Gravitation

So students, welcome to today's science lesson. Today we are going to study Chapter 9, which is about Gravitation. This is a very important chapter in your science curriculum, and I must tell you that the concepts you will learn here are fundamental to understanding how the universe works. So let's begin.

So students, let us start by thinking about something very common. You must have seen that when you drop any object from a height, it falls downwards towards the earth. Have you ever wondered why this happens? Also, you all know that the moon goes around the earth, and all the planets go around the sun. What is it that keeps the moon orbiting around the earth, or the planets orbiting around the sun? There must be some force acting on these objects. Today we are going to learn about this force, which is called the gravitational force.

Let us try to understand the motion of the moon. We have learnt about motion and force in previous chapters. We know that a force is needed to change the speed or the direction of motion of an object. Now, we observe that an object dropped from a height falls towards the earth. We also know that all the planets go around the sun, and the moon goes around the earth. In all these cases, there must be some force acting on the objects, the planets, and the moon. Isaac Newton was a great scientist who could understand that the same force is responsible for all these phenomena. This force is called the gravitational force.

Now students, let me tell you a very famous story. 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. He thought that if the earth can attract an apple, can it not attract the moon? Is the force the same in both cases? He conjectured that the same type of force is responsible in both the cases. He argued that 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 because it is moving sideways at the same time.

Now students, let us perform an activity to understand this concept better. This is Activity 9.1 from your textbook.

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. Now release the thread and again note the direction of motion of the stone.

Before the thread is released, the stone moves in a circular path with a certain speed and changes direction at every point. The change in direction involves change in velocity or acceleration. 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. The word centripetal means 'centre-seeking'. 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.

Now students, let me explain what a tangent to a circle is. A straight line that meets the circle at one and only one point is called a tangent to the circle. So when you release the stone, it flies off along a tangent to the circular path.

Now, this centripetal force is provided by the force of attraction of the earth on the moon. If there were no such force, the moon would pursue a uniform straight line motion.

Now students, let us think about another question. We have 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 does this happen?

According to Newton's third law of motion, the apple does attract the earth. But according to Newton's second law of motion, for a given force, acceleration is inversely proportional to the mass of an object. 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. The same argument explains why the earth does not move towards the moon, even though the moon attracts the earth.

Now students, from these facts, Newton concluded that 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. And this leads us to the Universal Law of Gravitation.

Now let me state the Universal Law of Gravitation for you. Every object in the universe attracts every other object with a force which is proportional to the product of their masses and inversely proportional to the square of the distance between them. The force is along the line joining the centres of two objects.

Now students, this is a very important law. Let me explain it in detail. 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, the force between two objects is directly proportional to the product of their masses. That is, F is proportional to M multiplied by m. And the force between two objects is inversely proportional to the square of the distance between them, that is, F is proportional to 1 divided by d squared.

Combining these two relationships, we get F is proportional to M multiplied by m divided by d squared. Or in other words, F equals G multiplied by M multiplied by m divided by d squared, where G is the constant of proportionality and is called the universal gravitation constant.

Now students, by rearranging this equation, we can find the SI unit of G. The SI unit of G can be obtained by substituting the units of force, distance and mass. The unit comes out as N m² kg⁻².

Now, who discovered the value of G? It was Henry Cavendish, who lived from 1731 to 1810, who found out the value of G by using a sensitive balance. The accepted value of G is 6.673 × 10⁻¹¹ N m² kg⁻².

Now students, let me tell you something interesting. We know that there exists a force of attraction between any two objects. You can compute the value of this force between you and your friend sitting close by. But you will find that this force is extremely small, and that is why you do not experience it!

Now, there is an important point about this law. It is called the inverse-square law. Saying that F is inversely proportional to the square of d means, for example, that if d gets bigger by a factor of 6, F becomes 1/36 times smaller. This is very important to remember.

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. This means that the same law that holds for an apple falling from a tree also holds for the planets revolving around the sun.

Now students, let me answer the questions given in your textbook after section 9.1.1.

