The universal law of gravity states that Every particle of matter in the universe attracts any other particle with a gravitational force directly proportional to the product of the masses and inversely proportional to the square of the distance between them.

Newton guessed that the acceleration of a body towards the earth is inversely proportional to the square of the distance of the body from the centre of the earth.
Thus, \(a \propto \frac{1}{r^2}\).
Also, the force is mass times acceleration and so it is proportional to the mass of the body. Hence,
\(
F \propto \frac{m}{r^2}
\)

By the third law of motion, the force on a body due to the earth must be equal to the force on the earth due to the body. Therefore, this force should also be proportional to the mass of the earth. Thus, the force between the earth (mass \(M\)) and a body (mass \(m\)) is
\(
\begin{aligned}
& F \propto \frac{M m}{r^2} \\
\text { or, } \quad & F=\frac{G M m}{r^2} \dots(1) 
\end{aligned}
\)

Newton further generalised the law by saying that not only the earth but all material bodies in the universe attract each other according to equation (1) with the same value of \(G\). The constant \(G\) is called the universal constant of gravitation and its value is found to be \(6.67 \times 10^{-11} \mathrm{~N}-\mathrm{m}^2 / \mathrm{kg}^2\). Equation (1) is known as the universal law of gravitation.

In general we can write \(F=\frac{G m_1 m_2}{r^2}\)

Where, \(F\)-Force acting on the object
\(m_1\) and \(m_2\) – mass of the objects
\(r\)-distance between the objects

Consider two bodies \(A\) and \(B\), each with a mass of \(m 1\) and \(m 2\), separated by \(r\), with the force of attraction acting on them, the law of gravitation gives the gravitational force between as illustrated in the diagram below:

The figure above shows how two bodies with attraction forces \(F1\) and \(F2\) tend to gravitate towards the center of gravity.

Example 1: Two particles of masses \(1.0 \mathrm{~kg}\) and \(2.0 \mathrm{~kg}\) are placed at a separation of \(50 \mathrm{~cm}\). Assuming that the only forces acting on the particles are their mutual gravitation, find the initial accelerations of the two particles.

Solution: The force of gravitation exerted by one particle on another is
\(
\begin{aligned}
F &=\frac{G m_1 m_2}{r^2} \\
&=\frac{6.67 \times 10^{-11} \frac{\mathrm{N}-\mathrm{m}^2}{\mathrm{~kg}^2} \times(1.0 \mathrm{~kg}) \times(2.0 \mathrm{~kg})}{(0.5 \mathrm{~m})^2} \\
&=5.3 \times 10^{-10} \mathrm{~N} .
\end{aligned}
\)
The acceleration of \(1.0 \mathrm{~kg}\) particle is
\(
\begin{aligned}
a_1 &=\frac{F}{m_1}=\frac{5 \cdot 3 \times 10^{-10} \mathrm{~N}}{1 \cdot 0 \mathrm{~kg}} \\
&=5 \cdot 3 \times 10^{-10} \mathrm{~m} \mathrm{~s}^{-2} .
\end{aligned}
\)
This acceleration is towards the \(2.0 \mathrm{~kg}\) particle. The acceleration of the \(2.0 \mathrm{~kg}\) particle is
\(
\begin{aligned}
a_2 &=\frac{F}{m_2}=\frac{5 \cdot 3 \times 10^{-10} \mathrm{~N}}{2 \cdot 0 \mathrm{~kg}} \\
&=2 \cdot 65 \times 10^{-10} \mathrm{~m} \mathrm{~s}^{-2} .
\end{aligned}
\)
This acceleration is towards the \(1.0 \mathrm{~kg}\) particle.

Gravitational Force expressed in Vector Form

Equation (1) can be expressed in vector form as
\(
\begin{aligned}
\mathbf{F} &=G \frac{m_1 m_2}{r^2}(-\hat{\mathbf{r}})=-G \frac{m_1 m_2}{r^2} \hat{\mathbf{r}} \\
&=-G \frac{m_1 m_2}{|\mathbf{r}|^3} \hat{\mathbf{r}}
\end{aligned}
\)
where \(\mathrm{G}\) is the universal gravitational constant, \(\hat{\mathbf{r}}\) 1s the unit vector from \(m_1\) to \(m_2\) and \(\mathbf{r}=\mathbf{r}_2-\mathbf{r}_1\) as shown in Fig below.

