Question

A diametrical tunnel is dug across the Earth. A ball is dropped into the tunnel from one side. Find the velocity of the ball when it reaches the centre of the Earth. (Given: gravitational potential at the centre of the earth is $-{\displaystyle \frac{3}{2}}{\displaystyle \frac{GM}{R}}$ ).
A) $\sqrt{R}$
B) $\sqrt{gR}$
C) $\sqrt{2.5gR}$
D) $\sqrt{7.1gR}$

Answer 1

Hint: As the ball is dropped from one side of the tunnel, its gravitational potential energy will be converted to kinetic energy so that the total energy of the ball remains constant. At the centre of the earth, the ball possesses gravitational potential energy as well as kinetic energy.

Formulas used:
The gravitational potential energy of a body on the surface of the earth is given by, $U={\displaystyle \frac{-GMm}{R}}$ where $G$ is the universal gravitational constant, $M$ is the mass of the earth, $m$ is the mass of the body and $R$ is the radius of the earth.
The kinetic energy of a body is given by, $K={\displaystyle \frac{1}{2}}m{v}^2$ where $m$ is the mass of the body and $v$ is the velocity of the body.

Complete step by step answer:
Step 1: Describe the problem at hand.
In the given problem, a ball is dropped from one end of a tunnel which is dug in such a way that it runs diametrically through the earth. We are to determine the velocity $v$ of the ball once it reaches the centre of the earth.
Let $M$ and $m$ be the masses of the earth and the ball respectively.
Let $R$ be the radius of the earth.
The gravitational potential at the centre of the earth is given to be $V=-{\displaystyle \frac{3}{2}}{\displaystyle \frac{GM}{R}}$.
Step 2: Express the energy of the ball before and after dropping.
Before dropping the ball, it only possesses gravitational potential energy
i.e., $E_{before}={\displaystyle \frac{-GMm}{R}}$ --------- (1)
Once the ball reaches the centre, or after dropping the ball, the ball possesses gravitational potential energy and kinetic energy.
The gravitational potential energy at the centre will be $U_{centre}=V\times m=-{\displaystyle \frac{3}{2}}{\displaystyle \frac{GMm}{R}}$ .
The kinetic energy of the ball is expressed as $K={\displaystyle \frac{1}{2}}m{v}^2$.
Then the energy after dropping will be $E_{after}=U_{centre}+K$ -------- (2)
Substituting for $U_{centre}=-{\displaystyle \frac{3}{2}}{\displaystyle \frac{GMm}{R}}$ and $K={\displaystyle \frac{1}{2}}m{v}^2$ in equation (2) we get, $E_{after}=-{\displaystyle \frac{3}{2}}{\displaystyle \frac{GMm}{R}}+{\displaystyle \frac{1}{2}}m{v}^2$ --------- (3)
Step 3: Based on the energy conservation principle, obtain the velocity of the ball at the centre of the earth.
We know that the total energy of the ball remains constant.
i.e., $E_{before}=E_{after}$ --------- (4)
Substituting equations (1) and (3) in (4) we get,
$\Rightarrow {\displaystyle \frac{-GMm}{R}}=-{\displaystyle \frac{3}{2}}{\displaystyle \frac{GMm}{R}}+{\displaystyle \frac{1}{2}}m{v}^2$
On simplifying the above expression becomes ${\displaystyle \frac{1}{2}}{v}^2=\left({\displaystyle \frac{3}{2}}{\displaystyle \frac{GM}{R}}\right)-\left({\displaystyle \frac{GM}{R}}\right)$
$\Rightarrow v^2={\displaystyle \frac{3GM}{R}}-{\displaystyle \frac{2GM}{R}}={\displaystyle \frac{GM}{R}}$
$\Rightarrow v=\sqrt{{\displaystyle \frac{GM}{R}}}$
Now as $GM=g{R}^2$ ; $g$ is the acceleration due to gravity, the velocity becomes
 $v=\sqrt{{\displaystyle \frac{g{R}^2}{R}}}=\sqrt{gR}$
Thus we finally have the velocity of the ball at the centre of the earth as $v=\sqrt{gR}$.

Hence the correct option is B.

Note:
The only force acting on the ball is the gravitational force of attraction between the ball and the earth. The gravitational force is a conserved force and so the total energy of the ball remains constant. Only a transfer of energy takes place. The acceleration due to gravity can be expressed as $g={\displaystyle \frac{GM}{R^2}}$ and it is from this relation we get, $GM=g{R}^2$ which was substituted to obtain the final expression of the velocity.