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Energy of Earth Orbiting Around the Sun

  1. Mar 20, 2015 #1
    1. The problem statement, all variables and given/known data
    The Earth's distance from the sun varies from 1.471 x 108 km to 1.521 x 108 km during the year. Determine the difference in (a) the potential energy, (b) the earth's kinetic energy, and (c) the total energy between these extreme points. Take the sun to be at rest.

    2. Relevant equations
    FGrav = (GMm)/r2

    U = integral(a→b)(GMm/r2)dr

    KE = .5mv2

    Law of the Conservation of energy: Ei = Ef

    ac = v2 / r

    3. The attempt at a solution

    I got (a) easily. I need help with b and c, though.

    I tried taking the average distance from the sun, which is just the length of the semi-major axis, which is

    1.521 x 108 km - 1.471 x 108 km = 5.00 x 109 m = rave

    Then I used the centripetal acceleration formula to get

    Fa = FG = (GMsun / (rave)2) = (v2)ave/rave.

    Solving for vave I get

    vave = 2.7 x 1010 m/s.

    I then calculated average kinetic energy to be

    .5 * mearth * (vave)2 = 2.1 * 1045 J.
    Average gravitational energy would be

    -GMm/rave = -1.6 * 1035 J.

    Average total energy would be average kinetic + average potential.

    Since energy is constant the average energy is constant for instantaneous energy.

    Therefore, at perihelion and aphelion, I would have average total energy - potential energy at the point, correct?

    Any help would be appreciated, thanks.
     
  2. jcsd
  3. Mar 20, 2015 #2

    SteamKing

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    This calculation makes no sense. How can you subtract two positive numbers and have the difference be greater than either number?

    Also, it's not even how you calculate an average distance.

    Really? What's the speed of light in a vacuum? Do you have the earth going faster or slower than light itself?

    http://en.wikipedia.org/wiki/Speed_of_light

    Correct your mistakes as pointed out above.
     
  4. Mar 21, 2015 #3
    OK.

    The mean distance is (1.521*108 km + 1.471*108 km)/2 = 1.496*108 km = 1.496*1011 m.

    Then the average velocity comes from the centripetal acceleration formula:

    Fa = FG = GM/(rave)2 = (vave)2 / rave

    Solve for vave

    vave = 30,000 m/s.

    So average KE = ,5mvave2 = 2.7*1033 J.

    Average potential energy is

    -GMm/rave = -5.3 * 1033 J.

    So total energy is U + KE = -2.7*1033 J.

    So at perihelion the KE would be

    Etotal - (-GMm/rclosest distance) = 2.8*1033 J.

    At aphelion the KE would be

    Etotal - (-GMm/rfar distance) = 2.6*1033 J.

    Is this correct?
     
  5. Mar 21, 2015 #4

    haruspex

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    I do not understand how an 'average' distance could be helpful in answering the question. Just find the PE and KE at the given extremes.
    Further, it is not clear what would be meant by an average distance from the sun. Taking the average of the two extreme distances is unlikely to produce anything useful. An integral would make more sense, but even then there are choices to be made. You could take an average over time, or an average over theta, etc.
     
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