# Newton's Law of Gravitation

When you see around, you'll find yourself and other objects around you being pulled by a force downwards. In fact the force pulls you and other objects around you towards the centre of the Earth.

The force a planet exerts on other bodies on its surface or inside its *gravitational field* is called gravitational force. The effect of the gravitational force of a planet goes beyond the surface of the planet and therefore it sets up a field around it called gravitational field.

Our planet Earth has a reasonable gravitational force inside its gravitational field which pulls every other bodies within its field towards its centre. In fact a neutral body exerts a force on every other neutral body near it.

Note that we are not talking about any force generated by induced charge. The effect of the force of a body of smaller mass on other bodies is negligible but if the body has very large mass in comparison to the other bodies, the effect of the force the body exerts on the other bodies is reasonable. That's the reason why planets show reasonable gravitational force and field.

The Newton's law of gravitation which tells us about the strength of force any two particles exert on each other and can be stated as

NEWTON'S LAW OF GRAVITATION: Every particle in the universe attracts every other particle and the force between any two particles is directly proportional to the product of their masses and inversely proportional to the square of distance between them.

So from the Newton's law of gravitation, if $F$ is the magnitude of force between the two particles and $m_1$ and $m_2$ are their masses, you'll get $F \propto m_1m_2/{{{r^2}}}$ and

\[F = G\frac{{{m_1}{m_2}}}{{{r^2}}} \tag{1} \label{1}\]

were $G$ is the universal constant called gravitational constant. Its value is $6.67\times{10^{-11}}{\rm{N}}{{\rm{m}}^{\rm{2}}}{\rm{/k}}{{\rm{g}}^{\rm{2}}}$ within three significant figures.

The above Newton's law of gravitation tells us the force between point particles of point masses but the law holds equally to larger bodies having spherically symmetric mass distributions, that is valid for both spherical shells and solid spheres but in this case the distance is measured between the centres of any two bodies. We'll prove this later.

On the other hand we use the same law given by Eq. \eqref{1} even if any two bodies have irregular shapes provided that the distance between the bodies is large enough in comparison to their sizes. For example, the gravitational force between the Earth of mass $M$ and an irregular body of mass $m$ on the Earth's surface is given by $F = G{{Mm}}/{{{R^2}}}$ where $R$ is the Earth's radius.

We used the Newton's law of gravitation for the irregular body because the body on the Earth's surface can be taken as a point particle in comparison to the distance $R$. Note that we have taken the Earth as a spherically symmetrical mass distribution here.

In Figure 1 the magnitude of gravitational force the Earth exerts on a body of mass $m$ at distance $r$ from the centre of the Earth is

\[F = G\frac{{M_Em}}{{{r^2}}} \tag{2} \label{2}\]

where $M_E$ is the mass of the Earth. Eq. \eqref{2} shows that the gravitational force decreases as the distance from the centre of the Earth increases.

According to the Newton's third law the body also exerts the same magnitude of force on the Earth. The body falls towards the centre of the Earth due to this gravitational force but in fact the Earth also falls towards the centre of the body.

The Earth has very large mass in comparison to the mass of the body so its acceleration due to the force the body exerts on it is negligible but you can see the acceleration of the body falling towards the centre of the Earth due to its very small mass in comparison to the mass of the Earth.

When a body falls under the action of a gravitational force, the body has an acceleration called gravitational acceleration or acceleration due to gravity which is denoted by $g$. Therefore, using Newton's second law, the magnitude of force the Earth exerts on the body is $F = mg$ and rewriting Eq. \eqref{2} gives

\[g = \frac{{GM_E}}{{{r^2}}} \tag{3} \label{3}\]

You can see in the above equation that the acceleration due to gravity of the body is independent of the mass of the body, that is all bodies either small or big fall with the same acceleration.

The above equation shows that the value of acceleration due to gravity changes as the distance from the centre of the Earth changes; this is the same for other planets as well.

In other words the magnitude of gravitational acceleration decreases as the distance from the centre of the Earth increases. If $R_E$ is the radius of the Earth, the magnitude of the acceleration due to gravity on the Earth's surface is

\[g = \frac{{GM_E}}{{{R_E^2}}} \tag{4} \label{4}\]

We had taken the Earth as spherically symmetrical mass distribution in the previous discussion but in fact the Earth is not spherically symmetrical. Due to this reason the magnitude of the acceleration due to gravity on the Earth's surface varies slightly from place to place.

If you lift an object on the Earth's surface, you'll feel a downward pull and you'll call the downward pull as the weight of the object. Newton's law of gravitation says that every particle in the universe attracts every other particle and this is also true for larger bodies not only for the particles. Therefore *the weight of a body can be defined as the net gravitational force on the body.*

On the Earth's surface the gravitational force exerted by other bodies near it is negligible and the downward pull is mainly due to the Earth's gravitational force. So we define the weight $w$ of a body of mass $m$ on the Earth's surface mainly by the gravitational force on the body due to the Earth's *gravity*.

You know from Eq. \eqref{4} that the magnitude of the gravitational acceleration on the Earth's surface is $g = {{G{M_E}}}/{{R_E^2}}$. So the magnitude of the weight of the body is $w=mg$ and hence,

\[w = \frac{{G{M_E}{\kern 1pt} m}}{{R_E^2}} \tag{5} \label{5}\]

But if the body is not on the surface and it is at a height $h$ from the surface of the Earth, the weight due to the Earth's gravitational force on the body is

\[w = \frac{{G{M_E}{\kern 1pt} m}}{{{{(R_E^{} + h)}^2}}} \tag{6} \label{6}\]

Therefore as the height from the surface of the Earth increases, the weight of the body decreases.