Center of gravity paradoxes

[pioneer]

If we were to begin with a hollow spherical shell of thinkness X, made of thick guage metal, like a metal basketball, the very center will define the center of gravity. If I was to put an object in the ball, off center, it would not go to the center of gravity, but would move to the closest wall. The reason this is so, is that we need mass to generate gravity and there is none in the center of gravity. One can do this with vector addition to get this same result.

 

Let us make this situation even more complicated. Inside the iron shell we place a flexible balloon, which is also spherical and centered at the center of gravity. The balloon is thin so the shell is still the dominant mass. What the balloon will do is stretch and move toward the perimeter. All the walls of the balloon willl be off center such that vector addition will cause each point on the balloon to attract toward the /closest and highest mass. The result will be a slight vacuum pulled at the center of gravity. This paradox occurs because the shell is rigid.

 

The term center of gravity was invented to help model the gravity attraction between two bodies. Each body is treated like all the mass is concentrated at the center and the results appear to work each time. If we took two such double shelled spheres with their internal vacuums, they would attract as though all the mass was concentrated at the center of gravity, even though there is little or no mass at the center of gravity but is undergoing a state of a partial vaccuum. The term center of gravity if there for the ease of math, but it doesn't always tell the entire story of what is actually occurring within the center of gravity.

 

In the case of these two situations attracting from a distance, their centers of gravity will create a false positive for each other, or make each other think this is where all the mass is concentrated. When they get close, they will realize the false positive, and rotate around a new center of gravity, which also contains no matter, which is also a false positive. The kinetic energy of the rotation, is the potential energy that can't be released due to the center of gravity played a joke of nature i.e., false positive, that will only look like a postive-positive at longer distances.

 

The earth is spinning with the solid iron core spinning a little faster. Because the crust and the core are both solid, the perpetual spin would suggest a situation sort of like the above, with a hollow at the center of gravity, which is creating a false positive for the rigid solids. The moon barely rotates suggesting a more continous solid. While its movement around the earth may be in response to the earth's hollow.

 

This is an interesting thing to ponder. I an not married to this but it does create some interesting food for thought that can explain things like the almost perpetual motion of planets around the sun. Gravity just can not close the deal, such that the potential energy becomes motion. The false positive suggests a reason why gravity can not close the deal.

[Sisyphus]

Your basic premise is false. An object inside a uniform hollow sphere experiences no gravitational attraction from that sphere, either towards the center or towards the walls. After that I don't know what you're talking about.

[Bignose]

Sisyphus is exactly right, it is a very classical result that a spherical shell of uniform density will have no gravity inside the shell. If your math says something different, you've made a mistake somewhere.

[pioneer]

I agree with you. The gravity is highest at the surface of the shell, which is why the internal balloon will not pressurize but will pull a vacuum. If we apply GR to this same situation, the space-time affects are highest at the surface and are lower at the center of gravity. The expansion of the balloon is analogous to space-time expanding at the center of gravity.

 

The shell analysis for gravity, does not change, irregardless of how thick the shell is. Even if there is only a small hole in the center of gravity space-time expands at the center of gravity. In this situation, the gravity trough within space-time should have a little hill in the middle. The fusion core of a star, causes an energy expansion of matter. Net mass convection is moving outward from the sun's core via the mass of the solar wind. So there is a hollow or GR hill in the center of the sun caused by the fusion.

 

This little hill at the bottom of the GR trough is the false positive. At a distance, the way the math works, this center of gravity is treated like the entire mass is concentrated at that point. Even with a bunch of hollow shells, the center of gravity is still mathematically modelled using center, even though, in reality, there is no matter there. Somewhere along the line this math simplification became a dogma that implies that the center of gravity is the zone of highest GR. The classical model would say there is no gravity in the center but GR says opposite. It is not GR saying the opposite, but a false assumption has made it that way.

 

So if we look at two separated bodies like the earth going around the sun. The hollow in the sun's center of gravity looks filled in from a distance since the mass will act using the mathematics of the center of gravity. It is possible that mass at a distance can be fooled by this false positive. But as the earth gets closer and sees the center of gravity is a false positive, i.e., it is not the bottom of a GR trough but it has a hill there, the left over potential energy becomes velocity or SR causing the orbit. It is sort of like a conservation of relativity where the system tries to minimize potential.

[Bignose]

I agree with you. The gravity is highest at the surface of the shell...

