Another Stupid Question. Warning: Science Content.

Another boring day of work led to more ridiculous conversation which brought up a question that caused much debate with no clear winner of an answer. After hours of arguing, we stood locked at four different hypothesis with no one side giving in, and no way to prove it one way or the other. I'm thinking those here would have the knowledge and the will to answer this, so I bring it to you. I'll save the lead up and get right to the facts and the four beliefs of what would be true. Pick one answer or supply your own, if you would.

Long story short, we imagined drilling a hole straight through the bottom of Lake Erie directly through the exact center of the Earth and out the other side, which would have you coming out at the bottom of the Indian Ocean. We assume the Earth is a perfect sphere, and that the "tunnel" is structurally sound and impervious to tectonic shift, pressure, heat, etc, forever. Once the initial rush of water filling the hole subsided, what would happen?

1. All that matters is the water "in the tunnel". Since there is equal water on each side of the center, the total effect would be 0. The water stays stagnant for eternity.

2. Salt water is heavier than frsh water. The Indian side would have more weight and push the fresh water out of the Erie side's tunnel.

3. The Indian Ocean is much deeper than Lake Erie and the pressure at the bottom of the Indian would overcome the pressure on the Erie side. The Indian Ocean would drain into Erie and flood the Great Lakes region.

4. Lake Erie is above sea level whereas the Indian Ocean is at sea level, so Lake Erie would drain into the Indian for eternity.

Hopefully this is more stimulating than my "5 unit fish" question =P

Editted to correct #2

Pretty sure you have a problem with #2. Salt water is heavier than fresh water that is why you float easier in salt water ie Great Salt Lake or Dead Sea.

Quote:

We assume the Earth is a perfect sphere

So 4 is out.

Quote:

The Indian Ocean is much deeper than Lake Erie

Not if you drill a hole through the center of an assumed perfectly spherical Earth. Then they are both equally deep. 3 is out.

Quote:

Fresh water is heavier than salt water.

Not so much heavier that it would push salt water out of the bottom of the Indian Ocean before mixing with it. 2 is out.

1 wins.

Quote: Face

Another boring day of work led to more ridiculous conversation which brought up a question that caused much debate with no clear winner of an answer. After hours of arguing, we stood locked at four different hypothesis with no one side giving in, and no way to prove it one way or the other. I'm thinking those here would have the knowledge and the will to answer this, so I bring it to you. I'll save the lead up and get right to the facts and the four beliefs of what would be true. Pick one answer or supply your own, if you would.

Long story short, we imagined drilling a hole straight through the bottom of Lake Erie directly through the exact center of the Earth and out the other side, which would have you coming out at the bottom of the Indian Ocean. We assume the Earth is a perfect sphere, and that the "tunnel" is structurally sound and impervious to tectonic shift, pressure, heat, etc, forever. Once the initial rush of water filling the hole subsided, what would happen?

1. All that matters is the water "in the tunnel". Since there is equal water on each side of the center, the total effect would be 0. The water stays stagnant for eternity.

2. Fresh water is heavier than salt water. The Erie side would have more weight and push the salt water out of the Indian side's tunnel.

3. The Indian Ocean is much deeper than Lake Erie and the pressure at the bottom of the Indian would overcome the pressure on the Erie side. The Indian Ocean would drain into Erie and flood the Great Lakes region.

4. Lake Erie is above sea level whereas the Indian Ocean is at sea level, so Lake Erie would drain into the Indian for eternity.

Hopefully this is more stimulating than my "5 unit fish" question =P

Assuming gravity does not have any effect or that effect is evened out, I feel 3 would be the closest to reality. Fresh water is heavier than salt water, but the volume of water in the Indian Ocean is much grater and would have much more weight. I do not think "sea level" would matter at all so 4 is out for me. As I said, volume is more important than unit weight so 2 is out for me. 1 is a nonstarter since the water on either side is not "equal".

Drill the hole and find out. Duh! :P

Quote: kenarman

Pretty sure you have a problem with #2. Salt water is heavier than fresh water that is why you float easier in salt water ie Great Salt Lake or Dead Sea.

Typo. The concept remains the same, though. One side's water is heavier and which makes that the deciding factor.

And a quick clarification to keep the questions the same, by "sphere" we didn't mean smooth like a marble. The geography is the same, we were just eliminating the possible oval/egg shape arguement that the Earth may have from detracting from the question at hand.

Quote: Nareed

Drill the hole and find out. Duh! :P

I tried when I was younger, Pops hollered because we could've been buried and killed =)

Physics is not my strong suit, but I go with #4. I see this as all the water in Lake Erie being siphoned into the Indian Ocean. There would also be stagnant water in the hole, from the Indian Ocean side, up to sea level on the US side.

I would await to hear from Doc, our resident physics expert, for the definitive answer. If he doesn't find this himself I'll send a reminder in a couple days.

If we accept sea level as being equal everywhere in the world than some variation of 4 is probably right. I am not sure that sea level is actually the same on all the oceans even ignoring tides. Once we bring tides into play we could have water surging back on forth as 5th option.

Since atmospheric pressure is lower at higher elevations, the Indian Ocean would push water into the tunnel and into the lake.

Quote: Face

we imagined drilling a hole straight through the bottom of Lake Erie directly through the exact center of the Earth and out the other side, which would have you coming out at the bottom of the Indian Ocean. We assume the Earth is a perfect sphere, and that the "tunnel" is structurally sound and impervious to tectonic shift, pressure, heat, etc, forever. Once the initial rush of water filling the hole subsided, what would happen?

My guess is that the earth's gravity would draw all of the water from both sides to the center of the sphere where it would remain. I imagine that anything dropped into that hole would simply fall due to the earth's gravity until it reached equilibrium - right in the middle. Yes? No??? Would that mean I've chosen answer #1?

Quote: matilda

Since atmospheric pressure is lower at higher elevations, the Indian Ocean would push water into the tunnel and into the lake.