The gravitational force is attractive, 1.e., the force \(\mathbf{F}\) is along – r. The force on point mass \(m_1\) due to \(m_2\) is of course- \(\mathbf{F}\) by Newton’s third law. Thus, the gravitational force \(\mathbf{F}_{12}\) on body 1 due to 2 and \(\mathbf{F}_{21}\) on body 2 due to 1 are related as \(\mathbf{F}_{12}=-\mathbf{F}_{21}\).

Gravitational force on a collection of point masses

Let’s consider the figure shown below. The gravitational force on point mass \(m_1\) is the vector sum of the gravitational forces exerted by \(m_2, m_3\), and \(m_4\).

The total force on \(m_1\) is
\(
\mathbf{F}_1=\frac{G m_2 m_1}{r_{21}^2} \hat{\mathbf{r}}_{21}+\frac{G m_3 m_1}{r_{31}^2} \hat{\mathbf{r}}_{31}+\frac{G m_4 m_1}{r_{41}^2} \hat{\mathbf{r}}_{41}
\)

Example 2: Three equal masses of \(m \mathrm{~kg}\) each are fixed at the vertices of an equilateral triangle ABC.
(a) What is the force acting on a mass \(2 \mathrm{~m}\) placed at the centroid \(G\) of the triangle?
(b) What is the force if the mass at the vertex A is doubled?

Take \(\mathrm{AG}=\mathrm{BG}=\mathrm{CG}=1 \mathrm{~m}\)

Solution:

(a) The angle between GC and the posittive \(x\)-axis is \(30^{\circ}\) and so is the angle between GB and the negative \(x\)-axis. The individual forces in vector notation are 

\(
\begin{aligned}
&\mathbf{F}_{\mathrm{GA}}=\frac{G m(2 m)}{1} \hat{\mathbf{j}} \\
&\mathbf{F}_{\mathrm{GB}}=\frac{G m(2 m)}{1}\left(-\hat{\mathbf{i}} \cos 30^{\circ}-\hat{\mathbf{j}} \sin 30^{\circ}\right) \\
&\mathbf{F}_{\mathrm{GC}}=\frac{G m(2 m)}{1}\left(+\hat{\mathbf{i}} \cos 30^{\circ}-\hat{\mathbf{j}} \sin 30^{\circ}\right)
\end{aligned}
\)
From the principle of superposition and the law of vector addition, the resultant gravitational force \(\mathbf{F}_R\) on \((2 m)\) is
\(
\begin{aligned}
\mathbf{F}_{\mathrm{R}}=\mathbf{F}_{\mathrm{GA}} &+\mathbf{F}_{\mathrm{GB}}+\mathbf{F}_{\mathrm{GC}} \\
\mathbf{F}_{\mathrm{R}}=2 G m^2 \hat{\mathbf{j}}+2 G m^2\left(-\hat{\mathbf{i}} \cos 30^{\circ}-\hat{\mathbf{j}} \sin 30^{\circ}\right) \\
&+2 G m^2\left(\hat{\mathbf{i}} \cos 30^{\circ}-\hat{\mathbf{j}} \sin 30^{\circ}\right)=0
\end{aligned}
\)
Alternatively, one expects on the basis of symmetry that the resultant force ought to be zero.

(b) Now if the mass at vertex A is doubled then
\(
\begin{aligned}
&\mathrm{F}_{G A}^{\prime}=\frac{\mathrm{G} 2 m \cdot 2 m}{1} \hat{\mathbf{j}}=4 \mathrm{Gm}^2 \hat{\mathrm{j}} \\
&\mathrm{F}_{G B}^{\prime}=\mathrm{F}_{G B} \text { and } \mathrm{F}_{G C}^{\prime}=\mathrm{F}_{G C} \\
&\mathrm{~F}_R^{\prime}=\mathrm{F}_{G A}^{\prime}+\mathrm{F}_{G B}^{\prime}+\mathrm{F}_{G C}^{\prime} \\
&\mathrm{F}_{\mathrm{R}}^{\prime}=2 \mathrm{Gm}^2 \hat{\mathrm{j}}
\end{aligned}
\)

Note: For the gravitational force between an extended object (ilke the earth) and a point mass, Eq. (1) is not directly applicable. Each point mass in the extended object will exert a force on the given point mass and these force will not all be in the same direction. We have to add up these forces vectorially for all the point masses in the extended object to get the total force.