The DYSONSPHERE: one of ST's, & Sci-Fi's, greatest mysteries.

[FordSVT]

The DYSONSPHERE:

How would such a *David Niven object come into being?


*Wrote a novel called Ringworld which inspired Freeman Dyson to invent the star-surrounding sphere. I am correct about Niven influencing Dyson yes If not

Freeman Dyson came up with the Dyson Sphere although he did not posit a solid shell as that would be unstable and more difficult to construct.

*Larry* Niven wrote Ringworld. His structure was a simpler, easier to build structure inspired by the Dyson Sphere.

*David* Niven, around the same time, was a famous actor.

As for why there is a Dyson sphere in TNG, I'm not really sure. It's pretty much a throwaway playing second fiddle to an implausible and silly crossover situation.

A real Dyson sphere should be a HUGE deal, no pun intended. It indicated a technology far in advance of any others the Feds had discovered.

There are countless reasons why a Dyson sphere would be abandoned if found. Honestly, the focus of the episode should have been the investigation of the artifact. Heck, you could have written a story arc about it.

Instead we got the Scotty episode. One of TNG's worst.

In the novels, at least, we know there are teams of scientists from many different Federation worlds exploring the sphere since that episode.

 

[Crazy Eddie]

No, this is wrong. For a perfect sphere, there is never going to be an 'up' gravity field. No matter how large, all perfect spherical shells MUST obey Newton's shell theory.

I agree with this. We just disagree on the implications of the theory. Anything on the inside of the sphere will always be attracted to the nearest part of the sphere, not the CENTER of the hollow sphere. And for precisely the reasons you described: only the part of the shell "below" you, as in closer to the center, will act on you. You only get "zero" gravitation exactly at the Earth's center because that mass is somewhat evenly distributed, but that zero is the result of those forces balancing out at equilibrium from the CENTER. If you're offset from equilibrium, it's a very different matter.

Experimental proof of this concept can be found in the fact that some geological features can be detected by gravitational anomalies; in some mountains, for example, the density of the crust directly beneath them is lower than the crust around their perimeter, which results--oddly enough--in a gravitational gradient; objects along the slope are gravitated in a direction slightly away from the base of the mountain.

As for the sphere: consider, for a moment, that for an object this size the inner surface is effectively flat for a radius several hundred times larger than the Earth. Assuming the sphere is made of an extremely dense material, that disk of sphere shell has greater gravitational influence than its surroundings, because the effect of gravity falls off at a distance. So a person on the inside is still attracted to the inner surface of the sphere for precisely the same reason that all of us are attracted to the Earth more than the sun or the moon. Even though the sun's gravity is more intense, Earth is that much closer.

 

[ancient]

No, this is wrong. For a perfect sphere, there is never going to be an 'up' gravity field. No matter how large, all perfect spherical shells MUST obey Newton's shell theory.

I agree with this. We just disagree on the implications of the theory. Anything on the inside of the sphere will always be attracted to the nearest part of the sphere, not the CENTER of the hollow sphere. And for precisely the reasons you described: only the part of the shell "below" you, as in closer to the center, will act on you. You only get "zero" gravitation exactly at the Earth's center because that mass is somewhat evenly distributed, but that zero is the result of those forces balancing out at equilibrium from the CENTER. If you're offset from equilibrium, it's a very different matter.

Experimental proof of this concept can be found in the fact that some geological features can be detected by gravitational anomalies; in some mountains, for example, the density of the crust directly beneath them is lower than the crust around their perimeter, which results--oddly enough--in a gravitational gradient; objects along the slope are gravitated in a direction slightly away from the base of the mountain.

As for the sphere: consider, for a moment, that for an object this size the inner surface is effectively flat for a radius several hundred times larger than the Earth. Assuming the sphere is made of an extremely dense material, that disk of sphere shell has greater gravitational influence than its surroundings, because the effect of gravity falls off at a distance. So a person on the inside is still attracted to the inner surface of the sphere for precisely the same reason that all of us are attracted to the Earth more than the sun or the moon. Even though the sun's gravity is more intense, Earth is that much closer.

