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# Gravitation

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Presentation on topic: Gravitation, included in physics syllabus for std XII (SYJC) of Maharashtra State Board

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### Gravitation

1. 1. Prof. Mukesh N. Tekwani Department of Physics Ismail Yusuf College, Mumbaimukeshtekwani@hotmail.com 1
2. 2. Isaac Newton Prof. Mukesh N Tekwani 2
3. 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. 4. Newton’s Law of GravitationG is called the Universal GravitationalConstantG = 6.67 x 10-11 N m2/kg2G is constant throughout the Universe and Gdoes not depend on the medium Mukesh N Tekwani the Prof. between 4
5. 5. Difference between G and g G gG is the Universal g is acceleration due toGravitational Constant gravityG = 6.67 x 10-11 N m2/kg2 Approx value g = 9.8 m / s2. Value of g varies from one place to another on the Earth.Constant throughout the Changes every place on aUniverse 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. 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. 7. Relation between G and g m M Prof. Mukesh N Tekwani 7
8. 8. Relation between G and gm Weight of the object is the gravitational force acting on M it. ………………………….(1) Prof. Mukesh N Tekwani 8
9. 9. Relation between G and g At height h from the surface of them Earth’s surface, acceleration due toh gravity is gh At height h, Weight of object = gravitational force M ………………………….(2) Prof. Mukesh N Tekwani 9
10. 10. Relation between G and g Dividing (2) by (1) we get,mh Thus, g is independent of the mass of M the object. Prof. Mukesh N Tekwani 10
11. 11. Projection of a SatelliteWhy is it necessary to have at least a two stagerocket to launch a satellite?A rocket with at least two stages is required tolaunch 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. 12. Critical Velocity of a SatelliteThe horizontal velocitywith which a satelliteshould be projected froma point above the earthssurface, so that it orbitsin a circular path aroundthe earth is called theorbital velocity or criticalvelocity (Vc) of thesatellite. Prof. Mukesh N Tekwani 13
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. 14. Kepler’s First Law – Law of OrbitEvery planet revolves in an ellipticalorbit around the Sun, with the Sunsituated at one focus of the ellipse. Prof. Mukesh N Tekwani 15
15. 15. Kepler’s Second Lawor Law of Equal AreasThe radius vector drawn from theSun to any planet sweeps out equalareas in equal intervals of time. Thislaw is called the law of areas.The areal velocity of the radius vectoris constant. Prof. Mukesh N Tekwani 16
16. 16. Kepler’s Law of Equal Areas Prof. Mukesh N Tekwani 17
17. 17. Kepler’s Laws Prof. Mukesh N Tekwani 18
18. 18. Kepler’s Third Law - Laws ofPeriodThe square of period of revolution ofthe planet around the Sun is directlyproportional to the cube of the semi-major axis of the elliptical orbit. T2 α r3According to this law, when the planet isclosest to the Sun, its speed is maximumand when it is farthest from the Sun, itsspeed is minimum. Prof. Mukesh N Tekwani 19
19. 19. Critical Velocity of a Satellitevc Let h M = mass of the Earth R = radius of Earth R m = mass of satellite h = height of the satellite above M Earth’s r = R + h, where r is r the distance of the satellite from the center of the Earth Vc = critical velocity Prof. Mukesh N Tekwani 20 of the satellite
20. 20. Critical Velocity of a SatelliteThe centripetal force necessary for the circularmotion of the satellite around the Earth is providedby the gravitational force of attraction between theEarth and the satellite.Centripetal force = gravitational force Prof. Mukesh N Tekwani 21
21. 21. Critical Velocity of a Satellite ……………………………………………….(1)Factors on which Critical Velocity of a satellitedepends:1. Mass of the planet2. Radius of the planet3. Height of the satelliteCritical velocity is not dependent on the mass of thesatellite as m does not appear in the aboveequation Prof. Mukesh N Tekwani 22
22. 22. Critical Velocity of a SatelliteBut we know thatSubstituting this value in eqn (1), we get, …………………………………..(2) Prof. Mukesh N Tekwani 23
23. 23. Critical Velocity of a SatelliteAssignment 1:Modify eqn (2) to find the critical velocity of asatellite orbiting very close to the surface of theEarth (h << R)Assignment 2:How does the critical velocity (or orbital velocity) ofa satellite vary with an increase in the height of thesatellite above the Earth’s surface? Prof. Mukesh N Tekwani 24
24. 24. Time Period of a SatelliteThe time taken by a satellite tocomplete one revolution around theearth is called its periodic time or timeperiod. Prof. Mukesh N Tekwani 25
25. 25. Time Period of a Satellitevc Let h M = mass of the Earth R = radius of Earth R m = mass of satellite h = height of the satellite above M Earth’s r = R + h, where r is r the distance of the satellite from the center of the Earth Vc = critical velocity Prof. Mukesh N Tekwani 26 of the satellite
26. 26. Time Period of a SatelliteDistance covered by the satellite in 1 revolution = Circumference of the circleTime taken to cover this distance is the time periodCritical speed Vc =Vc =But Prof. Mukesh N Tekwani 27
27. 27. Time Period of a Satellite As is a constant, so we get,Squaring both sides, 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. 28. Time Period of a SatelliteFactors on which Time Period of a satellite depends:1. Mass of the planet2. Radius of the planet, and3. Height of the satellite from the planet’s surfacePeriod of the satellite does not depend on the massof 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. 29. Binding Energy of a SatelliteDefinition:The minimum amount of energyrequired to remove a satellite from theearth’s gravitational influence is calledas binding energy of the satellite.A satellite revolving around the Earthhas Kinetic energy, andPotential energy Prof. Mukesh N Tekwani 30
