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

Question

Aicharya · Accepted Answer

Given,

Escape velocity from earth's surface $= v$

$v = v_{e a r t h} = \sqrt{2 g R}$

$= \sqrt{\frac{2 G M}{\frac{4}{3} π R^{3}} \frac{4}{3} π R^{2}} = \sqrt{\frac{2 G M}{V} \frac{4}{3} π R^{2}} = \sqrt{\frac{8 G ρ π R^{2}}{3}}$

$g =$ Gravitation acceleration on earth surface

$R =$ Radius of earth

$G =$ Universal gravitational constant

Escape velocity of another planet $= ?$

$∵ R_{p} = 4 R$
$&$

$ρ_{p l a n e t} = ρ_{e a r t h}$

So we can write,

$v_{e} \propto \sqrt{R^{2}} \propto R$

$\frac{{(v_{e})}_{e a r t h}}{{(v_{e})}_{p l a n e t}} = \frac{R_{e a r t h}}{R_{p l a n e t}} = \frac{R}{4 R} = \frac{1}{4}$

It means,

$4 {(v_{e})}_{e a r t h} = {(v_{e})}_{P l a n e t}$

${(v_{e})}_{P l a n e t} = 4 v$

The escape velocity from the Earth's surface is v . The escape velocity from the surface of another planet…

Solution