Another Stupid Question. Warning: Science Content.

Quote: weaselman

I don't understand what you mean by this. The gravity drops like g*(R-r)/R, the pressure rises as rho*g*r/2. (r is distance from surface, rho is density of water).
What I don't get is what good does comparing one rate to the other do you. If you know how the pressure rises, you have the answer. Who cares about gravity?

Unless I'm misreading, you're computing pressure as a function of the entire Earth on the column of water. But the premise is that the invincibilium tube is impervious to surrounding pressure, so the only pressure that matters is the pressure exerted by the water itself. Presumably, however, the tube does not shield its aqueous contents from the effects of gravity so that's why I left that in because, as I understand it, the gravity gradient on the water affects the pressure (doesn't it?) and, therefore, whether it changes phase. Maybe I'm mistaken somewhere -- I'm certainly no physicist -- but I thought the question was reduced to "does a column of water in a tube through the center of the Earth change phase if you discount the effect of the pressure on the tube itself because it's magically impervious to that pressure?"

Definately answer #1 and i think the person who wrote this question is not very intelligent. What if there was no water on the other half and the hole started a the Indian Ocea. Would water erupt from the hole or just settle as a small puddle. I believe the pressure would gradually get smaller as the water reached the core due to the center of the earth having no gravity. When the water reached the center it would slowly come to a halt in the center.

Quote: Face

In parsing through the number of responses, we have two Physics masters seemingly at odds (if I'm reading this right). Wiz's father appears to be saying there would be flow, but due only to tides, giving a net effect of 0. Doc, while pretty much stating there would be no grand explosion of water, seems to imply that there would still be a flow, immeasurably small but STILL THERE, out of Erie and into the Indian.

LOL! I pulled how many years of Master's level education out of retirement... just to get back to where I started!! =D
...
Feel free to derail into properties of ice, the nature of gravity, or whatever else has begun in the last few posts.

Face, I agree that in recent posts we seem to have wandered well afield of the original topic onto issues of only esoteric interest. I think I'll let all that drop, myself.

I don't really see that the Wizard's dad and I disagree substantially. We both think that a modest size tunnel will not do much: Wiz Senior said he would expect negligible effect while I expected the flow to be immeasurably small, later describing it as seeping through. Sounds like quite similar conclusions to me. I confess I didn't really follow the idea of twice-a-day tidal effects -- don't opposite sides of the globe have low tides at roughly the same time and high tides at roughly the same time? I suspect the flow through the tunnel due to tidal effects would be negligible, too, though I might have missed something.

And don't get too carried away with the level of Physics "mastery" you drew from this source. I never pursued pure sciences beyond the undergraduate level. Interestingly (amusingly?) Wiz Senior and I seem to have followed mirror-image (?) academic paths: he did a BS in Mechanical Engineering before turning to Physics, while I did the Physics degree first before turning to Mechanical Engineering.

After all this, I think I'll take another nap.

Quote: MathExtremist

Unless I'm misreading, you're computing pressure as a function of the entire Earth on the column of water. But the premise is that the invincibilium tube is impervious to surrounding pressure, so the only pressure that matters is the pressure exerted by the water itself.

Yes, you are misreading (unless, I am misreading you :)). I completely ignore the pressure exerted by the Earth insides, on the tunnel, and only compute the pressure of the water column.
Here is, once again, how I do it.
If gravity was constant, the pressure at the bottom of the column would be
P0 = rho*g*R,
where rho is the density of water, g is the gravitational acceleration, and R is the length of the column (coincidentally, the radius of the Earth).

Assuming that Earth has constant density, and using Gauss law, I conclude that the gravity (g in the P0 formula) should decrease linearly as r (distance from surface) grows, becoming 0 at r=R (using the boundary condition, it is obvious btw, that the actual function for gravity drop is g(r) = (R-r)*g/R, it is not really necessary for the answer though). This means, that the actual answer is P1 = P0/2 (I think, this is obvious from just common sense and basic geometry, but, if not, it can be found with a simple integral).