Only if we throw out the perfect sphere do we get a situation where you might be pulled 'up' I realize now. If you put a big, Earth-sized pimple on the dyson sphere somewhere, it would pull things towards itself. But only in a localized way, the same as a planet.

I understand what you're thinking, which is that the part of the sphere you're close to will pull you more than the rest of the sphere put together, because it's closer. But that's not true. A really big object far away, can have the same pull as a really small object close to you. With a perfect spherical shell this relation will always be balanced.

This isn't my own thing, something I came up with. I learned this specifically when I was learning about the shell theory in college. The 'no net gravity in a spherical shell (from the shell itself)' thing is something Newton came up with. Newton, the guy who invented calculus and discovered the universal law of gravitation. And they're still teaching it in physics class. And it has to be understood to calculate the gravitation in a solid sphere of varying density, such as Earth. Because with it, you know to ignore any mass from shells 'above' you.

So...I don't know what to tell you. Maybe I'm not explaining it very well, and it is something a little tricky to understand out of the blue this way, but it is something real.

 

[Santaman]

^^ I'm getting a headache from all that, I suggest using a Dyson Cube...

 

[Crazy Eddie]

I understand what you're thinking, which is that the part of the sphere you're close to will pull you more than the rest of the sphere put together, because it's closer. But that's not true. A really big object far away, can have the same pull as a really small object close to you. With a perfect spherical shell this relation will always be balanced.

I really don't think so. For it to be balanced out it would require the effect of gravity from the far side of the sphere to act with a greater effect than the mass on the near side. The point is gravitational pull falls off at a distance, and at the distances we're talking about they DO NOT balance out. If, for example, there was another planet of identical size and mass of Earth orbiting the sun directly opposite of us, we would still be attracted to Earth and not anti-Earth; local gravitation wins out in that case.

That's not to say there would be no tidal effect from the rest of the sphere, or that the pull would be uniformly "down" towards the inner shell of the sphere (actually, it would be more distributed, sort of a cone of pull that averages to a downward force). The only place forces balance out is in the very center of the sphere; the closer you are to the inner walls, the more influence that part of the wall has than its opposite number.

So...I don't know what to tell you.

You can plug in the numbers and tell me what you get back. Otherwise your interpretation of the theory is inconsistent with observations; just because the sphere is perfect doesn't mean it isn't subject to the same conditions of gravity-density parity as an imperfect sphere. Even in a multi-body system, the only reason you're attracted to the center of mass is because the sum of all gravitational pulls from all of the objects in that system average towards that point; if you're close enough to one of those objects, its gravity can overpower the influence of all the others. In this case--especially for an object as large as a dyson sphere--then the sphere walls (if they have enough density) can exert greater influence locally than the system as a whole. The center of mass is only the center of gravity for objects that are acted upon at a distance.

 

[Colonel Midnight]

^^ I'm getting a headache from all that, I suggest using a Dyson Cube...

Either that or we start building one -- by the time it gets completed, we may have come to a conclusion/answer.

Cheers,
-CM-

 

[JNG]

^^ I'm getting a headache from all that, I suggest using a Dyson Cube...

It would be awfully cool to live in the corner of a Dyson Cube.

 

[ancient]

I understand what you're thinking, which is that the part of the sphere you're close to will pull you more than the rest of the sphere put together, because it's closer. But that's not true. A really big object far away, can have the same pull as a really small object close to you. With a perfect spherical shell this relation will always be balanced.

I really don't think so. For it to be balanced out it would require the effect of gravity from the far side of the sphere to act with a greater effect than the mass on the near side.

Which is exactly what happens because of the shape of the sphere. The other side is further away, but there's a whole lot more of it.

