Hydrostatic pressure in a bottle of water on ISS

[Induana]

TL;DR

Hydrostatic pressure equals heightxdensityxg Does that mean that the bottle will have zero hydrostatic pressure when it is in a state of weightlessness?

Hello,
I am new in physics but I really like it so far. I have a question about hydrostatic pressure. I know that hydrostatic pressure equals (height)x(density)x(g). Am I right if I say that hydrostatic pressure in a bottle of water on ISS (state of weightlessness) will be zero ?
Thank you for your answers

 

[Orodruin]

No, $\rho g h$ is the pressure difference between two points at different heights in a gravitational field.

Being in free fall, the hydrostatic pressure inside a fluid at the ISS will be the same as the ambient pressure inside the ISS (which is non-zero - astronauts need to breathe).

 

[Lnewqban]

Welcome @Induana !

 

[russ_watters]

No, $\rho g h$ is the pressure difference between two points at different heights in a gravitational field.

Being in free fall, the hydrostatic pressure inside a fluid at the ISS will be the same as the ambient pressure inside the ISS (which is non-zero - astronauts need to breathe).

I think you're mixing together static pressure and hydrostatic pressure. Hydrostatic pressure is pressure due to height difference in gravity. In orbit you still have a height difference, but what you don't have is an apparent g, so hydrostatic pressure is zero.

But you do have static pressure, which is a more general term for pressure of a fluid not in motion. Static pressure is provided by the gas temperature, density and being an enclosed volume.

 

[russ_watters]

Welcome @Induana !

Why don't they use bags?

 

[Induana]

I think you're mixing together static pressure and hydrostatic pressure. Hydrostatic pressure is pressure due to height difference in gravity. In orbit you still have a height difference, but what you don't have is an apparent g, so hydrostatic pressure is zero.

But you do have static pressure, which is a more general term for pressure of a fluid not in motion. Static pressure is provided by the gas temperature, density and being an enclosed volume.

So the hydrostatic pressure will be zero as I said, right?

 

[Induana]

Welcome @Induana !

Thank you for the video and your welcome message :)

 

[russ_watters]

So the hydrostatic pressure will be zero as I said, right?

Yes.

 

[Andy Resnick]

Hydrostatic pressure equals heightxdensityxg Does that mean that the bottle will have zero hydrostatic pressure when it is in a state of weightlessness?

Yes, and this has a variety of ramifications besides being able to drink out of a bottle of water. It also impacts (fluid) fuel management in tanks, supplying water to the roots of plants, boiling and other multiphase fluid separation, including oil-water systems, and is possibly the reason for bone loss in astronauts.

 

[sophiecentaur]

So the hydrostatic pressure will be zero as I said, right?

Yes.

"Up to a point".

There is an effect called 'micro gravity' which reveals itself inside free orbiting spacecraft . The orbital speed and height of a craft are set by the Earth's gravity and its net effect on all the parts of the craft but what counts is the position of the craft's centre of mass. Parts of the craft are actually higher / further out (just a tiny bit) than the CM and the rigid structure craft is 'pulling them along' slightly faster than they would be going if they weren't attached to the rest of the craft. Likewise, the lower parts of the craft would be going a bit faster, to be in their natural orbital position so the craft is 'slowing them down' a bit. This results in a perceived gradient in the gravity inside. Things drift towards the walls and plant growth (so I believe) can show it. Same thing can happen to measured hydrostatic pressure (but this is a very tiny effect).

 

[Orodruin]

"Up to a point".

There is an effect called 'micro gravity' which reveals itself inside free orbiting spacecraft . The orbital speed and height of a craft are set by the Earth's gravity and its net effect on all the parts of the craft but what counts is the position of the craft's centre of mass. Parts of the craft are actually higher / further out (just a tiny bit) than the CM and the rigid structure craft is 'pulling them along' slightly faster than they would be going if they weren't attached to the rest of the craft. Likewise, the lower parts of the craft would be going a bit faster, to be in their natural orbital position so the craft is 'slowing them down' a bit. This results in a perceived gradient in the gravity inside. Things drift towards the walls and plant growth (so I believe) can show it. Same thing can happen to measured hydrostatic pressure (but this is a very tiny effect).

What you are describing are in essence tidal effects. These tend to stretch in the direction of the gravitational source and compress in directions orthogonal to the source.

