I've checked out related questions and didn't see an answer to my question.
Scenario: Liquid in open-top container, together accelerating downward with magnitude $g$. It was straightforward for me to surmise that $\nabla P = \rho (\vec{g} - \vec{a}) = \vec{0} \Rightarrow \frac{\partial P}{\partial y} = 0 \Rightarrow P(y) = \text{constant} $, so the pressure is uniform in the vertical (and all other) direction. No problem.
But what is the value of that uniform pressure? Since the container top is open to the atmosphere, does that guarantee the liquid's pressure is $P_{atm}$? Or, since the liquid is fleeing the atmosphere at exactly the acceleration that pulls the atmosphere down and generates atmospheric pressure, does that imply that a liquid in free-fall is free of atmospheric pressure? (I understand this would mean the liquid would vaporize, but perhaps that is beside the point.)
It's strange to me to conclude that $ P(y) = P_{atm} $, because I believe an object moving through air experiences a higher pressure on the leading surface and a lower pressure on the trailing surface. So shouldn't this liquid experience a lower pressure on its trailing (top) surface than if it were at rest? (And thus a lower pressure throughout, since its leading/bottom surface is shielded from the air by the container.)
Question: What is the value of the uniform pressure experienced by a liquid in free-fall (specifically in an open-top container)?