a weird question that puzzles the mind
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the question is: if we drilled a hole through the earth from one end to the other what would happen if you throw something down the hole. now this question can only be probable if the center wasn't filled with hot molten rock so let assume that there is no molten rock in the center.
I know this is a weird question but i want to know what people think the answer is.
I know this is a weird question but i want to know what people think the answer is.
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The rock would fall down, up, and down again, eventually coming to rest at the center. If it didn't hit a side on the way, the rock would be suspended. (The last part is a personal theory, the first is fact.)
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I think it would go past earth's center and sort-of sling shot back and forth untill it stops in the center. it was in a test I took once upon a time (and failed if I recall). oh i was second :(
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Flaser
OCD Hentai Collector
It all depends on what you include in the model.
If you sealed both ends of the tunnel, pumped out all the air, made sure the rock would never touch the side of the walls (maybe you drill a 1 km wide tunnel and the rock is 10 cm across) then the rock would fall down the hole accelerate until it hits the center of the Earth, then it would keep going but instead will slow down and come to a rest on the other end of the tunnel - where you could just grab it.
Then, if it's not grabbed it would fall once again, repeating the whole thing the other way around.
The reason is that once you're inside a spherical shell the pull of the shell cancels itself out. It's a bit complicated in both mathematics and writing the equations, but it comes out that even your position within the shell doesn't matter.
So if you go 30 kilometers below the surface that means that a 30 km sick shell no longer pulls you down.
So if he could make a pocket of the deep Earth (through magic!) inhabitable since the gravity would be "effectively lower" there all kinds of giant creatures could live. there.
Whee! Found it:
http://www.merlyn.demon.co.uk/gravity1.htm#GoSSh
Field Inside a Spherical Shell
It has been said that : "the field inside a uniform spherical shell is zero (which is indeed so for any 1/r2 type force); a non-obvious, non-trivial result needing calculus for proof."
This is not so; the cancellation of field is correct, but there is no need for calculus. This is relatively well known in the case of gravitation, but not for electrostatics. In the following, "charge" stands for "electric charge", "mass", or any other source of an inverse square field; and "small" means "infinitesimal".
Consider an arbitrary point X inside a uniformly-charged spherical surface, and consider both an arbitrary small element of "charged" surface area A and that second small element of area B which is marked out by straight lines from the edge of the first area through the point to meet the sphere again on the opposite side of the point.
The solid angles subtended by A and B at the point X are equal, and the directions XA and XB are opposite. Simple geometry makes it clear that the areas A and B (which have the same "tilt") and hence the charges are proportional to the squares of their distances from the point X. Their fields therefore cancel at that point; so there will be no net field from the whole set of such areas which covers the whole sphere, for any point X.
The argument is correct; I think I have expressed it correctly. Ramsey (s.3.2) says that Newton used it, in Principia Proposition LXX; indeed, it is in Hawking's cited book, p.880.
If you sealed both ends of the tunnel, pumped out all the air, made sure the rock would never touch the side of the walls (maybe you drill a 1 km wide tunnel and the rock is 10 cm across) then the rock would fall down the hole accelerate until it hits the center of the Earth, then it would keep going but instead will slow down and come to a rest on the other end of the tunnel - where you could just grab it.
Then, if it's not grabbed it would fall once again, repeating the whole thing the other way around.
The reason is that once you're inside a spherical shell the pull of the shell cancels itself out. It's a bit complicated in both mathematics and writing the equations, but it comes out that even your position within the shell doesn't matter.
So if you go 30 kilometers below the surface that means that a 30 km sick shell no longer pulls you down.
So if he could make a pocket of the deep Earth (through magic!) inhabitable since the gravity would be "effectively lower" there all kinds of giant creatures could live. there.
Whee! Found it:
http://www.merlyn.demon.co.uk/gravity1.htm#GoSSh
Field Inside a Spherical Shell
It has been said that : "the field inside a uniform spherical shell is zero (which is indeed so for any 1/r2 type force); a non-obvious, non-trivial result needing calculus for proof."
This is not so; the cancellation of field is correct, but there is no need for calculus. This is relatively well known in the case of gravitation, but not for electrostatics. In the following, "charge" stands for "electric charge", "mass", or any other source of an inverse square field; and "small" means "infinitesimal".
Consider an arbitrary point X inside a uniformly-charged spherical surface, and consider both an arbitrary small element of "charged" surface area A and that second small element of area B which is marked out by straight lines from the edge of the first area through the point to meet the sphere again on the opposite side of the point.
The solid angles subtended by A and B at the point X are equal, and the directions XA and XB are opposite. Simple geometry makes it clear that the areas A and B (which have the same "tilt") and hence the charges are proportional to the squares of their distances from the point X. Their fields therefore cancel at that point; so there will be no net field from the whole set of such areas which covers the whole sphere, for any point X.
The argument is correct; I think I have expressed it correctly. Ramsey (s.3.2) says that Newton used it, in Principia Proposition LXX; indeed, it is in Hawking's cited book, p.880.
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Well with no hot molten rock in the center...we'd probably freeze to death and die due to the ground beneath us falling in to fill the sudden gaping hole....also we'd probably lose a lot of spare change.
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What Flaser said.
I seems like a sound argument, though I have of cause not checked the assumptions on which he builds his arguments.
I seems like a sound argument, though I have of cause not checked the assumptions on which he builds his arguments.
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mibuchiha
Fakku Elder
Simply said, it'll undergo a simple harmonic motion as long as there's no disturbance to it.
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I learnt a lot today.... No teacher or lecturer would teach me this....
Thanks....
