Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 315 / 2 / d

Table of contents
- Solution d

Solution

For constant density

ρ_{c}

,

m (r) = \frac{4 π}{3} ρ_{c} r^{3}, \frac{d P}{d r} = - ρ_{c} \frac{G m ( r )}{r ^{2}} = - \frac{4 π G ρ _{c}^{2}}{3} r .

Imposing

P (R_{p}) = 0

and using

g = G M_{p} / R_{p}^{2} = 4 π G ρ_{c} R_{p} /3

gives

P (r) = \frac{2 π G ρ _{c}^{2}}{3} (R_{p}^{2} - r^{2}) = \frac{1}{2} ρ_{c} g R_{p} (1 - \frac{r ^{2}}{R _{p}^{2}}) .

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