The first question is: State the universal law of gravitation.

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

The second question is: Write the formula to find the magnitude of the gravitational force between the earth and an object on the surface of the earth.

The formula is F equals G multiplied by M multiplied by m divided by R squared, where M is the mass of the earth, m is the mass of the object, R is the radius of the earth, and G is the universal gravitation constant.

Now students, let me tell you about the importance of the universal law of gravitation. This law successfully explained several phenomena which were believed to be unconnected. These are: the force that binds us to the earth, the motion of the moon around the earth, the motion of planets around the sun, and the tides due to the moon and the sun. So you can see how important this law is!

Now let me work through Example 9.1 from your textbook.

The question is: 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.

Let me solve this step by step.

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

From the formula F = G M × m/d², we have:

F = 6.7 × 10⁻¹¹ × 6 × 10²⁴ × 7.4 × 10²² / (3.84 × 10⁸)²

Let me calculate this. First, (3.84 × 10⁸)² = 14.7456 × 10¹⁶ = 1.47456 × 10¹⁷

Now, 6.7 × 10⁻¹¹ × 6 × 10²⁴ = 40.2 × 10¹³ = 4.02 × 10¹⁴

Now, 4.02 × 10¹⁴ × 7.4 × 10²² = 29.748 × 10³⁶ = 2.9748 × 10³⁷

Now dividing by 1.47456 × 10¹⁷, we get approximately 2.02 × 10²⁰ N.

Thus, the force exerted by the earth on the moon is 2.02 × 10²⁰ N. This is a very large force, students! This force is what keeps the moon in its orbit around the earth.

Now students, let us move on to the next section, which is about Free Fall.

Let us try to understand the meaning of free fall by performing this activity. Take a stone. Throw it upwards. It reaches a certain height and then it starts falling down.

We have learnt that the earth attracts objects towards it. This is due to the gravitational force. Whenever objects fall towards the earth under this force alone, we say that the objects are in free fall.

Now, is there any change in the velocity of falling objects? While falling, there is no change in the direction of motion of the objects. But due to the earth's attraction, there will be a change in the magnitude of the velocity. Any change in velocity involves acceleration. Whenever an object falls towards the earth, an acceleration is involved. This acceleration is due to the earth's gravitational force. Therefore, this acceleration is called the acceleration due to the gravitational force of the earth, or acceleration due to gravity. It is denoted by g. The unit of g is the same as that of acceleration, that is, m s⁻².

Now students, we know from the second law of motion that force is the product of mass and acceleration. Let the mass of the stone be m. We already know that there is acceleration involved in falling objects due to the gravitational force and is denoted by g. Therefore the magnitude of the gravitational force F will be equal to the product of mass and acceleration due to the gravitational force, that is, F = m g.

From the universal law of gravitation, we have F = G M × m / d², where M is the mass of the earth and d is the distance between the object and the earth.

So, m g = G M × m / d²

Cancelling m from both sides, we get g = G M / d²

Now, let an object be on or near the surface of the earth. The distance d in this equation will be equal to R, the radius of the earth. Thus, for objects on or near the surface of the earth, m g = G M × m / R², or g = G M / R².

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 than at the equator. For most calculations, we can take g to be more or less constant on or near the earth. But for objects far from the earth, the acceleration due to gravitational force of earth is given by the equation g = G M / d².

Now students, let us calculate the value of g. To calculate the value of g, we should put the values of G, M and R in the equation g = G M / R². The values are: universal gravitational constant, G = 6.7 × 10⁻¹¹ N m² kg⁻², mass of the earth, M = 6 × 10²⁴ kg, and radius of the earth, R = 6.4 × 10⁶ m.

So, g = 6.7 × 10⁻¹¹ × 6 × 10²⁴ / (6.4 × 10⁶)²

(6.4 × 10⁶)² = 40.96 × 10¹² = 4.096 × 10¹³

6.7 × 10⁻¹¹ × 6 × 10²⁴ = 40.2 × 10¹³ = 4.02 × 10¹⁴

Dividing, we get approximately 9.8 m s⁻².

