Welcome dear students! Today we are going to learn about Gravitation from Class 9 Science. We have learnt about the motion of objects and force as the cause of motion. We know that a force is needed to change the speed or the direction of motion of an object. We always observe that an object dropped from a height falls towards the earth. We 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 on the moon. Isaac Newton could grasp that the same force is responsible for all these. This force is called the gravitational force. In this chapter we shall learn about gravitation and the universal law of gravitation. We shall discuss the motion of objects under the influence of gravitational force on the earth. We shall study how the weight of a body varies from place to place. We shall also discuss the conditions for objects to float in liquids.
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Let us begin with section nine point one, Gravitation. We know that the moon goes around the earth. An object when thrown upwards, reaches a certain height and then falls downwards. It is said that when Newton was sitting under a tree, an apple fell on him. The fall of the apple made Newton start thinking. 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. Let us try to understand the motion of the moon by recalling activity nine point one. Take a piece of thread. Tie a small stone at one end. Hold the other end of the thread and whirl it round, as shown in Figure nine point one. Note the motion of the stone. Release the thread. Again, note the direction of motion of the stone. In this diagram, we see a stone tied to a thread being whirled in a circular path. 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, meaning centre seeking force. 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. A tangent to a circle is defined as a straight line that meets the circle at one and only one point. The motion of the moon around the earth is due to the centripetal force. The centripetal force is provided by the force of attraction of the earth. If there were no such force, the moon would pursue a uniform straight line motion.
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It is seen that a falling apple is attracted towards the earth. Does the apple attract the earth? If so, we do not see the earth moving towards an apple. Why? According to the third law of motion, the apple does attract the earth. But according to the 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. Extend the same argument for why the earth does not move towards the moon. In our solar system, all the planets go around the Sun. By arguing the same way, we can say that there exists a force between the Sun and the planets. From the above 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. Now let us move on to the universal law of gravitation. 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. In Figure nine point two, we see two objects labelled A and B with masses M and m, separated by a distance d. The gravitational force F acts along the line joining their centres. According to the universal law of gravitation, the force between two objects is directly proportional to the product of their masses, so F is proportional to M times m. And the force between two objects is inversely proportional to the square of the distance between them, so F is proportional to one over d squared. Combining these, we get F is proportional to M times m divided by d squared, or F equals G times M times m divided by d squared, where G is the constant of proportionality and is called the universal gravitation constant. By multiplying crosswise, we get F times d squared equals G times M times m, or G equals F times d squared divided by M times m. The SI unit of G is N m squared kg to the power of minus two. The value of G was found out by Henry Cavendish by using a sensitive balance. The accepted value of G is six point six seven three times ten to the power of minus eleven N m squared kg to the power of minus two. We know that there exists a force of attraction between any two objects. Compute the value of this force between you and your friend sitting closeby. Conclude how you do not experience this force! 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. Saying that F is inversely proportional to the square of d means, for example, that if d gets bigger by a factor of six, F becomes one over thirty six times smaller.
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Let us solve Example nine point one. The mass of the earth is six times ten to the power of twenty four kg and that of the moon is seven point four times ten to the power of twenty two kg. If the distance between the earth and the moon is three point eight four times ten to the power of five km, calculate the force exerted by the earth on the moon. Take G equals six point seven times ten to the power of minus eleven N m squared kg to the power of minus two. Solution: The mass of the earth, M equals six times ten to the power of twenty four kg. The mass of the moon, m equals seven point four times ten to the power of twenty two kg. The distance between the earth and the moon, d equals three point eight four times ten to the power of five km, which converts to three point eight four times ten to the power of eight m. G equals six point seven times ten to the power of minus eleven N m squared kg to the power of minus two. From the equation F equals G times M times m divided by d squared, the force exerted by the earth on the moon is calculated as six point seven times ten to the power of minus eleven multiplied by six times ten to the power of twenty four multiplied by seven point four times ten to the power of twenty two, all divided by three point eight four times ten to the power of eight squared. This gives two point zero two times ten to the power of twenty N. Thus, the force exerted by the earth on the moon is two point zero two times ten to the power of twenty N. The universal law of gravitation successfully explained several phenomena which were believed to be unconnected: 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.