This was brought up. 10 feet of water is of equal pressure to 1 whole atmosphere. With such depths at work here, I figured atmosphere's impact was negligible, even if the ocean was in death valley and the lake sat atop Everest. I could be wrong...

Quote: Wizard

Physics is not my strong suit, but I go with #4. I see this as all the water in Lake Erie being siphoned into the Indian Ocean. There would also be stagnant water in the hole, from the Indian Ocean side, up to sea level on the US side.

I would await to hear from Doc, our resident physics expert, for the definitive answer. If he doesn't find this himself I'll send a reminder in a couple days.

I'm confused. If the Indian drains into the Erie side up to sea level and remains stagnant, how would Erie also drain into the Indian? Are you saying the Indian would keep some level of saltiness on the Erie side of the pipe, but not to the "surface", while Erie would at the same time sort of push through into the Indian?

Before Doc gets to put us all straight, Ill throw my official answer in the mix. I originally said #3 since the pressure at the bottom of the Indian was obviously quite a bit more than at the bottom of Erie's 300 foot deep basin. I switched to my final answer of #4 after thinking, of all things, college days and beer bongs =D. I equated Erie with the small mouth piece and the Indian with the large funnel, with the tube representing the hole we dug. With the entire volume of the device full of beer, if you slip up and lift the mouth piece too high, beer will overflow out of the funnel, no matter how big the funnel is or how heavy the beer is. Since Erie is most definately at a higher elevation than the ocean, I went with answer #4.

Quote: TheNightfly

My guess is that the earth's gravity would draw all of the water from both sides to the center of the sphere where it would remain. I imagine that anything dropped into that hole would simply fall due to the earth's gravity until it reached equilibrium - right in the middle. Yes? No??? Would that mean I've chosen answer #1?

From what I understand, if the hole was empty, say it was drilled, capped and had the water pumped out, yes, a rock dropped in would oscilate slower and slower until it stopped in the center. The water deal differs as it is still influenced by forces from outside of the pipe (the water still pushing from above), or at least it is my belief that it is so.

Quote: Face

From what I understand, if the hole was empty, say it was drilled, capped and had the water pumped out, yes, a rock dropped in would oscilate slower and slower until it stopped in the center. The water deal differs as it is still influenced by forces from outside of the pipe (the water still pushing from above), or at least it is my belief that it is so.

I also believe this is true. So, let's work from that.

Imagine all the water had somehow flowed to one side or the other. Wouldn't gravity (and the pressure above) pull some of it back into the tunnel? By the same argument, some of that water will go past the centre of the earth. There must be an equilibrium point where there is water on both sides. I'm leaning towards #1, with some initial changes: the actual levels would depend on complicated factors.

Quote: kenarman

If we accept sea level as being equal everywhere in the world than some variation of 4 is probably right. I am not sure that sea level is actually the same on all the oceans even ignoring tides. Once we bring tides into play we could have water surging back on forth as 5th option.

I read somewhere that sea level on both sides of the Panama Canal are about three feet apart. I'm ignoring that factor, as that effect is miniscule compared to the elevation of Lake Erie.

Quote: TheNightfly

My guess is that the earth's gravity would draw all of the water from both sides to the center of the sphere where it would remain. I imagine that anything dropped into that hole would simply fall due to the earth's gravity until it reached equilibrium - right in the middle. Yes? No??? Would that mean I've chosen answer #1?

My interpretation of the problem is that it is a narrow hole, like a foot in diameter, so it would not significantly lower the ocean level.

It is true that if you dug a hole straight through the earth, passing through the center, with dry land on both sides, and you fell in you would keep going back and forth until you finally settled in the middle of the earth, where you would experience weightlessness. I think I learned that from an episode of Cosmos.

Quote: Wizard

Physics is not my strong suit, but I go with #4. I see this as all the water in Lake Erie being siphoned into the Indian Ocean. There would also be stagnant water in the hole, from the Indian Ocean side, up to sea level on the US side.

I would await to hear from Doc, our resident physics expert, for the definitive answer. If he doesn't find this himself I'll send a reminder in a couple days.

The assumption is that the Earth is a perfect sphere thereby disqualifying #4 as a possibility.

Quote: Wizard

I read somewhere that sea level on both sides of the Panama Canal are about three feet apart. I'm ignoring that factor, as that effect is miniscule compared to the elevation of Lake Erie.

Yeah, we ignored conditional factors to keep from getting off topic and making it complicated. Hurricanes, Erie's ice-over, tides, all these things were ignored. Similar to gambling odds, we were thinking long haul. After a year for the initial rush to settle, what would be the net effect over the next million years, basically.

Quote: Wizard

My interpretation of the problem is that it is a narrow hole, like a foot in diameter, so it would not significantly lower the ocean level.

Lol, yes, you are correct. We have a guy here who always brings up derailers, which is where the weather, heat, tectonic shift conditions came from. As far as pipe size is concerned, I'm not sure it matters as long as it's not absurd. A hole 50 miles wide and thousands of miles deep may well drain Erie and kill the spirit of the thought experiment, for example

Quote: s2dbaker

The assumption is that the Earth is a perfect sphere thereby disqualifying #4 as a possibility.

Why does that kill it as a possibility? Water will still seek the lowest point.

Quote: s2dbaker

The assumption is that the Earth is a perfect sphere thereby disqualifying #4 as a possibility.

Quote: Face

And a quick clarification to keep the questions the same, by "sphere" we didn't mean smooth like a marble. The geography is the same, we were just eliminating the possible oval/egg shape arguement that the Earth may have from detracting from the question at hand.

To further clarify, the same derailer mentioned earlier kept throwing the "egg shaped Earth" debate at us. What the guy was saying was that the Earth was somewhat egg shaped and thus the distance from center Earth to sea level on one side could be a good deal different than center to sea level on the other. To kill the derailing debate we went with this - The distance from the direct center of the Earth to absolute sea level shall not change. (it is different, as the Wiz pointed out, but this was supposed to be fun, not a physics thesis ;)) The Earth will still retain it's geographic features; Death Valley will still be below sea level, Everest will still be the highest peak, Lake Erie will still be above sea level, but the Earth will be considered a sphere for this purpose. So, the "egg" debate was disallowed, Earth is a "sphere", and it still retains it's geography. =)

Quote: Nareed

Drill the hole and find out. Duh! :P

Thanks for the laugh of the day!

