# Gravity inside a massive hollow sphere.

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If I built a large hollow sphere from an extremely dense material (dense enough so that the shell has noticeable gravitational effects) what would the gravity inside it be like? Would objects fall towards the inner surface, or fall towards the empty center? Or would they fall towards the inner surface, while their acceleration decreased the closer they got to it?

I ask because visualizing this using the famous rubber sheet analogy (representing the sphere as a ring, since I'm removing a dimension in doing this...), it seems like the gravitational attraction should be towards the shell, regardless of whether you're inside or outside it - this makes sense intuitively, too, since the shell is what has mass... Visualizing it with gravity represented as "force lines", however, I'm not so sure.

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If I built a large hollow sphere from an extremely dense material (dense enough so that the shell has noticeable gravitational effects) what would the gravity inside it be like? Would objects fall towards the inner surface' date=' or fall towards the empty center? Or would they fall towards the inner surface, while their acceleration decreased the closer they got to it?

I ask because visualizing this using the famous rubber sheet analogy (representing the sphere as a ring, since I'm removing a dimension in doing this...), it seems like the gravitational attraction should be towards the shell, regardless of whether you're inside or outside it - this makes sense intuitively, too, since the shell is what has mass... Visualizing it with gravity represented as "force lines", however, I'm not so sure.[/quote']

Assuming a constant thickness and density of the shell it would be neutral. You would be pulled in no direction. You are correct about your ring example. You would be pulled more toward the closest part of the ring but a sphere is different.

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Depends on the mass of the object inside the sphere, but if you're dealing with an object of negligible mass inside the spere the smaller object will accelerate towards the center of mass between the object and the sphere, which would be towards its center.

If you're dealing with a small object with intense mass inside the sphere the sphere should accelerate to put its closest part physically against the intense mass if this intense mass is greater than that of the sphere, if the intense object starts off center because the part closest to it feels more acceleration from the object than the further parts and should over power it to determine its motion. But if the intense mass starts in the center of the hollow sphere it should just remain there as the sphere accelerates to keep the object at its center.

I haven't done any calculations, but I'm basing that on what I know about gravity. I hope I'm answering the point of this post, because this depends on many variables and the assumption that the system has the only objects in the universe.

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Assuming a constant thickness and density of the shell it would be neutral. You would be pulled in no direction. You are correct about your ring example. You would be pulled more toward the closest part of the ring but a sphere is different.

Umm, no? Why would a sphere be different? The only place where there would be no net attraction would be the exact center, same as a ring. In all other places you would pulled to the closest part of the sphere.

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wouldn't the object fall in slow motion since its being pulled from all directions?

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Exactly right and that's exactly what out world is like. Except our weather pattern swirls around causing the earth to spin and bulge in the middle. Heat gets sucked in at the polls and redistributed to the equator uniformly. this has allot to do with why there is global warming and it will happen regardless if you burn hydrocarbons (it happens faster with hydrocarbons burning in the mix).

If you look at the Weather Engine thread you should be able to figure out what's going on. Ill give you a hint high pressure on the surface low pressure in side.

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Um... no. The weather does not cause Earth to spin. The moon has no weather, and yet it spins. Also, your blatent self promotion of an unrelated thread, and your condecendance towards others are a surefire way to lose what little respect you have here.

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As Nevermore points out, you have reversed cause and effect. Weather patterns and the equatorial bulge are the result of the spin.

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'']Umm, no? Why would a sphere be different? The only place where there would be no net attraction would be the exact center, same as a ring. In all other places you would pulled to the closest part of the sphere.

A sphere has enough mass opposite to balance things out. Think of the vectors diverging as the wall approaches on the one hand and converging on the other. That and the distance squared rule makes for a balance. A ring does this also but not enough.

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Applying gauses law we can easily prove that the gravity inside a sphere would be 0 due to all the forces from each section of it balanancing all the forces from all the others.

I'll do the maths for this later if you so wish but I'm doing some work atm so don't have time.

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Cool thought, XYPH. I believe it is Einstein that said Large bodies curve space. It depends on hw big ur hollow sphere is. I think i agree wit the fact that if it is one massive sphere it should be able to form a form of magnetic vortex. THis i believe in turn would generate the gravitational pull in the sphere. As per Gauss' law sayin that the charge inside the sphere is zero, i agree but it refers to the net charge not that there is nt a charge in the sphere.

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The net pull is zero inside a sphere. It's not terribly difficult to show with calculus, if you realize that for any location within the sphere, there is an axis which passes through it, and that the force from a given point on the sphere can be divided into forces along the axis and perpendicular to it. The perpendicular forces obviously cancel out (for every point pulling you one way, there is a corresponding point pulling you exactly as much in the other direction), and so you are left with the forces along the axis, which vary as the sine of the angle between perpendicular and point divided by the square of its distance. The integral of this function will always equal zero in a given sphere

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I concur with Sisyphus. Its a simple application of Gauss' Law.

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A sphere has enough mass opposite to balance things out. Think of the vectors diverging as the wall approaches on the one hand and converging on the other. That and the distance squared rule makes for a balance. A ring does this also but not enough.

I think a ring balances out in the plane of the ring for an inverse linear field and a hypersphere balances out if the field is inverse cubed, etc. etc. (just don't ask me to do the math:D )

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OK, so the consensus seems to be that the inside of such a sphere would be pretty much a 0g environment, since the gravitational attraction from the rest of the sphere will always be balanced... That's interesting. Thanks to all who replied.

