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

The spacecraft clock measures proper time $\tau$, 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 giveIn Earth's inertial frame, the spacecraft's displacement at emission is $d=vt$. ThereforeHere $\tau =43200\mathrm{s}$ and the speed of light is $c=2.99792458\times 10^8\mathrm{m}{\mathrm{s}}^{-1}$. 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 $12\mathrm{h}$ in that frame, giving $c\tau /3=4$ light-hours. Specifying the reference frame is essential to the numerical answer.