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Old 2006-01-06, 14:38   #1
tha
 
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Default digging a hole through the earth.

Following my earlier puzzle (it is green) here is a new one:

Consider the earth to be a perfect sphere with a circumference of 25,000 miles. We start digging a hole at one point through the center of the earth and end up at exactly the other side of the globe. We ignore problems with heat, the core being fluid, etc. etc. We drop a stone from one side at ground level into the hole. How many feet does the stone fall?
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Old 2006-01-06, 14:49   #2
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assuming that you dug the hole straight through (a diameter)
and that the Earth is a sphere, approximately [b]12500/Pi[/b] miles,
to the center of the Earth

if we want to argue that it will go past the center due to its velocity
and then back towards the center due to gravity and so on, so forth
we will get a larger number, potentially infinite depending on the physics computations
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Old 2006-01-06, 15:34   #3
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Assuming the stone passes the centre of the Earth, and then reverses its direction of motion. Is this still called falling? It is surely now rising back towards the point from which it was dropped. It has negative velocity. Does this count as negative falling, and does the question ask us to subtract this from the positive falling to arrive at an answer equivalent only to the amount of positive distance the stone travelled?

Quote:
Originally Posted by tom11784
we will get a larger number, potentially infinite
I know nothing about physics, but it seems to me that the distance cannot be infinite. The stone can only lose energy on its journey, and since it didn't start with infinte energy it must stop at some point.
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Old 2006-01-06, 16:16   #4
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Quote:
Originally Posted by Numbers
The stone can only lose energy on its journey, and since it didn't start with infinte energy it must stop at some point.
I doubt it would be infinite, but this is a transfer of potential to kinetic energy
as it approaches the center and then kinetic to potential for the distance
travelled past the center, and not necessarily all a loss.

If the ratio of potential energies at each pause of the stone is sublinear
(like the famous k*(1+1/2+1/3+...) it can be infinite total travel,
despite eventually appearing to stop at the center
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Old 2006-01-06, 16:28   #5
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6 feet, then it is eaten by the green stone eating monster
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Old 2006-01-06, 16:58   #6
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Quote:
Originally Posted by Greenbank
6 feet, then it is eaten by the green stone eating monster
Fabulous answer. It deserves to be right!
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Old 2006-01-06, 17:07   #7
tha
 
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Quote:
Originally Posted by Numbers
Fabulous answer. It deserves to be right!
It is right. This puzzle is ideal for a party of teenagers or colleages at the office. Inbetween the first and the second question you wait for someone else to come up with another unrelated puzzle.
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Old 2006-01-06, 22:51   #8
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If I wanted to be strict in my physics, then I'd have to assume that a straight hole would NEVER allow the object to remain in freefall. The reason is that eventually the walls will touch whatever's falling. This would occour even without orbital considerations because the earth is not perfectly even in it's composition and thus density. It's not even a sphere! Of course this is all a bunch of nitpicking that really would take the fun out of a casual discussion.
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Old 2006-01-07, 00:14   #9
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Quote:
Originally Posted by nibble4bits
If I wanted to be strict in my physics, then I'd have to assume that a straight hole would NEVER allow the object to remain in freefall. The reason is that eventually the walls will touch whatever's falling.
Since it's already been decided that Greenbank wins the lifetime supply of cocoa at bedtime with extra squishy marshmallows on Sundays, would you care to expand on this a bit?

If we are allowed to suppose that we can dig a hole through the Earth then we are obviously not in the department of the physically possible, but I heard someone say something similar to your "the walls will touch whatever's falling" in a previous discussion and I just couldn't get what he was talking about. Surely, if the hole is straight, and the stone is dropped plumb centre, it just keeps going, doesn't it? Why do the walls get in the way?
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Old 2006-01-07, 01:21   #10
xilman
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Quote:
Originally Posted by Numbers
Since it's already been decided that Greenbank wins the lifetime supply of cocoa at bedtime with extra squishy marshmallows on Sundays, would you care to expand on this a bit?

If we are allowed to suppose that we can dig a hole through the Earth then we are obviously not in the department of the physically possible, but I heard someone say something similar to your "the walls will touch whatever's falling" in a previous discussion and I just couldn't get what he was talking about. Surely, if the hole is straight, and the stone is dropped plumb centre, it just keeps going, doesn't it? Why do the walls get in the way?
Unless the hole is dug between the poles, the earth's rotation will make the stone hit the walls.

Even if it does pass along the rotation axis, gravitational perturbations from the other bodies in the solar system will eventually make it hit the walls.

Paul
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Old 2006-01-07, 02:03   #11
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Quote:
Originally Posted by Xilman
gravitational perturbations from the other bodies in the solar system will eventually make it hit the walls.
I appreciate that the gravitational force of say Saturn is different to the gravitational force of say Pluto. And that as I move through the universe the difference between these forces will change dependent upon where I am relative to those two bodies. However, I would not have thought that moving such a short distance as the length of the Earth's axis would make any noticeable difference to an object the size of a stone.

There is also the fact (I don't like using words like "fact" in posts to you because it almost invariably turns out not to be a fact at all, but here goes) that the force of the Earth's gravity acting on the stone is many times stronger than the difference between the other two (keep it simple) forces. So it's a bit like trying to deflect a bullet by blowing on it. The force propelling it forwards is many times stronger than the attempted deflection which necessarily fails. No?
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