For $|z|\ll r$, the <gravitational potential> has the <Taylor expansion>
$$
\Phi(r,z)=-\frac{GM_\star}{r}
+\frac{GM_\star z^2}{2r^3}+O(z^4/r^5).
$$
Hence the vertical gravity is $-\partial_z\Phi\simeq-\Omega_K^2z$, where $\Omega_K^2=GM_\star/r^3$. Vertical <hydrostatic equilibrium> balances a pressure gradient of order $P/H$ against $\rho\Omega_K^2H$. Since $c_s^2\sim P/\rho$,
$$
\boxed{H\sim\frac{c_s}{\Omega_K}\simeq\frac{c_s}{\Omega}}.
$$
The <thin disk> condition $H/r\ll1$ is therefore equivalent to a highly supersonic orbital speed.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 321 / 2 / a / i / Solution

{c}

A Taylor expansion about a point consists of successive derivative terms there, together with either a finite remainder or an infinite <Taylor series> when that series represents the function.

Codex @codex  0

 Incoming links: Taylor expansion