This is my second approximation. In reality, the density is not constant though, and the gravity does not drop linearly. In fact it grows at first, and then starts dropping almost perfectly linearly from about r=R/2 to r=R.
My response to this is that this correction will make the final answer P2 > P1, and since P1 is already more than enough to turn water to ice, it is not necessary to look for the exact value of P2 in order to answer the original question. Simply knowing that it is greater than P1 (but less than P0) is enough to answer it.

Quote: Doc

After all this, I think I'll take another nap.

I aspire to have the lifestyle (and the environment) where I can take multiple naps per day. :)

Quote: Doc

Face, I agree that in recent posts we seem to have wandered well afield of the original topic onto issues of only esoteric interest. I think I'll let all that drop, myself.

Indeed, but for those enjoying it, I don't mind. Quite interesting, in fact.

Quote: Doc

I don't really see that the Wizard's dad and I disagree substantially. We both think that a modest size tunnel will not do much: Wiz Senior said he would expect negligible effect while I expected the flow to be immeasurably small, later describing it as seeping through. Sounds like quite similar conclusions to me. I confess I didn't really follow the idea of twice-a-day tidal effects -- don't opposite sides of the globe have low tides at roughly the same time and high tides at roughly the same time? I suspect the flow through the tunnel due to tidal effects would be negligible, too, though I might have missed something.

Oh, ho, ho! Well, I wasn't going to make that jump on my own, but after your interpretation and belief that the conclusions were "quite similar" I'm going to go ahead and declare workplace victory! Whether negligible, immeasurable, or infinitesimal, it's still THERE, and that was the spirit of the question. For what it's worth, tides were ignored. Whether the sides experienced the same tide simultaneously or not, I considered them a 0 net event. Whether or not that's true was a whole different day-long discussion and we had enough to work on at the time...

Quote: Doc

And don't get too carried away with the level of Physics "mastery" you drew from this source. I never pursued pure sciences beyond the undergraduate level. Interestingly (amusingly?) Wiz Senior and I seem to have followed mirror-image (?) academic paths: he did a BS in Mechanical Engineering before turning to Physics, while I did the Physics degree first before turning to Mechanical Engineering.

After all this, I think I'll take another nap.

Don't be so humble =). I eeked a H.S. diploma from the 3rd worst school in my region; I consider most people in this forum a "Master" in some form relative to me. Your physics is backed by accreditation from a University, mine is backed by reruns on The Discovery Channel =p. In any case, thanks for waking up and coming out of "forum retirement" for my silliness. Back to nappin' with ye!

Can water be a solid at very high temperatures?

If water were to fall down the "invincibilium" tube, its potential energy would be converted to kinetic energy. And when that kinetic energy is dissipated as internal energy, the temperature of the water near the center might be many thousands of degrees K turning it to steam, hydrogen and oxygen, or even a plasma.

To answer the original question, think of this model. A glass u-tube or even a clear plastic hose is partially filled with mercury. Into the right side pour some water onto the surface of the mercury. The level of the mercury on the right side would be lower than the mercury level on the left side. Do the calculation of levels by setting the weight of the fluid on the left side equal to the weight of the fluids on the right side and ignoring the weight of the air. The result is that the water-mercury border is below the mercury-air surface by a distance equal to (the difference in level of the water-air border and the mercury-air border) divided by (the difference between the density of mercury and the density of water.)

Think of the water-air surface as the surface of Lake Erie, and the mercury-air surface as sea level. Then the mercury-water surface corresponds to the border between the salt water and fresh water in the tube. From Wikipedia, Lake Erie's surface is 174 meters above sea level, and salt water has a density of about 1.025 g/ml. Make the assumption that after the system has come to rest that the fresh and salt water haven't mixed any and that no ice phase has formed in the tube to prevent equlibrium, then the salt water-fresh water border would be (174 meters)/(1.025 - 1.000) or about 7 kilometers below sea level. So, the invincibilium tube would be filled mostly with sea water, but with about 7 kilometers of fresh water on the Lake Erie side.

Quote: ChesterDog

Can water be a solid at very high temperatures?

If water were to fall down the "invincibilium" tube, its potential energy would be converted to kinetic energy. And when that kinetic energy is dissipated as internal energy, the temperature of the water near the center might be many thousands of degrees K turning it to steam, hydrogen and oxygen, or even a plasma.

