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Geosynchronous orbit derivation

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 revolution 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.

Constants and variables

$G = 6.67 \times 10^{-11}\; \mathrm{m}^3\, \mathrm{kg}^{-1}\, \mathrm{s}^{-2} \;$ gravitational constant

$M_e= 5.98 \times 10^{24}\; \mathrm{kg} \;$ mass of Earth

$\omega = \frac{2 \pi}{86164.09}\; \mathrm{rad s}^{-1} \;$ Earth's angular speed

$r \;$ radius of geosynchronous orbit

$v \;$ orbital speed

$m_s \;$ mass of satellite

Derivation

$F = \frac{m_s v^2}{r}$ (centripetal force required to maintain circular orbit)

$F = \frac{G M_e m_s}{r^2}$ (force of gravity from body

m e

acting on a body of mass

m s

)

$\frac{v^2}{r} = \frac{G M_e}{r^2}$ (equate and cancel previous formulae)

$\omega = \frac{v}{r}$ (rotational rate in radians per second as a function of v,r)

$\omega^2 = \frac{v^2}{r^2} = \frac{G M_e}{r^3}$ (from previous two formulae)

$r = \sqrt[3]\frac{G M_e}{\omega ^2} \approx 42,\!000 \mbox{ km }$ (rearrangement of the above formula)

This gives the distance of the circular geosynchronous (and hence geostationary) orbit from the centre of the Earth.

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 $m e$ .

Last updated: 10-20-2005 04:30:47

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10-26-2009 08:16:03
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