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Next, we will learn about free fall. Let us try to understand the meaning of free fall by performing activity nine point two. 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. 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 m s to the power of minus two. We know from the second law of motion that force is the product of mass and acceleration. Let the mass of the stone in activity nine point two 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 equals m times g. From the universal law equation and this equation, we have m times g equals G times M times m divided by d squared, or g equals G times M divided by d squared, where M is the mass of the earth, and d is the distance between the object and the earth. Let an object be on or near the surface of the earth. The distance d will be equal to R, the radius of the earth. Thus, for objects on or near the surface of the earth, g equals G times M divided by R squared. 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 equals G times M divided by d squared.
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To calculate the value of g, we should put the values of G, M and R in the equation. Universal gravitational constant, G equals six point seven times ten to the power of minus eleven N m squared kg to the power of minus two, mass of the earth, M equals six times ten to the power of twenty four kg, and radius of the earth, R equals six point four times ten to the power of six m. Calculating g equals G times M divided by R squared gives nine point eight m s to the power of minus two. Thus, the value of acceleration due to gravity of the earth, g equals nine point eight m s to the power of minus two. 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. Activity nine point three: 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. 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. 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 equals u plus g t, s equals u t plus one half g t squared, and v squared equals u squared plus two g 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.
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Let us solve Example nine point two. A car falls off a ledge and drops to the ground in zero point five s. Let g equals ten m s to the power of minus two for simplifying the calculations. What is its speed on striking the ground? What is its average speed during the zero point five s? How high is the ledge from the ground? Solution: Time, t equals zero point five second. Initial velocity, u equals zero m s to the power of minus one. Acceleration due to gravity, g equals ten m s to the power of minus two. Acceleration of the car, a equals plus ten m s to the power of minus two downward. For part one, speed v equals a t, so v equals ten multiplied by zero point five, which equals five m s to the power of minus one. For part two, average speed equals u plus v divided by two, which is zero plus five divided by two, equals two point five m s to the power of minus one. For part three, distance travelled, s equals one half a t squared, so s equals one half times ten times zero point two five, which equals one point two five m. Thus, its speed on striking the ground is five m s to the power of minus one, its average speed is two point five m s to the power of minus one, and the height of the ledge is one point two five m. Now Example nine point three. An object is thrown vertically upwards and rises to a height of ten m. Calculate the velocity with which the object was thrown upwards and the time taken by the object to reach the highest point. Solution: Distance travelled, s equals ten m. Final velocity, v equals zero m s to the power of minus one. Acceleration due to gravity, g equals nine point eight m s to the power of minus two. Acceleration of the object, a equals minus nine point eight m s to the power of minus two for upward motion. Using v squared equals u squared plus two a s, we get zero equals u squared plus two times minus nine point eight times ten. So minus u squared equals minus one hundred ninety six, giving u equals fourteen m s to the power of minus one. For time, using v equals u plus a t, we get zero equals fourteen minus nine point eight times t, giving t equals one point four three s. Thus, initial velocity is fourteen m s to the power of minus one, and time taken is one point four three s.
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Moving on to section nine point three, Mass. We have learnt that the mass of an object is the measure of its inertia. Greater the mass, the 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. Section nine point four, Weight. We know that the earth attracts every object with a certain force and this force depends on the mass of the object and the acceleration due to the gravity. The weight of an object is the force with which it is attracted towards the earth. We know that F equals m times a, that is, F equals m times g. The force of attraction of the earth on an object is known as the weight of the object. It is denoted by W. Substituting the same, we have W equals m times 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. 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. 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.
