Why is the value of acceleration due to gravity zero at the centre of the earth? Prove with mathematical calculations.

The value of acceleration due to gravity is zero at the center of the Earth because the gravitational forces from all parts of the Earth cancel each other out. Let's explain this in detail and also prove it mathematically.

### Explanation:
Gravitational force at any point on Earth is due to the mass of the Earth that is closer to that point. At the Earth's surface, the gravitational force is the result of the entire mass of the Earth pulling an object toward the center. However, as we move towards the center of the Earth, the situation changes.

Consider an object at a point inside the Earth. According to the shell theorem, a spherical shell of uniform mass does not exert any net gravitational force on an object located inside it. As the object moves closer to the center, the amount of mass pulling it from all directions decreases. At the center of the Earth, the masses are symmetrically distributed in all directions, and thus the gravitational forces from all parts of the Earth cancel each other out. This results in zero net gravitational force at the center.

### Mathematical Proof:

1. **Gravitational force at a distance r from the center of the Earth:**
The gravitational force on an object of mass \( m \) at a distance \( r \) from the center of the Earth (where \( r < R \), with \( R \) being the Earth's radius) is given by:

\[
F = \frac{G M_{\text{enc}} m}{r^2}
\]

where:
- \( G \) is the gravitational constant,
- \( M_{\text{enc}} \) is the mass enclosed within the radius \( r \),
- \( m \) is the mass of the object,
- \( r \) is the distance from the center of the Earth.

2. **Mass enclosed within a radius \( r \):**
The mass enclosed inside a sphere of radius \( r \) is proportional to the volume of the sphere, since the density of the Earth is approximately uniform. The enclosed mass is:

\[
M_{\text{enc}} = M \left( \frac{r}{R} \right)^3
\]

where \( M \) is the total mass of the Earth and \( R \) is the Earth's radius.

3. **Substitute the enclosed mass into the gravitational force formula:**

\[
F = \frac{G \cdot M \left( \frac{r}{R} \right)^3 \cdot m}{r^2}
\]

Simplifying:

\[
F = \frac{G M m r}{R^3}
\]

4. **Acceleration due to gravity:**
The acceleration due to gravity \( g \) at a distance \( r \) from the center is given by:

\[
g = \frac{F}{m} = \frac{G M r}{R^3}
\]

This shows that gravity decreases linearly with distance from the center of the Earth. When \( r = 0 \) (at the center of the Earth), the gravitational force is:

\[
g = \frac{G M \cdot 0}{R^3} = 0
\]

### Conclusion:
At the center of the Earth, the value of acceleration due to gravity is zero because the gravitational forces from all parts of the Earth cancel each other out symmetrically. The formula \( g = \frac{G M r}{R^3} \) clearly shows that as the distance \( r \) approaches zero, the acceleration due to gravity also becomes zero.