Question

The escape velocity for the earth is $11.2km/s$. The mass of another planet is $100$ times the mass of the earth and its radius is $4$ times the radius of the earth. The escape velocity for the planet is
$A)\text{ }280km/s$
$B)\text{ 56}.0km/s$
$C)\text{ 112}km/s$
$D)\text{ }24km/s$

Answer 1

Hint: This problem can be easily solved by using the direct formula for the escape velocity for a planet in terms of its mass and radius. By plugging in the values of the radius and the mass of the planet in terms of that of the earth, we can get the escape velocity for the planet in terms of the escape velocity of the earth and therefore, get the required numerical value.

Formula used:
${{v}_{esc}}=\sqrt{\dfrac{2GM}{R}}$

Complete step by step answer:
The escape velocity for a planet is the minimum velocity that has to be given to a body on the surface of the planet in the vertically upward direction so that it can escape the gravitational field of the planet and escape to infinity.
For a planet of mass $M$ and radius $R$, the escape velocity ${{v}_{esc}}$ is given by
${{v}_{esc}}=\sqrt{\dfrac{2GM}{R}}$ --(1)
Where $G=6.67\times {{10}^{-11}}{{m}^{3}}k{{g}^{-1}}{{s}^{-2}}$ is the universal gravitational constant.
Therefore, let us analyze the question.
Let the mass of the earth be ${{M}_{E}}$ and the radius be ${{R}_{E}}$.
Let the escape velocity for the earth be ${{v}_{E}}$.
Therefore, using (1), we get,
${{v}_{E}}=\sqrt{\dfrac{2G{{M}_{E}}}{{{R}_{E}}}}$ --(2)
Now, according to the question, the mass of the planet is ${{M}_{P}}=100{{M}_{E}}$.
The radius of the planet is ${{R}_{P}}=4{{R}_{E}}$.
Let the escape velocity for the planet be ${{v}_{p}}$.
Therefore, using (1), we get,
${{v}_{P}}=\sqrt{\dfrac{2G{{M}_{P}}}{{{R}_{P}}}}=\sqrt{\dfrac{2G\left( 100{{M}_{E}} \right)}{4{{R}_{E}}}}=\sqrt{\dfrac{2G{{M}_{E}}}{{{R}_{E}}}\left( 25 \right)}$ --(3)
Now, using (2) and (3), we get,
$\dfrac{{{v}_{P}}}{{{v}_{E}}}=\dfrac{\sqrt{\dfrac{2G{{M}_{E}}}{{{R}_{E}}}\left( 25 \right)}}{\sqrt{\dfrac{2G{{M}_{E}}}{{{R}_{E}}}}}=\sqrt{25}=5$
$\therefore {{v}_{P}}=5\times {{v}_{E}}$
Now, according to the question, ${{v}_{E}}=11.2km/s$
Putting this value in the above equation, we get,
${{v}_{P}}=5\times 11.2=56.0km/s$
Hence, the required escape velocity for the planet is $56.0km/s$.
Therefore, the correct option is $B)\text{ 56}.0km/s$.

Note: It is not necessary to memorize the formula for the escape velocity of a planet. It can be derived easily by using the concept that the mechanical energy of the body will remain constant. Hence, the decrease in the kinetic energy of the body will be equal to the increase in the gravitational potential energy. For escape velocity, the final potential energy (body at infinity) and kinetic energy of the body both are zero. Therefore, the magnitudes of the initial potential and kinetic energy will be equal. By equating them, we can easily get a relation for the escape velocity in terms of the mass and radius of the planet.