Nareed, you're coming up on 4,000 posts. Do you think you'll catch the Wizard?

Even if we allowed the squished-sphere shape of the earth to come into the question, I don't think it would matter. For those who don't know, the surface of the earth at the equator is further from the center of the earth than the poles. For this reason, the furthest point from the center of the earth is not Mount Everest, but Mount Aconcagua.

So even if we allowed for the squished earth shape to enter the discussion, I think water would not pour into Lake Erie for the same reason it doesn't drain out by the equator and cover the two poles. Why this doesn't happen, I'm not sure, to be honest with you. Maybe because the centrifugal force of the spinning earth drawing water to the equator is offset equally by gravity pulling water to the polls.

Doc, we need you here!

Quote: Wizard

Doc, we need you here!

Agreed. Physics aside, I kinda miss the ol' bugger. =) Seems he's had enough of you all since Vegas, what did you do to the poor man?! ;)

Quote: Wizard

It is true that if you dug a hole straight through the earth, passing through the center, with dry land on both sides, and you fell in you would keep going back and forth until you finally settled in the middle of the earth, where you would experience weightlessness. I think I learned that from an episode of Cosmos.

I remember from some geography class that you would fall in and speed up until you hit the core, then you would slow down until you were near the other side, if you did not grab onto something it would repeat. What would happen eventually, prepetual motion or settle, they did not say. I don't think you would be weightless. I don't think you would even stop. Some kind of gravity would keep moving you IMHO. It is the same reason the Universe must be expanding or contracting, supposedly in a cycle.

If it is contracting it will all collapse on itself sooner or later, somehting to consider if you want to put off mowing your lawn.

Quote: AZDuffman

I remember from some geography class that you would fall in and speed up until you hit the core, then you would slow down until you were near the other side, if you did not grab onto something it would repeat. What would happen eventually, prepetual motion or settle, they did not say. I don't think you would be weightless. I don't think you would even stop. Some kind of gravity would keep moving you IMHO. It is the same reason the Universe must be expanding or contracting, supposedly in a cycle.

If it is contracting it will all collapse on itself sooner or later, somehting to consider if you want to put off mowing your lawn.

Only in a vacuum, I'd reckon. The aero resistance would slow you eventually, although I think it'd take quite a long time to come to a complete stop.

5. No water would rush in at all, from either side. Since the Earth's core has a pressure about 3500x greater than the bottom of the ocean, you'd create volcanoes at both ends of the tube. This assumes that you're drilling into the actual Earth rather than a fictional Earth with a solid core.

Quote: MathExtremist

5. No water would rush in at all, from either side. Since the Earth's core has a pressure about 3500x greater than the bottom of the ocean, you'd create volcanoes at both ends of the tube. This assumes that you're drilling into the actual Earth rather than a fictional Earth with a solid core.

But you missed the part of this being ridiculous ;) The "tunnel" is impervious to heat, pressure, etc. Imagine a pipe made of invincibilium rather than a simple rough hole. And in case anyone missed this part of Chemistry, "invincibilium" reacts with nothing, is completely unbendable, unbreakable, unmaleable, exists only as a solid, is 100% non conductive of heat, and exists only in pipe form =p

It's between #2 and #4. The shape of the tunnel is irrelevant, as long as it is filled with water.
Imagine Indian Ocean sitting next to lake Erie, and connected with a tunnel.
If water on both ends was fresh, the flow would be from high level to lower level surface. But salt water is heavier, so, it could be that it is just heavy enough to compensate for the difference in heights, in which case there would be no flow at all. Or it could be even heavier, in which case the flow would go the opposite direction.
The exact answer depends on the exact densities of the water and exact heights of the surface. It is also possible that the water starts flowing from lake Erie, but as the level drops, the flow will stop and reverse itself, and repeat, and eventually stabilize at some level where the difference in densities exactly compensates for the difference in levels

Quote: AZDuffman

I remember from some geography class that you would fall in and speed up until you hit the core, then you would slow down until you were near the other side, if you did not grab onto something it would repeat. What would happen eventually, prepetual motion or settle, they did not say. I don't think you would be weightless. I don't think you would even stop. Some kind of gravity would keep moving you IMHO. It is the same reason the Universe must be expanding or contracting, supposedly in a cycle.

You would not stop if you were falling in a vacuum, that is correct. Not the same reason the universe has to be expanding or contracting, but rather same reason a pendulum would never stop oscillating if not for the resistance from the air, and its lever.
Yes, you'd be weightless the whole time. You don't have to fly to the center of the Earth to be weightless. Jumping down from your couch, you are weightless the whole time until you reach the floor.

Quote: weaselman

It's between #2 and #4. The shape of the tunnel is irrelevant, as long as it is filled with water.
Imagine Indian Ocean sitting next to lake Erie, and connected with a tunnel.
If water on both ends was fresh, the flow would be from high level to lower level surface. But salt water is heavier, so, it could be that it is just heavy enough to compensate for the difference in heights, in which case there would be no flow at all. Or it could be even heavier, in which case the flow would go the opposite direction.
The exact answer depends on the exact densities of the water and exact heights of the surface. It is also possible that the water starts flowing from lake Erie, but as the level drops, the flow will stop and reverse itself, and repeat, and eventually stabilize at some level where the difference in densities exactly compensates for the difference in levels.

The "water level" idea came up, we kind of ignored it figuring it was another derailer and assumed that which ever one drained would be at a rate slower than it was replenished, which, if the pipe was no bigger than, say, a highway-through-a-mountain sized tunnel, would be true. It would have to be a damn big hole to outrun all of the tributaries to either body, I'd think.

