Posers and Puzzles
07 Sep 07
24 Sep 07
Originally posted by sven1000How so? Gauss's law is about you can calculate the volume integral of the divergence of a potential field. Of course if the mass distribution has some special kind of symetry the result is more powerful and useful but you still can use it anyway.
Gauss's law is not being applied properly to this situation as stated so far.
Gauss's theorem applied to newtonian gravity tell us that the the surface integral is proportional to the mass inside the surface but it makes no mention whatsoever regrading the pecularities of the mass. Or am I remebering wrong. I'll hit the old books and see were I'm wrong in this.
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24 Sep 07
http://www.extinctionshift.com/SignificantFindings_Detail_15Aug2007.htm
just a little note from an expert,
" ... The gravitational effect that would be noted at the surface of an analytical Gaussian sphere does not care at all about the size or radius of the mass that is enclosed within the sphere. We can see from Gauss's Law that equal masses of different radii will have theoretically equal gravitational effects at an analytical Gaussian surface of radius R. The gravitational effect outside of an analytical Gaussian surface does not care at all about the size or radius of the mass that is enclosed within the Gaussian surface. We can see from Gauss's Law that equal masses of different radii will theoretically have equal gravitational effects at the surface of the Gaussian sphere of radius R. ... "
Dowdye, Jr., E.H.,
i understand that fighter aircraft also account for mountain ranges in their onboard flight calculations, as there is a definitive gravitational tug toward the mountain
so the peaks of said cubic planetoid have a definite effect that is stabilising for the low orbit when around the three orthogonal planes, but destabilising in orbits around the peaks, little or virtually no effect at distance
i like serigados' response " gonna try to solve it. ", and hope he can
suspect lag effects occur around corners, so squared ellipse, or squared circle would be my guess at a shape, but with some strange transitions around the corners
24 Sep 07
Originally posted by adam warlockdivergence of a potential field? Check that book again.
How so? Gauss's law is about you can calculate the volume integral of the divergence of a potential field. Of course if the mass distribution has some special kind of symetry the result is more powerful and useful but you still can use it anyway.
Gauss's theorem applied to newtonian gravity tell us that the the surface integral is proportional to the m ...[text shortened]... f the mass. Or am I remebering wrong. I'll hit the old books and see were I'm wrong in this.
But you are right. You can apply gauss' law wherever you want, as long as it's a closed integral on a vector field. The generalization you can make in gravity/electromagnetism are due to very nice symmetries (spherical ones) and because it's so easy to integrate a sphere in spherical coordinates , with direct application in gauss' law.
But I must check that book too 😛 I miss Calculus III and IV hehehe
25 Sep 07
1 edit
Originally posted by serigadoHey you!!! 😠😛
divergence of a potential field? Check that book again.
http://mathworld.wolfram.com/DivergenceTheorem.html
Normally F is called a potential field(vector field is a more mathematical name but even Chemetov used to say "campo potencial"... I think) in the first equation. Or isn't it? 😕 I gotta get back my memory...
Anyway see you tomorrow. 🙂
25 Sep 07
Originally posted by adam warlockehehe
Hey you!!! 😠😛
http://mathworld.wolfram.com/DivergenceTheorem.html
Normally F is called a potential field(vector field is a more mathematical name but even Chemetov used to say "campo potencial"... I think) in the first equation. Or isn't it? 😕 I gotta get back my memory...
Anyway see you tomorrow. 🙂
i always associate potential with something scalar.... unless it's explicitly called "vector potential", like the vector potencial A in Electrodynamics.
Chemetov is russian, they always call things different names.
25 Sep 07
Originally posted by adam warlockOne of the assumptions when using Gauss's Law is the spherical symmetry. This is necessary because any spherical shell of uniform density can be treated gravitationally as a point mass by an object outside it. So, the fact that the earth has stratified layers of density isn't important, because they are more or less uniform at a particular radius from the center (spherical symmetry). Therefore you can do the integral simply and voila.
How so? Gauss's law is about you can calculate the volume integral of the divergence of a potential field. Of course if the mass distribution has some special kind of symetry the result is more powerful and useful but you still can use it anyway.
Gauss's theorem applied to newtonian gravity tell us that the the surface integral is proportional to the m ...[text shortened]... f the mass. Or am I remebering wrong. I'll hit the old books and see were I'm wrong in this.
A cube is not spherically symmetric, so the assumption necessary for using Gauss's law in gravity fails.
Imagine this. A sphere with a diameter equal to the length of a side of the cube is removed from our cube. Now fly your spaceship in an orbit around the remaining pieces. Do you believe that you could treat this mass as a point in the empty center? This would have to follow if we can treat the entire cube as a point mass.
25 Sep 07
Originally posted by sven1000I never learned that way. The divegernce theorem only makes mention of the of the vector field and the result is independent of the density characteristics of the surface being considered. When applied to physics and potentials that goes as 1/r^2 we have Gauss's law, but again only the form of the potential matters.
One of the assumptions when using Gauss's Law is the spherical symmetry.
I'll check my book tomorrow but I'm pretty sure that that's how things are.
25 Sep 07
Originally posted by sven1000Thank you Sven.
A cube is not spherically symmetric, so the assumption necessary for using Gauss's law in gravity fails.
I'm not an expert using Gauss, but if the result of Gauss is that one can find an stable orbit around every shape by approximating with a sphere, then I have my doubts, it doesn't make sence.
I've said it before - if the math differ from reality, what is wrong, math or reality?
