Chapter 9 Gravitation PDF

## Chapter 9 Gravitation ### 9.1: Gravitation In Chapters 8 and 9, we have learnt about the motion of objects and force as the cause of motion. We have learnt that a force is needed to change the speed or the direction of motion of an object. We always observe that an object dropped from a heigh...

## Chapter 9 Gravitation ### 9.1: Gravitation In Chapters 8 and 9, we have learnt about the motion of objects and force as the cause of motion. We have learnt that a force is needed to change the speed or the direction of motion of an object. We always observe that an object dropped from a height falls towards the earth. We know that all the planets go around the Sun. The moon goes around the earth. In all these cases, there must be some force acting on the objects, the planets and on the moon. Isaac Newton could grasp that the same force is responsible for all these. This force is called the gravitational force. In this chapter we shall learn about gravitation and the universal law of gravitation. We shall discuss the motion of objects under the influence of gravitational force on the earth. We shall study how the weight of a body varies from place to place. We shall also discuss the conditions for objects to float in liquids. ### 9.1.1: Universal Law of Gravitation Every object in the universe attracts every other object with a force which is proportional to the product of their masses and inversely proportional to the square of the distance between them. The force is along the line joining the centres of two objects. - **More to know**: A straight line that meets the circle at one and only one point is called a tangent to the circle. Straight line ABC is a tangent to the circle at point B. The motion of the moon around the earth is due to the centripetal force. The centripetal force is provided by the force of attraction of the earth. If there were no such force, the moon would pursue a uniform straight line motion. It is seen that a falling apple is attracted towards the earth. Does the apple attract the earth? If so, we do not see the earth moving towards an apple. Why? According to the third law of motion, the apple does attract the earth. But according to the second law of motion, for a given force, acceleration is inversely proportional to the mass of an object (Eq. 9.4). The mass of an apple is negligibly small compared to that of the earth. So, we do not see the earth moving towards the apple. Extend the same argument for why the earth does not move towards the moon. In our solar system, all the planets go around the Sun. By arguing the same way, we can say that there exists a force between the Sun and the planets. From the above facts Newton concluded that not only does the earth attract an apple and the moon, but all objects in the universe attract each other. This force of attraction between objects is called the gravitational force. ### 9.1.2: Importance of The Universal Law of Gravitation The universal law of gravitation successfully explained several phenomena which were believed to be unconnected: - The force that binds us to the earth. - The motion of the moon around the earth. - The motion of planets around the sun. - The tides due to the moon and the sun. ### 9.2: Free Fall Let us try to understand the meaning of free fall by performing this activity. - **Activity**: Take a stone, throw it upwards, it reaches a certain height and then starts falling down. We have learnt that the earth attracts objects towards it. This is due to the gravitational force. Whenever objects move towards the earth under this force alone, we say that the objects are in free fall. Is there a change in the velocity of falling objects? While falling, there is change in the direction of motion of the objects. But due to the gravitational acceleration, a change in the direction of motion is involved. Any change in motion, there will be an acceleration. The gravitational acceleration is due to the earth's gravitational force. Therefore, the acceleration of the object towards the earth is called the acceleration due to gravity. It is denoted by g. The unit of g is the same as of that of acceleration, that is, m/s². We know from the second law of motion that force is the product of mass and acceleration. Let the mass of the stone in activity 9.2 be m. We already know that there is acceleration involved in falling objects due to the gravitational force and is denoted by g. Therefore the magnitude of the gravitational force F will be equal to the product of mass and acceleration due to the gravitational force, that is, F=mg From Eqs. (9.4) and (9.6) we have mg = G(Mxm)/d² or g=G(M)/d² Where M is the mass of the earth, and d is the distance between the object and the earth. Let an object be on or near the surface of the earth. The distance d in Eq. (9.7) will be equal to R, the radius of the earth. Thus, for objects on or near the surface of the earth, mg = G(Mxm)/R² or g=G(M)/R² The earth is not a perfect sphere. As the radius of the earth increases from the poles to the equator, the value of g becomes greater at the poles than at the equator. For most calculations, we can take g to be more or less constant on or near the earth. But for objects far from the earth, the acceleration due to gravity is calculated by the value of g given by Eq. (9.7). ### 9.2.1: To Calculate The Value of g To calculate the value of g, we should put the values of G, M and R in Eq. (9.9), namely, universal gravitational constant, G=6.7×10⁻¹¹ N m²/kg², mass of the earth, M=6×10²⁴ kg, and radius of the earth, R=6.4×10⁶ m. g=G(M)/R² = 6.7×10⁻¹¹ N m²/kg² × 6×10²⁴ kg/(6.4×10⁶ m)² = 9.8 m/s². Thus, the value of acceleration due to gravity of the earth, g=9.8 m/s². ### 9.2.2: Motion of Objects Under The Influence of Gravitational Force of The Earth Let us do an activity to understand whether all objects hollow or solid, big or small, will fall from a height at the same rate. - **Activity**: Take a sheet of paper and a stone. Drop them simultaneously from the first floor of a building. Observe whether both of them reach the ground simultaneously. We see that paper reaches the ground little later than the stone. This happens because of air resistance. The air offers resistance due to friction to the motion of the falling objects. The resistance offered by air to the paper is more than the resistance offered to the stone. If we do the experiment in a glass jar from which air has been sucked out, the paper and the stone would fall at the same rate. We know that an object experiences acceleration during free fall. From Eq. (9.9), this acceleration experienced by an object is independent of its mass. This means that all objects hollow or solid, big or small, should fall at the same rate. According to a story, Galileo dropped different objects from the top of the Leaning Tower of Pisa in Italy to prove the same. As g is constant near the earth, all the equations for the uniformly accelerated motion of objects become valid with acceleration a replaced by g……. The equations are: - v=u+at - s=ut+1/2at² - v²=u²+2as Where u and v are the initial and final velocities and s is the distance covered in time, t. In applying these equations, we will take acceleration, a to be positive when it is in the direction of the velocity, that is, in the direction of motion. The acceleration, a will be taken as negative when it opposes the motion. ### 9.3: Mass We have learnt in the previous chapter that the mass of an object is the measure of its inertia. We have also learnt that greater the mass, the greater is the Inertia. It remains the same whether the object is on the earth, the moon or even in outer space. Thus, the mass of an object is constant and does not change from place to place. ### 9.4: Weight We know that the earth attracts every object with a certain force and this force depends on the mass (m) of the object and the acceleration due to the gravity (g). The weight of an object is the force with which it is attracted towards the earth. We know that F=mxa, that is, F=mxg. The force of attraction of the earth on an object is known as the weight of the object. It is denoted by W. Substituting the same in Eq. (10.14), we have W=mxg As the weight of an object is the force with which it is attracted towards the earth, the SI unit of weight is the same as that of force, that is, newton (N). The weight is a force acting vertically downwards; it has both magnitude and direction.. We have learnt that the value of g is constant at a given place. Therefore at a given place, the weight of an object is directly proportional to the mass, say m, of the object, that is, Wo∝ m. It is due to this reason that at a given place, we can use the weight of an object as a measure of its mass. The mass of an object remains the same everywhere, that is, on the earth and on any planet whereas its weight depends on its location because g depends on location. From Eqs. (9.9) and (9.15) we have, W=G(Mxm)/R² Substituting the values from Table 9.1 in Eqs. (9.16) and (10.17), we get - W=G(m(7.36 × 1022 kg))/(1.74 × 10⁶ m)² - W=2.431×10¹⁰G × m and - W=G(m(5.98×10²⁴ kg))/(6.37 × 10⁶ m)² - W=1.474×10¹¹G × m Dividing Eq. (9.18a) by Eq. (9.18b), we get (W_m)/(W_e) = (2.431 × 10¹⁰)/(1.474 × 10¹¹) = 1/6 or (W_m)/(W_e) = 1/6 Weight of the object on the moon = (1/6) × weight of the object on the earth. Weight of the object on the moon = (1/6) × its weight on the earth. ### 9.4.1: Weight of An Object On The Moon We have learnt that the weight of an object on the earth is the force with which the earth attracts the object. In the same way, the weight of an object on the moon is the force with which the moon attracts that object. The mass of the moon is less than that of the earth. Due to this the moon exerts lesser force of attraction on objects.. Let the mass of an object be m. Let its weight on the moon be W. Let the mass of the moon be M and its radius be R. By applying the universal law of gravitation, the weight of the object on the moon will be W = G(Mxm)/R² Let the weight of the same object on the earth be W. The mass of the earth is M and its radius is R. | Celestial Body | Mass (Kg) | Radius (m) | | :--------------- | :---------- | :---------- | | Earth | 5.98 × 10²⁴ | 6.37 × 10⁶ | | Moon | 7.36 × 10²² | 1.74 × 10⁶ | From Eq. (9.9) and (9.15) we have, W = G(Mxm)/R² Substituting the values from Table 9.1 in Eqs. (9.16) and (10.17), we get - W = G(m(7.36 × 10²² kg))/(1.74 × 10⁶ m)² - W = 2.431×10¹⁰G × m and - W=G(m(5.98×10²⁴ kg))/(6.37 × 10⁶ m)² - W=1.474×10¹¹G × m Dividing Eq. (9.18a) by Eq. (9.18b), we get (W_m)/(W_e) = (2.431 × 10¹⁰)/(1.474 × 10¹¹) = 1/6 or (W_m)/(W_e) = 1/6 Weight of the object on the moon = (1/6) × weight of the object on the earth. Weight of the object on the moon = (1/6) × its weight on the earth. ### 9.5: Thrust and Pressure Have you ever wondered why a camel can run in a desert easily? Why an army tank weighing more than a thousand tonne rests upon a continuous chain? Why a truck or a motorbus has much wider tyres? Why cutting tools have sharp edges? In order to address these questions and understand the phenomena involved, it helps to introduce the concepts of the net force in a particular direction (thrust) and the force per unit area (pressure) acting on the object concerned. Let us try to understand the meaning of thrust and pressure by considering the following situations: - **Situation 1**: You wish to fix a poster or bulletin board, as shown in Fig 9.3. To do this task you will have to press drawing pins with your thumb. You apply a force on the surface area of the head of the pin. This force is directed perpendicular to the surface area of the board. This force acts on a smaller area at the tip of the pin. - **Situation 2**: You stand on loose sand. Your feet go deep into the sand. Now, lie down on the sand. You will find that your body will not go that deep in the sand. In both cases the force exerted on the sand is the weight of your body. You have learnt that weight is the force acting vertically downwards. Here the force acting perpendicular to the surface of the object is called thrust. When you stand on loose sand, the force acts on an area equal to area of your feet, When you lie down, the same force acts on an area equal to the contact area of your whole body, which is larger than the area of your feet. Thus, the effects of forces of the same magnitude on different areas are different. In the above cases, thrust is the same. But effects are different. Therefore the effect of thrust depends on the area on which it acts. The effect of thrust on sand is larger while standing than while lying. The thrust on unit area is called pressure. Thus, Pressure = Thrust/Area Substituting the SI unit of thrust and area in Eq. (9.20), we get the SI unit of pressure as N/m² or N m-2. In honour of scientist Blaise Pascal, the unit of pressure is called pascal, denoted as Pa. Let us consider a numerical example to understand the effects of thrust acting on different areas. ### 9.5.1: Pressure in Fluids All liquids and gases are fluids. A solid exerts pressure on a surface due to its weight. Similarly, fluids have weight, and they also exert pressure on the base and walls of the container in which they are enclosed. Pressure exerted in any confined mass of fluid is transmitted undiminished in all directions. ### 9.5.2: Buoyancy Have you ever had a swim in a pool and felt lighter? Have you ever drawn water from a well and felt that the bucket of water is heavier when it is out of the water? Have you ever wondered why a ship made of iron and steel does not sink in sea water, but while the same amount of iron and steel in the form of a sheet would sink? These questions can be answered by taking buoyancy in consideration. Let us understand the meaning of buoyancy by doing an activity. - **Activity**: