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ID: past-exam-of-the-general-relativity-course-of-the-university-of-oxford/2021/3/c/solution

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Past exam of the general relativity course of the University of Oxford / 2021 / 3 / c / Solution by

Codex  0 2026-10-25

For a timelike circular Schwarzschild geodesic, the coordinate orbital period and onboard proper time per orbit are

T_{\infty} = \frac{2 π}{Ω} = 2 π \frac{R ^{3}}{GM}, T_{crew} = \int_{0}^{T_{\infty}} \frac{d t}{u ^{0}} = T_{\infty} 1 - \frac{3 GM}{R c ^{2}} .

Using

G = 6.67430 \times 1 0^{- 11} m^{3} k g^{- 1} s^{- 2}

and

c = 299792458 m s^{- 1}

, the Schwarzschild radius and orbital areal radius are

R_{s} = 2 GM / c^{2} ≃ 1.21789 \times 1 0^{10} m

and

R = 200 R_{s} ≃ 2.43578 \times 1 0^{12} m

. Therefore

T_{\infty} T_{crew} ≃ 1.02100 \times 1 0^{6} s = 11.8172 days, = T_{\infty} \frac{397}{400} ≃ 1.01717 \times 1 0^{6} s = 11.7728 days .

The crew accumulates about

3836

fewer seconds, or

63.9

fewer minutes, per orbit. Both gravitational time dilation and kinematic time dilation contribute: a static clock at

R

has

d T_{static} = 1 - R_{s} / R d t

, while the orbiting clock additionally has

d τ = d T_{static} / γ

, where

γ = 398/397

. A distant observer measures the coordinate period; comparing arrivals emitted at the same orbital phase also removes the repeated light propagation delay.

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