(Edit May 2012-- my thoughts on this matter have now changed significantly-- I'm still convinced that the earth "feels" a downward push from the wing of an aircraft in flight, equal in magnitude to the weight of the aircraft, which is also equal to the upward gravitational attraction that the aircraft exerts on the earth, but I no longer believe that this downward force need involve any specific amount of downward momentum of the air (downwash). For more, see posts 58, 61, and 72. End edit.)
Originally Posted by HerkS
Momentum is conserved in a closed system - so if you fly your helicopter inside a box there is no net force on the system. Not so in an open system - an aircraft in flight is an open system
PS we would reach the same conclusions as I am suggesting, if we just considered in airplane in a closed box. Feel free to make the box as large as you want. Naturally there is a scale under the box. How does the airplanes weight get transmitted through the bottom of the box to the scale? By means of a nebulous pressure increase due to the kinetic stirring-up of the air by the wing, that is pressing equally on all sides of the box, pressing up on the top of the box just as hard as it is pressing down on the bottom of the box? That couldn't work. Or by means of the momentum of the downwash pushing down on the bottom of the box?
If the airplane pushes over into zero-G flight inside the box, does the box get lighter, as measured by the scale under the box? I say it does. Then the box gets heavier during the pull-out that follows.