25 Sep 07
Originally posted by Doctor RatMoving the corners would make the mass more spherelike of course.
The short answer is [b]yes.
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 it ...[text shortened]... Center of Mass constant. In this last case, the orbiter would experience no perturbation.[/b]
It would still have mascons that would change the shape of the orbit, it would deviate from a perfect ellipse but is still stable at least in the short run. I am not sure what you call the resultant shape around a cubical planet but it would approach a square shape with rounded corners. It would not be an ellipse for sure. The points of the cube would attract the satellite downward, giving a velocity kick and it would therefore try to fly away from the center with a slightly greater velocity where the local gravitational attraction is less, then it would be drawn again to the next point of the square where it would again pick up velocty devitating from an elliptical shape being drawn downward and so forth with the results to be a more squarish orbit with rounded corners. Thats my take on it.
25 Sep 07
3 edits
Originally posted by sonhouseAs the orbiter travels near to one point of the cube, this point would attract the orbiter downward, but at the same time, the other points of the cube have rotated farther away and the orbiter is feeling less attraction from them, and by definition this has to balance out if the orbiter is orbiting around the Center of Mass... and remember, the orbit must be still be far enough away that all the points are inside our orbital sphere. If we are looking at an "orbit" which has a point of the cube rising up like a mountain near to the spaceship's side, then absolutely yes, the original orbit will be perturbed. This is why the pilot of the spaceship would have to move the orbit farther out until no part of the cube rises out beyond the orbital shell. That's what the first post in this interesting thread is presenting to us.
Moving the corners would make the mass more spherelike of course.
It would still have mascons that would change the shape of the orbit, it would deviate from a perfect ellipse but is still stable at least in the short run. I am not sure what you call the resultant shape around a cubical planet but it would approach a square shape with rounded corners. It ...[text shortened]... so forth with the results to be a more squarish orbit with rounded corners. Thats my take on it.
25 Sep 07
Originally posted by Doctor RatIn my more extreme example, with the rod of equal mass to the earth, say we have two different orbits around a sphere which is larger than the rod. One orbit has the rod as its axis, the other passes close to the ends of the rod.
As the orbiter travels near to one point of the cube, this point would attract the orbiter downward, but at the same time, the other points of the cube have rotated farther away and the orbiter is feeling less attraction from them, and by definition this has to balance out if the orbiter is orbiting around the Center of Mass... and remember, the orbit mus ...[text shortened]... he orbital shell. That's what the first post in this interesting thread is presenting to us.
When the ship is above one end of the rod, ALL of the gravitational pull of the mass is directed straight down, because ALL of the mass is directly below the ship. When the ship is halfway between ends on its orbit, the gravitational pull of the mass is lessened, because there is some gravitational pull from each of the ends which cancels the other out, and some that adds together to pull downward. This is a result of vector addition. Therefore, you cannot have the same orbit around the center of mass, regardless of distribution of matter. A successful orbit around the "poles" of the rod would look different than one around the "equator".
A similar, but less obvious, reasoning is behind why a cube presents an orbital challenge.
25 Sep 07
2 edits
Originally posted by Doctor RatWhat does the rotation of the planet have to do with it? It could just as easily be spinning opposite to the movement of the satellite in which case the next corner point would be coming in that much faster not receding from the orbiting vehicle. Another way of looking at it would be this: suppose you have homogenous spheres, 4 of them orbiting around a common center. There you have the ultimate extension of the cube idea. Then our satellite comes into contact with these mutually orbiting planets and wants to study all four from a common orbit, therefore has to orbit them all in an orbit "higher" than the common radius. Can a stable orbit be made out of that arrangement? This gives the same kind of mass concentration as a cube but is more realizable in that each planet is already a sphere or spheroid. How could a common orbit be anything but a squashed square shape with rounded corners?
As the orbiter travels near to one point of the cube, this point would attract the orbiter downward, but at the same time, the other points of the cube have rotated farther away and the orbiter is feeling less attraction from them, and by definition this has to balance out if the orbiter is orbiting around the Center of Mass... and remember, the orbit mus he orbital shell. That's what the first post in this interesting thread is presenting to us.
