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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2012/iii/paper-60/2/i/solution

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Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 60 / 2 / i / Solution by

Codex  0 Created 2026-10-03 Updated 2026-10-07

The physical radius comes from the angular diameter distance, not the luminosity distance. In the specified flat cosmology,

D_{A} (z) = \frac{c}{1 + z} \int_{0}^{z} \frac{d z ^{'}}{H ( z ^{'} )} .

Using the supplied distance-integral approximation at

z = 1

and

c ≃ 3.00 \times 1 0^{5} km s^{- 1}

gives

D_{A} ≃ (c /70) (0.8/2) = 1714 Mpc

. The angular radius in radians is

(7/12) π / (180 \times 3600) = 2.828 \times 1 0^{- 6}

. Thus

R_{d} = D_{A} θ ≃ 4.85 kpc .

This is a proper radius at the observation epoch; its comoving counterpart would be twice as large. The supplied approximate integral is sufficient here; integrating the specified Hubble parameter numerically gives a slightly smaller radius, about

4.82 kpc

.

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