Universal Law of Gravitation

Newton’s Universal Law of Gravitation tells us that everything in the universe pulls on everything else. The bigger the masses, the stronger the pull — and the farther apart they are, the weaker the pull. This idea helps us understand how planets orbit the sun, why things fall to the ground, and how gravity works in space and on Earth. It’s an important concept for anyone studying physics, astronomy, or engineering.

1.0Newton's Law of Gravitation

This law states that every object in the universe pulls on every other object with a force that increases with their masses and decreases with the square of the distance between them.

It states that every particle in the universe attracts all other particles with a force which is directly proportional to the product of their masses and is inversely proportional to the square of the distance between them.

$\begin{array}{rl}& \left|{\vec{F}}_{12}\right|=\left|{\vec{F}}_{21}\right|=F_g\\ & F_g\propto m_1{m}_2.......(1)\\ & F_g\propto \frac{1}{r^2}........(2)\end{array}$

From equation (1) and (2)

$F_g\propto \frac{m_1{m}_2}{r^2}$

$F_g=G\frac{m_1{m}_2}{r^2}$

2.0Universal Gravitational Constant "G"

• The universal gravitational constant is a Calculate the magnitude of the gravitational force experienced by each particle. quantity.
• Value of Gravitational Constant in SI unit $G=6.67\times 10^{-11}N-m^2/k{g}^2$ AND in CGS unit $G=6.67\times 10^{-8}dyne-c{m}^2/g^2$
• Dimensional Formula = $\left[M^{-1}{L}^3{T}^{-2}\right]$
• The value of G is constant everywhere and does not depend on the objects’ properties or the medium between them.

Note: This formula is for point masses; for spheres like planets, use the distance between their centers.

3.0Vector Form of Newton's Law of Gravitation

$\begin{array}{rl}& {\vec{r}}_{12}=\text{ Displacement vector from }{m}_1\text{ to }{m}_2\\ & {\vec{r}}_{21}=\text{ Displacement vector from }{m}_2\text{ to }{m}_1\\ & {\vec{F}}_{21}=\text{ Gravitational Force exerted on }{m}_2\text{ by }{m}_1\\ & {\vec{F}}_{12}=\text{ Gravitational Force exerted on }{m}_1\text{ by }{m}_2\\ & \end{array}$

${\vec{F}}_{12}=-G\frac{m_1{m}_2}{r_{21}^2}{\hat{r}}_{21}=-G\frac{m_1{m}_2}{r_{21}^3}{\vec{r}}_{21}$

Negative sign shows that:

1. The direction of ${\vec{F}}_{12}$ is opposite to that ${\vec{r}}_{21}$
2. The gravitational force is attractive in nature

${\vec{F}}_{21}=-G\frac{m_1{m}_2}{r_{12}^2}{\hat{r}}_{12}$

or

${\vec{F}}_{21}=-G\frac{m_1{m}_2}{r_{12}^3}{\vec{r}}_{12}\Rightarrow {\vec{F}}_{12}=-{\vec{F}}_{21}$

The gravitational forces between two bodies are equal in size and opposite in direction.

4.0Characteristics of Gravitational Force

1. Gravitational force between two bodies forms an equal and opposite action-reaction pair.
2. It is a central force, acting along the line connecting their centers.
3. The force is unaffected by the medium between the bodies.
4. It does not depend on the presence of other bodies.
5. Negligible for light objects but significant for massive ones like stars and planets.
6. It is a long-range force, effective even over large distances.

For example, the gravitational force between the Sun and Earth is about 10²⁷ N, despite being 1.5 x 10⁷ km

5.0Force Due To Multiple Particles

• The total gravitational force on a particle is the vector sum of forces from all individual masses.

$\vec{F}={\vec{F}}_{01}+{\vec{F}}_{02}+\dots \dots \dots +{\vec{F}}_{0n}={\sum }_{i=1}^n{\vec{F}}_{0i}$

$\begin{array}{rl}& {\vec{F}}_{01}=\frac{G{m}_0{m}_1}{r^2}{\hat{r}}_{10}=\frac{G{m}_0{m}_1}{{\left|{\vec{r}}_{01}\right|}^2}{\hat{r}}_{01}\\ & {\vec{F}}_{02}=-\frac{G{m}_0{m}_2}{r^2}{\hat{r}}_{10}=\frac{G{m}_0{m}_2}{{\left|{\vec{r}}_{02}\right|}^2}{\hat{r}}_{02}\\ & {\vec{F}}_{0n}=-\frac{G{m}_0{m}_n}{r^2}{\hat{r}}_{10}=\frac{G{m}_0{m}_n}{{\left|{\vec{r}}_{0n}\right|}^2}{\hat{r}}_{0n}\end{array}$

$\vec{F}=\frac{G{m}_0{m}_1}{{|{\vec{r}}_{01}|}^2}{\hat{r}}_{01}+\frac{G{m}_0{m}_2}{{|{\vec{r}}_{02}|}^2}{\hat{r}}_{02}+\dots \dots \dots +\frac{G{m}_0{m}_n}{{|{\vec{r}}_{0n}|}^2}{\hat{r}}_{0n}$

$\vec{F}=G{m}_0\left[\frac{m_1{\hat{r}}_{01}}{{|{\vec{r}}_{01}|}^2}+\frac{m_2{\hat{r}}_{02}}{{|{\vec{r}}_{02}|}^2}\right]+\dots ..+\frac{m_n{\hat{r}}_{0n}}{{|{\vec{r}}_{0n}|}^2}$

6.0Solved Problems

Q1. Four point masses each of mass 'm' are placed on the corner of the square of side 'a'. Calculate the magnitude of the gravitational force experienced by each particle.

Solution:

F_r=resultant force on each particle = 2F cos 45°+F₁

$F_r=\frac{2G\cdot m^2}{a^2}\cdot \frac{1}{\sqrt{2}}+\frac{G\cdot m^2}{(\sqrt{2}a{)}^2}=\frac{G\cdot m^2}{2a}(2\sqrt{2}+1)$

Q2. Two particles of mass 2kg and 3kg are at points (1, 2) and (4, 6) respectively. Gravitational force on a 3kg particle due to a 2kg particle is:

Solution:

Force or mass 3kg due to 2kg in vector form is

$\vec{F}=-\left|F_g\right|\hat{r}$

$\hat{r}$ ->unit vector from 2 kg to 3 kg

$\hat{r}=\frac{\vec{r}}{r}=\frac{(4\hat{i}+6\hat{j})-(\hat{i}+2\hat{j})}{5}$

$\hat{r}=\frac{3\hat{i}+4\hat{j}}{5}$

$F_g=\frac{G{m}_1{m}_2}{r^2}=\frac{6.67\times 10^{-11}\times 2\times 3}{(5{)}^2}$

$\vec{F}=-\frac{1.60\times 10^{-11}}{5}(3\hat{i}+4\hat{j})N$

Q3. Find net force on 2kg due to 3kg  and 4kg particles.

Solution:

$F_g=\frac{G{m}_1{m}_2}{r^2}$

$F_1=\frac{6.67\times 10^{-11}\times 2\times 3}{(3{)}^2}=4.44\times 10^{-11}N$

Similarly 

$F_2=\frac{6.67\times 10^{-11}\times 2\times 4}{(2{)}^2}=13.34\times 10^{-11}N$

Here

F₁= Force on 2kg due to 3kg mass

F₂ = Force on 2kg due to 4kg mass

So, Net Force

$F_{net}=F_2-F_1=13.34\times 10^{-11}-4.44\times 10^{-11}=8.9\times 10^{-11}N$