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# Syjc chap2-gravitation

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gravitation, newtons law of gravitation satellite motion, critical velocity

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### Syjc chap2-gravitation

1. 1. Gravitation<br />Prof. Mukesh N. Tekwani<br />Department of Physics<br />Ismail Yusuf College,<br />Mumbai<br />mukeshtekwani@hotmail.com<br />1<br />
2. 2. Isaac Newton<br />Prof. Mukesh N Tekwani<br />2<br />
3. 3. Newton’s Law of Gravitation<br />Prof. Mukesh N Tekwani<br />3<br />Every particle of matter attracts every other particle with a force which is directlyproportional to the product of the masses and inversely proportional to the square of the distance between them.<br />
4. 4. Newton’s Law of Gravitation<br />Prof. Mukesh N Tekwani<br />4<br />G is called the Universal Gravitational Constant<br />G = 6.67 x 10-11 N m2/kg2<br />G is constant throughout the Universe and G does not depend on the medium between the masses <br />
5. 5. Difference between G and g<br />Prof. Mukesh N Tekwani<br />5<br />
6. 6. Relation between G and g<br />Let M = mass of the Earth<br />m = mass of an object on the surface of the Earth<br />g = acceleration due to gravity on the Earth’s surface<br />R = radius of the Earth<br />Prof. Mukesh N Tekwani<br />6<br />
7. 7. Relation between G and g<br />Prof. Mukesh N Tekwani<br />7<br />m<br />M<br />
8. 8. Relation between G and g<br />Prof. Mukesh N Tekwani<br />8<br />Weight of the object is the gravitational force acting on it.<br />m<br />M<br />………………………….(1)<br />
9. 9. Relation between G and g<br />Prof. Mukesh N Tekwani<br />9<br />At height h from the surface of the Earth’s surface, acceleration due to gravity is gh<br />At height h, <br />Weight of object = gravitational force<br />m<br />h<br />M<br />………………………….(2)<br />
10. 10. Relation between G and g<br />Prof. Mukesh N Tekwani<br />10<br />Dividing (2) by (1) we get, <br />m<br />h<br />Thus, g is independent of the mass of the object.<br />M<br />
11. 11. Projection of a Satellite<br />Prof. Mukesh N Tekwani<br />11<br />
12. 12. Projection of a Satellite<br />Prof. Mukesh N Tekwani<br />12<br />Why is it necessary to have at least a two stage rocket to launch a satellite?<br />A rocket with at least two stages is required to launch a satellite because<br />The first stage is used to carry the satellite up to the desired height.<br />In the second stage, rocket is turned horizontally (through 90 degrees) and the satellite is fired with the proper horizontal velocity to perform circular motion around the earth.<br />
13. 13. Critical Velocity of a Satellite<br />Prof. Mukesh N Tekwani<br />13<br />The horizontal velocity with which a satellite should be projected from a point above the earth's surface, so that it orbits in a circular path around the earth is called the orbital velocity or critical velocity (Vc) of the satellite.<br />
14. 14. Kepler’s Laws of Motion<br />Prof. Mukesh N Tekwani<br />14<br />Born: <br />December 27, 1571 <br />Died: <br />November 15, 1630 <br />German <br />Mathematician, <br />Astronomer<br />Astrologer.<br />
15. 15. Kepler’s First Law – Law of Orbit<br />Prof. Mukesh N Tekwani<br />15<br />Every planet revolves in an elliptical orbit around the Sun, with the Sun situated at one focus of the ellipse.<br />
16. 16. Kepler’s Second Lawor Law of Equal Areas<br />Prof. Mukesh N Tekwani<br />16<br />The radius vector drawn from the Sun to any planet sweeps out equal areas in equal intervals of time. This law is called the law of areas. <br />The areal velocity of the radius vector is constant.<br />
17. 17. Kepler’s Law of Equal Areas<br />Prof. Mukesh N Tekwani<br />17<br />
18. 18. Kepler’s Laws<br />Prof. Mukesh N Tekwani<br />18<br />
19. 19. Kepler’s Third Law - Laws of Period<br />Prof. Mukesh N Tekwani<br />19<br />The square of period of revolution of the planet around the Sun is directly proportional to the cube of the semi-major axis of the elliptical orbit.<br />T2 α r3<br />According to this law, when the planet is closest to the Sun, its speed is maximum and when it is farthest from the Sun, its speed is minimum.<br />
20. 20. Critical Velocity of a Satellite<br />Prof. Mukesh N Tekwani<br />20<br />Let<br />M = mass of the Earth<br />R = radius of Earth<br />m = mass of satellite<br />h = height of the satellite above Earth’s <br />r = R + h, where r is the distance of the satellite from the center of the Earth<br />Vc = critical velocity of the satellite<br />vc<br />h<br />R<br />M<br />r<br />