That's not to say there would be no tidal effect from the rest of the sphere, or that the pull would be uniformly "down" towards the inner shell of the sphere (actually, it would be more distributed, sort of a cone of pull that averages to a downward force). The only place forces balance out is in the very center of the sphere; the closer you are to the inner walls, the more influence that part of the wall has than its opposite number.

So...I don't know what to tell you.

You can plug in the numbers and tell me what you get back. Otherwise your interpretation of the theory is inconsistent with observations; just because the sphere is perfect doesn't mean it isn't subject to the same conditions of gravity-density parity as an imperfect sphere. Even in a multi-body system, the only reason you're attracted to the center of mass is because the sum of all gravitational pulls from all of the objects in that system average towards that point; if you're close enough to one of those objects, its gravity can overpower the influence of all the others. In this case--especially for an object as large as a dyson sphere--then the sphere walls (if they have enough density) can exert greater influence locally than the system as a whole. The center of mass is only the center of gravity for objects that are acted upon at a distance.

Once again, this is not something I came up with. This is not my theory, it is Newton's. And yes, I have confirmed this with my own calculations in the past because understanding this is required for any serious calculations using gravitation.

Ok, here are some links, found using a simple Google search:

http://cnx.org/content/m15104/latest/
I direct your attention to figure 6. And this quote (near the bottom):

This is yet another important result, which has been used to determine gravitational acceleration below the surface of Earth. The mass residing outside the sphere drawn to include the point below Earth’s surface, does not contribute to gravitational force at that point.
The mass outside the sphere is considered to be composed of infinite numbers of thin shells. The point within the Earth lies inside these larger shells. As gravitational intensity is zero within a shell, the outer shells do not contribute to the gravitational force on the particle at that point.

http://en.wikipedia.org/wiki/Shell_theory#Inside_a_shell

Therefore, the shell exerts no net force on particles anywhere within its volume. Reasoning intuitively using inverse-square law, the few pieces of the shell that are close to m exert a large force, but there aren't many of them. By contrast, there are many pieces of the shell far from m but their force contribution is smaller.

 

[Santaman]

Okay, last offer.. Dyson Pyramid?

 

[Crazy Eddie]

I understand what you're thinking, which is that the part of the sphere you're close to will pull you more than the rest of the sphere put together, because it's closer. But that's not true. A really big object far away, can have the same pull as a really small object close to you. With a perfect spherical shell this relation will always be balanced.

I really don't think so. For it to be balanced out it would require the effect of gravity from the far side of the sphere to act with a greater effect than the mass on the near side.

Which is exactly what happens because of the shape of the sphere. The other side is further away, but there's a whole lot more of it.

Not enough to overcome local attraction, as would be the case in a SOLID sphere. Even when our entire solar system is in an orbital position on the far side of the sun, we are still attracted to the Earth, and not to the rest of the solar system; hell, even in the case of the sun alone there is "a whole lot more" mass there than there is here, and yet--amazingly--we are still attracted to Earth more than the sun.

I direct your attention to figure 6. And this quote (near the bottom):

There's your problem. A dyson sphere isn't composed of "an infinite number of thin shells." It is composed of a single REALLY THICK shell. Earth is a solid sphere; rendering it as a series of shells is an abstract way of calculating gravitational attraction towards the center. Taken to its logical conclusion this would mean there is no gravitational attraction on ANY object inside the sphere, since ALL of the mass is concentrated in the outer shells.

Like I said, I'm not disputing the theory. Just the interpretation thereof, which is not applicable in this case. At the distances involved in an object this size, by the inverse square law it becomes the case that local gravitation of one section of the sphere wins out over distant gravitation from the rest of the sphere. This would be very weird gravity with some very extreme tidal forces (probably extremely variable, with effective weight dropping off quickly after only a few miles from the surface) but still gravitation towards the inner wall, not the center.

