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- 1. C hapter 10 GRAVITATIONIn Chapters 8 and 9, we have learnt about Let us try to understand the motion ofthe motion of objects and force as the cause the moon by recalling activity 8.11.of motion. We have learnt that a force isneeded to change the speed or the direction Activity _____________ 10.1of 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 thethe earth. We know that all the planets go other end of the thread and whirl itaround the Sun. The moon goes around the round, as shown in Fig. 10.1.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 ofthe same force is responsible for all these. the stone.This force is called the gravitational force. In this chapter we shall learn aboutgravitation and the universal law ofgravitation. We shall discuss the motion ofobjects under the influence of gravitationalforce on the earth. We shall study how theweight of a body varies from place to place.We shall also discuss the conditions forobjects to float in liquids.10.1 GravitationWe know that the moon goes around theearth. An object when thrown upwards,reaches a certain height and then fallsdownwards. It is said that when Newton wassitting under a tree, an apple fell on him. The Fig. 10.1: A stone describing a circular path with a velocity of constant magnitude.fall of the apple made Newton start thinking.He thought that: if the earth can attract anapple, can it not attract the moon? Is the force Before the thread is released, the stonethe same in both cases? He conjectured that moves in a circular path with a certain speedthe same type of force is responsible in both and changes direction at every point. Thethe cases. He argued that at each point of its change in direction involves change in velocityorbit, the moon falls towards the earth, or acceleration. The force that causes thisinstead of going off in a straight line. So, it acceleration and keeps the body moving alongmust 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 (meaningearth. ‘centre-seeking’) force. In the absence of this
- 2. force, the stone flies off along a straight line. 10.1.1 UNIVERSAL LAW OF GRAVITATIONThis straight line will be a tangent to thecircular 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. 10.2: 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 centripetalforce is provided by the force of attraction of Let two objects A and B of masses M andthe earth. If there were no such force, the m lie at a distance d from each other as shownmoon would pursue a uniform straight line in Fig. 10.2. Let the force of attraction betweenmotion. two objects be F. According to the universal It is seen that a falling apple is attracted law of gravitation, the force between twotowards the earth. Does the apple attract the objects is directly proportional to the productearth? If so, we do not see the earth moving 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 inverselyapple does attract the earth. But according proportional to the square of the distanceto the second law of motion, for a given force, between them, that is,acceleration is inversely proportional to themass of an object [Eq. (9.4)]. The mass of an 1apple is negligibly small compared to that of F∝ (10.2) d2the earth. So, we do not see the earth movingtowards the apple. Extend the same argument Combining Eqs. (10.1) and (10.2), we getfor why the earth does not move towards themoon. M ×m In our solar system, all the planets go F ∝ d2 (10.3)around the Sun. By arguing the same way,we can say that there exists a force between M×mthe Sun and the planets. From the above facts or, F = G 2 (10.4) dNewton concluded that not only does theearth attract an apple and the moon, but all where G is the constant of proportionality andobjects 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) givesthe gravitational force. F×d2=GM×m 132 SCIENCE
- 3. 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: 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 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 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. 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, to realise it. r3/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 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 Newton was also a great mathematician. He planetary motion is the gravitational 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 D C outside a sphere of uniform density, the sphere of Kepler to 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 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 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 ∝ v2/r. that time, there was hardly any doubt about If T denotes the period, then v = 2πr/T, its correctness. This is because Newton based so that v2 ∝ r2/T2. his theory on sound scientific reasoning and We can rewrite this as v2 ∝ (1/r) × backed it with mathematics. This made the ( r /T2). Since r3/T2 is constant by Kepler’s 3 theory simple and elegant. These qualities are third law, we have v2 ∝ 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
