Chapter 10 Gravitation (ISC Class 11 PDF)

C hapter 10 GRAVITATION In Chapters 8 and 9, we have learnt about Let us try to understand the motion of d the motion of objects...

C hapter 10 GRAVITATION In Chapters 8 and 9, we have learnt about Let us try to understand the motion of d the motion of objects and force as the cause the moon by recalling activity 8.11. of motion. We have learnt that a force is Activity _____________ 10.1 he needed to change the speed or the direction of motion of an object. We always observe that Take a piece of thread. an object dropped from a height falls towards Tie a small stone at one end. Hold the the earth. We know that all the planets go other end of the thread and whirl it around the Sun. The moon goes around the round, as shown in Fig. 10.1. is earth. In all these cases, there must be some Note the motion of the stone. force acting on the objects, the planets and Release the thread. on the moon. Isaac Newton could grasp that Again, note the direction of motion of bl the same force is responsible for all these. the stone. This force is called the gravitational force. In this chapter we shall learn about pu gravitation and the universal law of gravitation. We shall discuss the motion of objects under the influence of gravitational be T force on the earth. We shall study how the re weight of a body varies from place to place. o R We shall also discuss the conditions for objects to float in liquids. tt E 10.1 Gravitation We know that the moon goes around the C earth. An object when thrown upwards, reaches a certain height and then falls downwards. It is said that when Newton was no N Fig. 10.1: A stone describing a circular path with a sitting under a tree, an apple fell on him. The velocity of constant magnitude. 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 Before the thread is released, the stone © the same in both cases? He conjectured that moves in a circular path with a certain speed the same type of force is responsible in both and changes direction at every point. The the cases. He argued that at each point of its change in direction involves change in velocity orbit, the moon falls towards the earth, or acceleration. The force that causes this instead of going off in a straight line. So, it acceleration and keeps the body moving along must be attracted by the earth. But we do the circular path is acting towards the centre. not really see the moon falling towards the This force is called the centripetal (meaning earth. ‘centre-seeking’) force. In the absence of this  force, the stone flies off along a straight line. 10.1.1 UNIVERSAL LAW OF GRAVITATION 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 d he A straight line that meets the circle at one and only one point is called a Mm F = G tangent to the circle. Straight line d 2 ABC is a tangent to the circle at is point B. Fig. 10.2: The gravitational force between two uniform objects is directed along the line bl 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 pu 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. 10.2. Let the force of attraction between motion. two objects be F. According to the universal be T 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 re earth? If so, we do not see the earth moving o R of their masses. That is, towards an apple. Why? F M×m (10.1) According to the third law of motion, the And the force between two objects is inversely apple does attract the earth. But according tt E 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 C mass of an object [Eq. (9.4)]. The mass of an 1 apple is negligibly small compared to that of F 2 (10.2) d the earth. So, we do not see the earth moving no N 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 (10.3) d2 © 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 (10.4) 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. (10.4) gives the gravitational force. F×d 2=G M × m 132 SCIENCE  Isaac Newton was born How did Newton guess the in Woolsthorpe near inverse-square rule? Grantham, England. He is generally There has always been a great interest regarded as the most in the motion of planets. By the 16th original and century, a lot of data on the motion of influential theorist in planets had been collected by many the history of science. astronomers. Based on these data He was born in a poor Johannes Kepler derived three laws, farming family. But he which govern the motion of planets. was not good at These are called Kepler’s laws. These are: d farming. He was sent 1. The orbit of a planet is an ellipse with to study at Cambridge the Sun at one of the foci, as shown in Isaac Newton he University in 1661. In the figure given below. In this figure O (1642 – 1727) 1665 a plague broke is the position of