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

Hint: Gravity is the force which pulls the objects down towards the ground and acceleration produced by this force is called the acceleration due to gravity.Gravitational acceleration is a quantity of vector, that is it has both magnitude and direction.

Complete step by step answer:
Acceleration due to gravity at the surface of earth is approximately \[9.8\,m/{{s}^{2}}\]. It is not the same everywhere. It keeps on decreasing as we go above the surface of earth.
As we go below the surface of earth, the variation in g(acceleration due to gravity) is given by:
\[g'=g\left( 1-\dfrac{h}{{{R}_{e}}} \right)\]
where,
\[g'=\] Actual acceleration due to gravity at a depth ‘h’ from the surface of earth
\[g=\] Acceleration due to gravity at the surface of earth \[=9.8m/{{s}^{2}}\]
\[h=\] Distance from the surface of the earth
\[\operatorname{R_e}=\] Radius of the earth
Now, at the centre of earth, \[h=\operatorname{Re}\]
\[g'=g\left( 1-\dfrac{{{R}_{e}}}{{{R}_{e}}} \right)\]
\[\therefore g'=0\]
Therefore, acceleration due to gravity at the centre of the earth is \[0\] and it keeps on increasing till we reach at the surface of the earth and after that it keeps on decreasing.

Note: Logically, we can understand this by, when we move inside the earth, the mass that exerts gravitational force on us decreases and hence at the centre of the earth the acceleration due to gravity becomes zero. Variation of g as we go above the surface of the earth is given by \[g'=\dfrac{g}{{{\left( 1+\dfrac{h}{{{R}_{e}}} \right)}^{2}}}\].