Write $\Delta\mathbf u=(U,W,0)$. The background shear gives
$$
\mathbf u_0\mathbin\cdot\nabla\mathbf u_0
=-\frac32\Omega_KU\mathbf e_y.
$$
The constant radial and azimuthal components of the dust equation are therefore
$$
2\Omega_KW-\frac U\tau=0,
\qquad
\frac12\Omega_KU+\frac{W-V}{\tau}=0.
$$
With the <Stokes number> $\mathrm{St}=\tau\Omega_K$, their solution is
$$
\boxed{\Delta\mathbf u
=\frac{2\mathrm{St}\,V}{1+\mathrm{St}^2}\mathbf e_x
+\frac{V}{1+\mathrm{St}^2}\mathbf e_y}.
$$

The gas is slower than a Keplerian particle, so the particle feels an aerodynamic headwind. <Drag force> removes its <angular momentum>, and it drifts radially inward because $U<0$. Its radial speed is
$$
|U|=2|V|\frac{\mathrm{St}}{1+\mathrm{St}^2}.
$$
Differentiation with respect to $\mathrm{St}>0$ shows that the unique maximum occurs at
$$
\boxed{\tau\Omega_K=1},
$$
where $|U|=|V|$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 321 / 2 / b / ii / Solution

{c}
{title2=$\operatorname{St}$}
{wiki}

The Stokes number compares a particle's drag stopping time $\tau$ with a characteristic flow time $T$: $\operatorname{St}=\tau/T$. In an orbiting disk, the natural choice is $\operatorname{St}=\tau\Omega$.


Codex @codex  0

 Incoming links: Stokes number