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Period of Revolutions of a Planet

  1. Dec 8, 2009 #1
    1. The problem statement, all variables and given/known data
    A neutron star has a radius of 10,000 km, and takes a planet 30 days to complete one revolution around the star. When the star collapses, the new radius is 3 km. Find the new period of revolution of the newly formed neutron star.


    2. Relevant equations
    T=2(pi)/omega
    omega = (v cross r)/r^2

    3. The attempt at a solution
    Not sure even how to approach this. i'm so confused and i can't find the velocity to find omega!! this is my first post on PF, any help please?!
     
  2. jcsd
  3. Dec 8, 2009 #2
    Are you doing Kepler's Laws? The relationship between period and radius, one squared and the other cubed. Or is your class studying some other subject? And are you asking about the new period of the planet or the neutron star... something seems missing or stated slightly skewed...
     
  4. Dec 8, 2009 #3
    We're doing oscillatory motion and rotational motion and angular momentum. Oh and I'm assuming that he was asking for the period of the planet, the period of the star wouldn't make sense.
     
  5. Dec 8, 2009 #4

    mgb_phys

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    The collapse has no effect on the orbit of the planet (that will probably be the next question)
    For the star you just need conservation of angular momentum.
    Whats the angular momentum of a uniform sphere?
     
  6. Dec 8, 2009 #5
    Have you done the conservation of angular momentum?
     
  7. Dec 8, 2009 #6
    It's just 2pi/T. So would it be 2pi/2592000?
     
  8. Dec 8, 2009 #7

    mgb_phys

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    Angular momentum is = I * rotation rate
    Where I depends on the shape but for a sphere it is proportional to r^2
    (http://en.wikipedia.org/wiki/List_of_moments_of_inertia)

    So if you make r half as much the moment of intertia becomes 4x smaller, since the angular momentum is the same then it must go 4x faster.

    How much is r reduced in this case?
     
  9. Dec 8, 2009 #8
    Not following...

    From your link I= (2mr^2)/5 <----assuming you mean I= moment of inertia. But it would make more sense to me that I=r^2. So what do I do now with the rotational rate...? I'm so confused =X
     
  10. Dec 9, 2009 #9
    Ok I think I got it!!!

    I(initial)*Omega(initial)=I(final)*Omega(final)

    So I(initial)*(2pi/T(initial)) = I(final)*(2pi/T(final))

    If the distribution is symmetric then kMR(initial)^2*(I(initial))=kMR(final)^2*(I(final))

    So the kMR would cancel out on each side, so it would have Ri^2*(2pi/Ti)=Rf^2*(2pi/Tf)

    Divide 2pi/Ti on each side and then divide by Ri. Multiply Tf on the other side and you get:

    Tf = (Rf/Ri)^2 * Ti

    Plug in and solve.

    I feel smart now =)
     
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