- 4. 2 From Eq. (10.4), the force exerted by Fdor G= (10.5) the earth on the moon is M×m M ×m The SI unit of G can be obtained by F =G d2substituting the units of force, distance andmass in Eq. (10.5) 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)2Henry Cavendish (1731 – 1810) by using a = 2.01 × 1020 N.sensitive balance. The accepted value of G is6.673 × 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 2.01 × 1020 N. Qattraction between any two objects. Computethe value of this force between you and yourfriend sitting closeby. Conclude how you do uestionsnot 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 10.1.2 I MPORTANCE 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 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 the earth and the moon is 3.84×105 km, calculate the force exerted by the earth Let us try to understand the meaning of free 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. m = 7.4 × 1022 kg • It reaches a certain height and then it 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
- 5. any change in the velocity of falling objects? calculations, we can take g to be more or lessWhile falling, there is no change in the constant on or near the earth. But for objectsdirection of motion of the objects. But due to far from the earth, the acceleration due tothe earth’s attraction, there will be a change gravitational force of earth is given byin the magnitude of the velocity. Any change Eq. (10.7).in velocity involves acceleration. Whenever anobject falls towards the earth, an acceleration 10.2.1 TO CALCULATE THE VALUE OF gis involved. This acceleration is due to theearth’s gravitational force. Therefore, this To calculate the value of g, we should putacceleration 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,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,acceleration, that is, m s–2. R = 6.4 × 106 m. We know from the second law of motion Mthat force is the product of mass and g=G 2acceleration. Let the mass of the stone in Ractivity 10.2 be m. We already know that there 6.7 × 10-11 N m 2 kg -2 × 6 × 1024 kgis acceleration involved in falling objects due =to the gravitational force and is denoted by g. (6.4 × 106 m)2Therefore the magnitude of the gravitational = 9.8 m s–2.force F will be equal to the product of massand acceleration due to the gravitational Thus, the value of acceleration due to gravityforce, that is, of the earth, g = 9.8 m s–2. F=mg (10.6)From Eqs. (10.4) and (10.6) we have 10.2.2 MOTION OF OBJECTS UNDER THE INFLUENCE OF GRAVITATIONAL M ×m mg =G 2 FORCE OF THE EARTH d Let us do an activity to understand whether M all objects hollow or solid, big or small, will or g= G 2 (10.7) d fall from a height at the same rate.where M is the mass of the earth, and d isthe distance between the object and the earth. Activity _____________ 10.3 Let an object be on or near the surface of • Take a sheet of paper and a stone. Dropthe earth. The distance d in Eq. (10.7) will be them simultaneously from the first floorequal to R, the radius of the earth. Thus, for of a building. Observe whether both ofobjects 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 2 (10.9) R 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 fromradius of the earth increases from the poles which air has been sucked out, theto the equator, the value of g becomes greater paper and the stone would fall at theat the poles than at the equator. For most same rate.GRAVITATION 135
- 6. We know that an object experiences u +vacceleration during free fall. From Eq. (10.9), (ii) average speed = 2this acceleration experienced by an object is = (0 m s + 5 m s–1)/2 –1independent of its mass. This means that all = 2.5 m s–1objects hollow or solid, big or small, should (iii) distance travelled, s = ½ a t2fall at the same rate. According to a story, = ½ × 10 m s–2 × (0.5 s)2Galileo dropped different objects from the top = ½ × 10 m s–2 × 0.25 s2of the Leaning Tower of Pisa in Italy to prove = 1.25 mthe same. Thus, As g is constant near the earth, all the (i) its speed on striking the groundequations for the uniformly accelerated = 5 m s–1motion of objects become valid with (ii) its average speed during the 0.5 sacceleration a replaced by g (see section 8.5). = 2.5 m s–1The equations are: (iii) height of the ledge from the ground v = u + at (10.10) = 1.25 m. 1 s = ut + at2 (10.11) 2 v2 = u2 + 2as (10.12) Example 10.3 An object is thrown vertically upwards and rises to a heightwhere u and v are the initial and final of 10 m. Calculate (i) the velocity withvelocities and s is the distance covered in which the object was thrown upwardstime, 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 thedirection of the velocity, that is, in the Solution:direction of motion. The acceleration, a willbe taken as negative when it opposes the Distance travelled, s = 10 mmotion. Final velocity, v = 0 m s–1 Acceleration due to gravity, g = 9.8 m s–2 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 0 = u 2 + 2 × (–9.8 m s–2) × 10 m calculations). –u 2 = –2 × 9.8 × 10 m2 s–2 (i) What is its speed on striking the u = 196 m s-1 ground? (ii) What is its average speed during the u = 14 m s-1 0.5 s? (ii) v=u+at (iii) How high is the ledge from the 0 = 14 m s–1 – 9.8 m s–2 × t ground? t = 1.43 s. 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
