Gravitational Field Intensity – Definition, Formula

The space surrounding a material body in which its gravitational force of attraction can be measured is called its gravitational field. The gravitational force between two masses or the electrostatic force between two charges or the magnetic force between two magnetic poles does exist between them, even when they are at a distance from each other. For this reason, the gravitational, electrostatic, and magnetic interactions are sometimes called actions at a distance. Each of the three types of interactions can be successfully explained in terms of the field concept.

According to the field concept, a material body, when placed in space, sets up the gravitational field. In other words, it modifies the space around it in some manner. The space surrounding a material body in which its gravitational force of attraction can be measured is called its gravitational field.

It is because of the gravitational field of the body, due to which the body exerts force on any material body introduced in its field.

Gravitational Field Formula

The gravitational field due to a body of mass M at a distance r from the center of the body is given by

$ \displaystyle E=\dfrac{{GM}}{{{{r}^{2}}}}$

Intensity of the Gravitational Field

The intensity of the gravitational field of a body at any point in its field is defined as the force experienced by a unit mass (may be called test mass) placed at that point, provided the unit mass (test mass) itself does not produce any change in the field of the material body. It is a vector quantity and is denoted by $\vec{E}$.

The intensity of the gravitational field of a body is also known as the strength of the gravitational field or simply the gravitational field of the body.

S.I. Unit

Since the intensity of the gravitational field is the force experienced per unit mass, its unit is $N \, kg^{-1}$ in S.I. and $dyne \, g^{-1}$ in the cgs system.

Intensity of Gravitational Field of Earth

Let $M$ be the mass of the earth. Suppose that we have to find the intensity of the gravitational field due to the Earth at point $P$, distant $r$ from the center $O$ of the Earth. Place a test mass $m$ at the observation point $P$.

According to Newton’s law of gravitation, the force on the test mass $m$ at point $P$ is given by:

\[
F = \frac{G M m}{r^2}
\]

Therefore, by definition, the gravitational field of the Earth at the point $P$,

$ \displaystyle {{E}_{{\text{grav}}}}(\text{at }\text{r})=\dfrac{F}{m}$

$ \displaystyle {{E}_{{\text{grav}}}}(\text{at }\text{r})=\dfrac{{GMm}}{{{{r}^{2}}}}\times \dfrac{1}{m}$

$\displaystyle {{E}_{{\text{grav}}}}(\text{at}\text{ r})=\dfrac{{GM}}{{{{r}^{2}}}}$ $\displaystyle \text{ }\ldots \text{(1)}$

The direction of the gravitational field at the point \( P \) is along \( PO \), i.e. the gravitational field is directed towards the center of the Earth. It follows that the intensity of the gravitational field decreases as the distance \( r \) increases and becomes zero at infinity.

In equation (1), \( \frac{GM}{r^2} \) is equal to the acceleration due to gravity at the point \( P \).

Therefore, intensity of gravitational field at a point is equal to the acceleration due to gravity at that point.

In other words, when we say that at a point, the gravitational field is \( \frac{GM}{r^2} \), it implies that a test mass placed at that point will accelerate towards the Earth with acceleration \( \frac{GM}{r^2} \).

The gravitational field at a point on the surface of the Earth can be obtained by setting \( r = R \) (the radius of Earth) in equation (1). Therefore, the gravitational field at a point on the surface of the Earth is given by

$\displaystyle {{E}_{{\text{grav}}}}(\text{at}\text{ }r\text{ = R})=\dfrac{{GM}}{{{{R}^{2}}}}=g,$ $\displaystyle \text{ }\ldots \text{(2)}$

where \( g \) is the acceleration due to gravity at the surface of the Earth.

If the Earth is a sphere of material of uniform density \( \rho \), then

\[
M = \frac{4}{3} \pi R^3 \rho
\]

By substituting the value of \( M \) in equations (1) and (2), we have

$ \displaystyle {{E}_{{\text{grav}}}}(at\text{ }r)$ $\displaystyle =\frac{{G\times \frac{4}{3}\pi {{R}^{3}}\rho }}{{{{r}^{2}}}}$

$\displaystyle {{E}_{{\text{grav}}}}(at\text{ }r)=\dfrac{{4\pi G{{R}^{3}}\rho }}{{3{{r}^{2}}}}$

and

$\displaystyle {{E}_{{\text{grav}}}}(at\text{ }r=R)=\frac{{4\pi GR\rho }}{3}$

Remember: In case \( M \) represents the mass of a body (in place of the Earth), then equations (1) and (2) give the intensity of the gravitational field due to that body at a distance \( r \) from its center and at its surface respectively. In equation (2), \( g \) would mean acceleration due to gravity at the surface of that body.