What does a helium balloon do on the ISS vs in open space?

[DaveC426913]

In both cases, the air is *stationary*

The air is not stationary. Even if the density of the air itself is constant.
Because the air density is not constant throughout the vessel.

If you were to divide the vessel into two equal halves say, one cubic metre each, one half would have a cubic metre of air and the other half would have a cubic metre of air minus the volume of a balloon.

Under gravity or acceleration, the air will move so it displaces the balloon-shaped sphere of lighter than-air.

It is the air that moves in the direction the gravity/acceleration vector.

 

[russ_watters]

No, no, no, no, no. We're going to have to lock this thread if the misinformation persists. I'm at a bit of a loss as to how to fix it and get you guys to understand. Maybe draw a free body diagram? ...and then rotate it a little...?

I'm going to walk this back because evidently I misunderstood the thing I most objected to; the idea that the change in orientation is caused by wind. However, while I don't think the other things being said are technically wrong, I do think they are misleading/improper or unnecessary hairsplitting:

1. Density changes and gradients should only be mentioned for the purpose of dismissing them as irrelevant. Yes, they exist, but they are not included in the analysis.

2. Talk of "motion" in a completely static situation (once established) is misleading and unnecessary.

3. The balloon has three forces on it, and all of them are aligned with the apparent gravity vector. It's misleading or wrong depending on the characterization to say the apparent gravity vector is not involved or focus on one force and ignore the others. Note: it is possible to de-couple these vectors, and if you do that you might get a different angle for the string.

4. In addition to #3, saying the apparent gravity (acceleration) vector isn't acting on the balloon because it's the buoyant force buoyant force that acts on the balloon is an unnecessary hairsplit since the apparent gravitational acceleration vector is what causes the buoyancy vector.

 

[russ_watters]

It goes up because the air displaces it. The air is what provides the net upward force. No air, no upward force.

It's not just the air though:
Buoyancy + Weight = Net force
[and in this case, the string tension = - the net force]

A balloon without surrounding air still has the same gravitational vector, but the balloon does not go opposite to it.

Right: it goes down, in the direction of the apparent gravitational vector.

Gravity/acceleration alone cannot cause the balloon to move opposite the vector.

This is true. Adding the word "alone" makes the statement a lot better than your previous characterization.

The single factor that changes the direction of the balloon from 'down' to 'up' is the ambient presence of a heavier material.

This is true too...at least insofar as in that description it's the only thing you're changing in a particular scenario you are thinking of. I think it's misleading though because it implies the other factors don't matter at all. They do matter, you just assumed them to be already specified and unchanging. I could conversely say that if we change the gas in the balloon to air it will fall and that's the "single factor" making it fall. It's true, but it's not particularly relevant or complete; The buoyant force is still there, it's just pre-specified and unchanging.

The air is not stationary. Even if the density of the air itself is constant.
Because the air density is not constant throughout the vessel.

Please see the document I linked above. The air is both stationary and uniform density the way they modeled it, and most first-passes at buoyancy assume this to be true. The density gradient is correctly ignored.

If you were to divide the vessel into two equal halves say, one cubic metre each, one half would have a cubic metre of air and the other half would have a cubic metre of air minus the volume of a balloon.

That's not the same as saying the air has a density gradient, it means the air has a balloon-sized hole in it.

It is the air that moves in the direction the gravity/acceleration vector.

Again, the scenario presented is best analyzed static - nothing is moving (again; see the link). But if you want to release the balloon from some as yet unspecified position (say, the attachment point of the string), then yes, the balloon will move opposite the gravity vector and an equal volume of air will move toward it.

 

[zanick]

Could you quote the passage where it says that please. I'm not seeing it. It would be sloppy to say the air is moving because the buoyancy calculations assume the density is constant.

That was my mistake... I transposed "force" with "movement".

so it sounds like we are all on the same page now, so thanks for straightening a few things out, even though most of them were semantic in nature.

one of the things you can clarify is that when we talk about buoyancy of this helium balloon in a decelerating car...… the air doesn't get denser to any level to acknowledge , because we are ignoring that based on one of your points, buoyancy assumes a constant density...So, my first knee jerk reaction is to ask if the pressure gradient is shifting to the front vs straight down , does the pressure gradient change, without the density changing? how can we have higher pressure below the balloon and lower pressure above the balloon and not have a change in density? or is it the medium has a pressure gradient that changes directions based on the deceleration , and no density change? again, if it is a non compressible fluid, like water, then there is a pressure change, but no density change. maybe this density change is so small in both cases, its not worth adding to the discussion.

So, basically, it is the weight of the surrounding air displaced by the balloon that creates the buoyancy force , opposite the gravitational vector. ( and as you said, with normal gravity and a deceleration rate of 1g, that vector would be at a 45 degree angle. ) and if the weight of the object is higher, it sinks, if its lighter, it rises. (moves opposite to the vector force of gravity)

when I spoke of the" 2 pairs of forces". are they really just a pair, or are there two pairs? buoyancy obviously contains gravity , and gravity is obvious. But are they both providing the matched pressures acting on the balloon? balloon acting on the medium and the medium pushing on the balloon?

thanks!

 

[jbriggs444]

does the pressure gradient change, without the density changing? how can we have higher pressure below the balloon and lower pressure above the balloon and not have a change in density? or is it the medium has a pressure gradient that changes directions based on the deceleration , and no density change? again, if it is a non compressible fluid, like water

It is a question of degree. At typical sea level pressures and subject to such forces as are present from its own mass under the acceleration of gravity, the change in density of air over a distance of 1000 meters is about 10%. Over the span of a couple meters in a car, we are talking about something in the neighborhood of 0.02%. Maybe worth one or two tenths of a millimeter in balloon movement. Meanwhile buoyancy is shoving the balloon all the way from front to back or vice versa.

