Question

Given that mass of earth is $M$ and its radius is $R$. A body is dropped from a height equal to the radius of the earth above the surface of the earth. When it reaches the ground, its speed will be:
(A) $\dfrac{{GM}}{R}$
(B) $\sqrt {\dfrac{{GM}}{R}} $
(C) $\sqrt {\dfrac{{2GM}}{R}} $
(D) $\dfrac{{2GM}}{R}$

Answer 1

Hint: The initial velocity of the body is zero and the kinetic energy depends on the square of the velocity of a body. The law of conservation of mass should be put to use.
Formula Used: The formulae used in the solution are given here.
$K.E{._{initial}} + P.E{._{initial}} = K.E{._{final}} + P.E{._{final}}$ where $K.E.$ is the kinetic energy of a body, $P.E.$ is the potential energy and the subscripts denote the initial and final stages.
$K.E. = \dfrac{1}{2}m{v^2}$ where the mass of the body is $m$ and the final velocity of the body is $v$.
$P.E. = \dfrac{{G{m_1}{m_2}}}{r}$ where $G$ is the gravitational constant, ${m_1}$ and ${m_2}$ are the masses of two bodies and $r$ is the distance of separation.

Complete Step by Step Solution: The law of conservation of energy states that energy can neither be created nor be destroyed. Although, it may be transformed from one form to another. If you take all forms of energy into account, the total energy of an isolated system always remains constant. All the forms of energy follow the law of conservation of energy. In brief, the law of conservation of energy states that- ‘In a closed system, i.e., a system that is isolated from its surroundings, the total energy of the system is conserved.’
Therefore, by the law of conservation of energy, initial energy of a system is equal to the final energy of a system.
$K.E{._{initial}} + P.E{._{initial}} = K.E{._{final}} + P.E{._{final}}$ where $K.E.$ is the kinetic energy of a body, $P.E.$ is the potential energy and the subscripts denote the initial and final stages.
Given that, a body is dropped from a height equal to the radius $R$ of the earth above the surface of the earth.
Thus the initial kinetic energy of the body is zero, since initial velocity is zero.
$\therefore K.E{._{initial}} = 0$.
Let the final velocity of the body when it reaches the ground be $v$. Let the mass of the body dropped be $m$.
Thus the final kinetic energy of the body is $K.E{._{final}} = \dfrac{1}{2}m{v^2}$ where the mass of the body is $m$.
Gravitational potential energy is the energy stored in an object as the result of its vertical position or height. The energy is stored as the result of the gravitational attraction of the Earth for the object.
Gravitational potential energy is given by, $P.E{._{initial}} = - \dfrac{{GMm}}{{R + R}}$ where the radius of the earth is $R$.
At the final stage,
$K.E{._{final}} = \dfrac{1}{2}m{v^2}$ where the mass of the body is $m$ and $v$ is the final velocity of the body.
$P.E{._{final}} = - \dfrac{{GMm}}{R}$
Thus, the equation, $K.E{._{initial}} + P.E{._{initial}} = K.E{._{final}} + P.E{._{final}}$ can be rewritten as,
$0 + \dfrac{{ - GMm}}{{2R}} = \dfrac{1}{2}m{v^2} + \dfrac{{\left( { - GMm} \right)}}{R}$
Simplifying the equation, we get,
$\dfrac{1}{2}m{v^2} = \dfrac{{GMm}}{R}$
Thus, the value of final velocity is given by,
${v^2} = \dfrac{{GM}}{R}$
$ \Rightarrow v = \sqrt {\dfrac{{GM}}{R}} $
The velocity of the body when it reaches the ground, will be $v = \sqrt {\dfrac{{GM}}{R}} $.

Hence the correct answer is Option B.

Note: We can't apply Newton’s equation of motion as the height is comparable to earth's radius. We can apply the law of conservation here. Since the total energy remains constant so now we can equate the total energy at first point and the second point.