dtheriault

If you had a planet with a mantle that was more dense than the core could gravity move outward?

According to his theory gravity forces always pull inward. Why?

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Charlatan

Here is the question, why does gravity do what it does? I know we all exert a little pull of gravity on everything around us and vice versa... The greater an objects mass the greater the pull of gravity it exerts on the objects around it.

But what is gravity?

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teflonian

dtheriault-- I don't know exactly what "could gravity move outward" mean, but no... Gravity is only an attractive force. The fact that the mantle is more dense than the core could create some interesting scenarios for the planet, but any observer on or near the planet would still feel a pull towards the planet. It doesn't matter where the mass is located in the planet as long as it is there, it will pull "downwards". I don't think anyone can say why exactly it only pulls and doesn't push; there may be some theories on the edge of physics that may attempt to explain this, but I imagine it is still pretty unknown.

Charlatan-- gravity is the attraction of everything around us with mass to everything around us (and us) with mass. Sorry to get all tautological, but that is simply what "gravity" is defined to be. Gravity is one of the forces that act on bodies in the universe... To go any further (at least with what I perceive to be the intention of your question) would probably require either to go into metaphysics or more complicated theories like the gauge theory that I don't really understand anyway and probably wouldn't really get to the root of your question.

BadNick

This is oversimplified and I'm using common analogies but maybe try this: think of a mass floating in space and that mass distorts "space-time" around it - like a bowling ball sitting on a rubber sheet but in 3 dimensions not just a two dimensional surface. As you approach that area of "space-time" the local geometry is distorted so a "straight line" path to the traveller will actually be a curved path to an outside observer since the path follows this distortion in space ...errr space-time. Moving along that curved path produces forces on the moving object, similar to when you make any turn and feel acceleration forces pushing you toward the center of curvature. That force you feel is "gravity". As you get closer to the object causing the distortion, the curvature is greater and gravity forces are greater.

I'm no expert by any means, but I'm using "space-time" instead of just "space" since my limited understandinig is that space-time is a more accurate way for us to think of the "space" in our universe.

C4 Diesel

No, because the mantle is spherical. If you are inside the core, you would still be subjected to the gravity from all sides of the mantle, even if there is one that you are closer to. This would require one hell of a structurally sound core, though, to keep the mantle from collapsing in on it.

In theory, you could have a (man-made) structure with NO core, a la the halos in the video game Halo or a empty sphere or something. In this situation if you were "inside" it, then yes, you would get pulled outward if the radius of the object was sufficiently large, the object had a whole lot of mass, and was separated from other significant gravitational fields.

Excellent question. Gravity is the only force for which particles have not yet been discovered. Scientists theorize about "gravatons", which would be the particulate unit of gravity, but that would be one hell of a particle, since it would be powerful enough to connect everything in space (even to a minute degree) and totally ignore any forms of mass or other particles in its way. The notion is that gravity does not have a particle associated with it, but then that goes against everything else in the universe.

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Charlatan

C4 Diesel... you are describing a Dyson Sphere, no?

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Yakk

No, actually. If you are inside a spherical structure that is radially uniform, you feel no gravitational effects from it.

The calculus to show this was sort of fun. =)

To make it concrete:
Imagine you where inside a large, hollow, spherical shell. It is really really heavy.

Now, this shell 'pulls' at you.

If you are in the middle, in every direction there is the same amount of mass at the same distance. You feel no gravity.

Now, lets say you where nearer to one side.

Well, under the inverse square law, the gravitational force from that side is less attinuated by distance. However, there is more stuff on the other side.

The two forces (the closer, smaller mass on your right, and the farther, heavier mass on the left) cancel each other out perfectly.

Another fun structure is the heavy hollow tube. Things inside the tube are pulled towards a plane that cuts the tube into two equal-sized tubes. I believe the strength of the pull goes up as you move further away from the plane, to a limited extent.

An interesting question: going back to the heavy, hollow shell -- is there any effects when accellerate and move from point to point?