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Let us study the weight of an object on the moon. The weight of an object on the earth is the force with which the earth attracts the object. 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 W sub m. Let the mass of the moon be M sub m and its radius be R sub m. By applying the universal law of gravitation, the weight of the object on the moon will be W sub m equals G times M sub m times m divided by R sub m squared. Let the weight of the same object on the earth be W sub e. The mass of the earth is M and its radius is R. From Table nine point one, the Earth has a mass of five point nine eight times ten to the power of twenty four kg and a radius of six point three seven times ten to the power of six m. The Moon has a mass of seven point three six times ten to the power of twenty two kg and a radius of one point seven four times ten to the power of six m. From the equations, we have W sub e equals G times M times m divided by R squared. Substituting the values, we get W sub m equals two point four three one times ten to the power of ten G times m, and W sub e equals one point four seven four times ten to the power of eleven G times m. Dividing these, we get W sub m divided by W sub e equals zero point one six five, which is approximately one sixth. So, weight of the object on the moon equals one sixth times its weight on the earth. Let us solve Example nine point four. Mass of an object is ten kg. What is its weight on the earth? Solution: Mass, m equals ten kg. Acceleration due to gravity, g equals nine point eight m s to the power of minus two. W equals m times g, so W equals ten times nine point eight, which equals ninety eight N. Thus, the weight of the object is ninety eight N. Example nine point five. An object weighs ten N when measured on the surface of the earth. What would be its weight when measured on the surface of the moon? Solution: We know, weight of object on the moon equals one sixth times its weight on the earth. That is, W sub m equals W sub e divided by six, which is ten divided by six, equals one point six seven N. Thus, the weight of object on the surface of the moon would be one point six seven N.
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Now we will discuss section nine point five, 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, it helps to introduce the concepts of thrust and pressure. Let us try to understand them by considering situations. Situation one: You wish to fix a poster on a bulletin board, as shown in Figure nine point three. 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. In the diagram, we see a hand pressing a drawing pin into a board. Situation two: You stand on loose sand. Your feet go deep into the sand. Now, lie down on the sand. You will find that your body will not go that deep in the sand. In both cases the force exerted on the sand is the weight of your body. 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 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 per m squared or N m to the power of minus two. In honour of scientist Blaise Pascal, the SI unit of pressure is called pascal, denoted as Pa.
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Let us consider Example nine point six. A block of wood is kept on a tabletop. The mass of wooden block is five kg and its dimensions are forty cm times twenty cm times ten 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) twenty cm times ten cm and b) forty cm times twenty cm. In Figure nine point four, we see the wooden block in two orientations: first standing on the smaller base of twenty cm by ten cm, and second lying flat on the larger side of forty cm by twenty cm. Solution: The mass of the wooden block equals five kg. The dimensions are forty cm times twenty cm times ten cm. The weight of the wooden block applies a thrust on the table top. That is, Thrust equals F equals m times g equals five times nine point eight, which equals forty nine N. For part a, area of a side equals length times breadth equals twenty cm times ten cm, which is two hundred cm squared or zero point zero two m squared. Pressure equals forty nine N divided by zero point zero two m squared, which equals two thousand four hundred fifty N m to the power of minus two. When the block lies on its side of dimensions forty cm times twenty cm, it exerts the same thrust. Area equals forty cm times twenty cm, which is eight hundred cm squared or zero point zero eight m squared. Pressure equals forty nine N divided by zero point zero eight m squared, which equals six hundred twelve point five N m to the power of minus two. The pressure exerted by the side twenty cm times ten cm is two thousand four hundred fifty N m to the power of minus two and by the side forty cm times twenty cm is six hundred twelve point five N m to the power of minus two. 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.
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Section nine point five point one, 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. Section nine point five point two, 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 in consideration. Let us understand the meaning of buoyancy by doing activity nine point four. 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. 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.
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Section nine point five point three, why objects float or sink when placed on the surface of water? Let us do activity nine point five. 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. In Figure nine point five, we see an iron nail sinking and a cork floating when placed on the surface of water. Now activity nine point six. 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 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.
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Let us move to section nine point six, Archimedes' Principle. Activity nine point seven. 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 as shown in Figure nine point six part a. 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 as shown in Figure nine point six part b. In this diagram, part a shows the stone hanging in air with maximum elongation, and part b shows the stone partially submerged in water with reduced elongation. Observe what happens to 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. 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, 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 nine point seven, as the stone was fully immersed in water? Archimedes was a Greek scientist. He discovered the principle, subsequently named after him, after noticing that the water in a bathtub overflowed when he stepped into it. He ran through the streets shouting Eureka, which means I have got it. 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 Roman army. 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.
