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Orbital period of two moons in the same orbit

  1. Aug 24, 2014 #1
    1. The problem statement, all variables and given/known data

    Two identical moons of mass m maintain opposite positions in the same circular orbit of radius R around a planet of mass M. Find T2 the square of the orbital period.


    2. Relevant equations
    T2=(4*pi2*R3)/ ( G*M )







    3. The attempt at a solution
    Hi there, well so I tried just inputting that equation for the square of the orbital period as the anwser .And it marked me wrong.my logic behind that was well if one moon orbits with that period.And the two moons are opposite rather each other then they both are going top have the same orbital period.So I guess it would be the right anwser. But as I've said it's not.So I'm puzzled at what's wrong. Did I somehow miss read the question.or am I thinking of a wrong orbital period or something.could someone explain why my anwser is incorrect.And guide me along the right path , so I could improve my understanding of physics. Any help will be greatly appreciated. :)


    P.S. I don't know if it will work but here's the image I got with the problem. orbits-finding-error-find-trinary-period-multiple-guess-1.png
     
  2. jcsd
  3. Aug 24, 2014 #2

    BvU

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    Hello face, and welcome to PF.
    I wonder where you picked up the relevant formula you quote. I don't see m appearing in there (or 2m in this particular case). Perhaps you want to start your search with a few more fundamental equations.
     
  4. Aug 24, 2014 #3

    ehild

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    You cannot neglect the gravitational force from the other moon.


    ehild
     
    Last edited: Aug 24, 2014
  5. Aug 24, 2014 #4
    The formula is Kepler's 3rd law. And the big M there is the mass of the planet.
     
  6. Aug 24, 2014 #5
    I did think of that, but then dissmissed because I thought that newton's third law would cause them to cancel out.
    So in this case, the gravitational force exerted on one of the moons by the other is: (G*M*m)/(2*R). right? . But I have no idea how it will contribute to the period of the moon
     
  7. Aug 24, 2014 #6

    Orodruin

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    Yes, but Kepler's third law is only valid in a system where all masses except the central body are negligible. Since we are given the masses of the moons, this would not necessarily be the case (also, otherwise there really is no point in doing the problem with two moons).

    Do you know of some kinematical relations that have to hold true in a circular orbit?
     
  8. Aug 24, 2014 #7

    ehild

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    There is a third party, the other moon. You cannot apply Kepler's third law if the masses of the moons are comparable to the mass of the planet. Considering one of the moons, it is attracted both by the planet and the other moon. What is the total force it experiences? That is the centripetal force keeping it on circular orbit.

    ehild
     
  9. Aug 24, 2014 #8

    Orodruin

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    The forces of the moons on the planet cancel out. However, the forces of the moons on each other contribute to their centripetal acceleration ...
     
  10. Aug 24, 2014 #9
    Mhmm, So I calculated the centripetal acceleration to be: (G*M)/(R^2) + ( G*m )/((2*R)^2). am I right? and also by which equation could I relate the centripetal acceleration to the period ?

    p.s. sorry if some of what I'm writing doesn't make much sense , It's really late at night and I'm really tired. :)
     
  11. Aug 24, 2014 #10

    ehild

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    It is correct, and how is the centripetal acceleration acprelated to radius R and angular velocity ω?

    ehild
     
  12. Aug 24, 2014 #11
    since centripetal accelration is : v^2 / R. Where v = w*R. we get ((w*R)^2)/R . and that simplified is w*R^2. and angular velocity = (pi*2*R)/T since it is distance divided by time. putting all of that into one equatin and solving for T we get : T= (4*pi*R^(3/2))/( sqrt( G*( 4*M + n ) ) ). Then to get the anwser we square it .which we finally arrive at :( 16*pi^2*R^3)/( G*(4*M + m) ) . And WOOOAHOOOOAHAHHAHAOOOO!!!!! it's right . I've made a mistake in the final working above but wooooo I got it right :) . Thank you so much to all of you :)
     
  13. Aug 24, 2014 #12

    ehild

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    It is w^2R.


    Why did you take the square root and square again? But the result is correct at the end.

    ehild
     
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