According to the diagram, cited by Math Extremist, the temperature would need to be about 400 Celsius.
If the hole is being drilled directly at the bottom of the lake, it will be filled with water all the time, no sudden waterfalls.
However, even if the hole was first drilled dry, and then the water let into it, like Doc pointed out earlier, it would have a pretty modest terminal velocity due to air resistance, so, getting it sufficiently heated up this way is unlikely. Of course, the temperature near the center of the Earth could be used to heat the water and keep it liquid if necessary.

Quote:

From Wikipedia, Lake Erie's surface is 174 meters above sea level, and salt water has a density of about 1.025 g/ml. Make the assumption that after the system has come to rest that the fresh and salt water haven't mixed any and that no ice phase has formed in the tube to prevent equlibrium, then the salt water-fresh water border would be (174 meters)/(1.025 - 1.000) or about 7 kilometers below sea level. So, the invincibilium tube would be filled mostly with sea water, but with about 7 kilometers of fresh water on the Lake Erie side.

I have several problems with this formula. First, it turns to infinity when the densities on both sides are equal, which cannot be right.
Second, its units are length^4/mass, and the way you are using it implies length. And finally, you have meters in the nominator and cubic centimeters (milliliters) in the denominator, which can't be right either.

But the idea is right. The correct formula for two liquids is h1 = rho2 * h2 / rho1. So h1 = 174/1.025 = 169.76.
That is how high the Indian Ocean level would rise eventually if there was enough supply in lake Erie to feed all the necessary water, and not drop its own level at the same time. Also, this would be quite a lot of fresh water mixing into the Ocean, which would affect the density ... In two words, this is not as simple as it looks. All the dynamic problems are like this - you start looking at the initial condition, and it looks pretty simple to figure out how it will evolve, but then things start changing all at the same time, and it quickly gets out of hand.

Quote: weaselman

...If the hole is being drilled directly at the bottom of the lake, it will be filled with water all the time, no sudden waterfalls...

Good point! I like that idea. I wouldn't allow any heat transfer through the walls of the invincibilium tube, so I'm sure much of the tube would be filled with one or more of the solid phases of water.

Quote: weaselman

...Second, its units are length^4/mass, and the way you are using it implies length. And finally, you have meters in the nominator and cubic centimeters (milliliters) in the denominator, which can't be right either.

To correct the dimensional discrepancy, I'll revise my formula to this: depth to the salt water-fresh water border = (distance above sea level of the surface of Lake Erie) * (density of fresh water) / (density of sea water - density of fresh water).

Regarding the zero-denominator problem: if the sea water had the same density as fresh water, then Lake Erie would completely empty into the ocean through the tube. There would be no equilibrium depth for the border between the sea water and the water from Lake Erie.

Quote: ChesterDog

To correct the dimensional discrepancy, I'll revise my formula to this: depth to the salt water-fresh water border = (distance above sea level of the surface of Lake Erie) * (density of fresh water) / (density of sea water - density of fresh water).

Now, if the lake is at the sea level, you get zero for the border.

I am not sure what you mean by the "border" by the way. There is no "border" really, as long as one of the levels is sufficiently higher than the other to compensate for the difference in densities (like in this case) - the whole tunnel will be filled with the fresh water, that will be continuously flowing (however slowly) ), raising the level of the ocean and lowering the level of the lake, until h1/h2 = rho2/rho1, at which point the flow will stop.

Quote: Face

... Whether negligible, immeasurable, or infinitesimal, it's still THERE, and that was the spirit of the question.

...

Back to nappin' with ye!

I thought I had given up on this discussion, but it looks as if I got sucked back into the vortex.

I am reminded of an old discussion problem about a bouncing ball. The terms of the problem are that a semi-elastic ball is dropped from a modest height onto a semi-elastic, horizontal surface. There is no air resistance or other losses other than the in-elasticity in the collision/bounce of the materials. The elasticity of the materials is such that in each successive bounce, the ball rises to 90% of its height on the previous bounce. The question is this: how many times does the ball bounce before coming to rest on the surface (or does it ever come to rest)?

