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Homework Help: Triple star system

  1. Mar 16, 2008 #1
    1. The problem statement, all variables and given/known data
    A certain triple-star system consists of two stars, each of mass m = 7.00×1024kg, revolving about a central star of mass M = 1.69×1030kg in the same circular orbit of radius r = 2.00×1011m(see the figure). The two stars are always at opposite ends of a diameter of the circular orbit. What is the period of revolution of the stars in years(yr)?


    2. Relevant equations

    3. The attempt at a solution

    I know that the equation for period of motion is

    T=2*pi*r^3/2 / sqrt (GM)

    I'm just not sure if I can plug in values where M is the aggregate of masses M1 and M2..
    Last edited by a moderator: Apr 23, 2017
  2. jcsd
  3. Mar 16, 2008 #2
    I'm just a high school physics student, (so I could be wrong), but technically, if the stars are always at opposite ends, aren't the periods of the stars equal to each other? Which would mean using that equation and plugging in the 7E1024 and 1.69E1030 be correct?

    Like I said, I could be very very wrong, but this is just my thought on the problem.
  4. Mar 16, 2008 #3


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    Go back to the basics:

    [tex] F = m \frac{v^2}{R} = m \frac{4 \pi^2 R}{T^2} [/tex]
    where here m is the mass of the star that you are focusing on and R is the radius of its orbit.

    Now, the force will be the total force acting on that star so it will be the sum of teh force of gravity due to the other two stars. Can you go on from there?
    Last edited by a moderator: Apr 23, 2017
  5. Mar 16, 2008 #4
    ok so the force of gravity is Fg = (GM1M2)/R^2

    If total force is what you have above, should I set that equal to 2*(GM1M2)/R^2 -- or the sum of Fg of star1 on central star and star 2 on central star?
  6. Mar 16, 2008 #5


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    The equation I wrote is to be applied to one of the star in motion. So the force is the total force due to the other two stars. Note that the other two stars are AT DIFFERENT DISTANCES! So be careful with that
  7. Mar 16, 2008 #6
    but their orbital radius is the same though?
  8. Mar 16, 2008 #7
    i'm still confused, the force is the total force due to the other two stars - what does that mean? and how does gravitational force work into this equation?
  9. Mar 16, 2008 #8


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    Let's say we have stars A, B and C in that order. Star B is at rest. Stars A nd C are moving in a circle. The radius of the circle is the distance between star A and star B, right? (which is the same as between star B and star C) and is equal to 2 x 10^11 m.

    Calculate the total force on star A. It's the sum of the force of star B on star A PLUS the force of star C and star A.

    But I just noticed that the mass of stars A and C is really tiny compared to the central star (thos enumbers don't make much sense since the masses of stars A and C are comparable to the mass of a planet not of a star!)

    Anyway... because the mass of stars A and C is so small compared to mass B, you can neglect the effect of the gravitational pull of star C on star A. In that case, the force of gravity on star A is simply the force of star B on star A only.
  10. Mar 16, 2008 #9
    i think its supposed to be 10^24 and 10^30
  11. Mar 16, 2008 #10
    the total force on star A would be G(mass starB) / (R)^2 + G(mass starC) / (2R)^2 ?
  12. Mar 16, 2008 #11


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    Almost. You still have to multiply by the mass of star A. So it is

    [tex]\frac{G m_A m_B}{R^2} + \frac{G m_A m_C}{(2R)^2} [/tex]

    as I said, it turns out that the second term will be negligible compared to the first since they used such a small mass for "stars" A and C (the masses are actually close to the mass of our planet!).
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