Question

If the radius of the moon is $1.7\times {{10}^{6}}m$ and its mass is $7.34\times {{10}^{22}}kg$. Then its escape velocity is
A. $2.4\times {{10}^{3}}m{{s}^{-1}}$
B. $2.4\times {{10}^{2}}m{{s}^{-1}}$
C. $3.4\times {{10}^{3}}m{{s}^{-1}}$
D. $3.4\times {{10}^{2}}m{{s}^{-1}}$

Answer 1

Hint: It is a direct formula-based question so the student must know the formula to calculate the escape velocity which is given by:
${{v}_{e}}=\sqrt{\dfrac{2GM}{R}}$

Complete step by step answer:
The escape velocity on a planet or any celestial body is defined as the minimum velocity with which an object has to be projected vertically upwards so that it just crosses the planet’s gravitational force of attraction (and never to return or pull back).
The Escape velocity depends upon the radius and the mass of the planet from which a body must be projected.
The formula for escape velocity is given by:
 ${{v}_{e}}=\sqrt{\dfrac{2GM}{R}}$
Given that radius of moon is:
$R=1.7\times {{10}^{6}}m$
 Mass of moon is:
$M=7.34\times {{10}^{22}}kg$
G is the gravitational constant given by:
$G=6.67\times {{10}^{-11}}N{{m}^{2}}k{{g}^{-2}}$

Substituting these values in above equation we get:
${{v}_{e}}=\sqrt{\dfrac{2\times (6.67\times {{10}^{-11}})\times 7.34\times {{10}^{22}}}{1.74\times {{10}^{6}}}}m/s$
On solving we get:
${{v}_{e}}=\sqrt{5.627\times {{10}^{6}}}m/s$
${{v}_{e}}=2.37\times {{10}^{3}}m/s$
Which is approximately
${{v}_{e}}=2.4\times {{10}^{3}}m{{s}^{-1}}$
Hence, the correct option is option A. $2.4\times {{10}^{3}}m{{s}^{-1}}$

Additional Information:
The shortcut formula for calculating the escape velocity of earth is:
${{v}_{e}}=\sqrt{2gR}$
And the actual value of escape velocity on earth is $11.2km{{s}^{-1}}$.Although its value is still slightly greater than this as while deriving, $618km{{s}^{-1}}$ we have neglected the resistance to the body due to earth’s atmosphere.
Escape velocity on the Sun is $618km{{s}^{-1}}$ and that of the moon as we just calculated is $2.4km{{s}^{-1}}$.

Note: From the above discussion student must note that the escape velocity is independent of mass and radius of the body that has to be projected rather it depends upon the celestial body’s (planet or a star) mass and radius from which it has to be projected from.