The escape velocity from the Earth’s surface is v. The escape velocity from the surface of another planet having radius four times that of Earth and the same mass density is:

(1) v                     (2) 2v
(3) 3v                   (4) 4v

Solution

${v_e} = \sqrt {\frac{{2GM}}{R}}  \propto \sqrt {\frac{M}{R}} $

$\therefore {v_e} \propto \sqrt {\frac{{\rho .\frac{4}{3}\pi {R^3}}}{R}} $

$ \therefore {v_e} \propto R$ since mass density $\rho $ is same

$\frac{{{{({v_e})}_{Planet}}}}{{{{({v_e})}_{Earth}}}} = \frac{{{R_{Planet}}}}{{{R_{Earth}}}}$

$ \Rightarrow \frac{{{{({v_e})}_{Planet}}}}{v} = \frac{{4{R_{Earth}}}}{{{R_{Earth}}}} = 4$

$ \therefore {({v_e})_{Planet}} = 4v$

Answer: (4)