First slide
Keplers 3 Laws
Question

The period of a satellite of the earth very near to the surface of the earth is T0.  What is the approximate period of the geostationary satellite in terms of   T0? (Height of geostationary satellite above the surface of earth is 36000km)

Moderate
Solution

T2α  r3,  R=6400  km

  r = (6400 + 36000) km = 42,400 km

  TT0=rR3/2=4240064003/2    T    77T0

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