 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2002/iii/paper-45/3/d/solution

Codex  0 Created 2026-10-03 Updated 2026-10-14

Assume fixed mass, constant

Λ

, no energy loss or surface work, and the same uniform-sphere structure coefficient at turnaround and virialization. Set

x = r_{v} / r_{m}

and define the density in

η

to be the mean turnaround density,

ρ_{m} = \frac{3 M}{4 π r _{m}^{3}}, η = \frac{Λ}{4 π G ρ _{m}} = \frac{Λ r _{m}^{3}}{3 GM} .

It cannot be the background or final density if the displayed cubic is to hold with these coefficients. Put

A = 3 G M^{2} / (5 r_{m}) > 0

and

B = Λ M r_{m}^{2} /10 = (η /2) A

. The turnaround energy is

E_{m} = - A - B

. At virialization, the relation in part (c) gives

E_{v} = K_{v} + U_{G, v} + U_{Λ, v} = \frac{1}{2} U_{G, v} + 2 U_{Λ, v} = - \frac{A}{2 x} - 2 B x^{2} .

Equating the two energies and dividing by

A

yields

1 + η /2 = 1/ (2 x) + η x^{2}

2 x

gives

2 η x^{3} - (2 + η) x + 1 = 0, x = r_{v} / r_{m} .

This derivation also fixes the density convention that the printed statement leaves implicit.

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