Gravitation class 9 Notes, Solved Numerical and Work sheet PDF

 Gravitation class 9 Notes, Solved Numerical and Work sheet PDF 

Introduction:

What is Gravitation?

•             Gravitation is a natural phenomenon by which every object in the universe attracts every other object.

•             This attractive force is called the gravitational force.

•             It acts along the line joining the centers of the two objects.

 Key Points

1.          Gravitation is a universal force – it acts between any two objects, whether big or small, near or far.

–            Example:

•             Earth attracts an apple (falls down).

•             Earth and Moon attract each other (causing Moon’s revolution around Earth).

•             Planets revolve around the Sun due to gravitation.

2.          Gravitation is always an attractive force, never repulsive.

3.          The force of gravitation depends on:

–            The masses of the objects (heavier objects exert more force).

–            The distance between the objects (force decreases as distance increases).

Universal Law of Gravitation:

Introduction

In 1687, Sir Isaac Newton proposed the Universal Law of Gravitation. He stated that every object in the universe attracts every other object with a force.

This force:
- Is directly proportional to the product of their masses
- Is inversely proportional to the square of the distance between their centers

Statement of the Law

“Every particle of matter in the universe attracts every other particle with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.”

        F = G (m₁ m₂) / r²

Where,

     F = Gravitational Force

    m₁ = Mass of First object

    m₂ = Mass of another object

       r = Distance between two objects

      G = Universal Gravitational Constant

Value of Universal Gravitational Constant (G):

Experimental Determination of G

The value of G was first calculated by Henry Cavendish in 1798 using the torsion balance experiment.
- A horizontal rod with two small lead spheres was suspended.
- Two large lead spheres were placed nearby.
- The gravitational attraction between the large and small spheres caused a twist in the rod.
- By measuring this twist, Cavendish calculated the value of G.

- G = 6.67 × 10⁻¹¹ Nm²/kg²
- It is called universal because it is the same everywhere in the universe.

Examples in Daily Life:

•             Falling of fruits, raindrops, and objects towards the Earth.

•             Existence of atmosphere around Earth.

•             Tides in seas and oceans (caused by gravitational pull of Moon and Sun).

•             Planets revolving around the Sun.

Important Points

1. Gravitational force is a central force (acts along the line joining centers).

2. It is an attractive force.

3. It exists between all objects, but it is noticeable only when at least one mass is very large (like Earth, Sun).

4. This law explains why planets revolve around the Sun and why objects fall on Earth.

Solved Numericals

Numerical 1

Find the gravitational force between two objects of masses 10 kg and 20 kg kept 5 m apart.

Solution:
F = G (m₁ m₂) / r²
F = (6.67 × 10⁻¹¹) × (10 × 20) / 5²
F = 6.67 × 10⁻¹¹ × 200 / 25
F = 6.67 × 10⁻¹¹ × 8
F = 5.336 × 10⁻¹⁰ N

The gravitational force is 5.34 × 10⁻¹⁰ N (very small).

Numerical 2

Two objects, each of mass 1000 kg, are separated by 1 m. Find the force of attraction between them.

Solution:
F = G (m₁ m₂) / r²
F = (6.67 × 10⁻¹¹) × (1000 × 1000) / 1²
F = 6.67 × 10⁻¹¹ × 10⁶
F = 6.67 × 10⁻⁵ N

The force is 6.67 × 10⁻⁵ N (still very small).

Numerical 3 (Earth’s Attraction)

Find the gravitational force between the Earth (mass 6 × 10²⁴ kg) and a body of mass 10 kg placed on the Earth’s surface. Radius of Earth R = 6.4 × 10⁶ m.

Solution:
F = G (m₁ m₂) / r²
F = (6.67 × 10⁻¹¹) × (6 × 10²⁴ × 10) / (6.4 × 10⁶)²
F = (6.67 × 10⁻¹¹) × (6 × 10²⁵) / 4.096 × 10¹³
F = (6.67 × 10⁻¹¹) × (1.465 × 10¹²)
F ≈ 98 N

The force is 98 N, which is equal to the weight of the object.

