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 Incoming links: Past exam of the physics 1 paper of the University of Oxford / 2013 / 10 / c / i

Past exam of the physics 1 paper of the University of Oxford / 2013 / 10 / c / ii / Solution 2026-10-25

Launch and arrival occur on the capsule's own worldline, so its clock measures the proper time. In the ship frame its Lorentz factor is

γ_{c / s} = (1 - 0. 9^{2})^{- 1/2}

; therefore

Δ τ_{c} = \frac{Δ t _{s}}{γ _{c / s}} = Δ t_{s} 1 - 0. 9^{2} ≃ 0.131211 day ≃ 3.1491 h .

This uses the station-frame distance convention stated in i. With a one-light-day distance in the ship frame instead, it becomes

Δ τ_{c} = 0.19 /1.75 day ≃ 0.249080 day

. The uncertainty is a distance-frame convention in the wording, not a missing figure or an unanswered mathematical part.

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Past exam of the physics 1 paper of the University of Oxford / 2013 / 10 / d / Solution 2026-10-25

The velocity-addition formula combines the ship's station-frame velocity and the capsule's ship-frame velocity:

\frac{v _{c}}{c} = \frac{0.85 + 0.9}{1 + 0.85 ( 0.9 )} = \frac{350}{353} ≃ 0.9915014.

For the distance convention in i, station-frame arrival takes

D / v_{c} = (353/350) day = 1.008571 day

. Transforming its two endpoint events back to the ship frame gives

γ_{s} [D / v_{c} - 0.85 D / c]

, agreeing with i. The capsule remains below the speed of light in either frame.

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