Underwater Pressure on Sphere Dilemma

[joe_gunn]

Dear Friends,
I have a problem that has been driving me nuts for months now, and I am hoping you guys can help. I need to understand whether:
1. The Force with which a hollow, buoyant Steel Sphere is propelled upwards from under water at significant depth changes as the sphere ascends to higher elevations closer to the water surface? It seems counter-intuitive to me that the force exerted on the sphere 3 meters below the water surface, is identical to the force a 1000 meters below surface? My intuition tells me that due to higher pressures at the bottom, the force and therefore the rate of ascension should be greater at the bottom? If not, why not?
2. If I build a second hollow sphere of identical surface volume (yet different skin thickness – internal volume), but ensure that its internal gas pressure is say hundreds or thousands of times greater, will it ascend faster with greater force than the first sphere? After all, that extra pressure is not “visible” outside the skin of the sphere?

I understand that spheres must have a lower mass to volume ratio than water to be buoyant, however, to know the internal pressure exactly will demand a sensor, machining of a new opening, seals/gaskets (thus compromising further the integrity of the sphere and $$$). On the other hand, if I over-pressurize (for sufficient buoyancy) by filling it with say, dry ice, by the time all the ice sublimates to gas, the sphere will be at the bottom ready to surface; once manually triggered to do so. If my intuition is wrong, and if both spheres ascend at the same rate regardless of internal pressure, then all I need to do is machine the sphere to ensure minimal buoyancy, allow air to enter before closure and ignore the dry ice and pressure sensors.

Please disregard underwater temperature gradients, water densities, currents, and salinity issues.
Many thanks for your help.

 

[Bill_K]

The buoyancy force is equal to the weight of the water displaced. For a given volume, this is the weight an equal volume of water, and would vary very little with depth since water is incompressible. So as the object rises, the buoyancy force is essentially constant.

Opposing this buoyancy force is the downward force due to gravity of the object. If you add something, anything, be it pressurized gas or sand, the object will rise more slowly due to the added weight.

 

[Doc Al]

Dear Friends,
I have a problem that has been driving me nuts for months now, and I am hoping you guys can help. I need to understand whether:
1. The Force with which a hollow, buoyant Steel Sphere is propelled upwards from under water at significant depth changes as the sphere ascends to higher elevations closer to the water surface? It seems counter-intuitive to me that the force exerted on the sphere 3 meters below the water surface, is identical to the force a 1000 meters below surface? My intuition tells me that due to higher pressures at the bottom, the force and therefore the rate of ascension should be greater at the bottom? If not, why not?

The buoyant force is due to the difference in pressure with height, not the absolute pressure. That difference in pressure is ρgΔh, which is the same at any depth.

2. If I build a second hollow sphere of identical surface volume (yet different skin thickness – internal volume), but ensure that its internal gas pressure is say hundreds or thousands of times greater, will it ascend faster with greater force than the first sphere? After all, that extra pressure is not “visible” outside the skin of the sphere?

Internal forces are irrelevant--for one thing, they add to zero. Only external forces count, such as the buoyant force due to the surrounding water. The only purpose of having increased gas pressure would be to keep the sphere from being crushed by high pressures. But as long as it keeps its volume, the buoyant force on it will be the same. (Of course, its weight may well be different.)

 

[joe_gunn]

Thank you Bill & Doc,
So if I understand this correctly, the sphere can theoretically contain vacuum and still rise simply based on its m/v ratio with respect to water? It should rise even faster since gas has some added weight?

With regards to r*g*(delta h): Does this mean that if I had a cylinder instead of a sphere, it would rise faster the taller it would be?
Doc, I want to be sure I understand the (delta h) term correctly. It refers to the absolute size of the object (from top to bottom), not the depth of the ocean - correct?

 

[Doc Al]

So if I understand this correctly, the sphere can theoretically contain vacuum and still rise simply based on its m/v ratio with respect to water? It should rise even faster since gas has some added weight?

Yes. The upward buoyant force depends only on the volume. The less mass it has, the greater the acceleration.

With regards to r*g*(delta h): Does this mean that if I had a cylinder instead of a sphere, it would rise faster the taller it would be?

Again, the buoyant force depends only on the volume of water displaced. But the acceleration depends on the mass of the object as well.

Doc, I want to be sure I understand the (delta h) term correctly. It refers to the absolute size of the object (from top to bottom), not the depth of the ocean - correct?

That expression just describes the change in pressure with height. It leads to the buoyant force, which turns out to equal the weight of displaced fluid. The key point is that the change in pressure surrounding the sphere (or any object) depends on the dimensions of the object (its volume), but not on the depth beneath the surface it is (the actual height).

