A geostationary satellite orbits around the earth in a circular orbit of radius 36,000 km. then the time period of a spy satellite orbiting a few hundred km (600 km) above the earth's surface (R=6400 km) will approximately be

Options

A. `(1)/(2)` n
B. 1 n
C. 2 h
D. 4 h

Answer

c

Explanation

By Kepler’s law of planetary motion,
`T^(2)propr^(3)` , hence `T_(1)^(2)propr_(1)^(3)`, hence `T_(1)^(2)propr_(1)^(3)`
Simiarly `T_(2)^(2)propr_(2)^(3)`
therefore`T_(1)^(2)prop(36000)^(3)`
`T_(2)^(2)prop(6400+h)^(3)`
therefore ,
`(T_(2))^(2)=T_(1)^(2)((6400+h)/(36000))^(3)`
`gtT_(1)^(2)[(6400)/(3600)]^(3)`
`gt(24)^(2)[(8)/(45)]^(3)`
`thereforeT_(2)gt(24xx8xxsqrt(8))/(45xxsqrt(45))`
`gt1.8h`
So , `T_(2)~=2h`