CBSE Class 9 Science Revision Notes Chapter 10

CBSE Class 9 Science Revision Notes Chapter 10 – Gravitation

The Class 9 Science Chapter 10 Notes will give students an overview of gravitation and the universal law of gravitation. These notes cover various natural phenomena and how they happen. Class 9 Chapter 10 Science Notes will help students understand the concept of gravitation and the related things that happen around us. To understand the chapter better, students can download these CBSE revision notes from the Extramarks website. In Class 9 Science Notes Chapter 10, students will also learn about new concepts and different formulas to solve mathematical problems.

Access Class IX Science Chapter 10 – Gravitation Notes In 30 Minutes

The Universal Law of Gravitation Or Newton’s Law of Gravitation:

• The universal law of gravitation is the mathematical relationship that Sir Isaac Newton proposed for measuring gravitational force. 
• According to this law of gravitation, “every particle in the universe gets attracted to 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, the direction of the force being along the line joining the masses”. 
• Consider two objects with masses m1 and m2 separated by a distance d. According to Newton’s law, the product of the masses is proportional to the gravitational force F. The square of the distance between the masses is inversely proportional to the gravitational force.

F ∝ m1m2…… (1)

F ∝ 1d2 …….. (2)

Two objects with masses m1 and  m2 are separated by a distance d.

 F∝m1m2d2

F∝Gm1m2d2   …………..(3)

Here, G is proportionally constant, referred to as the universal gravitational constant, as its value will be constant throughout the cosmos and remains unaffected by object masses.

Universal Gravitational Constant:

The mathematical representation of Newton’s law of gravitation is:

F = Gm1m2d2

If m1=m2=1, andd = 1, then

F = G1112

F = G

We can derive from the above equation that the universal gravitational constant can be defined as the gravitational force between two unit masses separated by a unit distance. The SI unit of gravitational constant:

G = Fd2m1m2

The SI unit of force F is N, the SI unit of distance is metre, and the SI unit of mass is kg.

SI unit of G = Nm2kgkg

SI unit of g = Nm2kg2 or Nm²kg-2

The experimental value of G is 6.6734 1011Nm2kg², which Sir Henry Cavendish measured in 1798.

Dependence of Gravitational Force on Mass:

In Chapter 10 Science Class 9 Notes, students will learn about the dependence of gravitational force on mass. According to Newton’s law of gravitation, the force of attraction is directly proportional to the body mass.

• Two objects of mass m and 2m separated by a distance d:

When you double up the mass of one of the two given objects, the relation of the force of attraction will be as follows:

F2= Gm2md2

     = 2Gm2d2

    = 2F1

• Two objects of mass 2m separated by a distance d:

When you double the masses of both bodies, then the force of attraction will be as follows:

F3= G2m2md2

    = 4Gm2d2

    = 4F1 

Hence, whenever the mass of the body increases, the force of attraction also increases. 

Dependence of gravitational force on distance:

As per the universal law of gravitation, the force of attraction between two bodies will be inversely proportional to the square of their distance. 

1. Force of attraction between two bodies of mass m separated by a distance d:

F1= Gm1m2d2

    = Gm²d²

Hence, F1 = Gm²d²

2. The force of attraction when the distance is doubled:

F2 = Gmm (2d)²

      = Gm² 4d²

Here, the distance 2d separates two bodies of mass m.

F2=14Gm²d²

F2 = 14 F1

3. Force of attraction when the distance between the bodies is increased three times:

F3=Gmm(3d)²

     = Gm²9d²

Here, distance 3d separates two bodies of mass, and hence,

F3=19Gm²d²

F3 = 19F1

It Results That:

• When you double the distance, the force decreases to 1/4th of its original value.
• If you extend the distance three times, the force reduces to 1/9th of its original value. 
• As given in the preceding example, we can conclude that the force of attraction between the bodies varies inversely with the square of the distance between them. 

Gravitational Force Between Two Light Objects:

Let us calculate the gravitational force between two units of masses separated by a unit of distance. 

m1= 1kg

m2= 1kg

d = 1 m

G = 6.6734 x 10-11Nm²kg2

Therefore, the force will be 

F = Gm1m2d2

F = 6.673410-11111²

F = 6.6734 x 10-11 N

Hence, this is a very weak force. 

To find force existing between two objects of masses 60 kg and 100 kg separated by a distance of 1m:

F = Gm1m2d2

F = 6.673410-11601001²

F = 6.673410-86

F = 40.04 10-8 N

This is also a weak force.

It is evident from the previous two examples why we do not feel the force created by one object (on the Earth’s surface) on the other object.

