__What is meant by “Acceleration due to Gravity”?__

Earth attracts all bodies near its surface by a force of gravity called *gravitational force. *The acceleration produced by this force is called earth’s **acceleration due to gravity.**

- By
*Newton’s laws of gravitational force*, we have –

F = G \left ( \frac {m_1 m_2 }{ r ^2 } \right ) .

Consider that a body of mass ( m ) is kept on the earth’s surface. Radius and mass of the earth are ( R ) \ \text {and} \ ( M ) respectively.

- Then, earth’s gravitational pull on the body will be –

F = G \left ( \frac {M m }{ R ^2 } \right ) .

Let, earth’s gravitational pulling force produces an *acceleration* ( g ) on the body.

- Then by
*Newtons laws of motion,*we get –

F = ( mg )

Therefore, \quad F = G \left ( \frac {M m }{ R^2 } \right ) = ( mg )

- Hence, acceleration produced by the earth is –

g = G \left ( \frac {M}{ R^2 } \right )

*This acceleration is called acceleration due to gravity.*

__Variation of Acceleration due to Gravity__

Value of acceleration due to gravity acting on a body varies with the location of body on the earth’s surface, above the earth’s surface or below it.

This variation is due to the following effects –

- Effect of location on earth’s surface.
- Effect of earth’s rotation.
- Effect of height from earth surface.
- Effect of depth under earth surface.

### 1.Effect of location on Earth’s surface

*Acceleration due to gravity varies with the location on the earth’s surface.*

Consider that a point *mass* ( m ) * *is situated on the earth’s surface. Then, its distance from earth’s centre will be ( r = R ) .

- Therefore,
*gravitational force*on the point mass will be –

F = G \left ( \frac { M m }{ R^2 } \right )

Suppose the mass ( m ) experiences an acceleration of ( g ) .

- According to
*Newton’s Second Law of motion*, we have –

\text {Force} = \text {Mass} \ \times \ \text {Acceleration}

Therefore, \quad F = mg

Or, \quad mg = G \left ( \frac { M m }{ R^2 } \right )

Therefore, \quad g = G \left ( \frac { M }{ R^2 } \right )

- This equation gives the acceleration due to gravity on the surface of earth. We see that –

g \propto \left ( \frac {1}{R} \right )

Earth is not a perfect sphere but it is flattened at the poles. So, the equatorial radius ( R_e ) of earth is greater than the polar radius ( R_p ) . It has found that [ ( R_e - R_p ) = 21 \ km ] .

*This is the reason that, acceleration due to gravity at equator is less than acceleration due to gravity at poles.*

- Therefore, \quad ( g_e < g_p )

*Near the surface of earth, the standard value of acceleration due to gravity is taken as ( 9.8 \ ms^{ -2 } ) \quad or \quad ( 32 \ fts^{ -2 } ) *

__2.Effect of Earth Rotation__

Consider about the figure showing the rotation of earth.

As the earth rotates about its polar axis with angular velocity ( \omega ) , every particle laying on its surface also revolves along horizontal circles with the same angular velocity ( \omega ) .

Consider about a particle of mass ( m ) * *laying at point P whose latitude is ( \lambda ) .

With rotation of earth, the particle P describes a horizontal circle of radius ( r ) where –

r = PC = R \cos \lambda

- The
*centrifugal force*acting along PA on the particle is given by –

F_{cf} = ( mr \omega^2 )

This force has two components –

- Component force ( mr \omega^2 \cos \lambda ) – This force is acting opposite to the direction of weight ( mg ) of the particle.
- Component force ( mr \omega^2 \sin \lambda ) – This force is acting perpendicular to ( mg ) and has no effect on ( mg )
*.*

So, the apparent weight of the particle P * *is –

mg_{\lambda} = [ ( mg ) - ( mr \omega^2 \cos \lambda )

Or, \quad g_{\lambda} = ( g - r \omega^2 \cos \lambda )

But, \quad \left [ r = R \cos \lambda \right ]

- Therefore, \quad g_{\lambda} = ( g - R \omega^2 \cos^2 \lambda )

As ( \lambda ) increases, value of \left ( \cos \lambda \right ) decreases and ( g_{\lambda} ) increases.

*So, as we move from equator to pole, the acceleration due to gravity increases. **Therefore, acceleration due to gravity is minimum at the equator and maximum at the poles.*

__3.Effect of height from Earth surface__

Consider about a point mass ( m ) situated at an height of ( h ) above the earth’s surface as shown in figure.

- The effective distance between centres of point mass and earth is –

r = \left ( R + h \right )

Therefore, the *gravitational force* on mass ( m ) will be –

F = G \left [ \frac { M m }{ \left ( R + h \right )^2 } \right ]

If ( g' ) is the acceleration due to gravity at that point, then –

m g' = G \left [ \frac { M m }{ \left ( R + h \right )^2 } \right ]

Or, \quad g' = \left [ \frac { G M }{ \left ( R + h \right )^2 } \right ]

- Dividing numerator and denominator by ( R^2 ) we have –

g' = \left [ \left ( \frac { G M }{ R^2} \right ) \right] \times \left [ \frac {1}{\frac {( R + h )^2}{R^2}} \right ]

= \left [ \left ( \frac { G M }{ R^2} \right ) \right] \times \left [ \frac {( R + h )}{R} \right ]^{-2}

= \left [ \frac { G M }{ R^2 } \left (1 + \frac { h }{ R } \right )^{-2} \right ]

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

Since, ( h << R ) so we can neglect higher power terms of \left ( \frac {h}{R} \right ) .

- Thus, \quad g' = g \left [ 1 - \left ( \frac { 2h }{ R } \right ) \right ]

Therefore –

*Value of acceleration due to gravity decreases with increase in height above the earth’s surface.**When ( h ) is small i.e. \left ( h << R \right ) , then the fraction \left ( \frac { h }{ R } \right ) becomes negligible. Hence, the above equation reduces to ( g' = g )*

__4.Effect of depth under Earth surface__

Consider that a point mass ( m ) * *is situated at a depth of ( h ) below the earth’s surface as shown in figure.

Then, according to *Newton’s shell theorem*, the outer shell has no effect on the point mass m kept at point P . Only the inner sphere of radius ( r ) will exert a force. Where ( r = R - h )

Then effective mass of earth ( m_r ) is concentrated at the centre.

- Thus, the gravitational force on mass ( m ) is –

F = G \left [ \frac { m_r m }{ \left ( R - h \right )^2 } \right ]

But, \quad m_r = \left ( \frac { 4 }{ 3 } \right ) \pi \left ( R - h \right )^3 \rho = \left ( \frac { 4 }{ 3 } \right ) \pi R^3 \rho \left ( \frac { R - h }{ R } \right )^3 = M \left ( \frac { R - h }{ R } \right )^3

Or, \quad m_r = M \left ( \frac { R - h }{ R } \right )^3

Therefore, gravitational force acting on the mass will be –

F = \left [ \frac { G m_r m }{ \left ( R - h \right )^2 } \right ]

= \left [ \frac { G m }{ \left ( R - h \right )^2 } \right ] \left [ M \left ( \frac { R - h }{ R } \right )^3 \right ]

= \left [ \frac { G M m }{ R^2 } \right ] \left [ \frac { R - h }{ R } \right ]

But, \quad F = m g' \quad And \quad \left [ \frac { G M }{ R^2 } \right ] = g

Therefore, \quad m g'= \left [ \frac { G M m }{ R^2 } \right ] \left [ \frac { R - h }{ R } \right ]

= m g \left [ \frac { R - h }{ R } \right ]

Or, \quad g' = g \left ( \frac { R - h }{ R } \right ) = g \left (1 - \frac { h }{ R } \right )

Therefore –

*Value of acceleration due to gravity decreases while going in depth inside earth.**When ( h ) is small, i.e., \left ( h << R \right ) fraction \left ( \frac { h }{ R } \right ) becomes negligible. Hence, the above equation reduces to ( g' = g )**At the centre of the earth ( h = R ) . Therefore, \left [ 1 - \left ( \frac { h }{ R } \right ) = 0 \right ] . Hence, at the centre of earth ( g' = g \times 0 = 0 ) . This is the reason that, weight of a body at earth’s centre becomes zero.*

See numerical problems based on this article.