Question

The escape velocity of a body on an imaginary planet which has half the radius of earth and double the average density of earth, is (ve is escape velocity on earth)
​

Answer 1

Answer:

$Escape \: Velocity \:from \: imaginary \: planet \: (v_{p}) = \frac{v_{e}}{ \sqrt{2} }$

Explanation:

$= > \rho = \frac{M}{V} \\ \\ = > M = \rho V \\ \\ = > M = \rho \times\frac{4}{3} \pi {R}^{3} \\ \\ = > M = \frac{4}{3} \rho \pi {R}^{3}\\ \\ \\ Escape \: Velocity \: from \: earth \: (v_{e} )=\sqrt{ \frac{2GM}{R} } \\ \\ = > v_{e} = \sqrt{ \frac{2G \times\frac{4}{3} \rho \pi {R}^{3}}{R} } \\ \\ = > v_{e} = \sqrt{ \frac{2G \times4 \rho \pi {R}^{2}}{3} } \\ \\ = > v_{e} = \sqrt{ \frac{8G \rho \pi {R}^{2}}{3} } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: ....Eq_{1}\\ \\ \\ \rho = average \: density \: of \: earth \\ R = radius \: of \: earth \\ G = gravitational \: constant$

$According \: to \: the \: question \: on \: an \: imaginary \: planet \\ \\ R_{p} = \frac{1}{2} R \\ \rho_{p} = 2\rho \\ \\ Escape \: Velocity \:from \: imaginary \: planet \: (v_{p} ) = \sqrt{ \frac{8G \rho_{p} \pi {R_{p}}^{2}}{3} } \\ \\ = > v_{p} = \sqrt{ \frac{8G \times 2 \rho \pi { (\frac{1}{2} R)}^{2}}{3} } \\ \\ = > v_{p} = \sqrt{ \frac{8G \times 2 \rho \pi \times { \frac{1}{4} R}^{2}}{3} } \\ \\ = > v_{p} = \sqrt{ \frac{8G \times \rho \pi \times { \frac{1}{2} R}^{2}}{3} } \\ \\ = > v_{p} = \sqrt{ \frac{4G \rho \pi { R}^{2}}{3} } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: ....Eq_{2}$

$Dividing \: Eq_{1} \: by \: Eq_{2} , \: we\: get \\ \\ = > \frac{v_{e}}{v_{p}} = \frac{ \sqrt{ \frac{8G \rho \pi {R}^{2}}{3} }}{\sqrt{ \frac{4G \rho \pi { R}^{2}}{3} }} \\ \\ = > \frac{v_{e}}{v_{p}} = \sqrt{ \frac{8G \rho \pi {R}^{2}}{3} } \times \sqrt{ \frac{3}{4G \rho \pi { R}^{2}} } \\ \\ = > \frac{v_{e}}{v_{p}} = \sqrt{2} \\ \\ = > v_{p} = \frac{v_{e}}{ \sqrt{2} }$

Answer 2

Answer:

Given:

Planet has ½ radius and double density as that of Earth.

To find :

Escape Velocity in that planet in terms of escape Velocity of Earth.

Concept:

First we need to express Escape Velocity in terms of density and radius. This can be easily done by using mass as a product of volume and density .

$\boxed{ \red{ \huge{esc. \: v = \sqrt{ \dfrac{2Gm}{r} } }}}$

Considering planets to be spheres and having a constant density :

$\red{ \large{esc. \: v = \sqrt{ \dfrac{2G (\frac{4}{3} \pi {r}^{3} \times \rho)}{r} } }}$

$\red{ \large{ = > esc. \: v = \sqrt{ \dfrac{8 G \pi {r}^{2} \times \rho}{3} } }}$

Keeping all other quantities as constant , we can say that :

$\boxed{ \blue{ \huge{esc. \: v \: \propto \: (r \sqrt{ \rho} )}}}$

Calculation:

Taking ratio :

$= > \dfrac{v_{p}}{v_{e}} = \dfrac{ \dfrac{r}{2} \sqrt{2 \rho} }{r \sqrt{ \rho} }$

$= > v_{p} = \dfrac{v_{e}}{ \sqrt{2} }$

So final answer :

$\boxed{ \boxed{ \green{ \huge{ \sf{ \bold{ v_{p} = \dfrac{v_{e}}{ \sqrt{2} } }}}}}}$