 

[sophiecentaur]

The interesting thing about this is that the result is independent of the shape of the submerged object. It can be a sphere or a pointy thing (up or down) and the sums still give you an upwards force which is equal to the weight of the fluid displaced.
However, there may be a net torque on an odd shaped object that will make it rotate into an orientation with least Potential. So a vertical stick will end up floating horizontally. But that's 'second year work'.

 

[joe_gunn]

Gentlemen, forgive my stupidity, but I really want to understand this correctly. If the two objects have the same volume and mass - and one is wide and flat (like the surface dwelling fish) and the other is long and thin and vertical: according to r*g*(delta h) the long thin vertical one would surface faster? Yes? (ignoring turbulence and torque issues)

Secondly, when you speak of acceleration, if two objects have the same buoyancy, regardless of each object's mass to volume ratio, according to Bill's post above, there should be no acceleration at all. They should rise at exactly the same rate from the very bottom of the ocean to the very top. Yes?

This I think explains why bottom dwelling fish are horizontally so flat and thin, to minimize their buoyancy. But then, why are other fish thin and tall and vertical - to help them surface?

 

[Doc Al]

Gentlemen, forgive my stupidity, but I really want to understand this correctly. If the two objects have the same volume and mass - and one is wide and flat (like the surface dwelling fish) and the other is long and thin and vertical: according to r*g*(delta h) the long thin vertical one would surface faster? Yes? (ignoring turbulence and torque issues)

No. I brought up ρgΔh just to illustrate how the pressure difference is independent of depth, but that's not the formula for buoyant force. The buoyant force equals ρgv, where v is the volume of the object. The buoyant force is independent of shape.

Secondly, when you speak of acceleration, if two objects have the same buoyancy, regardless of each object's mass to volume ratio, according to Bill's post above, there should be no acceleration at all. They should rise at exactly the same rate from the very bottom of the ocean to the very top. Yes?

If they had no acceleration, why would they rise up? In any case, to determine the acceleration of an object, first find the net force on it. For objects at rest in a fluid, the forces are the upward buoyant force (which just depends on volume) and the downward force of gravity (equal to the object's weight). Once you have the net force, the acceleration is given by Newton's 2nd law: a = Fnet/mass.

 

[joe_gunn]

With all due respect,
But if gravity hasn't changed, and buoyancy hasn't changed and if mass and volume of the object hasn't changed then according to the math acceleration will also be constant. But there WILL BE ACCELERATION nevertheless. What I do not understand is: How is it possible for acceleration to be constant while the object is ascending through gradients of depths of different pressure. Its like poking a finger (with constant force) into a tube of butter that is frozen on one end and thawed out on the other. How is it possible that without changing the forces responsible for ascension, the rate of ascent is the same through the entire range of pressure differentials?

 

[russ_watters]

Constant force makes for constant acceleration. What will change the net force, however, is drag. An object ascending due to conatant buoyancy will quickly reach a terminal velocity and stop accelerating.

 

[Doc Al]

With all due respect,
But if gravity hasn't changed, and buoyancy hasn't changed and if mass and volume of the object hasn't changed then according to the math acceleration will also be constant. But there WILL BE ACCELERATION nevertheless.

You were the one who said the acceleration would be zero in your last post. I suppose that was just a typo.

In any case, of course there will be an acceleration. If we ignore the resistive force (fluid drag) as the object moves through the water, then the acceleration would be constant.

What I do not understand is: How is it possible for acceleration to be constant while the object is ascending through gradients of depths of different pressure. Its like poking a finger (with constant force) into a tube of butter that is frozen on one end and thawed out on the other. How is it possible that without changing the forces responsible for ascension, the rate of ascent is the same through the entire range of pressure differentials?

The gradient of pressure is the same everywhere. The buoyant force does not depend on depth. So the net force (ignoring fluid drag) remains constant, thus the acceleration remains constant. The rate of ascent would increase, of course.

(Edit: As Russ implies, ignoring drag is a bad assumption.)

 

[joe_gunn]

Doc,
You say that the:

The gradient of pressure is the same everywhere.

I assume you say this because water is incompressible, thus the pressure on one selected volume of water is the same as is on all the rest. How do you then explain the pressure difference due to the weight of the water at different depths. Surely the pressure of water on the object at the seafloor at 30 feet is different than it is at 1000 feet. Would the same object then ascend slower were it to begin its ascension at 1000 feet vs. at 30? I understand that the resulting ascending force is the same, but the volume of water pushing on the object is greater and thus heavier, causing the ascent at greater depth to initially be slower. Is this correct?

