Remix.run Logo
goku12 2 hours ago

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 5 minutes ago | parent | next [-]

>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 an hour ago | parent | prev [-]

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.