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NCERT Solutions for Class 9 Science Chapter 9 Gravitation

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NCERT Solutions for Gravitation Class 9 Questions and Answers FREE PDF Download

Class 9 Science Ch 9 explores the concept of gravitational force, understanding its effects on celestial bodies like planets and moons, and Archimedes’ Principle. Vedantu’s Class 9 Gravitation NCERT Solutions solves all the questions in the chapter and helps students navigate through complex concepts with clarity and precision.  Access Vedantu's Gravitation Class 9 solutions for step-by-step explanations and problem-solving strategies and enhance your learning experience.

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Download Vedantu's Science Class 9 Gravitation NCERT Solutions, revised to align with the Class 9 Science syllabus. Start your academic journey with Vedantu and pave your way towards academic excellence.


Quick Insights for NCERT Solutions for Class 9 Chapter 9 Science Gravitation

  • Class 9 Science Ch 9 comprehends the concept of gravitation, the force of attraction that exists between any two objects with mass, and elucidates Newton's Law of Universal Gravitation, a cornerstone of classical physics. 

  • Ch 9 Science Class 9 explores the effects of gravitation in maintaining the stability of celestial bodies such as planets, stars, and galaxies.

  • Class 9 Gravitation Question Answer delves into the concept of acceleration due to gravity, the distinction between weight and mass, and Archimedes’ Principle.

  • Gravitation Class 9 Questions And Answers develop proficiency in solving numerical problems related to the motion of objects under the influence of the earth's gravitational force, Pressure, and Thrust.

  • Vedantu offers additional resources such as class notes, important concepts, formulas, and exemplar solutions to reinforce learning and ensure a strong grasp of foundational scientific principles.

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NCERT Solutions for Class 9 Science Chapter 9 Gravitation
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Gravitation L9 | NCERT Exercises, Questions 16, 17 & 18 | CBSE Class 9 Physics | Science Chapter 10
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Gravitation L4 | Exercises, Questions 1, 2 & 3 | CBSE Class 9 Physics | Science Chapter 10 | Vedantu
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Access NCERT Solutions For Class 9 Science Chapter 9 – Gravitation

Intext Exercise 1

1. State the universal law of gravitation.

Ans: Every object in the universe attracts every other object with some force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. The force acts along the line joining the centres of two objects.

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


force of attraction between two objects be F


F=GMmr2

Where, 

Gis the universal gravitation constant which is given by:

G=6.67×1011Nm2kg2


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

Ans: Let the mass of the Earth be M and the mass of an object on its surface be m. If Ris the radius of the Earth, then according to the universal law of gravitation, the gravitational force (F) that acts between the Earth and the object can be given by the relation:

F=GMmR2.


Intext Exercise 2

1. What do you mean by free fall?

Ans: Each object is drawn towards the centre of the Earth by its gravity. When any object is released from a certain height, under the impact of gravitational force, it falls to the Earth's surface. The movement of the object is said to be in free fall.


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

Ans: When any object falls freely from a certain height towards the earth's surface, its velocity changes with respect to time. This change in velocity causes acceleration. This acceleration is known as the acceleration due to gravity (g). The value of acceleration due to gravity is 9.8ms2.


Intext Exercise 3

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

Ans: The difference between the mass of an object and its weight is given in the table below:

Sr. no.

Mass

Weight

1.

Mass can be defined as the quantity of matter

contained in the body.

Weight can be defined as the force of gravity acting on

the body.

2.

It is the quantity that is a measure of inertia of the body.

It is the quantity that is a measure of gravity.

3.

Mass is constant everywhere.

The value of weight varies at different places.

4.

It is a scalar quantity.

Weight is a vector quantity.

5.

SI unit of mass is kg.

SI unit of weight is N.

 

2. Why is the weight of an object on the moon 16th its weight on the earth?

Ans: Let the mass of the Earth be ME and the mass of an object on the surface of earth =m and the radius of earth RE.

