Question

The mass of the Earth is 81 times that of the Moon and the radius of the Earth is 3.5 times that of the moon. The ratio of the escape velocity on the surface of the Earth to that on the surface of the Moon will be:
A. 0.2
B. 2.57
C. 4.81
D. 0.39

Answer 1

Hint: Escape velocity is defined as the velocity required for a body to escape from the gravitational field of a planet or massive object. Take the ratio of escape velocity from the earth to the escape velocity from the surface of the moon. Simplify this ratio in terms of the mass of the earth and the mass of the moon. Substitute the mass of the earth in terms of the mass of the moon and hence find the required ratio.

Formula used:
The formula for escape velocity is given by:
${{v}_{e}}=\sqrt{\dfrac{2GM}{R}}$

Complete step by step answer:
Let ${{M}_{E}}$ be the mass of the Earth and ${{M}_{M}}$ be the mass of the Moon. Let \[{{R}_{E}}\] be the radius of the Earth and \[{{R}_{M}}\] be the radius of the Moon.
Given that the mass of the Earth is 81 times the radius of the Moon, therefore,

\[\begin{align}
  & {{M}_{E}}=81{{M}_{M}} \\
 & \dfrac{{{M}_{E}}}{{{M}_{M}}}=81 \\
\end{align}\]

Also, the radius of the Earth is 3.5 times the radius of the Moon, therefore,

\[\begin{align}
  & {{R}_{E}}=3.5{{R}_{M}} \\
 & \dfrac{{{R}_{E}}}{{{R}_{M}}}=3.5 \\
\end{align}\]

Escape velocity is defined as the velocity required for a body to escape from the gravitational field of a planet or massive object. It is given as

${{v}_{e}}=\sqrt{\dfrac{2GM}{R}}$
Where
G is the universal constant of gravitation.
M is the mass of the planet
R is the radius of the planet.
Thus, the escape velocity of an object from the Earth’s surface is given as
${{v}_{eE}}=\sqrt{\dfrac{2G{{M}_{E}}}{{{R}_{E}}}}$
And escape velocity of an object from the Moon's surface is given as
\[{{v}_{eM}}=\sqrt{\dfrac{2G{{M}_{M}}}{{{R}_{M}}}}\]
Taking the ratio of these two escape velocities, we get,
\[\begin{align}&\dfrac{{{v}_{eE}}}{{{v}_{eM}}}=\dfrac{\sqrt{\dfrac{2G{{M}_{E}}}{{{R}_{E}}}}}{\sqrt{\dfrac{2G{{M}_{M}}}{{{R}_{M}}}}} \\
&\therefore\dfrac{{{v}_{eE}}}{{{v}_{eM}}}=\dfrac{\sqrt{\dfrac{{{M}_{E}}}{{{R}_{E}}}}}{\sqrt{\dfrac{{{M}_{M}}}{{{R}_{M}}}}} \\
&\dfrac{{{v}_{eE}}}{{{v}_{eM}}}=\dfrac{\sqrt{{{M}_{E}}{{R}_{M}}}}{\sqrt{{{M}_{M}}{{R}_{E}}}}=\sqrt{\dfrac{{{M}_{E}}}{{{M}_{M}}}}\times \sqrt{\dfrac{{{R}_{M}}}{{{R}_{E}}}} \\
\end{align}\]

\[\begin{align}
  & \dfrac{{{v}_{eE}}}{{{v}_{eM}}}=\sqrt{81}\times \sqrt{\dfrac{1}{3.5}} \\
 & \dfrac{{{v}_{eE}}}{{{v}_{eM}}}=4.81 \\
\end{align}\]
Therefore, the ratio of the escape velocity on the surface of the Earth to that on the surface of the Moon will be 4.81.
Hence, the correct answer is option C.

Note:
Note that escape velocity depends on the mass and the radius of the planet or massive body. It does not depend on the mass of the body which is to be escaped i.e. a body of mass 10 kg and a body of mass 100 kg will have the same escape velocity from the surface of the Earth. The escape velocity from the surface of the Earth is 11.2 km/sec if we neglect the atmospheric friction.