The energy required to take a satellite to a height ‘h’ above Earth surface (radius of Earth`=6.4xx10^3` km ) is `E_1` and kinetic energy required for the satellite to be in a circular orbit at this height is `E_2`. The value of h for which `E_1` and `E_2` are equal, is:

Options

A. `1.28 xx 10^(4) km`
B. `6.4 xx 10^(3) km`
C. `3.2 xx 10^(3) km`
D. `1.6 xx 10^(6) km`

Answer

C

Explanation

Let m be the mass of satellite and orbital velocity of satellite at height h is
`v_(0) = sqrt((GM_(e))/((R+h)))`
Total energy of satellite at height h,
`T.E. = -(GmM_(e))/((R_(e) + h))`
Total energy of satellite on the earth surface is `-(GmM_(e))/((R_(e)))`
Applying law of conservation of mechanical energy for satellite
`E_(1) - (GM_(e)m)/(R_(e)) = -(GM_(e)m)/((R_(e) + h))`
`E_(1) = (GM_(e)m)/(R(R_(e)+h))`
Kinetic energy of satellite at height h
`E_(2) = (1)/(2) mv_(0)^(2) = (GM_(e)m)/(2(R_(e) + h))`
Since `E_(1) = E_(2)`
`(GM_(e)m)/(R(R_(e) + h)) = (GM_(e)m)/(2(R_(e) + h))`
`h = 3.2 xx 10^(3) km`