As $r\to0$, the right-hand side of the Mach-number equation tends to $GM$. Since $f(\mathcal M)\geq4$, a real positive solution at arbitrarily small $r$ requires
$$
4A\dot M^{1/2}\leq GM.
$$
Using the value of $A$ gives
$$
\dot M\leq\frac{(GM)^2}{16A^2}
=\boxed{\frac{\pi\rho_0(GM)^2}{v_{s0}^3}}
=\dot M_{\rm crit}.
$$
This is the <critical Bondi accretion rate for gamma equals five thirds>.

Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-314/1/d/solution