Thus, the value of acceleration due to gravity of the earth, g = 9.8 m s⁻². This is the value we use in most calculations.

Now students, let us understand the motion of objects under the influence of gravitational force of the earth. Let us do an activity to understand whether all objects, hollow or solid, big or small, will fall from a height at the same rate.

This is Activity 9.3. Take a sheet of paper and a stone. Drop them simultaneously from the first floor of a building. Observe whether both of them reach the ground simultaneously.

We see that paper reaches the ground little later than the stone. This happens because of air resistance. The air offers resistance due to friction to the motion of the falling objects. The resistance offered by air to the paper is more than the resistance offered to the stone. 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.

We know that an object experiences acceleration during free fall. From the equation g = G M / R², this acceleration experienced by an object is independent of its mass. This means that all objects, hollow or solid, big or small, should fall at the same rate. According to a story, Galileo dropped different objects from the top of the Leaning Tower of Pisa in Italy to prove the same. This is a famous experiment in the history of science.

Now students, as g is constant near the earth, all the equations for the uniformly accelerated motion of objects become valid with acceleration a replaced by g. The equations are:

v = u + a t s = u t + ½ a t² v² = u² + 2 a s

where u and v are the initial and final velocities and s is the distance covered in time, t.

In applying these equations, we will take acceleration a to be positive when it is in the direction of the velocity, that is, in the direction of motion. The acceleration a will be taken as negative when it opposes the motion. This is very important to remember, students!

Now let me work through Example 9.2.

The question is: 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?

Let me solve this step by step.

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

(i) We need to find speed v. Using the formula v = u + a t, v = 0 + 10 × 0.5 = 5 m s⁻¹

So the speed on striking the ground is 5 m s⁻¹.

(ii) Average speed = (initial speed + final speed) / 2 = (0 + 5) / 2 = 2.5 m s⁻¹

(iii) Distance travelled, s = u t + ½ a t² = 0 × 0.5 + ½ × 10 × (0.5)² = ½ × 10 × 0.25 = 1.25 m

Thus, the height of the ledge from the ground is 1.25 m.

Now students, let me work through Example 9.3.

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

(i) We use the formula v² = u² + 2 a s 0 = u² + 2 × (-9.8) × 10 -u² = -196 u² = 196 u = √196 = 14 m s⁻¹

So the initial velocity is 14 m s⁻¹.

(ii) We use the formula v = u + a t 0 = 14 - 9.8 × t t = 14 / 9.8 = 1.43 s

Thus, the time taken to reach the highest point is 1.43 seconds.

Now students, let me answer the questions given after section 9.2.

The first question is: What do you mean by free fall?

Free fall means when an object falls towards the earth under the influence of gravitational force alone, without any other force acting on it.

The second question is: What do you mean by acceleration due to gravity?

Acceleration due to gravity is the acceleration produced in a freely falling object due to the earth's gravitational pull. It is denoted by g and its value is approximately 9.8 m s⁻² on the surface of the earth.

Now students, let us move on to the next section, which is about Mass.

We have learnt in the previous chapter that the mass of an object is the measure of its inertia. We have also learnt that greater the mass, greater is the inertia. It remains the same whether the object is on the earth, the moon, or even in outer space. Thus, the mass of an object is constant and does not change from place to place. This is a very important point, students. Mass is an intrinsic property of an object.

Now students, let us learn about Weight.

We know that the earth attracts every object with a certain force, and this force depends on the mass (m) of the object and the acceleration due to gravity (g). The weight of an object is the force with which it is attracted towards the earth.

We know that force F = m × a, that is, F = m × g.

The force of attraction of the earth on an object is known as the weight of the object. It is denoted by W. So, W = m × g.

As the weight of an object is the force with which it is attracted towards the earth, the SI unit of weight is the same as that of force, that is, newton (N). The weight is a force acting vertically downwards; it has both magnitude and direction.

