Question

The escape velocity on a planet, four times the radius of the earth and having $9$ times acceleration due to gravity is
(A) $67.2\,km\,{s^{ - 1}}$
(B) $37.4\,km\,{s^{ - 1}}$
(C) $403.2\,km\,{s^{ - 1}}$
(D) $422.2\,km\,{s^{ - 1}}$

Answer 1

Hint: Escape velocity is the minimum speed needed for a body to escape the gravitational pull of the Earth
without any propulsion. We know, escape velocity of planet Earth ${v_e}$ is ${v_e} = \sqrt {2gR} $ .
Where, $g$ is the acceleration due to gravity at the surface of Earth. And $R$ is the radius of Earth.
The escape velocity for Earth is ${v_e} = 11.2\,km\,{s^{ - 1}}$ . Substitute values of the radius and acceleration due to
gravity of the other planet to get the answer.

Complete step by step solution:
As we know that the escape velocity of the Earth is given as:
${v_e} = \sqrt {2gR} $
Here, ${v_e}$ denotes the escape velocity of planet Earth.
$g$ is the acceleration due to gravity of Earth;
$R$ is the radius of Earth.
As, ${v_e} = 11.2\,km\,{s^{ - 1}}$
Therefore, we have:
\[\sqrt {2gR} = 11.2\,km\,{s^{ - 1}}\] …………………(1)
As per the given condition, for the given planet we have:
Radius of the planet, \[{R_0} = 4R\]
Acceleration due to gravity on the planet, \[{g_0} = 9g\]
Let the escape velocity of the planet be: \[{v_p}\] , substituting the values of radius and acceleration due to
gravity, we get
\[{v_p} = \sqrt {2 \times {g_0} \times {R_0}} \]
\[ \Rightarrow {v_p} = \sqrt {2 \times 9g \times 4R} \]
\[ \Rightarrow {v_p} = \sqrt {2gR\left( {36} \right)} \]
\[ \Rightarrow {v_p} = 6\sqrt {2gR} \]
But from equation \[1\] , we have \[\sqrt {2gR} = 11.2\,km\,{s^{ - 1}}\]
\[\therefore {v_p} = 6\left( {11.2} \right)\]
\[ \Rightarrow {v_p} = 67.2\,km\,{s^{ - 1}}\]
This is the escape velocity on the planet.

Therefore, option A is the correct option.

Note: The escape velocity on Earth is ${v_e} = 11.2\,km\,{s^{ - 1}}$ . This is the average escape velocity of the Earth as the radius of Earth and the acceleration due to gravity is variable on Earth at different locations. The escape velocity depends on the radius and acceleration due to gravity of the planet. A body having velocity less than escape velocity will not be able to be able to escape the surface of the planet.