Physics - Gravitation Question with Solution | TestHub

Escape velocity from the Earth,

${(v}_e)=11.2 \mathrm{k}\mathrm{m} {\mathrm{s}}^{-1}$

Let the mass, radius and density of Earth be $M$, $R$ and $\rho$, respectively, and for the given planet, mass, radius and density are $M^{\prime }$, $R^{\prime }$ and ${\rho }^{\text{'}}$, respectively.

$\therefore$ Escape velocity from the Earth,

$v_e=\sqrt{\frac{2G\times \left(\frac{4}{3}\pi R^3\mathrm{\rho }\right)}{R}}$

$v_e=\sqrt{\frac{8G\pi R^2\mathrm{\rho }}{3} } \dots .\left(\mathrm{i}\right)$

Similarly, escape velocity from the given planet,

${v\prime }_e=\sqrt{\frac{8G\pi {R\prime }^2\mathrm{\rho }}{3} } \dots .\left(\mathrm{i}\mathrm{i}\right)$

Dividing Eq. $\left(\mathrm{i}\right)$ by Eq. $(\mathrm{ii}),$ we get,

$\frac{v_e}{{v^{\prime }}_e}=\sqrt{\frac{8G\pi R^2\mathrm{\rho }}{3} }\times \sqrt{\frac{3}{8G\pi {R^{\prime }}^2\mathrm{\rho }} }=\sqrt{\frac{R^2}{{{\mathrm{4R}}^{\prime }}^2}}$

or $\frac{11.2}{{v^{\prime }}_{\mathrm{e}}}=\frac{R}{2R}$

$\therefore v{\prime }_{\mathrm{e}}=22.4\mathrm{ }\mathrm{k}\mathrm{m} {\mathrm{s}}^{-1}$