Question

Escape velocity at the surface of earth is $ 11.2km{s^{ - 1}} $ . Escape velocity from a planet whose mass is the same as that of earth and radius $ \dfrac{1}{4} $ that of earth, is?
(A) $ 2.8km{s^{ - 1}} $
(B) $ 15.6km{s^{ - 1}} $
(C) $ 22.4km{s^{ - 1}} $
(D) $ 44.8km{s^{ - 1}} $

Answer 1

Hint : We are given with the radius of the planet with respect to the radius of earth and are asked to find the escape velocity of the planet. Thus, we will use the formula of escape velocity.

Formulae used:
 $ {v_e} = \sqrt {\dfrac{{2G{M_p}}}{r}} $
Where, $ {v_e} $ is the escape velocity of a planet, $ G $ is the universal gravitational constant, $ {M_p} $ is the mass of the planet and $ r $ is the radius of the planet.

Complete step by step answer
For escape velocity of the earth, the formula turns out,
 $ {v_e}_{_e} = \sqrt {\dfrac{{2G{M_e}}}{{{r_e}}}} \cdot \cdot \cdot \cdot (1) $
Where, $ {M_e} $ is the mass of earth and $ {r_e} $ is the radius of earth.
Similarly, for the planet
 $ {v_{{e_p}}} = \sqrt {\dfrac{{2G{M_p}}}{{{r_p}}}} \cdot \cdot \cdot \cdot (2) $
Where, $ {M_p} $ is the mass of the planet and $ {r_p} $ is the radius of earth.
But,
According to the question,
 $ {M_p} = {M_e} $ And $ {r_p} = \dfrac{1}{4} \times {r_e} $
Substituting these values in equation $ (2) $ , we get
 $ {v_{{e_p}}} = \sqrt {\dfrac{{2G{M_e}}}{{\dfrac{1}{4} \times {r_e}}}} \cdot \cdot \cdot \cdot (3) $
Now,
Evaluating $ \dfrac{{(3)}}{{(1)}} $ , we get
 $ \dfrac{{{v_{{e_p}}}}}{{{v_{{e_e}}}}} = \sqrt 4 = 2 $
Thus,
 $ {v_{{e_p}}} = 2{v_{{e_e}}} $
We know,
 $ {v_{{e_e}}} = 11.2km{s^{ - 1}} $
Thus,
 $ {v_{{e_p}}} = 2 \times 11.2km{s^{ - 1}} = 22.4km{s^{ - 1}} $
Hence, the correct option is (C).

Note
The escape velocity of a planet is the velocity of a body which is required to escape the gravitational force of the planet. So for getting the formula of escape velocity, we can equate the energy of the body by virtue of its motion (kinetic energy) to the gravitational force multiplied by radius.
Thus,
 $ K.E. = {F_G} \times r $
Further, we get
 $ \dfrac{1}{2}m{v_e}^2 = \dfrac{{GMm}}{{{r^2}}} \times r $
After further evaluation, we get
 $ {v_e} = \sqrt {\dfrac{{2GM}}{r}} $ .
Here, we are given that the mass of the earth and the planet are the same. But if the masses were different, then the case will be something different.