Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 321 / 1 / c / ii

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- Solution ii

Solution

For a Keplerian orbit, write

h = K r^{1/2}

, so

h^{'} = h / (2 r)

. Substituting the stated torque laws into

\partial_{r} G - r T = \dot{M} h^{'}

gives

y^{'} + (\frac{1}{2 r} - \frac{β}{3 ν r ^{1/2}}) y = \frac{M ˙}{6 π r} .

With

x = \frac{r}{r _{in}}, λ = - \frac{2 β r _{in}}{3 ν},

the integrating factor is

x^{1/2} e^{λ x}

. The boundary condition

y (1) = 0

then gives

y (x) = \frac{M ˙}{3 πλ} x^{- 1/2} [1 - e^{λ (1 - x)}] .

Thus

f (x) = 1 - x

and

ν Σ = \frac{M ˙}{3 πλ} x^{- 1/2} [1 - exp (λ f (x))] .

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