## Force - Class 10 Science: Unit- 1 [Notes and Exercises]

## Overview

Gravitation is an attractive force existing in all the objects in the universe due to their masses. The force which attracts every object towards the centre of the Earth is called gravity. Sir Isaac Newton (1642-1727) (an English Mathematician and physicist) discovered gravity. Also, he proposed the universal law of Gravitation.

## What is Force?

Force is an external agency that changes or tries to change the position of an object. In general words, Force is the push or pull acting on an object.

The SI unit of Force is Newton (kgm/s²) and the C.G.S. unit of Force is Dyne (gcm/s²).

### Relation between Newton and Dyne

We know,

1 Newton = 1 kg × 1 m/s² [ F = ma ]

= 1× 1000g × 1 ×100cm/s²

= 1000g × 100cm/s²

= 100000 g cm/s²

= 10^5dynes

So, 1 Newton = 10^5dynes

### Gravitation (Gravitational Force)

Gravitational Force is the force of attraction between two bodies in the universe due to their masses. It is always the attracting force.

### Newton's universal law of gravitation:

According to Newton's universal law of gravitation, every object in the universe attracts every other object with a force which is called gravitational force. This gravitational force is (i) directly proportional to the product of their masses and (ii) inversely proportional to the square of the distance between their centre.

This law is called the universal law because it is applicable to every body in the universe regardless of their shape, size, mass, distance between the centres and time.

Prove: F = Gm1.m2 ⁄ r²

Let, the distance between centres of the two objects A and B of masses m1 and m2 respectively be 'r'. If F is the gravitational force between them then, according to Newton's universal law of gravitation:

F ∝ m1.m2 .................(1)

and

F ∝ 1/ r² ......................(2)

Combining equations (1) and (2), we get,

F ∝ m1.m2 / r²

Therefore, F = Gm1.m2 /r²

Where G is the universal gravitational constant.

#proved.

What happens to the force of gravitation if masses of the bodies are doubled keeping the distance between them constant?

Solution:

We have,

Gravitational force (F) = Gm1.m2 /r² ....(i)

According to the question,

Mass of body A (mA) = 2m1

Mass of body B (mB) = 2m2

Distance between A and B = r

Then,

F1 = GmA.mB /r²

[Where F1 is the new gravitational force]

or, F1 = G2m1.2m2 /r²

or, F1 = 4 Gm1.m2 /r²

So, F1 = 4F [ from equation (i)]

Hence, when the mass of both objects is doubled keeping the distance between them constant then the gravitational force will be increased by four times from the initial value.

Practice yourself:

What happens to the force of gravitation if the masses of the bodies are halved keeping the distance between them constant?

What happens to the force of gravitation if the masses of the bodies are doubled and the distance between them is also doubled?

Universal Gravitational Constant (G):

The Universal Gravitational Constant (G) is equal to the force of gravitation between two bodies of unit masses (i.e. of 1 kg each) separated by a unit distance (i.e. 1 m).

SI unit of G is Nm²/kg². Its value is 6.67×10^(-11) Nm²/kg². Its value is always constant because it does not depend upon the masses and distance of the bodies along with the variation in their (bodies and medium separating those bodies) characteristics.

It is a scalar quantity.

Gravity or Weight:

The force which attracts every object towards the centre of the Earth is called gravity or the weight of the body. The SI unit of gravity is Newton (N).

Effects of gravity:

Gravity makes human life very easy. Due to the earth's gravity, we can freely walk, stand, play and perform other activities. Any object that is thrown upwards again returns to the earth's surface. All types of construction of roads, bridges, and buildings are possible.

Mass:

The total amount of matter contained in a body is called its mass. Its SI unit is kilogram (kg). It is a scalar quantity. It is measured using a beam balance. Its value is constant all over the universe.

Acceleration due to Gravity (g):

Acceleration due to gravity (g) is the acceleration produced on a freely falling body due to the gravity present in the Earth.

Its SI unit is m/s² and its average value on the Earth's surface is 9.8 m/s². It is a vector quantity. Its value changes from place to place.

Relation between radius and acceleration due to gravity of the earth:

Let M be the mass of the earth with its radius R. Suppose the body on the surface of the earth has mass m..

Then,

According to Newton's universal law of gravitation, the force of attraction between them is given by:

F = GMm /R^2 .......... (a)

Here, F represents the force by which the body is attracted towards the earth. So F represents the weight(W) of the object. Then, equation a becomes:

W = GMm/R^2 ..........(b)

But we have, from the second law of motion,

W = mg ......................(c)

Hence, from (b) and (c), we get;

mg = GMm/ R^2

or, g = GMm/ R^2m

So, g = GM / R^2

So, you can calculate acceleration due to gravity with the above expression.

