Ungrouped Data: When a data collection is vast, a frequency distribution table is frequently used to arrange the data. A frequency distribution table provides the...
Ungrouped Data: Know Formulas, Definition, & Applications
December 11, 2024When you throw anything up, it again falls to the surface of the Earth. Why does this happen? Why is it easy to go downhill than going uphill? Why do the planets move around the sun? The answer to all of these questions is the gravitational force of attraction. Gravitational force is a universal force of attraction that acts between all the matter present in the universe. The idea of gravitational force was first explained by Issac Newton.
Newton was the first one to suggest that the gravitational force is universal and affects all objects in the universe. Newton’s Universal Law of Gravitation states that every point mass in the universe attracts every other point mass with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. It applies to all material bodies irrespective of their sizes. It applies to heavenly bodies like planets, stars, and it is also applicable to very small-sized particles like balls, pens, etc.
Now on combining the equations \((1)\) and \((2)\), we will get,
\(\vec F \propto \frac{{{M_1}{M_2}}}{{{r^2}}}\)
\( \Rightarrow \vec F = G\frac{{{M_1}{M_2}}}{{{r^2}}}\hat n\)
Where,
\(F = \) Gravitational force between two objects, measured in Newton \((N)\).
\(\hat n = \) It is vector in the direction of the line joining centres of the two masses.
\(G = \) Universal Gravitational Constant \( = 6.674 \times {10^{ – 11}}\;{\rm{N}}\;{{\rm{m}}^2}\;{\rm{k}}{{\rm{g}}^{ – 2}}\).
\({M_1} = \) Mass of first body measured in \({\rm{Kg}}\).
\({M_2} = \) Mass of Second body measured in \({\rm{Kg}}\).
\(R = \) Distance between two bodies measured in meter \((m)\).
Gravitational force is a universal force by which all the objects having mass are attracted towards each other. Gravitational force is very important to keep the universe functioning:-
1. The gravitational force between two bodies forms the action-reaction pair that means if the first body attracts the second body with the force of attraction \(F\) then the second body also attracts the first body with the same magnitude of force \(F\).
2. Gravitational force is a central force because the gravitational force between two masses is always acting along the line joining the centre of the two masses.
3. The gravitational force between the two masses is independent of the medium between the two masses and also independent of their sizes or distribution of their masses.
4. The work done by the gravitational force is independent of the path between the initial and the final position, so we can say the gravitational force is conservative.
5. It is a field force.
6. The magnitude of the gravitational force between two objects will not depend on the medium between them.
Unit of \(G\)
By Newton’s law of gravitation,
We have: \(\vec F = G\frac{{{M_1}{M_2}}}{{{r^2}}}\hat n\)
\( \Rightarrow G = \frac{F}{{{M_1}}}\frac{{{r^2}}}{{{M_2}}}\)
SI Units of \(G = \frac{{{\text{Unit}}\,{\text{of}}\,{\text{force}}\,{\text{(Unit}}\,{\text{of}}\,{\text{distance)}}{{\text{}}^2}}}{{{\text{(Unit}}\,{\text{of}}\,{\text{mass)(Unit}}\,{\text{of}}\,{\text{mass)}}}}\)
SI Units of \(G = \frac{{{\text{N}}\,{{\text{m}}^{\text{2}}}}}{{{\text{(kg)}}\,{\text{(kg)}}}}{\text{=N}}\,{{\text{m}}^{\text{2}}}\,{\text{k}}{{\text{g}}^{{\text{-2}}}}\)
Dimensions of \(G\)
By Newton’s law of gravitation,
We have: \(\vec F = G\frac{{{M_1}{M_2}}}{{{r^2}}}\hat n\)
\( \Rightarrow G = \frac{F}{{{M_1}}}\frac{{{r^2}}}{{{M_2}}}\)
\( \Rightarrow [G] = \frac{{[F]}}{{\left[ {{M_1}} \right.}}\frac{{{{[r]}^2}}}{{\left[ {{M_2}} \right]}} = \frac{{\left[ {{M^1}{L^1}{T^{ – 2}}} \right]}}{{\left[ {{M^1}{L^0}{T^0}} \right]}} \times \frac{{{{\left[ {{M^0}{L^1}{T^0}} \right]}^2}}}{{\left[ {{M^1}{L^0}{T^0}} \right]}}\)
\( \Rightarrow [G] = \left[ {{M^{ – 1}}{L^3}{T^{ – 2}}} \right]\)
Hence, the dimensional formula of \(G\) is \(\left[ {{M^{ – 1}}{L^3}{T^{ – 2}}} \right]\).
If we will compare the strength of gravitational force with other field forces (Electrostatic and Magnetic force), the order of gravitational force is much less. Let us consider an example:-
Why do we not feel the attraction force due to gravitation when we stand near our friends?
To get the answer to the above question, let us calculate the magnitude of the gravitational force acting between you and your colleague as you approach each other at a distance of one meter. We can do this quite simply by using newton’s universal law of gravitation.
\(\vec F = G\frac{{{M_1}{M_2}}}{{{r^2}}}\hat n\quad \ldots (4)\)
Suppose your mass \({M_1}\), is \(60\;{\rm{kg}}\) and the mass of your colleague \({M_2}\) is \(65\;{\rm{kg}}\). Your center-to-center separation \(^{\prime \prime }{r^{\prime \prime }}\) is \(1\;{\rm{m}}\) and \(G = 6.67 \times {10^{ – 11}}\;{\rm{N}}\;{{\rm{m}}^2}\;{\rm{k}}{{\rm{g}}^{ – 2}}\).
