Gravitation Class 9 Free Notes and Mind Map (Free PDF Download)

This chapter explores gravitation and the universal law of gravitation formulated by Isaac Newton. It covers the motion of objects under the influence of gravitational force on earth, free fall, acceleration due to gravity, the difference between mass and weight, how weight varies from place to place, thrust and pressure, buoyancy, and Archimedes’ principle. Understanding these concepts helps us explain phenomena like falling objects, planetary motion, and why objects float or sink in liquids.

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9.1 Gravitation

We know that the moon goes around the earth. An object when thrown upwards reaches a certain height and then falls downwards. It is said that when Newton was sitting under a tree, an apple fell on him. The fall of the apple made Newton start thinking.

Newton’s Thought Process:

If the earth can attract an apple, can it not attract the moon?
Is the force the same in both cases?

Newton’s Conclusion:

The same type of force is responsible in both cases.
At each point of its orbit, the moon falls towards the earth, instead of going off in a straight line.
So it must be attracted by the earth.
But we do not really see the moon falling towards the earth.

Understanding Moon’s Motion

Activity – Whirling a Stone:

Take a piece of thread.
Tie a small stone at one end.
Hold the other end of the thread and whirl it round.
Note the motion of the stone.
Release the thread.
Again, note the direction of motion of the stone.

Observations:

Before the thread is released:

The stone moves in a circular path with a certain speed.
It changes direction at every point.
The change in direction involves change in velocity or acceleration.

Force responsible:

The force that causes this acceleration and keeps the body moving along the circular path is acting towards the centre.
This force is called the centripetal force (meaning ‘centre-seeking’).

After the thread is released:

In the absence of this force, the stone flies off along a straight line.
This straight line will be a tangent to the circular path.

More to Know – Tangent to a Circle

A straight line that meets the circle at one and only one point is called a tangent to the circle.

Moon’s Motion and Centripetal Force

The motion of the moon around the earth is due to the centripetal force. The centripetal force is provided by the force of attraction of the earth. If there were no such force, the moon would pursue a uniform straight line motion.

Does the Apple Attract the Earth?

It is seen that a falling apple is attracted towards the earth. Does the apple attract the earth? If so, we do not see the earth moving towards an apple. Why?

Answer:

According to the Third Law of Motion:

The apple does attract the earth.

According to the Second Law of Motion:

For a given force, acceleration is inversely proportional to the mass of an object (F = ma).
The mass of an apple is negligibly small compared to that of the earth.
So, we do not see the earth moving towards the apple.

Same reasoning applies to the moon:

The moon attracts the earth.
But the earth’s mass is much larger than the moon’s mass.
So we don’t see the earth moving towards the moon.

Gravitational Force

In our solar system, all the planets go around the Sun. By arguing the same way, we can say that there exists a force between the Sun and the planets.

Newton’s Conclusion:

Not only does the earth attract an apple and the moon, but all objects in the universe attract each other.
This force of attraction between objects is called the gravitational force.

9.1.1 UNIVERSAL LAW OF GRAVITATION

Statement:

Every object in the universe attracts every other object with a force which is:

Directly proportional to the product of their masses
Inversely proportional to the square of the distance between them
The force is along the line joining the centres of the two objects

Mathematical Formulation

Let two objects A and B of masses M and m lie at a distance d from each other. Let the force of attraction between two objects be F.

According to the universal law of gravitation:

1. Force is directly proportional to the product of masses:

F ∝ M × m  ... (9.1)

2. Force is inversely proportional to the square of the distance:

F ∝ 1/d²  ... (9.2)

Combining both:

textF ∝ (M × m) / d²  ... (9.3)

Introducing constant of proportionality G:

F = G × (M × m) / d²  ... (9.4)

where G is the universal gravitation constant.

Finding the Value of G

Multiplying crosswise in Eq. (9.4):

F × d² = G × M × m

Therefore:

G = (F × d²) / (M × m)  ... (9.5)

SI Unit of G:

Substituting units of force, distance, and mass:

Unit of G = N m² kg⁻²

Value of G:

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

Accepted value:

G = 6.673 × 10⁻¹¹ N m² kg⁻²

More to Know

Universal Nature:

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

Inverse-Square Law:

Saying that F is inversely proportional to the square of d means:
If d gets bigger by a factor of 6, F becomes (1/36) times smaller.

Example 9.1

The mass of the earth is 6 × 10²⁴ kg and that of the moon is 7.4 × 10²² kg. If the distance between the earth and the moon is 3.84 × 10⁵ km, calculate the force exerted by the earth on the moon. (Take G = 6.7 × 10⁻¹¹ N m² kg⁻²)

Solution:

Given:

Mass of the earth, M = 6 × 10²⁴ kg
Mass of the moon, m = 7.4 × 10²² kg
Distance, d = 3.84 × 10⁵ km = 3.84 × 10⁵ × 1000 m = 3.84 × 10⁸ m
G = 6.7 × 10⁻¹¹ N m² kg⁻²

Using formula:

F = G × (M × m) / d²
F = (6.7 × 10⁻¹¹ × 6 × 10²⁴ × 7.4 × 10²²) / (3.84 × 10⁸)²
F = 2.02 × 10²⁰ N

Answer: The force exerted by the earth on the moon is 2.02 × 10²⁰ N.

Questions (9.1.1) and Solutions

Q1. State the universal law of gravitation.

Solution:

The universal law of gravitation states that:

Every object in the universe attracts every other object with a force which is:

Directly proportional to the product of their masses
Inversely proportional to the square of the distance between them
The force is along the line joining the centres of the two objects

Mathematical form:

textF = G × (M × m) / d²

where G is the universal gravitation constant.

Q2. Write the formula to find the magnitude of the gravitational force between the earth and an object on the surface of the earth.

Solution:

Formula:

F = G × (M × m) / R²

Where:

F = Gravitational force between earth and object
G = Universal gravitation constant (6.673 × 10⁻¹¹ N m² kg⁻²)
M = Mass of the earth (6 × 10²⁴ kg)
m = Mass of the object
R = Radius of the earth (6.4 × 10⁶ m)

9.1.2 IMPORTANCE OF THE UNIVERSAL LAW OF GRAVITATION

The universal law of gravitation successfully explained several phenomena which were believed to be unconnected:

1. The force that binds us to the earth

Explains why we stay on the ground and don’t float away.

