= Solution
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.
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