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Tube in the Earth


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If I had a perfectly straight tube running from the North pole of the Earth to the south pole and dropped a rock in it, would the rock fall to the other side or would ir stay in the center or the Earth? I guess it would depend on if the harmonic oscillator is over-damped, but I don't know for sure.

Edited by steevey
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If there was no friction or other factors that would interact with the rock, it would bounce back and forth indefinitely. At the least, this would require a perfect vacuum in the tunnel and a rock made of material that didn't interact with the earth's magnetic field. SM

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Forgetting for a moment the molten nature deep below the surface of the earth making such a tube improbable I would expect the rock to over shoot the centre and not reach the opposite surface before falling back toward the centre. It would overshoot again and again eventually coming to rest at the centre. However I would expect it to melt in the process and exist as a ball of lava.

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Such a tube could not exist on Earth because of the molten area and the pressure. However if you could build some sort of tube, then an item dropped would hit the wall of the tube due to conservation of angular momentum.

Sure about that?

The OP says it's a tube from pole to pole.

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Sure about that?

The OP says it's a tube from pole to pole.

 

Oh, I forgot that bit. There will still be some angular momentum issues based on the earth's orbit and tilt (so that if it is summer or winter dropping something down the tube would change its distance from the sun), but that would be a much smaller effect.

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Here is another factoid regarding this idea- The transit time from the north to the south pole, or the opposite, is slightly less than 42 minutes. Another question- In this transition, what is the region of maximum and minimum acceleration? SM

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Here is another factoid regarding this idea- The transit time from the north to the south pole, or the opposite, is slightly less than 42 minutes. Another question- In this transition, what is the region of maximum and minimum acceleration? SM

 

Well, if the Earth had a perfect distribution of mass, the region would be any place past the center of the Earth on the other end if a rock had been falling down one end, because once the rock got past the center of the Earth the first time, there would then be more gravity behind it then in front of it. And although there isn't a perfect distribution of mass, I think it's at least roughly equal all the way around since things such as water and air want to diffuse and deeper layers of rock and mantel also diffuse but also position themselves according to density.

Edited by steevey
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Here is another factoid regarding this idea- The transit time from the north to the south pole, or the opposite, is slightly less than 42 minutes. Another question- In this transition, what is the region of maximum and minimum acceleration? SM

 

Just like with a spring, there would be highest acceleration (but zero velocity) at either end, and it would be 9.8m/s^2. The lowest acceleration (but highest velocity) would be passing the center of mass, at which point the acceleration would be zero.

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It turns out that a frictionless object dropped into a theoretical tunnel connecting any two points on the surface of the earth would fall through in close to 42 minutes. This idea goes back to Newton, but was explored by Paul W. Cooper, "Through the Earth in Forty Minutes," American Journal of Physics, January 1966, Volume 34, Issue 1, pp. 68. The Time Magazine story about him is here: http://www.time.com/...,842469,00.html

Simple math: http://www.docstoc.c...y-Train-Project

More complex math for the shape of the fastest tunnel: www.physics.unlv.edu/~maxham/gravitytrain.pdf

SM

 

EDIT- I don't know why the PDF doesn't come up as a link, but it works as cut and paste.

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Just like with a spring, there would be highest acceleration (but zero velocity) at either end, and it would be 9.8m/s^2. The lowest acceleration (but highest velocity) would be passing the center of mass, at which point the acceleration would be zero.

Sine wave function and cosine wave function. (Which will both be damped and settle at zero)

Edited by TonyMcC
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Insane_alien, air resistance and possible interaction with the earth's magnetic field or the side of the tunnel, yes. But what if, as I mentioned above and in the "gravity train" references I cited, the tube contained a vacuum and was large enough (or whatever) to eliminate contact? Theoretically the rock should bounce back and forth between the north and south pole surfaces indefinitely. SM

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Insane_alien, air resistance and possible interaction with the earth's magnetic field or the side of the tunnel, yes. But what if, as I mentioned above and in the "gravity train" references I cited, the tube contained a vacuum and was large enough (or whatever) to eliminate contact? Theoretically the rock should bounce back and forth between the north and south pole surfaces indefinitely. SM

 

even then, no.

 

remember, indefinitely is an incredibly long time. even if we ignore the death of the sun and the gravity of the moon destabilizing the path and assume that this is a tube going through a homogenous non-rotating sphere and so on and do forth ideal conditions in other words, the answer is still no, it will not be indefinite.

 

 

it will eventually come to rest at the center as it loses energy to the emission of gravity waves.

 

all objects in orbit emit waves of gravity and at the scales we are considering, the effect is incredibly tiny. and this is an orbit we're talking about, albeit a very unusual one.

 

so it will eventually come to a standstill. it will take many trillions of years but it will happen. i suppose you could even work out exactly how long it would take if you wanted.

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What about radiating heat to atmoshere through the end caps of the evacuated tube? If a container of water sat on the end cap the water temperature would rise as the molten lump of rock came and went from it. This transfer of heat from the interior of the earth to atmosphere must (imo) involve losses.

If you consider a development of this idea with a very wide hole and a huge lump of molten rock transferring enough heat to boil water you could use the steam to turn a turbine. This could only work for a relatively short time otherwise you would have a perpetual motion machine.

Any other attempt to make the system do work for you is doomed to failure because it would damp out the oscillatory motion.

Edited by TonyMcC
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SMF - Not really sure myself! Would we just be cooling the earth at a faster rate? How about the temperature of the molten rock changing as it moves to and fro causing it to expand and contract. Also being molten there may be convection currents inside it causing internal friction. Would these produce losses ? I don't really know - what do you think?

 

Actually, it would keep oscillating for ever. If it stopped then you would know its position (at least approximately) and its momentum exactly which would be a breach of the uncertainty principle.

This is the origin of zero point energy.

Does this really apply to lumps of rock ( billions of atoms)? I thought this was only applicable to particles described in quantum physics. Would the oscillation you speak of be the same frequency as the original starting frequency?( apparently 1 cycle every 84 minutes)

Edited by TonyMcC
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SMF - Not really sure myself! Would we just be cooling the earth at a faster rate? How about the temperature of the molten rock changing as it moves to and fro causing it to expand and contract. Also being molten there may be convection currents inside it causing internal friction. Would these produce losses ? I don't really know - what do you think?

 

 

Does this really apply to lumps of rock ( billions of atoms)? I thought this was only applicable to particles described in quantum physics. Would the oscillation you speak of be the same frequency as the original starting frequency?( apparently 1 cycle every 84 minutes)

 

I think it applies.

In addition the rock would be "ringing" with all the vibrational modes of the atoms.

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