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Question: If \[{M_e}\] and \[{R_e}\] be the mass and radius of earth respectively, then the escape velocity wi...

If Me{M_e} and Re{R_e} be the mass and radius of earth respectively, then the escape velocity will be:
A. ve=2GMe2Re{v_e} = \sqrt {\dfrac{{2G{M_e}^2}}{{{R_e}}}}
B. ve=2GMe2Re2{v_e} = \sqrt {\dfrac{{2G{M_e}^2}}{{{R_e}^2}}}
C. ve=2GMeRe{v_e} = \sqrt {\dfrac{{2G{M_e}}}{{{R_e}}}}
D. ve=3GMe2Re{v_e} = \sqrt {\dfrac{{3G{M_e}}}{{2{R_e}}}}

Explanation

Solution

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 Me{M_e}.Radius of the earth is Re{R_e}. Let mm be the mass of a body on the surface of the earth or in earth’s gravitational force.Let vv 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,
Eb=GMmR{E_b} = \dfrac{{GMm}}{R}
where GG is gravitational constant, MM is the mass of the earth, mm is the mass of the body and RR is the distance from the center of the earth to the center of the body.
Here, M=MeM = {M_e} and R=ReR = {R_e}
So, gravitational energy is,
Eb=GMemRe{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=EbK.E = {E_b} (ii)
where K.EK.E is the kinetic energy of the body.
The kinetic energy of a body is given by,
K.E=12×(mass)×(velocity)2K.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=12mve2K.E = \dfrac{1}{2}m{v_e}^2 (iii)
Using equations (iii) and (i) in (ii), we get
12mve2=GMemRe\dfrac{1}{2}m{v_e}^2 = \dfrac{{G{M_e}m}}{{{R_e}}}
12ve2=GMeRe\Rightarrow \dfrac{1}{2}{v_e}^2 = \dfrac{{G{M_e}}}{{{R_e}}}
ve2=2GMeRe\Rightarrow {v_e}^2 = \dfrac{{2G{M_e}}}{{{R_e}}}
ve=2GMeRe\therefore {v_e} = \sqrt {\dfrac{{2G{M_e}}}{{{R_e}}}}
Therefore, escape velocity will be ve=2GMeRe{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 ve=2GMR{v_e} = \sqrt {\dfrac{{2GM}}{R}} where MM is the mass of the attracting body and RR 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.