Newton’s law of Universal Gravitation | Definition , Equation , Facts and Solved Examples.

Introduction

Before 1687, a large amount of data had been collected on the motions of the Moon and the planets, but a clear understanding of the forces related to these motions was not available. There was no answer to the questions that people asked about the night sky. Why doesn’t the moon fall to earth? Why do the planets move across the sky ? Why earth is remaining in orbit around the sun ? With the help of gravitation we can give the answer of all these questions and many more other questions . We know that there are four type of forces exist in the nature , gravitation is one of them .

In the 17th century Newton discovered that there must be some force that makes an apple fall out of a tree also keeps the planets in their orbits around the sun. This was the beginning of the study of the motion of objects in space. Newton analyzed astronomical data of the motion of the Moon around the Earth . In Mathematical Principles of Natural Philosophy , Newton published his work on the law of gravity .

Newton’s law of universal gravitation

According to Newton’s law of universal gravitation , every particle of the universe attract to each other with a force and the magnitude of the force is directly proportional to the product of their masses and inversely proportional to the square of the distance between them .

Thus , the magnitude of the gravitational force F that two particles of masses  m1 and m2 , separated by a distance r ,  there exists an attraction , which is proportional to the product of their masses and inversely proportional to the square of distance between the exert force on each other is given by 

F = Gm1m2 / r²

Where G is a constant , Called the universal gravitation constant . It’s value in SI units is

G = 6.674×10-11 N.m²/kg²

Applications of Gravitational  force ( Gravity )

  • Because of Gravitational attraction we are able to feet on the ground .
  • Due to gravitational force air is around the Earth.
  • Gravitational force makes the Earth roughly spherical.
  • Gravitational force keeps the Moon (and all those artificial satellites) in orbit around the Earth, the Earth in orbit around the Sun and the Sun in orbit around the galactic center.
  • Gravitational force creates the water pressure that runs hydroelectric turbines at the bottom of dams.
  • Gravitational force brings the rain back down after it evaporates into the atmosphere , etc .

Solved Examples ( Based on Universal Gravitational Law ) 

  • Problem 1 . Determine the gravitational force if the mass of two bodies are 30 kg and 60 kg and they are separated by a distance of 5m .

Solution :
Given:
m1 = 60 kg, m2 = 30 kg,
r = 5m and
G = 6.67 x 10–11 Nm²/kg².

Universal gravitation formula is given by,

F=Gm1m2/r²

F =6.67×10–11× 30×60 /25

F =4.28×10–9 N

  • Problem 2 . Calculate the magnitude of gravitational force of attraction between earth and sun due to their masses with help of data given below .

Take, the mass of the Sun = 1.99 x 10³⁰ Kg.
Earth’s mass = 6 x 10²⁴ Kg
G = 6.67 x 10-11 N-m²/kg²
and distance of Earth from sun =1.49 x 10¹¹ m

Solution :
Given , mass of sun ( m1 ) = 1.99 x 10³⁰ Kg.

mass of the Earth ( m2 ) = 61 x 10²⁴ Kg

G = 6.67 x 10-11 N-m²/kg²

and distance between the Sun and the Earth ( r ) =1.49 x 10¹¹ m

Force = Gm1m2/r²

= 6.67×10-11×( 1.99 × 10³⁰ )×( 6× 10²⁴ )/( 1.49 × 10¹¹ ) ²

F = 3.587×10²² N .

Gravitational force and The principle of Superposition .

Superposition principle help us to determine the net force ( or resultant) on a particle due to group of particles .

According to this principle the net effect on a body is the sum of the individual effects . Here , the principle means that first we will determine the individual gravitational forces that act on our selected particle due to the other particles then we find the net force by adding all the forces
vectorially .

F1,net = F1,2 + F1,2+ F1,3 + F1,4 ……..F1,n ( All the force are in their vector form )

Here F1,net is the net force on particle 1 due to the other particles and, for example, F1,2 is the force on particle 1 from particle 2 .

Solved Examples ( Based on Gravitation and Principle of Superposition )

Problem 1. There are two bodies A and B of masses m and 3m respectively are kept at a distance d from each other . For zero net gravitational force on a particle due to the bodies A and B , where should be small particle be placed ?

Solution – We know that gravitational force act along the central line . So , we must be placed the particle between bodies A and B to cancel out the gravitational force on the particle . Let the mass of the particle is m’. Suppose it’s distance from A is x , then it’s distance from B will be ( d-x )

The force due to A on m’ is

F1 = Gmm’/x² ( Towards A )

and force due to B on m’ is

F2 = 3Gmm’/(d-x)² ( Towards B )

Condition for zero net force is F1 = F2

Thus ,
Gmm’/x² = 3Gmm’/( d-x )²

( d-x )² = 3x²
d-x = ± ( 3 )1/2x

d = ( 1± 31/2 ) x

x = d/1+(3)1/2

or x = d/1- (3)1/2

As x cannot be negative
so , x = d/1+(3)1/2

Gravitational force between a uniform Spherical shell and a Particle

If we have to find the force on a object due to a uniform spherical shell . we can avoid the integration by assuming that the shell’s mass is concentrated at the center of the object and treating it as point mass . There are two type of possibilities ( as we discuss blow ) give us the force on a particle when the extended object is spherical shell .

Case I 

If a particle of mass m is located at P, a point outside a spherical shell of mass M , the shell attracts the particle as though the mass of the shell were concentrated at its center , i.e.,

F=GMm/r²

Fig. 20.2 Uniform spherical shell

Case II 

If the particle of mass m is located at Q, a point inside the shell, the force acting on it is zero, i.e ., F = 0.

NOTE – We can express these two important results as follows :

F=GMm/r² for r ≥ R

F=0 for r ≤ R

Gravitational force between a uniform Solid sphere and a Particle

As in case of Spherical shell we can assume in case of solid sphere that all the mass of the solid sphere is concentrated in the center of the sphere if the particle is out side the sphere . 

Fig. 20.3 Uniform solid sphere

Case I :

If a particle of mass m is located at P (figure), a point outside a homogeneous solid sphere of mass M, the sphere attracts
the particle as though the mass of the particle were concentrated at its center ,
i.e .,

F = GMm/r²

[This result follows from case-I (for a spherical shell), since a solid sphere can be considered as a collection of concentric spherical shells] .

Case II :

If a particle of mass m is located at Q, a point inside a homogeneous solid sphere of mass M ( figure 21.3 , the force on m is due only to the mass Mr contained within the sphere of radius r (represented by the dashed circle ) ,
i.e.,
F = GMrm/r² = GMrm/R³

NOTE – We can express these two important results as follows:

F = GMm/r² for r ≥R
or
F = GMmr/R³ for r ≤ R

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