Past exam of the physics 1 paper of the University of Oxford / 2016 / 10 / v

Use the supplied planet mass $5.9\times 10^{24}\mathrm{kg}$, $G=6.67430\times 10^{-11}{\mathrm{m}}^3\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{s}}^{-2}$, and adopt mean Earth radius $R_E=6371\mathrm{km}$. For a geostationary orbit, use one sidereal day, $T\simeq 86164\mathrm{s}$. The adopted radius and rotation period follow NASA's fact sheet.^[ref]The slightly smaller radius than the usual terrestrial value follows from using the question's smaller Earth mass. For coplanar prograde orbits, the preceding formulas giveThe escape transfer has $a\to \infty$, so its first tangential burn reaches the local escape speed $\sqrt{2\mu /r_0}$. No final circularization at infinity is needed:Thus the geostationary transfer requires about $0.687\mathrm{km}{\mathrm{s}}^{-1}$ more total impulse, or about 22 percent more, despite leaving the spacecraft gravitationally bound. Reaching the outer radius and circularizing there are separate requirements. Changing the starting orbit's inclination would add a plane-change cost absent from this calculation.