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 $p=K\rho^{1+1/m}$. <Hydrostatic equilibrium> in the uniform gravitational field gives
$$
\frac{dp}{dz}=-\rho g.
$$
Substitution and integration with $\rho=0$ at $z=0$ gives
$$
\rho=\rho_0\left(\frac{-z}{H}\right)^m,
\qquad z\leq0,
$$
where $H$ fixes the normalization. Integrating the hydrostatic equation from $z$ to the free surface gives the pressure directly:
$$
p(z)=g\int_z^0\rho(z')\,dz'
=\boxed{\frac{\rho_0gH}{m+1}
\left(\frac{-z}{H}\right)^{m+1}}.
$$
This indeed has $p\propto\rho^{1+1/m}$ and $p(0)=0$, as required for a <polytropic atmosphere>.

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 314 / 2 / a / Solution

{wiki=Polytrope}

A polytropic atmosphere obeys $p=K\rho^{1+1/m}$ for polytropic index $m$. Under uniform vertical gravity and a free surface, its density and pressure are powers of depth below that surface.


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