Gravitational Force Between Massive Objects:

Let us find the force of attraction between a 50 kg object and the Earth. The mass of Earth is nearly 6 1024kg, and the distance of Earth from the object is approximately 641024 m.

The force of attraction between the Earth and the object is,

F = Gm1m2d2

   = 6.673410-11 6 102450 641052

  = 6.67346510-111025 (64)21010

  = 6.67346510-11102510-10 6464

Therefore,

F = 6.673465104 6464

F = 6.67343105 6464

F = 488.7N

The force here is so strong that it cannot be ignored.

Now, let’s calculate the force of attraction between Earth and the Sun. The mass of Earth = 641024 kg

The mass of the Sun = 1.991030 kg

The distance between the Sun and Earth= 151010 m

Therefore,

F = Gm1m2d2

   = 6.673410-11 6 10241.991030 1510102

  = 6.673461.9910-111024103010-20 225

Hence,

F = 6.673461.991023 225

F = 3.541 1022N

As we can see, the force is very large, which keeps the planets in their respective orbits.

Conclusion:

When two objects of normal size are considered, the gravitational force is quite less, but when one of the items is huge, the force is quite large. 

Centre of Gravity:

• As we all know, every particle is attracted to the earth’s centre. A body consists of several particles. The gravitational attraction on these particles can be considered parallel to one another because the body is small compared to the earth. It can be understood from the following diagram.
• A single force acting in a downward direction can be replaced by a single force acting through a fixed location called the body’s centre of gravity.  
• The force that results is equivalent to the weight of the body.
• As a result, regardless of the body’s position, the centre of gravity is the point through which the weight of the body acts.
• The centre of gravity for bodies with uniform density and regular shapes is located at the body’s geometric centre. 
• For bodies with a regular shape and uniform density, the geometrical centre of gravity is where the body is placed.

Application Of Newton’s Law Of Gravitation:

• Newton’s law is mostly applied for calculating the masses of binary stars. A binary star refers to a system of two stars orbiting around their common centre of mass. 
• If the motion of a star is irregular, it would mean that another star or a planet revolves around the stars. The regularity in the motion of a star is known as a wobble. 

Mass And Weight:

• Mass and weight may sometimes create confusion, but they are two different concepts. Let’s find out what makes them different. 
• The mass of a body is the amount of matter contained in it. 
• Mass is a scalar quantity.
• The kilogram is the SI unit of mass (kg). 
• The amount of matter in a body will not change with time or location. The mass of a body will remain constant throughout the universe. However, the masses of the two bodies might be very different.
• A pan balance is used to calculate an individual’s mass.
• Weight is the force that pulls an object towards the Earth’s centre. 
• Weight of a body = force exerted by the Earth = mg (according to  Newton’s second law of motion)  
• W = mg

The SI unit of weight is Newton.

For example, a body with a mass of 1 kg weighs around

W = mg

W = 1 x 9.8 

W = 9.8 N

• Kg wt denotes the unit of weight. 
• 1kg weight is the force with which an object of mass of 1kg is pulled towards the Earth. 

W = mg 

1 kg wt = 1 x 9.8 

1 kg wt = 9.8 N 

• To determine the weight, spring balance is used. 
• Due to gravity’s acceleration, weight fluctuates from one place to another. 
• A body weighs more at the poles than it does at the equator, and it weighs nothing in the centre of the Earth because there is no acceleration of gravity.

Difference Between Mass And Weight:

To show that the weight of the body on the moon will be 16th of its weight on Earth:

Let’s imagine m represents a body’s weight on Earth. The equation that determines its weight on Earth is as follows:

We= mge

We=mGMeRe2 …..(1)

Since, 

ge=GMRe2

The same body’s weight on the moon (Wm) is given by

Wm= mgm

Wm=mGMmRm2  —-(2)

Since,

gm=GMmRm2

After dividing equation (2) by equation (1), we will get,

WmWc=mGMmRm2 X Rc2mGMc

Wm,Wc= Wm Rc2Rm2 Mc

WmWc=MmMcRcRm2 —– (3)

But we know that Me = 100 Mm and Re = 4 Rm , therefore

M m M e  = 1100

Re / Rm  = 4

By substituting above values in equation (3), we get

Wm W e = 1100( 4)2

 Wm W e = 16100

Wm W e = 16.25

Wm W e = 16

Wm =  16 We

So, the weight of a body on the moon will be 16th of its weight on Earth.

