Comment by TeMPOraL | Hacker News Reader

Comment by TeMPOraL | Hacker News Reader

Here's my rough impression of what we may expect to see if the collision happens. Note that the collision isn't going to be perfectly heads-on, and is statistically unlikely for the centers of mass to collide.

BEFORE:

       ^    /
       |   |
       |   o - CZ-4C R/B
       |  /
       | |
       | /
       |/
       |
       *
      /|
     | |
     / |
    /  o - COSMOS 2004
   |   |
   /   |
  v    |

AFTER:

     aaa^    /
     aaaa   |
      aaaa  |
       aa  /
     b aa b b
   bb b bb b b
       ccc  b b
      ccccc  b b
   b  ccccc  b b
 b b   ccc
  b b b b  b
   b  a |
     aa |
   aaa  |
   aaa  |
   aaa  |

Legend:

a - Shattered fragments of the parts of satellites that weren't involved in direct hypervelocity collision, but got torn off and spun. Essentially like a shotgun discharge, but in space. These will carry most of their original kinetic energy and follow close to their satellite's original trajectory, slowly spreading over time and becoming untrackable. These will stay around the longest.

b - Pieces that were close to or directly involved in the collision, retaining enough orbital velocity to stay in space. May be on highly eccentric orbits. Should decay faster than pieces labeled "a" (I think, I'm not sure).

c - Pieces that lost enough velocity to fall either straight down, or fly upwards and then fall straight down. These may cause some damage on their way down to ground, but should disappear quickly.

For reference, here are some (much prettier) diagrams of the 2009 collision, that happened at close to 90 degree angle: http://celestrak.com/events/collision/.

Question: This was posted half a day ago and was estimated to happen (or not) in less than 1.5 days from now. Now, or later today, when there is less time to the collision, can't they calculate with far more accuracy, whether the objects will collide?

Only if one of the two objects has passed over one of their tracker stations (and they got a good scan of it)

a 0.001% uncertainty in speed is about 7km uncertainty in position after a day

Then how can they say 1-20% risk of collision?

prob they can measure speed at a higher accuracy than one mm/s, also orbit have a relatively low inclination difference and objects have a size, so that the impact area is not a point but a larger ellipses of varying overlap between bodies.

Simulation of collision from 2009, i think you're pretty spot on

https://www.youtube.com/watch?v=BiHY5dR5Jsg

It’s interesting to see how much debris remained confined to the original orbit-although it makes sense because any fragments that go sideways would need to happen to pick up a fairly high tangential velocity as a result of the collision. Not impossible, but fairly unlikely for any given part I’d guess.

It's not easy to change the inclination.

The best way to get a feel for what a tangential component would do is to boot up KSP and start playing with the maneuver node tool in-game. I can't find any equivalent tool on-line, but there are some animated GIFs here:

https://t-neumann.github.io/space/OrbitalManeuvers/

Scroll down to "Normal and anti-normal maneuvers". Basically, adding sideways velocity will change inclination a little bit, but also make it more eccentric - as you're not cancelling your original velocity, so the speed along the new orbital direction is greater than it was for your current altitude.

Not easy for satellites but not necessarily hard for small fragments

Small fragments are just smaller satellites, and the rules are the same. Inclination changes are hard (since you still have all of your original velocity, which is quite high at this altitude).

Object/reaction mass fraction can be higher

Adding sideways velocity shouldn't change your eccentricity, because it doesn't change the kinetic or potential energy of the orbit.

But it does! Your orbit is defined by where your velocity vector is pointing at the moment. If you add a sideways component, you're a) increasing the magnitude, and b) rotating the vector.

Nope, but I see now why we're disagreeing. It's because the animation you linked to is misleading.

Work is the dot product of force and velocity. If the thrust is always orthogonal to the velocity, then the burn does no work on the rocket, and so it can't increase the magnitude of the velocity.

In the animation you linked, it's plotting out what would happen if you continue to burn due north for some period of time as your velocity vector changes. But you wouldn't do that--if you're trying to change your inclination, you burn perpendicular to your velocity, and as the direction of velocity changes, the direction of the burn changes too, such that it's always orthogonal.

It's easy to watch this happen in the game if you don't use the maneuver node, but just use SAS to burn in the normal direction. The direction of "normal" changes during the course of the burn, and (except for small errors), your apoapsis and periapsis stay constant.

I see your point, but we're talking about two different things.

Yes, you can rotate your orbit if you continuously burn in the direction orthogonal to your momentary velocity vector. In essence, this is the same thing the body below is doing to you already, with continuous acceleration in the radial direction. But we're talking about impact debris here, which do not accelerate under their own power. What happens is they get an impulse - a one-shot, near-instant change in velocity. If you just apply an impulse sideways, the eccentricity will increase, because your speed is now greater. The shape of your orbit around a spherical body is fully defined by your momentary position in 3D space and your speed. If your speed increases, the shape changes.

KSP maneuver node planner shows what would happen to your orbit if you applied an impulse to it, of a configured magnitude and direction. It's not accounting for the burn time. Which makes it perfect for the purpose of talking about debris propelled by a collision.

Ok, that's a good point.

I think the argument is: It’s normal to the direction of travel, so it does increase velocity, and eccentricity is what you get when you add velocity at one point in an orbit. Same as other burns. If you were to burn retrograde of normal (but still tangent to the orbit’s radius), there’s an orientation where the retrograde component of your thrust matches the component in the component that’s...prograde in a new orbit. Which now sounds to me like it would just be 45° from retrograde, tangent to the radius

Is this based on anything or just random musings and ascii-art?

Don't disparage ASCII art. The very internet you used to make this comment has most of its protocol's defined in ASCII art (See RFCs). Its the back-of-the-napkin of the engineers who created the internet age. ;-)

It's based on what happened in 2009 (pictures linked in the comment), some familiarity with orbital mechanics, as well as the rest of discussion in this thread.

There are a bunch of subthreads here that show people have a hard time imagining how a head-on collision in space would play out; this is an attempt to put it in a more visual form.

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