 Solution

For $|z|\ll r$, the gravitational potential has the Taylor expansionHence the vertical gravity is $-{\partial }_z\Phi \simeq -{\Omega }_K^{2}z$, where ${\Omega }_K^2=G{M}_{\star }/r^3$. Vertical hydrostatic equilibrium balances a pressure gradient of order $P/H$ against $\rho {\Omega }_K^{2}H$. Since $c_s^2\sim P/\rho$,The thin disk condition $H/r\ll 1$ is therefore equivalent to a highly supersonic orbital speed.