Past exam of the physics 1 paper of the University of Oxford / 2013 / 10 / c

Interpret the one-light-day separation $D$ as measured in the station frame at capsule launch, and retain the ship's initial inertial rest frame until reception at the station. The phrase does not explicitly name a frame; an alternative is given below. The ship-frame simultaneous separation is $D^{\prime }=D/{\gamma }_s$. The capsule moves at $0.9c$ towards a station approaching at $0.85c$, so their separation decreases at $1.75c$ in this frame. This is a separation rate, not either object's speed. Hence the elapsed ship-frame coordinate time isIt is the interval assigned using synchronized clocks in the ship frame to the launch and remote arrival. A single onboard clock does not directly mark the remote arrival event; observing its arrival signal would add a light-travel delay. If the stated one-light-day separation were instead measured in the ship frame, the answer would be $\Delta t_s=(1/1.75)\mathrm{day}=0.571429\mathrm{day}$.

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 ${\gamma }_{c/s}=(1-0.9^2{)}^{-1/2}$; thereforeThis uses the station-frame distance convention stated in i. With a one-light-day distance in the ship frame instead, it becomes $\Delta {\tau }_c=\sqrt{0.19}/1.75\mathrm{day}\simeq 0.249080\mathrm{day}$. The uncertainty is a distance-frame convention in the wording, not a missing figure or an unanswered mathematical part.