Q 11-07-074JEE MainJEE Main 2024 (30 Jan, Shift 1)Medium
The gravitational potential at a point above the surface of earth is $-5.12\times10^7\ \text{J kg}^{-1}$ and the acceleration due to gravity at that point is $6.4\ \text{m s}^{-2}$. Assume the mean radius of earth to be $6400\ \text{km}$. The height of this point above the earth's surface is:
Answer: (A) $1600\ \text{km}$
At distance $r$ from the centre, $|V| = \dfrac{GM}{r}$ and $g = \dfrac{GM}{r^2}$, so
$$r = \frac{|V|}{g} = \frac{5.12\times10^7}{6.4} = 8\times10^6\ \text{m} = 8000\ \text{km}$$
$$h = r - R = 8000 - 6400 = 1600\ \text{km}$$
Solution by Sreeraj P, M.Sc Physics