I was hoping it wouldn't come to specifics, but the more this is discussed the more it seems it might. I thought of it the same way, lose the "through the Earth" part and just imagine them side by side (my beer bong example). The height, I'm confident, is a key factor, and I'm starting to think the only way to come to a "true enough" answer is to have a vague knowledge of how high the lake is as well as a rough volume (one is a simple Google search, one is nigh impossible for this guy's mind to comprehend) I hope Doc can set us straight, if he can be found...

Quote: Face

Quote: Wizard

Doc, we need you here!

Agreed. Physics aside, I kinda miss the ol' bugger. =) Seems he's had enough of you all since Vegas, what did you do to the poor man?! ;)

Sorry to be so slow to respond, but I guess I must have been taking a nap lately. Actually, folks were quite kind to me during my last visit to Las Vegas. It was the dice that seemed to have a nasty attitude.

As for the "hole through the earth" topic, I'd bet most every kid who has ever heard that the earth is a sphere has contemplated some version of that, including those folks who have proposed the sub-oceanic, high-speed, inter-continental transport systems.

My first reaction when I read the question was very similar to what MathExtremist said, what with the molten core and such. Just saw a nice illustration of the moon's core in the August issue of National Geographic. The very center of that core is the hottest, but due to intense pressure it is a solid iron alloy, even though it is surrounded by a liquid core of iron alloy. I'm no geologist, so I don't know just how the inner layers of the earth are configured. I think the use of the invincibilium pipe is a good solution to the molten core issue though, provided you can get the drilling apparatus to work during construction. Well done, Face. :-)

I also would like to thank the Wizard for his comment about Mt. Aconcagua. I should read more about its status as farthest land point from the center of the earth. I had never heard that before.

As for the concept of a person falling into the empty tunnel, accelerating, going all the way through to the other side, and then oscillating, there is the complicating factor that the acceleration due to gravity varies with location and would be reduced at each greater depth in the tunnel. At the center, there would be gravitational attraction in all directions, with no net acceleration. Given the drag from air resistance, there is a terminal velocity of the fall (as experienced by sky divers), but with the reducing force of gravity as the fall progresses toward the center, this "terminal" velocity would decrease along the way. I suspect that the falling body will have a rather modest velocity as it passes through the center on its first oscillation and would come essentially to rest in the center rather quickly. This is probably a fairly common differential equations problem in viscous damping with variable driving force, but I doubt my atrophied brain could handle it any more.

So what about the original problem with water from Lake Erie and the Indian Ocean? I just re-read the proposed answers, and I don't really like any of them. If I really had to make some kind of choice, I think I would go mostly with #1, while leaning a bit to #4.

The way I view it, if the pipe were initially empty (or only contained air) and suddenly valves were opened at the bottom of Lake Erie and the bottom of the Indian Ocean, then the initial filling of the pipe would be quite an interesting process. Once the "dust" had settled, though, it would just be an invincibilium pipe filled with water. Yes, in the early days following the Great Flood, a portion of the pipe would have salt water from the ocean and part fresh water from the lake, but given just a little time, the combination of water mixing and salt diffusing through the water would lead to salt content essentially all the way through the pipe, regardless of where "that" water came from initially. It's not as if the individual water molecules are likely to be tracked.

Once stability is reached, water from Lake Huron will still flow through Lake Erie on its way to the Gulf of St. Lawrence and the prevailing currents will still circulate in the Indian Ocean. Neither body of water will drain, and neither body will exhibit a geyser of the wrong kind of water. But since the surface of the Indian Ocean is lower than the surface of Lake Erie, the tunnel will provide one more minor path for a tiny amount of the fresh water to reach the sea, mixing with the salt water in the tunnel instead of in the Gulf. I suspect the flow rate through the tunnel would be immeasurably small.

Edit: I just caught this comment from weaselman:

Quote: weaselman

Yes, you'd be weightless the whole time. You don't have to fly to the center of the Earth to be weightless. Jumping down from your couch, you are weightless the whole time until you reach the floor.

Well, this is a topic where the meaning of words could be argued for a long time (and probably are in dorm rooms or around water coolers), but I think I have to disagree with the way this is stated here.

The most common version of this argument is whether someone floating around the space station or in the midst of a dive into a pool is experiencing "zero g". In fact, both are free falling and experiencing exactly one g acceleration (local g) toward the earth's center. While I am sitting here in front of my computer, I don't seem to be falling at all, so apparently I am the one experiencing zero g, as well as most of you, perhaps. Of course, most of us use the term zero g to describe that condition where we aren't being pulled against something such as my desk chair, so we just have conflicting usage of terms.

Similarly, I disagree with the wording of the statement that when jumping down from your couch you are weightless until you reach the floor. If you were weightless, you would likely just float there, or even more likely be buoyed up to the ceiling by the air which does have weight. But you have weight, which is why you fall, and you still have the same weight after you crash onto the floor. There you just feel the resistance to your weight.

Quote: Doc

Neither body of water will drain, and neither body will exhibit a geyser of the wrong kind of water.

Thanks Doc. However, I still don't see why Lake Erie won't drain. The entire lake is above sea level (source). It seems to my amateur-in-science mind Lake Erie should drain completely? Much for the same reason if I siphoned water out of a bathtub it would stop when the water level of the bathtub equaled the end of the hose.

Quote: Face

But you missed the part of this being ridiculous ;) The "tunnel" is impervious to heat, pressure, etc. Imagine a pipe made of invincibilium rather than a simple rough hole. And in case anyone missed this part of Chemistry, "invincibilium" reacts with nothing, is completely unbendable, unbreakable, unmaleable, exists only as a solid, is 100% non conductive of heat, and exists only in pipe form =p

Okay -- no heat gets through. And no pressure from the surrounding molten rock (which is something like 3 million atmospheres in the inner core). However, what is the pressure exerted upon the water by itself as it descends? If it is great enough, and the heat does not increase, the water may turn into some non-standard form of ice as it descends. It may never reach the center of the Earth from either side, instead crystallizing inside the tube midway down.

See the Wikipedia article on the phases of ice. It has a nice chart.

Can anyone with more practical physics background than me determine what the pressure would be on the water column as it descends?

