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Colliding Binary Star (GPE Problem)

  1. Feb 15, 2013 #1
    I was looking over my old physics course problems, and I can't figure out how I'm doing this one wrong.

    1. The problem statement, all variables and given/known data
    Two identical stars, each having mass and radius M=2*10^29 kg and R = 7 *10^8 m are initially at rest in outer space. Their initial separation (between centers) is the same as the distance between our sun and the earth, D = 1.5*10^11 m. Their gravitational interaction causes the stars to be pulled toward one another. Find the speed of the stars just before they collide, i.e. when their centers are a distance 2R apart.


    2. Relevant equations
    [itex]GPE = \frac{- G m_1 m_2}{r}[/itex]
    [itex] KE = \frac{m v^2}{2}[/itex]

    3. The attempt at a solution
    I tried just doing conservation of energy, i.e.

    [itex]GPE + GPE = GPE + GPE + KE + KE[/itex], or more explicitly:
    [itex]\frac{-G M^2}{D} + \frac{-G M^2}{D} = \frac{-G M^2}{2R} + \frac{-G M^2}{2R} + \frac{1}{2} M v^2 + \frac{1}{2} M v^2[/itex]

    However solving this does NOT get the right answer, which is 9.7*10^4 m/s.

    Can anyone point out what I'm doing wrong? I can't find the flaw in my logic... Does it have to do with where I'm setting zero potential energy? I tried accounting for this by doing the problem another way:
    [itex] \Delta GPE = \Delta KE [/itex]
    But that seems to be equivalent to what I did above.
     
  2. jcsd
  3. Feb 15, 2013 #2

    Dick

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    Science Advisor
    Homework Helper

    You are overcounting the potential energy. The gravitational potential energy of two masses of mass m separated by a distance r is -G*m*m/r. It's not twice that. You are counting the same thing twice.
     
  4. Feb 15, 2013 #3
    *facepalm*. Thank you! I feel very silly now.
     
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