3 ms·

I am no expert either, but a paper-napkin calculation tells me yes, this can happen. Basically, just based on impulse and energy conservation, and some assumpt

I am no expert either, but a paper-napkin calculation tells me yes, this can happen. Basically, just based on impulse and energy conservation, and some assumptions on the size and number of fragments, a head-on collision of two satellites going opposite directions at the same orbit should send some pieces of debris into much higher orbit. How likely this is to happen in reality, and thus how big a problem this could be is a subject of modeling and much more complex calculations.

bdamm 6y ago
It turns out that intuitive reasoning for orbital mechanics is not easy. Two satellites having a head on collision become space junk because now you have many fragments that themselves can't control their altitude or deorbit. Fragments that are ejected from the collision cannot have more than energy than they had originally, nor can they change direction very much. If pieces are ejected vertically, they are always going to have less lateral velocity than they had going into the collision, therefore will end up in an elliptic orbit with a high probability of colliding with the atmosphere at perigree. See "can you throw a baseball from the ISS and hit the earth" (No, you can't.)
- jholman 6y ago
  I think you're claiming that it is impossible for two colliding objects to release fragments that have more energy than either of the original objects. That's wrong. If it helps, start by imagining two satellite-like machines colliding in interstellar space (i.e. not in orbit), where they were initially moving at 1m/s in opposite directions. Even ignoring the possibility of an explosion (unspent fuel, pressurized areas, I dunno), it's still very easy for interactions to violently fling individual parts outward at speeds higher than 1m/s. Of course, my guess is that the possible amounts of additional delta-v are pretty low, like 100 m/s, and as such the resulting orbit would not be much higher than the original orbit. But that's just wild conjecture.
  - lutorm 6y ago
    If you were to collide two LEO satellites and turn them into fragments, even if you give those fragments much higher velocity than the initial satellites, you _cannot_ launch them into a higher orbit. Fragments that are launched down obviously hit the Earth. But an orbit is a closed ellipse, so the fragments that are launched up will _also_ hit the Earth -- they'll just go up steeply, turn around, and come down. The only fragments that won't initially hit the Earth are the ones that are ejected tangentially, parallel to the Earth's surface. Those will go into an elliptical orbit with a high apogee and perigee at the altitude of the collision. Which means they will _still _come down to that altitude and gradually lose energy.
    - jholman 6y ago
      I agree that the best case (uh, best for people who like Kessler cascades?) is that the new fragment has a perigee equal to the location of the collision, and a higher apogee. That means, yes, that it will suffer whatever degree of drag was present at the altitude of the collision. However. The grandparent post claimed that "Fragments that are ejected from the collision cannot have more than energy than they had originally, nor can they change direction very much", which I think is clearly what I addressed in my comment, and I stand by my correction of that, and your reply doesn't address this at all. Also, with respect to "you _cannot_ launch them into a higher orbit", you just agreed that you can have a higher apogee than the collision. If the input satellites had perigees lower than the altitude of collision, you can also have a lower perigee. More velocity at this point on the orbit, higher apogee, higher perigee.... So how do you figure you cannot have a higher orbit?
  - bdamm 6y ago
    Parts could be squeezed between high pressure zones and get shot out of the collision, but even those parts won't gain a higher orbit because the perigee will be below the original orbit.
    - jholman 6y ago
      First of all, no, the perigee need not be "below the original orbit", where did you get that? The correct thing to say is that the new orbit's perigee cannot be any lower than the location of the collision. But more generally, how does this justify your claim that "Fragments that are ejected from the collision cannot have more than energy than they had originally"? Or were you conceding that point and making a different point? See also my more-detailed cousin of this comment.
- phreeza 6y ago
  The key insight there doesn't have that much to do with orbital mechanics, it's more that in a collision, small fragments can come out with vastly more momentum than they went in with, which in orbits translates to a higher apogee. If you have ever bounced a small ball off a large one, you know the effect. See this video for an intuitive example https://youtu.be/2UHS883_P60 https://youtu.be/2UHS883_P60
  - lutorm 6y ago
    Yes, it translates into a higher apogee, but not a higher perigee. So they will still come down to the altitudes with higher drag.
    - phreeza 6y ago
      Strictly speaking this is only true if you start with a perfectly circular orbit, right?
      - jamesmaniscalco 6y ago
        Nope, any orbit. Think of it this way: an orbiting particle (or piece of debris) with no independent control/thrust will stay along whatever orbital path it's following, round and round (modulo any further orbital decay). After a collision, the particle is on a new elliptical trajectory, but the new trajectory still passes through the location of the collision. So the lowest point of the new orbit cannot be any higher than the altitude of the collision. Likewise, if the particle is slowed down, the highest point of the new orbit cannot be any lower than than the altitude of the collision. An interesting feature of orbital mechanics is that a change in velocity at a given point in orbit will change the altitude of the orbit at every point except where the change took place.

