Most active commenters
  • ben_w(3)
  • randmeerkat(3)

←back to thread

Intelsat 33e breaks up in geostationary orbit

(spacenews.com)
157 points milgrim | 13 comments | 21 Oct 24 14:09 UTC | HN request time: 1.456s | source | bottom
Show context
nordsieck ◴[21 Oct 24 14:28 UTC] No.41904557[source]▶
It is particularly bad for a satellite in geostationary orbit to break up or fail. Satellites are packed as tightly as possible into that orbit due to its economic importance (it's very useful for a satellite, particularly communications satellites, to always be over the same part of the Earth), so there is a higher than normal likelihood that this could be seriously disruptive.
replies(7): >>41904586 #>>41904693 #>>41904725 #>>41905123 #>>41905207 #>>41905406 #>>41906037 #
perihelions ◴[21 Oct 24 14:41 UTC] No.41904693[source]▶
- "Satellites are packed as tightly as possible into that orbit due to its economic importance"

Note that that's in the sense of angular separation, as viewed from the ground. They're physically hundreds of kilometers apart.

edit: (Geostationary orbits are ~42,000 km from the Earth center-of-mass; each degree of angle is an arc of ~700 km).

replies(1): >>41905025 #
naikrovek ◴[21 Oct 24 15:09 UTC] No.41905025[source]▶
> They're physically hundreds of kilometers apart.

That’s pretty close when your neighbor just exploded and there’s almost exactly zero air resistance to prevent debris from reaching you.

replies(5): >>41905222 #>>41905231 #>>41905234 #>>41905245 #>>41905425 #
1. ben_w ◴[21 Oct 24 15:28 UTC] No.41905222[source]▶
Yes there's no air resistance, but also most of the fragments aren't going your way.

If you have a 25 m^2 cross section in the direction of the explosion, at that distance you have a roughly 1 in 246 billion chance of any given bit of debris hitting you.

replies(3): >>41905333 #>>41905589 #>>41905634 #
2. throw4950sh06 ◴[21 Oct 24 15:39 UTC] No.41905333[source]▶
What is the chance of getting hit by further broken pieces of that satellite and other satellites?

When calculating risk, you have to take into account how many are there and what is the chance that any will be hit. Then you have to calculate what's the chance this will happen again, etc - and only then you can calculate the risk to your own satellite.

It's true that the chance of getting hit by one broken satellite is small. But that assumes there are exactly 2 things on the orbit.

replies(1): >>41905614 #
3. anjel ◴[21 Oct 24 16:06 UTC] No.41905589[source]▶
It might not place neighbors at appreciable risk but wouldn't debris still prevent replacing the failed satellite with another one at the same precious original address?
replies(1): >>41905662 #
4. ethbr1 ◴[21 Oct 24 16:09 UTC] No.41905614[source]▶
Aka Kessler syndrome [0] or neutron flux / cross-section (and associated equations, if you want to model it that way) [1].

[0] https://en.m.wikipedia.org/wiki/Kessler_syndrome [1] https://en.m.wikipedia.org/wiki/Neutron_flux

replies(1): >>41905621 #
5. throw4950sh06 ◴[21 Oct 24 16:10 UTC] No.41905621{3}[source]▶
That's if the risk comes out at 100%, but there's some space below that.
6. randmeerkat ◴[21 Oct 24 16:11 UTC] No.41905634[source]▶
> If you have a 25 m^2 cross section in the direction of the explosion, at that distance you have a roughly 1 in 246 billion chance of any given bit of debris hitting you.

Source?

replies(1): >>41905714 #
7. ethbr1 ◴[21 Oct 24 16:14 UTC] No.41905662[source]▶
Wouldn't debris at the same address (after some time) therefore have zero relative motion?
replies(1): >>41905780 #
8. a1369209993 ◴[21 Oct 24 16:20 UTC] No.41905714[source]▶

  $ units
  You have: 25m2 / 2tau(700km)^2
  You want: /billion
    * 0.0040600751
    / 246.30086
replies(1): >>41908217 #
9. user32489318 ◴[21 Oct 24 16:28 UTC] No.41905780{3}[source]▶
No, because the debris = tiny pieces of aluminum, will be pushed around by solar radiation. Also, there’re tiny meteorites, and other pieces of debris colliding with it, which adds energy to the system, if you like. TLEs are not maintained for small debris, so you can’t really predict conclusively. But my hunch is that eventually, the orbit will become a bit more parabolic, precession of which could put it into a trajectory of a S/C and cause a collision.
10. randmeerkat ◴[21 Oct 24 20:30 UTC] No.41908217{3}[source]▶
LLM generated nonsense.
replies(1): >>41908820 #
11. ben_w ◴[21 Oct 24 21:41 UTC] No.41908820{4}[source]▶
The 'units' command line tool has been part of Unix since Bell Labs, and GNU Units came along in 1997.

Personally I used basic high school geometry knowledge of "what's the area of a sphere", and you could also have just asked WolframAlpha, which also predates LLMs.

replies(1): >>41909736 #
12. randmeerkat ◴[21 Oct 24 23:47 UTC] No.41909736{5}[source]▶
Yet a formula that actually calculates the probability of impact is nowhere to be found in your response. You don’t consider mass, density, velocity, orbits, or anything else for that matter.
replies(1): >>41911505 #
13. ben_w ◴[22 Oct 24 05:41 UTC] No.41911505{6}[source]▶
The formula is an excercise for the reader, even if my audience was a 14 year old learning about this for the first time. As is figuring out why "mass" and "density" are unimportant.

Might be a valuable lesson in "reading the question carefully" for them, though, as the scenario was: "That’s pretty close when your neighbor just exploded", which is why orbital mechanics can be disregarded in this instance.