Q 11-07-025NEETJEE MainMedium
At what height above the Earth's surface is the acceleration due to gravity equal to its value at a depth of $\dfrac{R}{2}$, where $R$ is the radius of the Earth?
Answer: (A) $(\sqrt{2} - 1)R$
Concept. $g_d = g\left(1 - \dfrac dR\right)$ and $g_h = g\dfrac{R^2}{(R + h)^2}$.
Step 1. Depth. $g_d = \dfrac g2$.
Step 2. Height. $(R + h)^2 = 2R^2 \Rightarrow h = (\sqrt2 - 1)R$.
Answer. $(\sqrt2 - 1)R$ (Option A).
Solution by Sreeraj P, M.Sc Physics