• question_answer           The value of acceleration due to gravity at Earth's surface is $9.8m{{s}^{-2}}$. The altitude above its surface at which the acceleration due to gravity decreases to $4.9\,m{{s}^{-2}}$, is close to : (Radius of earth $=6.4\times {{10}^{6}}m$) [JEE Main 10-4-2019 Morning] A) $1.6\times {{10}^{6}}m$   B) $6.4\times {{10}^{6}}m$C) $9.0\times {{10}^{6}}m$               D) $2.6\times {{10}^{6}}m$

$\frac{GM}{{{(R+h)}^{2}}}=\frac{GM}{2{{R}^{2}}}$                         $R+h=\sqrt{2}R$                         $h=(\sqrt{2}-1)R$                         $\simeq 2.6\times {{10}^{6}}m$