Class09 : Gravitation

In this tutorial we will study about ….

• Introduction of Gravitation.
• Universal Law of Gravitation.
• Free Fall.
• Expression for acceleration due to gravitation ‘g’.
• Calculation of value of g.
• Mass.
• Weight.

Introduction

Earth attracts everything towards it by an unseen force of attraction. This force of attraction is known as gravitation or gravitation pull.

Universal Law of Gravitation

Every object in the universe attracts other object by a force of attraction, called gravitation, which is directly proportional to the product of masses of the objects and inversely proportional to the square of distance between them. This is called Law of Gravitation or Universal Law of Gravitation.

The distance is considered between the centres of the objects.

Suppose there are two objects having mass M and m respectively.

The distance between their centres is equal to d.

The force of attraction is F.

Therefore, from Law of Gravitation which states that force of attraction by which an object attracts other object is directly proportional to the product of their masses,

Thus, $F\propto M\cdot m$ -----(i)

Now, Law of Gravitation also states that force of attraction by which an object attracts other object is inversely proportional to the square of distance between them.

Therefore, $F\propto \frac{1}{d^2}$ ----(ii)

Now from equation (i) and (ii) we get

$F\propto \frac{M\cdot m}{d^2}$
$\Rightarrow F=G\cdot \frac{M\cdot m}{d^2}$ -----(iii)

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

From equation (iii)

$F\times d^2=G\cdot M\cdot m$

$\Rightarrow G=\frac{F{d}^2}{M\cdot m}$ ----(iv)

The expression (iii) and (iv) are called expression for Universal Law of Gravitation.

This Law is applicable everywhere in universe, thus it is known as UNIVERSAL LAW OF GRAVIATION.

SI Unit of G

The SI unit of $G$ is $N{m}^{2}k{g}^{-2}$

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

The value of G was found out by Henry Cavendish, a British philosopher and scientist.

Importance of Universal Law of Gravitation

• This Law is applicable to every object in the universe.
• This law explains the cause of revolution of moon around earth and revolution of planets around sun.
• This law explains the cause of neap and tide due to moon and sun.

Free Fall

When an object falls from any height under the influence of gravitational force only, it is known as free fall. In the case of free fall no change of direction takes place but the magnitude of velocity changes because of acceleration.

This acceleration acts because of the force of gravitation and is denoted by ‘g’. This is called acceleration due to gravity.

Expression for acceleration due to gravitation ‘g’.

Let mass of the object put under free fall = m.

And acceleration due to gravity = g.

Therefore, according to Newton’s Second Law of Motion which states that Force is the product of mass and acceleration,

F = m x g -----------------(i)

Now, according to Universal Law of gravitation,

$F=G\cdot \frac{M\cdot m}{d^2}$ ---(ii)

Thus, from above two expressions, we get

$mg=G\cdot \frac{M\cdot m}{d^2}$
$\Rightarrow g=\frac{GMm}{d^2\cdot m}$
$\Rightarrow g=\frac{GM}{d^2}$ ---(iii)

Where, 

g is acceleration due to gravity,

G is the Universal Gravitational Constant.

M is the mass of earth.

And d is the distance between object and centre of earth.

When object is near the surface of earth

When an object is near the surface of earth, the distance between object and centre of the earth will be equal to the radius of earth because the distance of object is negligible in comparison of the radius of earth.

Let the radius of earth is equal to R.

Therefore, after substituting ‘R’ at the place of ‘d’ we get,

$g=\frac{GM}{R^2}$ ----(iv)

Since, earth is not a perfect sphere rather it has oblique shape. Therefore, radius at the equator is greater than at the poles.

Since, value of ‘g’ is reciprocal of the square of radius of earth, thus, the value of ‘g’ will be greater at the poles and less at the equator.

And the value of ‘g’ will decrease with increase of distance of object from earth.

Calculation of value of $g$

We know that,

👉 The accepted value of $G$ is $6.673\times {10}^{-11}N{m}^{2}k{g}^{-2}$

👉 The mass of earth, $M=6\times {10}^{24}kg$

👉 The radius of earth, $R=6.4\times {10}^{6}m$

Therefore, by using expression $g=\frac{GM}{R^2}$, the value of $g$ can be calculated.

Therefore, after substituting the value of G, M and R in the expression for ‘g’ we get.

Mass

Mass is the measurement of inertia and inertia is the property of any object which opposes the change in state of the object. It is inertia because of which an object in rest has tendency to remain in rest and an object in motion has tendency to remain in motion.

