In the chapter CIRCULAR MOTION it was shown that the force needed to keep an object moving in circular motion is called centripetal force and it is given by:
$$\begin{aligned} \mathrm{F = \frac {mv^2}{r}} \end{aligned}$$
The centripetal force can be exerted by friction of car tyres on the road, a string , magnetic and electric fields, etc... In this case the centripetal force is gravity, so that:
$$\begin{aligned} \mathrm{\frac {mv^2}{r} = \frac {GmM}{r^2}} \end{aligned}$$
Where m is the mass of the planet and M is the mass of the Sun
The smaller mass m cancels out and we get:
$$\begin{aligned} \mathrm{v = \sqrt\frac {GM}{r}} \end{aligned}$$
This is the condition for a circular orbit. If the velocity is different from the one determined by this formula, the orbit will be elliptic.
Example calculation
What is the velocity of the International Space Station (ISS), which orbits at a height of 400 km?
The mass of the Earth is 5.97 * 1024 kg.
The distance to the ISS is the radius of the Earth plus the height . Remember that for these calculations we consider all the mass of the Earth to be concentrated in a point at its centre. So,
r = 6.37 *106 m + 4*105 m = 6.77 *106 m
Assuming that it is a circular orbit, as it is the case for Earth´s artificial satellites, we have:
$$\begin{aligned} \mathrm{v = \sqrt\frac { 6.67 * 10^{-11}* 5.97 * 10^{24}}{6.77*10^6}=7.66*10^{3}m/s} \end{aligned}$$
Deducing Kepler´s third law from Newton gravity equation
Again referring to our circular motion lesson, we recall that the velocity of the circular motion is given by the length of the circumference (2πr) divided by the time taken to go through it, which is the period (T):
$$\begin{aligned} \mathrm{v = \frac{2\pi r}{T}} \end{aligned}$$
Equating to the formula above for speed:
$$\begin{aligned} \mathrm{v = \frac{2\pi r}{T}= \sqrt\frac {GM}{r}} \end{aligned}$$
If we square both sides so that we get T2 like in Kepler´s law , and solve for T, we get:
$$\begin{aligned} \mathrm{T^2= \left(\frac {4\pi^2}{GM}\right) r^3} \end{aligned}$$
This is Kepler´s third law. It is even better, because we have a value for the constatnt of proportionality.
This equation is very useful to relate the radius of orbits with their periods and , as a result, orbital speeds. Let´s look at this examples:
Example calculation
What is the height of a geostationary orbit?
Satellites in geostationary orbits move at the same speed that the Earth rotates so that they seem to be stationary to an observer in the surface of the Earth. this is very useful for communications because signals can be relayed and the satellite aerials on the ground always point in the same direction so that its position only have to be adjusted once.
The period (T) of the satellite is 24 hs which is 86400 s.
$$\begin{aligned} \mathrm{r=\sqrt[3]{ \left(\frac {GM}{4\pi^2}\right) T^2}} \end{aligned}$$
r = 4.2*107 m
To get the satellite height we need to subtract the radius of the Earth (spreadsheet with planetary data here), to get:
h = 35900 km
This is a very high orbit and it is expensive to place satelçlites in there. Another disadvantage is the time taken for the signals to travel back and forth. That is why companies like Starlink are placing communication satellites in LEO (Low Earth Orbit), which is around 550 km. The drawback is that you need lots of satellites to operate in LEO´s, in fact, thousand of them...