Through the use of re-arranging the above equation, we can come to the equation:
r³ = G (m2) T² / 4π²
We know that (m2) is the mass of the earth at 5.98×10^24 kg, T is the time period and G the universal gravitation constant at 6.67 x10^-11 kg^-2 .
Radius Of A Geostationary Orbit
We know every bit of information in the above equation to work out the radius of a geostationary orbit. The time period will be 24 hours which is 86400 seconds. Therefore, for a geostationary orbit,
r = 4.23×10^7 metres.
However this is the radius to from the center of the Earth. Therefore, we will need to deduct the radius of the Earth from this number: the height of the satellite from Earth = r – r(E) where r is the distance of the satellite from the center of the Earth and r(E) is the radius of the Earth.
From this, the height of a geostationary orbit above the earth is 3.6×10^7 meters, with a radius of orbit of 4.23×10^7 m (from the center of the earth).
Summary
- A geostationary orbit is an orbit which is fixed in respect to a position on the Earth. Therefore, the time period will always be 24 hours.
- From combining the centripetal force, gravitational force and basic velocity force equations, we can deduce that the radius required for a geostationary orbit is 3.6×10^7 meters from the surface of the earth.
