Q 11-07-176JEE MainJEE Main 2019 (9 Jan, Shift 2)Medium
The energy required to take a satellite to a height $h$ above the Earth surface (radius of Earth $= 6.4\times10^3\ \text{km}$) is $E_1$, and the 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
Answer: (C) $3.2\times10^3\ \text{km}$
Energy to lift the satellite (change in potential energy):
$$E_1 = -\frac{GMm}{R+h} + \frac{GMm}{R} = \frac{GMm\,h}{R(R+h)}$$
Kinetic energy in orbit at radius $R+h$:
$$E_2 = \frac{GMm}{2(R+h)}$$
Setting $E_1 = E_2$: $\dfrac{h}{R} = \dfrac12$, so $h = \dfrac{R}{2} = 3.2\times10^3\ \text{km}$.
Solution by Sreeraj P, M.Sc Physics