Question

The radius of the planet is $\dfrac{1}{4}th$ of the earth’s radius and acceleration due to gravity is double to that of the earth. What is the value of escape velocity at the planet as compared to its value on the earth?
A. $\dfrac{1}{{\sqrt 2 }}$
B. $\sqrt 2 $
C. $2$
D. $2\sqrt 2 $

Answer 1

Hint:Escape velocity is the amount of energy an object needs to escape the gravitational clutches from the body. Escape velocity depends on the acceleration due to gravity and radius of the earth. The formula for escape velocity used are ${v_e} = \sqrt {2{g_e}{R_e}} $.

Complete step by step answer:
The escape velocity on Earth is given by the
${v_e} = \sqrt {2{g_e}{R_e}} $ $ \ldots \left( 1 \right)$
where ${g_e}$ is the acceleration due to gravity on Earth and ${R_e}$ is the radius of earth.
According to the question we are given that radius of the planet is $\dfrac{1}{4}th$ to the earth’s radius i.e., ${R_p} = \dfrac{{{R_e}}}{4}$ where ${R_p}$is the radius of the planet.
And acceleration due to gravity of planet is double to the that of earth i.e.,
${g_p} = 2{g_e}$
Using $\left( 1 \right)$
Escape velocity of planet be given by
${v_p} = \sqrt {2{g_p}{R_P}} $
Using the values
${v_p} = \sqrt {2\left( {2{g_e}} \right)\left( {\dfrac{{{R_e}}}{4}} \right)} $
$\Rightarrow{v_p} = \sqrt {{g_e}{R_e}} $ $ \ldots \left( 2 \right)$
Comparing the above two marked equations
$\therefore{v_p} = \dfrac{{{v_e}}}{{\sqrt 2 }}$

So, the correct option is A.

Note:Minimum escape velocity assumes that there is no friction, such as atmospheric drag which would increase the needed instantaneous velocity to escape the gravitational pull. Other form of escape velocity is $v = \sqrt {\dfrac{{2GM}}{R}} $where $G$ is the gravitational constant, $M$ is the mass of the body from which object is trying to escape and $R$ is the distance between center of body and center of mass.