the Sun. out in Cambridge and so Newton took a year 2. The line joining the planet and the Sun off. It was during this year that the incident of sweep equal areas in equal intervals the apple falling on him is said to have of time. Thus, if the time of travel from is occurred. This incident prompted Newton to A to B is the same as that from C to D, explore the possibility of connecting gravity then the areas OAB and OCD are with the force that kept the moon in its orbit. equal. bl This led him to the universal law of 3. The cube of the mean distance of a gravitation. It is remarkable that many great planet from the Sun is proportional to scientists before him knew of gravity but failed the square of its orbital period T. Or, pu to realise it. r 3/T2 = constant. Newton formulated the well-known laws of It is important to note that Kepler motion. He worked on theories of light and could not give a theory to explain be T colour. He designed an astronomical telescope the motion of planets. It was Newton to carry out astronomical observations. who showed that the cause of the re Newton was also a great mathematician. He planetary motion is the gravitational o R invented a new branch of mathematics, called force that the Sun exerts on them. Newton calculus. He used it to prove that for objects used the third law outside a sphere of uniform density, the sphere D C of Kepler to tt E behaves as if the whole of its mass is calculate the concentrated at its centre. Newton gravitational force B transformed the structure of physical of attraction. The O C science with his three laws of motion and the gravitational force A universal law of gravitation. As the keystone of the earth is of the scientific revolution of the seventeenth weakened by distance. A simple argument no N century, Newton’s work combined the goes like this. We can assume that the contributions of Copernicus, Kepler, Galileo, planetary orbits are circular. Suppose the and others into a new powerful synthesis. orbital velocity is v and the radius of the It is remarkable that though the orbit is r. Then the force acting on an © gravitational theory could not be verified at orbiting planet is given by F v 2/r. that time, there was hardly any doubt about If T denotes the period, then v = 2 πr/T, its correctness. This is because Newton based so that v 2 r 2/T2. his theory on sound scientific reasoning and We can rewrite this as v 2 (1/r) × backed it with mathematics. This made the ( r /T2). Since r 3/T2 is constant by Kepler’s 3 theory simple and elegant. These qualities are third law, we have v 2 1/r. Combining now recognised as essential requirements of a this with F v2/ r, we get, F 1/ r2. good scientific theory. GRAVITATION 133  2 From Eq. (10.4), the force exerted by Fd or G (10.5) the 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. (10.5) as N m2 kg–2. 6.7 10 11 N m2 kg -2 6 1024 kg 7.4 1022 kg The value of G was found out by 8 (3.84 10 m ) 2 Henry Cavendish (1731 – 1810) by using a = 2.01 × 1020 N. sensitive balance. The accepted value of G is 6.673 × 10 –11 N m 2 kg–2. Thus, the force exerted by the earth on We know that there exists a force of the moon is 2.01 × 1020 N. d Q attraction between any two objects. Compute the value of this force between you and your uestions he friend sitting closeby. Conclude how you do 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 is 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. they are celestial or terrestrial. More to know bl Inverse-square 10.1.2 I MPORTANCE OF THE UNIVERSAL LAW OF GRAVITATION pu 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 be T bigger by a factor of 6, F becomes believed to be unconnected: 1 (i) the force that binds us to the earth; re times smaller. (ii) the motion of the moon around the o R 36 earth; (iii) the motion of planets around the Sun; and tt E Example 10.1 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 10.2 Free Fall C the earth and the moon is 3.84105 km, calculate the force exerted by the earth Let us try to understand the meaning of free no N on the moon. G = 6.7 10–11 N m2 kg-2. fall by performing this activity. Solution: Activity _____________ 10.2 The mass of the earth, M = 6 1024 kg Take a stone. © The mass of the moon, Throw it upwards. It reaches a certain height and then it m = 7.4 1022 kg starts falling down. The distance between the earth and the 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 134 SCIENCE  any change in the velocity of falling objects? calculations, we can take g to be more or less While falling, there is no change in the constant on or near the earth. But for objects direction of motion of the objects. But due to far from the earth, the acceleration due to the earth’s attraction, there will be