According to the Universal law of gravitation, weight WE of the object on the surface of the earth is given by,

WE=GMEmRE2

Let MM and RM be the mass and radius of the moon. Then, according to the universal law of gravitation, weight WM of the object on the surface of the moon is given by:

WM=GMMmRM2

So, ratio of weight of object on moon to weight on earth is

WMWE=MMRE2MERM2

Where, ME=5.98×1024kg 

MM=7.36×1022kg

RE=6.4×106m

RM=1.74×106m 

Substituting the values in the ratio,

WMWE=7.36×1022×(6.37×106)25.98×1024×(1.74×106)2

WMWE=0.16516

Hence, the weight of an object on the moon is 16th of its weight on the Earth.


Intext Exercise 4

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

Ans: Pressure can be given by the formula,

P=FA

Pressure is inversely proportional to the surface area on which the force is acting. The smaller is the surface area, the larger will be the pressure on the surface on which the force is being acted upon. In the case of a thin strap of the school bag, the contact surface area is very less. Hence, the pressure exerted on the shoulder is very high. Therefore, it becomes difficult to hold a school bag with a thin strap.


2. What do you mean by buoyancy?

Ans: The liquid exerts an upward force on any object when it is immersed in a liquid or fluid. The tendency of the liquid to exert such an upward force on the object is called buoyancy, and the upward force which is exerted on the object by the liquid is called the buoyant force.


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

Ans: If the density of an object is greater than the density of the liquid, it will sink into the liquid. This is due to the buoyant force which is acted by the object is less than the force of gravity. 


On the contrary, if the density of the object is less than the density of the liquid, it floats on the liquid's surface. This is because the force that is acting on the object is greater than the force of gravity.


Intext Exercise 5

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

Ans: An upward force acts on our body when we weigh our body while standing on a weighing machine. The buoyant force is which is a upward force that is acting. Consequently, the body is pushed up slightly, resulting in the weighing machine showing less reading than the real value.


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

Ans: Weight measured = Actual weight buoyant force

Therefore, Actual weight = Weight measured +buoyant force

As the surface area of the cotton, the bag is greater than the iron bar, more buoyant force acts on the bag than that on the iron bar. Hence, the mass of the cotton bag is more than that of the iron bar.


NCERT Exercise

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

Ans: According to the universal law of gravitation, the gravitational force (F) acting between two objects of mass m1and m2, separated by a distance ‘r’ is given by

F=Gm1m2r2

Where m1and m2are the masses of two bodies and ris the distance between them, G is the universal gravitational constant.

When the distance is reduced to half, i.e., r=r2

F=Gm1m2(r2)2

F=Gm1m2r24

F=4Gm1m2r2

Hence, if the distance is reduced to half, then the gravitational force becomes four times that of the previous value.


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

Ans: All the objects fall towards the ground with constant acceleration, called acceleration due to gravity (if there is no air resistance present). It is constant and independent of the mass of the object. Hence, heavy objects do not fall faster than light objects.


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

Ans: According to the Universal law of gravitation, the gravitational force exerted on an object of mass mis given by:

F=GMmr2

Where,

Mass of Earth, M=6×1024kg

Mass of object, m=1kg

Universal gravitational constant, G=6.7×1011Nm2kg2

Since the object is on the surface of the Earth, r=radius of the Earth (R)

r=R=6.4×106m

Gravitational force,

F=GMmr2

F=6.7×1011×6×1024×1(6.4×106)2=9.8N.

The magnitude of the gravitational force between the earth and a 1kgobject on its surface is 9.8N.


4. 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?

Ans: According to the Universal law of gravitation, two objects attract each other and according to Newton's third law of motion, the force of attraction between two objects is the same but acts in the opposite direction. Thus, the earth attracts the moon with the same force as the moon exerts on earth but the force acts in the opposite direction.


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

Ans: The Earth and the moon experience equal gravitational forces acting towards each other.