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Let us quickly review 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. Now let us solve the exercises to prepare for your exams.
Exercise one: How does the force of gravitation between two objects change when the distance between them is reduced to half? Solution: According to the universal law of gravitation, force is inversely proportional to the square of the distance. If distance d becomes d divided by two, the new force becomes proportional to one divided by (d/2) squared, which is four divided by d squared. Therefore, the force becomes four times greater.
Exercise two: Gravitational force acts on all objects in proportion to their masses. Why then, a heavy object does not fall faster than a light object? Solution: The acceleration due to gravity, g, is given by G times M divided by R squared, which is independent of the mass of the falling object. Since all objects experience the same acceleration g near the earth's surface, they fall at the same rate regardless of their mass, provided air resistance is negligible.
Exercise three: What is the magnitude of the gravitational force between the earth and a one kg object on its surface? Mass of the earth is six times ten to the power of twenty four kg and radius of the earth is six point four times ten to the power of six m. Solution: Using F equals m times g, where g equals nine point eight m s to the power of minus two, the force is one kg times nine point eight m s to the power of minus two, which equals nine point eight N.
Exercise four: The earth and the moon are attracted to each other by gravitational force. Does the earth attract the moon with a force that is greater or smaller or the same as the force with which the moon attracts the earth? Why? Solution: The force is exactly the same. According to Newton's third law of motion, every action has an equal and opposite reaction. The gravitational force between two bodies is mutual and equal in magnitude.
Exercise five: If the moon attracts the earth, why does the earth not move towards the moon? Solution: The earth does experience the same force, but because the earth's mass is enormously larger than the moon's mass, the resulting acceleration of the earth is extremely small and practically unnoticeable.
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Exercise six: What happens to the force between two objects, if (i) the mass of one object is doubled? (ii) the distance between the objects is doubled and tripled? (iii) the masses of both objects are doubled? Solution: (i) Force is directly proportional to the product of masses. If one mass doubles, the force doubles. (ii) Force is inversely proportional to the square of distance. If distance is doubled, force becomes one fourth. If distance is tripled, force becomes one ninth. (iii) If both masses are doubled, the product of masses becomes four times, so the force becomes four times greater.
Exercise seven: What is the importance of universal law of gravitation? Solution: It successfully explains 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.
Exercise eight: What is the acceleration of free fall? Solution: The acceleration of free fall is the acceleration due to gravity, denoted by g. Its value on the surface of the earth is approximately nine point eight m s to the power of minus two.
Exercise nine: What do we call the gravitational force between the earth and an object? Solution: It is called the weight of the object.
Exercise ten: 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? Solution: No, the friend will not agree. The value of g is greater at the poles than at the equator. Since weight equals mass times g, the gold will weigh less at the equator than at the poles, even though its mass remains the same.
Exercise eleven: Why will a sheet of paper fall slower than one that is crumpled into a ball? Solution: The flat sheet of paper has a larger surface area, which experiences greater air resistance compared to the crumpled ball. This increased air resistance slows down its fall.
Exercise twelve: Gravitational force on the surface of the moon is only one sixth as strong as gravitational force on the earth. What is the weight in newtons of a ten kg object on the moon and on the earth? Solution: On earth, weight equals mass times g equals ten kg times nine point eight m s to the power of minus two equals ninety eight N. On the moon, weight equals one sixth of earth weight, which is ninety eight divided by six, equals sixteen point three three N.
Exercise thirteen: A ball is thrown vertically upwards with a velocity of forty nine m per s. Calculate (i) the maximum height to which it rises, (ii) the total time it takes to return to the surface of the earth. Solution: (i) Using v squared equals u squared plus two a s, with v equals zero, u equals forty nine, and a equals minus nine point eight, we get zero equals forty nine squared minus two times nine point eight times s. Solving gives s equals one hundred twenty two point five m. (ii) Using v equals u plus a t, zero equals forty nine minus nine point eight times t, giving t equals five s to reach the top. Total time to return is twice this, so ten s.
