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.

The question that i.m.o. should be allowed on this site, is to ask about a more rigorous treatment of this problem.

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.

Vote up if you think this should be on topic under the new policy, or vote down if you think it should be off topic under the new policy.

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.

The question that i.m.o. should be allowed on this site, is to ask about a more rigorous treatment of this problem.

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.

The question that i.m.o. should be allowed on this site, is to ask about a more rigorous treatment of this problem.