Gravitation | Introduction and Universal Law of Gravitation

INTRODUCTION

When a body is held is hand and then released the body falls vertically downwards. Let us think as to why the body falls towards the earth? Why is it not going up? It was Sir Issac Newton posed this question and himself answer it.

It is said that when Newton was sitting under a tree, an apple fall in him. The fall the apple made Newton start thinking. He thought that if the earth can attract an apple can it not attract the moon? Is the force the same in both cases? He conjectured that the same type of force is responsible in both the cases

Newton’s Law of Gravitation

Newton gave a universal law that gave the relationship between the force of attraction between two bodies lying at certain distance.

Newton’s Universal Law of Gravitation

Every object is the universe attracts every other object with force which is proportional to the product of their masses and inversely proportional to the square of the distance between them. The force is along the line joining the centers of two objects indirect along the line joining their centers.

Let two objects A and B of masses M and m lie at a distance $\displaystyle d$ from each other as shown in figure. Let the force of attraction between two objects be F. According to the universal law of gravitation, the force between two objects is directly proportional to the product of their masses. That is,

$\displaystyle F\,\propto \,M\times m$ …(i)

And the force between two objects is inversely proportional to the square of the distance between them, that is,

$\displaystyle F\,\,\propto \,\,\frac{1}{{{d}^{2}}}$ …(ii)

Combining equation (i) and (ii)

$\displaystyle F\,\,\propto \,\,\frac{M\times m}{{{d}^{2}}}$

or $\displaystyle F\,=G\,\frac{M\times m}{{{d}^{2}}}$

Where G is the constant of proportionality and is called as Universal Gravitation Constant.

$\displaystyle F\,=G\,\frac{M\times m}{{{d}^{2}}}$

$\displaystyle F\times {{d}^{2}}=G\,M\times m$

$\displaystyle G=\frac{F\times {{d}^{2}}}{M\times m}$ …(iii)

The SI unit of G can be obtained by substituting the unit of force, distance and mass in equation (iii).

$\displaystyle G=\frac{N{{m}^{2}}}{\mathbf{kg}\times \mathbf{kg}}$

$\displaystyle G=N{{m}^{2}}/\mathbf{k}{{\mathbf{g}}^{2}}$ or $\displaystyle G=N{{m}^{2}}\,k{{g}^{-2}}$

The value of G was found out by Henry Cavendish (1731 – 1810) by using a sensitive balance.

The accepted value of G is $\displaystyle 6.673\times {{10}^{-11}}\,\,\mathbf{N}{{\mathbf{m}}^{\mathbf{2}}}\mathbf{/k}{{\mathbf{g}}^{\mathbf{2}}}$ .

Question: Why is Newton’s law of gravitation called as Universal Law of gravitation?

Solution: The law is universal is the sense that it is applicable to all the bodies, whether the bodies are big or small, whether they are celestial or terrestrial.

Question: If the distance between two objects is increased by a factor of 6, how is the force altered?

Solution: Saying that F is inversely proportional to the square of d means that if d gets bigger by a factor of 6, F becomes $\displaystyle \frac{1}{36}$ times smaller.

Question: The mass of the earth is $\displaystyle 6\times {{10}^{24}}kg$ and that of the moon is $\displaystyle 7.4\times {{10}^{22}}kg$ . If the distance between the earth and the moon is $\displaystyle 3.84\times {{10}^{5}}km$ , calculate the force exerted by the earth on the moon. [ $\displaystyle G=6.7\times {{10}^{-11}}N{{m}^{2}}\,k{{g}^{-2}}$ ]

Solution: The mass of the Earth, $\displaystyle M=6\times {{10}^{24}}\,F$

The mass of the Moon, $\displaystyle m=7.4\times {{10}^{22}}\,kg$

The distance between the earth and the moon,

$\displaystyle d=3.84\times {{10}^{5}}\,km$

$\displaystyle =3.84\times {{10}^{5}}\times 1000\,m$

$\displaystyle =3.84\times {{10}^{8}}\,m$

$\displaystyle G=6.7\times {{10}^{-11}}\,\,N{{m}^{2}}\,\,k{{g}^{-2}}$

From equation 10.4, the force exerted by the earth on the moon is

$\displaystyle F=G\frac{M\times m}{{{d}^{2}}}$

$\displaystyle =\frac{6.7\times {{10}^{-11}}N{{m}^{2}}/k{{g}^{2}}\times 6\times {{10}^{24}}kg\times 7.4\times {{10}^{22}}kg}{{{(3.84\times {{10}^{8}}m)}^{2}}}$ $\displaystyle =2.01\times {{10}^{20}}\,\,N$

Thus, the force exerted by the earth on the moon is $\displaystyle 2.01\times {{10}^{20}}N$ .

Question: Calculate the force of gravitation due to a child to a child of mass 25 kg on his fat mother of mass 75 kg if the distance between their centers is 1m from each other.

Given: $\displaystyle G=\frac{20}{3}\,\times {{10}^{-11}}N{{m}^{2}}kg$

Solution: Here, $\displaystyle {{m}_{1}}=25\,kg;\,\,\,\,{{m}_{2}}=75\,kg;\,\,\,\,d=1m;\,\,\,\,\,G=\frac{20}{3}\times {{10}^{-11}}N{{m}^{2}}/k{{g}^{2}}$

$\displaystyle F=\frac{2\times {{10}^{-11}}\times 75\times 25}{3\times {{(1)}^{2}}}$

$\displaystyle F=12,500\times {{10}^{-11}}$

$\displaystyle F=1.25\times {{10}^{-7}}N$

Since, this force is too small i.e., 0.000,000,125 N, hence the child or the mother is attracted by a negligible force due to gravitation and is not noticeable.

Question: A mass of 50 kg attracted by a mass of 20 kg lying at a distance of 2m with a force of $\displaystyle 1.67\times {{10}^{-8}}N$ . Find the value.

Solution: $\displaystyle F=1.67\times {{10}^{-8}}N$

$\displaystyle d=2m$

$\displaystyle {{m}_{1}}=50\,kg$

$\displaystyle {{m}_{2}}=20\,kg$

$\displaystyle G=?$

$\displaystyle F=\frac{G(50)\,(20)}{2\times 2}=\frac{G(1000)}{4}$

$\displaystyle 1.67\times {{10}^{-8}}=\frac{G(1000)}{4}$

$\displaystyle G=\frac{4\times 1.67\times {{10}^{-8}}}{1000}$

$\displaystyle G=6.68\times {{10}^{-11}}\,\,N{{m}^{2}}/k{{g}^{2}}$

Question: Calculate the force of gravitation due to earth on a child weighing 10 kg standing on the ground. (Mass of earth = $\displaystyle 6\times {{10}^{24}}kg$ ; Radius of earth = $\displaystyle 6.4\times {{10}^{3}}\,km$ & $\displaystyle G=6.7\times {{10}^{-11}}N{{m}^{2}}/k{{g}^{2}}$ )