21. 21. Critical Velocity of a Satellite<br />Prof. Mukesh N Tekwani<br />21<br />The centripetal force necessary for the circular motion of the satellite around the Earth is provided by the gravitational force of attraction between the Earth and the satellite.<br />Centripetal force = gravitational force<br />
22. 22. Critical Velocity of a Satellite<br />Prof. Mukesh N Tekwani<br />22<br />Factors on which Critical Velocity of a satellite depends:<br />Mass of the planet<br />Radius of the planet<br />Height of the satellite<br />Critical velocity is not dependent on the mass of the satellite as m does not appear in the above equation<br />……………………………………………….(1)<br />
23. 23. Critical Velocity of a Satellite<br />Prof. Mukesh N Tekwani<br />23<br />But we know that<br />Substituting this value in eqn (1), we get,<br />…………………………………..(2)<br />
24. 24. Critical Velocity of a Satellite<br />Prof. Mukesh N Tekwani<br />24<br />Assignment 1: <br />Modify eqn (2) to find the critical velocity of a satellite orbiting very close to the surface of the Earth (h << R)<br />Assignment 2:<br />How does the critical velocity (or orbital velocity) of a satellite vary with an increase in the height of the satellite above the Earth’s surface?<br />
25. 25. Time Period of a Satellite<br />The time taken by a satellite to complete one revolution around the earth is called its periodic time or time period.<br />Prof. Mukesh N Tekwani<br />25<br />
26. 26. Time Period of a Satellite<br />Prof. Mukesh N Tekwani<br />26<br />Let<br />M = mass of the Earth<br />R = radius of Earth<br />m = mass of satellite<br />h = height of the satellite above Earth’s <br />r = R + h, where r is the distance of the satellite from the center of the Earth<br />Vc = critical velocity of the satellite<br />vc<br />h<br />R<br />M<br />r<br />
27. 27. Time Period of a Satellite<br />Prof. Mukesh N Tekwani<br />27<br />Distance covered by the satellite in 1 revolution <br /> = Circumference of the circle<br />Time taken to cover this distance is the time period<br />Critical speed Vc = <br />Vc = <br />But <br />
28. 28. Prof. Mukesh N Tekwani<br />28<br />Time Period of a Satellite<br />As is a constant, <br />so we get,<br />T2α r3<br />Thus, the square of the period of revolution is directly proportional to the cube of the radius of its orbit.<br />Squaring both sides, we get<br />
29. 29. Prof. Mukesh N Tekwani<br />29<br />Time Period of a Satellite<br />Factors on which Time Period of a satellite depends:<br />Mass of the planet<br />Radius of the planet, and <br />Height of the satellite from the planet’s surface<br />Period of the satellite does not depend on the mass of the satellite.<br />Assignment:<br />Obtain an expression for the time period of a satellite in terms of gh.<br />For a satellite close to the earth, calculate the period of revolution in minutes.<br />
30. 30. Binding Energy of a Satellite<br />Definition: <br />The minimum amount of energy required to remove a satellite from the earth’s gravitational influence is called as bindingenergy of the satellite.<br />A satellite revolving around the Earth has <br /> Kinetic energy, and<br />Potential energy<br />Prof. Mukesh N Tekwani<br />30<br />
31. 31. What is kinetic energy?<br />The energy possessed by a body due to its motion is called its kinetic energy.<br />If m = mass of an object, and<br /> v = its velocity<br />K.E = (1/2) mv2<br />Prof. Mukesh N Tekwani<br />31<br />
32. 32. What is potential energy?<br />The energy possessed by a body due to its position is called its potential energy.<br />If m = mass of an object, and<br /> h = its height above the surface<br />P.E. = mgh<br />Quiz: Does a body in motion have potential energy?<br />Prof. Mukesh N Tekwani<br />32<br />
33. 33. Binding Energy of a Satellite<br />Prof. Mukesh N Tekwani<br />33<br />Let<br />M = mass of the Earth<br />R = radius of Earth<br />m = mass of satellite<br />h = height of the satellite above Earth’s <br />r = R + h, where r is the distance of the satellite from the center of the Earth<br />Vc = critical velocity of the satellite<br />vc<br />h<br />R<br />M<br />r<br />
34. 34. Prof. Mukesh N Tekwani<br />34<br />Binding Energy of a Satellite<br />The critical velocity is given by<br />Kinetic energy of motion KE = <br />Substituting (1) in (2), we get,<br />KE = <br />……………………………………………………...(1)<br />…..……...(2)<br />..………………………………………………..(3)<br />
35. 35. Prof. Mukesh N Tekwani<br />35<br />Binding Energy of a Satellite<br />The gravitational potential at a distance r from the centre of the Earth is given by:<br />GP = <br />Potential energy = gravitational potential x mass of the satellite<br />Therefore, PE = <br />..…….………………………..(4)<br />
