Past exam of the physics B course of the University of Cambridge / 2022 / Paper 1 / Section A / A1

The transfer is a Hohmann transfer orbit. Write $r_0$ for the original circular Kepler orbit radius and $a_t$ for the transfer ellipse's semi-major axis. The diagram measures both radii from Earth's centre, giving $r_1=1.2r_0$, and henceThe Hohmann transfer time is half the transfer orbital period, taking the satellite to the opposite end of the elliptic orbit. By Kepler's third law, relative to the original geostationary orbit's orbital period $T_0$,Using one sidereal day, $T_0=86164\mathrm{s}$, gives $t=4.9703\times 10^4\mathrm{s}=13.806\mathrm{h}$; the usual $24\mathrm{h}$ approximation gives $13.844\mathrm{h}$. The arrival impulse changes the velocity into that of the new circular Kepler orbit without adding coast time in the impulsive model.

There is a real discrepancy between the wording and the diagram: an increase in altitude above Earth's surface is different from the radius increase drawn. If altitude is interpreted literally, let $R_E$ be Earth's radius. Thenusing $R_E\simeq 6371\mathrm{km}$ and $r_0\simeq 42164\mathrm{km}$. Thus $13.8\mathrm{h}$ answers the explicitly labelled geometry; $13.5\mathrm{h}$ answers the literal altitude wording.