Syjc chap2-gravitation
Upcoming SlideShare
Loading in...5
×
 

Syjc chap2-gravitation

on

  • 1,691 views

gravitation, newtons law of gravitation satellite motion, critical velocity

gravitation, newtons law of gravitation satellite motion, critical velocity

Statistics

Views

Total Views
1,691
Views on SlideShare
1,691
Embed Views
0

Actions

Likes
0
Downloads
69
Comments
0

0 Embeds 0

No embeds

Accessibility

Categories

Upload Details

Uploaded via as Microsoft PowerPoint

Usage Rights

© All Rights Reserved

Report content

Flagged as inappropriate Flag as inappropriate
Flag as inappropriate

Select your reason for flagging this presentation as inappropriate.

Cancel
  • Full Name Full Name Comment goes here.
    Are you sure you want to
    Your message goes here
    Processing…
Post Comment
Edit your comment

    Syjc chap2-gravitation Syjc chap2-gravitation Presentation Transcript

    • Gravitation
      Prof. Mukesh N. Tekwani
      Department of Physics
      Ismail Yusuf College,
      Mumbai
      mukeshtekwani@hotmail.com
      1
    • Isaac Newton
      Prof. Mukesh N Tekwani
      2
    • 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.
    • 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
    • Difference between G and g
      Prof. Mukesh N Tekwani
      5
    • 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
    • Relation between G and g
      Prof. Mukesh N Tekwani
      7
      m
      M
    • Relation between G and g
      Prof. Mukesh N Tekwani
      8
      Weight of the object is the gravitational force acting on it.
      m
      M
      ………………………….(1)
    • 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)
    • 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
    • Projection of a Satellite
      Prof. Mukesh N Tekwani
      11
    • 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.
    • 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.
    • Kepler’s Laws of Motion
      Prof. Mukesh N Tekwani
      14
      Born:
      December 27, 1571
      Died:
      November 15, 1630
      German 
      Mathematician, 
      Astronomer
      Astrologer.
    • 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.
    • 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.
    • Kepler’s Law of Equal Areas
      Prof. Mukesh N Tekwani
      17
    • Kepler’s Laws
      Prof. Mukesh N Tekwani
      18
    • 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.
    • 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
    • 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
    • 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)
    • Critical Velocity of a Satellite
      Prof. Mukesh N Tekwani
      23
      But we know that
      Substituting this value in eqn (1), we get,
      …………………………………..(2)
    • 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?
    • 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
    • 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
    • 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
    • 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
    • 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.
    • 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
    • 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
    • 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
    • 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
    • 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)
    • 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)
    • 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)
    • 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.
    • 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.
    • 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.
    • 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.
    • 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.
    • 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
    • Escape Velocity of a Satellite
      Prof. Mukesh N Tekwani
      43
      Therefore, KE of satellite = BE
      KE =
      Therefore, =
      R
      R
    • 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
    • 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
    • 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
    • Prof. Mukesh N Tekwani
      47
      Variation of g with Altitude
      Weight of the object is the gravitational force acting on it.
      m
      M
      ………………………….(1)
    • 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)
    • 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
    • Prof. Mukesh N Tekwani
      50
      Variation of g with Altitude
      By Binomial expansion:
      Since the higher powers of h/R are neglected.
    • 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
    • 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
    • Variation of g due to depth
      Prof. Mukesh N Tekwani
      53
      Mass = Volume x Density
      Mass =
      …………………………………………. (1)
    • 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
    • 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
    • Variation of g due to depth
      Mass = volume x density
      Prof. Mukesh N Tekwani
      56
      ……………………………………………(2)
    • 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.
    • 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.
    • Prof. Mukesh N Tekwani
      59
      Variation of g due to depth
      g’
      Outside the earth
      Inside the earth
      depth
      altitude
    • 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
    • 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
    • Variation of g due to latitude
      Prof. Mukesh N Tekwani
      62
      NP-North Pole
      SP-South Pole
    • 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 = 
    • 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 =
    • 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
    • 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 
      • a = R ω2 cos2 ……………… (2)
      • 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
    • Variation of g due to latitude
      Prof. Mukesh N Tekwani
      67
      • g’ = g - R ω2 cos2
      • As latitude  increases, cos  decreases, so g’ will increase.
      • The value of g’ increases as we move from the equator to the pole due to rotation of the earth.
      a
    • 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.