Let $u=-v_r>0$ denote the inward speed. Steady spherical <mass conservation> and the radial <Euler momentum equation>[momentum equation] for an <Isothermal Bondi accretion>[isothermal gas] give
$$
\dot M=4\pi r^2\rho u,
\qquad
u\frac{du}{dr}=-c_s^2\frac{d\log\rho}{dr}-\frac{GM}{r^2}.
$$
Eliminating $d\log\rho/dr$ produces the <Isothermal Bondi equation>
$$
\left(u-\frac{c_s^2}{u}\right)\frac{du}{dr}
=\frac{2c_s^2}{r}-\frac{GM}{r^2}.
$$
A smooth flow can cross <Mach number> one only where both sides vanish. Its <Bondi sonic point> is therefore
$$
u_s=c_s,
\qquad
r_s=\frac{GM}{2c_s^2}.
$$
The integrated <Bernoulli equation> which approaches rest and density $\rho_\infty$ at infinity is
$$
\frac{u^2}{2}+c_s^2\log\frac{\rho}{\rho_\infty}-\frac{GM}{r}=0.
$$
At the <sonic point> this gives $\rho_s=e^{3/2}\rho_\infty$, and hence the unique regular <transonic branch> has
$$
\boxed{\dot M
=4\pi r_s^2\rho_s c_s
=\pi e^{3/2}\frac{G^2M^2\rho_\infty}{c_{s,\infty}^3}.}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 347 / 1 / a / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-347/1/a/solution