a change gravitational force of earth is given by in the magnitude of the velocity. Any change Eq. (10.7). in velocity involves acceleration. Whenever an object falls towards the earth, an acceleration 10.2.1 T O 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 acceleration is called the acceleration due to the values of G, M and R in Eq. (10.9), the gravitational force of the earth (or namely, universal gravitational constant, d acceleration due to gravity). It is denoted by G = 6.7 × 10 –11 N m2 kg-2, mass of the earth, g. The unit of g is the same as that of M = 6 × 1024 kg, and radius of the earth, he acceleration, that is, m s–2. R = 6.4 × 106 m. We know from the second law of motion M that force is the product of mass and g = G 2 acceleration. Let the mass of the stone in R activity 10.2 be m. We already know that there is -11 2 -2 24 6.7 1 0 N m kg 6 10 kg is acceleration involved in falling objects due = (6.4 1 06 m)2 to the gravitational force and is denoted by g. bl Therefore the magnitude of the gravitational = 9.8 m s–2. force F will be equal to the product of mass and acceleration due to the gravitational Thus, the value of acceleration due to gravity of the earth, g = 9.8 m s–2. pu force, that is, F=mg (10.6) From Eqs. (10.4) and (10.6) we have 10.2.2 M OTION OF OBJECTS UNDER THE be T INFLUENCE O F GRAVITATIONAL M m re mg =G 2 FORCE OF THE EARTH o R d Let us do an activity to understand whether M all objects hollow or solid, big or small, will or g= G (10.7) tt E d 2 fall from a height at the same rate. where M is the mass of the earth, and d is the distance between the object and the earth. Activity _____________ 10.3 C Let an object be on or near the surface of Take a sheet of paper and a stone. Drop the earth. The distance d in Eq. (10.7) will be them simultaneously from the first floor equal to R, the radius of the earth. Thus, for no N of a building. Observe whether both of objects on or near the surface of the earth, them reach the ground simultaneously. We see that paper reaches the ground M × m mg = G (10.8) 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 (10.9) 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 from radius of the earth increases from the poles which air has been sucked out, the to the equator, the value of g becomes greater paper and the stone would fall at the at the poles than at the equator. For most same rate. GRAVITATION 135  We know that an object experiences u +v acceleration during free fall. From Eq. (10.9), (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 d motion of objects become valid with (ii) its average speed during the 0.5 s acceleration a replaced by g (see section 8.5). = 2.5 m s–1 he The equations are: (iii) height of the ledge from the ground v = u + at (10.10) = 1.25 m. 1 2 s = ut + at (10.11) 2 Example 10.3 An object is thrown is v2 = u2 + 2as (10.12) vertically upwards and rises to a height where u and v are the initial and final of 10 m. Calculate (i) the velocity with velocities and s is the distance covered in which the object was thrown upwards bl time, t. and (ii) the time taken by the object to In applying these equations, we will take reach the highest point. acceleration, a to be positive when it is in the pu direction of the velocity, that is, in the Solution: direction of motion. The acceleration, a will be taken as negative when it opposes the Distance travelled, s = 10 m be T motion. Final velocity, v = 0 m s–1 Acceleration due to gravity, g = 9.8 m s–2 re o R Acceleration of the object, a = –9.8 m s–2 Example 10.2 A car falls off a ledge and (upward motion) drops to the ground in 0.5 s. Let (i) v 2 = u2 + 2a s g = 10 m s –2 (for simplifying the tt E 0 = u 2 + 2 × (–9.8 m s–2) × 10 m calculations). –u 2 = –2 × 9.8 × 10 m 2 s–2 (i) What is its speed on striking the u = 19 6 m s-1 C ground? (ii) What is its average speed during the u = 14 m s-1 0.5 s? (ii) v=u +a t 0 = 14 m s–1 – 9.8 m s–2 × t no N (iii) How high is the ledge from the t = 1.43 s. ground? Thus, Solution: (i) Initial velocity, u = 14 m s –1, and (ii) Time taken, t = 1.43 s. © Time, t = ½ second Q Initial velocity, u = 0 m s–1 Acceleration due to gravity, g = 10 m s–2 uestions Acceleration of the car, a = + 10 m s–2 1. What do you mean by free fall? (downward) 2. What do you mean by acceleration (i) speed v = at due to gravity? v = 10 m s–2 × 0.5 s = 5 m s–1 136 SCIENCE  10.3 Mass attracts the object. In the same way, the weight of an object on the moon is the force We have learnt in the previous chapter that with which the moon attracts that object. The the mass of an object is the measure of its mass of the moon is less than that of the inertia (section 9.3). We have also learnt that earth. Due to this the moon exerts lesser force