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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.


CONSTANTS AND VARIABLES

:
G = 6.67 imes 10^{-11}\; \mathrm{m}^3\, \mathrm{kg}^{-1}\, \mathrm{s}^{-2}
\;   Gravitational Constant

:
M_e= 5.98 imes 10^{24}\; \mathrm{kg}
\;   mass of Earth

:
\omega = rac{2 \pi}{86164.09}\; \mathrm{rad}\, \mathrm{s}^{-1}
\;   Earth's Angular Speed

:r \;   radius of geosynchronous orbit

:v \;   Orbital Speed

:
m_s \;   mass of satellite


DERIVATION

:
F = rac{m_s v^2}{r}
  ( Centrifugal Force required to maintain circular orbit)

:
F = rac{G M_e m_s}{r^2}
  (force of gravity from body m_e acting on a body of mass m_s)

:
rac{v^2}{r} = rac{G M_e}{r^2}
  (equate and cancel previous formulae)

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

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

:
r = \sqrt {Link without Title} rac{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. 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.