 

[Doc Al]

I assume you say this because water is incompressible,

Yes.

thus the pressure on one selected volume of water is the same as is on all the rest.

Pressure varies with depth.

How do you then explain the pressure difference due to the weight of the water at different depths. Surely the pressure of water on the object at the seafloor at 30 feet is different than it is at 1000 feet.

Absolutely. The pressure at 1000 feet is greater than the pressure at 30 feet. But the pressure difference between 1000 and 1001 feet is the same as the pressure difference between 30 and 31 feet. And it's only the pressure difference that contributes to the buoyant force.

Would the same object then ascend slower were it to begin its ascension at 1000 feet vs. at 30? I understand that the resulting ascending force is the same, but the volume of water pushing on the object is greater and thus heavier, causing the ascent at greater depth to initially be slower. Is this correct?

No, not correct. If the 'ascending force' is the same, how can they have different accelerations?

You need to realize that buoyant force is the net force due to the water pressure. A greater pressure will crush the object more, but will not increase the net force on it.

Let's take a very simple example and work it out. Say I have a cube of wood under the water at some depth. Further assume that it's aligned perfectly, so that the sides are vertical and the top and bottom horizontal. The length of each side is L. (So the volume is L³.) Let's calculate the buoyant force on this cube of wood directly by considering the water pressure on it. We can ignore the forces on the vertical sides of the cube, since the opposite sides will have equal and opposite force--thus those forces cancel out.

All we have to worry about is the force on the top and the bottom of the cube. Let's call the depth of the top of the cube h. (In other words, the top of the cube is a distance h below the water surface.) The depth of the bottom surface is a distance L deeper, thus h + L, and thus the water pressure on the bottom surface is a bit greater. Let's figure it out:
- top surface: Pressure = ρgh; total force on top surface = pressure*area = ρgh*L² downward
- bottom surface: Pressure = ρg(h+L); total force on bottom surface = pressure*area = ρg(h+L)*L² upward

Net force (the buoyant force):
Upward force - downward force =
ρg(h+L)*L² - ρgh*L² =
ρgh*L² + ρgL*L² - ρgh*L² = [STRIKE]ρgh*L²[/STRIKE] + ρgL*L² - [STRIKE]ρgh*L²[/STRIKE] = ρgL*L² = ρg*L³

That last term is simply ρg*volume, as expected. Note that the actual pressure due to the depth of the cube cancels out and thus does not matter. The final expression for the buoyant force does not include any mention of the depth of the water.

Although this calculation was for a special case, where the cube sides aligned perfectly for easy calculation, one can do the analysis for any shape and any orientation (using calculus) and get the same answer given by Archimedes' principle: The buoyant force equals the weight of the displaced fluid.

Let me know if this makes sense.

 

[sophiecentaur]

Actually, as the density of water will increase by a measurable amount, it could be argued that a larger weight of water is displaced for displaced a given volume at greater depth - so the upthrust would be greater.
But, on the other hand, if the submerged object has a lower modulus than the water, it will also be compressed at great depth and the volume of displaced water would be less so the upthrust would be less.
This is what can happen in discussions of 'basic principles' when they are taken to extremes. It can open more cans of worms and confuses the student.
Edit:
I see that I have repeated, in a few lines, more or less what the above post says except that I am saying that the difference in moduli will actually cause a difference in upthrust. After all, it applies to air bubbles so it must apply to everything.

 

[joe_gunn]

Doc, and Friends,

You are scholar/s and gentlemen! There is now one slightly less ignorant person walking/floating this earth. I am grateful for the fine explanations and your amazing quantity of your patience with me. It would take me months to go through all the books necessary to finally understand this. Don't get me wrong, I love books, it's just that my expertise is in telecommunications and optics - I never had this underwater stuff in undergrad or grad school. The cancellation of NET horizontal forces is what finally did it for me, leaving the top and bottom planes only as the relevant surfaces! The basic math of course is always beautiful icing on the cake. Russ, thanks for the critical velocity & drag explanation; once I correlated this to same effects in the atmosphere it made much more sense. Lastly, I suppose what Sophiecentaur could be referring to, is the experiment they conducted by sinking a Styrofoam mannequin head to a great depth only to retrieve the same head shrunken by like 40%. Many thanks again and if I can be of any help regarding fiber-optics or engineering, please holler!

 

[Integral]

A oceanographer friend once was on a cruise with a deep sea submersible. She had a 16oz Styrofoam cup which made a deep dive tied to the outside of the sub. It was the size of a shot glass.