If M is the mass of earth and R is the earth of radius the attracts a mass m on its surface with a force F given by F= GmM /R^{2}

This force imparts on acceleration to the mass m which is known as acceleration due to gravity (g)

By newton’s law we have

Acceleration = Force/Mass

Acceleration = \(\frac{F}{m}\frac{\frac{GmM}{{{R}^{2}}}}{m}=\frac{GM}{{{R}^{2}}}\).

g = GM/R^{2}

Substituting the values G, M, R we get g= 9.8 m/s^{2}

**VARIATION OF g WITH ALTIDUDE AND DEPTH:**

The acceleration due to gravity g = F/M

Where, F is the exerted force by the earth on an object of mass m. This force is affected by a number of factors, thus g depends on these factors.

**Variation in the values of g above the surface of earth: **

When an object is placed at a distance of h above the surface of the earth, the force of gravitation is

F = GmM/(R+h)^{ 2}

Acceleration due to gravity is g^{1} = GM/(R+h)^{ 2}

We see that the value of g^{1} decreases as one goes up. Thus,

\({{g}^{1}}=\frac{GM}{{{R}^{2}}{{\left( 1+\frac{h}{R} \right)}^{2}}}=\frac{g}{{{\left( 1+\frac{h}{R} \right)}^{2}}}\).

If h<<R

Then,

\({{g}^{1}}=g{{\left( 1+\frac{h}{R} \right)}^{-2}}=g\left( 1-\frac{2h}{R} \right)\).

**Variation in value of g below the surface of the earth:**

Acceleration due to gravity on the surface of the earth is g=GM/R^{2} = 4/3 πρGR

Where, ρ is the density of the earth

Acceleration due to gravity at depth d from the surface of the earth,

g^{1} = 4/3 πρG(R – d)

From above equations we get g^{1} = g [1- d/R]

**Variation on in the value of (g) due to rotation of the earth:**

Due to rotation of the earth, the value of g decreases as the speed of rotation of the earth increases. The value of acceleration due to gravity at a latitude is

g^{1}_{2} = g – ω^{2} cos^{2}λ

Following conclusions drawn from the above discussions

The effect of the centrifugal force due to rotation of the earth is to reduce the effective value of g.

The effective value of g is not truly in vertical direction

At the equators, λ = 0^{0}

Therefore g^{1} = g – Rω^{2}

At the poles λ = 90^{0}

Therefore g^{1} = g

At the equator, the rotation of the earth is maximum and value of g is minimum.

At the poles, effect of rotation of the earth is zero and value of g is maximum.