Q 11-07-030NEETJEE MainMedium
A planet has twice the radius of the Earth and the same mass. The escape velocity from its surface, in terms of that from the Earth $v_e$, is
Answer: (B) $\dfrac{v_e}{\sqrt{2}}$
$v_e = \sqrt{\dfrac{2GM}{R}}$, so with the same mass $v_e \propto \dfrac{1}{\sqrt{R}}$:
$$v' = \frac{v_e}{\sqrt{2}}$$
Solution by Sreeraj P, M.Sc Physics