Question

A tunnel is dug along the diameter of the earth (radius R and mass M). There is a particle of mass 'm' at the centre of the tunnel. Find the minimum velocity given to the particle so that it just reaches to the surface of the earth is
A. $\sqrt{\dfrac{GM}{R}}$
B. $\sqrt{\dfrac{GM}{2R}}$
C. $\sqrt{\dfrac{2GM}{R}}$
D. it will reach with the help of a negligible velocity

Answer 1

The particle is situated at the centre of the tunnel. We know that the value of acceleration due to gravity is given by $g=\dfrac{GM}{R_{e}^{2}}$. So, the value of acceleration is zero at the centre of the earth. The potential energy of the particle at the centre of the Earth is given as $-\dfrac{3GMm}{2R}$ and it has a kinetic energy due to the velocity given to reach at the surface of the earth. The potential energy of the particle at the surface is given as $-\dfrac{GMm}{R}$and here its kinetic energy is zero. We can make use of the law of conservation of energy to arrive at the solution.

Complete step by step answer:
Initially the particle is at the centre of the tunnel which is dug along the diameter of the earth. Therefore the particle is at the centre of the earth.Let us assume the velocity given to the particle be v.The potential energy for a particle at the centre of the Earth is given as $\dfrac{3GMm}{2R}$. So initially total energy of the particle can be written as,
$K.{{E}_{i}}+P.{{E}_{i}}=\dfrac{1}{2}m{{v}^{2}}+\left( -\dfrac{3GMm}{2R} \right)$
This velocity enables the particle to move to the surface of the earth. Therefore the velocity of the particle when it reaches the surface of the Earth is zero. The potential energy for a particle on the surface of the Earth is given as $-\dfrac{GMm}{R}$.

The final total energy of the particle can be written as,
$K.{{E}_{f}}+P.{{E}_{f}}=0+\left( -\dfrac{GMm}{R} \right)$
Considering the particle and the earth as a system and applied conservation of energy,
$\dfrac{1}{2}m{{v}^{2}}+\left( -\dfrac{3GMm}{2R} \right)$=$0+\left( -\dfrac{GMm}{R} \right)$
$\Rightarrow\dfrac{1}{2}m{{v}^{2}}=\dfrac{GMm}{2R} \\\ \therefore v=\sqrt{\dfrac{GM}{R}}$
Therefore, we get the velocity as $\sqrt{\dfrac{GM}{R}}$.

Hence, option A is correct.

Note: One should know the formula for gravitational potential energy of a particle at the centre of the earth$-\dfrac{3GMm}{2R}$ and on the surface of the earth $\dfrac{GMm}{R}$. We should know to apply the conservation of energy to solve such types of problems. Also, the value of g varies with the height and the depth of the earth and all such measurements are done from the centre of the earth. While doing such kind of problems we have to keep in mind the number of variables which needs to be introduced and what are the given quantities.