Variation in the value of acceleration due to gravity (g):

Variation because of the shape of the Earth: g ∝ 1/R²

This is the reason the value of 'g' is more than in the equatorial region.

Variation with the height from the surface of the Earth: g' = GM / (R+h)²

This is the reason the value of g at higher altitude is less than in the areas of lower altitude.

Variation with the depth of the Earth's surface: g' = (1 - d/R).g

From this relation, as the depth (d) increases, the value of 'g' also decreases.

Gravitational field and gravitational force intensity:

The area around a heavenly body up to which the influence of its gravitational force can be felt is called gravitational field.

Gravitational force intensity at any point is the amount of force experienced by a unit mass kept at that point in the gravitational field of the heavenly body.

Gravitational field intensity (I) = Gravitational Force Experienced (F) / Mass (m)

To calculate the gravitational field intensity, we use the formula,

Gravitational field intensity (I) = Universal Gravitational Constant (G)x Mass of body(M) / Radius of the body (R^2)

i.e. I = GM/ R^2

Therefore, this is numerically equal to acceleration due to gravity.

Free Fall:

When an object is falling towards the surface of the earth under the gravitation force only and without air resistance, then the body is said to be in free fall.

Weightlessness:

Weightlessness is the state in which the resulting weight of the body is zero due to the zero reaction force of the earth or other heavenly bodies. It occurs only when the acceleration due to gravity is zero.

Let us assume that we are ascending upwards in an elevator at an acceleration (g') that equals the downwards pull exerted by gravity (g). Since, these two forces are in opposite directions and equal, they can each other. So, reacting force (or resultant force) = g - g' = g - g = 0

We know,.weight = mass × acceleration due to gravity = mass × 0 = 0.

Thus, the body is said to be weightless. And, such a condition is called weightlessness. There are many more cases for weightlessness, as well.

Solved Numericals:

The mass of Jupiter is 1.9 x 10^27 kg and that of the sun is 2 x 10^30kg. If the distance between them is 78 x 10⁷ km, find the gravitational force between them.

Solution:

We have,

Jupiter's mass (m1) = 1.9 x 10^27 kg

mass of Sun (m2) = 2 x 10^30 kg

distance between them (r) = 78 x 10⁷ km = 78 x 10⁷ x 10³ m = 78 x 10^10 m

Now,

gravitational force (F) = Gm1m2 / r²

= 6.67 x 10^(-11) x 1.9 x 10^27 x 2 x 10^30 / (78 x 10^10)²

= 4.166 x 10^23 N

If the mass of a man is 70kg, what would be his weight on the moon? Find the mass of the same person on the Earth as well as on the moon.

Solution:

mass(m) = 70kg

acceleration due to gravity on moon(gm) = 1/6 * g = (1/6* 9.8)m/s^2 = 1.64m/s^2

So, weight of the man on the moon (W) = m * gm = 70* 1.64 = 114.3N

His man on the earth and the moon would be the same. i.e. 70kg.

What is the reason behind the weight of the body being more at poles than at the equator?

Why is it more difficult to lift a big stone than a small one?

Answer: Gravitational force is the weakest force. Its magnitude depends largely upon the mass of the substance (or object). A bigger stone has comparatively larger mass than the smaller one. Thus, the magnitude of the gravitational force acting on the bigger stone is greater than the smaller stone. This means the bigger stone is pulled more heavily by the Earth's gravity than the smaller one. When we try to lift both stones to the same height, more force has to be exerted on the bigger stone in order to tackle the larger magnitude of downwards pulling gravitational force. Hence, it is more difficult to lift a big stone than a small one.

Simply, due to the heavier mass of the bigger stone, it is pulled downwards by the Earth's gravity more than the smaller one. As a result, when we try to lift it up, the gravitational force acts as a barrier and makes it more difficult to lift a big stone than a small one.

Calculate the gravitational force between two bodies of masses 750 kg and 1500 kg if they are placed at 500 m distance?

(Ans: 3.0 x 10^(-10) N)

The masses of the sun and the earth are 2 x10^30 kg and 6 x10^24 kg respectively. Find the gravitational force between them if the distance between their centres is 1.5 x10^11 m. (Ans: 3.56 x10^22 N)

If you neglect the air resistance, on falling two objects with masses 50kg and 10kg respectively from equal heights, which one would touch the ground first? Give reason in support of your answer.