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

For a singular isothermal sphere host, ${\rho }_h=A_h{r}_o^{-2}$,Now $d\log M_h/d\log r_o=1$, so the circular tidal formula has a factor two:Using $M_s=2\pi A_s{r}_t^2$ givesThe frictional acceleration is now $a_{\mathrm{df}}=-G{\mu }_i\mathcal{L}{r}_o^2$. Since the specific angular momentum is $V{r}_o$,ThusNeither 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.