Solution
= Solution
The linear term $-(3/2)\Omega_Kx\mathbf e_y$ is the local <Keplerian shear> in the <shearing sheet>. The remaining constant
$$
V=r_0[\Omega(r_0)-\Omega_K(r_0)]
$$
is the gas's azimuthal velocity relative to the local Keplerian frame. The outwardly decreasing midplane pressure found in part (a)(iii) makes the gas <Sub-Keplerian>, so
$$
\boxed{V<0},
\qquad
|V|\sim\frac{c_s^2}{r_0\Omega_K}
\sim\Omega_Kr_0\left(\frac H{r_0}\right)^2.
$$