Free Fall:

Introduction

• When an object falls towards the Earth only under the influence of Earth’s gravitational force, it is said to be in free fall. 
• No other forces (like air resistance or push) act on it. 
• All objects, whether heavy or light, fall at the same rate in vacuum.

Definition of Free Fall

Free fall is the motion of a body where gravity is the only force acting on it.

Examples:
- An apple falling from a tree (ignoring air resistance).
- A stone dropped from the top of a building.
- A satellite orbiting the Earth is also in free fall.

Acceleration due to Gravity (g):

During free fall, objects accelerate towards the Earth with an acceleration called acceleration due to gravity (g).

Formula: g = GM / R²

Where:
G = 6.67 × 10⁻¹¹ Nm²/kg² (gravitational constant)
M = 6 × 10²⁴ kg (mass of Earth)
R = 6.4 × 10⁶ m (radius of Earth)

g ≈ 9.8 m/s²

This means the velocity of a freely falling body increases by 9.8 m/s every second.

Equations of Motion in Free Fall

The same equations of motion apply, but acceleration a is replaced with g:

1) v = u + gt  (Final velocity after time t)

2) s = ut + ½gt²  (Distance travelled in time t)

3) v² - u² = 2gs  (Relation between velocity and distance)

Important Points

1. In the absence of air, all objects fall with the same acceleration (g).

2. On Earth’s surface, g ≈ 9.8 m/s².

3. The value of g decreases with height and depth.

4. Weight of a body in free fall becomes zero (apparent weightlessness).

Differentiate between G and g :

Aspect	G (Gravitational Constant)	g (Acceleration due to gravity)
Definition	G is the universal constant of gravitation that appears in Newton’s law of gravitation.	G is the acceleration produced in a body when it falls freely under Earth’s gravity.
Formula	F = Gm1m2 / r2	g = GM / R2
Meaning	It gives the strength of gravitational force between two bodies.	It tells how fast objects accelerate towards Earth (or any planet)
Value	6.67 x 10-11 Nm2/kg2	On Earth = 9.8 m/s2 (varies slightly with location)
Depends on	Nothing – It is universal, same everywhere.	Depends on mass (M) and radius (R) of the planet.
Nature	Constant	Variable (changes from planet to planet, also with height and depth)
SI Unit	Nm2/kg2	m/s2

Solved Numericals

Numerical 1

A ball is dropped from a height of 20 m. Find the time taken to reach the ground. (Take g = 10 m/s²)

Solution:
s = ½gt²
20 = ½ × 10 × t²
20 = 5t²
t² = 4 → t = 2 s

Time taken = 2 seconds.

Numerical 2

A stone is thrown vertically downward with velocity 5 m/s from a height of 50 m. Find the time taken to reach the ground. (g = 10 m/s²)

Solution:
s = ut + ½gt²
50 = 5t + ½(10)t²
50 = 5t + 5t²
5t² + 5t - 50 = 0
t² + t - 10 = 0
t = (-1 + √41)/2 ≈ 2.7 s

Time taken = 2.7 seconds.

Numerical 3

A body falls freely from rest. Find its velocity after 3 s.

Solution:
v = u + gt
v = 0 + 9.8 × 3 = 29.4 m/s

 Final velocity = 29.4 m/s.

Mass and Weight:

Mass:

- Mass is the amount of matter contained in a body.
- It is constant everywhere in the universe.
- SI Unit: kilogram (kg).
- Mass is a scalar quantity.

Weight:

- Weight is the force with which Earth (or any planet) attracts a body.
- Formula: W = m × g
   where m = mass of body (kg), g = acceleration due to gravity (9.8 m/s² on Earth).
- Weight is a vector quantity.
- Weight varies from place to place because g is different on Earth, Moon, or other planets.

