[_factor]

Two objects colliding can send debris into different orbits. Combined kinetic energy and mass differences can send debris to many different orbits.

A golf ball hitting a bowling ball or basketball, both traveling at 30 units of speed can produce quite a fast golf ball. Not all of the debris will safely burn up.

[tlb]

At the speeds we're familiar with, basketballs and golf balls have elastic collisions. At orbital speeds, satellites are nearly inelastic. So fragment exit velocities lie between the two initial velocities, kv1 + (1-k)v2 for some k that depends on where each fragment came from. If they're colliding, the velocities must be somewhat different, so the weighted average speed has to be lower than orbital speed. So fragments usually don't survive many orbits.

[indoordin0saur]

Very well put. It also seems like there's a limit to how bad Kessler syndrome can get. The more debris there is the more collisions, but the more collisions the quicker the debris collides with itself and de-orbits.

[WithinReason]

That's what I was thinking, Kessler syndrome should be impossible for objects in LEO since all debris orbits decay rapidly (probably 99.9% enter the atmosphere and burn up in minutes, the rest in hours)

[perilunar]

I guess if a collision ruptures a pressurised tank, or causes an actual explosion then you could end up with a higher-than-orbit speed?

[indoordin0saur]

Possibly. But more likely the thrust from escaping gas will push it in a direction to either slow the orbit down or make it more eccentric and unstable.

[tlb]

Right, if there's something like a small hole in a pressure tank, it's very unlikely to be aligned exactly with the CG, so the tank will spin around and the net thrust will be near zero.

If a pressure tank splits in half, both halves will fly away but that's a very inefficient way of using the energy in the gas, so the added velocity will be a small fraction of the speed of sound in the gas, which is 1/6 of orbital speed for hydrogen, less for any other gas.

You can't really get much of a chemical explosion because the fuel and oxidizer both disperse very quickly in space.

[goku12]

Just to elaborate the correct reply given by the others, the perigee of all fragments will be less than or equal to the altitude at impact point. If that's low enough, they will all eventually decay and deorbit. Even the fragments in elongated high-eccentricity orbits will have their orbits circularized by lowering apogee (the perigee is never going to rise) due to air drag. It will eventually spiral into the atmosphere. Here is the best visualization for this phenomenon - the Gabbard plot.

[1] Gabbard Plot Discussion (NASA Orbital Debris Program Office): https://ntrs.nasa.gov/api/citations/20150009502/downloads/20...

[2] Satellite Breakup Analysis (Australian Space Academy): https://www.spaceacademy.net.au/watch/debris/collision.htm

[ViewTrick1002]

The periapsis will always pass through where the collision happened.

To circularize at a higher orbit you would need secondary collisions on the other side of the earth.

[FranOntanaya]

Solar pressure would be a small factor too, though I assume it's not a big deal compared with orbital speeds.

[goku12]

You're right that all the fragments will pass roughly through the impact point in orbit. But it's not always the periapsis.

1. The normal or anti-normal delta-v imparted by the explosion/fragmentation (i.e, the velocity imparted perpendicular the plane of initial orbit) will cause the orbital plane of the fragment to change. The new orbit will intersect the old orbit at the impact point. Meanwhile, the eccentricity (the stretch of the orbit), semi-major axis (the size of the orbit) and displacement of periapsis from the impact point (the orientation of the orbit) remains the same as the initial orbit.

2. The prograde and retrograde delta-v (velocity imparted tangential to the orbit) will cause the diametrically opposite side of the orbit to rise or fall respectively. Here too, the new orbit intersects the old orbit at the point of impact. But since the impact point isn't guaranteed to be the periapsis or apoapsis, the above mentioned diametrically-opposing point also cannot be guaranteed to be an apsis.

3. The radial and anti-radial delta-v (this is in the third perpendicular axis) will cause the orbit of the fragment to either dip or rise radially at the point of impact. Again the impact point remains the same for the new orbit. So the new orbit will intersect the old orbit either from the top or the bottom. The new orbit will look like the old orbit with one side lowered and the other side raised about the impact point.

So none of three components of delta-v shifts the orbit from the impact point. You can extrapolate this to all the fragments and you'll see that they will all pass through the impact point. The highest chance of recontact exists there. However the perturbation forces do disperse the crossing point (the original impact point) to a larger volume over time.

Edit: Reading the discussion again, I get what you were trying to say. And I agree. The lowest possible altitude of the fragments in orbit (i.e the periapsis) is the same that of the impact point. So if the impact point is low enough to cause drag, the orbit will decay for sure. There is nothing that demonstrates this better than a Gabbard plot [1][2] - the best tool for understanding satellite fragmentation.

[1] Gabbard Plot Discussion (NASA Orbital Debris Program Office): https://ntrs.nasa.gov/api/citations/20150009502/downloads/20...

[2] Satellite Breakup Analysis (Australian Space Academy): https://www.spaceacademy.net.au/watch/debris/collision.htm

[xoa]

>But it's not always the periapsis.

>But since the impact point isn't guaranteed to be the periapsis or apoapsis, the above mentioned diametrically-opposing point also cannot be guaranteed to be an apsis.

You're correct on the generalized case of the math here, no argument at all, but this also feels like it's getting a bit away from the specialized sub-case under discussion here: that of an existing functional LEO satellite getting hit by debris. Those aren't in wildly eccentric orbits but rather station-kept pretty circular ones (probably not perfectly of course but +/- a fraction of a percent isn't significant here). So by definition the high and low points are the same and which means we can say that the new low point of generated debris in eccentric orbits will be at worst no lower then the current orbit of the satellite (short of a second collision higher up, the probability of which is dramatically lower). All possible impact points on the path of a circular orbit are ~the same. And in turn if the satellite is at a point low enough to have significant atmospheric drag the debris will as well which is the goal.

[ViewTrick1002]

No worries. I think I could have been more precise in my wording. :)

My comment is based on the hunch concerning physical calculations and interactions from an engineering physics degree and way to many hours in kerbal space program a decade ago.

[goku12]

Thanks! I figured that you had a reasonable understanding in this subject. But I still couldn't help just laying it out. I have some background too - as a professional.