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# State the Newton law of gravitation. Hence define $G$ what are the units and dimensions of$G$? Why is it called a universal gravitational constant?

Last updated date: 13th Jun 2024
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Hint:-
- Newton’s law of gravitation is one of the fundamental laws in physics.
- This attractive force is very weak compared to other forces present in nature.
- The attractive force is still present even at a far distance.
- This is an inverse square law force.

Complete step by step solution:-
Newton’s universal law gravitation says that ‘there is an attractive force between particles which is directly proportional to their product of masses and inversely proportional to the square of the distance between them’.

Suppose two masses having masses ${m_1}$and ${m_2}$ respectively. The distance between them is$r$.
The gravitational force is given by $F \propto \dfrac{{{m_1}{m_2}}}{{{r^2}}}$
For removing the proportionality there is a constant introduced by Newton called universal gravitational constant. Denoted by the letter$G$.
$F = G\dfrac{{{m_1}{m_2}}}{{{r^2}}}$
This force is addictive and attractive in nature. Even though repulsive forces are present like in electrostatics there also this force is present , but compared to all other forces its very small.
$G$is numerically equal to the gravitational force experienced between two objects that have unit masses and are separated by unit distance.
$G$is a constant quantity everywhere in the universe. Under any physical condition it never changes. So, it is called a universal gravitational constant.

Now we want to find dimension and units.
Any force ($F$) can be written by Newton's second law, as a product of mass ($m$) and acceleration ($a$).
$F = ma$
By Gravitation law
$F = G\dfrac{{{m_1}{m_2}}}{{{r^2}}}$
Equate both equations,
$ma = G\dfrac{{{m_1}{m_2}}}{{{r^2}}}$
Re arrange,
$G = ma\dfrac{{{r^2}}}{{{m_1}{m_2}}}$
Dimensionally
$[a] \equiv {L^1}{T^{ - 2}}$
$[r] \equiv {L^1}$
$[m] \equiv [{m_1}] \equiv [{m_2}] \equiv {M^1}$
$[G] \equiv {M^1}L{T^{ - 2}}\dfrac{{{L^2}}}{{{M^1}{M^1}}} = {M^{ - 1}}{L^3}{T^{ - 2}}$
By this dimensional analysis easily can find the unit
$F = G\dfrac{{{m_1}{m_2}}}{{{r^2}}}$
Rearrange it
$G = F\dfrac{{{r^2}}}{{{m_1}{m_2}}}$
Unit of force ($F$) is Newton ($N$)
Unit of mass ($m$) is kilogram ($Kg$)
Unit of $r$ is meter ($m$)
So unit of $G = N{m^2}k{g^{ - 2}}$
More precisely $N = kgm/{s^2}$
So unit of $G = k{g^1}{m^1}{s^{ - 2}} \times {m^2}k{g^{ - 2}} = {m^3}k{g^{ - 1}}{s^{ - 2}}$

This constant has value $G = 6.674 \times {10^{ - 11}}{m^3}k{g^{ - 1}}{s^{ - 2}}$.

Note:-
- Acceleration due to gravity is explained by this law.
- The total planetary system is explained by this law.
- The earth gravitation is explained by this law.
- This gravitational force is playing the role in constant motion artificial satellites around the earth.
- If there is no gravitational force we can’t walk on the earth's surface.