greater the mass, the greater is the inertia. It of attraction on objects. remains the same whether the object is on Let the mass of an object be m. Let its the earth, the moon or even in outer space. weight on the moon be W m. Let the mass of Thus, the mass of an object is constant and the moon be Mm and its radius be Rm. does not change from place to place. By applying the universal law of gravitation, the weight of the object on the d moon will be 10.4 Weight Mm m Wm G he We know that the earth attracts every object R 2 (10.16) m with a certain force and this force depends Let the weight of the same object on the on the mass (m) of the object and the earth be We. The mass of the earth is M and acceleration due to the gravity (g). The weight its radius is R. of an object is the force with which it is is attracted towards the earth. We know that Table 10.1 F = m × a, (10.13) Celestial Mass (kg) Radius (m) bl that is, body F = m × g. (10.14) The force of attraction of the earth on an Earth 5.98 1024 6.37 106 pu object is known as the weight of the object. It is denoted by W. Substituting the same in Moon 7.36 1022 1.74 106 Eq. (10.14), we have be T W=m× g (10.15) From Eqs. (10.9) and (10.15) we have, re As the weight of an object is the force with M m We G (10.17) o R which it is attracted towards the earth, the R2 SI unit of weight is the same as that of force, Substituting the values from Table 10.1 in that is, newton (N). The weight is a force acting Eqs. (10.16) and (10.17), we get tt E vertically downwards; it has both magnitude and direction. 7.36 1022 kg m Wm G 2 We have learnt that the value of g is 1.74 10 m 6 C constant at a given place. Therefore at a given place, the weight of an object is directly Wm 2.431 1010 G × m (10.18a) proportional to the mass, say m, of the object, no N and W e 1.474 1011G × m (10.18b) that is, W m. It is due to this reason that Dividing Eq. (10.18a) by Eq. (10.18b), we get at a given place, we can use the weight of an object as a measure of its mass. The mass of Wm 2.431 1010 an object remains the same everywhere, that We 1.474 1011 © is, on the earth and on any planet whereas its weight depends on its location. Wm 1 or W = 0.165 ≈ 6 (10.19) e 10.4.1 WEIGHT OF AN OBJECT ON Weight of theobject on the moon 1 = THE MOON Weight of theobject on the earth 6 We have learnt that the weight of an object Weight of the object on the moon on the earth is the force with which the earth = (1/6) × its weight on the earth. GRAVITATION 137  of the net force in a particular direction Example 10.4 Mass of an object is 10 kg. (thrust) and the force per unit area (pressure) What is its weight on the earth? acting 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 10.3. To do Thus, the weight of the object is 98 N. 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 d Example 10.5 An object weighs 10 N when is directed perpendicular to the surface area measured on the surface of the earth. of the board. This force acts on a smaller area he What would be its weight when at the tip of the pin. measured on the surface of the moon? Solution: is We know, Weight of object on the moon = (1/6) × its weight on the earth. bl That is, We 10 Wm = = N. pu 6 6 = 1.67 N. Thus, the weight of object on the be T surface of the moon would be 1.67 N. Q re o R uestions 1. What are the differences between tt E the mass of an object and its weight? 2. Why is the weight of an object on C 1 the moon th its weight on the 6 no N earth? 10.5 Thrust and Pressure Fig. 10.3: 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 sharp edges? In order to address these not go that deep in the sand. In both cases questions and understand the phenomena the force exerted on the sand is the weight of involved, it helps to introduce the concepts your body. 138 SCIENCE  You have learnt that weight is the force by the wooden block on the table top if acting vertically downwards. Here the force it is made to lie on the table top with its is acting perpendicular to the surface of the sides of dimensions (a) 20 cm × 10 cm sand. The force acting on an object and (b) 40 cm × 20 cm. perpendicular to 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 lie down, the same force acts on an area equal The dimensions = 40 cm × 20 cm × 10 cm to the contact area of your whole body, which is larger than the area of your feet. Thus, the Here, the weight of the wooden block d effects of forces of the same magnitude on applies a thrust on the table top. different areas are different. In the above That is, Thrust = F = m × g he cases, thrust is the same. But effects are = 5 kg × 9.8 m s–2 different. Therefore the effect of thrust = 49 N depends on the area on which it acts. The effect of thrust on sand is larger while Area