We have learnt that the value of g is constant at a given place. Therefore, at a given place, the weight of an object is directly proportional to the mass, say m, of the object, that is, W is proportional to m. It is due to this reason that at a given place, we can use the weight of an object as a measure of its mass. However, students, you must remember that the mass of an object remains the same everywhere, that is, on the earth and on any planet, whereas its weight depends on its location because g depends on location.

Now let us learn about the weight of an object on the moon.

We have learnt that the weight of an object on the earth is the force with which the earth attracts the object. In the same way, the weight of an object on the moon is the force with which the moon attracts that object. The mass of the moon is less than that of the earth. Due to this, the moon exerts lesser force of attraction on objects.

Let the mass of an object be m. Let its weight on the moon be Wₘ. Let the mass of the moon be Mₘ and its radius be Rₘ.

By applying the universal law of gravitation, the weight of the object on the moon will be Wₘ = G Mₘ × m / Rₘ².

Let the weight of the same object on the earth be Wₑ. The mass of the earth is M and its radius is R.

Now look at Table 9.1 in your textbook. It gives the mass and radius of the earth and the moon.

From the table, the mass of the earth is 5.98 × 10²⁴ kg and its radius is 6.37 × 10⁶ m. The mass of the moon is 7.36 × 10²² kg and its radius is 1.74 × 10⁶ m.

Now, using the formulas, we can calculate the ratio of weight on moon to weight on earth.

Wₘ = G × 7.36 × 10²² × m / (1.74 × 10⁶)² Wₑ = G × 5.98 × 10²⁴ × m / (6.37 × 10⁶)²

After calculating, we get Wₘ / Wₑ = 0.165, which is approximately equal to 1/6.

So students, the weight of the object on the moon is one-sixth of its weight on the earth. This is why astronauts feel lighter on the moon. If you weigh 60 kg on earth, you would weigh only 10 kg on the moon!

Now let me work through Example 9.4.

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

Solution: Mass, m = 10 kg Acceleration due to gravity, g = 9.8 m s⁻² W = m × g = 10 × 9.8 = 98 N

Thus, the weight of the object is 98 N.

Now Example 9.5.

The question is: 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 that weight of object on the moon = (1/6) × its weight on the earth. That is, Wₘ = Wₑ / 6 = 10 / 6 = 1.67 N

Thus, the weight of the object on the surface of the moon would be 1.67 N.

Now students, let me answer the questions given after section 9.4.

The first question is: What are the differences between the mass of an object and its weight?

The differences are: 1. Mass is the quantity of matter contained in an object, while weight is the force with which the earth attracts the object. 2. Mass is a scalar quantity, while weight is a vector quantity. 3. Mass remains constant everywhere, while weight changes from place to place. 4. The SI unit of mass is kilogram (kg), while the SI unit of weight is newton (N).

The second question is: Why is the weight of an object on the moon 1/6th its weight on the earth?

This is because the mass of the moon is less than the mass of the earth, and the radius of the moon is also smaller than the radius of the earth. Due to this, the gravitational force exerted by the moon on objects is about 1/6th of the gravitational force exerted by the earth.

Now students, let us move on to the next section, which is about Thrust and Pressure.

Have you ever wondered why a camel can run in a desert easily? Why an army tank weighing more than a thousand tonne rests upon a continuous chain? Why a truck or a motorbus has much wider tyres? Why cutting tools have sharp edges? In order to address these questions and understand the phenomena involved, it helps to introduce the concepts of the net force in a particular direction, which is called thrust, and the force per unit area, which is called pressure.

Let us try to understand the meanings of thrust and pressure by considering the following situations.

Situation 1: You wish to fix a poster on a bulletin board. To do this task, you will have to press drawing pins 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.

You have learnt that weight is the force acting vertically downwards. Here the force is acting perpendicular to the surface of the sand. The force acting on an object perpendicular to the surface is called thrust.