Weightlessness:

• It is common knowledge that astronauts feel weightless in space. What does this actually mean? 
• Let’s conduct a little experiment to demonstrate weightlessness. When a stone is suspended from a spring balance, the weight of the stone is shown on the pointer of the balance.
• Allow the stone to fall freely, along with the spring balance.
• The spring balance reads “0 weight,” indicating that the stone has no weight.
• Does this suggest that the stone has no weight?
• The stone, however, is falling freely and appears to be weightless.
• The experimental value of G is 6.6734  x 1011 Nm2/kg2, which was measured by Sir Henry Cavendish in 1798.
• Now let’s try to explain why a spaceship astronaut feels weightless.
• Both the astronaut and the spaceship are in free fall towards the Earth when an astronaut is orbiting the Earth in a spaceship.
• During a free fall, both descend with the same acceleration, which is equal to the acceleration of gravity.
• Neither the astronaut exerts any force on the spaceship’s sides or floor, nor the spaceship’s sides and floor push the astronaut up. 
• So, the astronaut feels weightless while orbiting the Earth in a spaceship.

Density:

• Cotton takes up more space than iron. Hence, 0.5 kg of cotton takes up more space than 0.5 kg of iron.
• Cotton fibres are loosely packed, whereas iron fibres are densely packed. More iron is contained in a given volume.
• This explains why an equal volume of iron is heavier than cotton.
• The mass of a substance per unit volume is referred to as its density.

Density = Mass of the substanceVolume of the substance

D = Mv, where M is the mass, D is the density, and v is the volume.

SI unit of density is kgm³ 

• The density of a substance is constant under certain conditions.
• One of a material’s distinctive qualities is its density, which may be used to assess the purity of any substance.

Relative Density Of A Substance:

• We use the formula D=Mv to calculate the density of a substance or an item by figuring out the mass and value of the substance.
• This is only possible if the object has a regular shape. 
• It can be challenging to gauge the size of an object with an asymmetrical shape.
• In such cases, the object’s density is expressed in terms of water density. 
• The relative density of a substance is the ratio of its density to the density of water at 4 degrees Celsius. It is assumed that water has a relative density of one.
• What does it mean when someone claims gold’s relative density is 19.3?
• This indicates that gold has a density of 19.3 times that of water in the same volume.
• Objects with a relative density lower than one will float in water, and those with a higher density will sink.

Relative density of a substance = Density of the substanceDensity of water at 4° C

                                               = Mass of the substanceVolume of the substanceMass of waterVolume of water

                                               = Mass of the substanceVolume of the substanceVolume of waterMass of water

If we take an equal amount of water and material, then

Relative density of a substance = Mass of the substanceVolume of an equal volume of water

Thus, relative density has no unit as it is the ratio of two identical quantities.

Thrust and pressure:

• At the beginning of the chapter, we discussed that force is an external agent that alters the body’s direction of motion, speed, or shape. 
• We talked about the forces acting at a particular location on a body.
• Take into account the factors at work in a particular location.
• To hang a poster on the classroom bulletin board, you must exert force on the head of the drawing pin, which is perpendicular to the board’s surface.
• Thrust is the force which is acting perpendicular to the surface.
• The SI unit of thrust is Newton (N).
• Let’s check if the applied force and the area where it acts are related.
• Place a pin vertically in the centre of a stack of papers. Place another pin beside it, upside down, with its flat head on the stack.
• To press down both pins, place a flat object on top of them, such as a duster. We see that the upright pin has pierced the stack of papers.
• This is because the force acting on the erect pin is only applied over a small area, whereas the force acting on the second pin is applied over a large area.
• Your bag is held in place by a thin yet strong string strap.
• Now elevate the same bag using a wide cloth band as a strap. It is more comfortable to carry a school bag with a wide cloth band than one with a tiny strip.
• This is because the weight of the books is distributed over a larger area of the shoulder in the second example, which exerts less force.
• According to the above example, the efficiency of the applied force depends on the area on which it acts.
• It is now necessary to establish a new physical concept called pressure.
• Pressure is the force operating on a unit area.

Pressure = ForceArea

Pressure =ThrustArea

The SI unit of pressure is Nm².

Nm² is known as Pascal (Pa)

1Nm² = 1 Pascal

• Kilopascal is a commonly used small unit.  
• Pressure is determined by: the force applied and the area it affects.

