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Graph

spoilers:

$$\frac{T^{2}}{r^{3}}=\frac{4\pi^{2}}{GM} $$

and this is because

$$v^{2}=\frac{GM}{r} $$

Kepler's Law derivation (very useful)(ccp approved)

$$\begin{align} &F_{c}=\frac{m_{1}v^{2}}{r} \\ &F_{g}=G \frac{m_{1}m_{2}}{r^{2}} \\ &v=\frac{2\pi r}{T} \end{align} $$

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:

Kepler's Law

  1. Elliptical orbit

  2. Equal area swept in equal times - i.e. it goes faster when its closer

  3. $T^{2}\propto r^{3}$ (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

Earth's geostationary orbit

$$\begin{align} \frac{T^{2}}{r^{3}}&=\frac{4\pi^{2}}{GM} \\ \therefore r^{3}&=\frac{GMT^{2}}{4\pi^{2}} \\ r&=\sqrt[3]{\frac{GMT^{2}}{4\pi^{2}}} \\ &=\sqrt[3]{\frac{(6.67\times10^{-11})(5.97\times 10 ^{24})(24\times60\times 60)^{2}}{4\pi^{2}}} \\ &=4.22\times 10^{7}m \end{align} $$

If we want to find the distance above the surface, we need to subtract the radius of the orbiting body.