When you stand on loose sand, the force, that is, the weight of your body, is acting on an area equal to the area of your feet. When you lie down, the same force acts on an area equal to the contact area of your whole body, which is larger than the area of your feet. Thus, the effects of forces of the same magnitude on different areas are different. In the above cases, thrust is the same. But effects are different. Therefore, the effect of thrust depends on the area on which it acts.

The effect of thrust on sand is larger while standing than while lying. The thrust on unit area is called pressure. Thus, pressure equals thrust divided by area.

Substituting the SI unit of thrust and area, we get the SI unit of pressure as N/m² or N m⁻². In honour of scientist Blaise Pascal, the SI unit of pressure is called pascal, denoted as Pa.

Now students, let us consider a numerical example to understand the effects of thrust acting on different areas. This is Example 9.6.

The question is: A block of wood is kept on a tabletop. The mass of 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: The mass of the wooden block = 5 kg The dimensions = 40 cm × 20 cm × 10 cm Here, the weight of the wooden block applies a thrust on the table top. That is, Thrust = F = m × g = 5 kg × 9.8 m s⁻² = 49 N

(a) When the block lies on its side of dimensions 20 cm × 10 cm: Area = length × breadth = 20 cm × 10 cm = 200 cm² = 0.02 m² Pressure = thrust / area = 49 N / 0.02 m² = 2450 N m⁻²

(b) When the block lies on its side of dimensions 40 cm × 20 cm: Area = length × breadth = 40 cm × 20 cm = 800 cm² = 0.08 m² Pressure = 49 N / 0.08 m² = 612.5 N m⁻²

Thus, the same force acting on a smaller area exerts a larger pressure, and a smaller pressure on a larger area. This is the reason why a nail has a pointed tip, knives have sharp edges, and buildings have wide foundations.

Now students, let us learn about 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 and walls of the container in which they are enclosed. Pressure exerted in any confined mass of fluid is transmitted undiminished in all directions. This is known as Pascal's law, which has many applications in our daily life.

Now students, let us learn about Buoyancy.

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? Have you ever wondered why a ship made of iron and steel does not sink in sea water, but while 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. Let us understand the meaning of buoyancy by doing an activity.

This is Activity 9.4. Take an empty plastic bottle. Close the mouth of the bottle 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 exerted by the water goes on increasing as the bottle is pushed deeper till it is completely immersed. Now, release the bottle. It bounces back to the surface.

Now, does the force due to the 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?

The force due to the gravitational attraction of the earth acts on the bottle in the downward direction. So the bottle is pulled downwards. But the water exerts an upward force on the bottle. Thus, the bottle is pushed upwards. We have learnt that weight of an object is the force due to gravitational attraction of the earth. When the bottle is immersed, the upward force exerted by the water on the bottle is greater than its weight. Therefore, it rises up when released.

To keep the 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.

The upward force exerted by the water on the bottle is known as upthrust or buoyant force. In fact, 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.

Now students, let us understand why objects float or sink when placed on the surface of water. Let us do the following activities.

This is Activity 9.5. Take a beaker filled with water. Take an iron nail and place it on the surface of the water. Observe what happens.

The nail sinks. The force due to the 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 acting on the nail is greater than the upthrust of water on the nail. So it sinks.

Now Activity 9.6. 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.

The cork floats while the nail sinks. This happens because of the difference in their densities. The density of a substance is defined as the mass per unit volume. The density of cork is less than the density of water. This means that the upthrust of water on the cork is greater than the weight of the cork. So it floats.

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

Therefore, students, objects of density less than that of a liquid float on the liquid. The objects of density greater than that of a liquid sink in the liquid.

Now let me answer the questions given after section 9.5.3.

The first question is: Why is it difficult to hold a school bag having a strap made of a thin and strong string?

This is because the strap has a small surface area. When the bag is hung from the strap, the weight of the bag acts on a small area, creating more pressure on the shoulder. This makes it difficult to hold the bag.

The second question is: What do we mean by buoyancy?

