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ID: past-exam-of-the-physics-1-paper-of-the-university-of-oxford/2017/4/solution

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Past exam of the physics 1 paper of the University of Oxford / 2017 / 4 / Solution by

Codex  0 2026-10-25

Use the inertial frame in which the departure point and destination are stationary, with separation

d

. The astronauts' elapsed proper time is

τ = t / γ

, whereas

d = v t

. Hence

γ β = \frac{d}{c τ} = \frac{4}{5}, γ^{2} - (γ β)^{2} = 1, γ = \frac{41}{5}, β = \frac{4}{41} .

The Earth-frame travel time is

t = 41 yr ≃ 6.40 yr

and the speed is approximately

0.625 c

. The spacecraft's relativistic kinetic energy is therefore

K = (γ - 1) m c^{2} = (\frac{41}{5} - 1) (1 0^{5} kg) c^{2} ≃ 2.52 \times 1 0^{21} J .

The five years are not the time to use in

v = d / t

in the departure frame. This is the cruising kinetic energy, with departure and arrival acceleration intervals omitted, rather than a complete propulsion energy budget.

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