Geosynchronous orbit derivation
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This is the derivation of the geosynchronous orbital distance for a body in circular orbit around the Earth. It also applies to geostationary orbits. The geosynchronous orbits have the same period as the Earth's rotation.
The rotational period of the Earth is slightly shorter than a day (24 hours), because in one day the Earth does a complete rotation and a little extra due to it also moving round the Sun. Without this extra rotation speed the Sun would not quite appear in the same place at noon from day to day.
A sidereal day is 23 hours, 56 minutes, 4.09 seconds, or 86164.09 seconds.
[edit] Constants and variables
- mass of Earth
- Earth's angular speed
- radius of geosynchronous orbit
- mass of satellite
[edit] Derivation
- (centrifugal force required to maintain circular orbit)
- (force of gravity from body me acting on a body of mass ms)
- (equate and cancel previous formulae)
- (rotational rate in radians per second as a function of v,r)
- (from previous two formulae)
- (rearrangement of the above formula)
This gives the distance of the circular geosynchronous (and hence geostationary) orbit from the centre of the Earth.
[edit] Evaluation and approximation
Evaluation of the above formula yields r = 42 173 531 m (to the nearest meter), with an inaccuracy of about 0.05% arising mostly from the approximations to G and me. Note that this figure is measured from the centre of the Earth and not from the surface - to find the altitude of this orbit about the surface of the Earth we would simply deduct the Earth's radius at that point. This leaves an altitude of around 35 000 000 m (2sf) above the surface.