Physics - Gravitation Question with Solution | TestHub

Acceleration due to gravity is given by,

$\mathrm{g}=\frac{\mathrm{GM}}{{\mathrm{R}}^2}$

where$\mathrm{G}$is gravitational constant.

$\mathbf{For}\mathbf{ }\mathbf{earth}:\mathrm{ }{\mathrm{g}}_{\mathrm{e}}=\frac{{\mathrm{GM}}_{\mathrm{e}}}{{\mathrm{R}}_{\mathrm{e}}^2}$

$\mathbf{For}\mathbf{ }\mathbf{planet}:\mathrm{ }{\mathrm{g}}_{\mathrm{p}}=\frac{{\mathrm{GM}}_{\mathrm{p}}}{{\mathrm{R}}_{\mathrm{p}}^2}$

$\mathrm{or} \mathrm{ }\frac{{\mathrm{g}}_{\mathrm{e}}}{{\mathrm{g}}_{\mathrm{p}}}=\frac{{\mathrm{M}}_{\mathrm{e}}}{{\mathrm{M}}_{\mathrm{p}}}\mathrm{ }\times \frac{{\mathrm{R}}_{\mathrm{p}}^2}{{\mathrm{R}}_{\mathrm{e}}^2} \dots \left(\mathrm{i}\right)$

$\mathrm{Given}, {\mathrm{M}}_{\mathrm{p}}=2{\mathrm{M}}_{\mathrm{e}}, {\mathrm{R}}_{\mathrm{p}}=2{\mathrm{R}}_{\mathrm{e}}$

Putting the values in the Eq. (i), we obtain

$\mathrm{ }\frac{{\mathrm{g}}_{\mathrm{e}}}{{\mathrm{g}}_{\mathrm{p}}}=\frac{{\mathrm{M}}_{\mathrm{e}}}{2{\mathrm{M}}_{\mathrm{e}}}\times \frac{{\left(2{\mathrm{R}}_{\mathrm{e}}\right)}^2}{{\mathrm{R}}_{\mathrm{e}}^2}=\frac{1}{2}\times \frac{4}{1}=2$

$\therefore \mathrm{ }{\mathrm{g}}_{\mathrm{p}}=\frac{{\mathrm{g}}_{\mathrm{e}}}{2}$