# Pressure difference between two points on the Earth's surface

I'd like to find the pressure difference between two points $$X$$ and $$Y$$ on the Earth's surface given that wind blows steadily with speed $$U$$ from (say) West to East. Here's a picture: ($$z_i$$ is the height above sea-level in metres) I have seen

• The Bernoulli equation for potential flow:

$$\rho \frac{ \partial \phi}{\partial t} + \frac{1}{2} \rho \vec{u}^2 + p + \rho g z = f(t)$$

and

• the Euler equation for a rotating fluid:

$$\frac{D\vec{u}}{Dt} + 2 \rho \vec{\Omega} \times \vec{u} = -\nabla p + \rho \vec{g}$$

Attempt 1:

If I neglect the rotation of the Earth and use Bernoulli, then I get

$$p_X + \rho g z_1 = p_Y + \rho g z_2$$,

since

• if the flow is steady the potential function $$\phi$$ is independent of $$t$$, so $$\partial \phi/\partial t=0$$
• the $$\frac{1}{2} \rho \vec{u}^2$$ terms cancel.

This gives a pressure difference of $$\rho g(z_1-z_2)$$ (where $$\rho$$ is the air density).

This solution doesn't seem right as I haven't used the extra information about the wind speed.

Attempt 2:

If we further assume that the air is incompressible ($$\nabla U = 0$$) then $$Du/Dt = 0$$.

If I can write $$\vec{\Omega} \times \vec{u}$$ in terms of known quantities, then I can compute $$\nabla p$$ from the Euler equation and I'm essentially done.

However, I'm not sure how to handle the $$\vec{\Omega} \times \vec{u}$$ term.

Any ideas? I think I'm overcomplicating this.