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    Why is the pressure at the surface less than the atmospheric pressure.

    The effect makes sense if the pressure is less near the surface, but mathematically, I don't understand how is that the case.

    I think this equation is relevant to this:

    \frac{\partial p}{\partial n} + \rho g \frac{\partial z}{\partial n} = \rho \frac{V^2}{r_s}

    where V is the velocity, rs is the radius of curvature, and n is the distance from surface.
 
 
 
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Updated: May 17, 2013
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