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F= MA 2009 #5 (Gravitation)

  1. Jan 29, 2013 #1
    1. The problem statement, all variables and given/known data

    https://aapt.org/physicsteam/2010/upload/2009_F-ma.pdf

    2. Relevant equations
    L = mrv
    L = Iω


    3. The attempt at a solution
    For a circular orbit:
    Fc = Fg
    mv^2/r = Gmm/r^2
    v = √(GM/R)
    Thus:
    l = mR√(GM/R)
    l = m√(GMR)

    This means that LA > LC, eliminating choices B, C, and E.

    Now, to compare B, C
    I'm interested in finding a more rigorous approach, but here goes.
    The point of intersection between the Circlular path that C orbits on and the elliptical path that B orbits.
    We know that the velocity at the perihelion is greater than the aphelion, that is, the velocity of the intersection is the maximum velocity that B ever achieves. I then made an intelligent guess and postulated that thus B > C,
    leading to LA > LB > LC

    Could you suggest more rigor/principles to do this question?
     
  2. jcsd
  3. Jan 29, 2013 #2

    Simon Bridge

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    If choices B,C & E are eliminated - what is left are:

    (A) LA > LB > LC
    (E) The relationship between the magnitudes is different at various instants in time.

    Look at E.
    Consider: conservation of angular momentum.
     
  4. Jan 29, 2013 #3

    tms

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    You've already done that: for B, [itex]r[/itex] is never less than C's and never more than A's.
     
  5. Jan 29, 2013 #4
    Ya I see that but isn't it root(GM/a), where a is the semi major axis for object B. Doesn't that mean that the radius in the numerator won't cancel with the semi major axis on the denominator?
    That's where I was worried.
     
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