Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 346 / 3 / Solution

For a spherical orbit,so the galaxy rotation curve is flat. The isotropic spherical Jeans equation with constant dispersion isSince $\rho \propto r^{-2}$,

WriteFor $X\ll 1$,so Chandrasekhar dynamical friction is linear in $\mathbf{v}$ at low speed. At $X\gg 1$, $\mathcal{F}(X)\to 1$, so its acceleration magnitude decays as $v^{-2}$. At zero speed the wake is symmetric and the drag vanishes; at high speed the subhalo spends too little time deflecting each background particle efficiently.

For a circular orbit $v=v_0$, so $X=1$ and $\mathcal{F}(1)=0.428$. The tangential acceleration isThe specific angular momentum is $L=r{v}_0$, hence $v_0\dot{r}=r{a}_{\mathrm{df}}$ andIntegration from $r_i$ to zero gives

For a circular orbit of radius $r$,Solving for $r$ givesSince $\eta =L/L_{\mathrm{circ}}(E)$,

The friction acceleration is antiparallel to velocity, so locally $\dot{L}/L=(1/v)\dot{v}$ and $\dot{E}=v\dot{v}$. Applying the chain rule $\dot{e}=(de/d\eta )\dot{\eta }$ givesBecause $d\eta /de<0$ and $\dot{v}<0$, at pericentre $v>v_0$ gives $\dot{e}<0$: friction circularizes. At apocentre $v<v_0$ gives $\dot{e}>0$: it makes the orbit more eccentric. Orbit averaging produces substantial cancellation; the denser pericentre region generally gives modest net circularization, but the eccentricity changes much less dramatically than the orbital energy and radius.