Chapter 9 Gravitation PDF

C hapter 9 GRAVITATION Let us try to understand the motion of the We have learnt about the motion of objects and moon by recalling activity 7.11. force as the cause of...

C hapter 9 GRAVITATION Let us try to understand the motion of the We have learnt about the motion of objects and moon by recalling activity 7.11. force as the cause of motion. We have learnt that a force is needed to change the speed or Activity ______________ 9.1 the direction of motion of an object. We always observe that an object dropped from a height Take a piece of thread. Tie a small stone at one end. Hold the falls towards the earth. We know that all the other end of the thread and whirl it planets go around the Sun. The moon goes round, as shown in Fig. 9.1. around the earth. In all these cases, there must Note the motion of the stone. be some force acting on the objects, the planets Release the thread. and on the moon. Isaac Newton could grasp Again, note the direction of motion of that the same force is responsible for all these. the stone. 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. 9.1 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, Fig. 9.1: A stone describing a circular path with a an apple fell on him. The fall of the apple made velocity of constant magnitude. Newton start thinking. He thought that: if the earth can attract an apple, can it not attract Before the thread is released, the stone the moon? Is the force the same in both cases? moves in a circular path with a certain speed He conjectured that the same type of force is and changes direction at every point. responsible in both the cases. He argued that The change in direction involves change in at each point of its orbit, the moon falls velocity or acceleration. The force that causes towards the earth, instead of going off in a this acceleration and keeps the body moving straight line. So, it must be attracted by the along the circular path is acting towards earth. But we do not really see the moon falling the centre. This force is called the towards the earth. centripetal (meaning ‘centre-seeking’) force. 100 SCIENCE 2024-25 In the absence of this force, the stone flies off 9.1.1 UNIVERSAL LAW OF GRAVITATION along a straight line. This straight line will be a tangent to the circular path. Every object in the universe attracts every other object with a force which is proportional to the product of their masses and inversely Tangent to a circle proportional to the square of the distance between them. The force is along the line joining the centres of two objects. More to know A straight line that meets the circle at one and only one point is called a Mm F =G 2 tangent to the circle. Straight line d ABC is a tangent to the circle at point B. Fig. 9.2: The gravitational force between two uniform objects is directed along the line The motion of the moon around the earth joining their centres. is due to the centripetal force. The centripetal force is provided by the force of attraction of Let two objects A and B of masses M and the earth. If there were no such force, the m lie at a distance d from each other as shown moon would pursue a uniform straight line in Fig. 9.2. Let the force of attraction between motion. two objects be F. According to the universal It is seen that a falling apple is attracted law of gravitation, the force between two towards the earth. Does the apple attract the objects is directly proportional to the product earth? If so, we do not see the earth moving of their masses. That is, towards an apple. Why? F∝M × m (9.1) According to the third law of motion, the And the force between two objects is inversely apple does attract the earth. But according proportional to the square of the distance to the second law of motion, for a given force, between them, that is, acceleration is inversely proportional to the mass of an object [Eq. (8.4)]. The mass of an 1 apple is negligibly small compared to that of F∝ (9.2) d2 the earth. So, we do not see the earth moving towards the apple. Extend the same argument Combining Eqs. (10.1) and (10.2), we get for why the earth does not move towards the moon. M ×m In our solar system, all the planets go F ∝ d2 (9.3) around the Sun. By arguing the same way, we can say that there exists a force between M×m the Sun and the planets. From the above facts or, F = G 2 (9.4) d Newton concluded that not only does the earth attract an apple and the moon, but all where G is the constant of proportionality and objects in the universe attract each other. This is called the universal gravitation constant. force of attraction between objects is called By multiplying crosswise, Eq. (9.4) gives the gravitational force. F×d2=GM×m GRAVITATION 101 2024-25  2 From Eq. (9.4), the force exerted by the Fd or G= (9.5) earth on the moon is M×m M ×m The SI unit of G can be obtained by F =G d2 substituting the units of force, distance and mass in Eq. (9.5) as N m2 kg–2. 