Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 316 / 2 / v / Solution

Eliminate $a$ using the Poynting–Robertson drag invariant, $a=C{e}^{4/5}/(1-e^2)$. The eccentricity rate becomesSince the point-star limit corresponds to $a\to 0$ and $e\to 0$, the inspiral time under Poynting–Robertson drag isFor $e_0\to 0$, the integral is $(5/8)e_0^{8/5}+o(e_0^{8/5})$, recovering $a_0^2/(4A)$ despite the apparent singularity of $C$.

For $e_0\to 1$, use $C\simeq 2q_0$ and $1-e_0=2q_0/(Q_0+q_0)\simeq 2q_0/Q_0$. The endpoint asymptotic integral is $[2(1-e_0){]}^{-1/2}$, givingMost of this long lifetime is spent near the initially large apocentre distance, before substantial circularization. The result assumes weak drag and a valid orbital average; near-parabolic release can violate that assumption.