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Post by speakertoanimals on Apr 5, 2011 12:50:35 GMT 1
No. Why not?
simple. to be in a closed orbit, the force on the test body must always be towards the centre of its orbit. So, suppose we have body A, then one side of the orbit of the test body, interior of its orbital ellipse/circle whatever, then other side of orbit, then body B.
Lets call if A L (left) C R (right) and then B.
When at L, if force is to be inwards then grav effect of B must be greater than grav effect of A, which pulls in opposite direction. Hence when at R, we are closer to B, and further from A, hence if the pull of B won at L, it will win by even more at R. Hence we cannot set it up so net gravitational forces is towards the centre at both sides.
Hence we can't have an orbiot that insights wholly between A and B, we can only have an orbit where at least one of the bodies lyes WITHIN the orbit.
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Post by speakertoanimals on Apr 5, 2011 13:07:10 GMT 1
when it comes to the rugby ball, life gets a bit more complicated. We have to get away from the inverse square law for gravity, and use a different form.
this is Gauss law for gravity. It says that the gravitational flux through a closed surface is equal to the enclosed mass. Field strength is flux density, and the direction of the gravitational field is the direction of the flux lines (think of magnetic field lines drawn around a magnet, those are field lines, and flux means count the lines passing through a surface).
Okay, so inside a sphere, ANY closed surface lying wholly inside the sphere has zero enclosed mass. That doesn't say straight-off that field is zero though. We first need to say that since we have a sphere, ANY field that we do have will be spherically symmetric, either pointing away from the centre or towards it.
Hence for spherical surface inside the sphere, net flux is zero, and since either wholly outwards or inwards (symmetry), we can then see that flux density must instead be zero. Hence no field anywhere within the sphere.
Now take the rugby ball. We have symmetry about the long axis, if we rotate the ball about this axis things unchanged. We also have reflection symmetry, if we exchange the top point for the bottom, no change. But we don't have the total symmetry we did in the case of the sphere. In fact, I think the most we can say straight off is that the net field is going to be zero for any point on the axis. The point b out the sphere was that because of the special symmetry, we could argue that flux had to be the SAME at any point on our smaller enclised sphere, hence conclude it was actually zero. We can't do the same for the rugby ball, it doesn't have the same symmetry. So, the flux through a sphere inside the ball centered on it will have to be zero, but I can't see off the top of my head why the flux can't be outward at some points and inwards at others, to give zero net result.
Hence this way of doing both gravity (and electrostatics) can give you the answer is there is enough symmetry (sphere, infinite cyliner or infinite flat plate), but isn't enough when the situation isn't symmetric enough, like the rugby ball.
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Post by Progenitor A on Apr 5, 2011 13:12:11 GMT 1
No. Why not? simple. to be in a closed orbit, the force on the test body must always be towards the centre of its orbit. So, suppose we have body A, then one side of the orbit of the test body, interior of its orbital ellipse/circle whatever, then other side of orbit, then body B. Lets call if A L (left) C R (right) and then B. When at L, if force is to be inwards then grav effect of B must be greater than grav effect of A, which pulls in opposite direction. Hence when at R, we are closer to B, and further from A, hence if the pull of B won at L, it will win by even more at R. Hence we cannot set it up so net gravitational forces is towards the centre at both sides. Hence we can't have an orbiot that insights wholly between A and B, we can only have an orbit where at least one of the bodies lyes WITHIN the orbit. YES When the plane of the orbit is peropendicular to a line joining earth and moon and passing through the Cof G of the combined earth and moon masses ;D
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Post by Progenitor A on Apr 5, 2011 13:18:25 GMT 1
Okay, so inside a sphere, ANY closed surface lying wholly inside the sphere has zero enclosed mass. draw a circle on anywhere inside a sphere and the mass is proportional to the radius of the circle;D That doesn't say straight-off that field is zero though. We first need to say that since we have a sphere, ANY field that we do have will be spherically symmetric, either pointing away from the centre or towards it. Hence for spherical surface inside the sphere, net flux is zero, and since either wholly outwards or inwards (symmetry), we can then see that flux density must instead be zero. Hence no field anywhere within the sphere. Now take the rugby ball. We have symmetry about the long axis, if we rotate the ball about this axis things unchanged. We also have reflection symmetry, if we exchange the top point for the bottom, no change. But we don't have the total symmetry we did in the case of the sphere. In fact, I think the most we can say straight off is that the net field is going to be zero for any point on the axis. The point b out the sphere was that because of the special symmetry, we could argue that flux had to be the SAME at any point on our smaller enclised sphere, hence conclude it was actually zero. We can't do the same for the rugby ball, it doesn't have the same symmetry. So, the flux through a sphere inside the ball centered on it will have to be zero, but I can't see off the top of my head why the flux can't be outward at some points and inwards at others, to give zero net result. Hence this way of doing both gravity (and electrostatics) can give you the answer is there is enough symmetry (sphere, infinite cyliner or infinite flat plate), but isn't enough when the situation isn't symmetric enough, like the rugby ball. In other words she deserves a Nobel Prize ;D
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Post by speakertoanimals on Apr 5, 2011 13:25:28 GMT 1
We were talking about spherical shells, I thought that was OBVIOUS!
