Past exam of the mathematics course of the University of Cambridge / 2022 / ia / Paper 4 / 9C (Dynamics and Relativity)

The center of mass isBy Newton's third law, ${\mathbf{F}}_{21}=-{\mathbf{F}}_{12}$, soThus $\dot{\mathbf{R}}$ is constant. For $\mathbf{r}={\mathbf{x}}_1-{\mathbf{x}}_2$,Thereforewhere $\mu$ is the reduced mass.

For the proposed circular motion,The separation is ${\mathbf{x}}_1-{\mathbf{x}}_2=2a(\cos \omega t,\sin \omega t,0)$, so the inverse-square force on particle 1 isThe equation $m{\ddot{\mathbf{x}}}_1={\mathbf{F}}_{12}$ holds exactly whenThe equation for particle 2 follows by symmetry.

At a point $(0,0,z)$, the two particles are at equal distance $\sqrt{a^2+z^2}$. Their force components in the rotating $xy$-plane cancel, while their $z$-components add. Hence a particle initially moving along the $z$-axis remains on it, andCancelling $m_3$ gives

The equation has the one-dimensional effective potential per unit massThus the conserved specific energy isThe potential has its minimum $-2mk/a$ at $z=0$ and tends to $0$ from below as $|z|\to \infty$. Therefore $-2mk/a<\mathcal{E}<0$ gives bounded oscillation between two turning points; $\mathcal{E}=0$ gives marginal escape with asymptotic speed zero; and $\mathcal{E}>0$ gives escape with nonzero asymptotic speed. The minimum itself is the equilibrium $z=0$.

For $|z|\ll a$, linearization givesThe leading motion is simple harmonic motion with angular frequencyWith $z(0)=z_0$ and $\dot{z}(0)=0$,and its period is

The initial specific energy isEscape to infinity is possible exactly when $\mathcal{E}\ge 0$. Thus the escape velocity criterion isEquality corresponds to marginal escape.