For many practical purposes, one can treat air as an incompressible fluid.

Water is not incompressible either. That is just a working approximation we use because it is close to being accurate and it makes calculations easier.

 

[DaveC426913]

I think it's misleading though because it implies the other factors don't matter at all. They do matter, you just assumed them to be already specified and unchanging.

Agree. I didn't mean to suggest gravity/acceleration doesn't factor in. Really, my initial point was that it is easier to intuit what the balloon will do if you instead looks at what the air will do (in the presence of gravity/acceleration).

That's not the same as saying the air has a density gradient, it means the air has a balloon-sized hole in it.

Agree. I dropped the gradient part. I see that was in error.

 

[Zeke137]

But you can do it. Imagine you are plummeting feet-first, and you've deployed a skateboard ahead of you. The skateboard, when aligned properly, will be braked by the wind. You will (in theory) be able to stand on it, and keep it pressed to the bottom of your feet. If you replaced the skateboard with bathroom weigh scale, it would display a measurable weight.

Not so - the scale is only displaying a value corresponding to the difference in force exerted by the air on (1) the lower surface of the scale, and (2) on the person, as they both descend through the air. Remove the air from your second thought experiment, and the bathroom scale would then not register a force at all. So, "a measurable weight" is not what is being registered by the scale.

 

[DaveC426913]

Not so - the scale is only displaying a value corresponding to the difference in force exerted by the air on (1) the lower surface of the scale, and (2) on the person, as they both descend through the air. Remove the air from your second thought experiment, and the bathroom scale would then not register a force at all.

Just so.

The air - which is a medium (as opposed to a vacuum) - is exerting a force on the bottom of the skate board.

If that medium were, instead, solid ground, you'd call it weight. Ground just happens to take a lot longer to get out of the way of an object pressing down upon it. Remove the ground from your scenario and the bathroom scale would not register a force either.

But because it's air taking its time to get out of way you say its not the same thing?

Try this thought experiment: increase the density of the medium, bit by bit, from the density of air to the density of ground. At what point does it go from being "merely air" to being a surface dense enough that you consider it a bona fide weight on the scale?

 

[Zeke137]

Just so.

The air - which is a medium (as opposed to a vacuum) - is exerting a force on the bottom of the skate board.

If that medium were, instead, solid ground, you'd call it weight. Ground just happens to take a lot longer to get out of the way of an object pressing down upon it. Remove the ground from your scenario and the bathroom scale would not register a force either.

But because it's air taking its time to get out of way you say its not the same thing?

Try this thought experiment: increase the density of the medium, bit by bit, from the density of air to the density of ground. At what point does it go from being "merely air" to being a surface dense enough that you consider it a bona fide weight on the scale?

So, okay, keep the air, and let the person have a cross-sectional area normal to the direction of travel sufficient that the air imparts the same drag (upward force) on the person as the bathroom scale as they both fall through the air. What does the scale register?

 

[zanick]

So, okay, keep the air, and let the person have a cross-sectional area normal to the direction of travel sufficient that the air imparts the same drag (upward force) on the person as the bathroom scale as they both fall through the air. What does the scale register?

I would say, once you reach terminal velocity , (no acceleration) you then weight what you weigh on the surface of the earth, just like an elevator going down.

I think what was said, is that the buoyancy factor is the weight of the air around you . it makes you lighter by the volume and density (weight) of the air around you .

 

[A.T.]

So, "a measurable weight" is not what is being registered by the scale.

What does "measurable weight" mean here? Do you and Dave have the same thing in mind?

 

[A.T.]

So, okay, keep the air, and let the person have a cross-sectional area normal to the direction of travel sufficient that the air imparts the same drag (upward force) on the person as the bathroom scale as they both fall through the air. What does the scale register?

I would say, once you reach terminal velocity , (no acceleration) you then weight what you weigh on the surface of the earth, just like an elevator going down.

If part of the weight is balanced by the drag on the body, the scale will not show the full weight at terminal velocity.

 

[Zeke137]

What does "measurable weight" mean here? Do you and Dave have the same thing in mind?

No, I mean that the force registered by the bathroom scale in Dave's example is simply the difference in forces exerted by the air on the person and on the bathroom scale. If the drag on the person were the same as the drag on the scale as they both fall unhindered, then no net force would be registered by the scale. As in the Einstein elevator thought experiment.

 

[Zeke137]

Try this thought experiment: increase the density of the medium, bit by bit, from the density of air to the density of ground. At what point does it go from being "merely air" to being a surface dense enough that you consider it a bona fide weight on the scale?

Both the person and the scale are descending through the air, so they are surrounded by the air. Let's make the drag on the person and the bathroom scale equal: no net force is registered by the scale. Now, let's have the person and the scale fall through molasses: assuming the drag on both are still equal in this new fluid, which surrounds them, then again no net force will be registered. Replacing the molasses with soil or rock, they'll both be surrounded by this new medium and again, no net force will be registered by the scale.

 

[zanick]

If part of the weight is balanced by the drag on the body, the scale will not show the full weight at terminal velocity.

yes , i was assuming that the person was in the wake of the scale , not experiencing any drag... so it would be similar to the elevator thought experiment. but yes, the weight of the scale itself, matched by its drag would have measurable weight, but that would be measured by the drag forces required to stop its acceleration. same with the falling person ... there would be no measurement on the scale ...

 

[A.T.]

If the drag on the person were the same as the drag on the scale as they both fall unhindered, then no net force would be registered by the scale.

This is not true, unless the scale and the person have the same mass.

 

[zanick]

I was referring to their drag equaling their respective weights.