Q 11-07-042JEE MainJEE Main 2026 (4 Apr, Shift 2)Easy
The height in terms of radius of the earth $(R)$, at which the acceleration due to gravity becomes $\dfrac{g}{9}$, where $g$ is acceleration due to gravity on earth's surface, is ______.
Answer: (C) $2R$
Concept. Above the surface, $g$ falls with the square of the distance from the centre: $g_h=\dfrac{g}{(1+h/R)^2}$. This formula is exact; the shortcut $g(1-2h/R)$ only works for $h\ll R$.
Step 1. Set up. $\dfrac{g}{(1+h/R)^2}=\dfrac g9\Rightarrow\left(1+\dfrac hR\right)^2=9$.
Step 2. Solve. $1+\dfrac hR=3\Rightarrow h=2R$.
Answer. $2R$ (Option C). Distance from the centre is $3R$; $g$ falls by $3^2$.
Solution by Sreeraj P, M.Sc Physics
