In steady state, mass conservation makes the outward flux constant, $\mathcal F=\dot M$. Angular-momentum conservation then gives
$$
\dot Mh+\mathcal G=\text{constant}.
$$
The zero-torque condition at $r_{\rm out}$ fixes the constant to $\dot Mh_{\rm out}$, so
$$
\mathcal G=\dot M(h_{\rm out}-h).
$$
For a <Kepler orbit>, $h=\sqrt{GMr}$ and $d\Omega/dr=-3\Omega/(2r)$. The viscous torque is consequently $\mathcal G=3\pi\bar\nu\Sigma h$. Equating the two expressions yields the steady <decretion disk> structure
$$
\boxed{\bar\nu\Sigma=\frac{\dot M}{3\pi}
\left(\sqrt{\frac{r_{\rm out}}r}-1\right)}.
$$

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 321 / 1 / b / i / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-321/1/b/i/solution