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- 1. Prof. Mukesh N. Tekwani Department of Physics Ismail Yusuf College, Mumbai 1
- 2. Isaac Newton Prof. Mukesh N Tekwani 2
- 3. Newton’s Law of Gravitation Every particle of matter attracts every other particle with a force which is directly proportional to the product of the masses and inversely proportional to the square of the distance between them. Prof. Mukesh N Tekwani 3
- 4. Newton’s Law of Gravitation G is called the Universal Gravitational Constant G = 6.67 x 10-11 N m2/kg2 G is constant throughout the Universe and G does not depend on the medium Mukesh N Tekwani the between Prof. 4
- 5. Difference between G and g G G is the Universal Gravitational Constant G = 6.67 x 10-11 N m2/kg2 Constant throughout the Universe g g is acceleration due to gravity Approx value g = 9.8 m / s2. Value of g varies from one place to another on the Earth. Changes every place on a planet. E.g., on the Moon, the value of g is 1/6th of that on the Earth’s surface. Prof. Mukesh N Tekwani 5
- 6. Relation between G and g Let M = mass of the Earth m = mass of an object on the surface of the Earth g = acceleration due to gravity on the Earth’s surface R = radius of the Earth Prof. Mukesh N Tekwani 6
- 7. Relation between G and g m M Prof. Mukesh N Tekwani 7
- 8. Relation between G and g m M Weight of the object is the gravitational force acting on it. ………………………….(1) Prof. Mukesh N Tekwani 8
- 9. Relation between G and g At height h from the surface of the Earth’s surface, acceleration due to gravity is gh m h M At height h, Weight of object = gravitational force ………………………….(2) Prof. Mukesh N Tekwani 9
- 10. Relation between G and g Dividing (2) by (1) we get, m h M Thus, g is independent of the mass of the object. Prof. Mukesh N Tekwani 10
- 11. Projection of a Satellite Why is it necessary to have at least a two stage rocket to launch a satellite? A rocket with at least two stages is required to launch a satellite because The first stage is used to carry the satellite up to the desired height. 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. Prof. Mukesh N Tekwani 12
- 12. Critical Velocity of a Satellite 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. Prof. Mukesh N Tekwani 13
- 13. Kepler’s Laws of Motion Born: December 27, 1571 Died: November 15, 1630 German Mathematician, Astronomer Astrologer. Prof. Mukesh N Tekwani 14
- 14. Kepler’s First Law – Law of Orbit Every planet revolves in an elliptical orbit around the Sun, with the Sun situated at one focus of the ellipse. Prof. Mukesh N Tekwani 15
- 15. Kepler’s Second Law or Law of Equal Areas 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. The areal velocity of the radius vector is constant. Prof. Mukesh N Tekwani 16
- 16. Kepler’s Law of Equal Areas Prof. Mukesh N Tekwani 17
- 17. Kepler’s Laws Prof. Mukesh N Tekwani 18
- 18. Kepler’s Third Law - Laws of Period The square of period of revolution of the planet around the Sun is directly proportional to the cube of the semimajor axis of the elliptical orbit. T2 α r3 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. Prof. Mukesh N Tekwani 19
- 19. Critical Velocity of a Satellite Let vc h R M r M = mass of the Earth R = radius of Earth m = mass of satellite h = height of the satellite above Earth’s r = R + h, where r is the distance of the satellite from the center of the Earth Vc = critical velocity Prof. Mukesh N Tekwani of the satellite 20
- 20. Critical Velocity of a Satellite 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. Centripetal force = gravitational force Prof. Mukesh N Tekwani 21
- 21. Critical Velocity of a Satellite ……………………………………………….(1) Factors on which Critical Velocity of a satellite depends: 1. Mass of the planet 2. Radius of the planet 3. Height of the satellite Critical velocity is not dependent on the mass of the satellite as m does not appear in the above equation Prof. Mukesh N Tekwani 22
- 22. Critical Velocity of a Satellite But we know that Substituting this value in eqn (1), we get, …………………………………..(2) Prof. Mukesh N Tekwani 23
- 23. Critical Velocity of a Satellite Assignment 1: Modify eqn (2) to find the critical velocity of a satellite orbiting very close to the surface of the Earth (h << R) Assignment 2: 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? Prof. Mukesh N Tekwani 24
- 24. Time Period of a Satellite The time taken by a satellite to complete one revolution around the earth is called its periodic time or time period. Prof. Mukesh N Tekwani 25
