The escape velocity of a body from the surface of the earth is $11.2 \mathrm{~km} / \mathrm{s}$. The escape…
The escape velocity of a body from the surface of the earth is $11.2 \mathrm{~km} / \mathrm{s}$. The escape velocity of a body from a planet having same mean density as the earth but twice the radius of earth is
$5.5 \mathrm{~km} / \mathrm{s}$
$33.6 \mathrm{~km} / \mathrm{s}$
$22.4 \mathrm{~km} / \mathrm{s}$
$11.2 \mathrm{~km} / \mathrm{s}$
Solution
For escape velocity the formula is,
$v_e=\sqrt{\frac{2 G M}{R}}=\frac{\sqrt{2 G\left(\frac{4}{3} \pi R^3\right) \rho}}{R}=\sqrt{\frac{8}{3} G \pi R^2 \rho}$
Where, $\rho$ is density of the planet and $\mathrm{R}$ is its radius.
$\begin{aligned} & \therefore v_e=R \sqrt{\frac{8}{3} G \pi \rho} \\ & \therefore v_e \propto R\end{aligned}$
So, the escape velocity on the planet would be $22.4 \mathrm{~km} / \mathrm{s}$.