Answer
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Hint: Escape speed or velocity refers to the minimum velocity which is needed to leave a planet. To calculate the escape speed or velocity, the present elevation, or at what height the body is presently from sea level as gravity becomes lower and lower as we go away from the core of the earth. As far as we go from the core the gravity will be less effective at that point.
Complete step by step answer:
The gravitational potential energy of a body of mass m on the surface of the earth with mass M denoted by ${E_{{p_1}}}$ is,
$ \Rightarrow {E_{{p_1}}} = \dfrac{{GMm}}{R}$.
Here R= Radius of the Earth, M= Mass of Earth, m= Mass of body, and G= Gravitational force.
When the body escapes the earth’s gravity, the body’s potential energy will be,
$ \Rightarrow {E_{{p_2}}} = 0$.
Now, from the theory of energy conservation we have,
$ \Rightarrow {E_{{p_1}}} - {E_{{p_2}}} = \dfrac{1}{2}m{V_e}^2$.
Replacing the values of Ep1 and Ep2 in this equation we get,
$ \Rightarrow \dfrac{{GMm}}{R} = \dfrac{1}{2}m{V_e}^2$.
Simplifying this equation, we get,
$ \Rightarrow {V_e} = \sqrt {\dfrac{{2GM}}{R}} = \sqrt {2gR} $.
• Clearly from the equation of escape itself it shows that it does not depend on the mass of the body although it depends on the mass of the earth.
• There is no relation of escape velocity with the location from where it is projected. Hence it does not depend on the location from where it is projected.
• There is no relation between the angle of projection on the escape velocity. Hence it does not depend on it.
• We know that the acceleration due to gravity varies with the height from the surface of the earth. It can be shown by the following relation-
$ \Rightarrow g' = g\left( {1 - \dfrac{{2h}}{R}} \right)$
Where h=height from the surface of earth the new acceleration due to gravity is g’ and the normal acceleration due to gravity is $g$ the height above the surface of earth is h and the radius of the earth is R.
Thus the escape velocity is depending on the height from where the body is projected.
Hence, the correct answer is option (D).
Note: Mostly we get confused in the points about the location of the projection and the height of the projection. Always remember that, wherever the location, the main factor of importance here is the height or altitude of the body. The higher the altitude the lesser the escape velocity required by the body to escape the earth’s gravitational field.
Complete step by step answer:
The gravitational potential energy of a body of mass m on the surface of the earth with mass M denoted by ${E_{{p_1}}}$ is,
$ \Rightarrow {E_{{p_1}}} = \dfrac{{GMm}}{R}$.
Here R= Radius of the Earth, M= Mass of Earth, m= Mass of body, and G= Gravitational force.
When the body escapes the earth’s gravity, the body’s potential energy will be,
$ \Rightarrow {E_{{p_2}}} = 0$.
Now, from the theory of energy conservation we have,
$ \Rightarrow {E_{{p_1}}} - {E_{{p_2}}} = \dfrac{1}{2}m{V_e}^2$.
Replacing the values of Ep1 and Ep2 in this equation we get,
$ \Rightarrow \dfrac{{GMm}}{R} = \dfrac{1}{2}m{V_e}^2$.
Simplifying this equation, we get,
$ \Rightarrow {V_e} = \sqrt {\dfrac{{2GM}}{R}} = \sqrt {2gR} $.
• Clearly from the equation of escape itself it shows that it does not depend on the mass of the body although it depends on the mass of the earth.
• There is no relation of escape velocity with the location from where it is projected. Hence it does not depend on the location from where it is projected.
• There is no relation between the angle of projection on the escape velocity. Hence it does not depend on it.
• We know that the acceleration due to gravity varies with the height from the surface of the earth. It can be shown by the following relation-
$ \Rightarrow g' = g\left( {1 - \dfrac{{2h}}{R}} \right)$
Where h=height from the surface of earth the new acceleration due to gravity is g’ and the normal acceleration due to gravity is $g$ the height above the surface of earth is h and the radius of the earth is R.
Thus the escape velocity is depending on the height from where the body is projected.
Hence, the correct answer is option (D).
Note: Mostly we get confused in the points about the location of the projection and the height of the projection. Always remember that, wherever the location, the main factor of importance here is the height or altitude of the body. The higher the altitude the lesser the escape velocity required by the body to escape the earth’s gravitational field.
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