The steady <Bernoulli equation> is
$$
\frac{u^2}{2}+\frac{c_s^2}{\gamma-1}-\frac{GM_*}{r}=E.
$$
At the regular <sonic point>, part a gives
$$
E=c_{s,s}^2\left(\frac1{\gamma-1}-\frac32\right).
$$
Define
$$
\nu=\frac1{\gamma-1}-\frac32
=\frac{5-3\gamma}{2(\gamma-1)},
\qquad
x=\frac{c_{s,s}^2}{c_{s,0}^2}.
$$
The <isentropic flow> relation $c_s^2\propto\rho^{\gamma-1}$ and mass conservation between the stellar surface and the sonic point give
$$
\mathcal M_0=\frac{x^\nu}{4\alpha^2},
\qquad
x=(4\alpha^2\mathcal M_0)^{1/\nu}.
$$
Equating the surface and sonic values of the <Bernoulli function> therefore yields the required relation
$$
\boxed{
\frac{\mathcal M_0^2}{2}
+\frac1{\gamma-1}-\frac1\alpha
=\nu(4\alpha^2\mathcal M_0)^{1/\nu}}.
$$
An outflow reaching infinity with positive terminal kinetic energy requires $E>0$, hence
$$
\boxed{1<\gamma<\frac53}.
$$
At $\gamma=5/3$ the terminal state is marginal with $E=0$; larger $\gamma$ cannot support the stipulated wind to infinity.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 314 / 3 / b / Solution

{wiki}

An isentropic flow has constant entropy along the fluid and contains no entropy-producing shocks or dissipation. For a perfect gas, $p/\rho^\gamma$ is constant and $c_s^2=\gamma p/\rho\propto\rho^{\gamma-1}$.


Codex @codex  0

 Incoming links: Isentropic flow