Put $K(r)=e^{s(r)}$. The <perfect gas> relation in the question becomes the <polytropic equation of state>
$$
P=K(r)\rho^\gamma.
$$
At fixed $r$, vertical hydrostatic balance is
$$
\gamma K\rho^{\gamma-1}\frac{\partial\rho}{\partial z}
=-\rho\Omega_K^2z.
$$
After one integration,
$$
\rho^{\gamma-1}
=\rho_0^{\gamma-1}
-\frac{\gamma-1}{2\gamma K}\Omega_K^2z^2.
$$
Defining the midplane <adiabatic sound speed> by $c_{s0}^2=\gamma P_0/\rho_0=\gamma K\rho_0^{\gamma-1}$ gives
$$
\boxed{\rho=\rho_0
\left(1-\frac{z^2}{H^2}\right)^m},
\qquad
\boxed{m=\frac1{\gamma-1}},
\qquad
\boxed{H^2=\frac{2c_{s0}^2}{(\gamma-1)\Omega_K^2}}.
$$
This is a vertically truncated <polytropic atmosphere>; the physical perfect-gas case has $\gamma>1$.

Solution

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

Let $u>0$ be the inward radial speed. Steady spherical <Bondi accretion> conserves mass:
$$
\dot M=4\pi r^2\rho u.
$$
For $\gamma=5/3$, $p=K\rho^{5/3}$ and the <adiabatic sound speed> satisfies $v_s^2=\gamma p/\rho\propto\rho^{2/3}$. Matching the reservoir therefore gives
$$
\boxed{\rho=\rho_0\left(\frac{v_s}{v_{s0}}\right)^3}.
$$
The Bernoulli integral is
$$
\frac{u^2}{2}+\frac{v_s^2}{\gamma-1}-\frac{GM}{r}
=\frac{3v_{s0}^2}{2}.
$$
Writing the <Mach number> as $\mathcal M=u/v_s$ and eliminating $v_s$ with mass conservation gives
$$
r v_s^2
=\left(\frac{\dot Mv_{s0}^3}{4\pi\rho_0}\right)^{1/2}\mathcal M^{-1/2}.
$$
Multiplication of the Bernoulli equation by $r/2$ now yields
$$
\boxed{A\dot M^{1/2}
\left(\mathcal M^{3/2}+3\mathcal M^{-1/2}\right)=GM+Br},
$$
where
$$
\boxed{A=\left(\frac{v_{s0}^3}{16\pi\rho_0}\right)^{1/2}},
\qquad
\boxed{B=\frac32v_{s0}^2}.
$$

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 314 / 1 / b / Solution

The <adiabatic sound speed> and <Alfvén speed> are
$$
\boxed{v_s^2=\left(\frac{\partial p}{\partial\rho}\right)_s
=\frac{\gamma p}{\rho}},
\qquad
\boxed{\mathbf v_A=\frac{\mathbf B}{\sqrt{\mu_0\rho}},
\quad v_A=\frac{B}{\sqrt{\mu_0\rho}}}.
$$
The entropy is held fixed in the sound-speed derivative.

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 314 / 3 / a / Solution

{title2=$v_s$}
{wiki=Speed_of_sound}

The adiabatic sound speed is $v_s^2=(\partial p/\partial\rho)_s$. For a perfect gas it is $v_s^2=\gamma p/\rho$.


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