# How Does Buoyancy Behave in a Floating Water Sphere on the ISS?

• mishima
In summary: The absence of a buoyant force is justified by the fact that the only force vector is the gravitational force. There is no difference in pressure at all within the water sphere, so there would be no force to counteract the buoyant force.
mishima
If you push an air filled ball down to the bottom of a bucket of water an acceleration is caused by the buoyant force being greater than the force of gravity, so you have a net force up.

I was wondering if you tried this on the ISS where water becomes a floating sphere. How does the ball decide which direction to exit the sphere? Acceleration is in the direction of the net force, what would it be in this case? Thanks.

mishima said:
If you push an air filled ball down to the bottom of a bucket of water an acceleration is caused by the buoyant force being greater than the force of gravity, so you have a net force up.

I was wondering if you tried this on the ISS where water becomes a floating sphere. How does the ball decide which direction to exit the sphere? Acceleration is in the direction of the net force, what would it be in this case? Thanks.
Which way do you think is the direction of the buoyant force in these conditions?

There would still be a gravitational force due to the objects position in the gravitational field, there would still be a force due to gravity. So the buoyant force would still oppose this force, and could over power it. So it would move away from the gravitational center of the earth.

What about in a void in space, rather than free fall? A place where there isn't a gravitational field?

Would it just be that at the center of the sphere there is a region of higher pressure than outside (given the false atmosphere of the spaceship say)? Or is that indeed the case with a floating sphere of water even?

But then pressure is just a derived concept from force per unit area. What is really happening to oppose buoyancy on a small scale in this situation?

edit: what is the microscale explanation for buoyancy?

mishima said:
There would still be a gravitational force due to the objects position in the gravitational field, there would still be a force due to gravity. So the buoyant force would still oppose this force, and could over power it. So it would move away from the gravitational center of the earth.
Does that gravitational force also affect the ball of water? The people on the ISS? The ISS itself?

mishima said:
There would still be a gravitational force due to the objects position in the gravitational field, there would still be a force due to gravity.
There would still be a gravitational force, but would there be a buoyant force? What force will the water exert on the ball?

russ_waters: Yes, it affects all.

Doc_Al: Some kind of contact force between the ball and the water due to electromagnetic repulsion on the atomic scale? I don't know the explanation for buoyancy on this scale.

I'm assuming though that if at the center of the sphere there is a greater pressure than the outside, there could be a difference in constricting forces that would cause the ball to seek the outside. But I also don't know if that's the case for a sphere of water.

I would guess that small scale collisions of the atmosphere with the water sphere would lead to evaporation on the outside surface, and that those collisions could also transfer through to the center of the sphere, perhaps making a pressure difference throughout.

mishima said:
russ_waters: Yes, it affects all.
How? Does the person get pulled to the side of the ISS or does he float in the middle? Does the ball of water? If the ball of water isn't pulled to the side of the ISS, why would a bubble be pushed in the opposite direction?

russ_waters: As I understand it, everything is experiencing a non-zero acceleration towards the center of gravity (center of earth). This gravitational force is what keeps the ISS in orbit. If it were "switched off" the ISS would leave orbit in a straight line tangent to its current orbit. The person and water sphere float motionless to an observer inside the spaceship, but would appear to be traveling in a circle to an outside observer.

So, in the absence of a buoyant force, a ball of air inserted into the water sphere would stay where it is, since the only force vector it has is this gravitational force. It would fall along with everything else.

How is this absence of buoyant force justified? Is there no difference in pressure at all within the water sphere? Because I think that would create a buoyant force towards a lower pressure region (that's my normal understanding of it, anyways). Is there another explanation for buoyancy I'm missing? If not, and if there isn't a pressure difference, then it seems my answer is that the ball of air remains where you leave it in the water sphere.

mishima said:
russ_waters: As I understand it, everything is experiencing a non-zero acceleration towards the center of gravity (center of earth). This gravitational force is what keeps the ISS in orbit. If it were "switched off" the ISS would leave orbit in a straight line tangent to its current orbit. The person and water sphere float motionless to an observer inside the spaceship, but would appear to be traveling in a circle to an outside observer.

So, in the absence of a buoyant force, a ball of air inserted into the water sphere would stay where it is, since the only force vector it has is this gravitational force. It would fall along with everything else.

How is this absence of buoyant force justified? Is there no difference in pressure at all within the water sphere? Because I think that would create a buoyant force towards a lower pressure region (that's my normal understanding of it, anyways). Is there another explanation for buoyancy I'm missing? If not, and if there isn't a pressure difference, then it seems my answer is that the ball of air remains where you leave it in the water sphere.
There is a pressure difference but it is not caused by gravity.Surface tension causes a pressure difference towards the centre of the sphere.It is this that makes the water form a round sphere in space.This would probably result in the ball of air inserted into the sphere staying put towards the centre.

Here's a video of a bubble being inserted into a sphere of water in free fall (earth orbit). A ball filled with air would probably behave similarly.

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Thanks, that video is quite interesting. I was thinking of an air filled plastic ball, the kind in a kids ball pit. In the case of the video I assume it is the arrangement of water molecules on the inner surface which causes an attraction to the outer sides of the water sphere. With the plastic ball you wouldn't have that, right?