- 25. Time Period of a Satellite Let vc h R M r M = mass of the Earth R = radius of Earth m = mass of satellite h = height of the satellite above Earth’s r = R + h, where r is the distance of the satellite from the center of the Earth Vc = critical velocity Prof. Mukesh N Tekwani of the satellite 26
- 26. Time Period of a Satellite Distance covered by the satellite in 1 revolution = Circumference of the circle Time taken to cover this distance is the time period Critical speed Vc = Vc = But Prof. Mukesh N Tekwani 27
- 27. Time Period of a Satellite As Squaring both sides, we get is a constant, so we get, T2 α r3 Thus, the square of the period of revolution is directly proportional to the cube of the radius of its orbit. Prof. Mukesh N Tekwani 28
- 28. Time Period of a Satellite Factors on which Time Period of a satellite depends: 1. Mass of the planet 2. Radius of the planet, and 3. Height of the satellite from the planet’s surface Period of the satellite does not depend on the mass of the satellite. Assignment: (1)Obtain an expression for the time period of a satellite in terms of gh. (2)For a satellite close to the earth, calculate the period of revolution in minutes. Prof. Mukesh N Tekwani 29
- 29. Binding Energy of a Satellite Definition: The minimum amount of energy required to remove a satellite from the earth’s gravitational influence is called as binding energy of the satellite. A satellite revolving around the Earth has Kinetic energy, and Potential energy Prof. Mukesh N Tekwani 30
- 30. What is kinetic energy? The energy possessed by a body due to its motion is called its kinetic energy. If m = mass of an object, and v = its velocity K.E = (1/2) mv2 Prof. Mukesh N Tekwani 31
- 31. What is potential energy? The energy possessed by a body due to its position is called its potential energy. If m = mass of an object, and h = its height above the surface P.E. = mgh Quiz: Does a body in motion have potential energy? Prof. Mukesh N Tekwani 32
- 32. Binding Energy of a Satellite Let vc h R M r M = mass of the Earth R = radius of Earth m = mass of satellite h = height of the satellite above Earth’s r = R + h, where r is the distance of the satellite from the center of the Earth Vc = critical velocity Prof. Mukesh N Tekwani of the satellite 33
- 33. Binding Energy of a Satellite The critical velocity is given by ……………………………………………………...(1 ) Kinetic energy of motion KE = …..……...(2) Substituting (1) in (2), we get, KE = ..………………………………………………..(3 ) Prof. Mukesh N Tekwani 34
- 34. Binding Energy of a Satellite The gravitational potential at a distance r from the centre of the Earth is given by: GP = Potential energy = gravitational potential x mass of the satellite Therefore, PE = ..…….………………………..(4 ) Prof. Mukesh N Tekwani 35
- 35. Binding Energy of a Satellite The total energy of the satellite is given by TE = KE + PE TE = TE = + ..…….…………………………………………..(5) The negative sign indicates that the satellite is bound to the Earth due to the gravitational force of the Earth. Prof. Mukesh N Tekwani 36
- 36. Binding Energy of a Satellite 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. Therefore, BE = Where r = R + h Assignment: Calculate the BE of a satellite at rest on the surface of the Earth. N Tekwani Prof. Mukesh 37
- 37. Weightlessness in a Satellite 1. The weight of a body is the gravitational force exerted on it by the Earth. 2. 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. 3. This normal reaction is equal to the weight of the person. 4. The person has a feeling of weight due to this normal reaction. Prof. Mukesh N Tekwani 38
- 38. Weightlessness in a Satellite 5. Consider an astronaut of mass m, in a satellite that is moving around the Earth in a circular orbit. 6. 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. 7. 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 Prof. Mukesh N Tekwani astronaut. This causes the “feeling” of 39
- 39. Weightlessness in a Satellite 8. We must remember that the mass of the astronaut DOES NOT become zero. 9. This condition of weightlessness is also known (incorrectly) as zero gravity condition. 10.But, weightlessness does not mean the absence of gravity. Prof. Mukesh N Tekwani 40
- 40. Escape Velocity of a Satellite 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 Prof. Mukesh N Tekwani 41
- 41. Escape Velocity of a Satellite Consider a satellite of mass m, stationary on the surface of the Earth. The binding energy of the satellite, on the surface of the Earth, is given by BE = R To escape from the Earth’s influence, energy must be provided to the satellite in the form of kinetic energy. Prof. Mukesh N Tekwani 42
- 42. Escape Velocity of a Satellite Therefore, KE of satellite = BE KE = R Therefore, = R Prof. Mukesh N Tekwani 43