 

No, you can't agree and then say that the gravity is "highest at the surface of the shell." Inside a uniform shell, there is no gravity. Not in the center, not halfway between the center and the wall, not even on the inside of the wall. Zero gravity. Consider being on the surface of the shell. Despite being closer to the shell in one direction, there is more mass on the farther away side. The combination of farther away but with more mass completely balances the nearer but less mass. So everywhere inside a uniform shell has zero gravity.

[Sisyphus]

I don't know what you mean by "false positives." In classical physics, you just integrate for the whole mass to get a vector for gravitational pull. In relativity you're looking at the shape of space where you are. In neither case does it matter what the attracting body looks like. "Center of gravity" is an abstraction and a useful simplified approximation at long distances, but it has no physical being. I'm not sure where you're getting "leftover potential energy," or what you think that has to do with an orbit.

[BenTheMan]

Gauss' law. Spooky.

 

If you believe that there is not electric field inside a spherical shell, why is it so hard to believe there is no gravitaitonal field either.

[pioneer]

An easier way to see the gravity profile inside a spherical shell is to make the shell huge and place oneself inside. In this situation, the shell looks like a vertical wall directly in front of you and another vertical wall behind you, but at a very great distance. The operative word will be distance. The gravity force drops off with the square of distance. This means the strongest gravity will be directly in front of you. What is behind you will still exert gravity but being so far away it affect is very small. If we make the ball smaller and smaller, and you hold your relative position, the dynamics will not change. The only affect will be more vector cancelling. Only in the center of gravity are all distances exactly the same. Anything off center will cause you to be closer to one area of the wall and gravity will pull you in that direction with a net force.

 

Lets do the same thing with GR. We have our huge sphere and we are quite close to what appears to be a vertical wall in front of us. This area will exhibit the highest gravity affect on us so it has the max GR affect. The wall far away, but directly behind us, generates the same GR, but because it is so far away, its affect will diminish by the time it reaches us. The max GR, is at anywhere near these huge walls. It will get lower inside if we move away from this huge walls and drop off with square of d.

 

If we were to go outside the vertical wall of the shell the gravity will be even higher. At this point not only is that wall pulling us toward it, but the distance vertical wall far away is also pulling in the same direction. If we didn't know this was a spherical shell but thought it was a vertical wall, one might conclude the wall had to be thinker that it was. That is what I was referring to as a false positive. It adds up mathematically, so it is positive with respect to what we think it is. But it is negative in the sense that the wall, in reality, isn't as thick as we thought it was.

 

This is a joke of nature that may explain perpetual orbits. For example, the earth sees the sun's closest surface appearing to generate extra gravity. It sort of uses the math that works at great distances and gets fooled in believing the sun is such and such. Once it gets close, it isn't what it appeared to be. It can not do what it suggested from a long distance. The potential that can't be expressed remains but in the form of orbital velocity.

 

GR is an advanced version of Newtonian gravity, but because of its more abstract nature, somehow physics got fooled by the false positive. They just kept orbitting. One had to got back to the classical basics to set the record straight. Now GR has a better starting point to close the deal. Under the current system we can generate all types of false positives or theories that mathematically add up, but which don't reflect reality.

[Sisyphus]

An easier way to see the gravity profile inside a spherical shell is to make the shell huge and place oneself inside. In this situation, the shell looks like a vertical wall directly in front of you and another vertical wall behind you, but at a very great distance. The operative word will be distance. The gravity force drops off with the square of distance. This means the strongest gravity will be directly in front of you. What is behind you will still exert gravity but being so far away it affect is very small. If we make the ball smaller and smaller, and you hold your relative position, the dynamics will not change. The only affect will be more vector cancelling. Only in the center of gravity are all distances exactly the same. Anything off center will cause you to be closer to one area of the wall and gravity will pull you in that direction with a net force.

 

Again, I'm telling you that's incorrect. It's not just me saying this, it's a well known proof. Hell, Isaac Newton himself proved it in the Principia. It doesn't matter where you are inside in the hollow sphere, the forces always cancel. It is the exact same problem as being within a hollow sphere of electrical charge. Remember, when you're right next to one wall, the matter pulling you towards the wall is having more effect, but there is much more matter pulling you away.

 

Also, the false positive thing is meaningless.

 

1) You can't "fool" matter.

 

2) There is no force needed to explain circular orbits. It's just inertia and perpendicular gravity.

[BenTheMan]

It is the exact same problem as being within a hollow sphere of electrical charge.

 

SO MANY PEOPLE don't understand that, classically, the two forces are VERY similar.