Take an empty plastic bottle. Close the mouth of the bottle with an airtight stopper. Put it in a bucket filled with water. You see that the bottle floats. Push the bottle into the water. You feel an upward push. Try to push it further down. You will find it difficult to push deeper and deeper. This indicates that water exerts a force on the bottle in the upward direction. The upward force exerted by the water goes on increasing as the bottle is pushed deeper till it is completely immersed. - Now, release the bottle. It bounces back to the surface. Does the force due to the gravitational attraction of the earth act on this bottle? If so, why doesn't the bottle stay immersed in water after it is released? How can you immerse the bottle in water? The force due to the gravitational attraction of the earth acts on the bottle in the downward direction. So the bottle is pulled downwards. But the water exerts an upward force on the bottle. Thus, the bottle is pushed upwards. We have learnt that weight of an object is the force due to gravitational attraction of the earth. When the bottle is immersed, the upward force exerted by the water on the bottle is greater than its weight. Therefore it rises up when it is released. To keep the bottle completely immersed, an externally applied force acting downwards must be balanced. This can be achieved by an externally applied force acting downwards. This force must at least be equal to the difference between the upward force and weight of the bottle. The upward force exerted by the water on the bottle is known as upthrust or buoyancy force. In fact, all objects immerse in a fluid experience a buoyant force. The magnitude of this buoyant force depends on the density of the fluid. ### 9.5.3: Why Objects Float or Sink When Placed on The Surface of Water? Let us do the following activities to arrive an answer for the above question. - **Activity**: Take a beaker filled with water. Take an iron nail and place it on the surface of the water. Observe what happens. The nail sinks. The force due to the gravitational attraction of the earth on the iron nail pulls it downwards. There is an upthrust of water on the nail, which pushes it upwards. But the downward force acting on the nail is greater than the upthrust of water on the nail. So it sinks (Fig. 10.5). - **Activity**: Take a beaker filled with water. Take a piece of cork and an iron nail of equal mass. Place them on the surface of water. Observe what happens. The cork floats while the nail sinks. This happens because of the difference in their densities. The density of a substance is defined as the mass per unit volume. The density of cork is less than the density of water. This means that the upthrust of water on the cork is greater than the weight of the cork. So it floats (Fig. 9.5). The density of an iron nail is more than the density of water. This means that the upthrust of water on the iron nail is less than the weight of the nail. So it sinks. Therefore objects of density less than that of a liquid float on the liquid. The objects of density greater than that of a liquid sink in the liquid. ### 9.6: Archimedes' Principle - **Activity**: Take a piece of stone and tie it to one end of a rubber string or a spring balance. Suspend the stone by holding the balance or the string as shown in Fig. 9.6 (a). Note the elongation of the string or the reading on the spring balance due to the weight of the stone. Now, slowly dip the stone in the water in a container as shown in Fig. 9.6 (b). Observe what happens to elongation of the string or the reading on the balance. You will find that the elongation of the string or the reading of the balance decreases as the stone is gradually lowered in the water. However, no further change is observed once the stone gets fully immersed in the water. What do you infer from the decrease in the extension of the string or the reading of the spring balance? We know that the elongation produced in the string or the spring balance is due to the weight of the stone. Since the extension decreases once the stone is lowered in water, it means that some force acts on the stone in upward direction. As a result, the net force on the string decreases and hence the elongation also decreases. As discussed earlier, this upward force exerted by water is known as the force of buoyancy. What is the magnitude of the buoyant force experienced by a body? Is it the same in all fluids for a given body? Do all bodies in a given fluid experience the same buoyant force? The answer to these questions is contained in Archimedes' principle, stated as follows: