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

The specific angular momentum of a circular Keplerian orbit is $\sqrt{G{M}_{\ast }r}$. Therefore the disk angular momentum iswhereThe absence of an external or inner-boundary torque makes $J$, and hence $B$, constant.

From $\bar{\nu }=A{r}^{9/2}{\Sigma }^2$,The combination $A{B}^{2}t$ has dimension $L^{5/2}$. Dimensional analysis therefore gives

For the supplied similarity solution, put $\tau =t/t_0$ and $u=\sqrt{r/R(t)}$. Then $r=R{u}^2$ and $dr=2Rudu$. The total mass isSince $R^{1/2}=R_0^{1/2}{\tau }^{1/5}$,