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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2015/iii/paper-58/2/b/solution

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Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 58 / 2 / b / Solution by

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

Assemble the star from concentric shells. A shell of mass

d m

r

has interaction energy

- G m (r) d m / r

with the already assembled interior. This counts each gravitational pair once. Hence the Newtonian gravitational potential energy is

Ω = - \int_{0}^{R} \frac{G m ( r )}{r} 4 π ρ r^{2} d r = - \frac{16 π ^{2} G ρ ^{2}}{3} \int_{0}^{R} r^{4} d r .

Using

M = 4 π ρ R^{3} /3

gives

Ω = - \frac{3 G M ^{2}}{5 R} .

The negative sign expresses gravitational binding. A factor of one half should not be inserted again: the shell assembly already avoids double counting.

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