Just think, if the earth center has a very high tempereture and pressure. Until the carbon atom form a covalent bond with other carbon, the process repeated itself until it creates a new material - Diamond.... Would we be rich if we could drill to the center of the Earth and take the huge high quality diamond... Do you think it is ever possible
Thanks....
Just think, if the earth center has a very high tempereture and pressure. Until the carbon atom form a covalent bond with other carbon, the process repeated itself until it creates a new material - Diamond.... Would we be rich if we could drill to the center of the Earth and take the huge high quality diamond... Do you think it is ever possible
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Unlikely, as the temperatures go up pretty fast the lower we go, and because of the whole molten-thing. Though of course anything could be possible, if we had access to (practically) infinite amounts of energy and materials. I think it would be cheaper to just make a furnace that made diamonds, and definitely safer.
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Aud1o Blood wrote...
The rock would fall down, up, and down again, eventually coming to rest at the center. If it didn't hit a side on the way, the rock would be suspended. (The last part is a personal theory, the first is fact.)This is what I was going to say. Not verbatim of course, but you know what I mean.
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If there is no core, there would be no gravity thus all life as we knew it would end; thus, there would be no one to throw the coin down the hole, or even to drill the hole.... what a downer :P
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mibuchiha
Fakku Elder
rokushou wrote...
Funny question... of course it will stop on the center due to earth's gravity..if there's no disturbance to it (touching the tunnel wall, other force stops it etc) it won't stop actually. it will accelerate to the center, then decelerate until it reaches the same height it was dropped from but on the other side, then accelerate again to the center ad infinitum. in other words, a simple harmonic motion.
@soul: lol, not really. gravity is caused not by the core of the earth, but the mass of it as a whole. so even if the core is hollow, as long as the mass isn't significantly altered nothing will actually change.
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mibuchiha wrote...
rokushou wrote...
Funny question... of course it will stop on the center due to earth's gravity..if there's no disturbance to it (touching the tunnel wall, other force stops it etc) it won't stop actually. it will accelerate to the center, then decelerate until it reaches the same height it was dropped from but on the other side, then accelerate again to the center ad infinitum. in other words, a simple harmonic motion.
@soul: lol, not really. gravity is caused not by the core of the earth, but the mass of it as a whole. so even if the core is hollow, as long as the mass isn't significantly altered nothing will actually change.
You forgot the no air/resistance/friction argument... otherwise you are wrong, and it is always annoying when that happens one self :)
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mibuchiha
Fakku Elder
jenslyn wrote...
You forgot the no air/resistance/friction argument... otherwise you are wrong, and it is always annoying when that happens one self :)air resistance depends on the velocity of the of the object, and opposite in direction. so it's higher when it's falling to the center and becomes lower as it travel further from it.
i'm not really sure about this, but if we integrate the air resistance for the whole trip from one side to the other, I think it balances out, and so provides no real disturbance to the system.
btw, what does annoying got to do with this?
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It'd be broken up during its journy from point A to point B till it was merely dust.
Dust that would not get to the other side and would most likely end up on the sides of the hole.
Not much of a puzzle...
Not a puzzle actually.
Dust that would not get to the other side and would most likely end up on the sides of the hole.
Not much of a puzzle...
Not a puzzle actually.
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mibuchiha wrote...
jenslyn wrote...
You forgot the no air/resistance/friction argument... otherwise you are wrong, and it is always annoying when that happens one self :)air resistance depends on the velocity of the of the object, and opposite in direction. so it's higher when it's falling to the center and becomes lower as it travel further from it.
i'm not really sure about this, but if we integrate the air resistance for the whole trip from one side to the other, I think it balances out, and so provides no real disturbance to the system.
btw, what does annoying got to do with this?
adding air friction and x factor in Ftotal=Ffall potential (mgh)-Ffriction ,
then in the end (if the throwed material is not reduced by the friction), it'll still stop in the center, right?
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Flaser
OCD Hentai Collector
mibuchiha wrote...
jenslyn wrote...
You forgot the no air/resistance/friction argument... otherwise you are wrong, and it is always annoying when that happens one self :)air resistance depends on the velocity of the of the object, and opposite in direction. so it's higher when it's falling to the center and becomes lower as it travel further from it.
i'm not really sure about this, but if we integrate the air resistance for the whole trip from one side to the other, I think it balances out, and so provides no real disturbance to the system.
btw, what does annoying got to do with this?
That's wrong. Air resistance is not a conservative force. It will only take energy the whole way. As it has been said without resistance, you'd get an oscillating system where the coin/rock/whatever would swing between the two ends of the hole drilled through Earth. With resistance added it becomes a damped oscillating system. Each and every pass some of the velocity gained from "falling" will be bled off, so each pass the coin/rock would go less and less farther from the center of the Earth until eventually it looses all momentum and comes to a rest in the center.
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Flaser wrote...
mibuchiha wrote...
jenslyn wrote...
You forgot the no air/resistance/friction argument... otherwise you are wrong, and it is always annoying when that happens one self :)air resistance depends on the velocity of the of the object, and opposite in direction. so it's higher when it's falling to the center and becomes lower as it travel further from it.
i'm not really sure about this, but if we integrate the air resistance for the whole trip from one side to the other, I think it balances out, and so provides no real disturbance to the system.
btw, what does annoying got to do with this?
That's wrong. Air resistance is not a conservative force. It will only take energy the whole way. As it has been said without resistance, you'd get an oscillating system where the coin/rock/whatever would swing between the two ends of the hole drilled through Earth. With resistance added it becomes a damped oscillating system. Each and every pass some of the velocity gained from "falling" will be bled off, so each pass the coin/rock would go less and less farther from the center of the Earth until eventually it looses all momentum and comes to a rest in the center.
^this is what i mean....
