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B Angular Momentum of a Planet

  1. Jul 12, 2017 #1
    Hi,

    Consider a spherical planet of mass m and radius rp orbiting a star with a circular orbit of radius ro (distance from axis of orbit to the planet's center of mass). The planet has an angular velocity ω. Say we wanted to find the magnitude of the angular momentum of the planet. Going about that two different ways provides two different answers--which one is correct?

    1) p=mv
    v=ωro
    p=mωro
    L=ro×p
    |L|=|ro||p|sin(90)=|ro||p|
    |L|=romωro=mωro2
    Assuming that ω and ro are just magnitudes so you can leave off the absolute values.

    2) I=ICM+mro2
    ICM(sphere)=2mrp2/5
    I=2mrp2/5 + mro2)
    |L|=Iω
    |L|=2mωrp2/5 + mωro2
    Again, assuming ω is a magnitude.

    These two answers are off by 2mωrp2/5. Where's the mistake?

    Thanks in advance!
     
    Last edited by a moderator: Jul 12, 2017
  2. jcsd
  3. Jul 12, 2017 #2
    Why are you using the moment of inertia of a sphere? We are not considering a rotating sphere here if I understand your question correctly.
     
  4. Jul 12, 2017 #3
    In the first calculation, you assumed the planet to be a point particle rotating about the star. This is orbital angular momentum. In the second, you considered the spherical shape of the planet, and gave it a rotation about an axis through its center.
     
  5. Jul 12, 2017 #4
    It is a spherical planet rotating around a parallel axis a distance ro from its center of mass.

    However, many sources have the first result. That's why I am conflicted.
     
  6. Jul 12, 2017 #5
    Some more things have to be clarified here. In (2), you used the same angular velocity for the rotation around the sun and for the rotation about its own diameter. Was that intentional?
     
  7. Jul 12, 2017 #6
    Ok, I see. So it looks like in the second case you treat the planet as a distributed mass and in the first case it is treated as a point mass. The second case should yield a small correction over the first case.

    Generally speaking ##r_{p} << r_{o}## and ##r_{p}^{2}## will be many orders of magnitude smaller than ##r_{o}^{2}## so by considering the planet as a distributed mass, you gain a very small correction over the point mass case.
     
  8. Jul 12, 2017 #7
    There's no rotation around the planet's axis. The I term uses the parallel axis theorem but is the planet's angular momentum around the axis of the star.
     
  9. Jul 12, 2017 #8
    There is a rotation. The general case for a planet with homogeneous density is

    [itex]\left| L \right| = {\textstyle{2 \over 5}}m \cdot \omega _p \cdot r_p^2 + m \cdot \omega _0 \cdot r_0^2[/itex]

    where wp is the angular velocity of the rotation around the planet's axis and wo the angular velocity of the orbit. With wp=0 you get your first equation and with wp=wo (tidal locking) the second.
     
  10. Jul 12, 2017 #9
    Thank you!
     
  11. Jul 12, 2017 #10
    For number 2 in the original post the omegas are the same. I think you are talking about a different situation.

    I did not account for rotation around the planet's axis-- I only accounted for rotation around a central axis, in which I used the parallel axis theorem. I believe that if I did account for the planet's rotational angular momentum I would get
    ##|L|=\frac{2}{5}m\omega_or_p^2+mr_o^2+\frac{2}{5}m\omega_pr_p^2##
    Is that correct?
     
  12. Jul 13, 2017 #11
    Why you are including I (Moment of Inertia)?
     
  13. Jul 13, 2017 #12
    No that's exactly what I am talking about (wp=wo).

    That makes no sense. The paralle axis theorem includes the rotation around the planet's axis.
     
  14. Jul 13, 2017 #13
    Then what is the planet's moment of inertia about the central axis?
     
  15. Jul 14, 2017 #14

    jbriggs444

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    Science Advisor

    If you consider a planet rigidly rotating about a point at the center of the solar system (i.e. tidally locked) then you can use the parallel axis theorem to determine the moment of inertia of that planet about that axis and multiply by its angular velocity to get the angular momentum. There is only one relevant angular velocity, so the result is unambiguous.

    If the planet is not rigidly rotating about that center (i.e. is not tidally locked) but is instead, for instance, maintaining a fixed orientation with respect to the distant stars then the moment of inertia can not be multiplied by the angular velocity to obtain angular momentum. The formula ##L=I\omega## applies for rigid objects. For non-rigid objects, the formula for angular momentum is more complicated. One way of seeing this is to realize that there is no unique relevant angular velocity about the central axis. Different parts of a planet that is not tidally locked have different angular velocities about the center of the solar system.
     
  16. Jul 18, 2017 #15
    I have another thought on the general idea that applies to our very own solar system. It might be out of order but the moon strikes as having very similar momentum patterns in that it has to speed up and slow down as it orbits in the common Earth/Moon orbit around the sun. That really seems to complicate things.
     
  17. Jul 18, 2017 #16
    Hello,
    I think that by Steiner rule 2) is correct .
    1) should be correct in case if planet by assumed as matter point with no own inertia moment!- but you input planet diameter , so it is not.

    Hi
     
  18. Jul 18, 2017 #17
    if the planet is tidally locked

    or non-rotating
     
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