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Pressure inside a liquid planet in rotation

  1. Oct 20, 2013 #1
    Hi,

    I would like to calculate the pressure inside a liquid planet in rotation. How can I do ? Pressure depend of depth under gravity but it depend of rotational speed too. Is it gravity pressure less centripetal pressure ?

    Is it [itex]\frac{1}{2}ρω^2r^2 - ρgh[/itex] ?
     
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  3. Jan 16, 2014 #2

    Simon Bridge

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    No - much more complicated than that.

    To start with the gravity contribution needs to use ##GMm/r^2##

    if R is the radius of the planet/sphere, then ##g=GM/R^2##
    ##M=\frac{4}{3}\pi R^3 \rho##

    There would also be a coriolis contribution to the rotation though - so different layers of liquid could have different angular speeds. You could probably approximate small liquid spheres as "rigidly" rotating as in:
    http://www.physics.princeton.edu/~mcdonald/examples/bernoulli_rot.pdf
    http://en.wikipedia.org/wiki/Centrifugal_force_(rotating_reference_frame [Broken]

    You have to account for the different geometries of the forces - the centrifugal effect will follow a cylindrical geometry and the gravitational one is spherical.

    Then there's equilibrium to consider:
    High spins produce an oblate spheroid etc.

    You sure you want to pursue this?
     
    Last edited by a moderator: May 6, 2017
  4. Jan 16, 2014 #3

    D H

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    It's even more complicated than you make it out to be.

    That's for two point masses, or two objects with a spherically symmetric density. The planet in question is spinning. It's not going to be spherically symmetric.

    And it is not going to have a constant density.
     
  5. Jan 16, 2014 #4

    Simon Bridge

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    Good point - you'll notice that I did not say that was the solution, I said that the solution has to use that relation.
    Perhaps I should have been clearer?

    The gravity, alone, should exhibit spherical symmetry - i.e. we'd want to use spherical polar coordinates - or maybe cylindrical ... it's nasty at first glance - but the resulting object, already indicated, probably won't be a sphere. iirc we'd normally add the centrifugal effect as a perturbation on the overall shape of a planet sized body?
    I have a feeling OP will want a stronger effect.

    IRL the liquid won't be a constant density ... but the OP has a history of making approximations with incompressible liquids. You know - the usual classical idealizations.

    I don't want to do a lot of the math without hearing back from OP.
    More recent posts are using centrifugal gravity in cylindrical containers.
     
  6. Jan 17, 2014 #5

    Dale

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    From the symmetry you know that it is going to be an oblate ellipsoid. The different degrees of oblateness will correspond to different rotational speeds for a given body of liquid. But I cannot think of an easy way to calculate the rotational speed from a given oblateness or vice versa.
     
  7. Jan 17, 2014 #6

    Dale

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  8. Jan 17, 2014 #7

    Simon Bridge

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    iirc: there's is no general way to do this - but it is possible to come up with rules of thumb for classes of objects. It's been too long for me to recall now though.

    I think Gh778 is more interested in "antigravity" possibilities - part of his quest for perpetual motion - so there is a limit to the amount of math I'll do before the details come out.
     
  9. Jan 17, 2014 #8
    The pressure in the center is equal to any radial unit column whether it be polar, equitorial or in between. Gravity is a spherically centered accelaration. Centrifugal acceleration is cylindrically centered with a magnitude proportion to the cosine of the latitude. I think a first order approximation can be made rather simply.
     
  10. Jan 17, 2014 #9

    BruceW

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    a first-order approximation is not exactly what Gh778 was looking for. But yeah, that's a good idea to make the problem a bit simpler. Also, I'm not sure what you mean about pressure in center being equal to any radial unit column?
     
  11. Jan 17, 2014 #10

    BruceW

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    if the matter is not spherical, then the gravity won't be either. unless we're talking about first order perturbation of the matter. i.e. we set gravity as if the matter was spherically symmetric, and then do a perturbation on the matter (leaving gravity as 'zeroth order').
     
