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Describe various reason for perturbation of satellite orbit calculate maximum daily eclipse duration for GEO orbit.

This question appears in Mumbai University > Satellite Communication and Network subject

Marks: 10 M

Year: Dec 2013, May 2013

1 Answer
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  • A satellite moving in the equatorial orbit at a height of 35786 km above the earth in the same direction as the earth’s surface with a forward velocity of 3.07km/sec will complete one orbit in about 24 hours (precisely 23 hours, 56 min and 4 seconds). The satellite will therefore appear stationary with respect to earth and hence called a geostationary satellite and the orbit is referred to as geostationary orbit or Clarke orbit.
  • There exists only one geo stationary orbit.
  • Three satellites in a geostationary orbit can cover the entire globe except the Polar Regions.
  • A geosynchronous satellite is one which has a period of rotation equal to the sidereal period of rotation $(T_p)$ of its primary body which is Earth. For Earth the sidereal period is 23 hours 56 minutes and 4 seconds.
  • The following point differ geostationary and geosynchronous orbit
  • The geostationary orbit is a special case of geosynchronous orbit.
  • The geosynchronous orbit may be inclined, polar or may be elliptical but geostationary orbit is always circular.
  • Geostationary orbit is always stationary but geosynchronous need not be always stationary.

Maximum daily eclipse duration for GEO orbit:

  • The sun is considered to be at infinity with respect to earth. So the earth’s shadow becomes a cylinder of constant diameter. The maximum shadow angles at equinoxes is given by:

$∅_max=180°-2 cos^{(-1)} \frac{r_e}{r_s}$

$∅_max= 180°-2 cos^{(-1)} (\frac{6378.15}{42164.2})=17.4°$

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  • Geostationary satellite period is 24 hours, the maximum shadow angle is equivalent to maximum daily eclipse duration $(τ_{max})$.

$τ_{max}= \frac{17.4°}{360°} ×24=1.16 h$

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