I don't know if there would be a similar effect with plastic. It might have something to do with surface tension, as Buckleymanor mentioned in post #9, except with the opposite effect. I don't think there is an "attraction" from a distance, but rather a tendency to cling to the surface once it touches it. Here's another demonstration of surface tension aboard the ISS.

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The way i see it, the sphere of water won't be under pressure in space, so there would be no buoyant force, and the ball could quite happily remain within the sphere. If the sphere was for some reason under pressure, then the ball would move away from the point which is at the highest pressure.

The way i see it, the sphere of water won't be under pressure in space, so there would be no buoyant force, and the ball could quite happily remain within the sphere. If the sphere was for some reason under pressure, then the ball would move away from the point which is at the highest pressure.
If you watch the demonstration posted by TurtleMeister the water clearly is under pressure how would it form a perfect sphere if this was not the case.Now as for a ball of air or maybe a ball full of air moveing away from the point of highest pressure.The former would have two layers of surface tension, the outer skin of the sphere of water and the outer skin of the bubble of air which was inserted.
Both layers would form an area of highest pressure towards the centre of the mass of water between the outer and inner bubbles.
So you would imagine that depending on the size of the bubble of air inserted it would probably move towards the centre then again it might not.

Buckleymanor said:
If you watch the demonstration posted by TurtleMeister the water clearly is under pressure how would it form a perfect sphere if this was not the case.

This was answered in the video. It's because the molecules are clinging together (like little springs), and since a sphere has the smallest surface area of any shape, it's only natural that this would be the shape it would take (in the absence of any other forces).

TurtleMeister said:
This was answered in the video. It's because the molecules are clinging together (like little springs), and since a sphere has the smallest surface area of any shape, it's only natural that this would be the shape it would take (in the absence of any other forces).

Don't little springs put objects under pressure?

Buckleymanor said:
If you watch the demonstration posted by TurtleMeister the water clearly is under pressure how would it form a perfect sphere if this was not the case.Now as for a ball of air or maybe a ball full of air moveing away from the point of highest pressure.The former would have two layers of surface tension, the outer skin of the sphere of water and the outer skin of the bubble of air which was inserted.
Both layers would form an area of highest pressure towards the centre of the mass of water between the outer and inner bubbles.
So you would imagine that depending on the size of the bubble of air inserted it would probably move towards the centre then again it might not.

The fact that the bulk of the water is under pressure does not imply that there is a pressure gradient within the bulk of the water. Buoyancy requires a pressure gradient.

I would expect the bulk of the water within a weightless water globule to be under a fixed pressure. Accordingly, there would be no "point of highest pressure" and no tendency for a bubble within such a globule to migrate in any particular direction.

It is my understanding that "surface tension" is not really an effect of a thin layer skin under tension but is actually an attractive force between the bulk of molecules within the fluid that mimics the behavior that a skin under tension would have -- the fluid isn't "really" under pressure. It just behaves as if it were.

In the case at hand the point is moot. I don't see how either description of "surface tension" would give rise to a net buoyant force.

TurtleMeister said:
This was answered in the video. It's because the molecules are clinging together (like little springs), and since a sphere has the smallest surface area of any shape, it's only natural that this would be the shape it would take (in the absence of any other forces).

Some care is required in this reasoning.

Yes, a sphere has the smallest surface area of any shape of a given volume. And it follows that the potential energy associated with a stretched skin under surface tension is minimized when surface area is minimized. All things being equal, a weightless fluid glob will incrementally adjust itself so that surface area is minimized.

A key word is incremental. A "hill climbing" optimization process like this one may find a local optimum rather than a global optimum. And it may fail to gravitate to a global optimum if small incremental changes result in no change to the parameter being optimized.

The fact that the bulk of the water is under pressure does not imply that there is a pressure gradient within the bulk of the water. Buoyancy requires a pressure gradient.
There is a gradient between the bulk of the water and the atmosphere around the sphere of water and another between the air inserted and the waters surface inside the sphere.

Buckleymanor said:
There is a gradient between the bulk of the water and the atmosphere around the sphere of water and another between the air inserted and the waters surface inside the sphere.

That's not a gradient. That's a discontinuity.

Is there a way it can be proven that there is no gradient? If we are saying that the spaceship atmosphere is pushing in on all sides, then there must be some force per area on all points on the surface. So we are saying that this force per area is distributed throughout all locations in the sphere evenly? Why?

jbriggs444 said:
That's not a gradient. That's a discontinuity.

The fact that the bulk of the water is under pressure
So if it's under pressure it must be a different pressure than the surrounding atmosphere.So a gradient must exist between the atmosphere and the water.If you like to call it a discontinuity then it's ok but this won't remove the difference between the two pressures.
There does not seem to be any clear reason why a gradient must only exist within the bulk of water only, and not it's surroundings, maybe it's one of the reasons an inserted bubble when attached to the outer surface does seem to wan't to cling there.

mishima said:
Is there a way it can be proven that there is no gradient? If we are saying that the spaceship atmosphere is pushing in on all sides, then there must be some force per area on all points on the surface. So we are saying that this force per area is distributed throughout all locations in the sphere evenly? Why?
Look up Pascal's principle.