 

[Santaman]

^^ Hmm you have a point there, its stated that the sun inside the sphere isn't influenced by the sphere at all so you have to REALLY finetune your sphere's orbit to be exactly the same as that of the star, else there will be a big badaboom and so on, nasty hole burned through the shell, kinda not what you want...

 

[ancient]

I really don't think so. For it to be balanced out it would require the effect of gravity from the far side of the sphere to act with a greater effect than the mass on the near side.

Which is exactly what happens because of the shape of the sphere. The other side is further away, but there's a whole lot more of it.

Not enough to overcome local attraction, as would be the case in a SOLID sphere. Even when our entire solar system is in an orbital position on the far side of the sun, we are still attracted to the Earth, and not to the rest of the solar system; hell, even in the case of the sun alone there is "a whole lot more" mass there than there is here, and yet--amazingly--we are still attracted to Earth more than the sun.

You're not thinking about this in the right way. The Earth and the Sun are not concentric, nor are they part of a spherical shell system. What you say is right, but irrelevant.

I direct your attention to figure 6. And this quote (near the bottom):

There's your problem. A dyson sphere isn't composed of "an infinite number of thin shells." It is composed of a single REALLY THICK shell.

...two ways of saying the exact same thing. Many thin shells = one thicker shell.

Earth is a solid sphere; rendering it as a series of shells is an abstract way of calculating gravitational attraction towards the center.

Abstract? Maybe, but since Earth's density changes with depth it's the only real way to calculate the gravity at any point.

Taken to its logical conclusion this would mean there is no gravitational attraction on ANY object inside the sphere, since ALL of the mass is concentrated in the outer shells.

So, you think that the Earth's core is massless? That only the parts near the surface have mass?

There is always gravity inside the Earth (except at the center) because there are always massive shells below you. The gravity does decrease as you get deeper because there are fewer and fewer shells below you. In a dyson sphere, once you're inside, there are no shells below you.

Like I said, I'm not disputing the theory. Just the interpretation thereof, which is not applicable in this case. At the distances involved in an object this size, by the inverse square law it becomes the case that local gravitation of one section of the sphere wins out over distant gravitation from the rest of the sphere.

Every physics book ever written disagrees. The size of a sphere makes no difference. If you understand the shell theory than the null zone in a spherical shell is a given.

This would be very weird gravity with some very extreme tidal forces (probably extremely variable, with effective weight dropping off quickly after only a few miles from the surface) but still gravitation towards the inner wall, not the center.

No. Read about the shell theory. Read my links. You aren't ever going to get 'up' gravity in a sphere.

^^ Hmm you have a point there, its stated that the sun inside the sphere isn't influenced by the sphere at all so you have to REALLY finetune your sphere's orbit to be exactly the same as that of the star, else there will be a big badaboom and so on, nasty hole burned through the shell, kinda not what you want...

The dyson sphere would still feel the sun's gravity, but since the sphere is way more massive than the sun I'm not sure if it would be enough to all work out. It's obviously not something easy to build.

When an object falls towards the Earth, the Earth falls towards the object too, but since the Earth is way more massive, that's not taken into account. I have no idea how much more massive the Dyson sphere would be than the sun, but I imagine it could be by quite a bit.

 

[Santaman]

I guess its a matter of how far the gravity pull from either the sun and the sphere itself reach, I imagine the calculations to cause a massive headache..

 

[ancient]

Newton came up with calculus to make it less of a headache. A sphere is still the most easy to work with, since they can be treated as point-masses in relation to eachother, and as layered shells as individual masses.

Since Earth becomes more dense towards the center, you can't take the 'average density', multiply by the volume and get the mass. Well, you can, but there will be a pretty big error. So if you divide the Earth into 10 shells, each treated as having a uniform desity being the avg density at its range of depth, you get closer to the right answer. If you divide it into 50 shells, it's even closer to the right answer, if you divide it into 100 shells you get even closer, and so on. Using calculus you can divide it into an infinite number of infinitely thin shells, so you can get very close to the right answer. Without doing an inifinite number of equations.

