A geostationary satellite is a satellite in geostationary orbit, with an orbital period the same as the Earth’s rotation period. The geostationary orbit is a circular orbit directly above the Earth’s equator.

How high above the Earth’s surface must the geostationary satellite be placed into orbit?

###### Solution.

The gravitational force between the satellite and the Earth is in the radial direction and its magnitude is given by the Newton’s equation

F = GMm/r^{2} (1)

where G is the gravitational constant, M and m are the masses of the Earth and the satellite respectively and r is the radius of the orbit.

In case of the circular motion the net force equals mass times acceleration, where acceleration can be calculated by ω^{2}r, where ω is the angular rate of rotation also known as angular velocity.

Thereby,

F = ma = mω^{2}r. (2)

The angular velocity is given by

ω = 2π/T, (3)

where T is the period for one rotation.

We substitute (3) into the equation (2) and we get

F = m4π^{2}r/T^{2}. (4)

Now we can use the equations (4) and (1) to find the following formula

m4π^{2}r/T^{2} = GMm/r^{2}

or

r^{3} = GMT^{2}/4π^{2}. (5)

We substitute the values and we get

r^{3} = 6.67*10^{-11}*5.972*10^{24}*86400^{2}/4π^{2}

and finally,

r = 4.22*10^{7} m.

The radius of the Earth is 6.37*10^{6} m.

We can calculate the height h above the Earth’s surface by subtracting the radius of the Earth from the radius of the orbit.

h = 4.22*10^{7} – 6.37*10^{6} = 3.583*10^{7} m.

Image Satellites in geostationary orbit By Lookang, many thanks to author of original simulation = Francisco Esquembre author of Easy Java Simulation = Francisco Esquembre – Own work, CC BY-SA 3.0, Link