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Gravity inside a massive hollow sphere.


Xyph

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It's well known and easy to demonstrate that the gravity inside the sphere, exerted by the sphere, would be zero at any point whatsoever. You don't get pulled towards the center, you don't get pulled towards the nearest edge (even if you're right next to it), it doesn't matter how big the sphere is or how thick its walls are. It doesn't matter if it's 50 billion miles across with 10 thousand mile thick lead walls that you're right next to. The net force is exactly zero. Always.

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Would the inside of this sphere be truly anti-gravitational, or would outside influences that affect the sphere also affect it's contents?

 

for instance, if the sphere was near a black hole, would the gravity also pull objects inside the sphere to the edge that was nearest to the black hole, or would there still be zero gravity insde the sphere, no matter what gravitational forces were also working on the sphere from outside?

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Would the inside of this sphere be truly anti-gravitational' date=' or would [b']outside influences that affect the sphere also affect it's contents?[/b]

for instance, if the sphere was near a black hole, would the gravity also pull objects inside the sphere to the edge that was nearest to the black hole, or would there still be zero gravity insde the sphere, no matter what gravitational forces were also working on the sphere from outside?

 

 

The bold is correct

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I think that if you were directly in the center of a massive hollow sphere, which was symetrical in mass and area, that you would feel no gravity, but if you were off-center, you would slowly be pulled to the area of sphere you were closer to.

 

 

Even after reading all the posts explaining that you wouldn't? Why would you think that?

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Does this work for magnetism as well? I suppose it would, but I'm not familiar with this area of physics.

 

 

No, the laws for magnetism are different. There are no magnetic charges, so you don't have an inverse-square force that fits the form to follow Gauss's law.

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No, the laws for magnetism are different. There are no magnetic charges, so you don't have an inverse-square force that fits the form to follow Gauss's law.

 

The Bio Savart law (spelling) is the usefull *analogy with Gauss's law isn't it?

 

*I know it's an inoccernt term in ths case but I'm tired :|

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It's well known and easy to demonstrate that the gravity inside the sphere, exerted by the sphere, would be zero at any point whatsoever. You don't get pulled towards the center, you don't get pulled towards the nearest edge (even if you're right next to it), it doesn't matter how big the sphere is or how thick its walls are. It doesn't matter if it's 50 billion miles across with 10 thousand mile thick lead walls that you're right next to. The net force is exactly zero. Always.

 

The Earth is close to being a sphere.

 

If one dug a hole 100 feet deep and stood at the bottom of the hole, would he not, in effect, be inside a sphere?

 

Are you saying that there is no gravity at the bottom of a hole in the ground?:rolleyes:

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The Earth is close to being a sphere.

 

If one dug a hole 100 feet deep and stood at the bottom of the hole' date=' would he not, in effect, be inside a sphere?

 

Are you saying that there is no gravity at the bottom of a hole in the ground?:rolleyes:[/quote']

 

 

He'd be in the middle of a solid sphere. not inside a shell, which leads to a different result when you do the guassian analysis of the system (there is gravity but it gets less as you head downwards as there is less mass below you).

 

A HOLLOW sphere has NO NET GRAVITY INSIDE! That is not a subject of discussion is it a know mathmatical (and probaly experimental) FACT!

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That is not a subject of discussion is it a know mathmatical (and probaly experimental) FACT!

 

Yes, as anyone above who has taken a calculus-based physics class has said, this is fairly easy to prove mathematically.

 

I would suspect that it would be very difficult to do experimentally however -- from a few posts back, you would have to eliminate all outside forces. Then get a measuring device into the hollow sphere. All this assuming you have made a perfect sphere with completely uniform density.

 

But, I do not understand how people are still arguing over the theoretical result -- pull out a decent 1st semester university physics text and it would show you how to do the calculations for yourself. In fact, if you google it, ignoring all the pseudoscience stuff about the hollow earth theory, you can find a few websites that explain this even further.

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Would one experience the same result of zero net gravity when placed within a hollow torus?

