If $M_s$ is constant, the frictional acceleration instead scales as $r_o^{-2}$. The angular-momentum equation gives
$$
\boxed{
\frac{dr_o}{dt}=-K_c r_o^{-3/2},
\qquad
K_c=\frac{4GM_s\mathcal L}{3\sqrt C}}.
$$
Therefore
$$
\boxed{
r_o(t)=\left[
r_{o0}^{5/2}-\frac52K_ct
\right]^{2/5}}.
$$
The orbit again reaches the centre in finite time, but its inward speed accelerates without the strong tidal-mass suppression present in part b.


Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 320 / 3 / c / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-320/3/c/solution