"Outrunning pressure": Liquid in free-fall

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)?

• – cms
Commented Oct 1, 2022 at 21:00
• @cms Thank you. It's not a direct comparison, but if I understand correctly, that answer might suggest the pressure in this situation is 0. Commented Oct 2, 2022 at 0:45

It's strange to me to conclude that $$P(y)= P_{\text{atm}}$$ because I believe an object moving through air experiences a higher pressure on the leading surface and a lower pressure on the trailing surface

I believe you are conflating acceleration with velocity.

Consider when you drop a bucket of water from a height. If you release the bucket without giving it any initial velocity, then at the moment of release, it is accelerating downward at $$-g$$ but has zero velocity. This meets the premises of your question, and indeed without any velocity the absolute pressure in the fluid would be atmospheric. As no motion between the atmosphere and the fluid-bucket surfaces has been created, the same number of atoms strike the surfaces as when the bucket was supported, resulting in the same pressure.

Once some velocity is obtained, then certainly there may be less atoms striking the fluid surface than the bucket’s surface and this could result in a change from atmospheric pressure. However this becomes a fluid dynamics question and whether the pressure increases or decreases would depend on the geometry and many other physical parameters.

• Thank you for your answer. I agree I was a bit loose with my use of acceleration and velocity. Something still isn't clicking for me, but I can't put my finger on it. Commented Oct 2, 2022 at 0:43
• The diagram you shared in your answer to the related question (physics.stackexchange.com/a/520000/162611) suggests to me that $P=0$. Since the liquid and container in my scenario are in free-fall, then can we consider the liquid "unconstrained"? (since the container is falling as well it does not exert a force on it?). If the container has no opening, the pressure should be zero, correct? And if we take the top off, you say it has atmospheric pressure. What if it is all closed except for a pinhole? Commented Oct 2, 2022 at 0:52
• @electronpusher In the diagram that shows P = 0, that is "gauge pressure" (i.e. pressure above atmospheric pressure). In this discussion we are talking about absolute pressure (i.e. pressure compared to a total vacuum).
– cms
Commented Oct 2, 2022 at 1:08
• I see, I didn't realize that was gauge pressure. Thanks for your insight. Commented Oct 2, 2022 at 1:53