By Newton's Second Law, F=ma

a=Fm

For a certain force, acceleration is inversely proportional to the mass of an object.

aFm

Mass of the Earth >> Mass of the moon.

Hence, the acceleration experienced by earth due to the gravitational pull of the moon is very small when compared to that experienced by the moon due to earth. That is why the Earth does not move towards the moon.


6. What happens to the force between two objects, if

a) The mass of one object is doubled?

Ans: According to the universal law of gravitation, the force of gravitation between two objects is given by: F=GMmr2

Fis directly proportional to the product of masses of the two objects.

FMm

If the mass of one object is doubled, then the gravitational force will also change to double the original.


b) The distance between the objects is doubled and tripled?

Ans: Fis inversely proportional to the square of the distance between the objects.

If the distance between the objects is doubled, then the gravitational force becomes one-fourth of its original value. Also, if the distance is tripled, then the gravitational force becomes one-ninth of its original value.


c) The masses of both objects are doubled?

Ans: Fis directly proportional to the product of masses of the objects.

FMm

If the masses of both the objects are doubled, then the gravitational force becomes four times the original value.


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

Ans: The universal law of gravitation states that every object in the universe attracts every other object.

The force of gravitation binds us to the earth.

It is the cause for the motion of the moon around the earth and planets around the sun.

It results in the formation of tides due to the moon and the Sun. High tide occurs at the side where the moon pulls towards itself.


8. What is the acceleration of free fall?

Ans: A free-falling object is an object that is falling due to gravity without any air resistance. When it falls, there is a variation in velocity with respect to time that is associated with it.

Acceleration of free fall is denoted by gand its value on the surface of the earth is 9.8ms2, which is constant for all objects (irrespective of their masses).


9. What do we call the gravitational force between the Earth and an object?

Ans: The gravitational force between the earth and an object is called the weight of that object. It is equal to the product of acceleration due to the gravity and mass of the object.


10. Amit buys a 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).

Ans: Weight of a body on the Earth is given by:

W=mg

Where,

m=Mass of the body

g=Acceleration due to gravity

The shape of Earth is not a perfect sphere. As the radius of the earth increases from the poles to the equator, the value of gbecomes greater at the poles than at the equator. Since the value of g is greater at the poles than the equator.

Therefore, gold at the equator weighs less than at the poles. Hence, Amit’s friend will not agree with the weight of the gold bought.


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

Ans: When a sheet of paper is crumbled into a ball, then its surface area becomes much lesser than the surface area of a plain non-crumpled sheet of paper.

Hence, the upward force exerted by air on the sheet is greater as compared to the one exerted on the ball. Hence the sheet falls slower as compared to a paper ball.


12.Gravitational force on the surface of the moon is only 16 as strong as the gravitational force on the Earth. What is the weight in newtons of a 10kgobject on the moon and on the Earth?

Ans: It is provided that, Weight of an object on the moon=16×Weight of an object on the Earth

Also,

Weight=Mass×Acceleration

Acceleration due to gravity, g=9.8ms2

Therefore, the weight of a 10 kg object on the Earth =10×9.8N=98N

Weight of the same object on the moon =16×9.8N=16.3N


13. A ball is thrown vertically upwards with a velocity of 49ms1. Calculate

a) The maximum height to which it rises.

Ans: According to the equation of motion under gravity:

v2u2=2gh

Where,

u=Initial velocity of the ball

v=Final velocity of the ball

h=Height achieved by the ball

g=Acceleration due to gravity

At maximum height, final velocity of the ball is zero, i.e., v=0

u=49ms1

During upward motion, g=9.8ms2

Let h be the maximum height attained by the ball.

Hence,

(0)2(49)2=2×(9.8)×h

h=49×492×9.8=122.5


b) The total time it takes to return to the surface of the earth.

Ans: Let t be the time taken by the ball to reach the height 122.5m, then according to the equation of motion:

v=u+gt

Substituting the values and solving,

0=49+t×(9.8)

9.8t=49

t=499.8=5s

However,

Time of ascent = Time of descent

Therefore, the total time taken by the ball to return is 5+5=10s.