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Um... no. The weather does not cause Earth to spin. The moon has no weather, and yet it spins. Also, your blatent self promotion of an unrelated thread, and your condecendance towards others are a surefire way to lose what little respect you have here.

I don't want the respect of the close minded, and im not self serving. if i was i would be tring to sell somthing, but im not.

The point about the moon is irrelevant since it had water on it once and in fact it still does at its polls.

You can actually try to think for your self or you can quote people who seem smarter than you its your choice.

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To me this proposition appears to depend to some degree on the dimensions of the sphere.

If the sphere were 1 million miles in diameter and the thickness of the walls were 3000 miles, then wouldn't the walls have enough mass to attract any smaller mass to it?

Wouldn't a person be able to walk around on the inside of this sphere just as if he were on the outside of it?

Isn't the opposite side is far enough away so that it attracts less than the near side?

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If the sphere were 1 million miles in diameter and the thickness of the walls were 3000 miles' date=' then wouldn't the walls have enough mass to attract any smaller mass to it?

[/quote']

yes.

Wouldn't a person be able to walk around on the inside of this sphere just as if he were on the outside of it?

no. the net gravitational force is zero

Isn't the opposite side is far enough away so that it attracts less than the near side?

yes, but there is a lot more mass "above" you which is enough to cancel out what is "below" you.

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I don't want the respect of the close minded' date=' and im not self serving. if i was i would be tring to sell somthing, but im not.

The point about the moon is irrelevant since it had water on it once and in fact it still does at its polls.

You can actually try to think for your self or you can quote people who seem smarter than you its your choice.[/quote']

Just stop now. You posted nonsense in the wrong thread and now you're attacking the person who pointed it out.

And how exactly does believing that weather makes planets turn make you open-minded?

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Gauss' law:

$\oint_A \mathbf{g} \cdot d\mathbf{A} = 4 \pi G M$

If you do the integral over an area which is a sphere inside the sphere we are looking out we find that g must = 0... as M, the mass inside the guassian sphere is 0.

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Gauss' law:

$\oint_A \mathbf{g} \cdot d\mathbf{A} = 4 \pi G M$

If you do the integral over an area which is a sphere inside the sphere we are looking out we find that g must = 0... as M' date=' the mass inside the guassian sphere is 0.[/quote']

For those who don´t see why g=0 follows in above, I´d like to add the an explanation for it:

The problem has spherical symmetry. Therefore, the gravitational field also has to have spherical symmetry. Therefore, the field has to be perpedicular to the surface of the sphere integrated over with the same magnitude at each point. Therefore, the integral equals A*g® with r being the radius of the sphere. Now, since A*g®=0 (assuming we are still inside the massive hollow sphere) and A>0, it follows that g®=0 for all r smaller than the radius of the massive hollow sphere.

Also note that the spherical symmetry holds true outside the mass distribution, too. Only difference now being that 4*pi*G*M isn´t zero now. In fact, one can easily see here why outside a spherical symmetric mass distribution you have the same gravitational field as if all the mass was concentrated at the center.

EDIT: The spheres integrated over have to have the same origin as the massive sphere, of course. Otherwise, the symmetry argument won´t hold anymore and the calculation gets more tedious.

@Klaynos: Take my comment as an addition for people who accept statements like "therefore g=0" too quickly without really questioning themselves why and understanding it (and I can well speak of myself doing that from time to time). Your answer was great and I think it´s good that someone actually dares to post a mathematical solution in the physics forums - we should do that much more often.

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For those who don´t see why g=0 follows in above' date=' I´d like to add the an explanation for it:

The problem has spherical symmetry. Therefore, the gravitational field also has to have spherical symmetry. Therefore, the field has to be perpedicular to the surface of the sphere integrated over with the same magnitude at each point. Therefore, the integral equals A*g® with r being the radius of the sphere. Now, since A*g®=0 (assuming we are still inside the massive hollow sphere) and A>0, it follows that g®=0 for all r smaller than the radius of the massive hollow sphere.

Also note that the spherical symmetry holds true outside the mass distribution, too. Only difference now being that 4*pi*G*M isn´t zero now. In fact, one can easily see here why outside a spherical symmetric mass distribution you have the same gravitational field as if all the mass was concentrated at the center.[/quote']

Yes thanks I did mean to post one I just seem to have forgotton, wow I'm really not with it today :|

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My first thoughts would be that anything inside the sphere would just "fall" into the centre and be suspended there at the center of gravity. Most likely with a bemused look on it's face.

[edit: Of course with the rubber sheet/sphere=ring analogy it would make sense that gravity inside the sphere would be zero, with the sheet being stretched flat inside the ring of mass.]

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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.

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However even if you were at the edge of this sphere, you would only be "next to" one small area of sphere.

The rest of the sphere's mass and gravity could be strong enough to pull you away in the other direction - which is why I first imagined you'd end up in equilibrium in the centre of gravity, at the middle.

But then who's to say the rest of the sphere would have stronger influence than the area of sphere you're next to, and not an equal amout - thereby cancelling out any gravitational force, no matter where you were inside the sphere...

I hope that made sense. I'm typing on a half-concious brain here. *yawn*

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