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Period of stars rotating

  1. Oct 26, 2012 #1
    1. The problem statement, all variables and given/known data

    Three stars, each with the mass of our sun (1.9891E31 kg), form an equilateral triangle with sides long. The triangle has to rotate, because otherwise the stars would crash together in the center. What is the period of rotation?

    2. Relevant equations
    F1=GM2/r2
    ƩF=0
    F2=ma
    a=4π2r/T2

    3. The attempt at a solution
    F1=2.639E26
    Because the sum of the forces is equal to zero F1=F2
    2.639E26=(1.9891E31)*a
    a=.000133m/s2=4π2/T2
    T=5.45E8 s

    Do I have to do something else because there's three stars instead of two?
     
  2. jcsd
  3. Oct 26, 2012 #2

    gneill

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    Staff: Mentor

    What is the length of the sides of the triangle?

    It's difficult to judge what calculations you're performing when you only show results; show more of your work.

    Yes, you need to take all the masses into account. Think summation of vectors.
     
  4. Oct 26, 2012 #3
    The length of the sides of the triangle is 1E12m.
    F1=(6.67E-11)(1.9891E30kg)/(1E12)2=2.639E26
    Because it's a equilateral triangle all angles are 60 so I when summing the vectors I made one side at 0o and the other 60o above it so the coordinates of the first side is Fx=2.639E26 Fy=0 and the coordinates of the other side is Fx=2.639E26 cos(60)=1.32E26 Fy=2.639E26 sin(60)=2.29E26.

    the total Fx=3.959E26 and Fy=2.29E26 F=(Fx2+Fy2).5=4.57E26

    a=F/m so a=4.57E26/1.9891E30=2.3E-4
    a=4π2r/T2
    2.3E-4=4π2*1E12/T2
    T=4.14E8

    The correct answer: 3.15E8
     
  5. Oct 26, 2012 #4

    gneill

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    Staff: Mentor

    Ah. Well the radius of the circle around which the stars are rotating is not the same as the length of the triangle sides.
     
  6. Oct 26, 2012 #5
    Yes that was the problem thanks for the help
     
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