6.7 × 10 −11 N m 2 kg -2 × 6 × 1024 kg × 7.4 × 1022 kg = The value of G was found out by (3.84 × 108 m)2 Henry Cavendish (1731 – 1810) by using a = 2.02 × 1020 N. sensitive balance. The accepted value of G is 6.673 × 10–11 N m2 kg–2. Thus, the force exerted by the earth on We know that there exists a force of the moon is 2.02 × 1020 N. attraction between any two objects. Compute the value of this force between you and your Q friend sitting closeby. Conclude how you do uestions not experience this force! 1. State the universal law of gravitation. 2. Write the formula to find the The law is universal in the sense that magnitude of the gravitational it is applicable to all bodies, whether force between the earth and an the bodies are big or small, whether object on the surface of the earth. More to know they are celestial or terrestrial. Inverse-square 9.1.2 IMPORTANCE OF THE UNIVERSAL LAW OF GRAVITATION Saying that F is inversely proportional to the square of d The universal law of gravitation successfully means, for example, that if d gets explained several phenomena which were bigger by a factor of 6, F becomes believed to be unconnected: 1 (i) the force that binds us to the earth; times smaller. (ii) the motion of the moon around the 36 earth; (iii) the motion of planets around the Sun; and Example 9.1 The mass of the earth is (iv) the tides due to the moon and the Sun. 6 × 1024 kg and that of the moon is 7.4 × 1022 kg. If the distance between the 9.2 Free Fall earth and the moon is 3.84×105 km, calculate the force exerted by the earth on Let us try to understand the meaning of free the moon. (Take G = 6.7 × 10–11 N m2 kg-2) fall by performing this activity. Solution: Activity 9. 2 The mass of the earth, M = 6 × 1024 kg Take a stone. The mass of the moon, Throw it upwards. m = 7.4 × 1022 kg It reaches a certain height and then it The distance between the earth and the starts falling down. moon, We have learnt that the earth attracts d = 3.84 × 105 km objects towards it. This is due to the = 3.84 × 105 × 1000 m gravitational force. Whenever objects fall = 3.84 × 108 m towards the earth under this force alone, we G = 6.7 × 10–11 N m2 kg–2 say that the objects are in free fall. Is there any 102 SCIENCE 2024-25 change in the velocity of falling objects? While calculations, we can take g to be more or less falling, there is no change in the direction of constant on or near the earth. But for objects motion of the objects. But due to the earth’s far from the earth, the acceleration due to attraction, there will be a change in the gravitational force of earth is given by magnitude of the velocity. Any change in Eq. (9.7). velocity involves acceleration. Whenever an object falls towards the earth, an acceleration 9.2.1 TO CALCULATE THE VALUE OF g is involved. This acceleration is due to the earth’s gravitational force. Therefore, this To calculate the value of g, we should put the acceleration is called the acceleration due to values of G, M and R in Eq. (9.9), namely, the gravitational force of the earth (or universal gravitational constant, G = 6.7 × 10–11 acceleration due to gravity). It is denoted by N m2 kg-2, mass of the earth, M = 6 × 1024 kg, g. The unit of g is the same as that of and radius of the earth, R = 6.4 × 106 m. acceleration, that is, m s–2. M We know from the second law of motion g= G 2 that force is the product of mass and R acceleration. Let the mass of the stone in 6.7 × 10-11 N m 2 kg -2 × 6 × 1024 kg activity 9.2 be m. We already know that there = is acceleration involved in falling objects due (6.4 × 106 m)2 to the gravitational force and is denoted by g. = 9.8 m s–2. Therefore the magnitude of the gravitational force F will be equal to the product of mass Thus, the value of acceleration due to gravity and acceleration due to the gravitational of the earth, g = 9.8 m s–2. force, that is, F=mg (9.6) 9.2.2 MOTION OF OBJECTS UNDER THE From Eqs. (9.4) and (9.6) we have INFLUENCE OF GRAVITATIONAL M ×m FORCE OF THE EARTH mg =G 2 d Let us do an activity to understand whether all objects hollow or solid, big or small, will M fall from a height at the same rate. or g = G 2 (9.7) d where M is the mass of the earth, and d is the