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

Table of contents
- Solution e

Solution

For

r ≪ GM / B

, the Mach number tends to a constant

M_{1}

. Since

A \dot{M}_{crit}^{1/2} = GM /4

, its algebraic equation is

\overset{m}{˙}^{1/2} (M_{1}^{3/2} + 3 M_{1}^{- 1/2}) = 4,

or equivalently

\overset{m}{˙} \frac{( M _{1}^{2} + 3 ) ^{2}}{M _{1}} = 16 .

The central Bernoulli balance gives

v_{s}^{2} \sim \frac{2 GM}{( M _{1}^{2} + 3 ) r},

so

v_{s} \propto r^{- 1/2}, ρ \propto v_{s}^{3} \propto r^{- 3/2} .

Finally

u = M_{1} v_{s}

. Using the algebraic relation to rewrite its coefficient gives

u \sim (\frac{m ˙ M _{1}^{3}}{4})^{1/4} (\frac{GM}{r})^{1/2} .

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