After substituting these values into equation \((4)\)
We get:-
\( \Rightarrow F = 6.67 \times {10^{ – 11}}{\rm{N}}\,{{\rm{m}}^2}\;{\rm{k}}{{\rm{g}}^{ – 2}} \times \frac{{60\;{\rm{kg}} \times 65\;{\rm{kg}}}}{{1\;{{\rm{m}}^2}}}\)
\( \Rightarrow F = 2.6013 \times {10^{ – 7}}\;{\rm{N}}\)
This means that you exert a gravitational pulling force of \(0.26\) millionths of a newton on your colleague! The force exists, but it is too small to notice in practice.
From the number, it is clear that because the value of the Universal gravitational constant \((G)\) is so small, the magnitude of the gravitational force will be very small unless one or another of the objects has a very large mass.
Every matter in this universe is experiencing a force of attraction called gravitational force. Newton was the first person who performed experiments and proposed that magnitude of the gravitational force between two masses is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.
Gravitational force, \(\vec F = G\frac{{{M_1}{M_2}}}{{{r^2}}}\hat n\). Here \(\hat n\) represents the direction of Gravitational force and is along the line joining the centre of mass of the two masses.
Q.1. Calculate the gravitational force of attraction between two metal spheres, each of mass \(100\;{\rm{kg}}\), if the distance between their centres is \(50\;{\rm{cm}}\). Given \(G = 6.67 \times {10^{ – 11}}\;{\rm{N}}\;{{\rm{m}}^2}\;{\rm{k}}{{\rm{g}}^{ – 2}}\).
Ans: Given: – Mass of the first body \( = 100\;{\rm{kg}}\),
Mass of the second body \( = 100\;{\rm{kg}}\),
Distance between the masses \( = r = 50\;{\rm{cm}} = 50 \times {10^{ – 2}} = 0.5\;{\rm{m}}\)
Universal Gravitation Constant, \(G = 6.67 \times {10^{ – 11}}\;{\rm{N}}\;{{\rm{m}}^2}\;{\rm{k}}{{\rm{g}}^{ – 2}}\).
Force of attraction between the two masses \( = F = \)?
According to Newton’s law of gravitation,
We have: – \(\vec F = G\frac{{{M_1}{M_2}}}{{{r^2}}}\hat n\)
\( \Rightarrow F = 6.67 \times {10^{ – 11}}\;{\rm{N}}\;{{\rm{m}}^2}\;{\rm{k}}{{\rm{g}}^{ – 2}} \times \frac{{100\;{\rm{kg}} \times 100\;{\rm{kg}}}}{{{{0.5}^2}\;{{\rm{m}}^2}}}\)
\( \Rightarrow F = 2.668 \times {10^{ – 6}}\;{\rm{N}}\)
Ans: – Force of attraction between the two masses is \(2.668 \times {10^{ – 6}}\;{\rm{N}}\).
Q.2 A sphere of mass \({\rm{50}}\,{\rm{Kg}}\) is attracted by another spherical mass of \({\rm{20}}\,{\rm{Kg}}\) by force of \(9.8 \times {10^{ – 7}}\rm{N}\) when the distance between their centre is \(0.3\;{\rm{m}}\). Then find the value of \(G\).
Ans: Given: – Mass of the first body \({\rm{ = 50}}\,{\rm{Kg}}\),
Mass of the second body \({\rm{ = 20}}\,{\rm{Kg}}\),
Distance between the masses \( = r = 0.3\;{\rm{m}}\)
Force of attraction between the two masses \( = F = 9.8 \times {10^{ – 7}}\;{\rm{N}}\)
To find: Universal Gravitation Constant, \(G = \)?
According to Newton’s law of gravitation,
We have: \(\vec F = G\frac{{{M_1}{M_2}}}{{{r^2}}}\hat n\)
\( \Rightarrow G = \frac{F}{{{M_1}}}\frac{{{r^2}}}{{{M_2}}}\)
\( \Rightarrow G = \frac{{9.8 \times {{10}^{ – 7}}\rm{N} \times {{0.3}^2}\;{{\rm{m}}^2}}}{{50\;{\rm{kg}}}}\)
\( \Rightarrow G = 8.82 \times {10^{ – 11}}{\rm{N}}\,{{\rm{m}}^2}\;{\rm{k}}{{\rm{g}}^{ – 2}}\)
Ans: – The value of the Universal gravitation constant is \(8.82 \times {10^{ – 11}}{\rm{N}}\,{{\rm{m}}^2}\;{\rm{k}}{{\rm{g}}^{ – 2}}\).
Q.1. Are gravity and gravitation the same?
Ans: No, Gravity is the Earth’s gravitational pull on a body, lying near the surface of the Earth but Gravitation is the force of attraction acting between any two bodies of the universe.
Q.2. What causes spring tides?
Ans: When the moon, and sun are in one line, their gravity combines to produce a high spring tide.
Q.3. What is the use of Newton’s law of gravitation as applied to the planets?
Ans: The law of gravitation is applied to find the force of attraction exerted by planets on the other bodies and also to determine the motion of the planets around the sun.
Q.4. Why is the Earth’s motion not noticeable when an apple falls to the ground, does the Earth moves up due to the gravitational pull of the apple?
Ans: We know that the gravitational force acting on both Earth and apple will equal in magnitude because of the form of action and reaction pair. Since the mass of the Earth is much larger than the apple, the acceleration produced on the Earth is so small. That’s why it is not noticeable.