2. The motion of the moon around the earth

The gravitational force provides the centripetal force needed for circular motion.

3. The motion of planets around the Sun

The Sun’s gravitational force keeps planets in their orbits.

4. The tides due to the moon and the Sun

Gravitational pull of the moon and Sun on earth’s oceans causes tides.

9.2 Free Fall

Activity – Throwing a Stone:

Take a stone.
Throw it upwards.
It reaches a certain height and then it starts falling down.

Explanation:

We have learnt that the earth attracts objects towards it. This is due to the gravitational force.

Definition of Free Fall:

Whenever objects fall towards the earth under gravitational force alone, we say that the objects are in free fall.

Characteristics of Free Fall

Question: Is there any change in the velocity of falling objects?

Answer: Yes.

Details:

While falling, there is no change in the direction of motion.
But due to the earth’s attraction, there will be a change in the magnitude of velocity.
Any change in velocity involves acceleration.

Acceleration Due to Gravity:

Whenever an object falls towards the earth, an acceleration is involved. This acceleration is due to the earth’s gravitational force.

Definition: This acceleration is called the acceleration due to the gravitational force of the earth or acceleration due to gravity.

Symbol: g

Unit: m s⁻² (same as acceleration)

Relation Between Force, Mass, and Acceleration

From the second law of motion:

Force = Mass × Acceleration

For an object in free fall:

Mass = m
Acceleration due to gravity = g
Gravitational force F = mg … (9.6)

From universal law of gravitation:

F = G × (M × m) / d²  ... (9.4)

Equating (9.6) and (9.4):

mg = G × (M × m) / d²

Simplifying:

g = G × M / d²  ... (9.7)

where:

M = Mass of the earth
d = Distance between object and earth

For Objects on or Near Earth’s Surface

Let an object be on or near the surface of the earth. The distance d will be equal to R (radius of the earth).

Thus:

textmg = G × (M × m) / R²  ... (9.8)

Simplifying:

g = G × M / R²  ... (9.9)

Important Point:

The earth is not a perfect sphere. As the radius of the earth increases from the poles to the equator, the value of g becomes:

Greater at the poles
Smaller at the equator

For most calculations, we can take g to be constant on or near the earth (approximately 9.8 m s⁻²). But for objects far from the earth, acceleration due to gravity is given by Eq. (9.7).

9.2.1 TO CALCULATE THE VALUE OF g

To calculate the value of g, we should put the values of G, M, and R in Eq. (9.9):

Given:

Universal gravitational constant, G = 6.7 × 10⁻¹¹ N m² kg⁻²
Mass of the earth, M = 6 × 10²⁴ kg
Radius of the earth, R = 6.4 × 10⁶ m

Calculation:

g = G × M / R²
g = (6.7 × 10⁻¹¹ × 6 × 10²⁴) / (6.4 × 10⁶)²
g = 9.8 m s⁻²

Answer: The value of acceleration due to gravity of the earth is 9.8 m s⁻².

9.2.2 MOTION OF OBJECTS UNDER THE INFLUENCE OF GRAVITATIONAL FORCE OF THE EARTH

Question: Do all objects (hollow or solid, big or small) fall from a height at the same rate?

Activity – Paper and Stone:

Take a sheet of paper and a stone.
Drop them simultaneously from the first floor of a building.
Observe whether both reach the ground simultaneously.

Observation:

The paper reaches the ground later than the stone.

Reason:

This happens because of air resistance.
The air offers resistance due to friction to the motion of falling objects.
The resistance offered by air to the paper is more than the resistance offered to the stone.

In Vacuum:

If we do the experiment in a glass jar from which air has been sucked out, the paper and the stone would fall at the same rate.

Important Conclusion

We know that an object experiences acceleration during free fall. From Eq. (9.9):

g = G × M / R²

This acceleration g is independent of the mass of the object (m does not appear in the equation).

This means:

All objects (hollow or solid, big or small) should fall at the same rate.

Historical Note:

According to a story, Galileo dropped different objects from the top of the Leaning Tower of Pisa in Italy to prove the same.

Equations of Motion Under Gravity

As g is constant near the earth, all the equations for uniformly accelerated motion become valid with acceleration a replaced by g.

The equations are:

1. First Equation:

v = u + gt  ... (9.10)

2. Second Equation:

s = ut + (1/2)gt²  ... (9.11)

3. Third Equation:

v² = u² + 2gs  ... (9.12)

where:

u = Initial velocity
v = Final velocity
s = Distance covered
t = Time
g = Acceleration due to gravity

Sign Convention

Important:

Acceleration a is taken as positive when it is in the direction of velocity (direction of motion).
Acceleration a is taken as negative when it opposes the motion.

For objects thrown upwards:

Velocity is upward (positive direction)
Gravity acts downward (negative direction)
So, a = -g = -9.8 m s⁻²

For objects falling downwards:

Velocity is downward
Gravity acts downward
So, a = +g = +9.8 m s⁻²

Example 9.2

A car falls off a ledge and drops to the ground in 0.5 s. Let g = 10 m s⁻² (for simplifying the calculations).

(i) What is its speed on striking the ground?
(ii) What is its average speed during the 0.5 s?
(iii) How high is the ledge from the ground?

Solution:

Given:

Time, t = 0.5 s
Initial velocity, u = 0 m s⁻¹ (falls from rest)
Acceleration due to gravity, g = 10 m s⁻²
Acceleration of the car, a = +10 m s⁻² (downward motion)

(i) Speed on striking the ground:

Using v = u + at:

v = 0 + 10 × 0.5
v = 5 m s⁻¹

(ii) Average speed:

Average speed = (u + v) / 2
             = (0 + 5) / 2
             = 2.5 m s⁻¹

(iii) Height of the ledge:

Using s = ut + (1/2)at²:

s = 0 × 0.5 + (1/2) × 10 × (0.5)²
s = 0 + 5 × 0.25
s = 1.25 m

Answer:

Speed on striking ground = 5 m s⁻¹
Average speed = 2.5 m s⁻¹
Height of ledge = 1.25 m

Example 9.3

An object is thrown vertically upwards and rises to a height of 10 m. Calculate (i) the velocity with which the object was thrown upwards and (ii) the time taken by the object to reach the highest point.

