What is the radius of a geostationary satellite?  



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Posted on (Answer #1)

A satellite in a geostationary orbit moves directly above the Earth’s radius and takes exactly one day to complete one revolution. There, looking at them from the ground it appears as if they are stationary.

Now to calculate the distance of a geostationary orbit, we have to consider the force of gravity of the Earth that attracts the satellite. This is equal to G*Ms*Me*/R^2, where Ms is the mass of the satellite, Me is the mass of the Earth, R is the distance of the satellite from the Earth and G is the universal gravitational constant and the centripetal force. The centripetal force in terms of the angular speed of the satellite is Ms*w^2*R, where w is the angular speed and R is the orbital radius.

Now for a geostationary satellite, the orbital speed is 2*pi / 86164 rad/s. [Note: here we use the duration of 1 sidereal day, which is equal to 86164 s]

So substituting this, and equating the equations we have on top:

G*Me*/R^2 = w^2*R = (2*pi / 86164) ^2 * R

=> R^3 = G*Me/ w^2

=> R = [G*Me/w^2] ^ (1/3)

Simplifying this using the mass of the Earth, the universal gravitational constant and the angular speed we get R is approximately equal to 35786 km.

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