 

[sophiecentaur]

What you are describing are in essence tidal effects.

Yes. And could you not measure that in the form of hydrostatic pressure against the sides of the ship?

 

[Orodruin]

Yes. And could you not measure that in the form of hydrostatic pressure against the sides of the ship?

Yes and no. Depends on which sides.

 

[sophiecentaur]

Yes and no. Depends on which sides.

And not in the middle.

 

[256bits]

What you are describing are in essence tidal effects. These tend to stretch in the direction of the gravitational source and compress in directions orthogonal to the source.

Is he description from @sophiecentaur called a tidal effects?
He is attributing the difference in pressure due to the revolution of the ship around the earth.
Nowhere did he mention the gravitational gradient of the earth that exists throughout the ship.

 

[256bits]

And not in the middle.

In that regard, if the pressure in the "middle" is different than at the sides, so you would measure a difference in pressure between two points.
I suppose one could measure the stress/strain on a wall of the spacecraft and compare that to another and convert that to a pressure.
PS. I think objects are attracted to the walls through mutual gravitation rather than any pressure gradient that might exist throughout the ship.
PS_2 - one also has the friction drag of the spacecraft through the atmosphere, which would appear as an acceleration ( deceleration ). The 'front' end of the ship would have a slightly higher air pressure than the rear.

 

[sophiecentaur]

In that regard, if the pressure in the "middle" is different than at the sides, so you would measure a difference in pressure between two points.

I'd suggest that bubbles if air in a water column would 'rise' very slowly, away from the inner and outer walls. There are countless possible examples of microgravity in spacecraft but many of them would be too subtle to notice unless you were trying to find them.
I just did a search on "microgravity" but, even on Nasa pages, it's not discussed very fully. The effect that I have been calling "microgravity" doesn't seem to be mentioned (can anyone find it?) but it seems to be described as something that 'just happens', due to non-ideal free-fall conditions. This link shows that microgravity has been considered in the context of smouldering in a spacecraft - gulp! But the reasons we have discussed don't seem to be mentioned

Nowhere did he mention the gravitational gradient of the earth that exists throughout the ship.

I didn't want to write too much about it but I assumed that my basic statement that the 'natural' orbits of different parts of the craft would be due to different gravitational force and different velocities would be enough EDIT: [(but using the term 'gradient' would have been useful). ]The "gradient" you refer to would not be felt on disconnected items but it would be noticed once they are held rigidly together.

 

[sophiecentaur]

Also, I have a bone to pick with NASA when they describe the gravitational field in Low Earth Orbit as being 'less' than on the surface. They really should point out / stress that it is actually hardly different at all. The No Gravity idea should be squashed at every opportunity.

 

[jbriggs444]

Is he description from @sophiecentaur called a tidal effects?
He is attributing the difference in pressure due to the revolution of the ship around the earth.
Nowhere did he mention the gravitational gradient of the earth that exists throughout the ship.

The two effects are of similar magnitude, but somewhat different in character.

If I have not muffed the analysis, the net of the centrifugal effect and the tidal effect is a pseudo-tidal effect which has 3 times the magnitude of the centrifugal effect alone in the radial direction and has zero magnitude in the tangential direction.

Let me show the work:

To be clear, $r$ is position relative to the center of mass of the space craft. $R_0$ is the distance of the craft's center of mass from the center of the Earth. When appropriate, $R$ will be the distance from the center of the Earth to a particular point in the craft. The craft is (we assume without evidence) consistently oriented with the same face toward the earth, rotating at the same rate as its orbital period.

The centrifugal force from rotation has magnitude $\omega^2 r$ and is repulsive in both the radial and tangential directions.

Tidal gravity in the radial direction results from the gradient of a $\frac{1}{R^2}$ force field which is equal to $\omega^2 R_0$ near the position of interest. So we can normalize the gravitational force to $\frac{\omega^2{R_0}^3}{R^2}$ and take the derivative with respect to $R$. That will yield $\frac{-2\omega^2{R_0}^3}{R^3}$. Evaluated at $R = R_0$, this yields $-2 \omega^2\ dR$.

This is to say that the gravitational tidal gradient in the radial direction is twice the centrifugal component. Both are repulsive. So they add to a net effect that is three times that of centrifugal alone.

