 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-oxford/2017/prelims/mathematics-v/3/b/iii/solution

The core contributes no change in

φ

, since

φ^{'} = 0

there. Integrating the radial gravitational field across the shell, with

u = r / r_{0}

, gives

φ (2 r_{0}) - φ (0) = - 4 π G ρ_{0} r_{0}^{2} \int_{1}^{2} \frac{2/ e - ( u + 1 ) e ^{- u}}{u ^{2}} d u .

Using the supplied antiderivative,

\int_{1}^{2} \frac{2/ e - ( u + 1 ) e ^{- u}}{u ^{2}} d u = [- \frac{2}{e u} + \frac{e ^{- u}}{u}]_{1}^{2} = (- \frac{1}{e} + \frac{1}{2 e ^{2}}) - (- \frac{2}{e} + \frac{1}{e}) = \frac{1}{2 e ^{2}} .

Hence

φ (2 r_{0}) - φ (0) = - \frac{2 π}{e ^{2}} G ρ_{0} r_{0}^{2}, α = - \frac{2 π}{e ^{2}} .

The negative sign agrees with

φ^{'} (r) < 0

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