Solution:

Given:

Distance travelled, s = 10 m
Final velocity at highest point, v = 0 m s⁻¹ (object stops momentarily)
Acceleration due to gravity, g = 9.8 m s⁻²
Acceleration of the object, a = -9.8 m s⁻² (upward motion, gravity opposes)

(i) Initial velocity:

Using v² = u² + 2as:

0² = u² + 2 × (-9.8) × 10
0 = u² - 196
u² = 196
u = √196 = 14 m s⁻¹

(ii) Time taken:

Using v = u + at:

0 = 14 + (-9.8) × t
9.8t = 14
t = 14 / 9.8
t = 1.43 s

Answer:

Initial velocity = 14 m s⁻¹
Time taken = 1.43 s

Questions (9.2) and Solutions

Q1. What do you mean by free fall?

Solution:

Free fall is the motion of an object falling towards the earth under the influence of gravitational force alone, without any other force acting on it.

Key points:

Only gravitational force acts on the object.
Air resistance is negligible or absent.
All objects in free fall experience the same acceleration (g = 9.8 m s⁻²).
The acceleration is independent of the mass of the object.

Examples:

A stone thrown upwards (after leaving the hand)
An object dropped from a height

Q2. What do you mean by acceleration due to gravity?

Solution:

Acceleration due to gravity is the acceleration experienced by an object when it falls towards the earth under the influence of earth’s gravitational force alone.

Key points:

It is denoted by g.
Its value on or near the earth’s surface is approximately 9.8 m s⁻².
It is independent of the mass of the falling object.
Its value is greater at the poles and smaller at the equator.
Unit: m s⁻² (same as acceleration)

Formula:

g = G × M / R²

where M is the mass of the earth and R is the radius of the earth.

9.3 Mass

We have learnt in the previous chapter that the mass of an object is the measure of its inertia.

Important Properties of Mass:

Greater the mass, greater is the inertia.
Mass remains the same whether the object is on the earth, the moon, or even in outer space.
The mass of an object is constant and does not change from place to place.

9.4 Weight

We know that the earth attracts every object with a certain force and this force depends on:

The mass (m) of the object
The acceleration due to gravity (g)

Definition of Weight:

The weight of an object is the force with which it is attracted towards the earth.

Formula:

We know:

F = m × a  ... (9.13)

For gravitational force:

F = m × g  ... (9.14)

The force of attraction of the earth on an object is known as the weight of the object. It is denoted by W.

Weight Formula:

W = m × g  ... (9.15)

Characteristics of Weight

1. Unit:

Weight is a force.
SI unit of weight is newton (N) (same as force).

2. Nature:

Weight is a force acting vertically downwards.
It has both magnitude and direction (vector quantity).

3. Variation:

The value of g is constant at a given place.
At a given place, weight is directly proportional to mass: W ∝ m.
Due to this reason, at a given place, we can use the weight of an object as a measure of its mass.

4. Mass vs. Weight:

Mass remains the same everywhere (on earth, moon, or any planet).
Weight depends on location because g depends on location.

9.4.1 WEIGHT OF AN OBJECT ON THE MOON

The weight of an object on the earth is the force with which the earth attracts the object. Similarly, the weight of an object on the moon is the force with which the moon attracts that object.

Important: The mass of the moon is less than that of the earth. Due to this, the moon exerts lesser force of attraction on objects.

Deriving the Relation

Let:

Mass of an object = m
Weight on the moon = Wₘ
Mass of the moon = Mₘ
Radius of the moon = Rₘ

By applying the universal law of gravitation:

textWₘ = G × (Mₘ × m) / Rₘ²  ... (9.16)

Let:

Weight on the earth = Wₑ
Mass of the earth = M
Radius of the earth = R

From Eqs. (9.9) and (9.15):

Wₑ = G × (M × m) / R²  ... (9.17)

Table 9.1 – Celestial Body Data

Celestial Body	Mass (kg)	Radius (m)
Earth	5.98 × 10²⁴	6.37 × 10⁶
Moon	7.36 × 10²²	1.74 × 10⁶

Calculation

Substituting values from Table 9.1 in Eq. (9.16):

Wₘ = G × (7.36 × 10²² × m) / (1.74 × 10⁶)²
Wₘ = G × m × 2.431 × 10¹⁰  ... (9.18a)

Substituting values in Eq. (9.17):

Wₑ = G × (5.98 × 10²⁴ × m) / (6.37 × 10⁶)²
Wₑ = G × m × 1.474 × 10¹¹  ... (9.18b)

Dividing Eq. (9.18a) by Eq. (9.18b):

Wₘ / Wₑ = (2.431 × 10¹⁰) / (1.474 × 10¹¹)
Wₘ / Wₑ = 0.165 ≈ 1/6  ... (9.19)

Conclusion:

Weight of object on moon = (1/6) × Weight of object on earth

Example 9.4

Mass of an object is 10 kg. What is its weight on the earth?

Solution:

Given:

Mass, m = 10 kg
Acceleration due to gravity, g = 9.8 m s⁻²

Using formula:

textW = m × g
W = 10 kg × 9.8 m s⁻²
W = 98 N

Answer: The weight of the object is 98 N.

Example 9.5

An object weighs 10 N when measured on the surface of the earth. What would be its weight when measured on the surface of the moon?

Solution:

We know:

Weight on moon = (1/6) × Weight on earth

Calculation:

Wₘ = Wₑ / 6
Wₘ = 10 N / 6
Wₘ = 1.67 N

Answer: The weight of object on the surface of the moon would be 1.67 N.

Questions (9.4) and Solutions

Q1. What are the differences between the mass of an object and its weight?

Solution:

Mass	Weight
Measure of inertia	Force with which earth attracts the object
Scalar quantity (only magnitude)	Vector quantity (magnitude and direction)
Remains constant everywhere	Varies from place to place
Does not depend on gravity	Depends on gravity (g)
SI unit: kilogram (kg)	SI unit: newton (N)
Measured using beam balance	Measured using spring balance
Never zero	Can be zero (in space far from any celestial body)

Formula:

Weight (W) = Mass (m) × Acceleration due to gravity (g)

Q2. Why is the weight of an object on the moon (1/6)th its weight on the earth?

