The radius of the planet is double that of the earth, but their average densities are same.…
The radius of the planet is double that of the earth, but their average densities are same. $\mathrm{V}_{\mathrm{p}}$ and $V_E$ are the escape velocities of planet and earth respectively. If $\frac{V_p}{V_E}=x$, the value of ' $x$ ' is
$\frac{1}{4}$
$\frac{1}{2}$
2
4
Solution
Escape velocity is given by,
$\begin{aligned}
\mathrm{v}_{\mathrm{c}} & =\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}} \\
& =\sqrt{\frac{2 \mathrm{G}}{\mathrm{R}} \times \frac{4}{3} \pi \mathrm{R}^3 \rho}=\sqrt{\frac{8 \mathrm{G}}{3} \pi \mathrm{R}^2 \rho}=\sqrt{\frac{8 \mathrm{G} \pi \rho}{3}} \times \mathrm{R}
\end{aligned}$
As the planets have the same density;
$\begin{array}{ll}
& \mathrm{V}_{\mathrm{e}} \propto \mathrm{R} \\
\therefore \quad & \frac{\mathrm{~V}_{\mathrm{P}}}{\mathrm{~V}_{\mathrm{E}}}=\frac{\mathrm{R}_{\mathrm{P}}}{\mathrm{R}_{\mathrm{E}}}=\frac{2 \mathrm{R}}{\mathrm{R}}=2 \\
\therefore \quad & \frac{\mathrm{~V}_{\mathrm{P}}}{\mathrm{~V}_{\mathrm{E}}}=2 \\
\therefore \quad & \mathrm{X}=2
\end{array}$