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Total Angular Momentum of the Earth

  1. Nov 14, 2009 #1

    v3r

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    1. The problem statement, all variables and given/known data
    How long should the day be so that the total angular momentum of the Earth
    (including its rotation about its own axis and its (approximately) circular orbit around the
    sun) is zero (Note: the magnitude of the angular velocity is 2pi/T where T is the period of rotation?)

    2. Relevant equations
    Ltot = Ltrans + Lrot
    Lrot = Iw^2
    w = 2pi/T

    3. The attempt at a solution
    Ltot = 0

    I am really clueless. I don't know where to start..
     
  2. jcsd
  3. Nov 14, 2009 #2
    you're started. double check your equation
    Lrot = Iw^2

    There shouldn't be a square there. For the earth's orbital angular momentum, just treat the earth like a point mass at a distance r.

    L=mvr
     
  4. Nov 14, 2009 #3

    v3r

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    So it should be
    Ltot = Ltrans + Lrot
    Ltot = 0
    Ltrans = Lrot
    mvr = Iw
    mvr = 2pi/T * I
    mvr = 2pi/T * mr^2
    v/r = 2pi/T

    I'm still confused.
     
  5. Nov 14, 2009 #4
    You're getting there. Remember, you're looking for the period of one day on earth under your new conditions. what variable do you want to solve for?

    Also, be carefull with your variables. You've used "r" to stand in for two different things.

    What is the r in Ltrans = mvr?
    What is the r in I = mr^2?
     
  6. Nov 14, 2009 #5

    v3r

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    r in I would be the perpendicular distance which would be the radius of the earth.
    r in Ltrans would be the distance to the center of mass which would be the distance of earth from sun?

    I want to solve for v in order to get the time by dividing it by the radius?
     
  7. Nov 14, 2009 #6
    You're almost there. Now use another variable to re-name one of your r's. Eventually, you want to solve for period T. However, you have the earth's orbital velocity v to get rid of. For this, use univorm circular motion, and universal gravitation.

    F=m(v^2) / r


    F=GmM / (r^2)
     
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