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Post by speakertoanimals on Apr 8, 2011 12:11:06 GMT 1
You could try being a bit politer, rather than DEMANDING that I answer every little point you care to raise.
Why, because I persist in trying to get the right answer, and pointed out your many errors and mistaken assumptions?
I've got better things to do than try and answering some of your more boring queries...............
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Post by carnyx on Apr 8, 2011 12:42:18 GMT 1
@sta
Ah.. that little word 'right' ... You do realise the difference between righteousness and truth?
I suggest you read what the founder of 'science' actually meant by it .. and why he had to invent it ... And as for your motivation, which you have clearly exposed, do you regard it as having generally positive consequences?
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Post by speakertoanimals on Apr 8, 2011 13:24:39 GMT 1
And there was me thinking this was a SCIENCE board!
Cut the meaningless blather!
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Post by carnyx on Apr 8, 2011 13:34:46 GMT 1
Ah!
Expert on Science now, are we? Must be the quickest reading of Bacon, ever.
Gercha!
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Post by buckleymanor1 on Apr 8, 2011 14:47:48 GMT 1
Sorry STA this does not seem to be right. This would only happen if the sphere had no mass and there was no ballbearing inside the sphere.Its allready been established that the sphere has mass.So if you drilled a hole without the ballbearing inside admitedly the centre of mass would now be close to the centre and without the hole at the centre. Once you drill the hole with the ballbearing inside the sphere the centre of mass does not remain near the centre it shifts to the opposite side as the ball bearing and the now more massive opposite side are mutualy attracted. If the opposite side of the sphere was not more massive after drilling the hole the ballbearing would not move at all. Once the ball bearing is at the opposite side, this point becomes the centre of gravity to any outside smaller body orbiting the sphere. Try it with a large hole say half a sphere size and make it smaller and smaller the same allways applies once you introduce the ballbearing.
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Post by speakertoanimals on Apr 8, 2011 15:48:23 GMT 1
Standard in these problems to take the test mass (the ball bearing) as having negligible mass compared to anything else in the problem. Hence we just have to solve for the gravitational field of the drilled sphere, then we can predict how the ball bearing would move. SO, if we took a sphere about the size of the earth, say, made of rock, with a hole drilled that was about the mass of a reasonably sized asteroid, then our test mass was a ball-bearing....................
The point is that the ballbearing doesn't move towards the centre of the drilled hollow sphere, but is repelled by the missing mass taken out of the hole.
The repulsion, even when right up next to the opening of the hole, isn't that strong -- it is the gravity we would find on an asteroid, say, just in the opposite direction.
If we started to tunnel through the hole? Well, initial assumption was shell of negligible thickness compared to radius. If we have a sphere of finite thickness, then we still have zero gravity inside, because each infinitesimal shell still gives zero, and a thick shell is just obtained by adding a series of infinitesimal ones!
What about when we drill the hole? We just have negative the effect of the removed plug. So, if this was asteroid sized, then gravity would be like on an asteroid, just pushing up rather than pulling down.
When we burrow through the thick sphere (before hole drilled), we go from zero inside, to increasingly inwards 'normal gravity', until we have maximum normal gravity when we reach the outside of the sphere.
Hence to find result of removing plug, we just have to subtract asteroid contribution from this.
Hence inside we would have repelled from hole. If we move a bit inside the hole, we have inwards gravity as if we were inside complete shell, plus minus the gravity we would find just under the surface of the asteroid plug. When outside, we have normal downwards gravity, minus the downwards gravity on asteroid plug. We hence get a smooth transition from being repelled from hole inside, until we come to outside shell, where we have normal downwards gravity outside shell, but slightly reduced due to loss of mass of plug.
Which, if we took a route from point opposite the hole, to through the hole, and to the outside, would be gravity DOWN all the way (towards point opposite hole), but getting STRONGER as we approached the inner side of the hole -- but I'm not sure yet if would still increase or decrease as you moved through the hole?..........................
But down all the way!
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Post by buckleymanor1 on Apr 8, 2011 16:32:04 GMT 1
Have to get the cap on for that, might be some time.
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