4 ms·
Just to add to sibling comments who already answered this, that isn't how orbital mechanics work (or else a lot of things would be much easier!). A good place t
by xoa 6y ago
Just to add to sibling comments who already answered this, that isn't how orbital mechanics work (or else a lot of things would be much easier!). A good place to start reading is on Hohmann transfer orbits [1, 2]. They're commonly covered in the context of interplanetary travel, but the exact same principle applies to changing orbits around a single gravity well. The key take home as applied to here is that a single instantaneous addition of energy (delta-v) to an object in stable circular orbit just changes it to an elliptical orbit where the high point is determined by the added delta-v but the low point is the same as where it started. It takes a properly timed second instantaneous application of energy to stabilize at a different orbit.
So any ejecta from a single collision will still orbit through whatever altitude they started at, and in turn will be affected by atmospheric drag at that point regardless of how much farther out they get at the high point. In principle it's of course not impossible that they could collide with something else already in a higher stable orbit, but the odds of that get very, very low particularly in the short time frame they have before decay assuming they're starting in VLEO.
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1: https://ocw.mit.edu/courses/aeronautics-and-astronautics/16-07-dynamics-fall-2009/lecture-notes/MIT16_07F09_Lec17.pdf https://ocw.mit.edu/courses/aeronautics-and-astronautics/16-...
2: https://solarsystem.nasa.gov/basics/chapter4-1/ https://solarsystem.nasa.gov/basics/chapter4-1/
- SideburnsOfDoom 6y agoThat's a comprehensive answer, thank you. TIL.
- tgb 6y ago> high point is determined by the added delta-v but the low point is the same as where it started Is it the low point that is that same? I would have thought it was the current point (at time of delta-v addition), wherever that is, that must be part of the new orbit. Not that this is relevant for the topic at hand, just trying to check my understanding.
- omegaham 6y agoThis is correct. The simplification being made is that the original orbit was circular, so if you increase the debris' velocity, the current point is now the low point of the new orbit.
- tgb 6y agoEven for circular orbits, is that the case? The new low point would be lower if impulse is not parallel to the velocity.
- sand500 6y agoCorrect, so basically the lowest point(periapsis) of the new orbit cannot be any higher than the point where the collision happened.
- TeMPOraL 6y agoAnother way to think about it is, an orbit around a single body is entirely defined[0] by your current position and your instantaneous velocity. Six numbers. Where you are, which direction you're going, and how fast. Such orbits are closed, which means wherever you are right now, you'll return to that exact point later. So if you're at point P in your orbit and suddenly kinetic energy was added, your velocity changes in some arbitrary way. While you aren't sure where you'll go and when you'll get there, as long as the new (position, velocity) pair still defines a closed orbit around the body, you can be damn sure you'll return to the exact point where you are now. From this follows that the lowest point of your new orbit cannot be higher than where you are now, and the highest point of your orbit cannot be lower. -- [0] - Ignoring gravitational effects of other bodies, residual drag, magnetic fields, solar pressure, etc. None of these matter on the scale of days to months.
- xyzzyz 6y agoYes, if you start from roughly circular orbit, and the impulse is in the opposite direction to the direction of travel, it will slow you down and then the location of impulse is now the new highest point on orbit, and the lowest point after the impulse is even lower than before it.
- IlliOnato 6y agoThank you, this is really informative.