Question

A planet has twice the radius but the mean density is $\frac{1}{4}th$ as compared to earth. What is the ratio of escape velocity from to that from the planet?

• A. 3 : 1
• B. 1 : 2
• C. 1 : 1
• D. 2 : 1

Answer 1

The correct option is C 1 : 1

The minimum velocity with which a body must be projected in the gravitational pull is known as escape velocity.
Escape velocity from earth's surface is
$V_{es}=\sqrt{\frac{2G{M}_e}{R_e}}$

$=\sqrt{\frac{2G\cdot \frac{4}{3}\pi R_e^3{d}_e}{R_e}}(\because M=\frac{4}{3}\pi R_e^3{d}_e)$

or $v_{es}\propto \sqrt{d_e}\times R_e$ ....(i)
Similarly, for a planet
$v_{es}^{\prime }\propto \sqrt{d_p}\times R_p$ ...(ii)
so, $\frac{v_{es}}{v_{es}^{\prime }}=(\frac{d_e}{d_p}{)}^{1/2}\times \frac{R_e}{R_p}$

Given, $d_p=\frac{1}{4}{d}_e,R_p=2R_e$

$\frac{v_{es}}{v_{es}^{\prime }}=\left(\frac{d_e}{\frac{d_e}{4}}\right)\times \frac{R_e}{2R_e}$
$=(4{)}^{1/2}\times \frac{1}{2}$
$=2\times \frac{1}{2}=1$

So, $=\frac{v_{es}}{v_{es}^{\prime }}=1:1$