Thanks to dmckee for digging up an old deleted question asked by me a long time ago here. Here it is:

>Suppose a plane of mass $M$ flies at constant velocity at a height $h$ above the ground. If $h$ is much larger than the size of the aircraft, what is the pressure increase on the ground as a function of the position relative to the point directly underneath the aircraft?

>I then added a heuristic way to tackle the problem (some of it was in the comments):

>If the velocity field were (on average) to have a radial component that decays like 1/r^2, this amounts to a source, which violates conservation of mass. But you can then imagine a sink nearby, in the limit that they get infinitely close and the sources get infinitively strong, you should get a dipole like velocity field. Then this decays like 1/r^3 at large distances. At the ground we impose zero velocity boundary conditions, the pressure should thus behave like 1/r^6 = 1/(h^2 + d^2)^3 where d is the distance on the ground from the point directly underneath the aircraft.

>Integrating the total pressure over the surface should yield the weight of the aircraft. If we write the pressure as A/(h^2 + d^2)^3, then the integral of 2 pi A d/(h^2 + d^2)^3 over d from 0 to infinity should equal M g. Solving for A then yields the pressure:

>P(d) = 2 Mg h^4/pi 1/(h^2 + d^2)^3

>But all this is on the basis of purely intuitive reasoning.

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