  12. Jan 17, 2014 #11

    BruceW

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    haha, awesome. I wish I was on a quest.
     
  13. Jan 17, 2014 #12
    The pressure at any depth is equal to the weight of the column above it. For example the pressure of water at 10 feet of depth is 624 pounds per square foot. In this case variations in g due to altitude above the center of the earth and due to being within the mass was ignored.

    For this OP problem g decreases with depth for two reasons. The inverse square law wrt distance from the center of gravity of the earth and because only the mass below that element of the column is the cause of gravity and this mass decreases with the cube of the radius. The mass of the spherical shell outside that column element cannot contribute to g (it cancels out, there is no g inside an empty sphere of any mass or any size.) A variation of this is often given as a problem in physics courses.
     
    Last edited: Jan 17, 2014
  14. Jan 17, 2014 #13

    Dale

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    True, but it isn't a sphere. It is an ellipsoid.
     
  15. Jan 17, 2014 #14
    You can't even assume that the density of the liquid will remain unchanged.
    And the liquid will be flowing - so viscosity will affect the exact shape.

    Ignoring those affects, it will be the same as a column of the liquid along the rotational axis.
    The affect of gravity is proportional to the distance from the center. So compute the gravitational force at the pole then use the density of the liquid to compute the rate at which the pressure builds as you descend. That rate times the radius is what you would then integrate over the descent to the center.
     
  16. Jan 17, 2014 #15

    D H

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    No, you can't. It is an ellipsoid. You are implicitly assuming it's a sphere.
     
  17. Jan 17, 2014 #16
    Actually, I'm counting on it being exactly an ellipsoid. But I avoided part of the ellipsoid issue by having you start with the surface gravity at the pole. If you can come up with that number, the gravity at depth will still be proportional to the distance to the center.

    As you descend from one of the poles, the remaining gravity could be computed by integrating the gravitational effects of circular sections of the planet along the axis of rotation. What I am doing is comparing the circular cross sections that you integrate when its a sphere to the circular cross sections that you would integrate for an ellipsoid. They are off by a factor, but it's a constant factor through out the integration - so it all stays proportional.
     
  18. Jan 17, 2014 #17

    Simon Bridge

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    Spherical approximations would only hold for slow rotation so the centrifugal effect is small compared with gravity. How slow would the rotation have to be for the oblate deformation to be negligible? Depends right?

    If the liquid ball planet-like object is very big, with a slow rotation, that may give Gh the kind of pressure gradient that he's looking for ... it depends what he wants to know for. He probably only cares about the effect on the pressure along the "equatorial" radius for example.

    DH's link is a pretty good find - it has pretty much the sorts of restrictions Gh778 like to have - i.e. all parts of the liquid have the same angular velocity.

    Everything depends on what Gh wants the information for.
    What effect is he looking for?

    note:
    I keep getting pinged by my use of the term "spherical symmetry"... maybe I got this wrong?
    Something does not have to be an actual sphere to exhibit spherical symmetry right?

    You mean like a self-consistent calculation?
     
  19. Jan 18, 2014 #18

    Nugatory

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    I'd use the word "axial" to describe the symmetry of an ellipsoid (rotational symmetry around one axis), reserve "spherical" for shapes that have rotational symmetry about all axes.
     
  20. Jan 18, 2014 #19

    BruceW

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    yeah, I've seen this problem a few times before and I can't fully remember, but I think it was done by simply setting gravity as zeroth order, and using first order perturbation on the matter (due to the rotation), which gives an oblate spheroid. It's not really self-consistent because we're breaking Newton's laws of gravity. But I think the idea is that since we want only first order in the shape of the planet, we only care about zeroth order of gravity.
     
  21. Jan 18, 2014 #20
    No it isn't because the density is not constant (see Preliminary Reference Earth Model)

    And even with constant density a spherical symmetry might be a bad approximation. Trying to calculate the oblateness of Earth with this assumption leads to the half of the correct result.
     
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