Buoyancy is the upward force exerted by a fluid on any object placed in it. This force is also called upthrust or buoyant force.

The third question is: Why does an object float or sink when placed on the surface of water?

An object floats on water if its density is less than the density of water, because the upthrust of water is greater than the weight of the object. An object sinks in water if its density is greater than the density of water, because the upthrust of water is less than the weight of the object.

Now students, let us learn about Archimedes' Principle.

This is Activity 9.7. 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 the water in a container. Observe what happens to the elongation of the string or the reading on the balance.

You will find that the elongation of the string or the reading of the balance decreases as the stone is gradually lowered in the water. However, no further change is observed once the stone gets fully immersed in the water. What do you infer from the decrease in the extension of the string or the reading of the spring balance?

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

Now, what is the magnitude of the buoyant force experienced by a body? Is it the same in all fluids for a given body? Do all bodies in a given fluid experience the same buoyant force? The answer to these questions is contained in Archimedes' principle, which is stated as follows:

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.

Now, can you explain why a further decrease in the elongation of the string was not observed in activity 9.7, as the stone was fully immersed in water? This is because once the stone is fully immersed, the weight of the fluid displaced by it remains constant, so the buoyant force remains constant.

Archimedes' principle has many applications. It is used in designing ships and submarines. Lactometers, which are used to determine the purity of a sample of milk, and hydrometers used for determining density of liquids, are based on this principle.

Now students, let me tell you about Archimedes. He was a Greek scientist who lived around 250 BC. 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". This knowledge helped him to determine the purity of the 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.

Now let me answer the questions given after section 9.6.

The first question is: You find your mass to be 42 kg on a weighing machine. Is your mass more or less than 42 kg?

When you stand on a weighing machine, it measures your weight, not your mass. The weighing machine actually measures the force you exert on it. In air, there is a buoyant force acting on you, which reduces the reading slightly. So your actual mass is slightly more than 42 kg. However, this difference is very small and usually ignored.

The second question is: 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?

In reality, the iron bar is heavier than the bag of cotton. This is because the weighing machine measures weight, and it gives the same reading for both when measured in air. However, the buoyant force exerted by air on the cotton bag is more than that on the iron bar because cotton has a larger volume. So to get the same reading on the weighing machine, the actual weight of the cotton bag must be more than that of the iron bar. Therefore, the bag of cotton is actually heavier.

Now students, we have completed the theory portion of the chapter. Now let me solve all the exercises from the textbook.

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

According to the universal law of gravitation, force is inversely proportional to the square of the distance. So if the distance is reduced to half, then d becomes d/2. The new force F' = G M m / (d/2)² = G M m / (d²/4) = 4 G M m / d² = 4F. So the force becomes four times the original force.

Exercise 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?

This is because acceleration due to gravity is independent of the mass of the object. When we apply Newton's second law, F = ma, and for free fall, F = mg. So, mg = ma, which gives a = g. The acceleration is g, which is independent of mass. Therefore, all objects fall at the same rate in the absence of air resistance.

Exercise 3: 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.)

We use the formula F = G M m / R² G = 6.7 × 10⁻¹¹ N m² kg⁻² M = 6 × 10²⁴ kg m = 1 kg R = 6.4 × 10⁶ m

F = 6.7 × 10⁻¹¹ × 6 × 10²⁴ × 1 / (6.4 × 10⁶)² = 6.7 × 6 × 10¹³ / (40.96 × 10¹²) = 40.2 × 10¹³ / 40.96 × 10¹² = 9.82 N

So the gravitational force is approximately 9.8 N, which is equal to the weight of a 1 kg object.

Exercise 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?

According to Newton's third law of motion, every action has an equal and opposite reaction. So the earth attracts the moon with the same force as the moon attracts the earth. The forces are equal in magnitude but opposite in direction.

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

The earth does move towards the moon, but the movement is very small and not noticeable. This is because the mass of the earth is very large compared to the moon. According to Newton's second law, for the same force, a larger mass will have a smaller acceleration. So the acceleration of the earth towards the moon is very small.

