 Solution

ID: past-exam-of-the-physics-1-paper-of-the-university-of-oxford/2013/10/c/ii/solution

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.

 New to topics? Read the docs here!