 Solution

The sudden potential change leaves the star's position and velocity unchanged. Its new energy is thereforeThus $E_1>0$ precisely when $r<a$. Parametrize the orbit by eccentric anomaly $\eta$:A phase-mixed orbit is uniform in the mean anomaly ${\theta }_r$. For $e>0$, the condition $r<a$ is $-\pi /2<\eta <\pi /2$ modulo one period. The corresponding mean-anomaly interval has lengthHenceThe energy increase $G(M_0-M_1)/r$ is largest near periapsis, so stars are more readily unbound there than near apoapsis. A perfectly circular orbit is the measure-zero marginal case $E_1=0$ at every phase.