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?
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/r2 (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 ω2r, where ω is the angular rate of rotation also known as angular velocity.
F = ma = mω2r. (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 = m4π2r/T2. (4)
Now we can use the equations (4) and (1) to find the following formula
m4π2r/T2 = GMm/r2
r3 = GMT2/4π2. (5)
We substitute the values and we get
r3 = 6.67*10-11*5.972*1024*864002/4π2
r = 4.22*107 m.
The radius of the Earth is 6.37*106 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*107 – 6.37*106 = 3.583*107 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