Well, mathematically, you can calculate the ever-decreasing height of a series of bounces that approaches an infinite number. That viewpoint suggests one might consider the bounce "whether negligible, immeasurable, or infinitesimal, it's still THERE". Another viewpoint might suggest that once the height of the bounces is less than the surface imperfections (if even on a molecular level) of the ball, then there isn't any motion left that can be considered a bounce.

If you could never detect a bounce, do you think the ball is really still bouncing? If the flow through the tunnel were negligible, immeasurable or infinitesimal, would you still consider the water really was flowing?

On second thought, taking a nap is a better idea than discussing this....

Quote: weaselman

...The correct formula for two liquids is h1 = rho2 * h2 / rho1. So h1 = 174/1.025 = 169.76....

Here's a concrete example to help us come to closer agreement. Mercury (density = 13.5 g/ml) is placed in a glass u-tube filling both sides to the same level. (I hadn't made this point clear before--I didn't mean that the mercury column was balanced againt a water column.) Water (density = 1.0 g/ml) is poured into the right-hand side of the tube until the mercury level is 2.8 cm lower on the right than on the left. What is the length of the water column?

I get a water column length of 37.8 cm for a difference between the water surface and the left-hand side mercury surface of 35.0 cm. This agrees with my formula since 2.8 = 35.0*1.0/(13.5-1.0).

Ok, now I understand. Yes, this is correct in this case, just in an unusual form :)


It is easier to calculate the length of the water column by only considering the stuff that is above the division line:
h1*rho1 = h2*rho2, so h1 = h2 * rho2/rho1 = 2.8 * 13.5/1.0

I find it less confusing this way, because it lets you deal with any shape of the tube (like a tunnel through the center of Earth for example :)), and any combination of liquids (there is no division by zero), absence of a division line (when there is a constant flow from one reservoir to the other) and is generally simpler

Quote: weaselman

...two liquids is h1 = rho2 * h2 / rho1. So h1 = 174/1.025 = 169.76...

Thanks! I see that your formula is much clearer than mine. And finally I remember it and a diagram of it from science class.

Using Lake Erie's elevation of 174 m above sea level, and sea water's density at 1.025 g/ml, I would use your formula this way: Suppose the fresh water and sea water mixed negligibly, and that a column of fresh water floated above a column of sea water in the tunnel--that is, suppose the system reached an equilibrium with no additional water flow in either direction. [Edit: To prevent any mixing of fresh water and salt water, put a sphere, with density intermediate between fresh and salt water and with a diameter just slightly smaller than the diameter of the tunnel, between the fresh and salt water.] If the sea water/fresh water boundary were x below sea level, then h1=x and rho1=1.025. h2=x+174 and rho2=1.000. Then 1.025x = 1.000(x+174). So x = 174m/0.025 = 6.96 km. So, perhaps a column of fresh water of length 6.96 km + 0.17 km = 7.06 km could float on a column of sea water.

Do you think the system could be at this equilibrium, or do you think the fresh water would continue to flow out of Lake Erie?

First of all, I think, it is really confusing to think about this in terms of a boundary. The boundary can really be anywhere, depending on the volumes of salt water and fresh water relative to each other, and the shape of the tunnel. It is the level of the surface, that should be fixed, not the boundary.
Now, if you have two reservoirs, connected to each other, one with fresh water, and one with salt water, and the former is at 174 meters, then the surface of the other one is at about 168 meters (I have calculated it in one of earlier posts, using this formula). As long as the boundary is below the sea level, it can be anywhere in the tunnel, depending, like I said, on the volumes, the shape of the tunnel, and also on the dynamics of how exactly that tunnel is initially filled.
For example, if supply of fresh water is unlimited, and we drill the tunnel slowly, starting at the bottom of lake Erie, it will be completely filled with fresh water when it reaches the floor of the ocean. The fresh water will then continue pushing up until the surface of the ocean rises by 168 meters. Where will the boundary be then? Obviously, it will be 168 meters above the ocean floor. But to tell it's location relative to the sea level, one would need to know the initial depth of the ocean. If the ocean floor is H meters below the sea level, then the boundary will be H-168 below.