Gravitational intensity on the earth’s surface is g:

Let us consider the earth as a uniform sphere of radius \( R \) and mass \( M \).

We know that the intensity of the gravitational field at any point on the surface of a uniform solid sphere is \( -\frac{GM}{R^2} \). Where \( G \) is the gravitational constant.

Gravitational Intensity Due To A Solid Sphere At A Point (i) Outside The Sphere, (ii) On The Surface, And (iii) Inside A Sphere. Read In Detail.

The negative sign indicates that the field is directed towards the centre of the earth.

Now consider a body of mass \( m \) is lying on the earth’s surface. The attraction force on it by the earth is,

\( mg=G\frac{Mm}{R^2} \\or,\ g=\frac{GM}{R^2} \), where \( g \) is the acceleration due to gravity.

Hence the intensity of the gravitational field on the earth’s surface is \( g \) (numerically).

Gravitational potential on the earth’s surface is gR:

Assuming the earth to be a uniform solid sphere of radius \( R \) and mass \( M \),

We know that the potential at a point on the surface of the earth is \( V=-\frac{GM}{R} \).

Gravitational Potential Due To A Solid Homogeneous Sphere At A Point (i) Outside, (ii) On The Surface, And (iii) Inside A Point Of The Sphere. Read In Detail.

or, \( V=-R\frac{GM}{R^2} =-gR \)

Hence the gravitational potential on the earth’s surface is \( gR \) (numerically).