 

pioneer---I am telling you... This was a question on my Classical Mechanics final in my first year of grad school, and is a canonical upper level mechanics question. One can formulate gravity in terms of a potential and a field, just like Maxwell's equations. The only difference is that the constants are different.

 

If you STILL don't believe me, check out this article:

http://www.nytimes.com/library/national/science/health/011800hth-behavior-incompetents.html

[Bignose]

An easier way to see the gravity profile inside a spherical shell is to make the shell huge and place oneself inside. In this situation, the shell looks like a vertical wall directly in front of you and another vertical wall behind you, but at a very great distance. The operative word will be distance. The gravity force drops off with the square of distance. This means the strongest gravity will be directly in front of you. What is behind you will still exert gravity but being so far away it affect is very small.

 

Think if it this way, each unit square of the shell has a certain amount of gravity force, right? Let's get close enough to the wall so that only 1 unit square is in "front" of you. Sure, that unit square is going to exert a certain amount of force. But, with every single other unit squares behind you, they all exert a force the other direction. Some of them, the ones farthest away, will exert only a tiny amount, but because there are so many behind you, it adds up to be the same as the one in front of you.

 

I said it above, but the combination of small amount of mass close to you completely equals the combination of a large amount of mass far away.

 

It is exactly the same as being between 2 masses: a 1 kg mass 1 m away and a 4 kg mass 2 m away. The smaller but closer mass completely balances the farther away but larger mass.

 

pioneer, this is a result that has been known a very long time. The math really isn't all that hard... have you done any of the math? Once you do it, you should be able to convince yourself that what the three of us are saying is true. If you have any problems, I personally volunteer to help you work through the math, post it to this thread, and I'll do my best to help. But, this is a problem that has been worked by literally tens of thousands of students, and the answer is very well known. So, please do the math to convince yourself of the zero gravity answer.

 

Otherwise, this thread is just going to go on and on and on. But, even more meaningful, is that since your very first premise is flawed, nothing you derive or reason from it can be right, so you're doing nothing but building on a rotten foundation -- the house you build on a rotten foundation will not stand.

[pioneer]

I believe you and stand corrected. I think I went through the math many years ago and the results are sort of coming back. One can model the interaction of two spheres by assuming the gravity comes from the center of gravity of each mass.

 

If we had a bunch of hollow mass spheres, but you didn't know these were hollow, one could not deduce if they were hollow, or not, based on how the centers of gravity are interacting. We would know the mass by the interaction, but not if this mass was hollow or not. Or would you?

 

Addendum

 

If we compare a hollow mass shell to one that is solid there is one big difference. Both will have gravity fields beneath the surface, but only the solid sphere will have actual mass in the zone of the interior field. This mass is not only bathing in the field but helping to generate it. The hollow sphere has a field but is only empty space inside, i.e., virtual mass?

 

This is only meant as an analogy. The hollow mass sphere is sort of like an actor who plays a doctor on TV. He can talk and act like the ldeal doctor but his doctorability is on the surface. Most people meeting him might wish he was their doctor due his very refined bedside manner. That is the attraction. As you go to his office to interview him, as a potential doctor, you realize he is not too up on your specific medical needs. But at the same time, his charm and mannor makes him seem like he would be perfect. So you are in a conflict, due to an attraction and a gut feeling that makes you think your maybe should try elsewhere. But since there is a good balance of motivation, you continue the interview, i.e., orbit.

 

He scheduals an appointment for a physical, the next day, since you have remained indecisive as to whether you wish to go or leave and he is trying to build up his practice. You are sort of locked in now due to indecision. He is more than an actor but also a smart guy and does his research on the things that concern you, to fill in that hollow spot that is making you indecisive. The next day, he has his normal attractive charm, but has also added data that helps to settle some of your scepticism. The result of the fill-in is you begin to spiral even closer toward him, losing doubt. The more he fills in the gap, you eventually make your final decision to merge. The orbits of the planets and moons of our solar system may reflect the subtle differences between the gravity fields within solids and hollows. The hollows don't have to imply empty but may be something like the fusion core of a star having energy pressure that rarifies matter so there is more virtual.

[BenTheMan]

If we compare a hollow mass shell to one that is solid there is one big difference. Both will have gravity fields beneath the surface, but only the solid sphere will have actual mass in the zone of the interior field. This mass is not only bathing in the field but helping to generate it. The hollow sphere has a field but is only empty space inside, i.e., virtual mass?

 

NO!

 

There is no field inside the hollow sphere.