more clue than you will ever have sunshine! If you put as much effort into answering questions as you do into trying to have a go at me...................
I introduced a method that is more tractable than the brute-force integrate and see approach, and it does so far say that you can't prove straight-off (as you can for the sphere ), that it will be zero. Hence I suspect that it won't be.
Sphere: maximal rotational symmetry. Infinite Cylinder: rotational symmetry about the axis, and translational symmetry along the axis Rugby ball: just rotational symmetry about the axis. Same as for any other surface of revolution.
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Post by Progenitor A on Apr 5, 2011 13:34:54 GMT 1
Okay, so inside a sphere, ANY closed surface lying wholly inside the sphere has zero enclosed mass. We were talking about spherical shells, I thought that was OBVIOUS! If you draw a circle on the interior of the shell then the enclosed mass is proportional to the radius of the circle
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Post by speakertoanimals on Apr 5, 2011 14:16:12 GMT 1
A circle isn't a surface! And as Gauss law states, it consider the flux through a closed surface, and I specifically mentioned a closed surface lying wholly within the sphere (spherical shell).
Nice to know that you know think Gauss was talking tripe, as well as Newton......................
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Post by speakertoanimals on Apr 5, 2011 14:23:37 GMT 1
Gauss' Law is a useful approach, to problems in electrostatics as well as problems in gravity.
To give an example, lets take the case of an infinite straight cylindrical shell of uniform density. We then have rotational symmetry about the axis, and translational symmetry along the axis.
SO take a cylinder of finite length, lying wholly within the shell, and on the same axis. By symmetry, if there is a flux through the curved surface, it will be the same everywhere. By symmetry, there will not be a flux through the ends, because one direction along the axis is equivalent to the other (an INFINITE cylindrical shell). Hence the total flux through the test cylinder will just be the curved surface area multiplied by the flux. Since total enclosed mass is zero, the flux through the curved surface must also be zero.
Hence there is no radial component of any possible gravitational field inside the cydrical shell. Translational symmetry also says there can be no component along the axis. Hence we have ZERO gravitational field within an infinite hollow cylinder, as well as within a hollow sphere.
You can also use Gauss law to find the field outside the cylinder. note that we could only do it because of the additional translational symmetry. Hence why we get stuck with the rugby ball case -- we can't eliminate the possibility that gravity may be inwards for some part of the interior, and outwards at other points, which would still give a zero net flux.
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Post by Progenitor A on Apr 5, 2011 14:47:39 GMT 1
A circle isn't a surface! ;D Another for the STA Science!
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Post by carnyx on Apr 5, 2011 14:51:08 GMT 1
So for the 'Rugby ball' shell, we will have a varing internal gravitational field only in one axis. This field could be said to be anisotropic, i.e. polarised!
Looking end-on, we can see it as a a succession of slices or planes of concentric circles, and so we can say that in these flat planes the ball is free to wander about.
But if we look side-on, we can visualise a set of 'nested' ellipses, and so will see varying gravitational fields in these planes. The ball will therefore move to the null points in these planes.
I intuit that by resolving the 3D gravitational field for every point within the rugby ball, there will be two places where there will be zero gravity in all directions, and they will be at the two focal points of the ellipse. There wil also be one point, the 'centre point' where there is a null, but it will be unstable in one axis.
