Science Subjects Tutorials and Lectures: Acceleration due to Gravity 'g'

Bodies allowed to fall freely were found to fall at the same rate irrespective of their masses (air resistance being negligible). The velocity of a freely falling body increased at a steady rate i.e., the body had acceleration. This acceleration is called acceleration due to gravity - 'g'.

We know,
F = mg --------------- (2)

Where F is the force, m is the mass of the body, g is the acceleration due to gravity, M is the mass of the Earth, R is the radius of the Earth and G is the gravitational constant.
From equations (2) and (3),

'g' varies with (a) altitude (b) depth (c) latitude

Variation of 'g' with altitude

Let a body of mass m be placed on the surface of the Earth, whose mass is M and radius is R.
From equation (4)

Let the body be now placed at a height h above the Earth's surface. Let the acceleration due to gravity at that position be g^|.

For comparison, the ratio between g^| and g is taken

By binomial theorem,

h is assumed to be very small when compared to radius R of the Earth.

Hence, they can be neglected

This shows that acceleration due to gravity decreases with increase in altitude.
Loss in weight at height h(h<<R)
From equation (7)

__________(7)

Variation of 'g' with depth

Consider a body of mass m, lying on the surface of the Earth of radius R and mass M. Let g be the acceleration due to gravity at that place.

Let the body be taken to a depth d from the surface of the Earth. Then, the force due to gravity acting on this body is only due to the sphere of radius R.
(R - d). If g^| is the acceleration due to gravity at depth 'd'

Let the Earth be of uniform density r and its shape be a perfect sphere.

(Where r is the density of the Earth)

Comparing g^| and g

The acceleration due to gravity decreases with increase in depth.
If d = R, then g^| = 0.

Weight of a body at the centre of the Earth is zero.

Variation of 'g' with latitude

The value of g changes from place to place due to the elliptical shape of the Earth and the rotation of the Earth. Due to the shape of the Earth,
From equation (4)

Hence, it is inversely proportional to the square of the radius.

It is least at the equator and maximum at the poles, since the equatorial radius (6378.2 km) is more than the polar radius (6356.8 km)

Due to the rotation of the Earth

If 'w' is the angular velocity of the Earth and f is the latitude of the place,

Circular motion

Suppose a body is undergoing circular motion about a circle of radius r, with an angular velocity w rad/s as shown in the figure. From geometry, we can observe that the corresponding distance travelled in 1 second is equal to rw. Since the distance travelled in 1 second is velocity, v = rw. Every body undergoing circular motion with a constant angular velocity is said to be undergoing uniform circular motion. It experiences acceleration towards the centre of the circle of

Since it is directed towards the centre of the circle, it is called centripetal acceleration.
Therefore, centripetal acceleration is associated with uniform circular motion and directed towards the centre of the circle

Let us now consider earth as a sphere of radius R, undergoing uniform circular motion about its polar axis, connecting the north and south poles. Equator is the horizontal circle passing through the centre of this axis, P¹.

What is latitude?

Every point on the sphere lies on the same latitude, which lie on the base of the cone whose axis coincides with the polar axis and whose generators make an angle f with the horizontal or equatorial plane. The angle f is called the latitude of the place. Latitude of equator = 0^o and latitude of north pole = 90^o and latitude of south pole = -90^o.

We can observe that if the Earth is considered to be a sphere of radius R, the radius of the smaller circle in the latitudef=Rcosf.

A particle on the latitude f which is undergoing uniform circular motion with angular velocity 'w', experiences centripetal acceleration

directed towards the centre of the small circle O^I. This acceleration 'a' can be resolved into two components, tangential and vertical.

Gravitational acceleration 'g' acts on the body.

Since all the forces acting on the body at the latitude f result in uniform circular motion, the net of all forces should be equal to centripetal force.

At the equator,

At the pole,

Hence, the gravitation acceleration is maximum at the poles and minimum at the equator.
We should observe that the net of all the forces acting on the body results in uniform circular motion, which means that uniform circular motion is the result of all the forces acting on the body.