a level physics

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a level physics
2014-10-04 07:13:16
gravitational fields

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  1. a gravitational field is
    a region where any other mass will experience a force
  2. the gravitational field strength , g , at a point us defined as the
    • force that acts on a unit mass placed at that point 
    • g = F/m 
    • units Nkg-1
    • vector quantity
  3. a gravitational field can be represented by
    • sketching field lines around a mass 
    • the field lines show the direction of force that acts on another mass
  4. near the earths surface the field lines are approx
    parallel to each other and perpendicular to the surface of the earth , the field is said to be uniform - the gravitational field strength is the same magnitude and direction throughout the field
  5. on a larger scale the field lines around a spherical mass
    spread out radially
  6. force of gravity on a small mass near a much larger spherical mass is always directed to
    the centre of the larger mass
  7. the magnitude of g in a radial field .... with increased distance from the massive body
  8. over small distances which are much less than the earths radius the change in gravitational field strength is
    insignificant so the field is considered to be uniform
  9. for a radial field such as that due to a spherical mass , M , the force that acts on a small mass , m , placed at distance , r, is given by Newtons law as
    F = GMm/r2
  10. g = F/m therefor g also =
    • g = GMm/mr2
    • so 
    • g = GM/rr
  11. inside the earth as r becomes smaller the mass becomes smaller so g
  12. draw a graph to show how g changes with r
  13. as a mass , m , moves closer to the earth its potential energy changes , usually
    Ep = mgΔh
  14. however for large distances
    g isn't constant
  15. in order to be able to calculate energy changes in a gravitational field we define a a quantity called
    the gravitational potential
  16. the gravitational potential at a point in a gravitational field is defined as
    the work done in bringing a unit mass from infinity to that point
  17. the gravitational potential at a point X from the surface of a planet is
    equal to the work done in bringing a unit mass from infinity to point X
  18. work done =
    force x distance moved in direction of force
  19. the gravitational force on a unit mass at distance r from a mass , M , is F= GMm/r2 so the work done in moving the mass through a distance Δr
    -ΔrGMm/r2 . the negative sign shows that the force F acts in the opposite direction to displacement
  20. gravitation potential =
  21. draw a graph to show how gravitational potential varies with distance from the earth
    y axis should be force on a 1kg mass due to Earths gravitational field for the purpose of the next question only 
  22. the total work done in moving from infinity to a distance r is
    • the area under the graph between r and infinity 
    • each strip has an area FΔr = - GMmr/r2 =
    • FΔr = - GMm/r 
    • so V = FΔr/m = -GM/r
  23. gravitational potential is a ... qauntity
  24. units of gravitational potential are
  25. the negative sign indicates that
    this is abound system : energy would have to be transferred to the unit mass in order to pull it away
  26. http://www.vvc.edu/academic/chemistry/Unit%209R%20-%20Carboxylic%20Acids.pdf
    for chemistry
  27. V is proportional to
  28. the gradient of a gravitational potential graph vs distance is
  29. the work done in moving a mass in a gravitational field corresponds to
    change in gravitational potential energy
  30. change in gravitational potential energy can be thought of as
    change in potential energy per unit mass
  31. to calculate the potential energy we need to
    • multiply the gravitational potential by the mass that's in the field 
    • Ep=mV
  32. the magnitude of the gravitational potential at a point is
    the amount of energy needed to move a unit mass from a point to infinity
  33. hence the magnitude of the gravitational potential is
    the amount of energy required to pull the mass completely out of the gravitational field
  34. this energy could come from
    the initial kinetic energy
  35. this means we can calculate how fast a mass has to be thrown vertically up so that is never comes down again this is known as
    escape velocity
  36. if a mass has enough energy to escape then neglecting work done against air resistance 
    ΔEk =
    • -ΔEp 
    • 0.5mv2=mV (V = -GM/r)
    • 0.5mv2=mGM/r 
    • so the escape velocity is √mGm/r0.5m
    • √2GM/r
  37. gravitational field strength =
    force per unit mass
  38. gravitational potential =
    energy per unit mass
  39. in any gravitational field the greater the field strength the .... the gravitational potential changes with distance
  40. the rate of change of gravitational potential with distance is equal to
    the field strength g = -V/r
  41. what are the units of equipotentials
  42. equipotentials are similar to contour lines on a map they
    • connect points where a mass would have the same gravitational potential energy
  43. as a mass moves in a gravitational field its .... changes due to the force of gravity
  44. the exact shape of the path that it follows depends on the initial velocity as well as
    the strength of the field
  45. if we consider faster moving objects like a spacecraft the general shape of the path
    is a hyperbola if v > escape velocity
  46. for an object above the earth F =
    gravitational force
  47. at a particular speed the spacecraft can go into circular orbit around the planet . this happens when
    the gravitational force (force between two masses) equals the centripetal force necessary to make the spacecraft move in a circular path
  48. what is the equation for the velocity at which the spacecraft can go into circular orbit
    • F=GMm/r2 F=mv2/r (where r is the distance between orbit and centre earth)
    • GMm/r2 = mv2/r 
    • GM/r = V2
    • √GM/r = v 
  49. the orbital speed depends on
    the radius of the orbit and the mass of the planet but not the mass of the satellite
  50. for artificial satellites orbiting the earth those in a lower orbit have higher orbital speeds because
    v = √GM/r and at lower orbits r is less
  51. polar satellites
    • close to earth 
    • have an orbital period of 90 minutes
    • have a high orbital speed
    • are weather satellites
  52. GPS
    • global positioning satellite 
    • orbit between polar and geostationary satellites 
    • need at least 3 satellites to find your position
  53. geostationary satellites can also be called
    geosynchronous satellites
  54. geostationary satellites
    • have an orbital period of 24 hours
    • have a high equatorial orbit 
    • low orbital speed 
    • are satellites used for relaying TV pictures and therefor have to be in a fixed position relative to earth so the domestic satellite doesn't have to track the satellite across the sky
  55. calculate the radius of a ... orbit
    • F=GMm/r2 
    • F = mv2/r 
    • GMm/r2 = mv2/r
    • GM/r = v2
    • time = distance /speed 
    • T = 2pir/v 
    • so v = 2pir/T
    • GM/v2 = r 
    • GM/(2pir/T)2 = r 
    • GMT2/4pi2r2=r 
    • GMT2/4pi2=r3
    • where r is the distance from centre of earth so to find out how far above the earths surface the orbit is calculate r - radius of earth
  56. the final formula on the previous page tells us that
    T2 is proportional to r3 as G , M and pi are constants
  57. T2 for earth
    rfor earth 

