Past exam of the short options course of the University of Oxford / 2014 / 7 / c / ii / Solution

Put $a=\omega T$. At emission, $s=2T$, the middle segment givesThe planet is at $x_p=c/\omega$, so the inward light signal takes coordinate timeIts arrival time on the planet is therefore $t_r=2(e^a-1)/\omega$. Subtract this from the mission's return time $4\sinh a/\omega$:For large $T$ at fixed acceleration this tends to $2/\omega =2c/(c\omega )$. With the Earth-gravity value above, the wait is approximately two years, despite the millions of planet-years elapsed since departure in part (i).