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. 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. = mghQuiz: Does a body in motion havepotential energy? Prof. Mukesh N Tekwani 32
32. 32. Binding Energy of a Satellitevc Let h M = mass of the Earth R = radius of Earth R m = mass of satellite h = height of the satellite above M Earth’s r = R + h, where r is r the distance of the satellite from the center of the Earth Vc = critical velocity Prof. Mukesh N Tekwani 33 of the satellite
33. 33. Binding Energy of a SatelliteThe critical velocity is given by ……………………………………………………...(1 )Kinetic energy of motion KE = …..……...(2)Substituting (1) in (2), we get, ..………………………………………………..(3KE = ) Prof. Mukesh N Tekwani 34
34. 34. Binding Energy of a SatelliteThe gravitational potential at a distance rfrom the centre of the Earth is given by:GP =Potential energy = gravitational potential xmass of the satellite ..…….………………………..(4Therefore, PE = ) Prof. Mukesh N Tekwani 35
35. 35. Binding Energy of a SatelliteThe total energy of the satellite is given byTE = KE + PETE = +TE = ..…….…………………………………………..(5)The negative sign indicates that the satelliteis bound to the Earth due to the gravitationalforce of the Earth. Prof. Mukesh N Tekwani 36
36. 36. Binding Energy of a SatelliteTo free the satellite from the Earth’sgravitational influence, an amount of energyequal to its total energy must be supplied.This is called the binding energy of thesatellite.Therefore, BE =Where r = R + hAssignment: Calculate the BE of a satelliteat rest on the surface of the Earth. N Tekwani Prof. Mukesh 37
37. 37. Weightlessness in a Satellite1. 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. 38. Weightlessness in a Satellite5. 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 astronaut. This causes the “feeling” of Prof. Mukesh N Tekwani 39
39. 39. Weightlessness in a Satellite8. 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. 40. Escape Velocity of a SatelliteThe minimumvelocity with whicha body should beprojected from thesurface of theearth so that itescapes thegravitational fieldof the earth iscalled the escapevelocity of the Prof. Mukesh N Tekwani 41
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. 42. Escape Velocity of a SatelliteTherefore, KE of satellite = BEKE = R Therefore, = R Prof. Mukesh N Tekwani 43
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. 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. 45. Variation of g with Altitude Weight of the object is the m gravitational force acting on it. M ………………………….(1) Prof. Mukesh N Tekwani 47
46. 46. Variation of g with Altitude At height h from the surface of the m Earth’s surface, acceleration due to h gravity is gh At height h, Weight of object = gravitational force M ………………………….(2) Prof. Mukesh N Tekwani 48
47. 47. Variation of g with Altitude Dividing (2) by (1) we get, m h Thus, g is independent of the mass of the M 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. 48. Variation of g with Altitude By Binomial expansion: Since the higher powers of h/R are neglected. Prof. Mukesh N Tekwani 50
49. 49. Variation of g due to depthThe acceleration due to gravity on thesurface of the earth is given by:Consider the earth as a sphere ofdensity ρ Mass = Volume x Density Prof. Mukesh N Tekwani 52
50. 50. Variation of g due to depthMass = Volume x DensityMass = …………………………………………. (1) Prof. Mukesh N Tekwani 53
51. 51. Variation of g due to depthConsider a point P at the depth d belowthe surface of the earth. At this point, letthe acceleration due to gravity be gd.Distance of point P from centre of Earthis (R – d) Prof. Mukesh N Tekwani 54
52. 52. Variation of g due to depthThe acceleration due to gravity at pointP due to sphere of radius (R –d) isHere, M’ is the mass of inner solidsphere of radius (R - d) Prof. Mukesh N Tekwani 55
53. 53. Variation of g due to depthMass = volume x density ……………………………………………(2) Prof. Mukesh N Tekwani 56
54. 54. Variation of g due to depthDividing (2) by (1), we get,This is the expression for the acceleration due togravity at a depth d below the surface of the earth.Therefore, acceleration due to gravity decreases withdepth. Prof. Mukesh N Tekwani 57
55. 55. Variation of g due to depthAt the centre of the earth, d = R, thereforegd = 0So if a body of mass m is taken to the centre of theearth, its weight will be equal to zero (since w= mg). Butits mass will not become 0. Prof. Mukesh N Tekwani 58
56. 56. Variation of g due to depthg’ depth altitude Prof. Mukesh N Tekwani 59
57. 57. What is latitudeThe latitude of a location on the Earth isthe angular distance of that locationsouth or north of the Equator.Latitude of equator is 0oLatitude of North pole is : 90o north(+90o)Latitude of South pole is : 90o south (-90o) Prof. Mukesh N Tekwani 60
58. 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. 59. Variation of g due to latitudeNP-North PoleSP-South Pole Prof. Mukesh N Tekwani 62
60. 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. 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. 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. 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. 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. 65. Variation of g due to latitudeAssignment 1: Obtain an expression / value for theacceleration due to gravity at the equator.Assignment 2: Obtain an expression / value for theacceleration due to gravity at the poles. Prof. Mukesh N Tekwani 68
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