Time Period of a Satellite

The time period of a satellite is the time it takes to complete exactly one full revolution around the Earth. Consider a satellite of mass m, placed at height h and given the critical (orbital) velocity $v_c$, so it moves in a stable circular orbit of radius $r=R+h$. In one full revolution of period T, the satellite covers a distance equal to the CIRCUMFERENCE of its orbit, $2\pi r$, so its speed can also be written as $v_c=\dfrac{2\pi r}{T}.$

Combining this with the critical-velocity expression derived earlier, $v_c=\sqrt{GM/r}$, gives $\dfrac{2\pi r}{T}=\sqrt{\dfrac{GM}{r}}\quad\Longrightarrow\quad \dfrac{4\pi^2r^2}{T^2}=\dfrac{GM}{r}\quad\Longrightarrow\quad T^2=\dfrac{4\pi^2r^3}{GM}.$ Since $4\pi^2$, G and M are all constants (for orbits around a fixed body), this shows $T^2\propto r^3$ -- exactly Kepler's third law (law of periods), now derived from first principles for artificial satellites rather than merely observed empirically as it was for the planets in section 5.2. Writing $r=R+h$ explicitly gives the working formula $T=2\pi\sqrt{\dfrac{(R+h)^3}{GM}}.$

The period of a satellite does NOT depend on its own mass -- only on the Earth's mass, the Earth's radius, and the height of the orbit; a heavier or lighter satellite at the same height has exactly the same orbital period. As the height of the orbit increases, the period also increases (a higher orbit takes longer to complete). Using $GM=g_h(R+h)^2$ (from the definition of $g_h$ at that height), the period can equivalently be written as $T=2\pi\sqrt{(R+h)/g_h}$. …