Exercise 6: 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?

(i) If the mass of one object is doubled, then F' = G (2M) m / d² = 2F. So the force becomes twice.

(ii) If the distance is doubled, then F' = G M m / (2d)² = F/4. So the force becomes one-fourth. If the distance is tripled, then F' = G M m / (3d)² = F/9. So the force becomes one-ninth.

(iii) If the masses of both objects are doubled, then F' = G (2M) (2m) / d² = 4F. So the force becomes four times.

Exercise 7: What is the importance of universal law of gravitation?

The universal law of gravitation is important because it explains several phenomena: 1. It explains the force that binds us to the earth. 2. It explains the motion of the moon around the earth. 3. It explains the motion of planets around the sun. 4. It explains the tides due to the moon and the sun. 5. It is applicable to all bodies in the universe, whether big or small, celestial or terrestrial.

Exercise 8: What is the acceleration of free fall?

The acceleration of free fall on the surface of the earth is approximately 9.8 m s⁻². It is denoted by g.

Exercise 9: What do we call the gravitational force between the earth and an object?

The gravitational force between the earth and an object is called the weight of the object. It is denoted by W and is equal to mg.

Exercise 10: 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.)

No, the friend will not agree with the weight of gold bought. This is because the value of g is greater at the poles than at the equator. So the weight of the gold at the equator will be less than the weight at the poles. The mass remains the same, but the weight changes because g varies from place to place.

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

A sheet of paper has a larger surface area compared to a crumpled ball. So it experiences more air resistance. Therefore, it falls slower than a crumpled ball. If we remove the air resistance, both would fall at the same rate.

Exercise 12: 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?

Weight on earth = m × g = 10 × 9.8 = 98 N Weight on moon = (1/6) × weight on earth = 98/6 = 16.33 N

So the weight on earth is 98 N and on the moon is 16.33 N.

Exercise 13: 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.

Given: Initial velocity, u = 49 m/s Final velocity at highest point, v = 0 Acceleration due to gravity, g = 9.8 m s⁻² (taking g = 9.8)

(i) Using v² = u² + 2as 0 = 49² + 2 × (-9.8) × h 0 = 2401 - 19.6h 19.6h = 2401 h = 2401/19.6 = 122.5 m

So the maximum height is 122.5 m.

(ii) Time to reach maximum height: v = u + at 0 = 49 - 9.8t t = 49/9.8 = 5 s

Total time to return = 2 × 5 = 10 s

So the ball takes 10 seconds to return to the surface.

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

Given: Initial velocity, u = 0 Distance, s = 19.6 m Acceleration, g = 9.8 m s⁻²

Using v² = u² + 2as v² = 0 + 2 × 9.8 × 19.6 = 39.2 × 19.6 = 768.32 v = √768.32 = 19.6 m/s

So the final velocity is 19.6 m/s.

Exercise 15: 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?

Given: Initial velocity, u = 40 m/s Final velocity at highest point, v = 0 Acceleration, g = 10 m s⁻²

Using v² = u² + 2as 0 = 40² + 2 × (-10) × h 0 = 1600 - 20h 20h = 1600 h = 80 m

So the maximum height is 80 m.

Now, the stone goes up to 80 m and then falls back down to the ground. So the net displacement is zero (it returns to the starting point). The total distance covered is 80 m + 80 m = 160 m.

Exercise 16: 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.

Given: Mass of earth, M = 6 × 10²⁴ kg Mass of sun, m = 2 × 10³⁰ kg Distance, d = 1.5 × 10¹¹ m G = 6.7 × 10⁻¹¹ N m² kg⁻²

Using F = G M m / d² F = 6.7 × 10⁻¹¹ × 6 × 10²⁴ × 2 × 10³⁰ / (1.5 × 10¹¹)² = 6.7 × 12 × 10⁴³ / 2.25 × 10²² = 80.4 × 10⁴³ / 2.25 × 10²² = 35.73 × 10²¹ = 3.573 × 10²² N

So the force is approximately 3.57 × 10²² N.