of a side = length × breadth = 20 cm × 10 cm standing than while lying. The thrust on unit is = 200 cm 2 = 0.02 m2 area is called pressure. Thus, From Eq. (10.20), thrust Pressure = (10.20) 49N bl area Pressure = 0.02m 2 Substituting the SI unit of thrust and area in Eq. (10.20), we get the SI unit of pressure as = 2450 N m-2. pu N/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. be T as Pa. Area= length × breadth Let us consider a numerical example to = 40 cm × 20 cm re = 800 cm 2 = 0.08 m 2 o R understand the effects of thrust acting on different areas. From Eq. (10.20), 49 N tt E Example 10.6 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 C 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. no N 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. 10.5.1 PRESSURE IN FLUIDS All liquids and gases are fluids. A solid exerts pressure on a surface due to its weight. Fig. 10.4 Similarly, fluids have weight, and they also GRAVITATION 139  exert pressure on the base and walls of the by the water on the bottle is greater than its container in which they are enclosed. weight. Therefore it rises up when released. Pressure exerted in any confined mass of fluid To keep the bottle completely immersed, is transmitted undiminished in all directions. the upward force on the bottle due to water must be balanced. This can be achieved by 10.5.2 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 d 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. he 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 10.5.3 WHY OBJECTS FLOAT OR SINK WHEN doing an activity. PLACED ON THE SURFACE OF WATER? is Activity _____________ 10.4 Let us do the following activities to arrive at an answer for the above question. bl Take an empty plastic bottle. Close the mouth of the bottle with an airtight stopper. Put it in a bucket filled with Activity _____________ 10.5 water. You see that the bottle floats. pu Take a beaker filled with water. Push the bottle into the water. You feel Take an iron nail and place it on the an upward push. Try to push it further surface of the water. down. You will find it difficult to push be T Observe what happens. deeper and deeper. This indicates that water exerts a force on the bottle in the re The nail sinks. The force due to the o R upward direction. The upward force 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 tt E 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. 10.5). C attraction of the earth act on this bottle? If so, why doesn’t the bottle stay immersed in water after it is released? no N 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 Fig. 10.5: An iron nail sinks and a cork floats when bottle is immersed, the upward force exerted placed on the surface of water. 140 SCIENCE  Activity _____________ 10.6 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 d defined as the mass per unit volume. The density of cork is less than the density of (a) water. This means that the upthrust of water he on the cork is greater than the weight of the cork. So it floats (Fig. 10.5). (b) The density of an iron nail is more than the density of water. This means that the is Fig. 10.6: (a) Observe the elongation of the rubber upthrust of water on the iron nail is less than string due to the weight of a piece of stone the weight of the nail. So it sinks. suspended from it in air. (b) The elongation Therefore objects of density less than that decreases as the stone is immersed bl of a liquid float on the liquid. The objects of in water. density greater than that of a liquid sink in Observe what happens to elongation Q the liquid. pu of the string or the reading on the balance. uestions You will find that the elongation of the be T 1. Why is it difficult to hold a school string or the reading of the balance decreases bag having a strap made of a thin re as the stone is gradually lowered in the water. and strong string? o R However, no further change is observed once 2. What do you mean by buoyancy? the stone gets fully immersed in the water. 3. Why does an object float or sink What do you infer from the decrease in the when placed on the surface of tt E extension of the string or the reading of the water? spring balance? We know that the elongation produced in 10.6 Archimedes’ Principle C the string or the spring balance is due to the weight of the stone. Since the extension Activity _____________ 10.7 no N decreases once the stone is lowered in water, it means that some force acts on the stone in Take a piece of stone and tie it to one upward direction. As a result, the net force end of a rubber string or a spring balance. on the string decreases and hence the © Suspend the stone by holding the elongation also decreases. As discussed balance or the string as