 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ii/paper-1/15d/a/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

The radial equation is

\overset{r}{¨} = - \frac{GM}{r ^{2}} + \frac{Λ c ^{2}}{3} r .

Multiplying by

\overset{r}{˙}

and integrating gives

\frac{1}{2} \overset{r}{˙}^{2} - \frac{GM}{r} - \frac{Λ c ^{2}}{6} r^{2} = E .

Since

ρ

M = \frac{4 π r ^{3}}{3 c ^{2}} ρ .

r = a x_{0}

, divide the energy equation by

r^{2} /2

\frac{2 E}{x _{0}^{2}} = - k c^{2} .

Then

(\frac{a ˙}{a})^{2} = \frac{8 π G}{3 c ^{2}} ρ - \frac{k c ^{2}}{a ^{2}} + \frac{Λ c ^{2}}{3} .

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