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Syjc chap2-gravitation

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gravitation, newtons law of gravitation satellite motion, critical velocity

gravitation, newtons law of gravitation satellite motion, critical velocity

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  • 1. Gravitation
    Prof. Mukesh N. Tekwani
    Department of Physics
    Ismail Yusuf College,
    Mumbai
    mukeshtekwani@hotmail.com
    1
  • 2. Isaac Newton
    Prof. Mukesh N Tekwani
    2
  • 3. Newton’s Law of Gravitation
    Prof. Mukesh N Tekwani
    3
    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.
  • 4. Newton’s Law of Gravitation
    Prof. Mukesh N Tekwani
    4
    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 between the masses
  • 5. Difference between G and g
    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
    Prof. Mukesh N Tekwani
    7
    m
    M
  • 8. Relation between G and g
    Prof. Mukesh N Tekwani
    8
    Weight of the object is the gravitational force acting on it.
    m
    M
    ………………………….(1)
  • 9. Relation between G and g
    Prof. Mukesh N Tekwani
    9
    At height h from the surface of the Earth’s surface, acceleration due to gravity is gh
    At height h,
    Weight of object = gravitational force
    m
    h
    M
    ………………………….(2)
  • 10. Relation between G and g
    Prof. Mukesh N Tekwani
    10
    Dividing (2) by (1) we get,
    m
    h
    Thus, g is independent of the mass of the object.
    M
  • 11. Projection of a Satellite
    Prof. Mukesh N Tekwani
    11
  • 12. Projection of a Satellite
    Prof. Mukesh N Tekwani
    12
    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.
  • 13. Critical Velocity of a Satellite
    Prof. Mukesh N Tekwani
    13
    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.
  • 14. Kepler’s Laws of Motion
    Prof. Mukesh N Tekwani
    14
    Born:
    December 27, 1571
    Died:
    November 15, 1630
    German 
    Mathematician, 
    Astronomer
    Astrologer.
  • 15. Kepler’s First Law – Law of Orbit
    Prof. Mukesh N Tekwani
    15
    Every planet revolves in an elliptical orbit around the Sun, with the Sun situated at one focus of the ellipse.
  • 16. Kepler’s Second Lawor Law of Equal Areas
    Prof. Mukesh N Tekwani
    16
    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.
  • 17. Kepler’s Law of Equal Areas
    Prof. Mukesh N Tekwani
    17
  • 18. Kepler’s Laws
    Prof. Mukesh N Tekwani
    18
  • 19. Kepler’s Third Law - Laws of Period
    Prof. Mukesh N Tekwani
    19
    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.
    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.
  • 20. Critical Velocity of a Satellite
    Prof. Mukesh N Tekwani
    20
    Let
    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 of the satellite
    vc
    h
    R
    M
    r
  • 21. Critical Velocity of a Satellite
    Prof. Mukesh N Tekwani
    21
    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
  • 22. Critical Velocity of a Satellite
    Prof. Mukesh N Tekwani
    22
    Factors on which Critical Velocity of a satellite depends:
    Mass of the planet
    Radius of the planet
    Height of the satellite
    Critical velocity is not dependent on the mass of the satellite as m does not appear in the above equation
    ……………………………………………….(1)
  • 23. Critical Velocity of a Satellite
    Prof. Mukesh N Tekwani
    23
    But we know that
    Substituting this value in eqn (1), we get,
    …………………………………..(2)
  • 24. Critical Velocity of a Satellite
    Prof. Mukesh N Tekwani
    24
    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?
  • 25. 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
  • 26. Time Period of a Satellite
    Prof. Mukesh N Tekwani
    26
    Let
    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 of the satellite
    vc
    h
    R
    M
    r
  • 27. Time Period of a Satellite
    Prof. Mukesh N Tekwani
    27
    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
  • 28. Prof. Mukesh N Tekwani
    28
    Time Period of a Satellite
    As is a constant,
    so we get,
    T2α r3
    Thus, the square of the period of revolution is directly proportional to the cube of the radius of its orbit.
    Squaring both sides, we get
  • 29. Prof. Mukesh N Tekwani
    29
    Time Period of a Satellite
    Factors on which Time Period of a satellite depends:
    Mass of the planet
    Radius of the planet, and
    Height of the satellite from the planet’s surface
    Period of the satellite does not depend on the mass of the satellite.
    Assignment:
    Obtain an expression for the time period of a satellite in terms of gh.
    For a satellite close to the earth, calculate the period of revolution in minutes.
  • 30. 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 bindingenergy of the satellite.
    A satellite revolving around the Earth has
    Kinetic energy, and
    Potential energy
    Prof. Mukesh N Tekwani
    30
  • 31. 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
  • 32. 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
  • 33. Binding Energy of a Satellite
    Prof. Mukesh N Tekwani
    33
    Let
    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 of the satellite
    vc
    h
    R
    M
    r
  • 34. Prof. Mukesh N Tekwani
    34
    Binding Energy of a Satellite
    The critical velocity is given by
    Kinetic energy of motion KE =
    Substituting (1) in (2), we get,
    KE =
    ……………………………………………………...(1)
    …..……...(2)
    ..………………………………………………..(3)