Activity 9.3 distance between the object and the earth. Take a sheet of paper and a stone. Let an object be on or near the surface of Drop them simultaneously from the the earth. The distance d in Eq. (9.7) will be first floor of a building. Observe equal to R, the radius of the earth. Thus, for whether both of them reach the objects on or near the surface of the earth, ground simultaneously. We see that paper reaches the ground M ×m mg = G (9.8) little later than the stone. This happens 2 R because of air resistance. The air offers resistance due to friction to the motion M of the falling objects. The resistance g=G (9.9) R 2 offered by air to the paper is more than the resistance offered to the stone. If The earth is not a perfect sphere. As the we do the experiment in a glass jar radius of the earth increases from the poles to from which air has been sucked out, the equator, the value of g becomes greater at the paper and the stone would fall at the poles than at the equator. For most the same rate. GRAVITATION 103 2024-25  We know that an object experiences u +v acceleration during free fall. From Eq. (9.9), (ii) average speed = 2 this acceleration experienced by an object is = (0 m s–1+ 5 m s–1)/2 independent of its mass. This means that all = 2.5 m s–1 objects hollow or solid, big or small, should (iii) distance travelled, s = ½ a t2 fall at the same rate. According to a story, = ½ × 10 m s–2 × (0.5 s)2 Galileo dropped different objects from the top = ½ × 10 m s–2 × 0.25 s2 of the Leaning Tower of Pisa in Italy to prove = 1.25 m the same. Thus, As g is constant near the earth, all the (i) its speed on striking the ground equations for the uniformly accelerated = 5 m s–1 motion of objects become valid with (ii) its average speed during the 0.5 s acceleration a replaced by g. = 2.5 m s–1 The equations are: (iii) height of the ledge from the ground v = u + at (9.10) = 1.25 m. 1 s = ut + at2 (9.11) 2 v2 = u2 + 2as (9.12) Example 9.3 An object is thrown vertically upwards and rises to a height of 10 m. where u and v are the initial and final velocities Calculate (i) the velocity with which the and s is the distance covered in time, t. object was thrown upwards and (ii) the In applying these equations, we will take time taken by the object to reach the acceleration, a to be positive when it is in the highest point. direction of the velocity, that is, in the direction of motion. The acceleration, a will Solution: be taken as negative when it opposes the motion. Distance travelled, s = 10 m Final velocity, v = 0 m s–1 Acceleration due to gravity, g = 9.8 m s–2 Example 9.2 A car falls off a ledge and Acceleration of the object, a = –9.8 m s–2 drops to the ground in 0.5 s. Let (upward motion) g = 10 m s –2 (for simplifying the (i) v 2 = u2 + 2a s calculations). 0 = u 2 + 2 × (–9.8 m s–2) × 10 m (i) What is its speed on striking the –u 2 = –2 × 9.8 × 10 m2 s–2 ground? u = 196 m s-1 (ii) What is its average speed during the 0.5 s? u = 14 m s-1 (iii) How high is the ledge from the (ii) v=u+at ground? 0 = 14 m s–1 – 9.8 m s–2 × t t = 1.43 s. Solution: Thus, (i) Initial velocity, u = 14 m s–1, and Time, t = ½ second (ii) Time taken, t = 1.43 s. Initial velocity, u = 0 m s–1 Acceleration due to gravity, g = 10 m s–2 Q Acceleration of the car, a = + 10 m s–2 uestions (downward) 1. What do you mean by free fall? (i) speed v = at 2. What do you mean by acceleration v = 10 m s–2 × 0.5 s due to gravity? = 5 m s–1 104 SCIENCE 2024-25 9.3 Mass attracts the object. In the same way, the weight of an object on the moon is the force with We have learnt in the previous chapter that the which the moon attracts that object. The mass mass of an object is the measure of its inertia. of the moon is less than that of the earth. Due We have also learnt that greater the mass, the to this the moon exerts lesser force of attraction greater is the inertia. It remains the same on objects. whether the object is on the earth, the moon Let the mass of an object be m. Let its or even in outer space. Thus, the mass of an weight on the moon be Wm. Let the mass of object is constant and does not change from the moon be Mm and its radius be Rm. place to place. By applying the universal law of gravitation, the weight of the object on the moon will be 9.4 Weight Mm × m Wm = G (9.16) We know that the earth attracts every object Rm2 with a certain force and this force depends on Let the weight of the same object on the the mass (m) of the object and the acceleration earth be We. The mass of the earth is M and its due to the gravity (g). The weight of