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

[Crazy Eddie]

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

I could see that for an object in the center, sure. But it still remains the fact that local gravitation wins out against more massive distant bodies at very close ranges.

As someone who has an A.S. and took 4 physics courses along with the required calculus, this is not some nonsense.

Nobody said it was. Just that imperfect physical models are often used entirely because they will never be applied to cases where that imperfection would skew the results. This happened alot in the physics classes I took, like in normalization of electronics equations and--especially--in fluid dynamics.

Hell, even Newton's theory of gravitation was ultimately superseded by Einstein's field equations, wasn't it? Newton's equations are still considered valid in the general sense, but they're not always applicable or accurate, otherwise Einstein's equations would be entirely redundant.

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.

That's why I don't think it's applicable here. In this case, there are only two shells, the outside and the inside. Not that I don't absolutely suck at calculus, but I vividly remember that it turns out some screwy results when applied to a non-iterative process; i.e. it produces curves that become more accurate the more points you feed into it. If you only have two points, you get a flat result that cannot help but be erroneous.

I do not even know for sure this would be an erroneous result in the case of a dyson sphere, I am simply restating that this kind of situation is, it appears to me, NOT the kind of thing that is best handled by calculus equations in shell theory.

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.

In that case, I don't understand shell theory. As in the addition analogy you're basically telling that X + Y always equals 2X regardless of the values for X and Y. Exactly how the theorem could manage to arrive at accurate results without any accounting for variable density is beyond me (of course, when we ran the numbers in physics classes we never had a variable density scenario, so <shrugs>).

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

Relatively solid spheres of mostly uniform density. At least in this case, the effect of the "mass above" shells, though nonzero gravitationally, can be disregarded as miniscule.

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.

The way you calculate the inverse square law is WITH an expanding sphere. As distance increases, the surface area of that sphere increases by the square of the distance: in this case, the radius of that sphere.

In a dyson sphere, it is also true that the mass of the sphere increases by the square of the radius. What I've been trying to figure out is whether or not the "far side" mass also increases at the square of the distance off-center and--if not--whether the vector of attraction has anything to do with it. I mean, if you move .1R in any direction, thats a ring of material .1R wide and 2PiR long. That doesn't increase the far side mass by the square of the distance change, and the direction of that new mass is at an extremely steep angle. It does in a SOLID sphere, where a large amount of volume changes sides every time you move (equivalent to a large cylinder .1R thick) but not in hollow sphere where each movement adds an additional ring of material to the "far" side whose total contribution does NOT increase the net gravitation of the far side by a square of any value (that I can see, anyway. I might be missing something).

In some sources I've read it mentions a conic section change that sweeps out an area equal to the square of the distance which cancels out the second power, but I haven't seem the numbers on that so I'm not totally sure. It seems to me the sphere could easily be modeled as a ring since both hemispheres above and below the plane of movement would both cancel one another's components at all locations and only the material along the plane of motion is relevant; in that case, as in the ring, cancelation is only of the first power, not the second.

Before you denounce Newton as some sort of hack

I'm not denouncing Newton at all. Just that Newton's equations have never been given much of a rigorous test (until the last century or so, which is how Einstein got his fame) and could probably stand a bit of refinement. Just because the man's a supreme genius doesn't make him infallible.

 

[ancient]

Well, I think this whole conversation is at a stand still, so the Science forum should really get this opic, imo. You're spitting out all sorts of stuff that doesn't even really relate to the shell theory, or goes without saying, or doesn't even make sense. Let me boil this down to its essence, then you can go ahead and believe whatever, because I'm tired of this and we're getting nowhere.

The whole shell theory, which works as I've described and linked to, is a tool used to make a complicated equation easier. Using calculus to add up each tiny inifinitely small chunk of mass, and calculating the net result of their gravity, one gets a big old zero for their efforts when they do so for a point inside a spherical shell of any size or thickness.

So why bother doing all that work every time just to end up with a big fat zero every time? Just save yourself the trouble, do the proof once, and then never do that big pesky mess of calculations again. Justify it by describing why you don't need to do them and call it something appropriate - like the "shell theory". That's all the shell theory amounts to. It's a mathematical simplification which makes calculations for gravity much easier. It's not that complicated, or something mysterious that only Newton understood. It's real science.

I forget which physics class I learned this in, I recall that I took part of a "General Physics" sequence, then went back and took the full sequence again, that time the Engineering version, which I believe was called "Physics with Calculus for Scientists and Engineers" or something to that effect. If you didn't learn about the 'zero grav in a hollow sphere' thing, I'm not sure what you were learning. I admit it's been a while, but something like that is pretty hard to forget once it's learned.

 

[Meredith]

I would like to see a Super-Jupiter sometime, that would truly be impressive.