 

http://webphysics.davidson.edu/physlet_resources/bu_efield/EField_Gauss_Text.html

 

How many times do we have to say this before you believe us?

[Bignose]

NO!

How many times do we have to say this before you believe us?

 

If pioneer would actually do the math, he could prove this for himself. Until he does the math, personally, I've given up on this thread. Like you said, Ben, we've told him 9 times now, is he waiting for 20? 50? Whatever it is, I'm not going to keep repeating myself.

[BenTheMan]

For some reason, I think that pioneer doesn't read my posts. I've answered questions of his in other threads, and I never get a response.

[pioneer]

I read the link about the charges and hollow shells. Where I am confused is whether gravity acts the same way as charge. With magnetic fields we have sort of positive and minus, or north and south magnetic poles. But gravity doesn't exactly have a north and south pole.

[insane_alien]

no, but you will notice in the electrostatic examples there is only one type of charge involved.

[pioneer]

Here is a quote from the article suggested by BenTheMan.

 

http://webphysics.davidson.edu/physl...auss_Text.html

 

Something interesting to note is that when the inner sphere is introduced, the charge distribution on the outer sphere changes. The excess charge no longer lies only on the outside. The charge must redistribute itself so that E = 0 inside the conductor.

 

In other words, if we have a hollow mass shell, the field is zero inside. If I introduce another mass inside that shell, to fill in the hollow, the field is still zero, but the inner mass does not just evaporate due to zero gravity. It still has it own gravity. The surface now has to change.

 

What I originally was saying, at a distance center of gravity appears to apply, but as we get closer depending on how the hollow is filled in, will affect the surface we which will see, when we are up close.

[swansont]

Here is a quote from the article suggested by BenTheMan.

"Something interesting to note is that when the inner sphere is introduced, the charge distribution on the outer sphere changes. The excess charge no longer lies only on the outside. The charge must redistribute itself so that E = 0 inside the conductor."

 

In other words, if we have a hollow mass shell, the field is zero inside. If I introduce another mass inside that shell, to fill in the hollow, the field is still zero, but the inner mass does not just evaporate due to zero gravity. It still has it own gravity. The surface now has to change.

 

What I originally was saying, at a distance center of gravity appears to apply, but as we get closer depending on how the hollow is filled in, will affect the surface we which will see, when we are up close.

 

(URL removed because it's a bad link. You have to use the actual URL, not paste the shorthand displayed in the post)

 

The charge distribution changes, but that's because the charges are free to move and you have two types of charge.

 

If you have two shells, there will be gravity outside the inner shell, due only to the mass of the inner shell.

[BenTheMan]

The analogy between gravity and EM isn't perfect. But the calcualtion (using an anologue of Gauss' law) is more or less the same.

[alan2here]

Just so I get this straight, if i was in deep space and I had a huge very dense yet hollow sphere of matter and I was standing on the outside of it then I would be attracted to the surface due to gravity, it would be like standing on a planet, however if I was standing on the inside I would float like in space? yes that does seem rather odd.

[swansont]

Just so I get this straight, if i was in deep space and I had a huge very dense yet hollow sphere of matter and I was standing on the outside of it then I would be attracted to the surface due to gravity, it would be like standing on a planet, however if I was standing on the inside I would float like in space? yes that does seem rather odd.

 

That's the way it is, though.

[alan2here]

I believe you, I still find it odd. Like many things in science that are odd but true.

[Severian]

I believe you, I still find it odd. Like many things in science that are odd but true.

 

It is not really that odd. When you are outside the sphere, all of the sphere is off in one direction, 'below' your feet, so it all pulls at you in the same way and you feel the force.

 

When you are inside the sphere, some of the sphere is above you and some is below you, so different bits of the sphere pull you in different directions. Let's say your head is the closest bit of you to the sphere, and that you are quite near the edge, for a sec (to simplify the discussion). Then the amount of mass above you is less than the amount below you, but since the mass above you is closer its gravitational effect (per unit area) is greater.

 

The nice/weird/particular thing about gravity and electromagnetism, is that the extra gravitational pull (per unit area) of the mass above you is exactly compensated by the reduced area, so the forces from above and below exactly are exactly the same. This is because the force [math]\sim 1/r^2[/math] but the area [math] \sim r^2[/math].

 

So if gravity happened to be [math]\sim 1/r^3[/math] (or for that matter [math]\sim 1/r^{2+\epsilon}[/math] with [math]\epsilon[/math] very small) this wouldn't work.

[alan2here]

I now understand