If this is so, then the ball will gravitate to one or other of these foci.
And, we should have the inverse of an elliptical orbit, at least statically. (So, I also vaguely intuit that there may be some connection between polarisation, anisotropy, and the properties of the ellipse, which may have relevance to EM waves ...) But it will need a bit of work to plot the net gravitational vector at each of a 3D mesh of points in the rugby ball. Any takers?
Of course a cheap way of doing it would be to assume a zero sum at each of the foci. Then do a calculation for a point a small distance away, to see if the gradient is towards the local point. Then do one from a point near to the 'centrepoint' to check that there will be an unstable null at this point ( I.e the gradient will be saddle-shaped)
So, STA, any of your students up for it?
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Post by speakertoanimals on Apr 5, 2011 15:41:55 GMT 1
Nope, you're just playing word salad now! Polarised does not mean the same as anisotropic (not that I'm sure you know what anisotropic means in the first place). Plus em and ellipses was nonsense anyway, ditto the gravity and 'inverses of orbits' nonsense here.
Plus your ideas about stuff wandering freely in planes etc are just plain wrong! We don't KNOW yet what the gravitational field is, and you can't apply symmetry purely within a slice, since the net field could well have a component out of the plane as well as in the plane.
Guessing about foci is just that -- a pure guess, based on the only point you know about related to an ellipse! but could equally well be that the 'special' point is the centre, rather than a focus.
And gawd knows why NM thinks that pointing out that a circle isn't a surface qualifies as STA science. I note that NM hasn't tried to actually say anything intelligent about the actual problem.
Let's look at this 'stable point' hypothesis. Look at the ball points up and down, axis vertical, so horizontal sections are circles.
1) Points on the axis will have a symmetric matter distribution in their plane, BUT not in general wrt up and down. Hence any gravitational field will be directed purely along the axis.
2) FOr a point at the centre, we also have a symmetry between up and down, hence we can deduce that the centre of the ball has zero gravitational field. A oarticle suspended here will stay here. Although we can't say yet whether this will be stable or not.
3) At the foci, the best we can say is field will be along the axis, we can't say that it is zero.
Oblate spheroids are another matter.............
Seems you can solve Laplaces equation using spheroidal coordinates, but involves some weird stuff called prolate spheroidal harmonics (spherical harmonics are bad enough!).
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Post by buckleymanor1 on Apr 5, 2011 15:53:17 GMT 1
Imagine a spinning top with an axis going through it. At the bottom of the rotating axis you have the earth at the top of the rotating axis you have moon. In the middle "the horizontal plane" the body rotates or orbits horizontal to the earth and moon.
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Post by carnyx on Apr 5, 2011 15:53:17 GMT 1
But having reviewed a number of her posts now, I see that STA is incapable of any constructive comment.
DNFTT .. and she'll soon be gone.
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Post by Progenitor A on Apr 5, 2011 16:06:05 GMT 1
Imagine a spinning top with an axis going through it. At the bottom of the rotating axis you have the earth at the top of the rotating axis you have moon. In the middle "the horizontal plane" the body rotates or orbits horizontal to the earth and moon. Ah I see! Good description In fact just a different way of describing the same orbit that I described before So we agree! Excellent. There is a plane between the earth and moon where another mass can orbit their combined CofG As discussed earlier, it will be an unstable orbit - any perturbation - the slightest deviation from that plane will cause it to (eventually) go into orbit around the earth - I think
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Post by speakertoanimals on Apr 5, 2011 16:48:36 GMT 1
If claims of stupidity could get people banned, we'd have no posters left!
That's what you said. you only referred to grav fields in bit about sections in other direction. Which is just plain nonsense -- you CAN'T assume 'freedom of movement' (I assume you meant no grav force) in one slice because of symmetry. You can't analyse 3D motion in this way. Hence this:
is nonsense as well. You are the one who doesn't know how to do it, and is just trying to make it up as you go along.
Which is nonsense again, since the CofG in this case is within the EARTH. Hence we just have orbits about the earth, perturbed by the presence of the moon. Certainly not a plane BETWEEN the two, given any sensible definition of between......................
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