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Due to how orbital mechanics work, debris will move away from the initial trajectory and return to it an orbit later. Since all the pieces have different veloci

Due to how orbital mechanics work, debris will move away from the initial trajectory and return to it an orbit later. Since all the pieces have different velocities, by the time an orbit has passed they will have spread out significantly. The trajectories are not completely parallel, but the relative velocities away from the point of impact are insignificant relative to orbital velocity - thus they are all moving in roughly the same direction.

AnimalMuppet 5y ago
Do the pieces have the orbital velocity of the satellite? Or that of the interceptor? Or a combination of the two? And if it's a combination, why would they all have (approximately) the same combination?
- shadowgovt 5y ago
  For friable colliders, it'll be some combination of the two, but (assuming total disintegration, i.e. "pile of pieces" instead of two bodies breaking big rigid chunks off each other), the bulk of the material will end up going in the direction of the sum of the velocities of the colliding objects. Let me see if I can describe why. Blow up something in space far away from any gravity wells, and you'll get a more-or-less sphere radiating out in all directions, forever (well, not forever; it's possible the gravity of the pieces defeats the escape velocity over long enough distances if the explosion is small, but you know what, too much detail). Blow up something in a circular Earth orbit, and there is a property that must hold true (1) for every individual piece that doesn't achieve escape velocity: the piece's path must be a circle with the center at the Earth's center of mass. Since all those circles are starting from the same point, the only difference they can take in shape is to kind of imagine grabbing the circle at the point of the explosion and giving a "twist" around the original path. And for reasons I can't really explain with math but make a lot of intuitive sense when you play a bunch of Kerbal Space Program, those twists are tiny. In Earth orbit, it takes more change in velocity to go from an west-east orbit to a north-south orbit than it takes to go from launch pad to west-east orbit. So from afar, almost all the debris will following almost the same path as the original satellite because the velocity of the original satellite was going that way, momentum is conserved, and the distribution in the explosion will be kind of a normal distribution. If you were to ride along in the orbit from the original center of mass, what you'd observe is the debris spreads out (following its own individual circles), but halfway around the planet it reconverges and gets very near the original location, possibly self-colliding. (1) I simplified a bit, but not enough to matter much... The orbits are technically elliptical, and the fact that the debris will be pushed in prograde, retrograde, nadir, and zenith directions in addition to normal and antinormal directions means the ellipses will change shape, and therefore area and Kepler's law means the orbital phase will change. So the debris will eventually drift out of phase and therefore "apart," but it will still tend to be found in an area described by a cylinder around the original orbit, and it will still all intersect at the point of the original detonation until something interacts with all those debris pieces.
- mlindner 5y ago
  It's a combination. In this case the interceptor is basically standing still. The satellite hits the interceptor is a better way of thinking about it. So most debris will be lower or at the orbit of the satellite, but still this was much better than the Chinese satellite that was blown up as the low orbit means most of these debris should de-orbit in the coming few years rather than the Chinese satellite debris that will be up there for decades or centuries.
- deleted 5y ago
  [deleted]
- BHSPitMonkey 5y ago
  From what little I understand so far, the interceptor was moving in a similar direction to its target (i.e. imagine a car being rear-ended). The debris would mostly continue "forward" with a higher velocity than before the collision. Obviously a real impact would be messy and the debris would scatter, but the average velocity of all the remains (including the direction vector) would still need to work out to the sum of the two input vectors. A head-on collision would be a different story, since the overall velocity would be reduced instead.

AnimalMuppet 5y ago

Do the pieces have the orbital velocity of the satellite? Or that of the interceptor? Or a combination of the two? And if it's a combination, why would they all have (approximately) the same combination?

shadowgovt 5y ago
For friable colliders, it'll be some combination of the two, but (assuming total disintegration, i.e. "pile of pieces" instead of two bodies breaking big rigid chunks off each other), the bulk of the material will end up going in the direction of the sum of the velocities of the colliding objects. Let me see if I can describe why. Blow up something in space far away from any gravity wells, and you'll get a more-or-less sphere radiating out in all directions, forever (well, not forever; it's possible the gravity of the pieces defeats the escape velocity over long enough distances if the explosion is small, but you know what, too much detail). Blow up something in a circular Earth orbit, and there is a property that must hold true (1) for every individual piece that doesn't achieve escape velocity: the piece's path must be a circle with the center at the Earth's center of mass. Since all those circles are starting from the same point, the only difference they can take in shape is to kind of imagine grabbing the circle at the point of the explosion and giving a "twist" around the original path. And for reasons I can't really explain with math but make a lot of intuitive sense when you play a bunch of Kerbal Space Program, those twists are tiny. In Earth orbit, it takes more change in velocity to go from an west-east orbit to a north-south orbit than it takes to go from launch pad to west-east orbit. So from afar, almost all the debris will following almost the same path as the original satellite because the velocity of the original satellite was going that way, momentum is conserved, and the distribution in the explosion will be kind of a normal distribution. If you were to ride along in the orbit from the original center of mass, what you'd observe is the debris spreads out (following its own individual circles), but halfway around the planet it reconverges and gets very near the original location, possibly self-colliding. (1) I simplified a bit, but not enough to matter much... The orbits are technically elliptical, and the fact that the debris will be pushed in prograde, retrograde, nadir, and zenith directions in addition to normal and antinormal directions means the ellipses will change shape, and therefore area and Kepler's law means the orbital phase will change. So the debris will eventually drift out of phase and therefore "apart," but it will still tend to be found in an area described by a cylinder around the original orbit, and it will still all intersect at the point of the original detonation until something interacts with all those debris pieces.
mlindner 5y ago
It's a combination. In this case the interceptor is basically standing still. The satellite hits the interceptor is a better way of thinking about it. So most debris will be lower or at the orbit of the satellite, but still this was much better than the Chinese satellite that was blown up as the low orbit means most of these debris should de-orbit in the coming few years rather than the Chinese satellite debris that will be up there for decades or centuries.
deleted 5y ago
[deleted]
BHSPitMonkey 5y ago
From what little I understand so far, the interceptor was moving in a similar direction to its target (i.e. imagine a car being rear-ended). The debris would mostly continue "forward" with a higher velocity than before the collision. Obviously a real impact would be messy and the debris would scatter, but the average velocity of all the remains (including the direction vector) would still need to work out to the sum of the two input vectors. A head-on collision would be a different story, since the overall velocity would be reduced instead.