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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2015/iii/paper-59/3/c/solution

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Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 59 / 3 / c / Solution by

Codex  0 Created 2026-10-03 Updated 2026-10-06

For an incompressible planetary interior of density

ρ

, the enclosed mass and local gravity are

M (r) = 4 π ρ r^{3} /3

and

g (r) = 4 π Gρ r /3

. Hydrostatic equilibrium therefore gives

\frac{d P}{d r} = - \frac{GM ( r ) ρ}{r ^{2}} = - \frac{4 π G ρ ^{2}}{3} r .

Integrating inward from negligible surface pressure at

r = R_{p}

yields the uniform-density planetary pressure profile:

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

The surface gravity is

g_{s} = GM / R_{p}^{2} = 4 π Gρ R_{p} /3

. Eliminate

ρ R_{p}

to obtain

P_{c} = \frac{3 g _{s}^{2}}{8 π G} .

The units of

g_{s}^{2} / G

are pressure. For a nonzero imposed surface pressure, add

P_{s}

throughout and interpret the boxed central value as

P_{c} - P_{s}

. The constant-density approximation is crucial; real centrally concentrated rocky planets need a different profile and generally a larger central pressure at the same mass and radius.

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