spoilers:
$$\frac{T^{2}}{r^{3}}=\frac{4\pi^{2}}{GM} $$and this is because
$$v^{2}=\frac{GM}{r} $$Note that $m_{1}$ represents the mass of the orbiting object, $m_{2}$ represents the mass of the body the object orbits around
But when in circular orbit, $F_{c}=F_{g}$.
$\therefore,$
$$\begin{align} \frac{m_{1}v^{2}}{r}=G \frac{m_{1}m_{2}}{r^{2}} \\ \therefore v^{2}=\frac{Gm_{2}}{r} \\ \text{ Sub in (3).} \\ \left( \frac{2\pi r}{T} \right)^{2}=\frac{Gm_{2}}{r} \\ \frac{4\pi^{2}r^{2}}{T^{2}}=\frac{Gm_{2}}{r} \\ \frac{r^{3}}{T^{2}}=\frac{Gm_{2}}{4\pi^{2}} \\ \therefore \frac{T^{2}}{r^{3}}=\frac{4\pi^{2}}{Gm_{2}} \\ \frac{T^{2}}{r^{3}}=\frac{4\pi^{2}}{GM} \end{align} $$Where:
Elliptical orbit
Equal area swept in equal times - i.e. it goes faster when its closer
$T^{2}\propto r^{3}$ (i.e, $\frac{T^{2}}{r^{3}}=\frac{4\pi^{2}}{GM}$)
plot twists now we can find what the geostationary orbit is, i.e. when it will have a period equal to the rotation of the body
Hence, we get the terms geosynchronous or geostationary, which will be defined at another time
geostationary
If we want to find the distance above the surface, we need to subtract the radius of the orbiting body.