    would be the same as
    • T2 for planet y 
    • -----------------
    • r3 for planet y
  58. the total energy of a satellite is the
    sum of its potential energy and its kinetic energy
  59. if the satellite is moving at a velocity v in a circular orbit of radius r around the earth then
    • Etotal = EK + E= 0.5mv2 + mV 
    • this equals 0.5mv2 - GMm/r
  60. the moons orbit around the earth is assumed to be
  61. work done =
    force x distance moved in the direction of the force
  62. in circular motion force and direction are perpendicular so there's no
    travel in the direction of the force so the work done is zero
  63. at point A why doesn't the spacecraft have acceleration when the thrust motor is switched off
    the force of attraction between the spacecraft and earth is equal and opposite to the force of attraction between the spacecraft and the moon so there's no resultant force and hence no acceleration
  64. at point B without any motor thrust the speed of the spacecraft is constant but there's acceleration this is because
    the spacecraft is in circular orbit around the earth so there's always a change in direction and hence velocity and hence theres acceleration . there's no change in speed because circular orbits are in uniform circular motion
  65. at C the velocity is 6000ms-1 it moves through an angle of 1/100 rad without changing speed this means that
    thrust must be perpendicular to the direction of motion
  66. how long can the the thrust be maintained for at C if the if the thrust is 12000N
    • F = Δmv/t 
    • t = Δmv/F
    • t = 2000x 60000x1/100
    •      -----------------------
    •               12000
    • t = 10 seconds
  67. the spacecraft at D must be accelerating towards the centre of the earth if the thrust isn't working because
    the only force acting on the spacecraft is the gravitational force of attraction between the earth and the spacecraft and this force is the centripetal force for the circular orbit
  68. what is the magnitude change in momentum of the spacecraft over 100s
    • F = Δmv/t 
    • Δmv = Ft 
    • F = GMm/r2 x t
    • 4x1014x2000/(1000000x103)x 100 = 800Ns 
  69. why can we ignore the mass of the moon in our force calculations because
    the mass of the earth is much bigger than the mass of the moon . even if they were the same mass the moon is 3 times further away than the earth so g is 9 times smaller g is proportional to 1/r2 . this means the effect of the moon is negligible
  70. the thrust motors aren't in use between c and e calculate the change in gravitational potential energy
    • EP = mV 
    • mV - mV 
    • V = -GM/r 
    • (GMm/r - GMm/r) 
    • (2000x4x1014/2 x108) = 3.6x109J
    • the potential energy increases between E and C because the gravitational potential is less at E hence the gravitational potential energy is less at E
  71. the change in gravitational potential energy would be the same even if the motors were used because
    the velocity is still perpendicular to thrust so ne extra work done so no change in gravitational potential and hence gravitational potential energy
  72. at C what is the kinetic energy
    0.5mv2 = 0.5 x 2000x60002 = 3.6x1010J
  73. what is the kinetic energy at E
    3.6 x 1010 - 3.6 x 109 = 2.34 x 1010 J
  74. what would be the kinetic energy is E was in permanent orbit around the earth
    • when in circular orbit v2 = GM/r 
    • so Ek = 0.5mGM/r
    • 0.5 x 2000 x 4x1014 / 20000x103 = 2x1010J
  75. if v> escape velocity path is
  76. v < escape velocity
    path is elliptical
  77. v2 = GM/r
    circular orbit
  78. at some point between the earth and moon g on an object = 0
    • because GMm/r= GMm/r2
    • where GMm/r2 is equal to the gravitational force of planet 

  79. newtons law of gravitation
    the gravitational force between two point masses is proportional to product of the masses and inversely proportional to the square of the distance between them
  80. Ep = mV change in m
    has the same effect on Ep but no effect on V
  81. when using V = - GM/r
    ignore the negative
  82. Mega =
  83. Giga =
  84. centripetal acceleration =