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Binary stars

  1. Oct 24, 2003 #1
    Hello.
    Two binary stars (m1 and m2) orbit each other around their center of mass. They orbit in circles of radii r1 and r2. I am to show that the period is given by T2=4pi2(r1+r2)3/(G(m1+m2))

    The mutual force between m1 and m2 is F=Gm1m2/(r1+r2)2
    Considering m1, the acceleration of m1 is given by a1=Gm2/(r1+r2)2=v12/r1
    Solving for v and using T1=2(pi)r1/v1=T2, I get T2=4pi2r1(r1+r2)2/(Gm2)

    What did I do wrong?
     
  2. jcsd
  3. Oct 24, 2003 #2
    I would reconsider the definition of "center of mass."
     
  4. Oct 24, 2003 #3
    What precisely do you mean by "reconsider?" If I take the CM to be my reference frame (which I do), then m1r1=m2r2. I don't know how this helps. However, I noticed that if I take m2 to be my reference frame, then I get T2=4pi2(r1+r2)3/(Gm2). This is unexpected to me. I thought that omega was independent of reference frame so long as these reference frames have only translational motion wrt each other. Since T is indirectly proportional to omega, T should also be independent of reference frame. I'm clearly making some mistake here, but I can't figure out what it is.
     
  5. Oct 24, 2003 #4
    Stephen, your original approach was exactly on target. You simply didn't finish it.

    Now, remembering that m1r1 = m2r2
    divide numerator by r1 and denominator by m2r2 /m1 giving you
    T2=4(pi)2(r1+r2)2/(Gm1/r2)
    Now multiply numerator and denominator by (r1+r2) giving
    T2=4(pi)2(r1+r2)3/(G(m1r1 /r2+m1)

    But m1r1 /r2 = m2 which is exactly what you wanted.


    But I'm puzzled by your later statement which seemed to suggest that it would be OK to use m2 as a reference frame on the assumption that these frames have only translational motion wrt each other. How can you make this assertion when these frames are rotating about each other?
     
  6. Oct 24, 2003 #5
    The frames would not rotate with respect to each other. If their axes were originally parallel, they would always be parallel. In that way, omega would be the same in either frame.

    Edit:
    I learned it yesterday actually in my physics book. Perhaps my inexperience with the new concept has caused me to use it incorrectly. In the book, it says that if an object rotates in a plane, then it rotates about any axis perpendicular to that plane. And no matter which axis is chosen, the angular velocity omega is the same (however, as they are noninertial frames, linear velocity is affected).

    Edit:
    Thanks for your answer. I would have never gotten that. It still took me ten minutes to finish the problem after I read your answer.
     
    Last edited: Oct 24, 2003
  7. Oct 24, 2003 #6
    That's very interesting. I don't remember coming across that before. I imagine it will prove useful someday, but right now I'm having trouble visualizing it. In particular, does it still hold if you choose an axis that is outside of the objects orbit?


    By the way, the equation you wrote using m2 as the reference frame is Kepler's Second Law, which we learned is only valid to describe the orbit of a body wrt another body which is so much larger that the mass of the smaller body can be ignored.
     
  8. Oct 24, 2003 #7
    No it shouldn't matter where the axis lies, inside or outside of the orbit. I drew a picture to convince myself. The book actually used a rotating object as an example. It's easy to see then how the reference frame must move with the other particles in the object. When I was doodling with orbits it wasn't as clear how the RF should move.
     
  9. Oct 25, 2003 #8
    The parallel axis theorem (having to do with moments of inertia I) states:

    IR=0+MR2=IR=R
     
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