And as an added bonus you can ignore mass effects from shells above you, since they cancel away because of the inverse relation of distance and mass. When inside the Earth you can't treat it as a point mass anymore since some of the Earth is now 'above' you and is pulling you 'up', decreasing the 'down' force. But there will never be enough mass 'above' you to pull you back towards the surface. Not in a hollow or solid sphere.

A Dyson Cube would be a huge headache to calulate the gravity for. A sphere is much nicer and straight-forward. It can still be done, but it would require a triple intigration I think.

 

[backstept]

with regards to how the dyson sphere would look on the inside here's a thread I started a good while ago

http://www.trekbbs.com/showthread.php?t=63919

 

[Crazy Eddie]

Abstract? Maybe, but since Earth's density changes with depth it's the only real way to calculate the gravity at any point.

Right, but in a HOLLOW sphere, the density doesn't change with depth. Any depth below the outer layer, the density is exactly zero. You would have to take this into account when applying the theory that no matter how deep you get into the dyson sphere, all of the mass of the sphere is in outer shells (not including the star, of course).

You could say that "gravity" doesn't see the sphere as a single object, but as a massive collection of mass points that happen to be arranged in a spherical pattern. The configuration doesn't really matter, since you're dealing with a vast empty space, there is nothing within that space to attract objects towards it. All of your attraction will be towards the nearest massive gravitating body, and that means the wall of the sphere.

Mathematically, it is indeed true that you would be drawn towards the center if you are OUTSIDE the sphere; the sum of all gravitational effects produce downward motion towards the center, and shell theory applies here because the shells are still beneath you. But you need to find an application where you are not yet in the center of the sphere and yet 100% of the mass is STILL in shells above you.

So, you think that the Earth's core is massless? That only the parts near the surface have mass?

What are you talking about? We're discussing dyson spheres, not the Earth's core. Inside the dyson sphere, the core--actually the majority of its volume--is indeed massless. Your calculations would have to account for this somehow.

There is always gravity inside the Earth (except at the center) because there are always massive shells below you.

This is NOT the case in a dyson sphere; there are no shells below you, all the mass is in the outer shell.

Every physics book ever written disagrees.

I never never seen a physics book that discusses gravitation in a dyson sphere. Center of planets, yes, but never a massive hollow sphere with all the mass concentrated near the center. And I do not see that the null zone would encompass the entire internal volume thereof; at best it would be closer to null as you approach the center (and completely so within, say, a quarter AU of the star) but not near the inner surface.

No. Read about the shell theory. Read my links. You aren't ever going to get 'up' gravity in a sphere.

You say this like I haven't.

ancient, there is ALWAYS "up" gravity in a sphere. The theory works the way it does because "up" gravity is nowhere near as strong as "down" gravity because there is always more mass in the lower shells than there is in the upper shells. When you get to the center and this is no longer the case, then ALL of the gravity is "up" gravity and they all cancel each other out.

What none of the sources I have checked out in the past few days seem able to explain is what happens for a sphere whose center really IS massless and you reach a point where you are not yet at the center and yet there are no more shells below you to apply the "down" force. If, for example, Earth's core WAS massless, then as you get closer to the core the "up" force has far more influence than it should since there is less attractive mass in the lower shell.

It appears to me that the null zone at the center of the core is a consequence of that position being equidistant from all mass points within the planet. As this is NOT the case on the inside of a hollow sphere, the calculations have to take this difference into account.

 

[ancient]

Abstract? Maybe, but since Earth's density changes with depth it's the only real way to calculate the gravity at any point.

Right, but in a HOLLOW sphere, the density doesn't change with depth. Any depth below the outer layer, the density is exactly zero. You would have to take this into account when applying the theory that no matter how deep you get into the dyson sphere, all of the mass of the sphere is in outer shells (not including the star, of course).

Yes...

