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)
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a
T0/7
b
7T0
c
7T0
d
77T0
answer is D.
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Detailed Solution
T2 α r3, R=6400 km∴ r = (6400 + 36000) km = 42,400 km∴ TT0=rR3/2=4240064003/2 ⇒ T ≈ 77T0