A 3627 kg satellite is geosynchronous orbit around the Earth

A 362.7 kg satellite is geosynchronous orbit around the Earth stays over the same spot and takes 24 hours to make one revolution. Given the following data, determine the satellite’s height above the surface of the Earth. G = 6.67 x 10-11 Nm2/kg2, mass of Earth = 5.98 x 1024 kg, radius of Earth = 6.38 x 106 m.

Solution

A satellite of any mass M while revolving around the earth is under constant action of the gravitational pull which enables the circular motion of the satellite.

We will make use of the expression for the gravitational pull to find the velocity of the satellite for a distance, say r. We will then form a relation with the time taken to orbit the earth, which in turn will be equal to 24 hours and then resolve to find the value of R

So for a satellite going around the earth, we can write:

Mv^2 / R = GMe M/R^2

or, V = (GMe/R)

now the satellite covers a distane of 2R in 24 hours or 86400 seconds.

That is, 2R / 86400 = (GMe/R)

, or R³ = 86400³ GMe / 4²

Putting in the values we get:

R³ = 86400³ x 6.67 x 10^-11 x 5.98 x10^24 / 4 x 3.14 x 3.14 = 7.542 x 10²²

or, R = 4.23 x 10^7 Meters. Now the earth\'s radius is 6.38 x 10^6 meters

Therefore the height of the satellite above the earth\'s surface is 4.23 x 10^7 - 6.38 x 10^6 = 3.59 x 10⁷ Meters