Solution:

The weight of an object on the moon is (1/6)th its weight on the earth because:

Reason 1 – Mass Difference:

The mass of the moon is much less than the mass of the earth.
Moon’s mass = 7.36 × 10²² kg
Earth’s mass = 5.98 × 10²⁴ kg

Reason 2 – Gravitational Force:

Due to smaller mass, the moon exerts lesser gravitational force on objects.

Mathematical Proof:

Weight = m × g
For moon: Wₘ = m × gₘ
For earth: Wₑ = m × gₑ
When we calculate using the universal law of gravitation and substitute values of masses and radii of moon and earth, we get:

Wₘ / Wₑ = 1/6

Therefore:

Wₘ = (1/6) × Wₑ

9.5 Thrust and Pressure

Questions to Think About:

Why can a camel run in a desert easily?
Why does an army tank weighing more than a thousand tonnes rest upon a continuous chain?
Why does a truck or motorbus have much wider tyres?
Why do cutting tools have sharp edges?

To answer these questions, we need to understand the concepts of thrust and pressure.

Understanding Through Situations

Situation 1: Fixing a Poster with Drawing Pin

When you wish to fix a poster on a bulletin board:

You press the drawing pin with your thumb.
You apply a force on the surface area of the head of the pin.
This force is directed perpendicular to the surface area of the board.
This force acts on a smaller area at the tip of the pin.

Situation 2: Standing vs. Lying on Sand

When you stand on loose sand:

Your feet go deep into the sand.

When you lie down on the sand:

Your body will not go that deep in the sand.

In both cases:

The force exerted on the sand is the weight of your body (same in both cases).

Explanation:

When you stand, the weight acts on an area equal to the area of your feet.
When you lie down, the same weight acts on an area equal to the contact area of your whole body (larger than area of feet).
Thus, the effects of forces of the same magnitude on different areas are different.

Definitions

Thrust:

The force acting on an object perpendicular to the surface is called thrust.

Important: Weight is the force acting vertically downwards. When acting perpendicular to a surface, weight is the thrust.

Pressure:

The effect of thrust depends on the area on which it acts. The thrust on unit area is called pressure.

Formula:

Pressure = Thrust / Area  ... (9.20)

SI Unit:

Substituting units:

Unit of Pressure = N / m² = N m⁻²

In honour of scientist Blaise Pascal, the SI unit of pressure is called pascal, denoted as Pa.

1 Pa = 1 N/m² = 1 N m⁻²

Example 9.6

A block of wood is kept on a tabletop. The mass of the wooden block is 5 kg and its dimensions are 40 cm × 20 cm × 10 cm. Find the pressure exerted by the wooden block on the table top if it is made to lie on the table top with its sides of dimensions (a) 20 cm × 10 cm and (b) 40 cm × 20 cm.

Solution:

Given:

Mass of wooden block = 5 kg
Dimensions = 40 cm × 20 cm × 10 cm

The weight of the wooden block applies a thrust on the table top.

Thrust:

Thrust = Force = Weight
       = m × g
       = 5 kg × 9.8 m s⁻²
       = 49 N

(a) When block lies on side 20 cm × 10 cm:

Area:

Area = length × breadth
     = 20 cm × 10 cm
     = 200 cm²
     = 0.02 m²

Pressure:

Pressure = Thrust / Area
         = 49 N / 0.02 m²
         = 2450 N m⁻²
         = 2450 Pa

(b) When block lies on side 40 cm × 20 cm:

Area:

Area = length × breadth
     = 40 cm × 20 cm
     = 800 cm²
     = 0.08 m²

Pressure:

Pressure = Thrust / Area
         = 49 N / 0.08 m²
         = 612.5 N m⁻²
         = 612.5 Pa

Answer:

Pressure by side 20 cm × 10 cm = 2450 Pa
Pressure by side 40 cm × 20 cm = 612.5 Pa

Conclusion

The same force acting on:

A smaller area exerts a larger pressure
A larger area exerts a smaller pressure

This is the reason why:

A nail has a pointed tip (small area → large pressure → easily penetrates)
Knives have sharp edges (small area → large pressure → easily cuts)
Buildings have wide foundations (large area → small pressure → prevents sinking)

9.5.1 PRESSURE IN FLUIDS

All liquids and gases are fluids. A solid exerts pressure on a surface due to its weight. Similarly, fluids have weight, and they also exert pressure on:

The base of the container
The walls of the container in which they are enclosed

Important: Pressure exerted in any confined mass of fluid is transmitted undiminished in all directions.

9.5.2 BUOYANCY

Questions to Think About:

Have you ever had a swim in a pool and felt lighter?
Have you ever drawn water from a well and felt that the bucket of water is heavier when it is out of the water?
Why does a ship made of iron and steel not sink in sea water, but the same amount of iron and steel in the form of a sheet would sink?

These questions can be answered by taking buoyancy into consideration.

Activity – Bottle in Water

Activity:

Take an empty plastic bottle.
Close the mouth with an airtight stopper.
Put it in a bucket filled with water.
You see that the bottle floats.
Push the bottle into the water.
You feel an upward push.
Try to push it further down. You will find it difficult to push deeper and deeper.
This indicates that water exerts a force on the bottle in the upward direction.
The upward force goes on increasing as the bottle is pushed deeper till it is completely immersed.
Now, release the bottle. It bounces back to the surface.

Questions:

Does the force due to gravitational attraction of the earth act on this bottle?
If so, why doesn’t the bottle stay immersed in water after it is released?
How can you immerse the bottle in water?

Explanation

Force due to gravity:

The force due to gravitational attraction of the earth acts on the bottle in the downward direction (weight).
So the bottle is pulled downwards.

Force due to water:

The water exerts an upward force on the bottle.
The bottle is pushed upwards.

When bottle is immersed:

The upward force exerted by water on the bottle is greater than its weight.
Therefore it rises up when released.

To keep bottle completely immersed:

The upward force on the bottle due to water must be balanced.
This can be achieved by an externally applied force acting downwards.
This force must at least be equal to the difference between the upward force and the weight of the bottle.

