PREFACE Welcome to UDAAN- a program to give wings to the girl students! The UDAAN programme has been initiated with the primary objective of increasing the enrollment of girl students in leading engineering institutions. The program is designed to provide a platform to deserving girls who aspire to pursue higher education in engineering and assist them in preparing for the entrance examinations. The study material covers the syllabus of the Joint Entrance Examinations. It has been designed to ensure that it is easy to understand. The progression of the content moves from simpler to complex concepts as the Unit/Chapter proceeds. Each topic and subtopic is followed by some practice questions for you to attempt. Answers to these questions are given at the end. In case you have difficulty in solving any of them, please email us at [email protected] or contact at 011-23231070. Best of luck and happy learning! Chairperson, CBSE INDEX Unit No. Topic Page No. Unit–6 Gravitation 01 Unit–7 Properties of Solids and Liquids 145 Unit–8 Thermodynamics 333 Unit–9 Kinetic Theory of Gases 436 Unit–10 Oscillations and Waves 538 UNIT–6 : GRAVITATION Learning Objectives After going through this unit, you would be able to understand, appreciate and apply the following concepts:  Newton’s law or universal law of gravitation.  Gravitational field, field intensity and potential energy.  Acceleration due to gravity (g), its variation with height and depth, for earth.  Kepler’s laws of plantary motion.  Escape velocity.  Satellites; orbital velocity, height, period and energy of satellite in orbit.  Geostationary satellites. 1 2 GRAVITATION Introduction The force between material objects is called gravitational force. It is caused by the mass of the objects. Gravitational force (or gravity) is one of four fundamental forces is nature which makes the universe exist as we know now. Newton’s Law of Gravitation The law of gravitation was discovered by Sir Issac Newton, a british mathematician, in the year 1665. Newton’s law states that every body in this universe attracts every other body with a force which is directly proportional to the product of the masses of the two bodies and inversely proportional to the square of the distance between their centres. According to this law, if m and m are two point masses sepearted by a distance of r, then the gravitational 1 2 force between them will have magnitude ‘F’ given as: m m F 1 2 r2 Gm m or F 1 2 r2 Here, the proportionality constant ‘G’ is called the universal gravitational constant. The value of G is determined by the British scientist Henry Cavendish in 1798 and it is equal to 6.67×10–11 N–m2Kg–2. The dimensions of G are [M–1L3T–3]. The value of G is a universal constant which has same value every where irrespective of time, space and physical or chemical conditions. G is a scalar quantity. Because G is a universal constant, Newton’s law of gravitation is also called universal law of gravittation. Characterstics of Gravitational Force 1. It obeys Newton’s third law of motion. 2. Gravitational force is always attractive. 3. Gravitational force between two bodies does not depend on the presence of a third body. 4. It is a central force (This means that the line of action of the gravitational force pass through the centres of centre of gravity of the two bodies). 5. Being central force, gravitational force doesnot produce any torque. Hence the angular momentum (L) of the two attracting bodies will remain as a constant. Therefore gravitational forces do not cause any change in angular momentum (L). 3 6. It is conservative in nature. That is the work done by or gainst gravitational force does not depend on the path followed. 7. It is the weakest force in nature. Gravitational force is approximately 1038 times weaker than strong nuclear force and 1036 times weaker than electrostatic force. 8. Gravitational force does not depend on the intervening medium, where as electric and magnetic forces depend on the medium. 9. Gravitational force cannot be shielded. 10. It obeys inverse square law. 11. It obeys principles of superposition. 12. It is a long range force, which extend, theoretically, up to infinity. Newton’s Law of Gravitation in Vector Form F = Gravitational on man m due to m. Let r and r be position vector 21 2 1 1 2 of mass m and m respectively as shown in Fig. r is position vector of 1 2 21 mass m w.r.t. mass m. 2 2 Gm m  Then, F  1 2 r 21  r2  21 Similarly, F = Gravitational force on mass m due to mass m 12 1 2 Gm m   1 2 r  r2  12 Obvisouly r = –r ; therefore F = –F 12 21 12 21 Superposition Principle for Gravitational Attraction Since gravitational force is always attractive, it is comparitively easier to find the resultant force on one body due to several other bodies. Suppose there are several bodies of masses m, m, m etc. stituated at 1 2 3 specific locations defined by position vector r, r, r etc. 1 2 3 The resultant force on m due to all other masses can then be obtained as: 1 F = F + F + .......... 