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Inside the uniform sphere, $V_c^2=3{\sigma }^2{r}^2/r_0^2$. Thus, with $X=V_c/(\sqrt{2}\sigma )$, part (b) givesThe Gaussian integral needed for the lowered Maxwellian velocity distribution isSubstitution $v=\sqrt{2}\sigma y$ therefore yieldsSince $V_c<v_e$ throughout the core, the integrand is positive and so is $I(X)$ for $X>0$. Using $m{n}_0={\rho }_0$, part (a)'s massive-test-body drag magnitude isThe slow drift of a circular orbit is determined by torque, not by equating its changing circular speed to the tangential drag. Here $V_c=\omega r$, with constant $\omega =\sqrt{4\pi G{\rho }_0/3}$. The specific angular momentum $j=r{V}_c=\omega r^2$ obeysHence the positive local decay timescale isThere is a genuine factor-of-two discrepancy with the requested denominator $32{\pi }^2$. That denominator would give $V_c/(2D)$, which is the timescale for a flat galaxy rotation curve, not this uniform-density core. It also follows from the incorrect replacement ${\dot{V}}_c=-D$, which drops the $V_c\dot{r}$ term in $d(r{V}_c)/dt$. The literal signed definition $r/(2\dot{r})$ is negative for an inward drift; it equals $-{\tau }_{+}$. A positive decay time requires the absolute value or a minus sign.

This is the slow circular inspiral under tangential drag calculation. Its assumptions require $D$ small enough for nearly circular adjustment and a meaningful large Coulomb logarithm in stellar dynamics; the leading-log approximation is not uniform at $V_c\to 0$. Also, integrating the truncated distribution all the way to $v_e$ does not give exactly $n_0$. The model uses $n_0$ as the stated core normalization; exact self-consistency of a sharply truncated uniform core with that distribution is not implied.