an object radius is R. is the force with which it is attracted towards the earth. Table 9.1 We know that F = m × a, (9.13) Celestial Mass (kg) Radius (m) that is, body F = m × g. (9.14) The force of attraction of the earth on an Earth 5.98 × 1024 6.37 ××106 object is known as the weight of the object. It Moon 7.36 ××1022 1.74 ××106 is denoted by W. Substituting the same in Eq. (9.14), we have W=m×g (9.15) From Eqs. (9.9) and (9.15) we have, As the weight of an object is the force with M ×m We = G (9.17) which it is attracted towards the earth, the SI R2 unit of weight is the same as that of force, that Substituting the values from Table 10.1 in is, newton (N). The weight is a force acting Eqs. (9.16) and (9.17), we get vertically downwards; it has both magnitude and direction. 7.36 × 1022 kg × m Wm = G We have learnt that the value of g is (1.74 × 10 m ) 6 2 constant at a given place. Therefore at a given place, the weight of an object is directly Wm = 2.431 × 1010 G × m (9.18a) proportional to the mass, say m, of the object, and We = 1.474 × 10 G × m 11 (9.18b) that is, W ∝ m. It is due to this reason that at a given place, we can use the weight of an Dividing Eq. (9.18a) by Eq. (9.18b), we get object as a measure of its mass. The mass of Wm 2.431 × 1010 an object remains the same everywhere, that = is, on the earth and on any planet whereas its We 1.474 × 1011 weight depends on its location because g Wm 1 depends on location. or W = 0.165 ≈ 6 (9.19) e 9.4.1 W EIGHT OF AN OBJECT ON Weight of the object on the moon = 1 THE MOON Weight of the object on the earth 6 We have learnt that the weight of an object on Weight of the object on the moon the earth is the force with which the earth = (1/6) × its weight on the earth. GRAVITATION 105 2024-25  of the net force in a particular direction (thrust) Example 9.4 Mass of an object is 10 kg. and the force per unit area (pressure) acting What is its weight on the earth? on the object concerned. Solution: Let us try to understand the meanings of Mass, m = 10 kg thrust and pressure by considering the Acceleration due to gravity, g = 9.8 m s–2 following situations: W=m×g Situation 1: You wish to fix a poster on a W = 10 kg × 9.8 m s-2 = 98 N bulletin board, as shown in Fig 9.3. To do this Thus, the weight of the object is 98 N. 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 Example 9.5 An object weighs 10 N when perpendicular to the surface area of the board. measured on the surface of the earth. This force acts on a smaller area at the tip of What would be its weight when the pin. measured on the surface of the moon? Solution: We know, Weight of object on the moon = (1/6) × its weight on the earth. That is, W 10 Wm = e = N. 6 6 = 1.67 N. Thus, the weight of object on the surface of the moon would be 1.67 N. Q uestions 1. What are the differences between the mass of an object and its weight? 2. Why is the weight of an object on 1 th the moon its weight on the 6 earth? 9.5 Thrust and Pressure Fig. 9.3: To fix a poster, drawing pins are pressed with the thumb perpendicular to the board. 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 Situation 2: You stand on loose sand. Your continuous chain? Why a truck or a motorbus feet go deep into the sand. Now, lie down on has much wider tyres? Why cutting tools have the sand. You will find that your body will not sharp edges? In order to address these go that deep in the sand. In both cases the questions and understand the phenomena force exerted on the sand is the weight of your involved, it helps to introduce the concepts body. 106 SCIENCE 2024-25  You have learnt that weight is the force by the wooden block on the table top if acting vertically downwards. Here the force is it is made to lie on the table top with its acting perpendicular to the surface of the sand. sides of dimensions (a) 20 cm × 10 cm The force acting on an object perpendicular to and (b) 40 cm × 20 cm. the surface is called thrust. When you stand on loose sand, the force, Solution: that is, the weight of your body is acting on The mass of the wooden block = 5 kg an area equal to area of your feet. When you The dimensions lie down, the same force acts on an area equal = 40 cm × 20 cm × 10 cm to the contact area of your whole body, which Here, the weight of the wooden block is larger than the area of your feet. Thus, the effects of forces of the same magnitude on applies a thrust on the table top. different areas are different. In the above That is, cases, thrust is the same. But effects are Thrust = F = m × g different. Therefore the effect of thrust = 5 kg × 9.8 m s–2 depends on the area on which it acts. = 49 N The effect of thrust on sand is larger while Area of a side = length × breadth standing than while lying. The thrust on unit = 20 cm × 10 cm = 200 cm2 = 0.02 m2 area is called pressure. Thus, From Eq. (9.20), thrust Pressure = (9.20) 49 N area Pressure = 0.02 m 2 Substituting the SI unit of thrust and area in Eq. (9.20), we get the SI unit of pressure as N/ = 2450 N m-2. m2 or N m–2. When the block lies on its side of In honour of scientist Blaise Pascal, the dimensions 40 cm × 20 cm, it exerts SI unit of pressure is called pascal, denoted the same thrust. as Pa. Area= length × breadth Let us consider a numerical example to = 40 cm × 20 cm understand the effects of thrust acting on = 800 cm2 = 0.08 m2 different areas. From Eq. (9.20), 49 N Example 9.6 A block of wood is kept on a Pressure = 0.08 m 2 tabletop. The mass of wooden block is 5 kg and its dimensions are 40 cm × 20 = 612.5 N m–2 cm × 10 cm. Find the pressure exerted The pressure exerted by the side 20 cm × 10 cm is 2450 N m–2 and by the side 40 cm × 20 cm is 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. 9.5.1 PRESSURE IN FLUIDS All liquids and gases are fluids. A solid exerts pressure on a surface due to its weight. Fig. 9.4 Similarly, fluids have weight, and they also GRAVITATION 107 2024-25 exert pressure on the base and walls of the water on the bottle is greater than its weight. container in which they are enclosed. Pressure Therefore it rises up when released. exerted in any confined mass of fluid is To keep the bottle completely immersed, transmitted undiminished in all directions. the upward force on the bottle due to water must be balanced. This can be achieved by 9.5.2 BUOYANCY an externally applied force acting downwards. This force must at least be equal to the Have you ever had a swim in a pool and felt difference between the upward force and the lighter? Have you ever drawn water from a weight of the bottle. well and felt that the bucket of water is heavier The upward force exerted by the water on when it is out of the water? Have you ever the bottle is known as upthrust or buoyant wondered why a ship made of iron and steel force. In fact, all objects experience a force of does not sink in sea water, but while the same buoyancy when they are immersed in a fluid. amount of iron and steel in the form of a sheet The magnitude of this buoyant force depends would sink? These questions can be answered on the density of the fluid. by taking buoyancy in consideration. Let us understand the meaning of buoyancy by 9.5.3 W HY OBJECTS FLOAT OR SINK doing an activity. WHEN PLACED ON THE SURFACE OF Activity ______________ 9.4 WATER? Take an empty plastic bottle. Close Let us do the following activities to arrive at the mouth of the bottle with an an answer for the above question. airtight stopper. Put it in a bucket filled with water. You see that the Activity ______________ 9.5 bottle floats. Push the bottle into the water. You feel Take a beaker filled with water. an upward push. Try to push it further Take an iron nail and place it on the down. You will find it difficult to push surface of the water. deeper and deeper. This indicates that Observe what happens. water exerts a force on the bottle in the upward direction. The upward force The nail sinks. The force due to the exerted by the water goes on increasing gravitational attraction of the earth on the as the bottle is pushed deeper till it is iron nail pulls it downwards. There is an completely immersed. upthrust of water on the nail, which pushes Now, release the bottle. It bounces it upwards. But the downward force acting back to the surface. on the nail is greater than the upthrust of Does the force due to the gravitational water on the nail. So it sinks (Fig. 9.5). 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 Fig. 9.5: An iron nail sinks and a cork floats when immersed, the upward force exerted by the placed on the surface of water. 