Let us take the mass of the Earth as M and its radius as R. If a body of mass m is present at the surface of the Earth, it experiences a gravitational attraction from the Earth. It can be written as:

$$\rm F = G \frac{Mm}{R^{2}}$$

According to Newton's second law of motion, $\rm F = ma$, where m is the mass of the object that is experiencing the force F.

In our case, the smaller body of mass m is experiencing the force; hence, we can rewrite the expression for force as:

$$\rm F = G \frac{Mm}{R^{2}} = ma$$

This acceleration that is produced on a smaller body due to the gravitational attraction of the Earth is called acceleration due to gravity. So we put $\rm a = g$.

$$\rm F = G \frac{Mm}{R^{2}} = mg$$

Taking the last two terms, we get,

$$\rm G \frac{Mm}{R^{2}} = mg$$

$$\rm g = G \frac{M}{R^{2}}$$

The mass of the Earth and the magnitude of the Gravitational Constant remain constant. Hence, the above equation tells us that the acceleration due to gravity is inversely proportional to the square of the distance between the center of the Earth and the object.

$$\rm g \propto \frac{1}{R^{2}}$$