The viscous time is
$$
t\sim\frac{R^2}{\bar\nu}
\sim A^{-1}R^{2-m}\Sigma^{-n},
\qquad \Sigma\sim\frac M{R^2}.
$$
With zero stellar torque, total <angular momentum> is conserved and $J\sim M(GM_\star R)^{1/2}$, so $M\propto R^{-1/2}$. Hence
$$
t\propto R^{2-m+5n/2},
$$
and
$$
\boxed{R\propto t^{2/(4-2m+5n)}},
\qquad
\boxed{M\propto t^{-1/(4-2m+5n)}}.
$$
Accreting material loses specific angular momentum; conservation requires a diminishing fraction of the disk to spread outward and carry that angular momentum.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 321 / 2 / a / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-321/2/a/solution