Liquid pressure in microgravity

[2147483647]

Homework Statement

This is more of a conceptual question rather than a homework problem. This is my first post, so apologies if this is in the wrong section.

Consider a jar full of liquid in microgravity.

My book says:
According to the formula p = ρgh, p→0 as g→0. Thus there is no pressure in the jar when it's in microgravity.

My problem:
Isn't it true, though, that the molecules in the liquid are jiggling around? Wouldn't these molecules still inadvertently bump into the sides of the jar, creating pressure?

Homework Equations

p = ρgh

The Attempt at a Solution

My current thought is that the pressure due to this bumping is negligible. So the formula p = ρgh is a (good) approximation.

 

[barryj]

Here is my opinion. "the formula p = ρgh" is generally used in determining the pressure of liquids at a given depth, h, in a container on the Earth.
I would think that in no gravity, like in space, there will always be some pressure due to the vapor of the liquid. If you had a gas in space, then of course there would be pressure.

 

[Chestermiller]

Barryj is correct. In addition, the equation p=ρgh is not really correct. It should not be used to determine the absolute pressure. The equation should really read Δp=ρgΔz, or, even better dp = ρg dz. The equation gives the relative change in pressure with distance in the direction of the gravitational vector, rather than giving the absolute pressure.

 

[2147483647]

Brilliant, thank you barryj and Chestermiller.

Would it be appropriate to suggest p = p_{0} + \rho gh, where p_{0} is the initial pressure? So, as you were saying Chestermiller, p = \rho gh would be the change in pressure.

 

[Chestermiller]

Brilliant, thank you barryj and Chestermiller.

Would it be appropriate to suggest p = p_{0} + \rho gh, where p_{0} is the initial pressure? So, as you were saying Chestermiller, p = \rho gh would be the change in pressure.

Yes, with spatial position in a gravitational field.