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
$$
\boxed{A=\left(\frac{v_{s0}^3}{16\pi\rho_0}\right)^{1/2}},
\qquad
\boxed{B=\frac32v_{s0}^2}.
$$

Solution

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

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A sonic point is a point at which the flow speed equals the local <sound speed>, so the <Mach number> is one.

Sonic point

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{title2=$\mathcal M$}
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The Mach number is the ratio of flow speed to local sound speed.


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 Incoming links: Mach number