Orbital Velocity of a Satellite

Consider a satellite of mass $m$ moving in a circular orbit of radius $r$ (measured from the Earth's centre) around the Earth. For the orbit to be circular and steady, the net force acting on the satellite must be exactly the centripetal force required for circular motion at that radius and speed -- and here, that entire centripetal force is supplied by nothing but the Earth's own gravitational pull on the satellite:

Cancelling one factor of $m$ from both sides and solving for the orbital speed $v_o$:

For a satellite skimming very close to the Earth's surface, so that $r \approx R$, this becomes

using $g = GM/R^2$ exactly as before. Comparing this with the escape velocity of Section 7.10, $v_e = \sqrt{2GM/R}$, the two expressions differ by exactly a factor of $\sqrt{2}$:

a genuinely useful relation, since it means finding either speed immediately gives the other. Physically, it says a satellite already coasting in a near-surface circular orbit would need its speed boosted by a factor of only $\sqrt{2} \approx 1.414$ (not, say, doubled) to break free of Earth's gravity entirely and escape to infinity.

Orbital period. For a satellite moving at constant speed $v_o$ around a circle of circumference $2\pi r$, its time period is simply

$T = \frac{2\pi r}{v_o} = \frac{2\pi r}{\sqrt{GM/r}} = 2\pi\sqrt{\frac{r^3}{GM}}$ …