Inertia depends upon the mass of an object. Object having greater mass has greater inertia and vice versa. Mass of an object remains constant everywhere, i.e. mass will remain same whether that object is at the moon, at the earth or anywhere in the universe.

Weight:

Earth attracts every object towards it. We know that force is the product of mass and acceleration due to gravity.

This means, $F=m\times g$  ----(i)

The force by which earth attracts an object towards it is called the weight of the object, which is the product of mass (m) of the object and acceleration due to gravity (g).

Weight is denoted by ‘W’.

Therefore, by substituting in the expression ‘F = mg’ we get,

$W=m\times g$   ----(ii)

Since weight is the force which is acting vertically downwards, therefore, weight has both magnitude and direction and hence it is a vector quantity.

Since the value of ‘g’ is always constant at a given place,

Therefore, expression ‘W = m x g’ can be written as follows:

$W\propto m$  ------(iii)

This means weight of any object is directly proportional to its mass, i.e. weight will increase with the increase of mass and decrease with decrease in mass.

This is the cause that weight of any object is the measure of its mass.

The unit of weight

Since, weight of an object is equal to the force by which an object is attracted towards earth, therefore, unit of weight is same as the unit of force.

Therefore, Unit of weight is ‘newton (N)’.

Weight of an Object on the Surface of Moon

Since, weight of an object on the earth is the force by which earth attracts that very object towards it. In similar way, weight of an object on the surface of moon or any other planet is the force by which moon or any other planet will attract the object towards it.

We know that,

The Mass of Earth = 5.98 × 10 24 kg

Radius of earth = 6.37 × 106 m

Mass of moon = 7.36 × 1022 kg

Radius of moon = 1.76 × 106 m

Since, mass of the moon is less than that of earth, therefore, moon will exert less force of attraction on any object; in comparison to the earth.

Let mass of an object is ‘m’

• The weight of the object on earth is We
• The weight of the object on moon is Wm
• Mass of the earth is M
• Mass of the moon is Mm
• Radius of earth is R
• Radius of moon is Rm
• Acceleration due to gravity on earth is ‘g’
• Acceleration due to gravity on moon is ‘gm’.

Therefore,

Weight of the object on earth We = m × g

By substituting the value of ‘g’ from the expression of Universal Law of Gravitation we get

$W_e=m\cdot \frac{GM}{R^2}$

$\Rightarrow W_e=\frac{m\times G\times 5.98\times {10}^{24}kg}{{(6.37\times {10}^{6}m)}^2}$

$\Rightarrow W_e=\frac{m\times G\times 5.98\times {10}^{24}kg}{6.37\times 6.37\times {10}^6\times {10}^6{m}^2}$

$\Rightarrow W_e=\frac{m\times G\times 5.98\times {10}^{24}kg}{40.57\times {10}^{12}{m}^2}$

$\Rightarrow W_e=\frac{m\times G\times 5.98\times {10}^{12}kg}{40.57m^2}$

$\Rightarrow W_e=m\times G\times 0.1474\times {10}^{12}kg{m}^{-2}$

$\Rightarrow W_e=m\times G\times 1.474\times {10}^{11}kg{m}^{-2}$

Weight of the object on moon $W_m=m\times g_m$

By the expression of universal law of gravitation,

$W_m=m\frac{G{M}_m}{R_m^2}$

$\Rightarrow W_m=\frac{m\times G\times 7.36\times {10}^{22}kg}{{(1.74\times {10}^{6}m)}^2}$

$W_m=\frac{m\times G\times 7.36\times {10}^{22}kg}{1.74\times 1.74\times {10}^6\times {10}^6{m}^2}$

$\Rightarrow W_m=\frac{m\times G\times 7.36\times {10}^{22}kg}{3.0276\times {10}^{12}{m}^2}$

$\Rightarrow W_m=\frac{m\times G\times 7.36\times {10}^{10}kg}{3.0276m^2}$

$\Rightarrow W_m=m\times G\times 2.4309\times {10}^{10}kg{m}^{-2}$

Now, $\frac{W_m}{W_e}$ $=\frac{m\times G\times 2.4309\times {10}^{10}kg{m}^{-2}}{m\times G\times 1.474\times {10}^{11}kg{m}^{-2}}$

$\Rightarrow \frac{W_m}{W_e}=\frac{2.4309}{1.474\times 10}$

$\Rightarrow \frac{W_m}{W_e}=\frac{2.4309}{14.74}$

$\Rightarrow \frac{W_m}{W_e}==\frac{1}{6.030}\approx \frac{1}{6}$

Therefore, Weight of an object on moon / Weight of an object on earth = 1/6

Or, Weight of an object on the moon = 1/6th of the weight of the object on earth.

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