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. 4h
B. 2h
C. 1h
D. `1//2h`

Answer

B

Explanation

From Kepler's law, `T^(2)prop r^(3)`
`therefore ((T_(2))/(T_(1)))=((r_(2))/(r_(1)))^(3)`
`r_(1)=36000 km, T_(1)=24` hour
`r_(2)div 6400 km`, we assume that the satellite is orbiting close to the surface of the earth
`therefore (T_(2))/(24)=((6400)/(36000))^(3//2)=((64)/(360))^(3//2)`
`therefore T_(2)=24((8)/(45))^(3//2)div 2h`