bdamm 6y ago

It turns out that intuitive reasoning for orbital mechanics is not easy. Two satellites having a head on collision become space junk because now you have many fragments that themselves can't control their altitude or deorbit. Fragments that are ejected from the collision cannot have more than energy than they had originally, nor can they change direction very much. If pieces are ejected vertically, they are always going to have less lateral velocity than they had going into the collision, therefore will end up in an elliptic orbit with a high probability of colliding with the atmosphere at perigree. See "can you throw a baseball from the ISS and hit the earth" (No, you can't.)

jholman 6y ago
I think you're claiming that it is impossible for two colliding objects to release fragments that have more energy than either of the original objects. That's wrong. If it helps, start by imagining two satellite-like machines colliding in interstellar space (i.e. not in orbit), where they were initially moving at 1m/s in opposite directions. Even ignoring the possibility of an explosion (unspent fuel, pressurized areas, I dunno), it's still very easy for interactions to violently fling individual parts outward at speeds higher than 1m/s. Of course, my guess is that the possible amounts of additional delta-v are pretty low, like 100 m/s, and as such the resulting orbit would not be much higher than the original orbit. But that's just wild conjecture.
- lutorm 6y ago
  If you were to collide two LEO satellites and turn them into fragments, even if you give those fragments much higher velocity than the initial satellites, you _cannot_ launch them into a higher orbit. Fragments that are launched down obviously hit the Earth. But an orbit is a closed ellipse, so the fragments that are launched up will _also_ hit the Earth -- they'll just go up steeply, turn around, and come down. The only fragments that won't initially hit the Earth are the ones that are ejected tangentially, parallel to the Earth's surface. Those will go into an elliptical orbit with a high apogee and perigee at the altitude of the collision. Which means they will _still _come down to that altitude and gradually lose energy.
  - jholman 6y ago
    I agree that the best case (uh, best for people who like Kessler cascades?) is that the new fragment has a perigee equal to the location of the collision, and a higher apogee. That means, yes, that it will suffer whatever degree of drag was present at the altitude of the collision. However. The grandparent post claimed that "Fragments that are ejected from the collision cannot have more than energy than they had originally, nor can they change direction very much", which I think is clearly what I addressed in my comment, and I stand by my correction of that, and your reply doesn't address this at all. Also, with respect to "you _cannot_ launch them into a higher orbit", you just agreed that you can have a higher apogee than the collision. If the input satellites had perigees lower than the altitude of collision, you can also have a lower perigee. More velocity at this point on the orbit, higher apogee, higher perigee.... So how do you figure you cannot have a higher orbit?
- bdamm 6y ago
  Parts could be squeezed between high pressure zones and get shot out of the collision, but even those parts won't gain a higher orbit because the perigee will be below the original orbit.
  - jholman 6y ago
    First of all, no, the perigee need not be "below the original orbit", where did you get that? The correct thing to say is that the new orbit's perigee cannot be any lower than the location of the collision. But more generally, how does this justify your claim that "Fragments that are ejected from the collision cannot have more than energy than they had originally"? Or were you conceding that point and making a different point? See also my more-detailed cousin of this comment.
phreeza 6y ago
The key insight there doesn't have that much to do with orbital mechanics, it's more that in a collision, small fragments can come out with vastly more momentum than they went in with, which in orbits translates to a higher apogee. If you have ever bounced a small ball off a large one, you know the effect. See this video for an intuitive example https://youtu.be/2UHS883_P60 https://youtu.be/2UHS883_P60
- lutorm 6y ago
  Yes, it translates into a higher apogee, but not a higher perigee. So they will still come down to the altitudes with higher drag.
  - phreeza 6y ago
    Strictly speaking this is only true if you start with a perfectly circular orbit, right?
    - jamesmaniscalco 6y ago
      Nope, any orbit. Think of it this way: an orbiting particle (or piece of debris) with no independent control/thrust will stay along whatever orbital path it's following, round and round (modulo any further orbital decay). After a collision, the particle is on a new elliptical trajectory, but the new trajectory still passes through the location of the collision. So the lowest point of the new orbit cannot be any higher than the altitude of the collision. Likewise, if the particle is slowed down, the highest point of the new orbit cannot be any lower than than the altitude of the collision. An interesting feature of orbital mechanics is that a change in velocity at a given point in orbit will change the altitude of the orbit at every point except where the change took place.