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Inside a Spherical Chamber

Towlie

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Towlie
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May 22, 2009
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(One thread inspires another. ;) )

If a large spherical hollow chamber, perhaps 1000 miles in diameter, was created with its center exactly at the center of the Earth, what gravitational force would be felt at various points within the sphere?
 
Anything within the sphere would likely be drawn towards the center.

Which means that all those dinosaurs, inner-space aliens, and lost Atlanteans would be unable to climb out of the hole at the North Pole and wander around in their Bigfoot suits, looking for goats to suck the blood out of.

;)
 
Towlie said:
(One thread inspires another. ;) )

If a large spherical hollow chamber, perhaps 1000 miles in diameter, was created with its center exactly at the center of the Earth, what gravitational force would be felt at various points within the sphere?
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A spherically symmetric shell has no gravitational field inside the shell. This is part of the famous Shell Theorem (the other part is that outside the shell, you can treat the gravitational field as if all the mass were at a point at the center of the shell).
 
A spherically symmetric shell has no gravitational field inside the shell.
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And earths rotation provide next to no artificial gravity?

So any dinosaurs etc. would need wings or really big flippers to get around.
 
Good question. On the one hand, there is an equal amount of bulk all around you, so you should weigh nothing, and not drift anywhere.

On the other hand, that would suggest the "gravity well" of a planet wasn't really an ever deepening well, but rather went downward until some point under the surface, then went back upward as you approached the center. That would seem to make a bunch of that mass be weightless, or nearly so, and would also suggest the center is precisely not where the heaviest elements would be. On the gripping hand, smaller grains of sand work their ways under larger with vibrations, and thus smaller atoms would work their way under larger, heavier ones, pushing them "up", which, in this case, would be towards the center of the Earth.


Stop me, I'm hurting myself. :(
 
Beerina said:
Good question. On the one hand, there is an equal amount of bulk all around you, so you should weigh nothing, and not drift anywhere.

On the other hand, that would suggest the "gravity well" of a planet wasn't really an ever deepening well, but rather went downward until some point under the surface, then went back upward as you approached the center. That would seem to make a bunch of that mass be weightless, or nearly so, and would also suggest the center is precisely not where the heaviest elements would be. On the gripping hand, smaller grains of sand work their ways under larger with vibrations, and thus smaller atoms would work their way under larger, heavier ones, pushing them "up", which, in this case, would be towards the center of the Earth.


Stop me, I'm hurting myself. :(
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Well, assuming that the density of the Earth remained fairly constant, and the center wasn't hollow, gravity's pull on you would decrease linearly from 1 G to Zero G as you went from the surface to the center. Putting a 500 mile radius hole in the center would seem to make that linear decrease go from 1 G to Zero G from the surface to the shell of the inner hollow. And then anywhere inside that hollow you would be weightless.
 
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Both Zig and JustThinking are correct. If the Earth is like a hollow shell, there is no gravitational field inside the shell; yet if the Earth has uniform density, the gravitational field will increase linearly with distance from the center of the planet.

All of this can be summarized by application of Gauss's Law to the inverse-square law of gravity.
 
Just thinking said:
I estimate that to be about 7.4 x 10-6 G's.

Ain't much.
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Throught so, I have not heard of any difference between the poles and equator. It must be in the same range.

It just got me thinking of a Sci-Fi novel taking place in a gas ring orbiting its sun. There is no gravity so the seas and lakes are big globules of water. The human colonists get around with flippers, or just stay in their scrubbery or on the huge trees. S-shaped trees with leaves in both ends.
 
Toke said:
It just got me thinking of a Sci-Fi novel taking place in a gas ring orbiting its sun. There is no gravity so the seas and lakes are big globules of water. The human colonists get around with flippers, or just stay in their scrubbery or on the huge trees. S-shaped trees with leaves in both ends.
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Hmmm . . . S-shaped trees, but it would be a stretched-out S, so more of an integral sign than an S. Might call them "integral trees" .. .
 
I believe that you wouldn't fall to the cetre point, the sphere would orbit you with you inside it, or rather the sphere's centre of gravity would orbit your centre of gravity (or the other way round, whichever way you want to look at it)

EDIT: If it was perfectly spherical inside and out then if there was a vacuum you'd barely touch the inside of the sphere each orbit, if there was atmosphere then you might eventually come to rest right at the centre.

EDIT AGAIN: thinking some more, if you were off centre then while there's more mass to one side, the mass on the lesser side is closer, wierd. If you were halfway between two equal size stars and move slightly towards one then that one will pull you towards it.

I dunno, I'll have a look at that link MM
 
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dasmiller said:
Hmmm . . . S-shaped trees, but it would be a stretched-out S, so more of an integral sign than an S. Might call them "integral trees" .. .
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Thank you :)
I could not remember the title.
I have some art for it somewhere with a cute woman floating in midair with harpoon and flippers.
 
Towlie said:
And if you're floating within the sphere while it's rotating around you, would the rotation make any difference at all?
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For the amount I calculated ... not likely. However, if one was trying to simulate 1 G by spinning the sphere quickly enough, it might start to induce enough drag on the air within the sphere to start it moving as well ... which in turn would start you moving, and eventually drag you to the inner edge of the sphere. Just how closely your speed will match up to that of the sphere upon contact ... that may be interesting to watch.
 
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alexi_drago said:
I believe that you wouldn't fall to the cetre point, the sphere would orbit you with you inside it, or rather the sphere's centre of gravity would orbit your centre of gravity (or the other way round, whichever way you want to look at it)
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No. There is no net gravitational interaction between a spherical shell and anything within the shell. The shell will feel a compressive stress because of anything inside, but that will produce no motion as long as the shell retains its shape.

EDIT AGAIN: thinking some more, if you were off centre then while there's more mass to one side, the mass on the lesser side is closer
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Indeed, and these effects exactly cancel.

If you were halfway between two equal size stars and move slightly towards one then that one will pull you towards it.
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Yes, because unlike inside a shell, the amount of mass pulling one way vs. the other doesn't change as you move.
 
Funny, I read The Integral Trees ages ago (probably when it came out in 1984). I just got an old copy of it and the sequel The Smoke Ring which I've never read, so I'm just now re-reading The Integral Trees.
 
Ziggurat said:
Indeed, and these effects exactly cancel.
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So if you had a bubble with a large mass the gravitational fields on the outside of the bubble would be affected by that mass but the gravitational fields inside the bubble would be as if there was no bubble at all?
Would there be some sort of step as you cross the boundary?
 
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