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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.

Chandrasekhar dynamical friction is the drag exerted by the overdense gravitational wake that a massive body creates in a background of lighter particles. It transfers orbital energy and angular momentum to the host, causing satellites, star clusters, and massive black holes to spiral inward and promoting galaxy mergers. In a homogeneous isotropic Maxwellian background its force iswhere $\log \Lambda$ is the Coulomb logarithm in stellar dynamics. Its scaling can be reconstructed from the strong-deflection impact parameter $b_{\mathrm{cr}}\sim GM/v^2$: the encountered mass rate is $\sim \rho v\pi b_{\mathrm{cr}}^2$, and multiplying by momentum change $\sim v$ gives $F_{\mathrm{df}}\sim G^2{M}^2\rho /v^2$.

For a spherical host with a flat galaxy rotation curve,This is a singular isothermal sphere with one-dimensional dispersion $\sigma =v_c/\sqrt{2}$, so $X=1$. DefineFor a constant-mass satellite on a circular orbit, the drag magnitude and its torque areIt follows thatThe quadratic radius dependence and inverse mass dependence explain why massive nearby satellites merge much faster than light or distant ones.

Let the satellite also have a flat internal rotation curve of speed $v_s$. Equating its edge density to the host density gives the tidal radiusBecause $M_s(<R_t)=v_s^2{R}_t/G$, the bound mass decreases linearly:Under the question's literal closure that the derivative of the remaining satellite's total orbital angular momentum equals the frictional torque,and henceIf stripped material is explicitly assigned the satellite's instantaneous specific orbital angular momentum, the balance for the bound remnant is instead $M_{s}d(v_{c}r)/dt=-r{F}_{\mathrm{df}}$; that convention gives $t_{\mathrm{df},\mathrm{strip}}=2t_{\mathrm{df},0}$. Both treatments show the robust point: tidal stripping weakens the drag as the orbit shrinks and substantially delays coalescence. In less idealized profiles the mass can fall faster than linearly, producing dynamical-friction stalling by tidal stripping.

For either component $j\in \{h,s\}$ and $0<\gamma <3$, direct integration of the power law density givesLet $B_j=4\pi G{A}_j/(3-\gamma )$. Since $d{\Phi }_j/dr=G{M}_j/r^2=B_j{r}^{1-\gamma }$,The additive constants depend on the chosen reference and cannot generally be set by requiring $\Phi \to 0$ at infinity for an untruncated power law.

For a satellite on a circular orbit, the linearized effective radial force gives the tidal radiusHere ${\Omega }^2=G{M}_h(<r_o)/r_o^3=B_h{r}_o^{-\gamma }$ and ${\Phi }_h^{\prime \prime }=(1-\gamma ){\Omega }^2$, soSubstitution of the two mass profiles yieldsand equivalently

Now set $\gamma =1$. ThenThe host circular speed is $v=B\sqrt{r_o}$ with $B=\sqrt{2\pi G{A}_h}$. The magnitude of Chandrasekhar dynamical friction becomesa constant. The specific angular momentum is $j=r_{o}v=B{r}_o^{3/2}$. Since the drag torque gives $dj/dt=-r_o{a}_{\mathrm{df}}$,and thereforeIntegration giveswhich 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 $a_{\mathrm{df}}=a_0{r}_{o0}^2/r_o^2$, where $a_0$ is the constant acceleration in the stripped calculation. The same torque equation givesThusWithout 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, ${\rho }_h=A_h{r}^{-2}$, $M_h\propto r$, and $v=B$ is constant. The tidal formula with a satellite ${\rho }_s=A_s{r}^{-1}$ givesConsequently $a_{\mathrm{df}}\propto M_s{\rho }_h/v^2\propto r_o^2$. Since now $j=B{r}_o$, the torque equation yieldsIts solution satisfies $r_o^{-2}=r_{o0}^{-2}+2C_{2}t$, 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.