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

Ans: According to the equation of motion under gravity:

v2u2=2gh

Where,

u=Initial velocity of the stone =0

v=Final velocity of the stone

s=Height of the stone =9.6m

g = Acceleration due to gravity =9.8ms2

v202=2×9.8×19.62

v2=2×9.8×19.6=(19.6)2

v=19.6ms1

Hence, the velocity of the stone just before touching the ground is 19.6ms1.


15. A stone is thrown vertically upward with an initial velocity of 40ms1. Taking g=10ms2, find the maximum height reached by the stone. What is the net displacement and the total distance covered by the stone?

Ans: According to the equation of motion under gravity:

v2u2=2gh

Where,

u=Initial velocity of the stone =40ms1

v=Final velocity of the stone=0

s=Height of the stone 

g = Acceleration due to gravity =10ms2

Let h be the maximum height attained by the stone.

Therefore,

0(40)2=2×h×(10)

h=40×4020=80m

Therefore, the total distance covered by the stone during its upward and downward journey is 80+80=160m.

The net displacement of the stone during its upward and downward

journey is 80+(80)=0m.


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

Ans: According to the Universal l law of gravitation, the force of attraction between the Earth and the Sun is given by:

Given,

MSun=Mass of the Sun =2×1030kg

MEarth=Mass of the Earth =6×1024kg

R= Average distance between the Earth and the Sun =1.5×1011m

G=Universal gravitational constant =6.7×1011Nm2kg2.

F=GMSunMEarthR2

F=6.7×1011×2×1030×6×1024(1.5×1011)2

F=3.57×1022N

Hence, the force of gravitation between the earth and the sun is 3.57×1022N.


17. A stone is allowed to fall from the top of a tower 100m high and at the same time another stone is projected vertically upwards from the ground with a velocity of 25ms1. Calculate when and where the two stones will meet.

Ans: Let the two stones meet after time tfrom the start.

a) For the stone dropped from the tower:

Initial velocity, u=0.

Let the displacement of the stone in time t from the top of the tower be s.

Acceleration due to gravity, g=9.8ms2

From the equation of motion,

s=ut+12gt2

s=0×t+12×9.8×t2

s=4.9t2……. (1)

b) For the stone thrown upwards:

Initial velocity, u=25ms1

Let the displacement of the stone from the ground in time tbe s.

Acceleration due to gravity, g=9.8ms2

Equation of motion,

s=ut+12gt2

s=25t12×9.8×t2

s=25t4.9t2…… (2)

The combined displacement of both the stones at the meeting point is equal to the height of the tower 100m.

s+s=100…… (3)

Substituting equation (1) and (2) in (3),

4.9t2+25t4.9t2=100

25t=100

t=10025=4s

In 4s, the falling stone has covered a distance given by equation (1) as

s=12×9.8×42=78.4m

Therefore, the stones will meet after 4s at a height (10078.4)=21.6m from the ground.


18. A ball thrown up vertically returns to the thrower after 6s. Find

a) The velocity with which it was thrown up,

Ans: Time of ascent is equal to the time of descent. The ball takes a total of 6sfor its upward and downward journey.

Hence, time taken for upward journey, t=62=3s

Final velocity of the ball at the maximum height, v=0

Acceleration due to gravity, g=9.8ms2

Equation of motion, v=u+gtwill give,

0=u+(9.8×3)

u=9.8×3=29.4ms1

Hence, the ball was thrown upwards with a velocity of 29.4ms1.


b) The maximum height it reaches 

Ans: Let the maximum height attained by the ball be h.

Initial velocity during the upward journey, u=29.4ms1 

Final velocity, v=0

Acceleration due to gravity, g=9.8ms2

s=ut+12at2

From the equation of motion,

h=29.4×3+12×(9.8)×(3)2=44.1m 

c) Its position after 4s.

Ans: Ball attains the maximum height after 3s. After attaining this height, it will start falling downwards.

