What is the height from the surface of earth, where acceleration due to gravity will be $\frac{1}{4}$ of…
What is the height from the surface of earth, where acceleration due to gravity will be $\frac{1}{4}$ of that of the earth? $\left(R_E=6400 \mathrm{~km}\right)$
6400 km
3200 km
1600 km
640 km
Solution
Acceleration due to gravity at height $h$ is
$\begin{aligned}
& g_h=\frac{g}{(R+h)^2} \Rightarrow g_h \propto \frac{1}{(R+h)^2} \\
& \frac{g_{h_2}}{g_{h_1}}=\left(\frac{\mathrm{R}+\mathrm{h}_1}{R+h_2}\right)^2 \Rightarrow \frac{\frac{g}{4}}{g}=\left(\frac{6400+0}{6400+h_2}\right)^2 \\
& \Rightarrow \frac{1}{2}=\frac{6400}{6400+h_2} \Rightarrow h_2=6400 \mathrm{~km}
\end{aligned}$