  • 35. Prof. Mukesh N Tekwani
    35
    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)
  • 36. Prof. Mukesh N Tekwani
    36
    Binding Energy of a Satellite
    The total energy of the satellite is given by
    TE = KE + PE
    TE = +
    TE =
    The negative sign indicates that the satellite is bound to the Earth due to the gravitational force of the Earth.
    ..…….…………………………………………..(5)
  • 37. Prof. Mukesh N Tekwani
    37
    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.
  • 38. Prof. Mukesh N Tekwani
    38
    Weightlessness in a Satellite
    The weight of a body is the gravitational force exerted on it by the Earth.
    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.
    This normal reaction is equal to the weight of the person.
    The person has a feeling of weight due to this normal reaction.
  • 39. Prof. Mukesh N Tekwani
    39
    Weightlessness in a Satellite
    Consider an astronaut of mass m, in a satellite that is moving around the Earth in a circular orbit.
    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.
    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.
  • 40. Prof. Mukesh N Tekwani
    40
    Weightlessness in a Satellite
    We must remember that the mass of the astronaut DOES NOT become zero.
    This condition of weightlessness is also known (incorrectly) as zero gravity condition.
    But, weightlessness does not mean the absence of gravity.
  • 41. Escape Velocity of a Satellite
    Prof. Mukesh N Tekwani
    41
    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.
  • 42. Escape Velocity of a Satellite
    Prof. Mukesh N Tekwani
    42
    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 =
    To escape from the Earth’s influence, energy must be provided to the satellite in the form of kinetic energy.
    R
  • 43. Escape Velocity of a Satellite
    Prof. Mukesh N Tekwani
    43
    Therefore, KE of satellite = BE
    KE =
    Therefore, =
    R
    R
  • 44. 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
  • 45. 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
  • 46. Variation of g with Altitude
    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
    46
  • 47. Prof. Mukesh N Tekwani
    47
    Variation of g with Altitude
    Weight of the object is the gravitational force acting on it.
    m
    M
    ………………………….(1)
  • 48. Prof. Mukesh N Tekwani
    48
    Variation of g with Altitude
    At height h from the surface of the Earth’s surface, acceleration due to gravity is gh
    At height h,
    Weight of object = gravitational force
    m
    h
    M
    ………………………….(2)
  • 49. Prof. Mukesh N Tekwani
    49
    Variation of g with Altitude
    Dividing (2) by (1) we get,
    m
    h
    Thus, g 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.
    M
  • 50. Prof. Mukesh N Tekwani
    50
    Variation of g with Altitude
    By Binomial expansion:
    Since the higher powers of h/R are neglected.
  • 51. Variation of g due to depth
    Prof. Mukesh N Tekwani
    51
    Let
    M = mass of the Earth
    R = radius of Earth
    g = acceleration due to
    gravity on the surface of
    earth
    ρ = density of earth
    P
    R-d
  • 52. 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
  • 53. Variation of g due to depth
    Prof. Mukesh N Tekwani
    53
    Mass = Volume x Density
    Mass =
    …………………………………………. (1)
  • 54. 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
  • 55. 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
  • 56. Variation of g due to depth
    Mass = volume x density
    Prof. Mukesh N Tekwani
    56
    ……………………………………………(2)
  • 57. Prof. Mukesh N Tekwani
    57
    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.
  • 58. Variation of g due to depth
    Prof. Mukesh N Tekwani
    58
    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.
  • 59. Prof. Mukesh N Tekwani
    59
    Variation of g due to depth
    g’
    Outside the earth
    Inside the earth
    depth
    altitude
  • 60. 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
  • 61. 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
  • 62. Variation of g due to latitude
    Prof. Mukesh N Tekwani
    62
    NP-North Pole
    SP-South Pole
  • 63. Variation of g due to latitude
    Prof. Mukesh N Tekwani
    63
    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 = 
  • 64. Variation of g due to latitude
    Prof. Mukesh N Tekwani
    64
    Therefore, OPQ =  (alternate angles – transversal cutting two parallel lines)
    Centripetal acceleration needed for a body at point P is
    ar = rω2
    Consider  OPQ
    cos =
  • 65. Variation of g due to latitude
    Prof. Mukesh N Tekwani
    65
    cos =
    cos =
    r = R cos 
    Therefore, ar = R cos  x ω2
    ar = R ω2cos  ………………. (1)
    a
  • 66. Variation of g due to latitude
    Prof. Mukesh N Tekwani
    66
    From  OPQ, the radial component of centripetal acceleration is
    a = arcos
    • a = R ω2cos cos 
    • 67. a = R ω2 cos2 ……………… (2)
    • 68. The effective acceleration g’ due to gravity at point P is directed towards the centre of the earth and is given by:
    g’ = g - a
    a
  • 69. Variation of g due to latitude
    Prof. Mukesh N Tekwani
    67
    • g’ = g - R ω2 cos2
    • 70. As latitude  increases, cos  decreases, so g’ will increase.
    • 71. The value of g’ increases as we move from the equator to the pole due to rotation of the earth.
    a
  • 72. Variation of g due to latitude
    Prof. Mukesh N Tekwani
    68
    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.