You could say that "gravity" doesn't see the sphere as a single object, but as a massive collection of mass points that happen to be arranged in a spherical pattern. The configuration doesn't really matter, since you're dealing with a vast empty space, there is nothing within that space to attract objects towards it. All of your attraction will be towards the nearest massive gravitating body, and that means the wall of the sphere.

No. Larger masses that are farther away can pull stronger because of the inverse relationship of mass and distance. The further parts of the sphere will always be more massive (much MUCH more massive) than the part near you.

Mathematically, it is indeed true that you would be drawn towards the center if you are OUTSIDE the sphere; the sum of all gravitational effects produce downward motion towards the center, and shell theory applies here because the shells are still beneath you. But you need to find an application where you are not yet in the center of the sphere and yet 100% of the mass is STILL in shells above you.

The shell theory exists specifically to make calculations within a sphere easier. From outside a sphere applying it is useless, since from outside you can just treat the whole thing as a point mass.

So, you think that the Earth's core is massless? That only the parts near the surface have mass?

What are you talking about? We're discussing dyson spheres, not the Earth's core. Inside the dyson sphere, the core--actually the majority of its volume--is indeed massless. Your calculations would have to account for this somehow.

I was replying to your assertions about the Earth.

This is NOT the case in a dyson sphere; there are no shells below you, all the mass is in the outer shell.

I'm sorry, but you don't seem to have followed me at all, Newton's calculations work for either a hollow shell or solid sphere. The results are different. The result for a hollow sphere is that gravity drops to zero as soon as you are inside. The result for a solid sphere is that gravity gradually decreases until you reach the center. At which point it is zero also.

Every physics book ever written disagrees.

I never never seen a physics book that discusses gravitation in a dyson sphere. Center of planets, yes, but never a massive hollow sphere with all the mass concentrated near the center.

A dyson sphere is a hollow shell. The shell theory is used to simplify calculations by cancelling all mass at a greater distance from the center of a sphere than the point of interest. All physics books do mention that all shells above you are irrelevant and can be disregarded. They may not mention the Dyson Sphere by name (though some do) but a dyson sphere and a hollow spherical shell are the same thing. They are equal.

And I do not see that the null zone would encompass the entire internal volume thereof; at best it would be closer to null as you approach the center (and completely so within, say, a quarter AU of the star) but not near the inner surface.

If that were true the shell theory would not exist because it would, in fact, be totally useless.

No. Read about the shell theory. Read my links. You aren't ever going to get 'up' gravity in a sphere.

You say this like I haven't.

You certainly haven't understood it.

ancient, there is ALWAYS "up" gravity in a sphere. The theory works the way it does because "up" gravity is nowhere near as strong as "down" gravity because there is always more mass in the lower shells than there is in the upper shells. When you get to the center and this is no longer the case, then ALL of the gravity is "up" gravity and they all cancel each other out.

So if you're 20 miles from the center of the Earth, that little core part is more massive than the entire Earth? That's absurd, the real reason is because only the part of the Earth below you is relevant. The rest, the parts above you, will always cancel eachother out and do nothing.

What none of the sources I have checked out in the past few days seem able to explain is what happens for a sphere whose center really IS massless and you reach a point where you are not yet at the center and yet there are no more shells below you to apply the "down" force. If, for example, Earth's core WAS massless, then as you get closer to the core the "up" force has far more influence than it should since there is less attractive mass in the lower shell.

...Try reading them again.

It appears to me that the null zone at the center of the core is a consequence of that position being equidistant from all mass points within the planet.

That's true, but only explains the gravity field at one point, and is pretty useless for finding the gravity at any other point within the sphere.

As this is NOT the case on the inside of a hollow sphere, the calculations have to take this difference into account.

They do. When you are off center within the shell the further side of the shell always has more mass, cancelling the effect of it being further away. This is also true for a solid sphere, except that there is always mass below you to pull down.