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Exercise fourteen: A stone is released from the top of a tower of height nineteen point six m. Calculate its final velocity just before touching the ground. Solution: Using v squared equals u squared plus two g s, with u equals zero, g equals nine point eight, s equals nineteen point six. v squared equals zero plus two times nine point eight times nineteen point six, which equals three hundred eighty four point one six. Taking square root, v equals nineteen point six m per s.
Exercise fifteen: A stone is thrown vertically upward with an initial velocity of forty m per s. Taking g equals ten m per s squared, find the maximum height reached by the stone. What is the net displacement and the total distance covered by the stone? Solution: Using v squared equals u squared plus two a s, zero equals forty squared minus two times ten times s. So s equals one hundred sixty divided by twenty, which equals eighty m. Net displacement is zero because it returns to the starting point. Total distance covered is eighty m up plus eighty m down, which equals one hundred sixty m.
Exercise sixteen: Calculate the force of gravitation between the earth and the Sun, given that the mass of the earth equals six times ten to the power of twenty four kg and of the Sun equals two times ten to the power of thirty kg. The average distance between the two is one point five times ten to the power of eleven m. Solution: Using F equals G times M times m divided by d squared, with G equals six point six seven three times ten to the power of minus eleven. F equals six point six seven three times ten to the power of minus eleven times six times ten to the power of twenty four times two times ten to the power of thirty, divided by one point five times ten to the power of eleven squared. This calculates to approximately three point five six times ten to the power of twenty two N.
Exercise seventeen: A stone is allowed to fall from the top of a tower one hundred m high and at the same time another stone is projected vertically upwards from the ground with a velocity of twenty five m per s. Calculate when and where the two stones will meet. Solution: Let them meet after time t at height h from ground. For falling stone: distance covered is one hundred minus h equals zero times t plus one half g t squared, so one hundred minus h equals five t squared. For rising stone: h equals twenty five t minus one half g t squared, so h equals twenty five t minus five t squared. Adding both equations: one hundred equals twenty five t, so t equals four s. Substituting t in second equation: h equals twenty five times four minus five times sixteen, which equals one hundred minus eighty, equals twenty m. They meet after four seconds at twenty m above the ground.
Exercise eighteen: A ball thrown up vertically returns to the thrower after six s. Find (a) the velocity with which it was thrown up, (b) the maximum height it reaches, and (c) its position after four s. Solution: (a) Time to reach top is half of total, so three s. Using v equals u plus a t, zero equals u minus nine point eight times three, so u equals twenty nine point four m per s. (b) Using v squared equals u squared plus two a s, zero equals twenty nine point four squared minus two times nine point eight times s, giving s equals forty four point one m. (c) After four s, it has reached top at three s and is falling for one s. Distance fallen in one s equals one half g t squared equals zero point five times nine point eight times one, equals four point nine m. Position from ground equals maximum height minus distance fallen, which is forty four point one minus four point nine, equals thirty nine point two m.
Exercise nineteen: In what direction does the buoyant force on an object immersed in a liquid act? Solution: The buoyant force acts vertically upwards.
Exercise twenty: Why does a block of plastic released under water come up to the surface of water? Solution: The density of the plastic block is less than the density of water. Therefore, the upward buoyant force is greater than the downward gravitational force (weight), causing it to rise to the surface.
[CHECKPOINT]
Exercise twenty one: The volume of fifty g of a substance is twenty cm cubed. If the density of water is one g per cm cubed, will the substance float or sink? Solution: Density of substance equals mass divided by volume, which is fifty divided by twenty, equals two point five g per cm cubed. Since its density is greater than the density of water, it will sink.
Exercise twenty two: The volume of a five hundred g sealed packet is three hundred fifty cm cubed. Will the packet float or sink in water if the density of water is one g per cm cubed? What will be the mass of the water displaced by this packet? Solution: Density of packet equals mass divided by volume, which is five hundred divided by three hundred fifty, equals one point four three g per cm cubed. Since its density is greater than water, it will sink. The mass of water displaced equals the volume of the packet times the density of water, which is three hundred fifty cm cubed times one g per cm cubed, equals three hundred fifty g.
Thank you for listening! Keep revising and practicing. Goodbye! [CHAPTER_COMPLETE]