Put
$$
q=\frac{A_s}{A_h},
\qquad
\mu_s=\frac{2\pi A_s^3}{A_h^2},
\qquad
C=2\pi GA_h,
\qquad
\mathcal L=\mathcal K\log\Lambda.
$$
Part a gives
$$
r_t=qr_o,
\qquad
M_s=\mu_s r_o^2,
$$
while the host <circular speed> is $v_c=\sqrt{Cr_o}$. The <Chandrasekhar dynamical friction> acceleration reduces to the radius-independent value
$$
a_{\rm df}
=-\frac{4\pi G^2M_s(A_h/r_o)\mathcal L}{v_c^2}
=-2G\mu_s\mathcal L.
$$
Assume stripped material leaves with the satellite's instantaneous specific angular momentum. The remaining orbit then obeys
$$
\frac d{dt}(r_ov_c)=r_oa_{\rm df}.
$$
Since $r_ov_c=\sqrt C,r_o^{3/2}$,
$$
\boxed{
\frac{dr_o}{dt}=-K_s\sqrt{r_o},
\qquad
K_s=\frac{4G\mu_s\mathcal L}{3\sqrt C}}.
$$
For initial radius $r_{o0}$,
$$
\boxed{
r_o(t)=r_{o0}\left(1-\frac{t}{t_d}\right)^2,
\qquad
t_d=\frac{2\sqrt{r_{o0}}}{K_s}},
$$
and
$$
\boxed{
M_s(t)=M_{s0}\left(1-\frac{t}{t_d}\right)^4,
\qquad
r_t(t)=r_{t0}\left(1-\frac{t}{t_d}\right)^2}.
$$
In this ideal cusp, radius and remaining mass both reach zero at the finite time $t_d$.

Solution

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

 Solution

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