" 

 (Zapletal Writes ->) Mark, Your last sentence pretty much nails it (in terms of this simplified ""billiard ball"" model...). 

 For an even simpler version, consider an aerofoil moving at constant horizontal velocity through a vacuum, and above a large massive body such as the Earth. The gravity force between Earth and foil pulls them together, and gives them both equal time-rate-of-change of inwards momenta (ie. d(m.V)). Of course, the lighter foil has the higher dV here, so follows a more curved path (roughly parabolic, concave down). Now picture something like a football bouncing up and down ""elastically"" (!) between the two bodies. Each time it hits the Earth and bounces back up, it ""exchanges"" twice its vertical linear momentum with the Earth, and gives it some downward dP (but of very tiny dV). Similarly, each time it hits the underside of the foil and bounces back down, it gives the foil the same upward momentum dP. So answer to your original question is, indeed, YES. End result is that the foil travels along a path a bit like ""mmm"". 

 The bouncing ball in this model is roughly equivalent to Lanchester's ""pillar"" from page 9 of his book, ie.,  ""As a whole, the fluid, in the previous section, does not gain or lose momentum any more than does a cast-iron pillar supporting a load."" . 

 Adding more detail to the model has many more balls bouncing every which way, which add, in Continuum Mechanics terms, an ""ambient pressure"" everywhere. But, for this BB model to work there MUST be a time-averaged increase in the number of these balls bouncing between each other in the zone between the Earth and foil. This ""increased pressure zone"" MUST ALSO travel steadily with the foil, much as the single football does above, or as does Lanchester's ""cast-iron pillar"". (End Zapletal)  "