Differentiate between Mass and Weight:

Aspect	Mass	Weight
Definition	Mass is the amount of matter contained in a body.	Weight is the force with which a body is attracted towards the center of the Earth (or any planet) due to gravity.
Formula	m = W/g	W = m x g
Nature	Scalar quantity (only magnitude)	Vector quantity (magnitude + direction)
SI unit	Kilogram (kg)	Newton (N)
Constant or Variable	Constant everywhere in the universe.	Varies with the value of g (gravity)
Measuring Device	Measured by a beam balance or electronic balance.	Measured by a spring balance (or weighing machine)
Effect of Location	Same on Earth, Moon, or in space.	Changes from Earth to Moon (Moon weight = 1/6th of Earth weight)
Example	A body has 10 kg mass everywhere.	The same body has a weight of 98 N on Earth but only 16.3 N on Moon

Solved Numerical 1:

Q: A stone has a mass of 5 kg. Find its weight on Earth (take g = 9.8 m/s²).

Solution:

W = m × g = 5 × 9.8 = 49 N
Mass of stone = 5 kg (same everywhere).
Weight of stone on Earth = 49 N.

Weight of an Object on the Surface of Moon:

- The Moon’s gravity is 1/6th of Earth’s gravity.
                           g_moon = (1/6) g_earth
- So, 

                          W_moon  = (1/6) W_earth

Solved Numerical 2: Weight on Moon

Q: A boy has a mass of 60 kg.
(i) Find his weight on Earth.
(ii) Find his weight on Moon. (Take g = 9.8 m/s²).

Solution:
(i) On Earth: W_earth = m × g = 60 × 9.8 = 588 N
(ii) On Moon: W_moon = (1/6) × 588 = 98 N
Mass of boy = 60 kg (constant).
Weight on Earth = 588 N.
Weight on Moon = 98 N.

Thrust and Pressure:

Thrust:

- Force acting perpendicularly on a surface.
- SI Unit: Newton (N).

Pressure:

- Pressure is thrust per unit area.
- Formula: P = F / A
   where F = perpendicular force (N), A = area (m²).
- SI Unit: Pascal (Pa) = 1 N/m².
- Pressure is inversely proportional to area.

Difference between Thrust and Pressure:

Aspect	Thrust	Pressure
Definition	Thrust is the force acting perpendicularly on a surface.	Pressure is the thrust (force perpendicularly) acting per unit area of the surface.
Formula	F = P x A	P = F/A
Nature	Vector quantity (has magnitude + direction)	Scalar quantity (only magnitude)
SI unit	Newton (N)	Pascal (Pa) = N/m2
Depends on	Only on the force applied.	Depends on both force applied and the area of contact.
Effect of Area	Independent of area.	Inversely proportional to area.
Example	The total force a girl exerts on the ground while standing is thrust.	The force per unit area exerted by the girl’s shoes on the ground is pressure.

 

Solved Numerical 3: Thrust and Pressure

Q: A girl weighing 300 N stands on the ground wearing shoes. Each shoe has an area of 200 cm² in contact with the ground. Find the pressure exerted on the ground.

Solution:
Total area = 2 × 200 = 400 cm² = 0.04 m²
Force = 300 N
Pressure = F / A = 300 / 0.04 = 7500 Pa
Pressure on ground = 7500 Pascal.

Solved Numerical 4: Effect of Area on Pressure

Q: A sharp knife has a blade area of 0.5 cm², while a blunt knife has a blade area of 10 cm². If a force of 50 N is applied on both, calculate the pressure exerted in each case.

Solution:
Sharp knife: A = 0.5 cm² = 5 × 10⁻⁵ m²
P = 50 / (5 × 10⁻⁵) = 1 × 10⁶ Pa

Blunt knife: A = 10 cm² = 1 × 10⁻³ m²
P = 50 / (10⁻³) = 5 × 10⁴ Pa

Pressure with sharp knife = 1,000,000 Pa
Pressure with blunt knife = 50,000 Pa
This is why sharp knives cut better!

Buoyancy (Upthrust):

• When an object is immersed in a fluid (liquid or gas), the fluid exerts an upward force on it. This upward force is called Buoyant Force or Upthrust.
• It arises because fluid pressure increases with depth, so the bottom of the object experiences more pressure than the top.
• Formula ,                 F_b = ρ × V × g

Numerical -1:Buoyancy

A wooden block of volume 0.02 m³ is fully immersed in water. Find the buoyant force on it (density of water = 1000 kg/m³, g = 9.8 m/s²).
Solution:

 F_b = ρ × V × g

     = 1000 × 0.02 × 9.8

F_b  = 196 N

Factors affecting on buoyant force:

• Volume of fluid displaced
• Density of fluid
• Acceleration due to gravity (g)

Why Objects Sink or Float?

