Physics - Gravitation Question with Solution | TestHub

The value of acceleration due to gravity changes with height $(ie, \mathrm{a}\mathrm{l}\mathrm{t}\mathrm{i}\mathrm{t}\mathrm{u}\mathrm{d}\mathrm{e})$. If $g\prime$ is the acceleration due to gravity at a point, at height $h$ above the surface of earth, then

$g^{\prime }=\frac{GM}{{\left(R+h\right)}^2}$

$\mathrm{b}\mathrm{u}\mathrm{t},\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }g=\frac{\mathrm{G}\mathrm{M}}{R^2}$

$\therefore \frac{g^{\prime }}{g}=\frac{GM}{{\left(R+h\right)}^2}\times \frac{R^2}{GM}=\frac{R^2}{{\left(R+h\right)}^2}$

$\mathrm{H}\mathrm{e}\mathrm{r}\mathrm{e},\mathrm{ }\mathrm{ }\mathrm{ }{\mathrm{g}}^{\mathrm{\prime }}=\frac{GM}{{\left(R+h\right)}^2}=\frac{GM}{{\left(R+3R\right)}^2}$

$=\frac{GM}{{\left(4R\right)}^2}=\frac{GM}{{16R}^2}=\frac{g_e}{16}$

           $\Rightarrow \frac{g_e}{g^{\text{'}}}=16$