= Solution
For either component $j\in\{h,s\}$ and $0<\gamma<3$, direct integration of the <power law> density gives
$$
\boxed{M_j(<r)=\frac{4\pi A_j}{3-\gamma}r^{3-\gamma}},
\qquad
\boxed{v_{c,j}^2(r)=\frac{GM_j(<r)}r
=\frac{4\pi GA_j}{3-\gamma}r^{2-\gamma}}.
$$
Let $B_j=4\pi GA_j/(3-\gamma)$. Since $d\Phi_j/dr=GM_j/r^2=B_jr^{1-\gamma}$,
$$
\boxed{
\Phi_j(r)=
\begin{cases}
B_jr^{2-\gamma}/(2-\gamma)+C_j,&\gamma\ne2,\\
B_j\log r+C_j,&\gamma=2.
\end{cases}}
$$
The additive constants depend on the chosen reference and cannot generally be set by requiring $\Phi\to0$ at infinity for an untruncated power law.
For a satellite on a circular orbit, the linearized effective radial force gives the <tidal radius>
$$
r_t^3=\frac{GM_s(<r_t)}{\Omega^2-\Phi_h''(r_o)}.
$$
Here $\Omega^2=GM_h(<r_o)/r_o^3=B_hr_o^{-\gamma}$ and $\Phi_h''=(1-\gamma)\Omega^2$, so
$$
r_t^3=\frac{GM_s(<r_t)}{\gamma\Omega^2}.
$$
Substitution of the two mass profiles yields
$$
\boxed{\frac{r_t}{r_o}=
\left(\frac{A_s}{\gamma A_h}\right)^{1/\gamma}},
$$
and equivalently
$$
\boxed{\frac{r_t}{r_o}
=\left[\frac{M_s(<r_t)}{\gamma M_h(<r_o)}\right]^{1/3}
\sim\left[\frac{M_s(<r_t)}{M_h(<r_o)}\right]^{1/3}}.
$$
Now set $\gamma=1$. Then
$$
r_t=\frac{A_s}{A_h}r_o,
\qquad
M_s(<r_t)=2\pi A_sr_t^2
=\mu r_o^2,
\qquad
\mu=\frac{2\pi A_s^3}{A_h^2}.
$$
The host circular speed is $v=B\sqrt{r_o}$ with $B=\sqrt{2\pi GA_h}$. The magnitude of <Chandrasekhar dynamical friction> becomes
$$
a_{\rm df}
=4\pi G^2M_s\frac{A_h}{r_o}
\frac{\log\Lambda}{B^2r_o}K
=2G\mu K\log\Lambda,
$$
a constant. The specific angular momentum is $j=r_ov=Br_o^{3/2}$. Since the drag torque gives $dj/dt=-r_oa_{\rm df}$,
$$
\frac32B\sqrt{r_o}\frac{dr_o}{dt}
=-r_oa_{\rm df},
$$
and therefore
$$
\boxed{\frac{dr_o}{dt}=-C\sqrt{r_o}},
\qquad
C=\frac{2a_{\rm df}}{3B}>0.
$$
Integration gives
$$
2\sqrt{r_o(t)}=2\sqrt{r_{o0}}-Ct,
\qquad
\boxed{t_{\rm merge}=\frac{2\sqrt{r_{o0}}}{C}},
$$
which is finite.
If stripping is switched off, hold the satellite mass at its initial value $M_0=\mu r_{o0}^2$. The frictional acceleration is then $a_{\rm df}=a_0r_{o0}^2/r_o^2$, where $a_0$ is the constant acceleration in the stripped calculation. The same torque equation gives
$$
\frac{dr_o}{dt}=-Cr_{o0}^2r_o^{-3/2}.
$$
Thus
$$
t_{\rm no\ strip}
=\frac1{Cr_{o0}^2}\int_0^{r_{o0}}r^{3/2}\,dr
=\frac{2\sqrt{r_{o0}}}{5C}
=\boxed{\frac{t_{\rm merge}}5}.
$$
Without <tidal stripping>, the satellite retains its mass while the background density increases inward, so dynamical friction strengthens rapidly. Stripping instead gives $M_s\propto r_o^2$ and removes the very mass that creates the gravitational wake.
For a singular isothermal host, $\rho_h=A_hr^{-2}$, $M_h\propto r$, and $v=B$ is constant. The tidal formula with a satellite $\rho_s=A_sr^{-1}$ gives
$$
r_t\propto r_o^2,
\qquad
M_s\propto r_t^2\propto r_o^4.
$$
Consequently $a_{\rm df}\propto M_s\rho_h/v^2\propto r_o^2$. Since now $j=Br_o$, the torque equation yields
$$
\boxed{\frac{dr_o}{dt}=-C_2r_o^3}.
$$
Its solution satisfies $r_o^{-2}=r_{o0}^{-2}+2C_2t$, so $r_o$ approaches zero only as $t^{-1/2}$ and never arrives in finite time. This is <dynamical-friction stalling by tidal stripping>: the inward tidal field strips the satellite so aggressively that its wake and drag disappear. Observationally, disrupted satellites should deposit stars in streams and the stellar halo, surviving low-mass remnants can remain at finite radii for very long times, and merger times inferred from a constant satellite mass can be severe underestimates.
Solved by gpt-5.6-sol high.
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