Quote: Doc

Well, this is a topic where the meaning of words could be argued for a long time (and probably are in dorm rooms or around water coolers), but I think I have to disagree with the way this is stated here.

The most common version of this argument is whether someone floating around the space station or in the midst of a dive into a pool is experiencing "zero g". In fact, both are free falling and experiencing exactly one g acceleration (local g) toward the earth's center. While I am sitting here in front of my computer, I don't seem to be falling at all, so apparently I am the one experiencing zero g, as well as most of you, perhaps. Of course, most of us use the term zero g to describe that condition where we aren't being pulled against something such as my desk chair, so we just have conflicting usage of terms.

Hm. I don't know about that. Up until now, every time I heard a term "so many Gs" used, it was always in the context of the force (weight) you are feeling. By the same logic, if you are being spun in a centrifuge, you have to conclude you are experiencing "zero g" even while your body is being flattened along the wall.

I am sure, you know this ... If you are sitting in front of your computer in a falling elevator with no windows, you have no way to tell if you are falling down towards the center of the Earth or floating around in space.
It was very surprising for me to hear this kind of objection from you of all people on this board (no offense intended to the other folks :) It's just that Doc being considered our resident physics expert and all ...)

Quote:

But you have weight, which is why you fall,

Come on, Doc. Weight is not the same thing as mass. Mass is why you are falling, not weight.
Weight is the force that you exert on the elevator floor due to gravity. If you are in a free fall, there is no force, you are weightless.
Think about astronauts orbiting the Earth in a space station. I hope nobody will argue that they are not "weightless" in any reasonable sense of the word. At the same time, they are not free from the Earth's gravitational field (if they were, they would fly away from Earth), they are essentially in the same state of the free fall as you would be inside an elevator falling down towards the center of the Earth.

weaselman:

This is why I said this is a matter of conflicting usage of terms. First, with regard to "so many Gs" being a description of the force, yes, that is one interpretation of the term. But since "g" also is used to refer to a rate of acceleration, there is the potential for conflicting interpretations.

If I were being spun in a centrifuge and flattened against the wall, I would certainly not be experiencing zero g acceleration. The rotational motion of the centrifuge can create enormous accelerations. I think the simple formula for the centripetal acceleration is the radius multiplied by the square of the angular velocity. While sitting in my chair, I actually am experiencing a centripetal acceleration due to the rotation of the earth, but it is fairly small and isn't obvious to me as I sit here, unlike the more obvious acceleration during a jump off a couch.

To me, "weight" means the force that a body experiences due to the fact it has mass and the fact that it is exposed to a gravitational field. If you are in free fall, there indeed is a force -- you are accelerating because the force of your weight is not counteracted by any other force. I do not consider the free falling person to be weightless at all. But there is a sensation during free fall that is commonly referred to as weightlessness. It is something that most of us never experience for extended periods. We are much more accustomed to the situation where a resisting force (such as from my chair) acts against our weight and keeps us from accelerating. When that resisting force is removed, it feels as if we have no weight holding us in the chair, but the only way to do that in the gravitational field is to allow for acceleration and free fall. Our weight is still there, but without the resisting force we experience a very different sensation.

Quote: Wizard

Thanks Doc. However, I still don't see why Lake Erie won't drain. The entire lake is above sea level. It seems to my amateur-in-science mind Lake Erie should drain completely? Much for the same reason if I siphoned water out of a bathtub it would stop when the water level of the bathtub equaled the end of the hose.

Because Lake Erie is already draining right now... into the next lake and the next... but also being filled up from the other end at the same time. Forgetting the fact the hole has a link to the Ocean, if the outflow down our imaginary hole is bigger than the inflow, it will eventually drain away.

The weight of the water column at the centre of the earth doesn't depend on the size of the lake/ocean... only the height and area of the column (something I learnt/relearnt when we had the discussion about the force on a dam). Unless I misunderstood that.

I Emailed my dad this problem, who agrees with Doc and others saying the effect would be negligible. My dad has an undergraduate degree in mechanical engineering from Yale, and Ph.D. in physics from Cal Tech, so I think his credentials are good.

Quote: Wizard's dad

That is an interesting question. Bear in mind that Lake Erie and the oceans of the world are already connected through the St. Lawrence and Lake Ontario, and water flows out continously (except in very cold winter) at a rate determined by the width and depth of that river, and the rate that rivers feed into the Great Lakes.

I doubt that the "very wide pipe" (a mile or less) would add much to the drain rate beyond what the St. Lawrence already provides. Its long length would also provide significant flow resistance. Therefore I would expect a negligible effect, unless the pipe were hundreds of miles wide. In that case, the lake level would drop somewhat and the Great Lakes would significant experience tidal effects as water flows back and forth through the pipe twice a day.

It is more an engineering type fluid mechanical calculation than a this-or-that physics question.

I think I'm finally convinced that my original #4 answer is wrong.

Quote: Wizard

Thanks Doc. However, I still don't see why Lake Erie won't drain. The entire lake is above sea level (source). It seems to my amateur-in-science mind Lake Erie should drain completely? Much for the same reason if I siphoned water out of a bathtub it would stop when the water level of the bathtub equaled the end of the hose.

Look at it this way. I have said that because Lake Erie is at higher elevation than the Indian Ocean, the tunnel would provide an alternate path for the water to reach the sea. I have also guestimated that the rate of flow through that path would be immeasurably small. In contrast, I have stood and watched in amazement the tremendous flow rate of the Niagara River over the falls, and even that flow rate doesn't seem to be having much effect on emptying Lake Erie. In Face's post last evening he said that they had agreed (as terms of the problem, I think) that the tributaries of both the lake and the ocean would continue to supply them.

If the water supply were somehow cut off, yes, the lake would eventually empty. The flow down through the falls and through Lake Ontario would be much more significant factors until the level of Lake Erie got below the bed of the Niagara River. Then the rest of the water could seep through the tunnel. I still think the flow rate would be quite small due to the combination of the modest hydraulic head pushing the flow (approx 570 ft of water head in the beginning) and the frictional resistance to flow through an 8,000 mile long invincibilium pipe. Try siphoning out your tub with a 1/4" tube a few thousand feet long and see how quickly you get it empty.