36. 36. Prof. Mukesh N Tekwani<br />36<br />Binding Energy of a Satellite<br />The total energy of the satellite is given by<br />TE = KE + PE<br />TE = + <br />TE = <br />The negative sign indicates that the satellite is bound to the Earth due to the gravitational force of the Earth.<br />..…….…………………………………………..(5)<br />
37. 37. Prof. Mukesh N Tekwani<br />37<br />Binding Energy of a Satellite<br />To free the satellite from the Earth’s gravitational influence, an amount of energy equal to its total energy must be supplied. This is called the binding energy of the satellite.<br />Therefore, BE = <br />Where r = R + h<br />Assignment: Calculate the BE of a satellite at rest on the surface of the Earth.<br />
38. 38. Prof. Mukesh N Tekwani<br />38<br />Weightlessness in a Satellite<br /> The weight of a body is the gravitational force exerted on it by the Earth.<br />When a person stands on a floor, he exerts a force on the floor. The floor in turn exerts a force (normal reaction) on the person. <br />This normal reaction is equal to the weight of the person. <br />The person has a feeling of weight due to this normal reaction.<br />
39. 39. Prof. Mukesh N Tekwani<br />39<br />Weightlessness in a Satellite<br />Consider an astronaut of mass m, in a satellite that is moving around the Earth in a circular orbit.<br />There is a centripetal force on the satellite and the astronaut. Thus, both are attracted towards the Earth with the same acceleration, due to the Earth’s gravitational force.<br />So the astronaut is not able to exert a force on the floor of the satellite & the satellite in turn cannot exert normal reaction on the astronaut. This causes the “feeling” of weightlessness.<br />
40. 40. Prof. Mukesh N Tekwani<br />40<br />Weightlessness in a Satellite<br />We must remember that the mass of the astronaut DOES NOT become zero. <br />This condition of weightlessness is also known (incorrectly) as zero gravity condition.<br />But, weightlessness does not mean the absence of gravity. <br />
41. 41. Escape Velocity of a Satellite<br />Prof. Mukesh N Tekwani<br />41<br />The minimum velocity with which a body should be projected from the surface of the earth so that it escapes the gravitational field of the earth is called the escape velocity of the body.<br />
42. 42. Escape Velocity of a Satellite<br />Prof. Mukesh N Tekwani<br />42<br />Consider a satellite of mass m, stationary on the surface of the Earth.<br />The binding energy of the satellite, on the surface of the Earth, is given by<br /> BE = <br />To escape from the Earth’s influence, energy must be provided to the satellite in the form of kinetic energy.<br />R<br />
43. 43. Escape Velocity of a Satellite<br />Prof. Mukesh N Tekwani<br />43<br />Therefore, KE of satellite = BE<br />KE = <br />Therefore, = <br />R<br />R<br />
44. 44. Numerical Problems<br />Obtain an equation for the escape velocity of a body from the surface of a planet of radius R and mean density ρ.<br />What would be the duration of the year if the distance between the Earth and Sun gets doubled?<br />Prof. Mukesh N Tekwani<br />44<br />
45. 45. Numerical Problems<br />Calculate the height of a communications satellite from the surface of the Earth. Values of G, M and R are as given in the text book. (These values will also be provided in the question paper)<br />A body weighs 4.5 kgwt on the surface of Earth. How much will it weigh on the surface of a planet whose mass is (1/9)th that of the Earth’s mass and radius is half that of the Earth?<br />Prof. Mukesh N Tekwani<br />45<br />
46. 46. Variation of g with Altitude<br />Let M = mass of the Earth<br />m = mass of an object on the surface of the Earth<br />g = acceleration due to gravity on the Earth’s surface<br />R = radius of the Earth<br />Prof. Mukesh N Tekwani<br />46<br />
47. 47. Prof. Mukesh N Tekwani<br />47<br />Variation of g with Altitude<br />Weight of the object is the gravitational force acting on it.<br />m<br />M<br />………………………….(1)<br />
48. 48. Prof. Mukesh N Tekwani<br />48<br />Variation of g with Altitude<br />At height h from the surface of the Earth’s surface, acceleration due to gravity is gh<br />At height h, <br />Weight of object = gravitational force<br />m<br />h<br />M<br />………………………….(2)<br />
49. 49. Prof. Mukesh N Tekwani<br />49<br />Variation of g with Altitude<br />Dividing (2) by (1) we get, <br />m<br />h<br />Thus, g is independent of the mass of the object.<br />From this eqn. it is clear that the acceleration due to gravity decreases as altitude of the body from the earth’s surface increases.<br />M<br />
