Spherical symmetry reduces the <Poisson equation> for gravity to
$$
\frac1{r^2}\frac d{dr}\left(r^2\Phi_{\rm dm}'\right)
=4\pi G\Psi_0r^{-5/2}.
$$
After one integration,
$$
r^2\Phi_{\rm dm}'=8\pi G\Psi_0r^{1/2}.
$$
The omitted integration constant would represent a point mass and is absent for the dark-matter configuration alone. Choosing the additive constant so that the potential tends to zero at large radius gives
$$
\boxed{\Phi_{\rm dm}(r)=-\frac{16\pi G\Psi_0}{\sqrt r}}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 314 / 3 / a / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-314/3/a/solution