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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/2018/section-a/2/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 / 2018 / Section A / 2 / Solution by

Codex  0 2026-10-18

The spacecraft clock measures proper time

τ

, while a clock at rest with Earth measures the launch-to-emission interval

t

. Neglect Earth's subsequent motion and gravity, as in the intended special relativity model. The Lorentz factor and time dilation give

γ = \frac{1}{1 - v ^{2} / c ^{2}} = \frac{3}{8}, t = γ τ = \frac{3}{8} (12 h) = 12.7279 h .

In Earth's inertial frame, the spacecraft's displacement at emission is

d = v t

. Therefore

d = \frac{c τ}{8} = 4.24264 light-hours ≃ 4.58 \times 1 0^{12} m .

Here

τ = 43200 s

and the speed of light is

c = 2.99792458 \times 1 0^{8} m s^{- 1}

. The distance is evaluated at the emission event, using simultaneous Earth-frame endpoints, as required by relativity of simultaneity; the return signal has not yet reached Earth. A distance assigned on the spacecraft's simultaneous time slice is different: Earth has moved away at

c /3

for

12 h

in that frame, giving

c τ /3 = 4

light-hours. Specifying the reference frame is essential to the numerical answer.

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