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

The enclosed masses of the $r^{-1}$ host and truncated satellite areFor a circular orbit, the Jacobi tidal radius isSince $d\log M_h/d\log r_o=2$, substitution givesand henceThe assumption $A_s\ll A_h$ makes $r_t/r_o\ll 1$, precisely the scale separation required by the local tidal approximation.

PutPart a giveswhile the host circular speed is $v_c=\sqrt{C{r}_o}$. The Chandrasekhar dynamical friction acceleration reduces to the radius-independent valueAssume stripped material leaves with the satellite's instantaneous specific angular momentum. The remaining orbit then obeysSince $r_o{v}_c=\sqrt{C},r_o^{3/2}$,For initial radius $r_{o0}$,andIn this ideal cusp, radius and remaining mass both reach zero at the finite time $t_d$.

If $M_s$ is constant, the frictional acceleration instead scales as $r_o^{-2}$. The angular-momentum equation givesThereforeThe orbit again reaches the centre in finite time, but its inward speed accelerates without the strong tidal-mass suppression present in part b.

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.