The acceleration produced in a freely falling body by the gravitational pull of the earth is called the acceleration due to gravitation.

We know that,

Force = Mass × Acceleration

F = m × a

a = \(\frac{F}{m}\)........................1

Where F is the force on the object of mass m dropped from a distance r from the centre of earth of mass M.

So, force exerted by the earth on the object is

F = G\(\frac{M\times m}{r^2}\)......................2

M = Mass of Earth

m = mass of object

r = distance of object from centre of earth

Now, from equation 1 and 2,

a = G\(\frac{M\times m}{r^2\times m}\)

a = G\(\frac{M}{r^2}\)

Now, from above,

a = g = Acceleration due to gravity

We also see that, although force is depending on the mass of the object,

F = G\(\frac{M\times m}{r^2}\)

But acceleration due to gravity is independent of the mass.

g = G\(\frac{M}{r^2}\)

Factors on which g depends are:

(i) Value of gravitational constant (G)

(ii) Mass of Earth (M)

(iii) Radius of Earth (r)

As gravitational constant G and mass of earth M are always constant, so the value f acceleration due to gravity g is constant as long as the radius of earth remains constant.

At the surface of earth also, the value of g is not constant.  \((g\,∝ \frac{1}{r^2})\)

At the poles, radius of earth is minimum, hence the g is maximum. Similarly, at the equator, the radius of earth is maximum, and hence the value of g is minimum. Also, as we go up from the surface of the earth, distance from the centre of the earth increases and hence value of g decreases.

The value of g also decreases as we go inside the surface of earth and g is zero at the centre of earth, as at the centre of the earth the object has mass around it, so net force cancels and thus net acceleration becomes zero.