An object will either sink or float depending on the comparison between its weight (W) and the buoyant force (F_b):

If W > F_b → Object sinks

If W < F_b → Object floats

If W = F_b → Object remains fully immersed but does not sink

Numerical 2: Sink or Float

An iron cube of volume 0.001 m³ and density 7800 kg/m³ is placed in water. Will it sink or float?

Solution:

Mass = 7800 × 0.001 = 7.8 kg

Weight (W) = 7.8 × 9.8 = 76.44 N

Buoyant force (F_b) = 1000 × 0.001 × 9.8 = 9.8 N
Since W > F_b, the cube sinks.
Answer: Cube sinks

Archimedes’ Principle:

Archimedes’ Principle states:  "When a body is immersed fully or partially in a fluid, it experiences an upward force (upthrust) equal to the weight of the fluid displaced by it".

• Formula, F_b = Weight of displaced fluid

Applications:
- Designing ships and submarines
- Hydrometer
- Explaining floating of bodies

Numerical 3: Archimedes’ Principle

An object of weight 50 N is immersed in water. It displaces water of weight 20 N.

Find (i) upthrust, (ii) apparent weight.
Solution:

Upthrust = 20 N

Apparent weight = 50 - 20 = 30 N

Relative Density (R.D.)

• Relative density is the ratio of the density of a substance to the density of water.
• Formula, R.D. = ρ_substance / ρ_water

Key points:
- It has no unit (ratio)
- RD > 1 → sinks in water
- RD < 1 → floats in water

Numerical 4: Relative Density

A solid weighs 200 g in air and 150 g in water. Find its relative density.

Solution:

Loss of weight = 200 - 150 = 50 g = weight of displaced water
RD = W_air / (W_air – W_water)

       = 200 / 50

RD = 4

Practice Numerical 

1)      An object thrown vertically upwards reaches a height of 500 m. What was its initial velocity? How long will the object take to come back to the earth? Assume g = 9.8 m/s² (Answer: u = 98.99 m/s and 20.20 sec)

2)      With what speed must an object be thrown vertically upwards to reach a height of 125 m? (Take g = 10 m/s²)                                                                                                        (Answer: 50 m/s )

3)      An object is thrown vertically upward to a height of 10 m. calculate its,

                   i) Velocity with which it was thrown upward. (Answer: 14 m/s)

                  ii) Time taken by the object to reach the highest point. (Answer: 1.43 sec)      

4)      A stone is thrown vertically upwards with an initial velocity of 30 m/s. The time taken for the stone to rise to its maximum height is (Take g = 10 m/s²) (Answer : 3sec)

5)      What will be weight of object on the moon if its weight on Earth is 120 N? (Answer : 20 N )

6)      The weight of an object on earth is 800 N. If it were taken to the moon, it would weight 100 N. What is its mass on the moon? (Acceleration due to gravity on earth = 10 m/s²)

                                                                                                                              (Answer : 80 kg)

7)      A block of wood is kept on a table top. The mass of the wooden block is 5 kg and its dimension 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 with its sides of dimensions,

         (a) 20 cm × 10 cm (Answer: 2450 Pa)                     

        (b) 40 cm × 20 cm. (Answer: 612.5 Pa)                                               (Given g = 9.8 m/s²)

8)      How much force should be applied on an area of 1 cm² to get a pressure of 15 Pa?                                                                                                                                        (Answer: 1.5 x 10^-3 N)   

9) Calculate the pressure developed by a force of 50 N acting on an area of 10 m². (Answer: 5 Pa)  

10)  Find the pressure exerted by block (given in figure below) on a table top, if mass of block is 3 kg.                                                                                                               (Answer : 3266.66 Pa)                                                                         

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