25 Sep 07
Originally posted by sven1000Absolutely you can. That's the property of a center of mass. When an olympic high-jumper curves backwards over the bar, their center of mass actually passes below the bar, as an example of a center of mass being in empty space. As long as the orbital sphere encompasses the entirety of the object (or objects) one is orbiting, and that orbital sphere has as its center the Center of Mass of the object(s), everything is fine. This is the beauty of finding the center of mass. Even as our spacecraft passes over one of the points, and the mass of the point begins to have a stronger effect, at the same time, the other points have rotated much farther away, and their gravitational attraction is weaker, and it is weaker in exactly the same magnitude as the nearer point is stronger, because that is what the Center of Mass defines.
...Imagine this. A sphere with a diameter equal to the length of a side of the cube is removed from our cube. Now fly your spaceship in an orbit around the remaining pieces. Do you believe that you could treat this mass as a point in the empty center?...
25 Sep 07
Originally posted by Doctor RatSo you believe that an orbit will remain the same as long as the center of mass is in the same location, and the masses are equal? If I have a rod of one meter in thickness, but very long, that weighs the same as the earth, the orbit of my spaceship should be the same around the rod's center of mass as around the earth's?
Absolutely you can. That's the property of a center of mass. When an olympic high-jumper curves backwards over the bar, their center of mass actually passes below the bar, as an example of a center of mass being in empty space. As long as the orbital sphere encompasses the entirety of the object (or objects) one is orbiting, and that orbital sphere has as ...[text shortened]... ame magnitude as the nearer point is stronger, because that is what the Center of Mass defines.
The center of mass is the fulcrum around which a solid body turns, but it isn't the deciding factor for an orbit. These are two different scenarios.
25 Sep 07
1 edit
Take the case which IS in reality, those asteroids we have very good closeups of, the ones that have resulted from two more or less spherical objects colliding at a relatively slow velocity and the resultant shape is more like a peanut, two spheres more or less just barely together. If you put that inside a Gaussian sphere, how could it possibly have a uniform gravitational potential at almost any distance and especially close to the surface of the sphere? There would HAVE to be a differance in gravitaional potential where the two halves meet V the gravitational potential on the ends. Also the same thing happens of earth due to Mascons, there are satellites in orbit now, pairs of them linked by lasers, that measure these Mascons, measurable because of the distortion of the orbits as they glide overhead. That is real and measurable. BTW, Mascon is short for Mass Concentration.
25 Sep 07
5 edits
Originally posted by sven1000The short answer is yes.
So you believe that an orbit will remain the same as long as the center of mass is in the same location, and the masses are equal? If I have a rod of one meter in thickness, but very long, that weighs the same as the earth, the orbit of my spaceship should be the same around the rod's center of mass as around the earth's?
The center of mass is the ful ...[text shortened]... ody turns, but it isn't the deciding factor for an orbit. These are two different scenarios.
As long as the original center of mass does not change its position, and the value for the mass doesn't change, the original massive object (the orbitee, if you will) can change into any wild shape you can think of (with curves or lobes or perforations or hollow spaces) and an orbiting object will merrily continue in its original stable orbit -- as long as one doesn't stretch the original massive object so that a piece of it physically intersects the orbit of the spacecraft itself, and one doesn't stretch any of the object past the orbital shell of our spacecraft (our original Gaussian Sphere) because in almost every case this last point would cause a shift in the value of mass or an additional center of mass. That's the point of finding the Center of Mass, that's the cool thing about a Center of Mass, that's where all the gravity can be considered to be concentrated when you're far enough away. So as long as the mass remains constant and the we restrict any topological transformations of shape and density such that the position of the Center of Mass does not change, and the orbiter remains far enough away, then the orbiter will continue upon its orignal course with no changes even if the orbitee is undulating wildly in shape and size from moment to moment beneath the orbiter.
(The original post in this thread was stressing this last bit -- how far away is "far enough away" when trying to find an orbit around a square planet)
Given a spaceship that has a stable orbit around the Earth (and this could be a close orbit such as the case of a geosynchronous satellite, or a more distant orbit such as the distance of the moon, or whatever orbit you already have as an intitial condition), if one were to suddenly topographically crush the shape of the earth into a point, or compress the Earth into a cube the size of a car, or transform the Earth into a very long rod (whose ends do not extend past the original orbital shell of the orbiter), then the orbiter would see no shift in its orbit.
In the case of Earth transformed into a very long rod whose Center of Mass does not positionally change (so the orbiter is still orbiting around the same position in space) and whose ends of the rod do not extend past the original orbital shell, the rod may be formed in any positional orientation with respect to the orbiter with no perturbation in the orbit. The orbiter may have an orbital plane parallel to the long axis of the rod, and the orbiter may pass right over one end of the rod, but there will be no change in orbit because at the same time, the other end of the rod is now so much further away, the stronger attraction of one now exactly balancing the weaker attraction of the other because that is the definition of our Center of Mass.
In the case of a rod 1 meter in diameter whose ends now extend past the original orbital shell (because lets face it, this would be a very very long rod and I think this is the case you are wanting to examine) there is still a single case where the orbital path will not change. If the rod is created such that the orbital plane of the orbiter is perpendicular to the long axis of the rod, we have a case of symmetry with the far ends of the rod throughout the orbiters orbit, so again, the orbital path will not be perturbed. However, in this "very long rod" case, if the Earth is suddenly transformed into a rod with a tilt with respect to the orbiters orbital plane, symmetry fails, and the orbit will be perturbed.
In the case earlier of our constant-density cube, when you cut out a sphere from inside the cube leaving a symmetrical collection of corners, if you actually made that mass disappear, then of course the orbit would change because even though the position of the center of mass would be the same, the value of the mass would suddenly be different! I think we were both talking about the case where the mass removed from the spherical cut was symmetrically moved into the remaining corners, keeping the mass and the Center of Mass constant. In this last case, the orbiter would experience no perturbation.