Buoyant Force

Definition: The upward force exerted by a fluid (water) on an object (bottle) is known as upthrust or buoyant force.

Important Points:

All objects experience a force of buoyancy when they are immersed in a fluid.
The magnitude of this buoyant force depends on the density of the fluid.

9.5.3 WHY OBJECTS FLOAT OR SINK WHEN PLACED ON THE SURFACE OF WATER?

Activity – Iron Nail in Water

Activity:

Take a beaker filled with water.
Take an iron nail and place it on the surface of the water.
Observe what happens.

Observation: The nail sinks.

Explanation:

The force due to gravitational attraction of the earth on the iron nail pulls it downwards.
There is an upthrust of water on the nail, which pushes it upwards.
But the downward force (weight) acting on the nail is greater than the upthrust of water.
So it sinks.

Activity – Cork and Iron Nail of Equal Mass

Activity:

Take a beaker filled with water.
Take a piece of cork and an iron nail of equal mass.
Place them on the surface of water.
Observe what happens.

Observation:

The cork floats.
The nail sinks.

Explanation:

This happens because of the difference in their densities.

Density:

The density of a substance is defined as the mass per unit volume.

For Cork:

The density of cork is less than the density of water.
This means the upthrust of water on the cork is greater than the weight of the cork.
So it floats.

For Iron Nail:

The density of iron nail is more than the density of water.
This means the upthrust of water on the iron nail is less than the weight of the nail.
So it sinks.

General Rule

Objects of density less than that of a liquid → Float on the liquid

Objects of density greater than that of a liquid → Sink in the liquid

Questions (9.5) and Solutions

Q1. Why is it difficult to hold a school bag having a strap made of a thin and strong string?

Solution:

It is difficult to hold a school bag having a strap made of a thin and strong string because:

Reason:

The weight of the bag acts as thrust on the shoulder.
A thin string has a very small area of contact with the shoulder.
Pressure = Thrust / Area
Since area is small, the pressure is very large.
This large pressure causes pain and discomfort to the shoulder.

Solution:

School bags have broad straps (large area).
This reduces pressure on the shoulder.
Makes it comfortable to carry.

Q2. What do you mean by buoyancy?

Solution:

Buoyancy is the upward force exerted by a fluid on an object immersed in it.

Key points:

Also called upthrust or buoyant force.
Acts in the upward direction (opposite to weight).
Depends on the density of the fluid.
All objects experience buoyancy when immersed in a fluid (liquid or gas).
The magnitude of buoyant force is equal to the weight of the fluid displaced by the object (Archimedes’ Principle).

Example:

When you put a plastic bottle in water, you feel an upward push. This upward push is buoyancy.

Q3. Why does an object float or sink when placed on the surface of water?

Solution:

Whether an object floats or sinks depends on the comparison between two forces:

1. Weight of the object (downward force due to gravity)
2. Buoyant force (upward force exerted by water)

And also depends on the density of the object compared to water.

Case 1: Object Floats

Condition: Density of object < Density of water
Force comparison: Buoyant force > Weight
Result: Object floats
Example: Cork, wood, plastic bottle

Case 2: Object Sinks

Condition: Density of object > Density of water
Force comparison: Buoyant force < Weight
Result: Object sinks
Example: Iron nail, stone, metal objects

General Rule:

If density of object < density of liquid → Object floats
If density of object > density of liquid → Object sinks

9.6 Archimedes’ Principle

Activity – Stone and Spring Balance

Activity:

Take a piece of stone and tie it to one end of a rubber string or a spring balance.
Suspend the stone by holding the balance or the string.
Note the elongation of the string or the reading on the spring balance due to the weight of the stone.
Now, slowly dip the stone in water in a container.
Observe what happens to the elongation of the string or the reading on the balance.

Observations:

In Air:

The elongation of the string or reading on balance shows the full weight of the stone.

When lowered in water:

The elongation of the string or the reading on the balance decreases as the stone is gradually lowered in the water.

When fully immersed:

No further change is observed once the stone gets fully immersed in the water.

Explanation

What do you infer from the decrease in extension?

The elongation produced in the string or spring balance is due to the weight of the stone.
Since the extension decreases once the stone is lowered in water, it means some force acts on the stone in the upward direction.
As a result, the net force on the string decreases and hence the elongation also decreases.
This upward force exerted by water is the force of buoyancy.

Archimedes’ Principle

Statement:

When a body is immersed fully or partially in a fluid, it experiences an upward force (buoyant force) that is equal to the weight of the fluid displaced by it.

Why No Further Decrease After Full Immersion?

In the activity, a further decrease in the elongation of the string was not observed as the stone was fully immersed in water because:

Once the stone is fully immersed, the volume of water displaced remains constant.
Since volume displaced is constant, the weight of water displaced is constant.
Therefore, the buoyant force remains constant.
Hence, no further decrease in elongation.

About Archimedes (287 BC – 212 BC)

Archimedes was a Greek scientist. He discovered the principle subsequently named after him, after noticing that the water in a bathtub overflowed when he stepped into it. He ran through the streets shouting “Eureka!”, which means “I have got it”.

Contributions:

This knowledge helped him determine the purity of gold in the crown made for the king.
His work in the field of Geometry and Mechanics made him famous.
His understanding of levers, pulleys, wheels-and-axle helped the Greek army in its war with the Roman army.

Applications of Archimedes’ Principle

1. Designing Ships and Submarines

Used to calculate buoyant force and ensure ships float.

2. Lactometer

Used to determine the purity of milk.
Based on density and buoyancy.

3. Hydrometer

Used for determining the density of liquids.
Works on the principle of buoyancy.

Questions (9.6) and Solutions

Q1. You find your mass to be 42 kg on a weighing machine. Is your mass more or less than 42 kg?

Solution:

Your actual mass is slightly more than 42 kg.

Reason:

The weighing machine measures your apparent weight.
When you stand on the weighing machine, you displace air around you.
According to Archimedes’ principle, air exerts an upward buoyant force on your body.
This buoyant force reduces your apparent weight.
Therefore, the weighing machine shows a reading slightly less than your actual mass.

However:

The buoyant force due to air is very small (because density of air is very small).
So the difference is negligible for practical purposes.
For all practical purposes, your mass is 42 kg.

