= Solution
For a spherical orbit,
$$
v_c^2=r\frac{d\phi}{dr}=v_0^2,
$$
so the <galaxy rotation curve> is flat. The isotropic spherical <Jeans equation> with constant dispersion is
$$
\frac d{dr}(\rho\sigma^2)=-\rho\frac{d\phi}{dr}.
$$
Since $\rho\propto r^{-2}$,
$$
-\frac{2\rho\sigma^2}{r}
=-\frac{\rho v_0^2}{r},
\qquad
\boxed{\sigma=\frac{v_0}{\sqrt2}}.
$$
Write
$$
\mathcal F(X)=\operatorname{erf}X
-\frac{2X}{\sqrt\pi}e^{-X^2}.
$$
For $X\ll1$,
$$
\mathcal F(X)=\frac{4X^3}{3\sqrt\pi}+O(X^5),
$$
so <Chandrasekhar dynamical friction> is linear in $\mathbf v$ at low speed. At $X\gg1$, $\mathcal F(X)\to1$, 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 is
$$
a_{\rm df}
=-0.428\,\frac{4\pi G^2M\rho\log\Lambda}{v_0^2}
=-0.428\,\frac{GM\log\Lambda}{r^2}.
$$
The specific angular momentum is $L=rv_0$, hence $v_0\dot r=ra_{\rm df}$ and
$$
\boxed{
r\dot r=-0.428\,\log\Lambda\,\frac{GM}{v_0}
}.
$$
Integration from $r_i$ to zero gives
$$
\boxed{
t_{\rm df}
=\frac{r_i^2v_0}{2(0.428)GM\log\Lambda}
=\frac{1.17\,r_i^2v_0}{GM\log\Lambda}
}.
$$
For a circular orbit of radius $r$,
$$
E=\frac{v_0^2}{2}+v_0^2\log(r/r_0).
$$
Solving for $r$ gives
$$
\boxed{
L_{\rm circ}(E)
=v_0r_0
\exp\left(\frac{E-v_0^2/2}{v_0^2}\right)
}.
$$
Since $\eta=L/L_{\rm circ}(E)$,
$$
\boxed{
\dot\eta
=\eta\left(\frac{\dot L}{L}
-\frac{\dot E}{v_0^2}\right)
}.
$$
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 $\dot e=(de/d\eta)\dot\eta$ gives
$$
\boxed{
\dot e
=\frac{\eta}{v}\frac{de}{d\eta}
\left(1-\frac{v^2}{v_0^2}\right)\dot v
}.
$$
Because $d\eta/de<0$ 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.
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