 

[Crazy Eddie]

No. Larger masses that are farther away can pull stronger because of the inverse relationship of mass and distance. The further parts of the sphere will always be more massive (much MUCH more massive) than the part near you.

I do not believe this holds true in an inverse-square relationship.

The shell theory exists specifically to make calculations within a sphere easier. From outside a sphere applying it is useless, since from outside you can just treat the whole thing as a point mass.

True as that is, it is only applicable for a solid sphere for that reason. If all of the mass of the sphere is contained in an infinitesimally thin shell, none of the calculations would be valid except for directly at the outside surface.

I'm sorry, but you don't seem to have followed me at all, Newton's calculations work for either a hollow shell or solid sphere. The results are different. The result for a hollow sphere is that gravity drops to zero as soon as you are inside.

I follow that just fine. I simply disagree with you (or maybe with Newton, however the case may be) that this is a valid application of the theory. It works well for calculating the gravitational attraction below the surface of a planet, but I do not think it would be accurate for a dyson sphere. Again because of that pesky inverse square relationship, and the fact that (in this case, for example) from any given point, 2/3rds of the mass of the total sphere is at least than 75 million kilometers away.

A dyson sphere is a hollow shell. The shell theory is used to simplify calculations by cancelling all mass at a greater distance from the center of a sphere than the point of interest. All physics books do mention that all shells above you are irrelevant and can be disregarded.

Yes they do, but they take this as a feature of the calculations, which they do not appear to be. This appears, instead, to be an imperfect model treated as gospel just because it is simple and convenient, where the reality would be much more complicated.

If that were true the shell theory would not exist because it would, in fact, be totally useless.

Hardly. For what it's used for (calculating gravitation below the surface of a planet) it is perfectly adequate to give rough approximations. But it doesn't really have to be precise since, firstly, there is no such thing as a "perfect" sphere in nature and, in any case, there's never been (and probably never will be) an opportunity to put it to the test close to the center core of a planet.

I say this theory is good for approximations, because in this condition there is a bit of a discrepancy in application. If the theory assumes that the mass distribution of a planet is uniform throughout (constant density) then it is indeed ALREADY useless. If on the other hand the theory assumes that the lower shells are more dense than the upper shells, then it is indeed already useless with respect to the dyson sphere which violates this assumption and makes the theory inapplicable here.

It appears to me that the null zone at the center of the core is a consequence of that position being equidistant from all mass points within the planet.

That's true, but only explains the gravity field at one point, and is pretty useless for finding the gravity at any other point within the sphere.[/quote]Precisely. Which is the main reason I do not think shell theory is applicable here, since at other points not equidistant from the center, the forces would no longer cancel out.

They do. When you are off center within the shell the further side of the shell always has more mass, cancelling the effect of it being further away. This is also true for a solid sphere, except that there is always mass below you to pull down.

But the effect of that motion has to be to increase the mass of the further side by a factor proportional to the square of the movement. A movement of, say, 4% of the radius reduces the gravity of the far side by 16% and increases the force of the near side by 16%.

 

[ancient]

Maybe we should just dump this topic into the Science forum.

No. Larger masses that are farther away can pull stronger because of the inverse relationship of mass and distance. The further parts of the sphere will always be more massive (much MUCH more massive) than the part near you.

I do not believe this holds true in an inverse-square relationship.

Yes, that's what I was referring to.

The shell theory exists specifically to make calculations within a sphere easier. From outside a sphere applying it is useless, since from outside you can just treat the whole thing as a point mass.

True as that is, it is only applicable for a solid sphere for that reason. If all of the mass of the sphere is contained in an infinitesimally thin shell, none of the calculations would be valid except for directly at the outside surface.

Where did you read this?

I follow that just fine. I simply disagree with you (or maybe with Newton, however the case may be) that this is a valid application of the theory. It works well for calculating the gravitational attraction below the surface of a planet, but I do not think it would be accurate for a dyson sphere. Again because of that pesky inverse square relationship, and the fact that (in this case, for example) from any given point, 2/3rds of the mass of the total sphere is at least than 75 million kilometers away.

