# Why neglect gravity in Kutta-Joukowski theorem?

In an example I was asked to calculate the lift force using Bernoulli's equation

$$u^2/2 \ + p/\rho +gz \ = \ constant$$

and show that its consistent with Kutta-Joukowski theorem, but seems like gravity is neglected in the theorem? why neglect gravity?

The theorem computes the lift force, which by definition is a non-gravitational contribution weighed against gravity to determine whether there is a net upward acceleration. We "neglect" gravity (i.e. regard it separately) as we would when computing upthrust.

• but isn't the pressure actually dependant on g? $$p \ = \ constant -\rho (g \ + \ u^2/2)$$ and the pressure is what is actually causing the lift force, so if you say the lift force without gravity is 0 there may be a non 0 acceleration still due to the gravity term in the pressure... – physicsnoob1000 Jun 3 '17 at 11:21