Quote: Doc

If I were being spun in a centrifuge and flattened against the wall, I would certainly not be experiencing zero g acceleration.

You would towards the center of the Earth :)

Quote:

While sitting in my chair, I actually am experiencing a centripetal acceleration due to the rotation of the earth, but it is fairly small and isn't obvious to me as I sit here, unlike the more obvious acceleration during a jump off a couch.

But when you are falling in an elevator, it should be as obvious to you, that you experience no acceleration whatsoever, isn't it?

Quote:

To me, "weight" means the force that a body experiences due to the fact it has mass and the fact that it is exposed to a gravitational field.

This is not exactly correct. Weight is not a force experienced by you, it is the force exerted by you on something else (like a scale or a chair). When you are at rest, and in absence of other forces, it is the same force by magnitude (and direction) that the gravity exerts on yourself, but that's a special scenario - for those two forces to be equal, you have to be at rest with respect to the local gravitational field. In all other cases, these forces are different. For example, if an elevator accelerates upward, your weight (the force you exert on the floor) increases, while the gravitational force exerted on you by the Earth is constant. If you float in water, you are more or less weightless, because your buoyancy compensates for gravity almost perfectly.

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If you are in free fall, there indeed is a force -- you are accelerating because the force of your weight is not counteracted by any other force.

So, you are telling me that the astronauts in the orbit are not weightless? Seriously?

Quote:

I do not consider the free falling person to be weightless at all.

The problem is that there is no physical experiment that could locally distinguish an object free falling in a gravitational field from one just sitting at rest inertially. This is called "equivalence principle" and is the basic postulate of the General Relativity.

Now, you personally may or may not consider a free falling object weightless. But, speaking physically, you would have to, because in physics it makes no sense whatsoever to define states that cannot be determined by some sort of a physical experiment. Which means that your personal definition of weightlessness has no physical sense.

Physics isn't my strongest card in the deck, but I am under the impression that you don't really have a single tunnel as much as you do two tunnels that just happened to be joined to each other. The center of the Earth is also Earth's center of gravity; it will pull water in from both ends, and they will "meet in the middle". (1) sounds like the closest answer to me.

Quote: MathExtremist

Okay -- no heat gets through. And no pressure from the surrounding molten rock (which is something like 3 million atmospheres in the inner core). However, what is the pressure exerted upon the water by itself as it descends? If it is great enough, and the heat does not increase, the water may turn into some non-standard form of ice as it descends. It may never reach the center of the Earth from either side, instead crystallizing inside the tube midway down.

See the Wikipedia article on the phases of ice. It has a nice chart.

Can anyone with more practical physics background than me determine what the pressure would be on the water column as it descends?

Now this could be a really interesting side discussion, though I don't really want to distract things too far from the more basic issues. I hadn't even considered the possibility of phase change of the water, and my initial reaction was (perhaps quite incorrectly) the exact opposite of what MathExtremist suggested. In more normally experienced conditions, increasing the pressure on ice will cause it to melt rather than causing the water to turn solid. This is because of the extremely complex phase diagram for water and specifically relates to that little blip on the left side of the green zone in the chart that MathExtremist linked.

However, at extremely high pressures we would enter a realm of ice structure that I have never considered. If the water supply were 300K (a bit warm for Lake Erie but easy to read in the chart), a pressure of 1 GPa would be required to form ice. That's 10,000 times atmospheric pressure. Would that much pressure be generated by a 4,000 mile high water column that exists in a gravitational field that gets weaker and weaker toward the bottom? Interesting question, but one that I am too lazy to try to solve. One of the benefits of retirement is that you can just refuse to do most work you don't want to take on. (Assignments from the wife not included.)

Maybe it's easiest to change the rules and say that the invincibilium would allow just enough heat in to keep the water liquid.

Quote: Doc

However, at extremely high pressures we would enter a realm of ice structure that I have never considered. If the water supply were 300K (a bit warm for Lake Erie but easy to read in the chart), a pressure of 1 GPa would be required to form ice. That's 10,000 times atmospheric pressure. Would that much pressure be generated by a 4,000 mile high water column that exists in a gravitational field that gets weaker and weaker toward the bottom? Interesting question, but one that I am too lazy to try to solve

If the gravity was constant, and did not become weaker towards the center, the pressure at the center of Earth would be (radius of Earth * water density * g) = 6,400,000 * 1,000 * 10 = 64,000,000,000 = 64GPa
Assuming that the density of Earth is constant, the gravity decreases linearly with distance as you get deeper, the real pressure at the center will be 32GPa - way more than enough ...

Quoting just a few of the comments so that this doesn't get completely out of hand:

Quote: weaselman

Weight is not a force experienced by you, it is the force exerted by you on something else (like a scale or a chair). When you are at rest, and in absence of other forces, it is the same force by magnitude (and direction) that the gravity exerts on yourself, but that's a special scenario - for those two forces to be equal, you have to be at rest with respect to the local gravitational field. In all other cases, these forces are different. For example, if an elevator accelerates upward, your weight (the force you exert on the floor) increases, while the gravitational force exerted on you by the Earth is constant. If you float in water, you are more or less weightless, because your buoyancy compensates for gravity almost perfectly.

Well, this is probably a continuing disagreement in usage of terms, but here goes.

I believe that weight is the force imposed by the gravitational field upon a mass. My weight is indeed a force experienced by me. If there is no counterbalancing force, this weight force causes my mass to accelerate in a free fall. If I am supported by a stationary, non-accelerating device such as my chair, I impose a force on it equal to my weight. By Newton's third law, the chair imposes an equal and opposite force on me, counterbalancing the gravitational force and keeping me from accelerating. In all of these states, my weight is unchanged, assuming I skipped lunch.