- 7. 10.3 Mass attracts the object. In the same way, the weight of an object on the moon is the forceWe have learnt in the previous chapter that with which the moon attracts that object. Thethe mass of an object is the measure of its mass of the moon is less than that of theinertia (section 9.3). We have also learnt that earth. Due to this the moon exerts lesser forcegreater 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 itsthe earth, the moon or even in outer space. weight on the moon be Wm. Let the mass ofThus, 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 moon will be10.4 Weight Mm × m Wm = G (10.16)We know that the earth attracts every object 2 Rmwith a certain force and this force depends Let the weight of the same object on theon the mass (m) of the object and the earth be We. The mass of the earth is M andacceleration due to the gravity (g). The weight its radius is R.of an object is the force with which it isattracted towards the earth.We know that Table 10.1 F = m × a, (10.13) Celestial Mass (kg) Radius (m)that is, body F = m × g. (10.14) The force of attraction of the earth on an Earth 5.98 × 1024 6.37 × 106object is known as the weight of the object. Itis denoted by W. Substituting the same in Moon 7.36 × 1022 1.74 × 106Eq. (10.14), we have From Eqs. (10.9) and (10.15) we have, W=m×g (10.15) As the weight of an object is the force with M ×m We = G (10.17)which it is attracted towards the earth, the R2SI unit of weight is the same as that of force, Substituting the values from Table 10.1 inthat is, newton (N). The weight is a force acting Eqs. (10.16) and (10.17), we getvertically downwards; it has both magnitudeand direction. 7.36 × 1022 kg × m Wm = G (1.74 × 10 m ) 2 We have learnt that the value of g is 6constant at a given place. Therefore at a givenplace, the weight of an object is directly Wm = 2.431 × 1010 G × m (10.18a)proportional to the mass, say m, of the object, and We = 1.474 × 10 G × m 11 (10.18b)that is, W ∝ m. It is due to this reason that Dividing Eq. (10.18a) by Eq. (10.18b), we getat a given place, we can use the weight of anobject as a measure of its mass. The mass of Wm 2.431 × 1010 =an object remains the same everywhere, that We 1.474 × 1011is, on the earth and on any planet whereasits weight depends on its location. Wm 1 or W = 0.165 ≈ 6 (10.19) e10.4.1 WEIGHT OF AN OBJECT ON Weight of the object on the moon 1 = THE MOON Weight of the object on the earth 6We have learnt that the weight of an object Weight of the object on the moonon the earth is the force with which the earth = (1/6) × its weight on the earth.GRAVITATION 137
- 8. 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 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 What would be its weight when at the tip of 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, We 10 Wm = = 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?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 runin a desert easily? Why an army tank weighingmore than a thousand tonne rests upon a Situation 2: You stand on loose sand. Yourcontinuous chain? Why a truck or a motorbus feet go deep into the sand. Now, lie down onhas much wider tyres? Why cutting tools have the sand. You will find that your body willsharp edges? In order to address these not go that deep in the sand. In both casesquestions and understand the phenomena the force exerted on the sand is the weight ofinvolved, it helps to introduce the concepts your body. 138 SCIENCE
- 9. You have learnt that weight is the force by the wooden block on the table top ifacting vertically downwards. Here the force it is made to lie on the table top with itsis acting perpendicular to the surface of the sides of dimensions (a) 20 cm × 10 cmsand. 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 kgan area equal to area of your feet. When you The dimensionslie down, the same force acts on an area equal = 40 cm × 20 cm × 10 cmto the contact area of your whole body, which Here, the weight of the wooden blockis larger than the area of your feet. Thus, theeffects 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 × gdifferent. Therefore the effect of thrust = 5 kg × 9.8 m s–2depends on the area on which it acts. = 49 N The effect of thrust on sand is larger while Area of a side = length × breadthstanding than while lying. The thrust on unit = 20 cm × 10 cm = 200 cm2 = 0.02 m2area is called pressure. Thus, From Eq. (10.20), thrust Pressure = (10.20) 49 N area Pressure = 0.02 m 2Substituting the SI unit of thrust and area inEq. (10.20), we get the SI unit of pressure as = 2450 N m-2.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 exertsSI 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 cmunderstand the effects of thrust acting on = 800 cm2 = 0.08 m2different areas. From Eq. (10.20), 49 N 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 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. 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 alsoGRAVITATION 139
