Question

The kinetic energy needed to project a body of mass $m$ from the earth’s surface to infinity is:
A) $\dfrac{1}{4}mgR$
B) $\dfrac{1}{2}mgR$
C) $mgR$
D) $2mgR$

Answer 1

Hint: To find out the kinetic energy of the mass, first write the formula of the kinetic energy and substitute the formula for the escaping velocity of the mass to move it out of gravitational force in it. The simplification of the equation provides the answer.

Formula used:
(1) The formula of the kinetic energy is given by
$KE = \dfrac{1}{2}m{V^2}$
Where $KE$ is the kinetic energy of the mass, $m$ is the mass and $V$ is the velocity at which the mass is thrown.

(2) The escape velocity is given by
${V_{esc}} = \sqrt {2gR} $
Where ${V_{esc}}$ is the escaping velocity of the mass, $g$ is the acceleration due to gravity and the $R$ is the radius of the earth.

Complete step by step solution:
It is given that the mass projects to infinity from the earth surface. It means the mass destination is known but it is not returning back to the point of the projection. Hence the mass escapes the earth’s gravitational force that causes the mass to return back to the earth, after the force of the mass is zero. Hence this velocity that makes escape of the mass from the gravity is called escaping velocity.

The formula of the kinetic energy is considered.
$KE = \dfrac{1}{2}m{V^2}$
$KE = \dfrac{1}{2}m{V_{esc}}^2$ ------------------(1)
Substituting the formula (2) in the equation (1).
${V_{esc}} = \sqrt {2gR} $
$KE = \dfrac{1}{2}m{\left( {\sqrt {2gR} } \right)^2}$
By simplifying the above equation,
$KE = \dfrac{2}{2}mgR$
By the further simplification, we get
$KE = mgR$
Hence the kinetic energy of the mass is $mgR$ .

Thus the option (C) is correct.

Note: If the mass has the normal velocity, when the velocity at which the mass is projected up becomes zero, the mass again returns back to the original surface due to the gravity. This gravitational force is limited up to a certain distance, after this the objects can move with escape velocity.