Ok, so I guess my question has to do with how Pascal's principle operates in both free fall and zero gravity situations. I will try and find some info there, thanks.

Buckleymanor said:
So if it's under pressure it must be a different pressure than the surrounding atmosphere.So a gradient must exist between the atmosphere and the water.If you like to call it a discontinuity then it's ok but this won't remove the difference between the two pressures.

But it does mean that it's not a gradient.

At an introductory level, a "scalar field" (such as pressure) is an assignment of a scalar value to each point in a three dimensional space.

A "vector field" is an assignment of a vector value to each point in a three dimensional space.

The "gradient" of a scalar field S is a vector field V that is formed by taking the partial derivitive of S with respect to x, the partial derivitive of S with respect to y and the partial derivitive of S with respect to z as the three components of V.

Symbolically: V(x,y,z) = ( dS/dx (x,y,z), dS/dy (x,y,z), dS/dz (x,y,z) )

Another way of stating this is that the gradient vector points in the direction in which the scalar field increases most rapidly and that its magnitude indicates the rate at which the scalar field increases in that direction.

There does not seem to be any clear reason why a gradient must only exist within the bulk of water only, and not it's surroundings, maybe it's one of the reasons an inserted bubble when attached to the outer surface does seem to wan't to cling there.

The pressure within the bulk of the water is constant. Accordingly, the pressure gradient is zero within the bulk of the water. That is what I mean when I say that there is no gradient within the bulk of the water. It is not clear whether you disagree with this.

The pressure within the bulk of the water is constant. Accordingly, the pressure gradient is zero within the bulk of the water. That is what I mean when I say that there is no gradient within the bulk of the water. It is not clear whether you disagree with this.
That depends, if it's just the bulk of the water aboard the ISS as a floating sphere and in the absence of forces that may create a gradient then I agree. Once a bubble of air is introduced into the sphere then to be honest I am not so sure the parameters remain the same.
You will end up haveing two different substances with different properties and densities one on the inside of the sphere of water as air and another on the outside of the sphere also as air.
Also two layers of surface tension.

Buckleymanor said:
That depends, if it's just the bulk of the water aboard the ISS as a floating sphere and in the absence of forces that may create a gradient then I agree. Once a bubble of air is introduced into the sphere then to be honest I am not so sure the parameters remain the same.

You will end up haveing two different substances with different properties and densities one on the inside of the sphere of water as air and another on the outside of the sphere also as air.
Also two layers of surface tension.

Now consider a small parcel of water somewhere between the outer surface where water meets air and the inner surface where water meets air. Further suppose that this parcel of water is under a pressure higher than a neighboring parcel.

Under this scenario, the water between those two regions will be under a net force away from the high pressure region and toward the low pressure region, yes?

And this water will undergo an acceleration away from the high pressure region and toward the low pressure region, yes?

And the resulting influx of water will increase the pressure in the low pressure region and decrease the pressure in the high pressure region, yes?

So the only stable static arrangment must be one in which the pressure throughout the bulk of the water is constant.

One quibble is that there can be stable dynamic arrangements, e.g. a whirlpool or a compression wave, but those will tend to decay over time, so let's agree to consider only a static equilibrium.

The same reasoning applies for the air in the cabin and for the air in the embedded bubble.

## 1. How does the ball behave when dropped in water?

When a ball is dropped in water, it experiences three main forces: buoyancy, drag, and gravity. Initially, the ball will sink due to the force of gravity pulling it down. However, as it sinks, the water exerts an upward force on the ball known as buoyancy. This force increases as more water is displaced by the ball, eventually balancing out the force of gravity. The drag force also acts on the ball, slowing it down as it falls through the water. As a result, the ball will reach a terminal velocity and continue to sink at a constant speed.

## 2. Why does the ball float back to the surface after sinking?

The ball floats back to the surface due to the force of buoyancy. As the ball sinks, it displaces water equal to its own weight. When the buoyancy force equals the weight of the ball, the ball will stop sinking and start to float back up to the surface.

## 3. How is the speed of the ball affected by the water?

The speed of the ball is affected by the water due to the force of drag. As the ball falls through the water, the water molecules collide with it, exerting a force in the opposite direction of motion. This force increases with the speed of the ball, eventually balancing out the force of gravity and causing the ball to reach a constant speed known as terminal velocity.

## 4. What factors can affect the behavior of the ball in water?

Several factors can affect the behavior of the ball in water, including the density and shape of the ball, the density and temperature of the water, and the depth of the water. These factors can impact the forces acting on the ball, altering its speed and buoyancy.

## 5. How does the ball in water (in free fall) differ from a ball in air?

The behavior of a ball in water is significantly different from a ball in air due to the presence of buoyancy and drag forces. In air, the only force acting on the ball is gravity, causing it to fall at an increasing speed until it reaches a terminal velocity. In water, the forces of buoyancy and drag act on the ball, balancing out the force of gravity and resulting in a constant sinking speed.

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