Angular Momentum of a Planet

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  • Thread starter Isaac0427
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  • #1
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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!
 
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Answers and Replies

  • #2
I=ICM+mro2
ICM(sphere)=2mrp2/5
I=2mrp2/5 + mro2)
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.
 
  • #3
Chandra Prayaga
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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.
 
  • #4
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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.
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.
 
  • #5
Chandra Prayaga
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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?
 
  • #6
It is a spherical planet rotating around a parallel axis a distance ro from its center of mass.
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.
 
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  • #7
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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?
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.
 
  • #8
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There's no rotation around the planet's axis.
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.
 
  • #9
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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.
Thank you!
 
  • #10
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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.
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?
 
  • #11
rrthakur75
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!
Why you are including I (Moment of Inertia)?
 
  • #12
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For number 2 in the original post the omegas are the same. I think you are talking about a different situation.
No that's exactly what I am talking about (wp=wo).

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.
That makes no sense. The paralle axis theorem includes the rotation around the planet's axis.
 
  • #13
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That makes no sense. The paralle axis theorem includes the rotation around the planet's axis.
Then what is the planet's moment of inertia about the central axis?
 
  • #14
jbriggs444
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Then what is the planet's moment of inertia about the central axis?
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.
 
  • #15
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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.
 
  • #16
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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
 
  • #17
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I think that by Steiner rule 2) is correct .
if the planet is tidally locked

1) should be correct in case if planet by assumed as matter point with no own inertia moment!
or non-rotating
 

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