Order-of-magnitude vertical balance gives
$$
P\sim\rho\Omega^2H^2,\qquad
F\sim\alpha P\Omega H,\qquad
F\sim\frac{\sigma T^4}{\kappa\rho H},\qquad
\rho\sim\frac{\Sigma}{H}.
$$
Magnetic support also gives $\Omega^2H^2\sim r\Omega T^{1/2}$, so $T\sim\Omega^2H^4/r^2$. Eliminating $P,F,\rho,T$ gives
$$
H^{14}\propto\Sigma^2r^8\Omega^{-5}.
$$
For Keplerian $\Omega\propto r^{-3/2}$,
$$
\boxed{H\propto r^{31/28}\Sigma^{1/7}}.
$$
The <alpha disk> stress implies $\bar\nu\sim\alpha P/(\rho\Omega)\sim\alpha\Omega H^2$, and therefore
$$
\boxed{\bar\nu\propto r^{5/7}\Sigma^{2/7}},
\qquad
\boxed{a=\frac57,\quad b=\frac27}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 321 / 1 / c / ii / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-321/1/c/ii/solution