In this case, Initial velocity, u=0

Position of the ball after 4s of the throw is given by the distance

travelled by it during its downward journey in 4s3s=1s

Equation of motion,

s=ut+12at2

will give,

s=0×t+12×9.8×12=4.9m

Total height =44.1m

This means that the ball is 44.1m4.9m=39.2m above the ground

after 4seconds.


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

Ans: An object immersed in a liquid is acted upon by the buoyant force in the vertically upward direction.


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

Ans: The number of forces acting on a certain item in water are two. The first one is the gravitational force pulling down the object, and the other is the buoyant force pushing up the object. If the buoyant force acting in the upward direction is higher than the gravitational force that is acting downward, then the object goes up to the water's surface as quickly as it is released into water. That is why a block of plastic released under the water comes up to the surface of the water.


21. The volume of 50gof a substance is 20cm3 . If the density of water is 1gcm3, will the substance float or sink?

Ans: If the density of an object is more than the density of a liquid, then it sinks in the liquid. If the density of an object is less than the density of a liquid, then it floats

Density of the substance=Mass of the substanceVolume of the substance

Density of the substance=5020

Density of the substance=2.5gcm3.

The density of the substance > The density of water (1gcm3).

Hence, the substance will sink in water.     


22. The volume of a 500g sealed packet is 350cm3. Will the packet float or sink in water if the density of water is 1gcm3? What will be the volume of the water displaced by this packet?

Ans: If the density of an object is greater than the density of a liquid, then the object will sink in the liquid. If the density of an object is less than the density of a liquid, then it will float on the surface of the liquid.

Density of the 500 g sealed packet=Mass of the packetVolume of the packet

Density of the 500 g sealed packet=500350

Density of the 500 g sealed packet=1.428gcm3

The density of the substance is more than the density of water (1gcm3).

Hence, the object will sink in water.

Clearly, the mass of water displaced by the packet can be considered equal to the volume of the packet=0.350g.


Class 9th Science Gravitation Class 9 - Quick Overview of Detailed Structure of Topics

Topic 

Subtopics

Gravitation

  • Universal Law of Gravitation

  • Importance Of The Universal Law Of Gravitation

Free Fall

  • To Calculate the Value Of ‘g’

  • Motion of Objects Under the Influence Of Gravitational Force of the Earth

Mass

  • Understanding Mass and Weight

Weight

  • Weight of an Object on the Moon

Thrust And Pressure

  • Pressure In Fluids

  • Buoyancy

  • Why Objects Float or Sink When Placed on the Surface of Water?

Archimedes’ Principle

  • Understanding the Archimedes’ Principle


Class 9 Science Ch 9 Gravitation - Important Formula and Concepts

  • The universal law of gravitation: The force of attraction between any two objects is proportional to the product of their masses and inversely proportional to the square of the distance between them. 

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

  • F=GMmr2

Where Gis the universal gravitation constant, which is given by:

G=6.67×1011Nm2kg2

  • Weight of the object on the moon = (1/6) × its weight on the earth.

  • Pressure= Thrust/Area


Benefits of Vedantu’s NCERT Class 9 Science Gravitation Question Answer

  • Vedantu’s solutions explain all key concepts covered in Ch 9 Science Class 9, including the Universal Law of Gravitation, acceleration due to gravity, the distinction between mass and weight, free fall, and gravitational fields.

  • Class 9 Science Gravitation Question Answers are presented in a step-by-step format, making it easier for students to follow and understand the problem related to the motion of objects under the influence of the gravitational force of the earth, pressure, and Thrust. 

  • Vedantu’s NCERT Gravitation Class 9 Questions And Answers help students prepare effectively for their exams. 

  • The solutions include various types of questions, from multiple-choice to descriptive, ensuring comprehensive exam readiness.

  • Class 9 Gravitation Question Answers are prepared by Vedantu Master Teachers with a deep understanding of the curriculum and examination patterns. This ensures that the content is accurate, reliable, and aligned with the latest syllabus.