shown in earlier, this upward force exerted by water is Fig. 10.6 (a). known as the force of buoyancy. Note the elongation of the string or the What is the magnitude of the buoyant reading on the spring balance due to force experienced by a body? Is it the same the weight of the stone. Now, slowly dip the stone in the water in all fluids for a given body? Do all bodies in in a container as shown in a given fluid experience the same buoyant Fig. 10.6 (b). force? The answer to these questions is GRAVITATION 141  contained in Archimedes’ principle, stated as 10.7 Relative Density follows: When a body is immersed fully or partially As you know, the density of a substance is in a fluid, it experiences an upward force that defined as mass of a unit volume. The unit of is equal to the weight of the fluid displaced density is kilogram per metre cube (kg m–3). by it. The density of a given substance, under Now, can you explain why a further specified conditions, remains the same. decrease in the elongation of the string was not observed in activity 10.7, as the stone Therefore the density of a substance is one was fully immersed in water? of its characteristic properties. It is different for different substances. For example, the d Archimedes was a Greek density of gold is 19300 kg m-3 while that of scientist. He discovered the water is 1000 kg m-3. The density of a given he principle, subsequently sample of a substance can help us to named after him, after determine its purity. noticing that the water in a It is often convenient to express density bathtub overflowed when he of a substance in comparison with that of stepped into it. He ran is through the streets shouting water. The relative density of a substance is Archimedes the ratio of its density to that of water: “Eureka!”, which means “I have got it”. This knowledge helped him to bl determine the purity of the gold in the crown Density of a substance Relativedensity = made for the king. Density of water His work in the field of Geometry and pu Mechanics made him famous. His Since the relative density is a ratio of understanding of levers, pulleys, wheels- similar quantities, it has no unit. and-axle helped the Greek army in its war be T with Roman army. re Example 10.7 Relative density of silver is o R Archimedes’ principle has many applications. It is used in designing ships and 10.8. The density of water is 103 kg m–3. submarines. Lactometers, which are used to What is the density of silver in SI unit? tt E determine the purity of a sample of milk and hydrometers used for determining density of Solution: liquids, are based on this principle. Q Relative density of silver = 10.8 C uestions Relative density no N 1. You find your mass to be 42 kg Density of silver on a weighing machine. Is your = Density of water mass more or less than 42 kg? Density of silver 2. You have a bag of cotton and an = Relative density of silver © iron bar, each indicating a mass of 100 kg when measured on a × density of water weighing machine. In reality, one is heavier than other. Can you = 10.8 × 103 kg m –3. say which one is heavier and why? 142 SCIENCE  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. d Gravitation is a weak force unless large masses are involved. he Force of gravitation due to the earth is called gravity. 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 is attracts it. The weight is equal to the product of mass and acceleration due to gravity. bl The weight may vary from place to place but the mass stays constant. pu 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 be T 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 re o R which it is immersed then it sinks in the liquid. E xercises tt E C 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 no N 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? GRAVITATION 143  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 d 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 he 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? is 1 12. Gravitational force on the surface of the moon is only as 6 bl 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? 13. A ball is thrown vertically upwards with a velocity of 49 m/s. pu Calculate (i) the maximum height to which it rises, be T (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. re o R 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 tt E by the stone. What is the net displacement and the total distance covered by the stone? C 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 × 10 30 kg. The average distance between the two is no N 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. 144 SCIENCE  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? d he is bl pu be T re o R tt E C no N © GRAVITATION 145