108 SCIENCE 2024-25  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 (a) 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 (Fig. 9.5). (b) The density of an iron nail is more than the density of water. This means that the Fig. 9.6: (a) Observe the elongation of the rubber string due to the weight of a piece of stone upthrust of water on the iron nail is less than suspended from it in air. (b) The elongation the weight of the nail. So it sinks. decreases as the stone is immersed Therefore objects of density less than that in water. of a liquid float on the liquid. The objects of density greater than that of a liquid sink in Observe what happens to elongation the liquid. of the string or the reading on the balance. Q uestions You will find that the elongation of the string 1. Why is it difficult to hold a school or the reading of the balance decreases as the bag having a strap made of a thin stone is gradually lowered in the water. However, and strong string? no further change is observed once the stone 2. What do you mean by buoyancy? gets fully immersed in the water. What do you 3. Why does an object float or sink infer from the decrease in the extension of the when placed on the surface of string or the reading of the spring balance? water? We know that the elongation produced in the string or the spring balance is due to the 9.6 Archimedes’ Principle 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 Activity ______________ 9.7 upward direction. As a result, the net force on Take a piece of stone and tie it to one the string decreases and hence the elongation end of a rubber string or a spring also decreases. As discussed earlier, this balance. upward force exerted by water is known as Suspend the stone by holding the the force of buoyancy. balance or the string as shown in What is the magnitude of the buoyant Fig. 9.6 (a). Note the elongation of the string or force experienced by a body? Is it the same the reading on the spring balance due in all fluids for a given body? Do all bodies to the weight of the stone. in a given fluid experience the same buoyant Now, slowly dip the stone in the water force? The answer to these questions is in a container as shown in contained in Archimedes’ principle, stated as Fig. 9.6 (b). follows: GRAVITATION 109 2024-25  When a body is immersed fully or partially Archimedes’ principle has many in a fluid, it experiences an upward force that applications. It is used in designing ships and is equal to the weight of the fluid displaced submarines. Lactometers, which are used to by it. determine the purity of a sample of milk and Now, can you explain why a further hydrometers used for determining density of decrease in the elongation of the string was liquids, are based on this principle. not observed in activity 9.7, as the stone was Q fully immersed in water? uestions Archimedes was a Greek scientist. He 1. You find your mass to be 42 kg discovered the principle, subsequently named on a weighing machine. Is your after him, after noticing that mass more or less than 42 kg? the water in a bathtub 2. You have a bag of cotton and an overflowed when he stepped iron bar, each indicating a mass into it. He ran through the of 100 kg when measured on a streets shouting “Eureka!”, weighing machine. In reality, which means “I have got it”. one is heavier than other. Can This knowledge helped him to you say which one is heavier determine the purity of the and why? Archimedes 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. 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. 110 SCIENCE 2024-25  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. Exercises 1. How does the force of gravitation between two objects change when the distance between them is reduced to half ? 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? 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.) 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? 5. If the moon attracts the earth, why does the earth not move towards the moon? 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? 7. What is the importance of universal law of gravitation? 8. What is the acceleration of free fall? 9. What do we call the gravitational force between the earth and an object? 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.] 11. Why will a sheet of paper fall slower than one that is crumpled into a ball? 1 12. Gravitational force on the surface of the moon is only as 6 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? GRAVITATION 111 2024-25  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. 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. 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? 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. 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. 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. 19. In what direction does the buoyant force on an object immersed in a liquid act? 20. Why does a block of plastic released under water come up to the surface of water? 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? 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? 112 SCIENCE 2024-25