The specific angular momentum of a circular <Keplerian orbit> is $\sqrt{GM_*r}$. Therefore the disk angular momentum is
$$
J=\int_0^R\sqrt{GM_*r}\,\Sigma\,2\pi r\,dr
=\sqrt{GM_*}\,B,
$$
where
$$
\boxed{B=\int_0^Rr^{1/2}\Sigma\,2\pi r\,dr}.
$$
The absence of an external or inner-boundary torque makes $J$, and hence $B$, constant.

From $\bar\nu=Ar^{9/2}\Sigma^2$,
$$
[A]=L^{3/2}M^{-2}T^{-1},
\qquad
[B]=ML^{1/2}.
$$
The combination $AB^2t$ has dimension $L^{5/2}$. <Dimensional analysis> therefore gives
$$
\boxed{R\propto(AB^2t)^{2/5}\propto t^{2/5}}.
$$

For the supplied <similarity solution>, put $\tau=t/t_0$ and $u=\sqrt{r/R(t)}$. Then $r=Ru^2$ and $dr=2Ru\,du$. The total mass is
$$
\begin{aligned}
M_D
&=2\pi\int_0^R\Sigma r\,dr\\
&=4\pi\Sigma_0R_0^{3/2}\tau^{-2/5}R^{1/2}
\int_0^1(1-u)^{1/2}\,du\\
&=\frac{8\pi}{3}\Sigma_0R_0^{3/2}
\tau^{-2/5}R^{1/2}.
\end{aligned}
$$
Since $R^{1/2}=R_0^{1/2}\tau^{1/5}$,
$$
\boxed{M_D=\frac{8\pi}{3}\Sigma_0R_0^2
\left(\frac t{t_0}\right)^{-1/5}}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 321 / 2 / e / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/iii/paper-321/2/e/solution