At a height ' $R$ ' above the earth's surface the gravitational acceleration is ( $\mathrm{R}$ = radius of…
At a height ' $R$ ' above the earth's surface the gravitational acceleration is ( $\mathrm{R}$ = radius of earth, $\mathrm{g}$ = acceleration due to gravity on earth's surface)
g
$\frac{\mathrm{g}}{8}$
$\frac{\mathrm{g}}{4}$
$\frac{\mathrm{g}}{2}$
Solution
Acceleration due to gravity is given by
$\begin{aligned}
& \mathrm{g}^{\prime}=\frac{\mathrm{GM}}{\mathrm{r}^2}=\frac{\mathrm{GM}}{(\mathrm{R}+\mathrm{h})^2}=\frac{\mathrm{GM}}{(2 \mathrm{R})^2} \\
& =\frac{\mathrm{GM}}{4 \mathrm{R}^2}=\frac{\mathrm{g}}{4}
\end{aligned}$