= Slow circular inspiral under tangential drag
{title2=$\dot r=-\frac{rD}{(1+q)V_c},\quad q=d\log V_c/d\log r$}
Let a weak tangential deceleration $D>0$ act on a test body moving along a slowly evolving family of <circular orbits>. Write its <circular speed> as $V_c(r)>0$ and its <specific angular momentum> as $j=rV_c(r)$. The specific <torque> is $\dot j=-rD$. Defining $q=d\log V_c/d\log r$ gives
$$
\frac{dj}{dr}=(1+q)V_c,\qquad
\dot r=-\frac{rD}{(1+q)V_c}.
$$
This derivation follows angular-momentum balance; simply setting $\dot V_c=-D$ would omit the <angular momentum> changed by radial migration.
An adiabatic inward drift along radially stable circular orbits requires $1+q>0$. Indeed, with $\Omega=V_c/r$, the <radial epicyclic frequency> satisfies
$$
\kappa^2=\frac1{r^3}\frac{d(j^2)}{dr}=2(1+q)\Omega^2.
$$
For this stable branch, the positive local timescale defined as $r/(2|\dot r|)$ is $(1+q)V_c/(2D)$. It is $V_c/D$ in a uniform-<mass density> core with $q=1$, and $V_c/(2D)$ for a flat <galaxy rotation curve> with $q=0$.
At $q=-1$, the circular-orbit <angular momentum> is locally independent of <radius>, so division by $1+q$ fails. For $q<-1$, the algebraic drift formula points outward, but those circular orbits are radially unstable; it is not a stable slow-inspiral solution. The quasistatic approximation also requires the drift time to exceed an orbital time. Finally, this local timescale is not automatically the remaining time to the center, which requires integrating the radius-dependent drift.
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