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3 stars in an equilateral triangle

  1. Oct 30, 2008 #1
    Three identical stars of mass M form an equilateral triangle that rotates around the triangle's center as the stars move in a common circle about that center. The triangle has edge length L. What is the speed of the stars?


    All I've been able to come up with is they rotate around the center. I took Kepler's law, T^2 = (4pi^2*r^3)/(GM), and replaced T with 2pi/angvelocity. Then, I replaced angvelocity = velocity/radius, and put the equation for radius (radius = (L*sqrt(3))/3) back in....

    v=sqrt (GM/R) = sqrt (3GM/(L*sqr(3))) - - - the answer given is sqrt (GM/L) .... not sure where I went wrong, because I know that R isn't equal to L. Does anyone have any direction, or do you think this is just a typo in the book's answer key?
     
  2. jcsd
  3. Oct 30, 2008 #2

    tiny-tim

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    Welcome to PF!

    Hi burianek! Welcome to PF! :smile:

    Calculate the forces from each of the two other two stars separately, add them, and then equate to the centripetal force.
     
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