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 Be the first pioneers to continue the Astronomy Discussions at our new Astronomy meeting place...The Space and Astronomy Agora Re: Making Sure I'm Not Misunderstood. Forum List | Follow Ups | Post Message | Back to Thread Topics | In Response ToPosted by Zephram Cochrane/">Zephram Cochrane on November 30, 1999 22:14:26 UTC

: : The hole itself is at the center, but then the object in it can be anywhere in the hole, not just at the center of the hole, and then no matter where the object is in the hole it will undergo no gravitational acceleration. For instance it is not attracted to a side wall no matter how close it gets to one.

: why?

There's a general relativistic answer, and a Newtonian answer. I'll give the Newtonian answer. The answer is superposition. Consider a point off center in a thin spherical shell of mass of density s. Using this off center point as the origin for a small solid angle dWin one direction we find the amount of mass enclosed by the element of area dA = r2dW on the surface by the solid angle to be dm = sdA = sr2dW The element of the gravitational acceleration due to this element of mass is: g = Gdm/r2 = GsdW Now consider the same solid angle in the opposite direction. This results in: gopp = GsdW These are now obviously equal in magnitude even though r and dm are different due to the point being off center, but acting in opposite directions and so they add to zero. Since we didn't specify where in the spherical shell this point is the net field must also be zero everywhere inside it. Now add shell layers until the thin shell is a ball with a spherical hole in the center and the net gravitational acceleration is then still zero everywhere inside the cavity. You are not pulled apart, you are weightless everywhere inside the hole.