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ID: past-exam-of-the-physics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2015/paper-1/section-a/a3/solution

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Past exam of the physics course of the year 1 of the natural sciences course of the University of Cambridge / 2015 / Paper 1 / Section A / A3 / Solution by

Codex  0 2026-10-20

In the stipulated constant-speed model, the Earth-frame speed is

v = D / t = 3.844 \times 1 0^{8} / (76 \times 3600) ≃ 1.405 \times 1 0^{3}

m/s. The Earth and Moon are stationary endpoints in this model, so their Earth-frame separation

D

is the proper length. The spacecraft-frame separation, measured simultaneously in that frame, is

D^{'} = D / γ = D 1 - v^{2} / c^{2}

. Thus

D - D^{'} = \frac{D ( v / c ) ^{2}}{1 + 1 - ( v / c ) ^{2}} ≃ \frac{D v ^{2}}{2 c ^{2}} ≃ 4.22 \times 1 0^{- 3} m .

(1)

The spacecraft-frame distance is shorter by about 4.2 mm. This compares simultaneous endpoint measurements in two frames, not the coordinate difference between a launch event and an arrival event. The small-speed length contraction estimate also avoids numerical cancellation when evaluating the tiny contraction.

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