1 12 13 where F is resultant force on m 1 1 F is gravitational attraction on m by m 12 1 2 F is gravitationl attraction on m by m and so an. 13 1 3 4 Gm m Gm m Also, F  1 2 r  1 3 r ..... 1 r2 12 r2 13 12 13 N m   Gm  j  1  r2  j2 1j  Example–1: Two heavy bodies of 200 kg and 500 kg are seperated by a distance of 10m. What is gravitational force between then? Solution: Given: m = 200 Kg 1 m = 500 Kg 2 r = 10 m G = 6.67×10–11 N–m2 Kg–2 Gm m 6.671011200  F 1 2  6.67108N r2 102 (Please note that this force is negligible in magnitude, even though the masses are heavey. Our feeling that they are heavy is because of their weight which is none other than the gravitational attraction on then by earth). Example–2: Earth, considered spherical in shape, has a radius of 6400 Km. Its mass is 6.0×1024 Kg. A body of mass 50 Kg is situated on the surface of earth. What is the gravitational force on the body due to earth? Solution: The figure represents the situation given inthe question. The distance between the centres of earth and the body = R.  r R 6400103m m = m = 50 Kg 1 m = M = 6×1024 Kg 2 Gm m Now, F 1 2 r2 5 Gmm  R2 6.6710115061024  64001032 6.673   6464 = 488.5 N (Note that this attractive force on 50 Kg is practically large and it is directed towards the centre of earth. This is called the weight of the body. We all feel this gravitational attraction or ‘weight’ which enables us to walk and move on the surface of the earth). Example–3: Calculate the gravitational force between a proton and electron separated by a distance of 1 femtometer. Given mass of proton = 1.67×10–27 Kg Mass of electron = 9.1×10–31 Kg and 1 femtometer = 10–15 m Solution: Gm m F 1 2 r2 6.6710111.6710279.11031  10152 = 101.36399×10–39 N = 1.01×10–37 N (Note that this force is extremely negligible. In comparision, the electrostatic force between the same proton and electron is: 1 q q 1 F  . 1 2 9109 SI units [ e 4 r2 4   6 91.610191.61019  10152 q1 = + 1.6×10–19 Coulumb = 23.04×10 q = – 1.6×10–19 Coulumb] 2 = 230.4 N (Attractive) This is very large compared to the gravitational force between them). Example–4: Three identical particles, each of mass ‘m’, are placed at the vertices of an equilateral triangle of side ‘a’. Find the gravitational force exerted by this system on an identical particle, of mass m placed at (i) the mid- point of one side of the triangle and (ii) the centroid of the triangle. Solution: (i) Figure represents the position of the fourth particle at D. It is clear that two forces on m at D; due to mass at B and C, are equal and opposite. So they cancel out.  The third force, directed from D to A is the resultant force on m at D. Gmm Gm2 4 Gm2  F    D AD2 3  3 a2  a2 4  (ii) When the fourth particle is placed at the centroid O, the three forces on the mass m at O due to the other three have equal magnitude of F each. The angle between any two of these forces is 1200. Therefore the third force (i.e. along OA) will become the equilibrant of the other two. Hence the net force would be zero. Example–5: Two variable masses m and m are placed at a variable distance r from each other. Plot graphs showing 1 2 the variation of the gravitational force (F) between then with: (i) product of the masses (m m) 1 2 (ii) r Solution: Gm m We have, F 1 2 r2 7 or Fm m 1 2 1 and F r2 (i) (ii) Gravitational Field at a Point Gm m From Newton’s law of gravitation, F 1 2 , it follows that the magnitude of gravitational force becomes r2 zero, only when r . But, as the distance increases to large value, F becomes negligible. Hence, practically the force F is pexperiened up to a limited distance. The region around a body in which the gravitational attractive force of it can be experienced by other bodies is called its gravitational field (or field region). It is a three dimensional space around a body (or source mass). At different points of this field region the strength of the force experienced by another body / mass will be different. In order to predict how much force would a body or mass experience when placed at a particular point in the field region of a source mass, it would be convenient if we instroduce a physcial concept called ‘gravitational field intensity’ at a point. 8