Q2. You have a bag of cotton and an iron bar, each indicating a mass of 100 kg when measured on a weighing machine. In reality, one is heavier than the other. Can you say which one is heavier and why?

Solution:

The bag of cotton is actually heavier (has more mass) than the iron bar.

Reason:

For Bag of Cotton:

Cotton has very low density.
To have a mass of 100 kg, the volume of cotton must be very large.
Large volume displaces a large volume of air.
According to Archimedes’ principle, cotton experiences a large buoyant force from air.
This large buoyant force reduces the apparent weight significantly.

For Iron Bar:

Iron has high density.
To have a mass of 100 kg, the volume of iron is relatively small.
Small volume displaces a small volume of air.
Iron experiences a small buoyant force from air.
The reduction in apparent weight is small.

Conclusion:

Both show 100 kg on weighing machine (apparent weight is same).
But cotton experiences more buoyant force than iron.
Therefore, the actual mass of cotton must be more than that of iron to show the same reading on the weighing machine.

Formula:

textActual weight = Apparent weight + Buoyant force

Since buoyant force on cotton is more, its actual weight (and mass) is more.

Exercises – Questions and Solutions

Q1. How does the force of gravitation between two objects change when the distance between them is reduced to half?

Solution:

The force of gravitation becomes four times when the distance between them is reduced to half.

Explanation:

From universal law of gravitation:

textF = G × (M × m) / d²

This shows that force is inversely proportional to the square of distance:

F ∝ 1/d²

When distance is reduced to half:

Original distance = d
New distance = d/2

New force:

F' = G × (M × m) / (d/2)²
F' = G × (M × m) / (d²/4)
F' = 4 × [G × (M × m) / d²]
F' = 4F

Answer: Force becomes 4 times the original force.

Q2. Gravitational force acts on all objects in proportion to their masses. Why then, does a heavy object not fall faster than a light object?

Solution:

A heavy object does not fall faster than a light object because the acceleration due to gravity is independent of mass.

Explanation:

From Newton’s Second Law:

textForce = Mass × Acceleration
F = m × a

For gravitational force:

F = m × g

From Universal Law of Gravitation:

F = G × (M × m) / R²

Equating both:

m × g = G × (M × m) / R²

Simplifying:

g = G × M / R²

Observation:

The mass of the object (m) gets cancelled.
g depends only on M (mass of earth) and R (radius of earth).
g is independent of the mass of the falling object.

Conclusion:

All objects (heavy or light) experience the same acceleration (g = 9.8 m s⁻²) when falling.
Therefore, they fall at the same rate.
Even though a heavy object experiences more gravitational force, it also has more inertia (more mass to move), so the acceleration remains the same.

Q3. What is the magnitude of the gravitational force between the earth and a 1 kg object on its surface? (Mass of the earth is 6 × 10²⁴ kg and radius of the earth is 6.4 × 10⁶ m.)

Solution:

Given:

Mass of earth, M = 6 × 10²⁴ kg
Mass of object, m = 1 kg
Radius of earth, R = 6.4 × 10⁶ m
G = 6.7 × 10⁻¹¹ N m² kg⁻²

Using universal law of gravitation:

F = G × (M × m) / R²
F = (6.7 × 10⁻¹¹ × 6 × 10²⁴ × 1) / (6.4 × 10⁶)²
F = (6.7 × 6 × 10¹³) / (40.96 × 10¹²)
F = (40.2 × 10¹³) / (40.96 × 10¹²)
F = 40.2 / 4.096
F ≈ 9.8 N

Alternative method:

F = m × g
F = 1 kg × 9.8 m s⁻²
F = 9.8 N

Answer: The magnitude of gravitational force is approximately 9.8 N.

Q4. The earth and the moon are attracted to each other by gravitational force. Does the earth attract the moon with a force that is greater or smaller or the same as the force with which the moon attracts the earth? Why?

Solution:

The earth attracts the moon with a force that is exactly the same as the force with which the moon attracts the earth.

Reason:

According to Newton’s Third Law of Motion:

To every action, there is an equal and opposite reaction.
These forces act on different objects.

Application:

The earth exerts a gravitational force on the moon (action).
The moon exerts an equal and opposite gravitational force on the earth (reaction).
These two forces are equal in magnitude but opposite in direction.

From Universal Law of Gravitation:

Force on moon by earth = G × (Mₑ × Mₘ) / d²
Force on earth by moon = G × (Mₘ × Mₑ) / d²

Both expressions are identical, confirming that the forces are equal.

Answer: The forces are equal (Newton’s Third Law).

Q5. If the moon attracts the earth, why does the earth not move towards the moon?

Solution:

The earth does move towards the moon, but the movement is so small that it is not noticeable.

Explanation:

From Newton’s Second Law:

Force = Mass × Acceleration
F = m × a
Therefore, a = F/m

For the Earth:

Mass of earth (Mₑ) = 6 × 10²⁴ kg (very large)
Force on earth = F
Acceleration of earth = F/Mₑ (very small because mass is very large)

For the Moon:

Mass of moon (Mₘ) = 7.4 × 10²² kg (much smaller than earth)
Force on moon = F (same as force on earth)
Acceleration of moon = F/Mₘ (relatively larger)

Conclusion:

The same force acts on both earth and moon (Newton’s Third Law).
But the acceleration is inversely proportional to mass.
Earth has much larger mass, so it has much smaller acceleration.
Moon has smaller mass, so it has larger acceleration.
The earth does move towards the moon, but the displacement is extremely small and negligible.
The moon’s movement is more noticeable, so it appears that only the moon moves around the earth.

Q6. What happens to the force between two objects, if:
(i) the mass of one object is doubled?
(ii) the distance between the objects is doubled and tripled?
(iii) the masses of both objects are doubled?

Solution:

From universal law of gravitation:

F = G × (M × m) / d²

(i) If the mass of one object is doubled:

Let original mass = M, New mass = 2M

F' = G × (2M × m) / d²
F' = 2 × [G × (M × m) / d²]
F' = 2F

Answer: Force becomes 2 times (doubles).

(ii) If the distance between objects is doubled and tripled:

When distance is doubled:

Original distance = d
New distance = 2d

F' = G × (M × m) / (2d)²
F' = G × (M × m) / (4d²)
F' = (1/4) × [G × (M × m) / d²]
F' = F/4

Answer: Force becomes (1/4)th (one-fourth).

When distance is tripled:

Original distance = d
New distance = 3d

F'' = G × (M × m) / (3d)²
F'' = G × (M × m) / (9d²)
F'' = (1/9) × [G × (M × m) / d²]
F'' = F/9

Answer: Force becomes (1/9)th (one-ninth).

(iii) If the masses of both objects are doubled:

Original masses = M and m
New masses = 2M and 2m

F' = G × (2M × 2m) / d²
F' = 4 × [G × (M × m) / d²]
F' = 4F

Answer: Force becomes 4 times.

Q7. What is the importance of universal law of gravitation?

Solution:

The importance of the universal law of gravitation includes:

1. Explains Force Binding Us to Earth:

Explains why we stay on the ground and don’t float away.
Explains why objects fall towards the earth.

2. Explains Motion of Moon Around Earth:

The gravitational force provides the centripetal force needed for the moon’s circular motion around earth.

3. Explains Motion of Planets Around Sun:

The Sun’s gravitational force keeps planets in their orbits.
Explains the elliptical orbits of planets.

4. Explains Tides:

The gravitational pull of the moon and Sun on earth’s oceans causes tides.

5. Universal Application:

Applicable to all bodies in the universe (big or small, celestial or terrestrial).
Helps us understand the structure and behavior of the universe.

6. Satellite Motion:

Used to calculate the motion of artificial satellites.
Helps in launching and maintaining satellites in orbit.

7. Discovery of New Celestial Bodies:

Has led to the discovery of new planets and stars based on gravitational effects.

Q8. What is the acceleration of free fall?

Solution:

The acceleration of free fall is the acceleration experienced by an object when it falls towards the earth under the influence of gravitational force alone.

Key Points:

Also called acceleration due to gravity.
Denoted by g.
Value on earth’s surface = 9.8 m s⁻² (or approximately 10 m s⁻² for calculations).
It is independent of the mass of the falling object.
All objects (heavy or light) experience the same acceleration during free fall.
Its value is greater at poles and smaller at equator.

Formula:

g = G × M / R²

where M is mass of earth and R is radius of earth.

Q9. What do we call the gravitational force between the earth and an object?

Solution:

The gravitational force between the earth and an object is called the weight of the object.

Key Points:

Weight is a force (not mass).
Symbol: W
SI Unit: Newton (N)
Formula: W = m × g
It acts vertically downward towards the center of the earth.
Weight is a vector quantity (has both magnitude and direction).
Weight varies from place to place depending on the value of g.

Q10. Amit buys few grams of gold at the poles as per the instruction of one of his friends. He hands over the same when he meets him at the equator. Will the friend agree with the weight of gold bought? If not, why? [Hint: The value of g is greater at the poles than at the equator.]

Solution:

No, the friend will not agree with the weight of gold bought.

Reason:

Value of g:

The value of g is greater at the poles than at the equator.
This is because the earth is not a perfect sphere; its radius is smaller at poles and larger at equator.

Weight depends on g:

Weight (W) = Mass (m) × g

At Poles:

g is greater
Therefore, weight is greater
Gold weighs more at poles

At Equator:

g is smaller
Therefore, weight is smaller
The same gold weighs less at equator

Conclusion:

When Amit bought gold at the poles, its weight was measured (not mass).
When he hands it over at the equator, the weight will be less because g is less.
The friend will feel he received less gold.

However:

If measured by mass (using beam balance), both will agree because mass remains constant everywhere.
But if measured by weight (using spring balance), there will be a difference.

Q11. Why will a sheet of paper fall slower than one that is crumpled into a ball?

Solution:

A sheet of paper falls slower than a crumpled paper ball because of air resistance.

Explanation:

For Sheet of Paper:

Has a large surface area.
Experiences more air resistance (friction with air).
Air resistance opposes the downward motion.
Net downward force = Weight – Air resistance (smaller)
Falls with smaller acceleration.
Takes more time to reach the ground.

For Crumpled Paper Ball:

Has a smaller surface area.
Experiences less air resistance.
Net downward force = Weight – Air resistance (larger)
Falls with larger acceleration.
Takes less time to reach the ground.

In Vacuum:

If the experiment is done in vacuum (no air), both will fall at the same rate because there is no air resistance.

Note: Both have the same mass, so gravitational force is the same. The difference is only due to air resistance.

Q12. Gravitational force on the surface of the moon is only (1/6) as strong as gravitational force on the earth. What is the weight in newtons of a 10 kg object on the moon and on the earth?

Solution:

Given:

Mass of object, m = 10 kg
g on earth = 10 m s⁻² (or 9.8 m s⁻²)
g on moon = (1/6) × g on earth

Weight on Earth:

Wₑ = m × gₑ
Wₑ = 10 kg × 10 m s⁻²
Wₑ = 100 N

Weight on Moon:

Wₘ = (1/6) × Wₑ
Wₘ = (1/6) × 100 N
Wₘ = 16.67 N

Or, alternatively:

gₘ = (1/6) × 10 = 1.67 m s⁻²
Wₘ = m × gₘ
Wₘ = 10 kg × 1.67 m s⁻²
Wₘ = 16.7 N

Answer:

Weight on earth = 100 N
Weight on moon = 16.67 N (approximately 16.7 N)

Q13. A ball is thrown vertically upwards with a velocity of 49 m/s. Calculate (i) the maximum height to which it rises, (ii) the total time it takes to return to the surface of the earth.

Solution:

Given:

Initial velocity, u = 49 m/s (upward)
Final velocity at maximum height, v = 0 m/s
Acceleration, a = -g = -9.8 m s⁻² (gravity opposes upward motion)

(i) Maximum height:

Using v² = u² + 2as:

0² = 49² + 2 × (-9.8) × s
0 = 2401 - 19.6s
19.6s = 2401
s = 2401 / 19.6
s = 122.5 m

(ii) Total time to return:

Time to reach maximum height:

Using v = u + at:

0 = 49 + (-9.8) × t
9.8t = 49
t = 49 / 9.8
t = 5 s

Time to come down from maximum height:

By symmetry, time to come down = time to go up = 5 s

Total time:

textTotal time = Time to go up + Time to come down
Total time = 5 s + 5 s
Total time = 10 s

Answer:

Maximum height = 122.5 m
Total time = 10 s

Q14. A stone is released from the top of a tower of height 19.6 m. Calculate its final velocity just before touching the ground.

Solution:

Given:

Height of tower, s = 19.6 m
Initial velocity, u = 0 m/s (released from rest)
Acceleration, a = g = 9.8 m s⁻² (downward)

Using v² = u² + 2as:

v² = 0² + 2 × 9.8 × 19.6
v² = 2 × 9.8 × 19.6
v² = 384.16
v = √384.16
v = 19.6 m/s

Answer: Final velocity = 19.6 m/s

Q15. A stone is thrown vertically upward with an initial velocity of 40 m/s. Taking g = 10 m/s², find the maximum height reached by the stone. What is the net displacement and the total distance covered by the stone?

Solution:

Given:

Initial velocity, u = 40 m/s (upward)
Final velocity at maximum height, v = 0 m/s
Acceleration, a = -g = -10 m s⁻² (opposes motion)

Maximum height:

Using v² = u² + 2as:

0² = 40² + 2 × (-10) × s
0 = 1600 - 20s
20s = 1600
s = 1600 / 20
s = 80 m

Net Displacement:

The stone returns to the starting point (thrower’s hand).
Net displacement = Final position – Initial position = 0 m

Total Distance Covered:

Distance while going up = 80 m
Distance while coming down = 80 m
Total distance = 80 + 80 = 160 m

Answer:

Maximum height = 80 m
Net displacement = 0 m
Total distance covered = 160 m

Q16. Calculate the force of gravitation between the earth and the Sun, given that the mass of the earth = 6 × 10²⁴ kg and of the Sun = 2 × 10³⁰ kg. The average distance between the two is 1.5 × 10¹¹ m.

Solution:

Given:

Mass of earth, Mₑ = 6 × 10²⁴ kg
Mass of Sun, Mₛ = 2 × 10³⁰ kg
Distance, d = 1.5 × 10¹¹ m
G = 6.7 × 10⁻¹¹ N m² kg⁻²

Using universal law of gravitation:

F = G × (Mₑ × Mₛ) / d²
F = (6.7 × 10⁻¹¹ × 6 × 10²⁴ × 2 × 10³⁰) / (1.5 × 10¹¹)²
F = (6.7 × 6 × 2 × 10⁴³) / (2.25 × 10²²)
F = (80.4 × 10⁴³) / (2.25 × 10²²)
F = 35.73 × 10²¹
F = 3.573 × 10²² N
F ≈ 3.6 × 10²² N

Answer: Force of gravitation = 3.6 × 10²² N (approximately)

Q17. A stone is allowed to fall from the top of a tower 100 m high and at the same time another stone is projected vertically upwards from the ground with a velocity of 25 m/s. Calculate when and where the two stones will meet.

Solution:

Let the stones meet after time t at a height h from the ground.

For Stone 1 (falling from top):

Initial velocity, u₁ = 0 m/s
Acceleration, a₁ = +g = 10 m s⁻²
Distance travelled = 100 – h (downward from top)

Using s = ut + (1/2)at²:

100 - h = 0 × t + (1/2) × 10 × t²
100 - h = 5t²  ... (1)

For Stone 2 (thrown upward from ground):

Initial velocity, u₂ = 25 m/s
Acceleration, a₂ = -g = -10 m s⁻²
Distance travelled = h (upward from ground)

Using s = ut + (1/2)at²:

h = 25t + (1/2) × (-10) × t²
h = 25t - 5t²  ... (2)

Adding equations (1) and (2):

100 - h + h = 5t² + 25t - 5t²
100 = 25t
t = 100/25
t = 4 s

Finding height h:

Substituting t = 4 in equation (2):

h = 25 × 4 - 5 × 4²
h = 100 - 5 × 16
h = 100 - 80
h = 20 m

Answer:

The stones will meet after 4 seconds
They will meet at a height of 20 m from the ground

Q18. A ball thrown up vertically returns to the thrower after 6 s. Find (a) the velocity with which it was thrown up, (b) the maximum height it reaches, and (c) its position after 4 s.

Solution:

Given:

Total time of flight = 6 s
Time to reach maximum height = 6/2 = 3 s
g = 10 m s⁻² (taking for easier calculation)

(a) Initial velocity:

At maximum height, v = 0

Using v = u + at:

0 = u + (-10) × 3
0 = u - 30
u = 30 m/s

(b) Maximum height:

Using v² = u² + 2as:

0² = 30² + 2 × (-10) × h
0 = 900 - 20h
20h = 900
h = 45 m

Or using s = ut + (1/2)at²:

h = 30 × 3 + (1/2) × (-10) × 3²
h = 90 - 5 × 9
h = 90 - 45
h = 45 m

(c) Position after 4 s:

After 3 s: Ball reaches maximum height of 45 m

After 4 s: Ball has been falling for (4 – 3) = 1 s

Distance fallen in 1 s:

Using s = ut + (1/2)gt² (where u = 0 for falling):

s = 0 + (1/2) × 10 × 1²
s = 5 m

Position after 4 s:

Position = 45 - 5 = 40 m from ground

Answer:

(a) Initial velocity = 30 m/s
(b) Maximum height = 45 m
(c) Position after 4 s = 40 m from ground

Q19. In what direction does the buoyant force on an object immersed in a liquid act?

Solution:

The buoyant force on an object immersed in a liquid acts in the upward direction (vertically upward).

Explanation:

Buoyant force always acts opposite to the weight of the object.
Weight acts downward, so buoyant force acts upward.
It acts along the line of action of weight but in opposite direction.

Q20. Why does a block of plastic released under water come up to the surface of water?

Solution:

A block of plastic released under water comes up to the surface because the buoyant force acting on it is greater than its weight.

Explanation:

Density Comparison:

Density of plastic is less than the density of water.

Forces Acting:

Weight of plastic block acts downward.
Buoyant force (upthrust by water) acts upward.

According to Archimedes’ Principle:

Buoyant force = Weight of water displaced by plastic

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