Q 11-07-137JEE MainJEE Main 2021 (31 Aug, Shift 2)Medium
If $R_E$ be the radius of Earth, then the ratio between the acceleration due to gravity at a depth $r$ below and a height $r$ above the earth surface is: (Given: $r < R_E$)
Answer: (A) $1 + \dfrac{r}{R_E} - \dfrac{r^2}{R_E^2} - \dfrac{r^3}{R_E^3}$
Let $x = \dfrac{r}{R_E}$. Depth: $g_d = g(1 - x)$. Height: $g_h = \dfrac{g}{(1 + x)^2}$.
$$\frac{g_d}{g_h} = (1 - x)(1 + x)^2 = (1 - x)(1 + 2x + x^2) = 1 + x - x^2 - x^3$$
Solution by Sreeraj P, M.Sc Physics