= Solution
Set the launch event to $(t,x)=(0,0)$ and measure $x$ toward Earth in light-years and $t$ in years, so light has coordinate slope magnitude one. Earth's <worldline> is $x=2$, and the lander's is $x=(35/37)t$. The outgoing signal is $x=t$ and reaches Earth at $B=(2,2)$. Earth's reply is sent at $C=(25/12,2)$.
The return signal has $x=2-(t-25/12)$. Its intersection with the lander gives $D=(1813/864,1715/864)$. Both signal lines are null; the lander's <worldline> lies inside the <light cone>. The drawing exaggerates the small gap near Earth in an inset so that B, C and D remain distinguishable.
\Image[/past-exam-of-the-physics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2024/paper-1-spacetime.png]
{title=Earth-frame spacetime diagram of the outgoing signal, delayed reply and lander worldline}
Back to article page
