Past exam of the astrophysics course of the University of Oxford / 2022 / 2 / a

There are two different physical interpretations. The supplied logarithmic integral signals free fall in a fixed singular isothermal sphere, with a test particle moving through a frozen background. Write $C={\rho }_0{r}_0^2$. By the shell theorem, the enclosed mass is $M(r)=4\pi Cr$ and the radial equation of motion isThe second equality follows by multiplying by $\dot{r}$ and using the initial rest condition. The free fall trajectory is given implicitly byThus the intended fixed-profile answer isA collapsing material shell gives a different answer. If the entire cloud undergoes pressureless gravitational collapse from rest, the edge follows the fixed enclosed mass $M_0=4\pi {\rho }_0{r}_0^3$, not the instantaneous fixed-profile value $4\pi Cr$. With no shell crossing, $\ddot{r}=-G{M}_0/r^2$, soThis is a concrete counterexample to treating the fixed-profile result as the exact collapse time of the whole nebula. The initial mean density inside $r_0$ is $3{\rho }_0$, consistent with the usual free-fall time of a uniform sphere expressed using that mean density. The logarithmic integral is valid for the explicitly frozen background approximation.