A geostationary satellite orbites around the earth in a circular orbit of radius `36000` km Then, the time period of a spy satellite orbiting a few hundred km above the earth's surface `(R_(e)=6400km)` will approximately be

Options

A. 1h
B. 2h
C. 24h
D. 36 h

Answer

B

Explanation

By kepler's law of planetary motion
`T^(2) alpha r^(3)`
`T^(2) alpha (36000)^(3) and T^(2) alpha (6400 + h)^(3)`
Therefore `(T')^(2) = T^(2)((6400 +h)/(36000))`
`gt T^(2)((6400)/(36000))^(3)" "(because h lt lt R)`
`gt(24)^(2) ((8)/(45))^(3)T gt (24 xx 8 xx sqrt(8))/(45sqrt(45)) gt 1.8 h`
So, ` T ~~ 2h`