For a body of mas ' $\mathrm{m}$ ', the acceleration due to gravity at a distance ' $R$ ' from the surface…
For a body of mas ' $\mathrm{m}$ ', the acceleration due to gravity at a distance ' $R$ ' from the surface of the earth is $\left(\frac{g}{4}\right)$. Its value at a distance $\left(\frac{R}{2}\right)$ from the surface of the earth is $(R=$ radius of the earth, $g$ = acceleration due to gravity)
$\left(\frac{\mathrm{g}}{8}\right)$
$\left(\frac{9 \mathrm{~g}}{4}\right)$
$\left(\frac{4 \mathrm{~g}}{9}\right)$
$\left(\frac{\mathrm{g}}{2}\right)$
Solution
$\mathrm{g}=\frac{\mathrm{GM}}{\mathrm{r}^2}$ where $\mathrm{r}$ is the center of the earth
In the first case, $r_1=R+R=2 R$
In the second case, $r_2=R+\frac{R}{2}=\frac{3}{2} R$
$\begin{aligned}
& \therefore \frac{\mathrm{g}_2}{\mathrm{~g}_1}=\left(\frac{\mathrm{r}_1}{\mathrm{r}_2}\right)^2=\frac{16}{9} \\
& \mathrm{~g}_2=\frac{16}{9} \mathrm{~g}_1=\frac{16}{9} \times \frac{\mathrm{g}}{4}=\frac{4}{9} \mathrm{~g}
\end{aligned}$