And for a sphere that big, the immense mass of the sphere makes up for the distance.

Yes they do, but they take this as a feature of the calculations, which they do not appear to be. This appears, instead, to be an imperfect model treated as gospel just because it is simple and convenient, where the reality would be much more complicated.

As someone who has an A.S. and took 4 physics courses along with the required calculus, this is not some nonsense. I've done the proofs and it works. The model is perfect and if you actually understand it, quite intuitive. Of course this was a while back that I learned this, so I did have to look up the specifics. But if you can come up with an eqation that can disprove the shell theory, I'd sure like to hear it.

Hardly. For what it's used for (calculating gravitation below the surface of a planet) it is perfectly adequate to give rough approximations. But it doesn't really have to be precise since, firstly, there is no such thing as a "perfect" sphere in nature and, in any case, there's never been (and probably never will be) an opportunity to put it to the test close to the center core of a planet.

The theory works for 'almost' perfect spheres as well. In engineering, and I assume other fields, it is important to keep in mind that 'small' errors that arise from a non-ideal shapes or measurements can be ignored if they are small enough. This is why reputable science papers aimed at other scientists include predicted error for most calculations.

I say this theory is good for approximations, because in this condition there is a bit of a discrepancy in application. If the theory assumes that the mass distribution of a planet is uniform throughout (constant density) then it is indeed ALREADY useless.

No no no, here again you've not understood what I said. The shell theory is applied to a sphere of varying density because you can treat each change in density as a new shell. This is why using lots of shells - ideally an infinite amount - works the best. Each shell represents the density at its depth.

If on the other hand the theory assumes that the lower shells are more dense than the upper shells, then it is indeed already useless with respect to the dyson sphere which violates this assumption and makes the theory inapplicable here.

...no, again, you really expose a lack of understanding of what the shell theory actually does and how it works. The shell theory can be applied to more than one special situation. It applies to ALL spheres, solid, hollow, uniform, it says nothing about which shells are more dense or less dense. That will change your final result to be sure. But just as addition works with any two numbers, the shell theory works for any concentric arrangement of shells, becaue each shell works fine on its own.

It appears to me that the null zone at the center of the core is a consequence of that position being equidistant from all mass points within the planet.

That's true, but only explains the gravity field at one point, and is pretty useless for finding the gravity at any other point within the sphere.

Precisely. Which is the main reason I do not think shell theory is applicable here, since at other points not equidistant from the center, the forces would no longer cancel out.

I'm curious to know what you think it is applicable to then.

They do. When you are off center within the shell the further side of the shell always has more mass, cancelling the effect of it being further away. This is also true for a solid sphere, except that there is always mass below you to pull down.

But the effect of that motion has to be to increase the mass of the further side by a factor proportional to the square of the movement. A movement of, say, 4% of the radius reduces the gravity of the far side by 16% and increases the force of the near side by 16%.

A bigger sphere has more mass. Its increase in mass balances with its increase in size. Its increase in radius will never outpace its mass. A little sphere acts the same a bigger sphere and when you're using the inverse square law, they remain proportional regardless of sphere size, thickness, or density.

Before you denounce Newton as some sort of hack, you should really try to understand his findings better. There is a reason that the shell theory is still being taught in schools. A lot of people don't realize just how smart Newton was.

 

[Neopeius]

This gravity in a hollow sphere problem was the one of the first things we worked on in second semester college physics.

Mathematically, inside a uniform hollow sphere, there is no gravity anywhere. All objects inside, wherever they are, are uniformly attracted from all sides.

Now, our problems did not include the giant sun in the middle. Since the Dyson's sphere is likely not rotating, I would imagine anything on the inside surface of the sphere would be gently tugged by the sun (.0006g), eventually to burn.