Exercise 17: 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.

Let the stone from the top fall with initial velocity u1 = 0, acceleration g = 10 m/s². Let the stone from ground be projected with initial velocity u2 = 25 m/s, acceleration g = -10 m/s² (opposite direction).

Let them meet after time t at a distance x from the top (or from the ground).

For the stone falling from top: Distance covered = x = u1 t + ½ g t² = 0 + ½ × 10 × t² = 5t²

For the stone going up: Distance covered = 100 - x = u2 t + ½ g t² = 25t + ½ × (-10) × t² = 25t - 5t²

So we have: x + (100 - x) = 5t² + 25t - 5t² = 25t 100 = 25t t = 4 s

Now, x = 5t² = 5 × 16 = 80 m

So they meet after 4 seconds, at a height of 80 m from the top (or 20 m from the ground).

Exercise 18: 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.

Since the ball returns to the thrower after 6 s, the time to reach maximum height is 3 s.

(a) Using v = u + at 0 = u - g × 3 u = 3g = 3 × 10 = 30 m/s

So the velocity with which it was thrown up is 30 m/s.

(b) Maximum height: using v² = u² + 2as 0 = (30)² - 2 × 10 × h h = (30)² / (2 × 10) = 900 / 20 = 45 m

(c) Position after 4 s: Total time is 6 s, so after 4 s, the ball is on its way down. Time from maximum height = 4 - 3 = 1 s.

Using s = ut + ½ at² for downward motion from maximum height: s = 0 × 1 + ½ × 10 × 1² = 5 m

So the ball is 5 m below the maximum height, i.e., at height = 45 - 5 = 40 m from the ground.

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

The buoyant force acts in the upward direction, perpendicular to the surface of the liquid.

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

This is because the density of plastic is less than the density of water. So the buoyant force acting on the plastic block is greater than its weight. Hence, it rises up to the surface.

Exercise 21: The volume of 50 g of a substance is 20 cm³. If the density of water is 1 g cm⁻³, will the substance float or sink?

Density of substance = mass / volume = 50 g / 20 cm³ = 2.5 g cm⁻³

Density of water = 1 g cm⁻³

Since the density of the substance is greater than the density of water, the substance will sink.

Exercise 22: The volume of a 500 g sealed packet is 350 cm³. Will the packet float or sink in water if the density of water is 1 g cm⁻³? What will be the mass of the water displaced by this packet?

Density of packet = mass / volume = 500 g / 350 cm³ = 1.43 g cm⁻³

Since the density of the packet is greater than the density of water (1 g cm⁻³), the packet will sink.

The mass of water displaced = volume of packet × density of water = 350 cm³ × 1 g cm⁻³ = 350 g.

Now students, we have completed all the exercises. Let me now give you a summary of everything we have learned in this chapter.

What you have learnt:

- The law of gravitation states that 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. The law applies to objects anywhere in the universe. Such a law is said to be universal.

- Gravitation is a weak force unless large masses are involved.

- The force of gravity decreases with altitude. It also varies on the surface of the earth, decreasing from poles to the equator.

- The weight of a body is the force with which the earth attracts it.

- The weight is equal to the product of mass and acceleration due to gravity.

- The weight may vary from place to place, but the mass stays constant.

- All objects experience a force of buoyancy when they are immersed in a fluid.

- Objects having density less than that of the liquid in which they are immersed float on the surface of the liquid. If the density of the object is more than the density of the liquid in which it is immersed, then it sinks in the liquid.

Students, this is the end of our lesson on Gravitation. I hope you have understood all the concepts clearly. Remember, gravitation is a fundamental force that governs the motion of all objects in the universe. The concepts of mass, weight, thrust, pressure, buoyancy, and Archimedes' principle are very important and have many practical applications. Make sure you practice all the numerical problems and remember the formulas. Thank you for your attention, and see you in the next lesson!