Question

The kinetic energy required to make a body move to infinity from earth surface is
(A) Infinity
(B) 2mgR
(C) $\dfrac{1}{{{\kern 1pt} 2mgR}}$
(D) mgR

Answer 1

Hint
We will first use the equation of binding energy on a particle on the earth’s surface, as if a body wants to escape its energy has to be greater than or equal to the binding energy.
This binding energy must be equal to kinetic energy of the body, so after equating this equation we get the escape velocity of the body.
Putting this value of escape velocity in the kinetic energy equation we get the kinetic energy required to make a body move to infinity from earth surface.

Complete step by step answer
The kinetic energy needed to move a body from earth's surface to infinity is known as escape energy of the body.
We know that the binding energy on a particle on the earth’s surface kept at rest is $\dfrac{{GMm}}{R}$ . So, if this much energy in the form of kinetic energy is supplied to the particle, then it leaves the earth gravitational field.
So, let ${v_e}$ be the escape velocity of the particle, then kinetic energy should be equal to the binding energy.
So we can write the equation as $\dfrac{1}{2}m{v_e}^2 = \dfrac{{GMm}}{R}$ , where m is the mass of the object , G is the gravitational constant, M and m are the masses of the particles and R is the distance between the particles.
So, after equating the equation we get ${v_e} = \sqrt {\dfrac{{2GM}}{R}}$ .
Now we know that $g = \dfrac{{GM}}{{{R^2}}}$ , substituting the value of g in the above equation we get ${v_e} = \sqrt {2gR}$ .
We substitute this value of ${v_e}$ in the kinetic energy equation to find the kinetic energy of the body.
Hence $KE = \dfrac{1}{2}m{v_e}^2 = \dfrac{1}{2}m \times 2gR$ , after solving we get $KE = mgR$
So option (D) is correct.

Note
At the surface of earth escape velocity of a body is $\sqrt 2$ times its orbital velocity. Also, after substituting the value of acceleration due to gravity and distance i.e. $g = 9.8\dfrac{m}{{{s^2}}}$ and $R = 6.4 \times {10^6}m$ , we get the value of escape velocity ${v_e} = 11.2\dfrac{{km}}{s}$ (approx.).