- 10. exert pressure on the base and walls of the by the water on the bottle is greater than itscontainer 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 by10.5.2 BUOYANCY an externally applied force acting downwards. This force must at least be equal to theHave you ever had a swim in a pool and felt difference between the upward force and thelighter? 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 onwhen it is out of the water? Have you ever the bottle is known as upthrust or buoyantwondered why a ship made of iron and steel force. In fact, all objects experience a force ofdoes not sink in sea water, but whether the buoyancy when they are immersed in a fluid.same amount of iron and steel in the form of The magnitude of this buoyant force dependsa sheet would sink? These questions can be on the density of the fluid.answered by taking buoyancy inconsideration. Let us understand the 10.5.3 WHY OBJECTS FLOAT OR SINK WHENmeaning of buoyancy by doing an activity. PLACED ON THE SURFACE OF WATER? Activity _____________ 10.4 Let us do the following activities to arrive at • Take an empty plastic bottle. Close the an answer for the above question. 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. • 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 • Observe what happens. deeper and deeper. This indicates that water exerts a force on the bottle in the The nail sinks. The force due to the 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 • 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). 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 gravitationalattraction of the earth acts on the bottle inthe downward direction. So the bottle ispulled downwards. But the water exerts anupward force on the bottle. Thus, the bottleis pushed upwards. We have learnt thatweight of an object is the force due togravitational attraction of the earth. When the Fig. 10.5: An iron nail sinks and a cork floats whenbottle is immersed, the upward force exerted placed on the surface of water. 140 SCIENCE
- 11. 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. Thishappens because of the difference in theirdensities. The density of a substance isdefined as the mass per unit volume. Thedensity of cork is less than the density of (a)water. This means that the upthrust of wateron the cork is greater than the weight of thecork. So it floats (Fig. 10.5). (b) The density of an iron nail is more thanthe density of water. This means that the Fig. 10.6: (a) Observe the elongation of the rubberupthrust of water on the iron nail is less than string due to the weight of a piece of stonethe 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 immersedof a liquid float on the liquid. The objects of in water.density greater than that of a liquid sink in • Observe what happens to elongationQthe liquid. of the string or the reading on the balance. uestions You will find that the elongation of the 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 as the stone is gradually lowered in the water. and strong string? 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 extension of the string or the reading of the water? spring balance? We know that the elongation produced in10.6 Archimedes’ Principle the string or the spring balance is due to the weight of the stone. Since the extension Activity _____________ 10.7 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 end of a rubber string or a spring upward direction. As a result, the net force 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 isGRAVITATION 141
- 12. contained in Archimedes’ principle, stated asfollows: 10.7 Relative Density When a body is immersed fully or partially As you know, the density of a substance isin a fluid, it experiences an upward force that defined as mass of a unit volume. The unit ofis 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 wasnot observed in activity 10.7, as the stone Therefore the density of a substance is onewas fully immersed in water? of its characteristic properties. It is different for different substances. For example, the 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 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 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 determine the purity of the gold in the crown Density of a substance Relative density = made for the king. Density of water His work in the field of Geometry and 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 with Roman army. Archimedes’ principle has many Example 10.7 Relative density of silver isapplications. 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?determine the purity of a sample of milk andhydrometers used for determining density of Solution:Qliquids, are based on this principle. Relative density of silver = 10.8 uestions Relative density 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? 2. You have a bag of cotton and an Density of silver iron bar, each indicating a mass = Relative density of silver 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
- 13. 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. • 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 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. 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?GRAVITATION 143
- 14. 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? 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. 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.144 SCIENCE
- 15. 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?GRAVITATION 145

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