the flood, mammoths, elphants, and food.
by Crazyguy

Viviane
in other words no acceleratin in the center of earth and less acceleration in a mineshaft. Also no acceleration inside a hollow sphere. Hope that clears something up.
That's not true. You yourself just proved it by saying "less acceleration ina mineshaft".
bohm: thank you and we remember that acceleration is equivalent to the pull of gravity.
Finding someone to agree that you are wrong doesn't make you right. You've still a ton of math you said you could show that, as far as you shown, you don't even comprehend.
Unless, of course, you can show that math.
Follow the the graph line of the strength of the gravitational pull from the maximum at the surface (if any) to the zero at the center, one realises there is a steady or incremental decline.
Case in point, you keept mixing "gravitational pull" with claiming "no gravity". Very wildly different things that you can't seem to grasp. Also, quantum physics pretty makes your "no gravity in the center" idea impossible no matter how you word it.
Unless, of course, you can show that math.
This means that the layer above, like layers down a mine shaft have CRASED to contribute to the strength of the downward pull. how otherwise could the force have declined? It follows that if the inner layers of your choice could be removed, there would be no gravity force left.
Or, it means you don't understand math, gravity or arithmetic, as you've shown by using every excuse in the book to avoid showing the math you claimed you could to.
What gravity is, a property of mass, a tensioning of space, an exchange of gravitonradiation, the action of the Hicks Boson, does not matter here.
And you just proved, with that statement, you understand absolutely none of the things you claim. Well, that statement and dozens of others.
Unless you can, as you claimed, show the math.
These ideas are fine in an ideal scenario, like the Laws of motion of Kepler, working only for pointmasses. You do not want to be OVERMASS (you can not be overweight) floating inside that gravityfree cavity and pull the fragile ceiling down when coming too close.
Seriously? You're argument is that because star (or galaxial or galaxial superclusters, also) aren't spherical, that's your proof that gravity in a sphere cancels itself out? Using that very very very very very bad logic, you would float off the planet, the star wouldn't be shining and we wouldn't exist.
I say again to EVERYTHING you've said....show your math. I loathe and despise the spreading of bad, lazy thiking and pseudoscience and that is EXACTLY what you are selling. I have children that I have worked very hard to educate outside of the nonesense of the WT, I will be dead before I let charlatans spread idiot ideas in the name of science to dumb them or anyone else back down.
Show your math. Put up or shut up.
These ideas are fine in an ideal scenario, like the Laws of motion of Kepler, working only for pointmasses. You do not want to be OVERMASS (you can not be overweight) floating inside that gravityfree cavity and pull the fragile ceiling down when coming too close.
Weight is not the same as mass. At ALL. You clearly have no idea what you are talking about. Also, Kepler was a good starting point but ultimately wrong about several things, as we will all eventually be. You are wrong on several things now, easily proven.
Unless, of couse, you can show your math, which, strangely, you have argued against doing again and again.

Caedes
Caedes, point well taken but it is a language issue, in short terse sentences, only the bare essentials of a thought are expresses. When I mean gravityneutral, I talked about the effect inside an onionlike layered entity. If you
Follow the the graph line of the strength of the gravitational pull from the maximum at the surface (if any) to the zero at the center, one realises there is a steady or incremental decline. This means that the layer above, like layers down a mine shaft have CRASED to contribute to the strength of the downward pull. how otherwise could the force have declined? It follows that if the inner layers of your choice could be removed, there would be no gravity force left. (sic)
in other words no acceleratin in the center of earth and less acceleration in a mineshaft. Also no acceleration inside a hollow sphere. Hope that clears something up. (sic)
If you were in a shaft leading to the centre of the earth your acceleration would be at a maximum at the surface and would gradually reduce to zero at the centre. Your velocity would start at zero at the surface and increase until you reached terminal velocity (assuming there is air in your shaft of course), you would then travel at that velocity until you got to the centre. Once past the centre your velocity would decrease as gravity started to decelerate you. Eventually the gravity would reduce your velocity to zero and then would start accelerating you back towards the centre. Hence your velocity would still be at it's maximum when you go through the centre.
So Bohm is correct, however Prologos, you are still forgetting the scenario you are actually discussing, the earth is still at the centre. Your water canopy is not gravity neutral because it can't be considered on it's own. The fact that the earth wouldn't feel any gravity effects from the water canopy does not make the canopy gravity neutral, in fact the water canopy would very much feel the effects of it's own gravity along with that from earth.

bohm
Bohm: in other words no acceleratin in the center of earth and less acceleration in a mineshaft. Also no acceleration inside a hollow sphere. Hope that clears something up.
Viviane: That's not true. You yourself just proved it by saying "less acceleration ina mineshaft".
Well actually it is true (It is known as the "shell theorem" which I was not aware of when I wrote my post) and you can find a proof in any elementary book on classical mechanics, c.f. wikipedia. I will be happy to help you with the required integration if neccesary:
http://en.wikipedia.org/wiki/Shell_theorem
Isaac Newton proved the shell theorem [1] and said that:
 A spherically symmetric body affects external objects gravitationally as though all of its mass were concentrated at a point at its centre.
 If the body is a spherically symmetric shell (i.e., a hollow ball), no net gravitational force is exerted by the shell on any object inside, regardless of the object's location within the shell.
Now, so returning to the mineshaft, if we assume the earth is a body of uniform density the gravitational pull at a radius r will (per 1 of the shell theorem) scale as the mass (proportional to r^3) divided by square of distance (newtons low of gravitation) and so scale as r. Ofcourse if you make assumptions on the density of earth this may affect the result, however my comment was discussing the idealized situation and at any rate I wont look up the density of the mantle compared with the inner parts of the earth now.

bohm
Update: I tried to work it out and it does seem like the difference in density of the earth when one compares the mantle with the inner parts. So the correct answer is the acceleration of an object will properly increase slightly as one pass down a mineshaft and then begin to decrease at some depth; This does not affect my two previous comments since they were made on the assumption of uniform density, specifically, the shell theorem still holds which was the point of my comment.
This is not to say the shell theorem support the idea of a "water canopy" for several reasons:
1) If we assume the canopy is hard sphere around the earth it would need to be very strong indeed to withstand the gravitational pull and the tidal effects from the moon and sun. Assuming this, it would be in the unfortunate situation not to be gravitationally stable, that is, any small change in velocity of the shell (by e.g. a meteor impact) would cause it to drift into the earth.
2) If it was instead not rigid and rotating, it would eventually collapse to something resembling the rings around saturn or a new small moon, depending on it's height above the surface.

Viviane
Well actually it is true (It is known as the "shell theorem" which I was not aware of when I wrote my post) and you can find a proof in any elementary book on classical mechanics, c.f. wikipedia. I will be happy to help you with the required integration if neccesary:
It's still not true. Please do show the math. I would suggest you pick the exact center of the sphere and, for a good distribution of test points, five random places inside the hollow sphere and run the math. Oh, and please be sure to use relativity when describing the gravity. Classical mechanics is so hamfisted at it.
Now, so returning to the mineshaft, if we assume the earth is a body of uniform density the gravitational pull at a radius r will (per 1 of the shell theorem) scale as the mass (proportional to r^3) divided by square of distance (newtons low of gravitation) and so scale as r.
So, even using classical mechanics, it's not, as you said, "no acceleration in a hollow sphere". Yeah, I did some gravitational math earlier on the thread. It's not very complicated.
Ofcourse if you make assumptions on the density of earth this may affect the result, however my comment was discussing the idealized situation and at any rate I wont look up the density of the mantle compared with the inner parts of the earth now.
Yeah, I did that the other day. It's a tougher number to find tham I thought it would be, lots of assumptions have to be made.
Anyway, at the exact center of a perfect sphere of perfectly uniform density in a nonmoving system (that bit's important) with no other external forces whatsoever, there isn't "no gravity". There is equal acceleration in all directions. Any particle or body placed in and only in the exact center of a sphere under such conditions would experience the exact same gravitational acceleration in all directions and be held in place.
No gravity would mean it could float around and experience zero gravitational forces at all in any direction.
Hope that clears it up! Looking forward to seeing your math!

Viviane
Once past the centre your velocity would decrease as gravity started to decelerate you. Eventually the gravity would reduce your velocity to zero and then would start accelerating you back towards the centre. Hence your velocity would still be at it's maximum when you go through the centre.
So... how is there no gravity at the center if gravity is decelerating the object? See my previous post.

Viviane
BTW, for those about to say "Bbbbut Shell Theorem!", it's got a LOT to do with the radius of the objet under questions. It's not as simple as reading Wikipedia.

prologos
Caedes, I appreciated your previous point that a massive object possibly would feel and exert a self gravitaional effect on and inside a shell.
My calling an object gravitational 'NEUTRAL' can not be correct in a TOTAL sense. I stand corrected and always love to learn.
My point about the canopy shell was, that being high up, it could not have had any gravitational effect on the surface gravity or the gravity on top of Everest UNTIL it rained and settled below there. Then its full weight was felt, and it then INCREASED the gravity measurable at that new surface of the planet ( not by much) Of course that misterious high water was always part of the total attractor, when measured from the outside.
This effect what I called 'gravity neutrality' is only true of a shell, assuming the water canopy was a shell, and not a dome that rested on the flat earth and "turtles all the way down".

bohm
Viv: So, even using classical mechanics, it's not, as you said, "no acceleration in a hollow sphere". Yeah, I did some gravitational math earlier on the thread. It's not very complicated.
we agree the shell theorem is correct right? We also agree the shell theorem sayes there is no acceleration of an object inside a hollow sphere? ( point 2)
I got no idea what you object to to be honest....

bohm
Viv: BTW, for those about to say "Bbbbut Shell Theorem!", it's got a LOT to do with the radius of the objet under questions. It's not as simple as reading Wikipedia.
the radius of which object? what part of wikipedia is wrong?
Please dont shift the goal post by introducing qm or gr.