Intensity of Gravitational Field

Intensity of Gravitational Field

The intensity of the gravitational field of a body at a point in the field is defined as the force experienced by a body of unit mass placed at that point provided the presence of unit mass does not disturb the original gravitational field. It is always directed towards the centre of gravity of the body whose gravitational field is considered. Intensity of the gravitational field at a point is a vector quantity and is denoted by \(\overrightarrow{E}\).

Let, M be the mass of a body with centre of gravity at O. Let, F be the gravitational force of attraction experienced by a test mass m₀ when placed at P in the gravitational field of the body, such that OP = x. According to Newton’s law of gravitation:


Intensity of gravitational field at P will be:

\(E=\frac{F}{{{m}_{0}}}=\frac{\frac{GM{{m}_{0}}}{{{x}^{2}}}}{{{m}_{0}}}=\frac{GM}{{{x}^{2}}}\) … (1)

In vector form:

\(\overrightarrow{E}=-\frac{Gm}{{{x}^{2}}}\widehat{x}\) … (2)

Here, the negative sign shows that the gravitational intensity is of attractive force. From equation (2), we note that as x increases, E decreases and when x = ∞, E = 0 it means that the intensity of gravitational field is zero only at infinite distance from the body.

If the test mass is free to move at P, it will be accelerated due to force of attraction F, by the body of mass m₀. Let, a be the acceleration produced in the test mass due to force F, then F = m₀a.

F/m = a … (3)

From equations (2) and (3), E = a.

It means that the intensity of gravitational field at a point in the gravitational field is equal to the acceleration of test mass placed at that point. If the gravitational field is due to earth and point P lies on the surface of the earth then x = R. Now



g = Acceleration due to gravity at the surface of earth.

Unit of intensity of gravitational field in SI system is N/kg or m/ sec² and in CGS system, it is dyne/ g (or) cm/ sec².

Dimensional formula of gravitational intensity: