It's because the pull from every direction is exactly the same. This is obvious from the center, but as you move to one side, you're closer to that side, which increases its pull, but this is exactly offset by the fact that there's now MORE mass on the other side. The proof is extremely lengthy; see this link:
http://en.wikipedia.org/wiki/Shell_theorem
2006-09-11 05:09:15
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answer #1
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answered by hslayer 3
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hslayer misinterprets the Shell Theorem. Others may be confusing this with the fact that the electric field within a perfectly conducting shell is zero (the electrical potential is uniform). Since matter is neither an insulator nor a conductor of gravity, the gravitational force on a mass inside the shell is almost identical to the force outside the mass. The equivalence to a point-mass is valid as long as the distance to that point isn't so small that it drives the force to infinity. What the Shell Theorem actually says for the case within the shell is that the contribution to the total gravitational field from the shell is zero. So it's as if the shell wasn't there.
If there's a Gauss law relevant to gravitational fields, that would support this interpretation of the Shell Theorem.
2006-09-12 01:53:54
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answer #2
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answered by Frank N 7
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Can't be done using standard procedures. Gravitational waves move toward the core point of a sphere. There is no place within a hollow sphere where g = 0. There is no location within a solid mass where g = 0, even the very core point of a mass does not equal zero, as is supposed.
But, were a person to remove all the heat energy from within a hollow sphere, then the g value could equal zero. The reason this is true is that the equation for a gravitational field is c2 = E/m. The "E" value in this instance is that of the heat energy within a mass. Were our planet to have all of its heat energy removed, our planet would not have a gravitational field.
There is a writing at http://360.yahoo.com/noddarc entitled "The Problem and Repair of Relativity." It is short and easy to read, and should answer your question.
2006-09-11 13:20:45
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answer #3
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answered by Anonymous
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According to Gauss's law the surface integral of gravitational force is equal to the 4piG times the mass enclosed within the closed surface. Now for a point inside the hollow sphere, imagine that it lies on a closed surface inside the hollow sphere. Since this closed surface does not enclose any mass, the gravitational force on every point on this closed surface is zero. Hence the gravitational force at the given point inside the hollow sphere is zero.
2006-09-11 14:45:05
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answer #4
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answered by rabi k 2
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The gravitation field an electrical field folow basically the same rule, and both fall under Gauss law.
The description of Electric field is E=force per unit energy which boils down to an inverse meter.
The description of the gravity field is GF=Force per unit mass which boils down to meter per second squared.
So both mass and energy content ,whether its a holow sphere or solid sphere can be represnted as if it was a point in the center of mass of the sphere.
There is nothing inside a mass that has been proven to cause a gravitational pull( attraction.)
2006-09-11 12:30:17
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answer #5
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answered by goring 6
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It is zero only at the center. Non zero everywhere else.
2006-09-11 12:32:54
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answer #6
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answered by Dr M 5
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i have never heard of such an concept
any how
2006-09-11 12:09:41
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answer #7
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answered by Ravitej 2
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