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Consider a plate moving frontally through a gas with velocity +v. A molecule moving vertically towards the plate with velocity -u in the lab frame has a velocity -u-v in the reference frame of the plate and hence bounces back from the front side with the velocity u+v (as the plate is much heavier than the molecule), i.e. in the lab frame the velocity of the molecule is now u+2v. Correspondingly, at the backside of the plate, a molecule with velocity u in the lab frame has a velocity u-v in the reference frame of the plate and thus after reflection a velocity -u+v, i.e. in the lab frame the velocity is now -u+2v.

This means in the lab frame the kinetic energy of the molecules being reflected from the front has increased from m/2*u² to m/2*(u+2v)² =m/2*u² +2*m*u*v + 2*m*v², i.e. an increase of

**+2*m*u*v + 2*m*v²**.

Molecules reflected from the back side on the other hand have changed their energy from m/2*u² to m/2*(-u+2v)² = m/2*u² -2*m*u*v + 2*m*v² i.e. a change of

**-2*m*u*v + 2*m*v²**.

Adding and averaging the two contributions one obtains therefore the average increase of the kinetic energy of the molecules with mass m hitting the plate with velocity v as

**ΔK = 2*m*v²**. (Note: this value will actually be smaller by about factor 1/2 as the molecules won't be reflected straight back but will be scattered into the whole half-space due to the roughness of the plate's surface).

(One can derive this result also from the energy and momentum conservation equations applied an elastic collision, with the same result if one assumes the plate mass as large compared to the molecular mass).

Let's apply this result to air (although the latter is not strictly inviscid in the above defined sense):

if the plate has a cross section of 1m² and a velocity of 10 m/sec it will collide with 3*10^26 molecules per second, which, assuming a molecule mass of 4.5*10^-26 kg, amounts to a total mass of M=13 kg per second. Replacing m with M in the above result for the energy gain of a molecule, one sees that the kinetic energy gained by the air molecules per second corresponds to the kinetic energy of a plate of about 26 kg (taking the above mentioned correction into account), i.e. a plate of with a mass of 26 kg, a cross section of 1m² and a velocity of 10m/sec would lose its energy within about one second. This result looks actually quite realistic and shows that the assumption of a non-viscous gas can account for observed drag in air, and this obviously suggests also that the same may hold for the aerodynamic lift.