Variation in Value of g

Variation with Height

The magnitude of g decreases as square of the distance from the centre of the earth increases. However, if the distance h above the surface of the earth, called altitude, is small compared with the radius of the earth, the value of g, denoted by g_h, is given by

g_h = GM/(R+h)²

Variation with Depth

Consider a point P at a depth d inside the earth. Assume that the earth is a sphere of uniform density ρ. The distance of the point P from the center of the earth is r = (R – d). A mass placed at P will experience gravitational force from particles in (i) the shell of thickness d, and (ii) the sphere of radius r. It can be shown that the forces due to all the particles in the shell cancel each other. That is, the net force on the particle at P due to the matter in the shell is zero. Therefore, in calculating the acceleration due to gravity at P, you have to consider only the mass of the sphere of radius (r – d).

As d increases, (R – d) decreases. This means that the value of g decreases as you go below the earth. At d = R, that is, at the centre of the earth, the acceleration due to gravity will vanish.

Variation with Latitude

The earth rotates about its axis. Due to this, every particle on the earth’s surface executes circular motion. In the absence of gravity, all these particles would be flying off the earth along the tangents to their circular orbits.

To keep a particle in circular motion, it must be supplied centripetal force. A small part of the gravity force is used in supplying this centripetal force. As a result, the force of attraction of the earth on objects on its surface is slightly reduced. The maximum effect of the rotation of the earth is felt at the equator. At poles, the effect vanishes completely.

If g_λ denotes the value of g at latitude λ and g is the value at the poles, then

g_λ = g – Rω² cosλ

where ω is the angular velocity of the earth and R is its radius.

At the poles, λ = 90°, and hence gλ = g.