Past exam of the physics course of the year 1 of the natural sciences course of the University of Cambridge / 2015 / Paper 1 / Section A / A2

For a circular orbit, $v^2/r=GM/r^2$. The relation for circular orbital speed from surface gravity eliminates the planet's gravitational parameter: $GM=g{R}^2$. The orbital radius is $r=R+h=7.4\times 10^6$ m, soThis is about 7.4 km/s, under the circular-orbit and spherical-Earth approximation. Using surface $g$ unchanged at orbital height would overestimate the centripetal acceleration; its role here is to determine $GM$.