Buoyancy And Archimedes’ Principle:

• Bodies appear lighter when they are submerged in water or any other liquid.
• When taking a bath, we notice that the mug of water becomes substantially heavier as soon as it crosses the water’s surface.
• When a fish is lifted out of the water, it appears heavier in the air than it was in the water.
• Let’s examine why this happens.
• When objects are submerged in water or any other liquid, they appear lighter because the liquid or water exerts an upward force on them.
• Let’s experiment to check if there is a loss of weight when submerged in water.
• Attach a stone to one end of the spring balance. Suspend the spring balance as shown in the diagram.

Experimental set up to prove Archimedes’ Principle:

• Take note of the spring balance reading and call it W1.
• Place the stone in a water container and record the reading on the spring balance.
• The spring balance reading will continue to drop until it is completely submerged in water.
• The reading on the spring balance will determine the weight of the stone.
• The reading keeps going down; thus, we can conclude that the object’s weight diminishes as it is dropped into water.
• The apparent weight loss proves that a force acting on an object in the upward direction is causing it to lose weight.
• The upward force that is applied to an object submerged in a liquid is the buoyant force that causes the object to appear to lose weight.
• Buoyancy is the tendency of the liquid to exert an upward force on an object placed in it, causing it to float or rise.

Factors Affecting The Buoyant Force:

• We know that an iron ship floats on the surface of the water, whereas an iron nail sinks. This is because the ship is bigger or has more volume.
• When an iron nail and a cork of equal mass are submerged in water, the iron nail sinks because it has a higher density than the water and a lower density than the cork.
• A body’s buoyant force causes it to float when the liquid’s density is higher than that of the body’s material and vice versa.
• As seen in the examples above, the buoyant force that a body experiences while submerged in a liquid depends on the volume of the body and the density of the liquid.

Archimedes’ Principle:

• Archimedes performed several experiments to study the upward thrust that a body experiences when it is partially or completely submerged in fluid before formulating Archimedes’ Principle. 
• This principle states when a body is partially or entirely submerged in a fluid, it experiences a buoyant force equal to the weight of the liquid displaced.

Experiment To Verify Archimedes’ Principle:

• Calculate the mass of a clean, dry beaker using a physical balance.
• To calculate a solid’s weight, suspend it from a spring balance. Add water to a Eureka can until it reaches the spout and place the mass (m) beaker under it.
• Gently drop the stone into the Eureka can while it is suspended from a spring balance. Continue doing this until the stone is completely submerged in water.
• When immersed in water, the stone displaces a specific amount of water.
• The spring balance gives a lower value that indicates the solid is upthrust. The displaced water is collected in the beaker.
• The physical balance is used to compute the mass of the water and beaker. Let’s call it m1.
• Hence, the amount of water displaced will be m1– m
• When compared to the volume of water displaced, the apparent weight loss of the solid in the water is found to be equal. This experiment validates the Archimedes Principle.

Application Of Archimedes’ Principle:

• It is used in the development of ships and submarines. 
• Based on this principle, lactometers and hydrometers devices that assess the purity of a sample of milk and the density of liquids, respectively, function.

Gravitation: Summary Of Class 9 Chapter 10 Revision Notes

Class 9 Science Chapter 10 Notes will describe the universal law of gravitation. The law states that it is possible to determine the gravitational force between two bodies. The terms included in the law are properly explained in the notes so that students can relate the expressions of the law with its concepts. Students are advised to pay close attention to each term so they can later use it as a formula to solve problems.

The Class 9 Science Chapter 10 Notes are prepared by a team of experts according to the latest CBSE syllabus to help students prepare for board exams. Apart from CBSE revision notes, students can also access CBSE sample papers and CBSE past years’ question papers from the Extramarks website. Practising sample papers, past years’ question papers, and CBSE extra questions will help students understand the exam pattern and learn time-management skills to perform better in exams.

After understanding the law of gravitation, students learn to calculate the speed of a freely falling body. You learn how the earth attracts all bodies towards its centre. Newton’s laws of motion can also take advantage of this gravitational acceleration. You learn the definition and calculation of the acceleration caused by gravity.

After understanding the law of gravitation and acceleration, you’d study the concept of the moon’s revolution around the Earth without falling into the Earth. According to this section, all bodies travel straight when there is no disturbance or force. Proceeding further, you’d  learn how a centripetal force operates to cause a body to rotate around the axis of another body despite the presence of a gravitational force between them. You will understand how centripetal force acts.

Additionally, in this chapter you’d study the concepts of motion and free fall. You’d learn how these terms are expressed by the signs. The equations related to the laws of motion can be used to compute free fall and gravitational motion. After going through these notes, you can define thrust and pressure and explain Archimedes’ Principle. You’d learn about relative density and its applications, Kepler’s Law and how a planet moves in its orbit around the sun.