I've been looking at this article. This question is about the motion of fragments in the immediate aftermath of the collision. I've drawn the following figures and quotes from the link.

Figure 9 shows the evolution of the debris clouds 180 minutes post-collision, almost two revolutions later. The spread of each debris cloud around its respective orbit is already becoming apparent.

Figure 9: Figure 9. View of Iridium 33 and Cosmos 2251 orbits and debris 180 minutes post-collision

The next extract relates to a snapshot of about 30 seconds after the collision.

Examination of the interactive 3D scenario (provided below Figure 10) shows large out-of-plane relative velocity components, the apparent result of coupling of the two masses, despite the hypervelocity nature of the collision. There are also a large number of pieces of Cosmos 2251 debris with significant radial (downward) relative velocities, although it is not apparent why this situation would be the case. It is hoped that the availability of this data set will help researchers with expertise in hypervelocity impacts develop a more complete description of the collision geometry for this event.

Figure 10: enter image description here

This question is not about the mentioned radial (downward) velocities, though I grant it is curious.

The implication of the two screenshots, where figure 9 is three hours after figure 10, is that the spectrum of relative velocities is very small compared to original satellite velocities. The question is, why are there not more fragments immediately generated with a range of high velocity headings between the two original trajectories?

If it helps, think of the impact of two billiard-balls, in such a case neither ball retains its original heading and both depart in very different directions, according to the normal expectation of a near elastic collision.

I can see a possible explanation, that the collision was a glancing blow, the main bodies continued as they were and the spectrum of ejected fragments is near inelastic, in the sense that much collision energy is lost and so the relative velocities are low.

Does anyone have any more insight into this?


2 Answers 2


With the satellites in LEO, relatively few debris will end up with velocity higher than at the impact moment, while moving at roughly similar trajectory as initially. That means most debris scattered in random directions would enter elliptical orbits with velocity roughly similar to initial at the point of impact. And that means apoapsis going significantly up, and periapsis - down. And lowering the periapsis significantly in LEO means one thing: reentry.

About all the debris that were knocked out of circular orbits, by the time of two revolutions later were already burned up, whether going directly down from the impact, or going up towards the new apoapsis, and then heading down almost a revolution later.

What remained were debris that didn't participate in the core part of the collision, torn away and shattered by the vibration of the stressed structure, not gaining or losing much of the initial velocity, plus scarce few that achieved enough velocity at just the right angle not to fall - apoapsis rising, periapsis not - certainly not many of them, because hypervelocity collisions tend to be of inflexible nature, metal splashing instead of shattering, meaning gaining speed vs losing it is unlikely, - lower periapsis much more likely.

In short, putting a satellite in a stable LEO requires quite a bit of mathematics, and if you change its velocity randomly by a large factor, the chance it will remain in orbit is quite small. The wiggle room isn't much.

It isn't that most debris entered these orbits shown by the "clouds". Simply, most debris that didn't continue along these orbits burned up on reentry.

  • $\begingroup$ Until now I was following the naive view that the outcome of the out of plane collision (and we could abstract it further to say starting in a circular orbit, 90 deg out of plane collision, no radial component, no along track component) would be a rotation of the line of nodes if it occurred at the poles, equally it would be an inclination change if it happened at the equator. Am I right that you are saying instead, the new velocity vector being simply different, regardless of orientation, means that there will be a new opposite apsis created thus making an ellipse? $\endgroup$
    – Puffin
    Commented Feb 25, 2016 at 1:55
  • $\begingroup$ Assuming so, this seems to make sense but do you really mean that all the fragments with the most diverse headings apparent in fig 10 have re-entered by fig 9, three hours later? Just to see if I follow you well, surely the initial height is unchanged, its just the opposite apsis that has either gone up or done, so we should be left with all the higher ones if the lower ones have de-orbited. Do you follow my reasoning there? $\endgroup$
    – Puffin
    Commented Feb 25, 2016 at 2:01
  • $\begingroup$ @Puffin: your first question: Yes. The orbit really doesn't care about poles or rotation of the planet below, it's an arbitrary correlation we care about only because we are down there and use the satellite. If the velocity vector is merely rotated in a plane tangent to Earth, then the ellipse (circle) will just rotate but remain an orbit. If it rotates and the speed grows, you're creating a higher apoapsis but periapsis stays. Anything else will almost certainly increase the eccentricity - meaning most likely lowering the periapsis. $\endgroup$
    – SF.
    Commented Feb 25, 2016 at 3:27
  • $\begingroup$ For your second question: When you're changing the direction of velocity in radial/antiradial direction, you're changing the eccentricity, not the average radius. That means apoapsis rises about as much as periapsis drops (not exactly, but that's a way to imagine it). Another inaccurate way to imagine it is if you're in a circular orbit and perform a radial/antiradial burn, you're rotating the circle of the orbit around the point you're in, without changing its plane, so it stops being coaxial with the planet, one side dips below the surface while the other rises. $\endgroup$
    – SF.
    Commented Feb 25, 2016 at 3:32
  • $\begingroup$ ...so with any change of direction out of the tangent plane, you need a lot of positive change of velocity to overcome the loss of altitude of the periapsis. In the crash you're not getting it - you're mostly losing speed. All the debris that went up will fall down on the other side of Earth, when the potential energy borrowed to rise to the higher apoapsis needs to be paid back by falling towards the lower periapsis. $\endgroup$
    – SF.
    Commented Feb 25, 2016 at 3:34

Satellites are not billiard balls, they are complex structures designed to withstand the expected stresses in their lifetimes. As an example, imagine the impact was between the bodies of the satellites. The arms holding the solar arrays are not very strong, so would snap before imparting much velocity change to the arrays. This would cause the arrays (or maybe pieces of them) to leave the impact with similar velocity to what the satellite had before the impact.

  • $\begingroup$ I can see that the example you have given makes sense, it says what could be happening if we accept the outcome as the starting point, but it doesn't actually address the question and attempt to say why it is this way. $\endgroup$
    – Puffin
    Commented Feb 25, 2016 at 1:48
  • $\begingroup$ Or perhaps have I missed something there? $\endgroup$
    – Puffin
    Commented Feb 25, 2016 at 2:02
  • $\begingroup$ I think it does. If we view a satellite as a collection of loosely bound pieces, most of them will exit the collision with just about the same velocity as they went in with. This explains the number along the old orbit. A few of them that have direct impact on pieces of the other satellite will have large velocity changes, but those bits will scatter widely. $\endgroup$ Commented Feb 25, 2016 at 4:07

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