 

 

Not for a general point; the equations work for spherical symmetry only. If you were at the center of a symmetric distribution it would be zero, but would not in general stay zero away from that point.

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I think that if you were directly in the center of a massive hollow sphere, which was symetrical in mass and area, that you would feel no gravity, but if you were off-center, you would slowly be pulled to the area of sphere you were closer to.

I kind of concur, it is said that the earth's gravitational field attracts things to its centre. So this sphere should act in like manners.

Concernin the charges, Gauss's applies, but it is not that there isnt a charge witin the sphere.

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Not really, there's an equal about of gravitational attraction above you and to the sides of you and in every other direction as much as there is directly below you.

 

On Earth there's ONLY gravitational attraction below you.

 

However, with regard to being inside the ring of a torus (I presume the question was about being inside the ring itself and not just in the hole of the doughnut?) my thoughts are that you would be attracted to the side that forms the inside edge of this ring/hole, because I imagine the gravity to be stronger on the inside of the torus (the edge of the hole) than the gravity on the outside of the torus' ring, at the edge next to outside space.

 

This would be because there is a lot more Torus on the other side from where you are, as you're only inside one section of ring. In a sphere there is an equal amount of sphere all around you.

 

I hope this makes sense, perhaps if it's confusing - someone else who knows what I'm trying to say can translate...?

 

I'm inclined to believe that gravity inside any hollow but regular shape, be it cube, tetrahedron or sphere will always be zero as there is an equal amount of shell around you in all directions...?

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Originally Posted by Euclid

I think that if you were directly in the center of a massive hollow sphere, which was symetrical in mass and area, that you would feel no gravity, but if you were off-center, you would slowly be pulled to the area of sphere you were closer to.

 

I kind of concur' date=' it is said that the earth's gravitational field attracts things to its centre. So this sphere should act in like manners.

Concernin the charges, Gauss's applies, but it is not that there isnt a charge witin the sphere.[/quote']

 

 

Oh, Good Grief!

 

Gauss's law says that there is no gravity inside the hollow sphere. This has been said by several people, and three people with "physics expert" titles have concurred! What, exactly, is your objection?

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Gauss's law says that there is no gravity inside the hollow sphere. This has been said by several people, and three people with "physics expert" titles have concurred! What, exactly, is your objection?

 

I have a proof of this in a physics text. You should definitely believe it.

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Even without using any maths it is pretty obvious.

 

If you are in the centre of the sphere, you will be pulled in all directions at the smae time, so no net force. (I don't think anyone is disputing that.)

 

Now, if you move closer to one of the edges, then you are closer to the side in front of you and it will therefore pull you more strongly towards it (gravity gets stronger as you approach the source). But now there is more mass behind you (most of the sphere is behind you now), and since gravity increases with mass, the part of sphere behind you will also be pulling you more strongly. Since the area of a sphere increases as r2 and the gravitational force behaves as 1/r2 these two effects are exactly balanced, and you feel zero net force.

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So what about the gravity in a cube like Transdecimal was saying? truthfully, I dont think it would function like a sphere at all.

 

If you sat in the center of a cube that was much like the sphere (thickness size etc.) you'd just float in the center.

 

however unlike the sphere if you moved off center... well I'm having a tough time visualizing what would happen, but I don't think it'd be the same as the sphere...

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I've had some time to think, If you drew lines connecting the center of the faces of the cube, those lines would have zero gravity.

 

most regular 3d shapes would do the same. The sphere just has an infinite amount of faces though so there are lines connecting sides at all points inside the sphere so it would be zero gravity

 

EDIT: Please correct me if this is wrong :)

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I've had some time to think' date=' If you drew lines connecting the center of the faces of the cube, those lines would have zero gravity.

 

most regular 3d shapes would do the same. The sphere just has an infinite amount of faces though so there are lines connecting sides at all points inside the sphere so it would be zero gravity

 

EDIT: Please correct me if this is wrong :)[/quote']

 

It's right. I already said so in post 42.

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