Question #1905

Obtain an expression for the time period of a satellite orbiting the earth. At what altitude should a satellite be placed for its orbit to be geosynchronous?

Expert's answer

When satellite is moving over Earth with circular orbit of radius r, the gravitation force treating it is

** GmM/r**^{2}, where G - gravity constant, m - satellite mass, M - Earth mass.

Due to second Newton's Law gravitation force = centripetal force**mv**^{2}/r.

Hence we obtain the expression for the velocity of satellite:

**v=(G M/r)**^{1/2}

Period T_{сп} = circumference of the orbit, **2π r,** divided by the satellite velocity **v**:

**Tсп=2π r/v=2π (r**^{3}/GM)^{1/2}

A geosynchronous orbit (sometimes abbreviated GSO) is an orbit around the Earth with an orbital period that matches the Earth's sidereal rotation period.The synchronization of rotation and orbital period means that for an observer on the surface of the Earth, the satellite appears to constantly hover over the same meridian (north-south line) on the surface, moving in a slow oscillation alternately north and south with a period of one day, so it returns to exactly the same place in the sky at exactly the same time each day

Earth's sidereal rotation period = 23 hours 56 min 4.1 s=86164.1 s

**Altitude H=(T*R*√(G)/2π)**^{2/3} - R

where R-Earth radius, R=6370 km

H=35 786km

Due to second Newton's Law gravitation force = centripetal force

Hence we obtain the expression for the velocity of satellite:

Period T

A geosynchronous orbit (sometimes abbreviated GSO) is an orbit around the Earth with an orbital period that matches the Earth's sidereal rotation period.The synchronization of rotation and orbital period means that for an observer on the surface of the Earth, the satellite appears to constantly hover over the same meridian (north-south line) on the surface, moving in a slow oscillation alternately north and south with a period of one day, so it returns to exactly the same place in the sky at exactly the same time each day

Earth's sidereal rotation period = 23 hours 56 min 4.1 s=86164.1 s

where R-Earth radius, R=6370 km

H=35 786km

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