50. 50. Prof. Mukesh N Tekwani<br />50<br />Variation of g with Altitude<br />By Binomial expansion:<br />Since the higher powers of h/R are neglected.<br />
51. 51. Variation of g due to depth<br />Prof. Mukesh N Tekwani<br />51<br />Let<br />M = mass of the Earth<br />R = radius of Earth<br />g = acceleration due to <br /> gravity on the surface of <br /> earth<br />ρ = density of earth<br />P<br />R-d<br />
52. 52. Variation of g due to depth<br />The acceleration due to gravity on the surface of the earth is given by:<br />Consider the earth as a sphere of density ρ<br /> Mass = Volume x Density<br />Prof. Mukesh N Tekwani<br />52<br />
53. 53. Variation of g due to depth<br />Prof. Mukesh N Tekwani<br />53<br />Mass = Volume x Density<br />Mass = <br />…………………………………………. (1)<br />
54. 54. Variation of g due to depth<br />Consider a point P at the depth d below the surface of the earth. At this point, let the acceleration due to gravity be gd.<br />Distance of point P from centre of Earth is (R – d)<br />Prof. Mukesh N Tekwani<br />54<br />
55. 55. Variation of g due to depth<br />The acceleration due to gravity at point P due to sphere of radius (R –d) is<br />Here, M’is the mass of inner solid sphere of radius (R - d)<br />Prof. Mukesh N Tekwani<br />55<br />
56. 56. Variation of g due to depth<br />Mass = volume x density<br />Prof. Mukesh N Tekwani<br />56<br />……………………………………………(2)<br />
57. 57. Prof. Mukesh N Tekwani<br />57<br />Variation of g due to depth<br />Dividing (2) by (1), we get,<br />This is the expression for the acceleration due to gravity at a depth d below the surface of the earth.<br />Therefore, acceleration due to gravity decreases with depth.<br />
58. 58. Variation of g due to depth<br />Prof. Mukesh N Tekwani<br />58<br />At the centre of the earth, d = R, therefore <br />gd = 0<br />So if a body of mass m is taken to the centre of the earth, its weight will be equal to zero (since w= mg). But its mass will not become 0.<br />
59. 59. Prof. Mukesh N Tekwani<br />59<br />Variation of g due to depth<br />g’<br />Outside the earth<br />Inside the earth<br />depth<br />altitude<br />
60. 60. What is latitude<br />The latitude of a location on the Earth is the angular distance of that location south or north of the Equator.<br />Latitude of equator is 0o<br />Latitude of North pole is : 90o north (+90o)<br />Latitude of South pole is : 90o south (-90o)<br />Prof. Mukesh N Tekwani<br />60<br />
61. 61. Variation of g due to latitude<br />The Earth is rotating from west to east and the axis of rotation passes through the poles.<br />Let angular velocity of earth be ω.<br />Every point on the surface of the earth is moving in a circle, i.e. every point is in an accelerated motion<br />Prof. Mukesh N Tekwani<br />61<br />
62. 62. Variation of g due to latitude<br />Prof. Mukesh N Tekwani<br />62<br />NP-North Pole<br />SP-South Pole<br />
63. 63. Variation of g due to latitude<br />Prof. Mukesh N Tekwani<br />63<br />Consider a body at point P on the surface of the earth.<br />Let the latitude of point P be <br />The body at point P moves in a circular path whose center is at Q and radius is PQ.<br />PQ = r<br />XOP = <br />
64. 64. Variation of g due to latitude<br />Prof. Mukesh N Tekwani<br />64<br />Therefore, OPQ =  (alternate angles – transversal cutting two parallel lines)<br />Centripetal acceleration needed for a body at point P is <br />ar = rω2<br />Consider  OPQ<br />cos =<br />
65. 65. Variation of g due to latitude<br />Prof. Mukesh N Tekwani<br />65<br />cos =<br />cos =<br />r = R cos <br />Therefore, ar = R cos  x ω2<br />ar = R ω2cos  ………………. (1)<br />a<br />
66. 66. Variation of g due to latitude<br />Prof. Mukesh N Tekwani<br />66<br />From  OPQ, the radial component of centripetal acceleration is <br />a = arcos<br /><ul><li>a = R ω2cos cos 
67. 67. a = R ω2 cos2 ……………… (2)
68. 68. The effective acceleration g’ due to gravity at point P is directed towards the centre of the earth and is given by:</li></ul>g’ = g - a<br />a<br />
69. 69. Variation of g due to latitude<br />Prof. Mukesh N Tekwani<br />67<br /><ul><li> g’ = g - R ω2 cos2
70. 70. As latitude  increases, cos  decreases, so g’ will increase.
71. 71. The value of g’ increases as we move from the equator to the pole due to rotation of the earth.</li></ul>a<br />
72. 72. Variation of g due to latitude<br />Prof. Mukesh N Tekwani<br />68<br />Assignment 1: Obtain an expression / value for the acceleration due to gravity at the equator.<br />Assignment 2: Obtain an expression / value for the acceleration due to gravity at the poles.<br />