Variation of the value of ‘g’

What is acceleration due to gravity?

Gravitational force is one of four fundamental forces. This force is attractive in nature and is very weak in magnitude for bodies that we encounter in our day-to-day life. Everybody on Earth experiences a strong gravitational force of attraction from the centre of Earth. This force is responsible for keeping bodies stuck on the surface of Earth. 

According to Newton’s law of gravitation, if two bodies have mass m and M and if they are separated by a distance x then they will experience a force of attraction according to the following formula:

F=GMm/x2

Where,

G = Universal gravitational constant = 6.6710-11 Nm2/Kg2

The above expression gives the gravitational force of attraction between two masses.

In the case of Earth, when a body is sitting on the surface, it will experience an attractive force from the Earth. 

If mass of body is m and mass of Earth is M and the radius of Earth is X  then the force on that body is,

F=GMm/X2 …………….(1)

By comparing this equation with Newton’s second law, 

F=ma ………………(2)

We will get the value of gravitational acceleration of a body on the surface as,

g=a=GM/X2  …………….(3)

Variation of the value of gravitational acceleration due to rotation of Earth 

Equation (3) gives us the value of acceleration caused by gravity on the surface of Earth. This value is around 9.8 m/s2. But this value is not constant. This value varies with change in the latitude from the Earth’s surface. 

In this section, we will derive an expression for the variation of acceleration caused by gravitational force with latitude. We can assume our Earth has spherical shape. In the case of a spherical body (for instance, the Earth), the entire mass of the body is considered to be situated at the centre of Earth.

Consider the diagram given below,

Latitude is an angle made by a radius vector of any point from the centre of the Earth with the equatorial plane. Obviously, it ranges from 0 at the equator to 90 at the poles.

The Earth rotates about its polar axis from west to east with uniform angular velocity .

Hence every point on the surface of the Earth (except the poles) moves in a circle parallel to the equator. 

The motion of a mass m at point B on the Earth is shown by the smaller circle with centre at O’. Let the latitude of B be and radius of the circle be r.

BO’=r

∠AOB= Θ, where, A is a point on the equator

∴ ∠OBO’=Θ.

In  OBO’, cos(Θ)= BO’/BO=r/R

∴ r=Rcos(Θ)

The centripetal acceleration for the mass k, directed along BO’ is:

a=r2

∴ a=R2cos(Θ)

The component of this centripetal acceleration along BO, i.e., towards the centre

of the Earth is,

ar=acos(Θ)

∴ ar=r2cos2(Θ)

Part of the gravitational force of attraction on B acting towards BO is utilised in providing these components of centripetal acceleration.

Thus the effective force of gravitational attraction on k at B can be written as

mg’=mg-mR2cos2(Θ)

g’ being the effective acceleration due to gravity at B i.e., at latitude . This is thus given by,

g’=g-R2cos2(Θ)

As the latitude increases, cos() decreases. Therefore g’ will increase towards the pole. Thus, acceleration due to gravity is large at the pole as compared to the equator.

Conclusion 

This article explains the variation of ‘g’ due to rotation of the earth. The value of acceleration caused by gravity on the surface of Earth. This value is around 9.8 m/s2. But this value is not constant. This value varies with change in the latitude from the Earth’s surface. The acceleration due to gravity (g) is large at the pole as compared to the equator.