For this thought experiment, I am assuming the spins of the planet lining up with the common orbital plane and the craft is going to such an orbit as to see each spinning equator in turn, thus after several orbits depending on the mutual orbit velocity, all the surface of all the planets would come into view eventually, except for the poles which may be at too shallow of an angle to see surface features.
25 Sep 07
Originally posted by sven1000All of the mass is always pulling from the Center of Mass, whether it is a rod or a sphere. All of the gravitational attraction has already been accounted for by determining the Center of Mass, whether the rod is lined up below the orbiting ship, or whether it isn't. That's the beauty of the physics of Center of Mass! Do you see it now? When the ship is halfway between the ends, there is still a summation of components of gravity toward the Center of Mass, just as there is when the rod is lined up below the ship, and the summations are equal. The vector addition MUST give the same Center of Mass point-gravitational field because that is the very definition of the Center of Mass!
In my more extreme example, with the rod of equal mass to the earth, say we have two different orbits around a sphere which is larger than the rod. One orbit has the rod as its axis, the other passes close to the ends of the rod.
When the ship is above one end of the rod, ALL of the gravitational pull of the mass is directed straight down, because ALL ...[text shortened]...
A similar, but less obvious, reasoning is behind why a cube presents an orbital challenge.
25 Sep 07
Originally posted by Doctor RatNo, it's not. Here's another example. Say you have a spherical shell, which you are inside. The spherical shell has a center of mass at the center of the shell, of course. But you feel NO gravitational attraction toward that center, or indeed at all, because the gravitational attraction of different points on the shell balance out. This is a direct case where the vector addition of the gravitational effects contradicts the calculation given by the center of mass.
All of the mass is always pulling from the Center of Mass, whether it is a rod or a sphere. All of the gravitational attraction has already been accounted for by determining the Center of Mass, whether the rod is lined up below the orbiting ship, or whether it isn't. That's the beauty of the physics of Center of Mass! Do you see it now? When the ship is h ...[text shortened]... r of Mass point-gravitational field because that is the very definition of the Center of Mass!
25 Sep 07
Originally posted by Doctor RatAnd another way. My rod, which you say creates all of its gravitational pull as if the all the mass were at the center of mass, now breaks in two pieces. Does each individual piece exert its own pull from its own, new, center of mass, or from the center of mass of the original rod?
All of the mass is always pulling from the Center of Mass, whether it is a rod or a sphere. All of the gravitational attraction has already been accounted for by determining the Center of Mass, whether the rod is lined up below the orbiting ship, or whether it isn't. That's the beauty of the physics of Center of Mass! Do you see it now? When the ship is h ...[text shortened]... r of Mass point-gravitational field because that is the very definition of the Center of Mass!
25 Sep 07
Originally posted by sonhouseThe rotation has to do with it because the center of mass tells us that even though the orbiting ship may be coming very close to one of the pointy cube bits during a pass of its orbit, then the other parts of the cube must be getting farther away, and they are getting farther away in exactly a manner such that the additional nearby gravitational strength of the close pointy bit is perfectly balanced by the weakening distant gravitational strength of the far away pointy bits, and it must be perfectly balanced because that's what orbiting the Center of Mass means. If an imbalance was detected, that means by definition that the orbiting spaceship must not be orbiting around the Center of Mass, and that is a case we aren't considering because we already know that would cause problems.
What does the rotation of the planet have to do with it? It could just as easily be spinning opposite to the movement of the satellite in which case the next corner point would be coming in that much faster not receding from the orbiting vehicle. Another way of looking at it would be this: suppose you have homogenous spheres, 4 of them orbiting around a com ...[text shortened]... ntually, except for the poles which may be at too shallow of an angle to see surface features.
In your example of mutually orbiting bodies, you have introduced an entirely different problem. With the cube we are considering a fixed body, and in the case of a cube with a spherical section removed, we should only consider the case where the remains of the cube continue to hold fixed points in space because only in that way can you have a fixed Center of Mass, which holds true with the original example of a cube planet in our discussion. With your example, your mutually orbiting bodies may have the same mass concentration as the cube, but you've suddenly introduced the possibility of a spatially non-fixed Center of Mass. That would be an example of the classic N-Body Problem in orbital mechanics, and that is not the same problem proposal as the original post in this thread. So, if for example, the spaceship was in a stable circular orbit around the cube, and the cube suddenly transformed into multiple mutually orbiting bodies, that system would not have a stable fixed Center of Mass and the dynamics of this new system would mean most orbits by a relatively close observing spaceship would be unstable/chaotic ... BUT, if you were saying that hypothetically these mutually orbiting bodies somehow kept a constant fixed Center of Mass that was fixed and equal to the original Center of Mass of the cube, then the orbiting spaceship would have no special concerns; it would simply orbit the fixed Center of Mass of the system at a reasonable distance to avoid collisions with the mutually orbiting bodies, and observe a fascinating dynamic system.
