Question:
a satellite is orbiting earth in a circular orbit of radius R completes 1 revolution in 3 hrs . If orbital radius of geostationary satellite is 42000 km then the value of R is
Answer:
According to Keplers law, we know that (T1/T2)² = (R1/R2)³
Let T1 and R1 represent the time period and radius of the geostationary satellite.
Similarly, T2 and R2 represents the time period and radius of earth.
Hence, T1 = 3 hrs, T2 = 24 hrs, R2 = 42,000 km
(3/24)2 = (R1/ 42000)3
(1/64 )1/3= R1/42000
R1 = 42000/4
R1 = 10500 km
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