\displaystyle = (\xi_1,\,\eta_1,\,0)=\left(-\mu_2\,\cos (\omega\, t),\,-\mu_2\,\sin(\omega \,t),\,0\right),$ (9.3)
$\displaystyle {\bf r}_2$ $\displaystyle =(\xi_2,\,\eta_2,\,0)= \left(\mu_1\,\cos( \omega\, t),\,\mu_1\,(\sin\omega \,t),\,0\right),$ (9.4)

respectively. See Figure 9.1. 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

$\displaystyle \skew{3}\ddot{\xi}$ $\displaystyle = -\mu_1\,\frac{(\xi-\xi_1)}{\rho_1^{\,3}} - \mu_2\,\frac{(\xi-\xi_2)}{\rho_2^{\,3}},$ (9.5)
$\displaystyle \skew{3}\ddot{\eta}$ $\displaystyle = -\mu_1\,\frac{(\eta-\eta_1)}{\rho_1^{\,3}} - \mu_2\,\frac{(\eta-\eta_2)}{\rho_2^{\,3}},$ (9.6)
$\displaystyle \skew{3}\ddot{\zeta}$ $\displaystyle = -\mu_1\,\frac{\zeta}{\rho_1^{\,3}} - \mu_2\,\frac{\zeta}{\rho_2^{\,3}},$ (9.7)


$\displaystyle \rho_1^{\,2}$ $\displaystyle = (\xi-\xi_1)^2+(\eta-\eta_1)^2 + \zeta^{\,2},$ (9.8)
$\displaystyle \rho_2^{\,2}$ $\displaystyle = (\xi-\xi_2)^2+(\eta-\eta_2)^2 + \zeta^{\,2}.$ (9.9)