25 Sep 07
Originally posted by sven1000You've changed the original conditions so the argument does not hold. We cannot consider being inside a spherical shell because that is inside the Gaussian Sphere, and we have already agreed that our gravity would change should the orbit be too small. Your example is correct, but it does not apply to our consideration.
No, it's not. Here's another example. Say you have a spherical shell, which you are inside. The spherical shell has a center of mass at the center of the shell, of course. But you feel NO gravitational attraction toward that center, or indeed at all, because the gravitational attraction of different points on the shell balance out. This is a direct cas ...[text shortened]... addition of the gravitational effects contradicts the calculation given by the center of mass.
25 Sep 07
Originally posted by sven1000I don't want to say that a rod creates its gravitational pull from one point. An outside object will be affected as if all the gravitational field were centered at the point of the Center of Mass. Every individual piece of matter within the rod has its own gravitational field, and we can calculate their cumulative effect and discover the point of the Center of Mass. I think we already both agree on that, just wanted to re-state that.
And another way. My rod, which you say creates all of its gravitational pull as if the all the mass were at the center of mass, now breaks in two pieces. Does each individual piece exert its own pull from its own, new, center of mass, or from the center of mass of the original rod?
Even if the rod breaks into two pieces, and even though each rod can have its own calculated center of mass, the system as seen by an outside observer would still have its own unique Center of Mass. If you took your massive rod, split it into two, and moved the piece on the left straight to the left a light year away, and then you took the piece on the right, and moved it straight to the right also a light year away, each piece would have their own center of mass, but if the question was asked, "Where is the Center of Mass for the original rod, for the system now comprised of the two pieces?" the answer would be that the Center of Mass of the system never moved. It is still right where it was when the rod was one piece (because both pieces were moved symmetrically, there was no change in where the Center of Mass for the system is).
25 Sep 07
Originally posted by FabianFnasI withdraw from this thread. I did not want this to be a battle. I did not want this to be a contest of who was right and who was wrong. Knowing the answer but being unable to clearly illuminate the ideas behind it is just as bad as not knowing the answer at all. I am sure there are a few readers who feel I have only muddied the waters of the conversation, and to them, I apologize. If I wanted to teach physics, I should have committed to that choice a long time ago.
I just sit and enjoy this battle of the brains.
But one of them is only cerebellum, the other cerebrum majoris.
Keep using your imaginations and keep exploring this fascinating universe of ours!
Question everything! Experience everything!
It's a wonderful life.
25 Sep 07
Originally posted by Doctor RatWell I tire of this as well, but if you like, here are a few links about why you need to consider the shape, and not just the mass and center of mass.
I withdraw from this thread. I did not want this to be a battle. I did not want this to be a contest of who was right and who was wrong. Knowing the answer but being unable to clearly illuminate the ideas behind it is just as bad as not knowing the answer at all. I am sure there are a few readers who feel I have only muddied the waters of the conversation ...[text shortened]... cinating universe of ours!
Question everything! Experience everything!
It's a wonderful life.
http://scienceworld.wolfram.com/physics/CylinderGravitationalPotential.html
http://scienceworld.wolfram.com/physics/OblateSpheroidGravitationalPotential.html
http://scienceworld.wolfram.com/physics/MacCullaghsFormula.html
and especially
http://scienceworld.wolfram.com/physics/OrbitalPerturbation.html
25 Sep 07
So the original poster, sonhouse, doesn't actually have an answer? Good discussion, but it would be nice to understand the answer. I can't even figure out for certain how to apply Gauss's theorem correctly in this. If it doesn't apply in the way that I understood, I'm strongly guessing that there's only a numerical/iterative solution and/or no stable orbit.