Newtons Law of Gravitation – Definition, Derivation and Formula

1. Statement of the Law

Every particle in the universe attracts every other particle with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers.
This force acts along the line joining the two particles.

[
F = G \dfrac{m_1 m_2}{r^2}
]

Where:

• [ F ] = Gravitational force between two bodies
• [ m_1, m_2 ] = Masses of the two bodies
• [ r ] = Distance between their centers
• [ G ] = Universal Gravitational Constant

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2. Explanation and Mathematical Derivation

Consider two point masses [ m_1 ] and [ m_2 ] separated by a distance [ r ].

According to Newton, the gravitational attraction between them is proportional to the product of their masses:

[
F \propto m_1 m_2
]

and inversely proportional to the square of the distance between them:

[
F \propto \dfrac{1}{r^2}
]

Combining these relations:

[
F \propto \dfrac{m_1 m_2}{r^2}
]

Introducing the constant of proportionality ( G ):

[
F = G \dfrac{m_1 m_2}{r^2}
]

Where ( G ) is the Universal Gravitational Constant:

Its value is the same throughout the universe and determined experimentally by Cavendish using the torsion balance method.

[
G = 6.674 \times 10^{-11} \text{N·m}^2/\text{kg}^2
]

The negative sign (if vector form is used) indicates that the force is attractive in nature.

Vector Form:

[
\vec{F}_{12} = -G \dfrac{m_1 m_2}{r^2} \hat{r}_{12}
]

Here, [ \hat{r}_{12} ] is the unit vector from mass [ m_1 ] to [ m_2 ].

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Example of Newtons Law of Gravitation

The rotation of Earth around the Sun is a great example to understand the concept of gravitational attraction. The Earth and the Sun are attracted to each other by the gravitational force of attraction. Due to this gravitational force the Earth would have gone towards the Sun, however the Earth remains in its orbit rotating around the Sun.

This is because of the centrifugal force that acts on the Earth in outward direction. Therefore, the force of gravitation attraction towards the Sun gets balanced by the centrifugal force acting outwards i.e. away from the Sun.

The formation of high tides and low tides are also due to the gravitation attraction of the Moon. The gravitational force of attraction between the Earth and the Moon results the ocean water to rise resulting to high tide. When the Moon is on the other side of the Earth there is a low tide on the ocean.

The artificial satellites in their orbit around the Earth is due to the gravitational force of attraction between Earth and satellite. The Earth’s gravity tries to pull the satellite towards  its center, however the centrifugal force of satellite while rotating in a circular orbit around Earth tries to pull outwards i.e. away from the center of the Earth.

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Principle of Superposition of Gravitation

It states that the resultant gravitational force $\vec{F}$ acting on a particle due to a number of masses is equal to the vector sum of the forces exerted by the individual masses on the given particle, i.e.,

[ \vec{F}={{\vec{F}}_{{01}}}+{{\vec{F}}_{{02}}}+\ldots +{{\vec{F}}_{{0n}}}] [=\sum\limits_{{i=1}}^{n}{{{{{\vec{F}}}_{{0i}}}}}]

where $\vec{F}_{01}, \vec{F}_{02}, \dots, \vec{F}_{0n}$ are the gravitational forces on a particle of mass $m_0$ due to particles of masses $m_1, m_2, m_3, \dots, m_n$, respectively.

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3. Dimensions and Units

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4. Key Features

1. The gravitational force is mutual — both bodies experience equal and opposite forces.
2. It is a central force, acting along the line joining the centers of the two masses.
3. It is an inverse-square law — if the distance doubles, force becomes one-fourth.
4. It is a long-range force — acts even at astronomical distances.
5. It is an always attractive force (never repulsive).
6. It depends only on mass and distance, not on the medium.
7. It provides the basis for planetary motion, satellites, and orbital mechanics.

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5. Important Formulas to Remember

(Refer to Differentiation for how inverse-square dependence leads to radial acceleration in orbital motion.)

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6. Conceptual Questions with Solutions

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7. FAQ / Common Misconceptions

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8. Practice Questions (Step-by-Step Solutions)

Q1. Find the gravitational force between two bodies of masses [ 20 \text{kg} ] and [ 5 \text{kg} ] separated by [ 10 \text{m} ].
Solution:
[F = G \dfrac{m_1 m_2}{r^2}] [= (6.67 \times 10^{-11}) \dfrac{(20)(5)}{(10)^2}] [= 6.67 \times 10^{-10} \text{N}]

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Q2. By what factor does the gravitational force change if distance becomes three times?
Solution:
[F’ = \dfrac{F}{3^2} = \dfrac{F}{9}]
Force becomes one-ninth of original.

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Q3. The gravitational force between two masses is [ 2 \text{N} ] at a distance [ 3 \text{m} ]. What will be the force at [ 6 \text{m}] ?
Solution:
[F’ = 2 \times \left(\dfrac{3}{6}\right)^2] [= 2 \times \dfrac{1}{4} = 0.5 \text{N}]

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Q4. Find the gravitational constant [ G ] if two bodies of [ 10 \text{kg} ] each attract with force [ 6.67 \times 10^{-9} \text{N} ] at [ 1 \text{m} ].
Solution:
[F = G \dfrac{m_1 m_2}{r^2} \Rightarrow G] [= \dfrac{F r^2}{m_1 m_2}] [= \dfrac{6.67 \times 10^{-9} (1)^2}{10 \times 10}] [= 6.67 \times 10^{-11} \text{N·m}^2/\text{kg}^2]

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Q5. The mass of Earth is [ 6 \times 10^{24} \text{kg} ] and radius is [ 6.4 \times 10^6 \text{m} ]. Find [ g ] using Newton’s law.
Solution:
[g = G \dfrac{M_E}{R_E^2}] [= 6.67 \times 10^{-11} \times \dfrac{6 \times 10^{24}}{(6.4 \times 10^6)^2}] [= 9.8 \text{m/s}^2]