For a <singular isothermal sphere> host, $\rho_h=A_hr_o^{-2}$,
$$
M_h=4\pi A_hr_o,
\qquad
v_c^2=4\pi GA_h\equiv V^2.
$$
Now $d\log M_h/d\log r_o=1$, so the circular tidal formula has a factor two:
$$
r_t=r_o\left(\frac{M_s}{2M_h}\right)^{1/3}.
$$
Using $M_s=2\pi A_sr_t^2$ gives
$$
\boxed{
r_t=\frac{A_s}{4A_h}r_o^2,
\qquad
M_s=\mu_i r_o^4,
\qquad
\mu_i=\frac{\pi A_s^3}{8A_h^2}}.
$$
The frictional acceleration is now $a_{\rm df}=-G\mu_i\mathcal L r_o^2$. Since the specific angular momentum is $Vr_o$,
$$
\boxed{
\frac{dr_o}{dt}=-K_i r_o^3,
\qquad
K_i=\frac{G\mu_i\mathcal L}{V}}.
$$
Thus
$$
\boxed{
r_o(t)=\frac{r_{o0}}
{\sqrt{1+2K_ir_{o0}^2t}},
\qquad
M_s(t)\propto[1+2K_ir_{o0}^2t]^{-2}}.
$$
Neither radius nor mass reaches zero at finite time. Compared with the $r^{-1}$ host, the steeper isothermal cusp shrinks the tidal radius as $r_o^2$, so stripping suppresses the friction rapidly enough to produce <dynamical-friction stalling by tidal stripping>.

Solution

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

 Solution

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