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Homework Help: Two stars orbiting a common center

  1. May 8, 2014 #1
    1. The problem statement, all variables and given/known data
    Two stars, each of mass M, move in a circular orbit of radius R around their common center of mass C.
    a) what is the gravitational force on each star?

    b) what is the period of each star's orbit?

    c) what is the total mechanical energy (potential + kinetic) of this system?

    2. Relevant equations

    [itex]F = \frac{GMm}{r^2} [/itex]

    [itex] a_R = \frac{v^2}{r} [/itex]

    [itex] v = \frac{2piR}{T} [/itex]

    3. The attempt at a solution

    a) [itex]F = \frac{GCM}{R^2} [/itex]

    b) [itex] \frac{GCM}{R^2} = \frac{Mv^2}{R}[/itex]

    [itex] \frac{GC}{R} = v^2[/itex]

    [itex] \frac{GC}{R} = (\frac{2piR}{T})^2[/itex]

    [itex] \frac{GC}{R} = \frac{4pi^2R^2}{T^2}[/itex]

    [itex] GC= \frac{4pi^2R^3}{T^2}[/itex]

    [itex] \frac{GC}{4pi^2R^3}= \frac{1}{T^2}[/itex]

    [itex] \frac{4pi^2R^3}{GC}= T^2[/itex]

    [itex] T = (\frac{4pi^2R^3}{GC})^\frac{1}{2}[/itex]

    [itex] E = \frac{2GCM}{R^2} + \frac{2v^2}{R} [/itex]

    Am i doing these right?
    Last edited: May 9, 2014
  2. jcsd
  3. May 8, 2014 #2
    The thread title says two planets around a star. The problem says two planets around each other. The questions ask about orbits of stars. Make up your mind will you?
  4. May 9, 2014 #3
    oops just edited it and fixed it. It's two starts orbiting around C. Picture C as a point and one star is a distance R to the left of it and the other star is a distance R to the right of it
  5. May 9, 2014 #4


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    What do you mean on GC?

  6. May 9, 2014 #5
    G is the gravitational constant and C is the center of mass
  7. May 9, 2014 #6
    OH. I see the problem now. Give me a second I will fix my equation
  8. May 9, 2014 #7
    ##C## initially was just a label of a point, the centre of mass. It does not mean anything else and it should not enter Newton's gravitation law as if it were mass. The only mass you have is ##M##, it is the mass of each star.

    The radius ##R## is distance between a star and ##C##, not between the two stars.

    Take the above into account and you should be able to solve this problem.
  9. May 9, 2014 #8
    Nevermind I think I have confused myself. I do see now that i cant use C as a mass that they are orbiting around but I don't know where to go from here.
  10. May 9, 2014 #9
    [itex] F = \frac{GM^2}{4R^2} [/itex]

    is that correct for the force on each star?
  11. May 9, 2014 #10
    Yes, that is correct.
  12. May 9, 2014 #11
    so the starting equation to find T would be:

    [itex] \frac{GM^2}{4R^2} = \frac{Mv^2}{R} [/itex]
  13. May 9, 2014 #12


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  14. May 9, 2014 #13


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    Staff: Mentor

    [note: I have modified the thread title to reflect the clarified problem statement -- gneill]
  15. May 9, 2014 #14
    and for the last part E = K + U
    [itex] \frac{GM^2}{4R^2} + \frac{Mv^2}{R} + \frac{GM^2}{4R^2} + \frac{Mv^2}{R} [/itex]

    [itex] E = \frac{GM^2}{2R^2} + \frac{2Mv^2}{R} [/itex]
  16. May 9, 2014 #15


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    Do you know how the potential energy and kinetic energy are defined?
    Are they forces? Or something else?

  17. May 9, 2014 #16


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    what dimensions does [itex] \frac{GM^{2}}{R^{2}}[/itex] have?
    let's see...
    [itex] \frac{Nm^{2}}{kg^{2}}|_{G} \times kg^{2}|_{M^2} \times \frac{1}{m^{2}}|_{1/R^{2}}[/itex]
    I don't see energy..... (Nm)

    Neither for [itex]\frac{Mu^{2}}{R}[/itex] which obviously has force units.... kg m/s^2
  18. May 9, 2014 #17
    oh right they are not forces the force of gravity would be the gravitational potential energy times the mass i think. so remove one M from the force of gravity equation and it would be the potential energy? and the kinetic energy is (1/2)mv^2
    so for one of the stars:

    [itex] U =\frac{GM}{4R^2} [/itex]

    [itex] K = \frac{1}{2}mv^2 = \frac{1}{2}M\frac{4pi^2R^2}{T^2} = \frac{2Mpi^2R^2}{T^2}[/itex]
  19. May 9, 2014 #18


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    Let's check it again :P
    [itex] \frac{Nm^2}{kg^{2}} kg \frac{1}{m^{2}}\ne Nm=J[/itex]
    By the previous (Post 16) dimensional analysis I gave you, you can see you are missing a [lenght] power on the nominator for the potential energy....
    Also this can be seen by the fact that the force is (- gradient of a potential). So since forces are 1/r^2 how should the potential energy be?
    by [itex]m[/itex] I mean meters (because I am trying to show that you are dimensionally wrong, and what you have is not energy, energy is [itex]J=N*m[/itex])
    Last edited: May 9, 2014
  20. May 9, 2014 #19


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    wrong... How the potential energy is defined? Is not it connected to work somehow?

  21. May 9, 2014 #20


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    Force is the negative gradient of the potential energy.

  22. May 9, 2014 #21


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    I corrected it, although I didn't care about the +/- sign since dimensionally it wouldn't play a role....since the OP shows some weakness in seeing the dimensions of what he's writting.
  23. May 9, 2014 #22
    [itex] \frac{M^2G}{R} [/itex]

    this would make the units fit and it makes sense because if you look at U as mgy since g is for earth this version of it would be [itex] \frac{GM}{R^2}[/itex] m would be M and y would be R so mgy would be [itex] \frac{M^2G}{R} [/itex]
  24. May 9, 2014 #23


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    That's fine (for the potential)... and the result is general, every force law which goes as ~1/r2 corresponds to a potential with ~1/r behavior.
  25. May 9, 2014 #24
    and my result for kinetic energy was also wrong?
  26. May 9, 2014 #25


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    If the stars move on circles (fixed R), then your kinetic energy result is also correct (I guess)...
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