A satellite is moving in a circular orbit at a certain height above the earth's surface. It takes `5.26xx10^3` s to complete one revolution with a centripetal acceleration equal to 9.32 m `s^(-2)`. The height of the satellite orbit above the earth's surface is (Take radius of earth =`6.37xx10^6` m )

Options

A. 70 km
B. 170 km
C. 190 km
D. 220 km

Answer

B

Explanation

As, `T=2pisqrt(((R+h)^3)/(GM))`
`T^2/(4pi^2)=(R+h)^3/(GM)` …(i)
Centripetal acceleration , `a=(GM)/(R+h)^2`
`(R+h)^2/(GM)=1/a`
`(R+h)=T^2/(4pi^2)xxa` [Using (i)]
`=(5.26xx10^3 // 2pi)^2 xx 9.32=6.54xx10^6` m
`therefore h=6.54xx10^6 - R = 6.54xx10^6 - 6.37xx10^6`
`= 0.17xx10^6` m = 170 km