= Solution
Steady spherical <mass conservation> gives $4\pi r^2\rho u=\dot M$. For the <polytropic equation of state>, $c_s^2=dp/d\rho$, so
$$
\frac{\rho'}\rho=-\frac{u'}u-\frac2r,
\qquad
\frac{p'}\rho=c_s^2\frac{\rho'}\rho.
$$
Substitution into the radial <Euler equations for an inviscid fluid> gives the <Parker wind equation>
$$
\boxed{
\left(u-\frac{c_s^2}{u}\right)\frac{du}{dr}
=\frac{2c_s^2}{r}-\frac{GM_*}{r^2}}.
$$
At a <sonic point>, the coefficient of $du/dr$ vanishes. A smooth <transonic branch> can pass it only if the right-hand side vanishes simultaneously. Hence
$$
\boxed{u_s=c_{s,s},
\qquad
r_s=\frac{GM_*}{2c_{s,s}^2}}.
$$
Back to article page
