Question

If \[{M_e}\] and \[{R_e}\] be the mass and radius of earth respectively, then the escape velocity will be:
A. \[{v_e} = \sqrt {\dfrac{{2G{M_e}^2}}{{{R_e}}}} \]
B. \[{v_e} = \sqrt {\dfrac{{2G{M_e}^2}}{{{R_e}^2}}} \]
C. \[{v_e} = \sqrt {\dfrac{{2G{M_e}}}{{{R_e}}}} \]
D. \[{v_e} = \sqrt {\dfrac{{3G{M_e}}}{{2{R_e}}}} \]

Answer 1

Hint: We are asked to find out the escape velocity from earth. First, recall the formula for gravitational energy which binds a body to earth’s gravitational field. Find the kinetic energy of the body. Check for the condition in which a body can escape the earth’s gravitational field, use this condition to find the escape velocity.

Complete step by step answer:
Given, the mass of the earth is \[{M_e}\].Radius of the earth is \[{R_e}\]. Let \[m\] be the mass of a body on the surface of the earth or in earth’s gravitational force.Let \[v\] be the velocity of the body.

The earth will pull the body towards itself through gravitational force. This gravitational energy of the earth to keep the body in this gravitational field is given by,
\[{E_b} = \dfrac{{GMm}}{R}\]
where \[G\] is gravitational constant, \[M\] is the mass of the earth, \[m\] is the mass of the body and \[R\] is the distance from the center of the earth to the center of the body.
Here, \[M = {M_e}\] and \[R = {R_e}\]
So, gravitational energy is,
\[{E_b} = \dfrac{{G{M_e}m}}{{{R_e}}}\] (i)

For the body to escape from the earth’s gravitational field, it must have a kinetic energy equal to the gravitational energy. That is,
\[K.E = {E_b}\] (ii)
where \[K.E\] is the kinetic energy of the body.
The kinetic energy of a body is given by,
\[K.E = \dfrac{1}{2} \times \left( {{\text{mass}}} \right) \times {\left( {{\text{velocity}}} \right)^2}\]
So, here the kinetic energy of the body will be,
\[K.E = \dfrac{1}{2}m{v_e}^2\] (iii)
Using equations (iii) and (i) in (ii), we get
\[\dfrac{1}{2}m{v_e}^2 = \dfrac{{G{M_e}m}}{{{R_e}}}\]
\[ \Rightarrow \dfrac{1}{2}{v_e}^2 = \dfrac{{G{M_e}}}{{{R_e}}}\]
\[ \Rightarrow {v_e}^2 = \dfrac{{2G{M_e}}}{{{R_e}}}\]
\[ \therefore {v_e} = \sqrt {\dfrac{{2G{M_e}}}{{{R_e}}}} \]
Therefore, escape velocity will be \[{v_e} = \sqrt {\dfrac{{2G{M_e}}}{{{R_e}}}} \]

Hence, option C is the correct answer.

Note: Escape velocity of a body is given by \[{v_e} = \sqrt {\dfrac{{2GM}}{R}} \] where \[M\] is the mass of the attracting body and \[R\] is the radius of the attracting body. We can see that escape velocity is independent of the mass of the body which is attracted by gravitational force and depends only on the mass and radius of the attracting body.