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Calculate the ratio of the radii of the orbits

  1. Jul 12, 2003 #1
    Two objects of masses m and M move in circular orbits that have the same center. The force that gives rise to this motion is a force of attraction between the two objects acting along the line joining them. Use conservation of momentum and N's 2nd Law to show that they move with the same angular speed. Calculate the ratio of the radii of the orbits.

    My attempt:
    Conservation of Momentum

    N's 2nd Law

    For uniform (I assume, because I know no better, rather than show that the motion is uniform - feel free to give me a lesson on that topic) circular motion: F=mv2/r


    So, F=mrw2, for uniform circular motion

    But since v=wr
    mwr + MWR = K

    dP/dt= mrw2 + MRW2=F1+F2=0


    mr(dw/dt)+MR(dW/dt)= mrw2 + MRW2=0
    And you may have thought I was going somewhere with that. But I'm not. No matter what I do to these equations I can't get w=W!
    A little hint would be nice. Thank you.
  2. jcsd
  3. Jul 13, 2003 #2
    Seems no one likes my problem :( - Well, here's another anyway. A billard ball collides head on with a second billard ball that is touching a third billard ball (I'm guessing this means they form a straight line). Assume all collisions are elastic. Describe the motion of the balls.

    Not much I can do with this problem. Since they all have approximately the same mass, p conservation yields
    v1i= v1f+v2f+v3f
    energy conservation yields
    v1i2= v1f2+v2f2+v3f2
    There are three unknowns (I consider v1i a known), but two equations. So at best I can give some sort of relation between the variables, but there appears to be so many possible outcomes that this would be of little use.
    Anyway, I came up with
    through various calculations. Can anyone do any better?
  4. Jul 13, 2003 #3
    Correct. Now since the attraction is mutual, you know F1 = -F2. Now you use a reference frame where the center-of-mass is at rest. This means K=0. And there you are.

    As for the 2nd problem:
    Do you really not know what happens in this case ? If so, you should try a little experiment - it's very instructive. Well, I can help you with the math...
  5. Jul 13, 2003 #4
    You mean that since
    mwr + MWR=0
    that Psystem=0=mVsystem?
    I can agree with that if that means the CM has v=0. I guess I didn't consider that the constant is arbitrary depending on your reference frame.
    But the idea is to show that the angular velocities are equal (ie. w=W).
    And given that K=0 doesn't help that matter.
    w/W = -(MR)/(mr)
    Depending on which direction I define to be negative, either r or R will be negative. If w=W, then w/W=1. Or, |(MR)/mr|=1, which is only true if MR=mr. We don't know that to be true.
    As for the second question, I'm sorry I don't know what happens. I could speculate, but that's not physics. I thought I'd do some math instead. I would appreciate some help with that matter, anyway. If you can describe what happens that would be nice, but I like to see the math behind it all. Actually, I was playing pool today (ironically), but it didn't occur to me do that experiment. I was busy paying attention to whether they really depart at right angles or not. I don't know. I guess they do. I was wondering how one would treat a problem when the velocity of the billard ball is not collinear with the line joining the point of contact and the CM.... Still wondering.
  6. Jul 13, 2003 #5
    Fine. Please show me a more direct way to get to this result. I've done it, but it's excessively complicated I think. Say there are two balls, and one is initially at rest.
    v1i=v1f+ v2f
    v1i2= v1f2+ v2f2
    = (v1f+ v2f)2
    v1f2+ 2v1f*v2f + v2f2= v1f2+ v2f2
    v1f*v2f =0
    so either v_1f=0, v_2f=0 or both. Clearly it can't be both, since that would violate conservation of momentum.
    Using the first two equations,
    v1i2-v1f2= v2f2
    Divide each by each to get,
    multiply by v_2f on each side
    thus, v_1f=0
    Alternatively, I could have multiplied by v_1f on each side. I would have found,
    and v_2f=0
    Unlikely. Strange that the math supports that possibility. My guess is the complicated nature of the solution made the last one feasible. There must be a way to show that v_2f[x=]0. But anyway, that's only the solution for two balls. Could I find the solution where two are at rest by considering the collision which immediately occurs when the second ball begins to move. It would (ideally) be the same as the first collision with v_2f=v_1i, and thus I would expect to see the last ball move with v_3f=v_1i, and the other two balls at rest, in contact. But again, at this point, pure speculation.
    Last edited: Jul 13, 2003
  7. Jul 14, 2003 #6
    1) actio = reactio =>
    F1 + F2 = 0
    mrw^2 + MRW^2 = 0.
    2) conservation of momentum =>
    mrw + MRW = 0

    2) => mrw = -MRW
    2) into 1) -MRWw + MRW^2 = 0
    => -w + W = 0
    => W = w

    into 1) mr + MR = 0
    => mr = -MR
    => r/R = -M/m
  8. Jul 14, 2003 #7
    Right. And that's all that math can do for you here. Now, we know that a collison happens, and that means, ball #2 gets some momentum, and from the math we know that it gets all. And that's it.
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