Now if my chair happens to be in an elevator that is accelerating upward, that doesn't change my weight. It does, though, change the force that I impose on the chair and the force the chair imposes on me. Those forces will equal my mass times the sum of the acceleration of gravity and the acceleration of the elevator. It will "feel" as if I got heavier, as I sink deeper into the cushion, but my weight will not have changed. No more than I would gain some sort of "horizontal weight" when my car accelerates.

Quote: weaselman

So, you are telling me that the astronauts in the orbit are not weightless? Seriously?

Seriously. They are, indeed, experiencing that sensation we call "weightlessness" and which would cause me some abdominal queasiness. But they still have weight (since they have mass and are in a gravitational field), and that weight force is causing them to accelerate in free fall toward the center of the earth. This acceleration is obvious in that they maintain their orbit rather than flying off on a tangential straight line (in a state of uniform, non-accelerating motion as described in Newton's first law). If they had no tangential velocity in orbiting the earth, this acceleration would give them a (negatively) increasing radial velocity and bring them plummeting directly toward the surface. And yet, in spite of always having this weight, they would feel that sensation we call weightlessness right up until contacting the atmosphere.

I didn't follow the rest of you comment. It has been a long time since I studied anything at all about relativity, and I never studied it seriously. All of my comments here are based on Newtonian mechanics. If an object has mass and is in a gravitational field, then it has weight -- that is the name for the force of gravity on the mass -- and the weight does not depend upon whether the object is stationary, in uniform motion, or accelerating.

Quote: Doc

I didn't follow the rest of you comment. It has been a long time since I studied anything at all about relativity, and I never studied it seriously.

That was the key though :) You don't need to know relativity to understand it. The argument is really pretty simple.
Imagine yourself inside an elevator, without windows. The elevator is falling down in a (uniform) gravitational field.

The simple fact, referred to as the Equivalence Principle, is that there is no way in principle by observing any physical phenomena to distinguish that state from an elevator floating somewhere in an intergalactic space, free from any gravitational influence whatsoever.

This is akin to the principle of relativity, which says that if you experience no acceleration, there is no way to tell if you are moving with a constant speed or are completely at rest.

The latter principle is the basis of Special Relativity. The former is the beginning of General Relativity.

You don't need the knowledge of tensor algebra to understand it though. It is just as intuitive and obvious as the relativity principle, known since Galileo.

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All of my comments here are based on Newtonian mechanics. If an object has mass and is in a gravitational field, then it has weight

This is not true. Even in Newtonian mechanics, weight is the force, exerted by the body on its support, not the force between two gravitating masses. The latter is called gravitational force.

Quote: weaselman

If the gravity was constant, and did not become weaker towards the center, the pressure at the center of Earth would be (radius of Earth * water density * g) = 6,400,000 * 1,000 * 10 = 64,000,000,000 = 64GPa
Assuming that the density of Earth is constant, the gravity decreases linearly with distance as you get deeper, the real pressure at the center will be 32GPa - way more than enough ...

Excellent start, weaselman!

Are you sure that the change is linear? I was thinking that within the body of the earth, the gravitational attraction was dependent upon the mass that is closer to the center than the point of interest. If that thought was correct (and I haven't pondered it thoroughly), then I think you need an adjustment in your figures. The mass of the material closer to the center than a point at some radius varies with radius cubed (assuming constant density), while the distance to the center of mass is radius to the 1st power. Wouldn't that lead to a quadratic relationship for gravity variation with distance to the center?

Whichever of those is correct, the only other thing I see possibly lacking here is an adjustment for the variation in the water density under different pressures. We typically approximate water as incompressible, but we don't often deal with 32GPa pressures. I still suspect that incompressibility is a reasonable approximation.

So MathExtremist's point would be an important one if the tunnel were ever constructed. Face should keep this in mind when working out the details.

Quote: Doc

Excellent start, weaselman!

Are you sure that the change is linear?

I am sure it would be linear if the density of the Earth was constant :)
In reality, it of course, isn't constant, increasing towards the center, so the drop in gravity is really not linear, it drops much faster near the center than it does near the surface. An even closer to the reality view is that the gravity actually keeps increasing near the surface until about half the radius, and then falls linearly from there.

These considerations would of course affect the final number, but they would only make it higher, so qualitatively the answer would remain the same - it'll freeze.

Quote:

I was thinking that within the body of the earth, the gravitational attraction was dependent upon the mass that is closer to the center than the point of interest. If that thought was correct (and I haven't pondered it thoroughly), then I think you need an adjustment in your figures. The mass of the material closer to the center than a point at some radius varies with radius cubed (assuming constant density), while the distance to the center of mass is radius to the 1st power. Wouldn't that lead to a quadratic relationship for gravity variation with distance to the center?

Right, the mass closer to the center decreases as r^3, while it's gravitational effect grows as r^2, so the result is linear.

Edit: fixed way over complicated (and actually incorrect) explanation of why the drop is linear

Quote: weaselman

...

Quote:

All of my comments here are based on Newtonian mechanics. If an object has mass and is in a gravitational field, then it has weight

This is not true. Even in Newtonian mechanics, weight is the force, exerted by the body on its support, not the force between two gravitating masses. The latter is called gravitational force.

I guess we're just going to have to disagree on this one and perhaps attribute it to different uses of terminology. Do you think of weight as a force that can act in arbitrary directions? I think of it as only acting in the direction of the gravitational field. To me, weight and gravitational force are essentially identical terms, at least in the context of the earth and a small object.

I think there are a heck of a lot of ways for a body to exert forces on its supports other than something I would refer to as its weight -- think of a body exerting forces during impact or a body with a propulsion system. Does a rocket trying to blast off but still being restrained have a negative weight? I don't think I have ever seen things described that way. To me, the weight is the force due to the body's mass and the gravitational field, and there can be a lot of other interactions (or none at all) between the body and any support that may or may not exist.

Quote: weaselman

The final formula (assuming the constant density - rho) is F(r) = m*G*4/3*pi*r*rho. It can be deduced from the Gauss law (if you want, I can show you how, it's pretty simple, but may be too involved for this board).