Let us turn our attention to the gravitational tidal gradient in the tangential direction. Offhand, I do not see a convenient way to attack this. So I will be floundering a bit. However, I expect the result that is obtained will be proportional $\omega^2 r$ with a small integer as the constant of proportionality.

The tangential gradient will be given by $\omega^2 R_0 \sin \theta$ where $\theta$ is the angle subtended by an arc of length $dR$ on a circle of radius $R_0$. The small angle approximation is appropriate, so we can rewrite the gradient as $\frac{\omega^2 R_0\ dR}{R_0}$. Which simplifies to $\omega^2\ dR$. [Which was surprisingly straightforward after the initial panic had worn off].

So the gravitational gradient in the tangential direction is equal to the centrifugal effect. But the two act in opposite directions, so the net effect is zero tidal gradient in the tangential direction (to first order).

 

[256bits]

The two effects are of similar magnitude, but somewhat different in character.

If I have not muffed the analysis, the net of the centrifugal effect and the tidal effect is a pseudo-tidal effect which has 3 times the magnitude of the centrifugal effect alone in the radial direction and has zero magnitude in the tangential direction.

Let me show the work:

To be clear, $r$ is position relative to the center of mass of the space craft. $R_0$ is the distance of the craft's center of mass from the center of the Earth. When appropriate, $R$ will be the distance from the center of the Earth to a particular point in the craft.

The centrifugal force from rotation has magnitude $\omega^2 r$ and is repulsive in both the radial and tangential directions.

Tidal gravity in the radial direction results from the gradient of a $\frac{1}{R^2}$ force field which is equal to $\omega^2 R_0$ near the position of interest. So we can normalize the gravitational force to $\frac{\omega^2{R_0}^3}{R^2}$ and take the derivative with respect to $R$. That will yield $\frac{-2\omega^2{R_0}^3}{R^3}$. Evaluated at $R = R_0$, this yields $-2 \omega^2\ dR$.

This is to say that the gravitational tidal gradient in the radial direction is twice the centrifugal component. Both are repulsive. So they add to a net effect that is three times that of centrifugal alone.

Let us turn our attention to the gravitational tidal gradient in the tangential direction. Offhand, I do not see a convenient way to attack this. So I will be floundering a bit. However, I expect the result that is obtained will be proportional $\omega^2 r$ with a small integer as the constant of proportionality.

The tangential gradient will be given by $\omega^2 R_0 \sin \theta$ where $\theta$ is the angle subtended by an arc of length $dR$ on a circle of radius $R_0$. The small angle approximation is appropriate, so we can rewrite the gradient as $\frac{\omega^2 R_0\ dR}{R_0}$. Which simplifies to $\omega^2\ dR$. [Which was surprisingly straightforward after the initial panic had worn off].

So the gravitational gradient in the tangential direction is equal to the centrifugal effect. But the two act in opposite directions, so the net effect is zero tidal gradient in the tangential direction (to first order).

Cool.
I guess ( not guessing, just figure of speech ) it would make a second order difference if the ship was as curved as its orbit ( circular orbit assumption ), versus a straight line of sight ship. Since the ISS is small compared to its orbit, that should be quite tiny as the deviation from the circular orbit for the straight line of sight IIS is a function of its length 218 feet to the curvature of an orbit on the order of 24, 000 miles.
Kind of a word salad there.

 

[sophiecentaur]

Having thought again, I think that the ‘tidal’ distortion is when gravity and centrifugal force are in equilibrium. But the rigid hull adds another force to restrain actual displacement. Doesn’t that introduce another factor?

 

[Orodruin]

Having thought again, I think that the ‘tidal’ distortion is when gravity and centrifugal force are in equilibrium. But the rigid hull adds another force to restrain actual displacement. Doesn’t that introduce another factor?

The tidal effect is always present. It is just the gradient of the gravitational field (a rank 2 tensor) acting on the displacement. Whether or not tidal forces represent the most apparent manifestation of gravity or not depends on the situation. What you are discussing here is an object in free fall, which is precisely when tidal forces tend to dominate because the leading order gravitational effects are cancelled by going to the accelerated rest frame of the center of gravity.

The rigid hull plays the same role a bath tub. It contains the fluid and prevents it from spreading out.