  • Using Class 9 Gravitation NCERT Solutions, students can save time by quickly finding answers and explanations for their doubts and questions. This allows them to allocate more time to practice and revision.

  • Vedantu’s Class 9 Chapter 9 Science solutions are available online, making them accessible anytime and anywhere. This flexibility supports continuous learning and allows students to study independently.


Related Study Materials for Class 9 Science Ch 9 Gravitation


Conclusion

Vedantu’s NCERT Class 9 Gravitation NCERT Solutions is an important study material. They provide clear explanations and step-by-step solutions, helping students grasp important concepts like the laws of gravitation, mass, weight, and gravitational force. Focusing on these areas is essential, as they form the chapter's foundation. In previous years, around 7-8 questions on gravitation have been asked in exams, highlighting its importance. By using Gravitation Class 9 Questions And Answers, students can effectively prepare, clear their doubts, and perform well in their exams, ensuring a strong understanding of gravitation.



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FAQs on NCERT Solutions for Class 9 Science Chapter 9 Gravitation

1. How can students understand the features of Gravitational Force Properly?

Chapter 9 Science Class 9 is an important part of the Science syllabus. Focus on the classroom sessions and concentrate on what the teachers are explaining. Study the chapter unit-wise and clear your doubts by using the Science Class 9 NCERT Solutions provided by Vedantu. You will surely understand these newfound concepts well.

2. How can I solve Gravitation problems quickly?

You must practise regularly using the NCERT Solutions Class 9 Science Chapter 9 as a reference and become more efficient. Your speed will automatically increase as you can remember the formulas properly.

3. Why do students prefer using NCERT Solutions for Class 9 Science Chapter 9?

By using the NCERT Solutions for Class 9 Science Gravitation, a student can save time in finding the right answers. They can focus better in preparing the chapter and score higher in the exams by following the ideal answering format recommended by the experts.

4. Why does the Earth not move towards objects due to Gravitation according to Chapter 9 Gravitation of Class 9 Science? 

Newton's third law states, “Every action has its equal and opposite reaction”. It means the force applied by an object on the Earth is equal to the force applied by the earth on the object, but we know that acceleration is inversely proportional to mass. This means when the acceleration is increased, the mass is decreased, or when the mass is increased, the acceleration is decreased. As the mass of the earth is large, the acceleration due to an object is small or negligible. Therefore, it's not noticeable. And the Earth doesn’t seem to be moving.

5. What are the different applications of Archimedes' principle?

The different applications of Archimedes' principle include the following:

  • It is used in designing ships and submarines.

  • Lactometers used to determine the purity of a milk sample and hydrometers used to determine the density of a liquid are based on this principle. 

6. Why does the Earth doesn’t move towards objects due to Gravitation according to Chapter 9 Gravitation of Class 9 Science? 

As per Newton's third law, “Every action has its equal and opposite reaction”. It means the force applied by an object on the Earth is equal to the force applied by the earth on the object, but we know that acceleration is inversely proportional to mass. This means when the acceleration is increased, the mass is decreased or when the mass is increased, the acceleration is decreased. As the mass of the earth is large, the acceleration due to an object is small or negligible. Therefore, it's not noticeable. And the Earth doesn’t seem to be moving.

7. What are topics covered in Vedantu’s Class 9 Gravitation NCERT Solutions?

Vedantu’s NCERT Class 9 Science Gravitation Question Answer covers all key concepts such as Newton’s Law of Gravitation, the universal law of gravitation, the relationship between gravitational force, mass, and distance, free fall, mass, weight, and the concept of acceleration due to gravity.

8. Why should I refer to Vedantu’s NCERT Solutions for Gravitation Class 9?

Vedantu’s NCERT Solutions for Gravitation Class 9 offers detailed explanations and step-by-step answers to textbook problems, making it easier to understand complex concepts. They also provide diagrams and illustrative examples that aid in better comprehension.