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If a satellite is destroyed and blown to multiple small pieces of debris, each one of these pieces has tiny probability of destroying another satellite, creating more debris which now will have even higher probability of destroying another satellite and so on (Kessler syndrome).

I'm wondering, how many satellites humanity would have to deploy to LEO (let's say 350-600km altitude, where large number of Internet satellites will likely reside in near future) before probability of chain reaction after even a single catastrophic failure becomes non-trivial?

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    $\begingroup$ Related (GEO) and this question $\endgroup$ – user10509 May 14 '19 at 14:18
  • $\begingroup$ This would depend on how big the satellites were, their sturdiness of construction, the exact orbits followed and other things. So it is not a well-defined number. As an extreme example, if a single satellite packed with small ball bearings was detonated in low Earth orbit I think it would be enough to lay waste to a wide range of low Earth orbits for some time. $\endgroup$ – Slarty May 15 '19 at 18:02
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Well I actually came here because I had a very similar question. But I have had a few thoughts about it and this is my answer to the question "What is the critical satellite density to cause a severe Kessler syndrome?". It is not complete, not the direct answer to your question and not necessarily correct.


So if you want to determine the satellite density in a given orbital layer you must first know the available area. The earth has a mean radius of 6371 km, to which we add the orbit height of 500km. That would give us a sphere with a surface area of 593 266 412 km².

Even if you have a satellite network with 1 million satellites that are spread across this area evenly each satellite has over 593 km² of area to itself.


Next you can determine the capacity of a given orbit. Since all objects that are at a given height in orbit have the same velocity there won't be any collisions if all satellites in a given orbit travel in the same direction (i.e. all of them at 0° inclination).

So if your satellites have an average length of 30m (which is actually big for a satellite) and you want to have the same distance between the satellites you can fit about 720 000 satellites in a 500km orbit with a circumference of 43 200 km.


But since the satellites actually travel in different directions their orbits intersect sometimes. This is where it gets difficult. My approach is to check how many satellites can pass a given single spot without colliding with each other.

So we need to know the orbital speed: $$ v=\sqrt{\frac{398600.5}{6871}}(km/s)=7.617km/s$$

At this speed a 30m satellite with it's safety buffer of another 30m passes a given spot in about 8 milliseconds. With an orbital period of about 90 minutes that means that about 675 000 satellites can pass this same spot before the first satellite reaches the spot again.


Now all those are very abstract numbers and it doesn't take eccentric orbits, different orbit directions or even orbital maneuvering into account, but keep in mind that this is only for the single orbital layer at 500 km. If you consider every 100 m increase in height as a new orbital layer that gives a massive amount of space for the satellites to be in. On the other hand of course you need to factor in intersections between those layers and so on. This is where I stopped because at this point all the calculations are getting way too complex to be made by hand or maybe even by my computer.


My conclusion so far is, that the space in LEO (~200 - 1000km) will surely be enough to handle at least 100 000 satellites if the orbits are planned with respect to each other and most of the satellites have maneuvering capabilities to avoid crashes in the rare case of a direct collision course. Even only the "small" space directly around earth is still insanely big after all.

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  • $\begingroup$ Feel free to complement some formulas if you can, otherwise I will add them myself over time. $\endgroup$ – Paedow Jun 17 at 22:44
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    $\begingroup$ The surface area of the circular orbit is a bit misleading. All orbits with different inclinations and the same height will cross over the equator. All polar orbits will cross over the poles. $\endgroup$ – Uwe Jun 17 at 23:01
  • $\begingroup$ @Uwe that's exactly what I am saying in the second-to-last paragraph $\endgroup$ – Paedow Jun 19 at 13:40

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