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. `(2h)/((R+2h))`
B. `(2h)/((2R+3h))`
C. `(R)/(R+h)`
D. `(2R)/(2h+R)`

Answer

A

Explanation

`E=Delta U=(mgh)/(1+h//R)=(mghR)/((R+h))`
E' = energy of satellite - energy of satellite on surface of earth
`=-(GMm)/(2(R+h))-(-(GMm)/(R))`
`=mgR^(2)[(1)/(R)-(1)/(2(R+h))]=(mg(R+2h)R)/(2(R+h))`
Now, `(E)/(E')=(2h)/((R+2h))`.