The longest unstable disturbance has mixing length $\ell\sim k_1^{-1}$ and grows on the orbital timescale $\Omega^{-1}$. Its characteristic turbulent velocity is therefore $u_{\rm turb}\sim\Omega\ell$, so the <gravitoturbulent viscosity> estimate is
$$
\bar\nu\sim u_{\rm turb}\ell
\sim k_1^{-2}\Omega.
$$
For $Q\ll1$,
$$
k_1\sim\frac QH
=\frac{\Omega^2}{\pi G\Sigma},
$$
up to an unimportant numerical factor. Hence
$$
\bar\nu\sim\frac{G^2\Sigma^2}{\Omega^3}
\propto r^{9/2}\Sigma^2
$$
in a <Keplerian disk>.

For a disk of characteristic radius $R$ and mass $M_D$, $\Sigma\sim M_D/R^2$. The <viscous timescale> is
$$
t_\nu\sim\frac{R^2}{\bar\nu}
\sim\frac{R^6\Omega^3}{G^2M_D^2}.
$$
Using $\Omega^2=GM_*/R^3$ gives
$$
\boxed{t_\nu\sim
\left(\frac{M_*}{M_D}\right)^2\Omega^{-1}}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 321 / 2 / d / Solution

{title2=$t_\nu$}

The viscous timescale is the radial diffusion time $t_\nu\sim R^2/\nu$ associated with a kinematic viscosity $\nu$ on length scale $R$.


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 Incoming links: Viscous timescale