Originally posted by sven1000
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.
The short answer is
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 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.