Set
$$
D(r)=b+(\sqrt b+\sqrt r)^2
=r+2\sqrt{br}+2b,
\qquad
\Psi(r)=-\Phi(r)=\frac{GM}{D(r)}.
$$
The spherical <Poisson equation> gives
$$
\rho(r)=\frac1{4\pi G r^2}
\frac d{dr}\left(r^2\frac{d\Phi}{dr}\right)
=\frac{M[6b^{3/2}\sqrt r+10br+3\sqrt b\,r^{3/2}]}
{8\pi r^2D^3}.
$$
The numerator identity
$$
6b^{3/2}\sqrt r+10br+3\sqrt b\,r^{3/2}
=3\sqrt{br}\,D+4br
$$
splits this into two <scale-free density-potential components>:
$$
\boxed{
\rho(r)=
\frac{3\sqrt b}{8\pi G^2M}r^{-3/2}\Psi^2
+\frac{b}{2\pi G^3M^2}r^{-1}\Psi^3}.
$$
Thus
$$
(\gamma_1,p_1,A_1)
=\left(\frac32,2,\frac{3\sqrt b}{8\pi G^2M}\right),
$$
$$
(\gamma_2,p_2,A_2)
=\left(1,3,\frac{b}{2\pi G^3M^2}\right).
$$

At small radius, $D\sim2b$ and the first term dominates:
$$
\rho(r)\sim\frac{3M}{32\pi b^{3/2}}r^{-3/2}.
$$
The cusp has finite enclosed mass because $r^2\rho\sim r^{1/2}$. Moreover,
$$
v_c^2=r\frac{d\Phi}{dr}
=\frac{GMr[1+\sqrt{b/r}]}{D^2}
\sim\frac{GM}{4b^{3/2}}\sqrt r,
$$
so the <circular speed> tends to zero as $r^{1/4}$. At large radius, $D\sim r$ and
$$
\rho(r)\sim\frac{3M\sqrt b}{8\pi}r^{-7/2}.
$$
Since $r^2\rho\sim r^{-3/2}$ is integrable at infinity and $\Phi\sim-GM/r$, the total mass is finite and equals $M$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 320 / 2 / a / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-320/2/a/solution