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Thread: Q: How to calculate internal pressure of a planetoid?

  1. #1
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    Q: How to calculate internal pressure of a planetoid?

    How does one calculate the pressure within a massive body at a certain depth? I've tried to look up some information on this and found this entry for "Lithostatic pressure" on wikipedia:

    http://en.wikipedia.org/wiki/Lithostatic_pressure

    Unfortunately, I do not understand the terms in that article well enough to apply the formula it provides.

    Would anyone in this forum be willing to elaborate on relationship or provide links to other sources? How could this pressure be calculate for a simple example... the pressure at a specified depth for a body with a uniform density throughout?

  2. #2
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    With uniform density, the equation is:

    P=(density)(depth)g

    Where (density) is the density of the planetoid in kg/m3
    (depth) is depth in meters
    and g is the acceleration due to gravity of the planetoid.

    Say a density of 2500 kg/m3 and a depth of 3km on a planetoid with an acceleration due to gravity of 1m/s2

    that gives a pressure of of 7.5 million Pa

    Most densitys are given in g/cc instead of kg/m3, so multiply by 1000 in that case

  3. #3
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    On a planetary scale, neither density (ρ) nor g would be constant. The value of g, if I got my algebra correct, would be:

    g(r) = (4/3)*π ρ r

    Note that it, too, depends on ρ.

    This means that the integral on Wikipedia would have to be rearranged so g is inside the integral. Assuming ρ is constant, you can pull it out of the integral. Just remember z = R0 - r.

    I did some planetary hydrostatics in one of my introductory fluids class. Alas, it was about 40 years ago and I remember little of it. I'll dig through the book and find out what the equation should be.
    Information about American English usage here and here. Floating point issues? Please read this before posting.

  4. #4
    Quote Originally Posted by swampyankee View Post
    On a planetary scale, neither density (ρ) nor g would be constant. The value of g, if I got my algebra correct, would be:

    g(r) = (4/3)*π ρ r

    Note that it, too, depends on ρ.

    This means that the integral on Wikipedia would have to be rearranged so g is inside the integral. Assuming ρ is constant, you can pull it out of the integral. Just remember z = R0 - r.

    I did some planetary hydrostatics in one of my introductory fluids class. Alas, it was about 40 years ago and I remember little of it. I'll dig through the book and find out what the equation should be.

    I have heard it stated that in an extremely rough approximation----> ideal gas law may provide some quick and dirty answers---->

    the product of Pressure and Volume is proportional to Temperature----I apologize to the purists and to the "real astronomers" in the forum but in a pinch ---if you were to look upon the ideal gas law ---as opposed to the more accurate hydrostatic approach---you may argue on the grounds of Freshman level calculus that there is a "rough" similarity in terms----

    I am sure I will have my detractors--

  5. #5
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    Quote Originally Posted by John Jaksich View Post
    I have heard it stated that in an extremely rough approximation----> ideal gas law may provide some quick and dirty answers---->


    I am sure I will have my detractors--
    We are looking at solid or liquid, where compressibility is small compared to pressure (and thus can be neglected) and stratification can also be neglected. In which case, the gravitational acceleration is proportional to distance from centre, and central pressure is one half the product of density, surface acceleration and radius.

  6. #6
    Quote Originally Posted by chornedsnorkack View Post
    We are looking at solid or liquid, where compressibility is small compared to pressure (and thus can be neglected) and stratification can also be neglected. In which case, the gravitational acceleration is proportional to distance from centre, and central pressure is one half the product of density, surface acceleration and radius.
    May i be so bold as to ask you what the exact equation looks like?---or if it's posted what it is?

  7. #7
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    Quote Originally Posted by chornedsnorkack View Post
    central pressure is one half the product of density, surface acceleration and radius.
    surface acceleration? What about planets that have no surface, such as gas giants?

  8. #8
    Baric---here is a differential equation describing pressure variation with height---I hope it is understandable?

    I retrieved the equation from text Fundamental Astronomy-5th ed

    Karttunen, et al. editors

    Springer Publications 2007


    Differential equation for gas pressure varying with height:

    dP / P = -g ( μ / k T) dh




    k----> Boltzman's constant
    T----> Temperature in K
    μ----> mass of 1 molecule
    g-----> acceleration of gravity
    h-----> height
    P-----> Pressure

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