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Particles in a circular orbit

  1. Oct 17, 2007 #1
    1. The problem statement, all variables and given/known data
    If two particles are orbiting each other in a circular orbit with a period of tau, show that if you stop them both in an instant, they will fall into each other after a time tau/ 4 sqrt2 .


    2. Relevant equations



    3. The attempt at a solution
    So, we know that they must be along a diameter and we can use f = m v^2/ r_0 to express f in terms of tau and then V in terms of tau. I get:

    [tex] f = m \tau^2 4 \pi^2 r_0 [/tex]

    V = -f * r

    But don't we know that V = -GM1M2/r so why is my expression for V different than that?
     
  2. jcsd
  3. Oct 18, 2007 #2

    nrqed

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    I don't quite understand the statement of the problem. "If you stop them in an instant"?? That means that their velocity is set to zero?? But then they start falling toward one another right away! Or do you mean how long before they hit each other??
     
  4. Oct 18, 2007 #3
    Yes I mean how long before they hit each other.
     
  5. Oct 18, 2007 #4

    nrqed

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    I don't know what your equation V = -f r comes from.

    You only need to find the initial distance before starting the problem. Just set the centripetal force equal to GmM/r^2. Becareful about the fact that "r" here is different than the r_0 in your centripetal force equation. (I am assuming they were moving along a circular orbit around their common center of mass, right?)
     
  6. Oct 18, 2007 #5
    You're right. I think that expression for V is wrong.

    Do you agree that tau = v 2 pi r_0?

    Do you agree that r_0 = tau^2/(8 pi^2 G) ?

    So now I just need to integrate something w.r.t r from r_0 to 0 and I am trying to figure out what the best way to do that is?

    Should I get the equations of motion from the EL equations or what?
     
  7. Oct 18, 2007 #6

    nrqed

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    No, I don't agree.

    [tex] v = \frac{2 \pi R}{\tau} = \frac{ \pi r_0}{\tau} [/tex]

    where R is the radius of the circle and is equal to half the initial distance between the objects.
     
  8. Oct 18, 2007 #7
    Let's try that again.

    Do you agree that R^3 = G/(2 pi^2 tau^2)

    with your definition of R?
     
  9. Oct 18, 2007 #8

    nrqed

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    No. The units don't work in your expression. I get R^3 = MG tau^2/(2 pi^2) with R being the distance between the two objects.
     
  10. Oct 18, 2007 #9
    I get R^3 = MG tau^2/(4 pi^2).

    I guess we should assume that the masses are the same even though the problem does not say so explicitly. Otherwise, one would accelerate more than the other and a circular orbit could not be maintained.

    So, now my question is still:

    I just need to integrate something w.r.t r from R to 0 and I am trying to figure out what the best way to do that is?

    Should I get the equations of motion from the EL equations or what?
     
  11. Nov 7, 2007 #10

    D H

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    Note: The OP may no longer be interested in this problem (its a nearly month old), but this thread has been referenced elsewhere.

    There is an easy way to solve this problem. Stopping the particles changes the semi-major axis from [itex]a[/itex] to [itex]a_0[/itex], where the subscript zero denotes the semi-major axis after stopping the particles. Using Kepler's third third law,

    [tex] \left(\frac{T_0}{\tau}\right)^2 = \left(\frac{a_0}{a}\right)^3[/tex]

    or

    [tex] T_0 = \left(\frac{a_0}{a}\right)^{3/2}\tau[/tex]

    where [itex]T_0[/itex] is the period after stopping the particles. Intuitively, stopping the particles halves the semi-major axis, and thus

    [tex]T_0 = \frac{1}{2^{3/2}}\;\tau = \frac{\tau}{2\surd2}[/tex]

    To make this intuitive approach a bit more rigorous, suppose that the attempt to completely stop the two particles doesn't quite work, leaving the two particles with a tiny tangential velocity [itex]\epsilon_v[/itex]. The particles are still in some orbit and are at apofocus. Half an orbit later they will be at perifocus. By the vis-viva equation, the semi-major axis is

    [tex]\frac 1 a = \frac 2 r - \left(\frac{\epsilon_v}{\mu}\right)^2[/tex]

    In the limit [itex]\epsilon\to 0[/itex], the semi-major axis becomes half the initial separation distance. The initial separation distance is the initial circular orbit semi-major axis, so stopping the particles does indeed halve the semi-major axis.

    Again looking at the case where stopping the particles doesn't quite work, the particles reach perifocus half a period after the failed stopping attempt. The half period to perifocus becomes the time to collision in the limit that [itex]\epsilon_v \to 0[/itex]:

    [tex]T_{\text{collision}} = \frac 1 2 T_0 = \frac{\tau}{4\surd2}[/tex]
     
    Last edited: Nov 7, 2007
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