## Pages

### 2.6 Gravitational field

We already discussed in the previous notes that there are four fundamental forces in nature and that gravity is one of them. Here we'll elaborate a little more on the Newtons theory of Universal Gravitation.

Firstly we'll discuss the concept of field. In Physics we can define "field of force as the region of the space within which any object of concern experience a force."

In case of the gravitational field "it is the region of space within which any object that has mass experience a gravitational force". That is to say that any object in the vicinity of the earth as shown below gets attracted towards the center of the earth, and that is what we mean in physics as the field of force.
Gravitational field strength is a vector quantity. Its direction is towards the mass that creates the field. "Mathematically gravitational field strength is defined as gravitational force acting per unit mass." So,

$$g=\frac{F_{G}}{m}$$
Where, $g$=gravitational field strength, $F_{G}$=Gravitational force and $m$=unit of mass
Its S.I units is newtons per kilogram (Nkg$^{-1}$).

Newtons Universal Gravitational forces: In 1687, English mathematician Sir Isaac Newton published Principia, which hypothesizes the inverse-square law of universal gravitation. It states that "a particle attracts every other particle in the universe using a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them."
$$F_{G}=\frac{GMm}{r^{2}}$$
where, $M$=mass of earth,
$m$=mass of the moon,
$G$= universal gravitational constant and
$r$=distance between the center of an object.

Now that we have the definition of the gravitational force, we can now find the gravitational field strength, substituting the definition of the gravitational force in the definition of the gravitational field strength we have as follows,

$$g=\frac{GM}{r^{2}}$$
It is evident from this formula that gravitational field strength at the surface of the earth can be calculated given that, $M_{earth}=6\times 10^{24}$kg and $r_{earth}=6.4\times10^{6}$m with $G=6.67\times10^{-11}$Nm$^{2}$kg$^{-2}$ is 9.77Nkg$^{-1}$.

This magnitude is closely equal to the gravitational field strength at the surface of the earth and this is exactly the reason why we take the gravitational field strength near the earth surface to be a constant 9.8Nkg$^{-1}$.

Here's an easy to understand video of gravitational field strength.

Exercise: Find the gravitational field strength at 2500km above the earth surface.
Soln:  5.05Nkg$^{-1}$.