The escape velocity from the Earth's surface is v . The escape...

Given,

Escape velocity from earth's surface $= v$

$v = {v}_{earth} = \sqrt{2gR}$

$= \sqrt{\dfrac{2GM}{\dfrac{4}{3}\pi{R}^{3}}\dfrac{4}{3}\pi{R}^{2}} = \sqrt{\dfrac{2GM}{V}\dfrac{4}{3}\pi{R}^{2}} = \sqrt{\dfrac{8Gρ\pi{R}^{2}}{3}}$

$g =$ Gravitation acceleration on earth surface

$R =$ Radius of earth

$G =$Universal gravitational constant

Escape velocity of another planet $= ?$

$∵ {R}_{p} = 4R \text{ and }$

${ρ}_{planet} = {\text{ρ}}_{earth}$

So we can write,

${v}_{e} ∝ \sqrt{{R}^{2}} ∝ R$

$\dfrac{{\left({v}_{e}\right)}_{earth}}{{\left({v}_{e}\right)}_{planet}} = \dfrac{{R}_{earth}}{{R}_{planet}} = \dfrac{R}{4R} = \dfrac{1}{4}$

It means,

$4{\left({v}_{e}\right)}_{earth} = {\left({v}_{e}\right)}_{Planet}$

${\left({v}_{e}\right)}_{Planet} = 4v$