Whoops! I screwed up on this one! I now think that you are correct that it is linear and that my error is that I was treating it as if the denominator in the gravitational law equation was the radius instead of the radius squared. The (r^3)/(r^2) do indeed lead to a linear relationship. Just glad that I said up front that I hadn't pondered the issue thoroughly -- makes it easier to clean the egg off my face.

I still haven't thought it all the way through, but I think that all of the mass at a greater radius does indeed become irrelevant. I think the gravity field inside a hollow shell is zero, or rather that there is no gravitational field due to the shell.

Quote: Doc

I guess we're just going to have to disagree on this one and perhaps attribute it to different uses of terminology. Do you think of weight as a force that can act in arbitrary directions?

No, weight s collinear with the gravity, that causes it.

Quote:

I think of it as only acting in the direction of the gravitational field. To me, weight and gravitational force are essentially identical terms, at least in the context of the earth and a small object.

Imagine that you are hitting a big fence with your car. There are several forces at work here.
The car engine applies the force to the car, making it move forward. The car in turn applies force to the fence, making it fall down. The latter force is caused by the former, but it is not the same force. If there was no fence in your way, there would not be second force.

Quote:

I think there are a heck of a lot of ways for a body to exert forces on its supports other than something I would refer to as its weight -- think of a body exerting forces during impact or a body with a propulsion system. Does a rocket trying to blast off but still being restrained have a negative weight?

Negative - no. Zero - yes. It is not exerting negative force on the support.

Quote:

I don't think I have ever seen things described that way.

Hmmm ... They are described exactly this way in every physics text book I have seen ... I have seen lots of physics textbooks :)


BTW, I fixed the explanation of why the force drop is linear in my previous post. What I wrote there at first was way overcomplicated, and actually wrong too :) The only thing correct there was the formula and the reference to the Gauss law :)

Quote: Doc

I still haven't thought it all the way through, but I think that all of the mass at a greater radius does indeed become irrelevant. I think the gravity field inside a hollow shell is zero, or rather that there is no gravitational field due to the shell.

Yes, exactly. This is exactly what Gauss law tells us.

Quote: weaselman

Yes, exactly. This is exactly what Gauss law tells us.

The gravity at the center of a shell might be zero, but not just inside the edge (or on the surface). That's how a sun can live at the center of a Dyson Sphere (assuming you could build one) but the people on the surface wouldn't be floating around.

This is obviously getting way more complicated than was initially intended, but if you really wanted to hack it out, you'd need to take into effect the density of each layer of the Earth's core as it surrounds the invincibilium tube so you could calculate the gravity gradient on the water passing through it. Assuming the tube intersects the Earth's center of mass, that spot will have no gravity (I think). But the rest will, just to a lesser degree. What I don't know is whether the pressure on the water column increases at a rate sufficiently greater than the decrease in gravity so as to turn the water into some ice phase or whether the water will stay liquid the whole way through. I don't know the physics to provide the answer, but I think it'd make a good question on a college physics exam.

Edit: I just looked up Dyson Sphere on Wikipedia to make sure I was talking about the right construct, and the article says "Not to be confused with Dyson Ball" (i.e. the vacuum cleaner). That gave me a chuckle.

Quote: MathExtremist

The gravity at the center of a shell might be zero, but not just inside the edge (or on the surface).

Huh? Of course the gravity on the surface is not zero. That's the reason we do not fall off the surface of the Earth ... But thanks for reiterating that important point! :D

Quote:

This is obviously getting way more complicated than was initially intended, but if you really wanted to hack it out, you'd need to take into effect the density of each layer of the Earth's core as it surrounds the invincibilium tube so you could calculate the gravity gradient on the water passing through it.

I did :)
Like I said, I have the exact answer for the constant density.
In reality, density is not constant, in fact, the gravity even raises a little all the way up to the half way to the center (because the density is way lower in the mantle than it is in the core), and then falls linearly from there. But these corrections can only make my answer larger, not smaller, so qualitatively, the answer does not change - the water in the center (as well as well before the center, actually) will be frozen.

Quote:

What I don't know is whether the pressure on the water column increases at a rate sufficiently greater than the decrease in gravity so as to turn the water into some ice phase or whether the water will stay liquid the whole way through.

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?

=D!! I'm absolutely tickled pink that this gained so much attention. Thanks to all for giving my stupid obsessions so much of your valued time.

Quote: Daddy Wiz

That is an interesting question. Bear in mind that Lake Erie and the oceans of the world are already connected through the St. Lawrence and Lake Ontario, and water flows out continously (except in very cold winter) at a rate determined by the width and depth of that river, and the rate that rivers feed into the Great Lakes.

I doubt that the "very wide pipe" (a mile or less) would add much to the drain rate beyond what the St. Lawrence already provides. Its long length would also provide significant flow resistance. Therefore I would expect a negligible effect, unless the pipe were hundreds of miles wide. In that case, the lake level would drop somewhat and the Great Lakes would significant experience tidal effects as water flows back and forth through the pipe twice a day.

It is more an engineering type fluid mechanical calculation than a this-or-that physics question.

Quote: Doc

But since the surface of the Indian Ocean is lower than the surface of Lake Erie, the tunnel will provide one more minor path for a tiny amount of the fresh water to reach the sea, mixing with the salt water in the tunnel instead of in the Gulf. I suspect the flow rate through the tunnel would be immeasurably small.

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 That's awesome on so many levels. In all seriousness (if that word can be used in a discussion such as this) I really enjoyed this experiment and again, thanks to all for taking part. It seems for an absolute answer, I'd have to either save for the dig (invincibilium isn't cheap, ya know) or get some serious Physics group to waste a good amount of their time and resources for no reason whatsoever. Still, finding that I CAN'T know something is comforting, and I think I'll finally be able to silence the nagging this topic has had on my brain since it was first discussed over a month ago. (for the record, I still say the pipe would be no different than any of the multiple rivers, Erie will drain into the Indian, no matter how minute the flow. (Sorry Father Wiz))

All this from a fishing discussion gone awry...

Feel free to derail into properties of ice, the nature of gravity, or whatever else has begun in the last few posts.