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Hydrostatic equilibrium in sphere

  1. Jul 19, 2012 #1
    I want to derive a formula for pressure at depth in a constant density planet, which sounds pretty simple.

    Setting up a force balance,

    d(4∏*r2P) = -4/3∏r3*ρ*4∏ρ*r2*dr*G/r2

    I'm too lazy to write down a thorough derivation, but by pretty simple calculus and integrating from the planet surface (where P~0), I get:

    r2P = ∏/3*ρ*ρ*(R4-r4)

    Then dividing by the surface area I get something like P = C *(R4/r2-r2)

    At the center of the planet, this goes to infinity. Every peer reviewed paper etc. has an equation which is similar near R (the radius) but very much smaller and flatter approaching r=0. There's no asymptote.

    Their version is easy to get assuming dP/dr=ρ*g. But this implies equal areas (a cylinder or something) on top and bottom. I can't see how they can neglect the surface area change. I tried my same derivation as a balance of forces in the z direction (z=r sin(θ)) and got the same thing.

    So obviously, I am misunderstanding something basic about the problem. Is pressure just defined in a different way for some reason? Even if you posit a tiny cube at the center of the planet, it's not just the column above which presses down. It's the whole cone, and the area will not scale with volume as it will for a column.

    So...what am I not getting?

    Edit: Thinking about it, there is some net effect upwards from lateral forces in the shell. Is assuming equal areas just an approximation for that effect? Or is the common answer really the rigorous solution?
    Last edited: Jul 19, 2012
  2. jcsd
  3. Jul 19, 2012 #2


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    If you are counting the mass of a given incremental shell as having to be entirely supported by the next incremental shell down then you may be ignoring the supporting force of the shell on itself due to curvature.
  4. Jul 19, 2012 #3
    So I've realized. But can the lateral contributions of the shells be proved exactly to negate the area effect? Seems pretty convenient at first glance.
  5. Jul 19, 2012 #4
    gravity ~ mass
    mass ~ diameter of earth

    [f(sphere_Pressure(massUpper)) - f(sphere_Pressure(massLower))] * gravitational_acceleration-earth f(diameter) = pressure

    edited: (gravity -> gravitational_acceleration-earth f(diameter))

    formula = (infinite sum of) [(delta)mass(as a function of (diameter_earth_upper)) * acceleration (as a function of (diameter_earth_upper))]
    Last edited: Jul 19, 2012
  6. Jul 20, 2012 #5


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    G/r2? The gravity at radius r inside a uniform solid sphere is proportional to r. But I didn't understand the rest of that equation, so maybe that isn't the problem.
    The gravitational force on an element dAdr of the shell at radius r is 4Gρ2π(r/3)dAdr. The pressure that exerts on the layer below is 4Gρ2π(r/3)dr. The pressure at radius r is the sum of these pressures from there upwards, 2Gρ2π(R2-r2)/3.
  7. Jul 20, 2012 #6
    I've researched and found relevant papers.

    Does the Surface Pressure Equal the Weight per Unit Area of a Hydrostatic Atmosphere?
    Peter R. Bannon, Craig H. Bishop, and James B. Kerr

    General relationships between pressure, weight and mass of a hydrostatic fluid
    Maarten H.P Ambaum

    To extremely summarize, the answer to the first paper's title is no in non-Cartesian geometries.
  8. Jul 20, 2012 #7


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  9. Jul 20, 2012 #8
    Below is a write-up I made on the correct derivation. The lateral pressure contribution cancels out conveniently with the area effect, making the usually used equation valid.

  10. Jul 20, 2012 #9


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    The 'lateral pressure' issue is a bit of a confusion. Your analysis is fine, using forces rather than pressure, and finding that the tangential forces at the sides cancel the concentration effect of the difference in areas above and below. But you can work quite validly entirely in terms of pressure, as I did, and get that cancellation automatically.
    You can see the same going on if you consider the forces and pressures in a closed region (at constant gravity) that is not uniform in horizontal section.
  11. Jul 20, 2012 #10
    If you start from the equation of hydrostatic equilibrium, grad P=-rho*g, sure. I haven't seen a rigorous proof of that though, except by assuming cylindrical or Cartesian geometry in the beginning.

    That said, for how convenient the proof turns out to be, it seems like there may be a profound (not geometry-based) proof of grad P=-rho*g. I've just never seen one.

    Edit: The point about a constant pressure situation or one without gravity is one I'd never thought about. I guess it does make sense geometrically, that lateral pressure would cancel out any area effect you impose by considering a geometry.
    Last edited: Jul 20, 2012
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