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Last edited by JHVH : 10-29-4004 BC at 09:00 PM. Reason: Time for a rest.

Paradise Lost

The best way to show that it doesn't matter where alot of the weight of a planet is, as
long as it's mostly uniform (and it will be), the combined mass of all the atoms from you,
drawn straight down to the core of the planet, will, roughly, be equivalent.

So, whether the decent amount of mass is in the core, or in the mantle (impossible, the
pressure of the weight would force it to become a core, but still,) the total mass is
still the same and therefore pulling on you the same no matter where the
mass is located in the planet.

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Coppertop

Ok, I don't I can't be the only dork who thought of this:

Hain

I see good clean potential for some fun with this. Stand on one end, jump in... and just keep bouncing back and forth from one end to the next.

Do you mean accelerate the sphere you are in? If I remember right you will remain in the same location as the sphere, until you collide into the side. Since the net forces on you inside are zero, you should remain still.

Someone else read J. Richard Gott's Time Travel in Einstein's Universe. He proposes if you take the mass of about 6 Jupiter's, collapse it into a spherical shell 10m diameter of the hollow and 12m overall diameter, you will bend space enough that your time will slow down significantly.

Just some fun effects of gravity.

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Hain

Why even begin with a Dyson Sphere? Niven's Ringworld is much better. Sure scrith is a "hand waved" material that requires nuclear forces to be occurring constantly within it. But that's a minor detail to overcome when you are on a ring 95 million miles away from the sun, nearly a million miles laterally, and only 100 feet thick!

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Yakk

No, I mean accellerate inside the sphere.

Relatavistic effects should make the sphere no longer spherical, to a small degree. Would that change the math at all?

If the sphere weighed ALOT (like, ALOT), these effects might be detectable even at small amounts of accelleration or velocity. You may be able to avoid black-hole problems by making the sphere larger. . .

Random: might the effects behave alot like inertia?

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Last edited by JHVH : 10-29-4004 BC at 09:00 PM. Reason: Time for a rest.

fckm

That's an interesting question. It'll probably break symmetry. I don't know, it would depend on how you'd do the divergence thereom in that case.

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C4 Diesel

I cecede to Yakk... Sorry, man... Forgot my calculus in the midst of my ramblings.

Charlatan... A Dyson sphere would do the trick, although many other things would do as well.

Coppertop... My limitied stint with D&D only started at the very tail end of 2nd ed and barely made it to 3.5, so yes, you probably are the only dork who thought of that. Haha...

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fckm

More thoughts on the relativistic sphere. I talked to my roomate about it, she has no idea either.
The problem I have with the length contraction causing a field is that in the rest frame of the sphere, there is no field. I have no idea what acceleration would do to it. I'll go ask a prof on friday. This is interesting.

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stingc

If the sphere isn't rotating, then there is no gravitational field inside - even relativistically. If it is rotating, then you have a "frame dragging" (Lense-Thirring) effect.

C4 Diesel

Frame Dragging, for those like me who had no idea what this was.

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fckm

So the proper frame is always in the rest frame of the object?

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stingc

I'm not sure what you mean. If you have something which is spherically symmetric (this can be defined in an invariant way), then it is possible to prove that the spacetime inside of it is flat (this is another invariant statement). The motion of a particle inside the sphere would therefore be the same as it would be if the sphere weren't there.

This result follows from Birkhoff's theorem if you want to look it up. As far as I know, there isn't any simple, exact generalization of this that could be called an equivalent of Gauss' form for Newtonian gravity. It is only in this special case of spherical symmetry that things work out nicely.

My point is that you shouldn't try to think of this as an application of an inverse square law with length contractions built in. That procedure isn't correct, and it's a nontrivial coincidence that GR happens to predict something so simple in this case.

For those wondering why this result doesn't apply to a rotating sphere, gravitational fields depend on the sources' internal momenta and stresses as well as their masses. Rotation therefore breaks spherical symmetry.

fckm

Does the same thing apply to uniformed spheres of charge?

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And if you say to me tomorrow, oh what fun it all would be.
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