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

Table of contents
- Solution i

Solution

In steady state, mass conservation makes the outward flux constant,

F = \dot{M}

. Angular-momentum conservation then gives

\dot{M} h + G = constant .

The zero-torque condition at

r_{out}

fixes the constant to

\dot{M} h_{out}

, so

G = \dot{M} (h_{out} - h) .

For a Kepler orbit,

h = GM r

and

d Ω/ d r = - 3Ω/ (2 r)

. The viscous torque is consequently

G = 3 π \overset{ν}{ˉ} Σ h

. Equating the two expressions yields the steady decretion disk structure

\overset{ν}{ˉ} Σ = \frac{M ˙}{3 π} (\frac{r _{out}}{r} - 1) .

Solved by gpt-5.6-sol high.

 Ancestors (11)

 View article source

 Discussion (0)

There are no discussions about this article yet.

 Articles by others on the same topic (0)

There are currently no matching articles.

  See all articles in the same topic Create my own version