- 43. Numerical Problems Obtain an equation for the escape velocity of a body from the surface of a planet of radius R and mean density ρ. What would be the duration of the year if the distance between the Earth and Sun gets doubled? Prof. Mukesh N Tekwani 44
- 44. Numerical Problems 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) 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? Prof. Mukesh N Tekwani 45
- 45. Variation of g with Altitude Weight of the object is the gravitational force acting on it. m M ………………………….(1) Prof. Mukesh N Tekwani 47
- 46. Variation of g with Altitude At height h from the surface of the Earth’s surface, acceleration due to gravity is gh m h M At height h, Weight of object = gravitational force ………………………….(2) Prof. Mukesh N Tekwani 48
- 47. Variation of g with Altitude Dividing (2) by (1) we get, m h M Thus, gh is independent of the mass of the object. From this eqn. it is clear that the acceleration due to gravity decreases as altitude of the body from the earth’s surface increases. Prof. Mukesh N Tekwani 49
- 48. Variation of g with Altitude By Binomial expansion: Since the higher powers of h/R are neglected. Prof. Mukesh N Tekwani 50
- 49. Variation of g due to depth The acceleration due to gravity on the surface of the earth is given by: Consider the earth as a sphere of density ρ Mass = Volume x Density Prof. Mukesh N Tekwani 52
- 50. Variation of g due to depth Mass = Volume x Density Mass = …………………………………………. Prof. Mukesh N Tekwani (1) 53
- 51. Variation of g due to depth 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. Distance of point P from centre of Earth is (R – d) Prof. Mukesh N Tekwani 54
- 52. Variation of g due to depth The acceleration due to gravity at point P due to sphere of radius (R –d) is Here, M’ is the mass of inner solid sphere of radius (R - d) Prof. Mukesh N Tekwani 55
- 53. Variation of g due to depth Mass = volume x density ……………………………………………(2) Prof. Mukesh N Tekwani 56
- 54. Variation of g due to depth Dividing (2) by (1), we get, This is the expression for the acceleration due to gravity at a depth d below the surface of the earth. Therefore, acceleration due to gravity decreases with depth. Prof. Mukesh N Tekwani 57
- 55. Variation of g due to depth At the centre of the earth, d = R, therefore gd = 0 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. Prof. Mukesh N Tekwani 58
- 56. Variation of g due to depth g’ depth altitude Prof. Mukesh N Tekwani 59
- 57. What is latitude The latitude of a location on the Earth is the angular distance of that location south or north of the Equator. Latitude of equator is 0o Latitude of North pole is : 90o north (+90o) Latitude of South pole is : 90o south (90o) Prof. Mukesh N Tekwani 60
- 58. Variation of g due to latitude The Earth is rotating from west to east and the axis of rotation passes through the poles. Let angular velocity of earth be ω. Every point on the surface of the earth is moving in a circle, i.e. every point is in an accelerated motion Prof. Mukesh N Tekwani 61
- 59. Variation of g due to latitude NP-North Pole SP-South Pole Prof. Mukesh N Tekwani 62
- 60. Variation of g due to latitude Consider a body at point P on the surface of the earth. Let the latitude of point P be The body at point P moves in a circular path whose center is at Q and radius is PQ. PQ = r XOP = Prof. Mukesh N Tekwani 63
- 61. Variation of g due to latitude Therefore, OPQ = (alternate angles – transversal cutting two parallel lines) Centripetal acceleration needed for a body at point P is ar = rω2 Consider OPQ cos = Prof. Mukesh N Tekwani 64
- 62. Variation of g due to latitude cos = cos = r = R cos Therefore, ar = R cos x ω2 ar = R ω2 cos ………………. (1) Prof. Mukesh N Tekwani 65
- 63. Variation of g due to latitude From OPQ, the radial component of centripetal acceleration is a = ar cos = R ω2 cos cos a = R ω2 cos2 ……………… (2) a effective acceleration g’ due to The gravity at point P is directed towards the centre of the earth and is given by: Prof. Mukesh N Tekwani 66
- 64. Variation of g due to latitude = g - R ω2 cos2 g’ latitude increases, cos As decreases, so g’ will increase. value of g’ increases as we The move from the equator to the pole due to rotation of the earth. Prof. Mukesh N Tekwani 67
- 65. Variation of g due to latitude Assignment 1: Obtain an expression / value for the acceleration due to gravity at the equator. Assignment 2: Obtain an expression / value for the acceleration due to gravity at the poles. Prof. Mukesh N Tekwani 68

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