Another thread about sat collision probability 🛰️💥🛰️

This time let's talk about regulations, as the FCC has just solicited input about how to regulate collision risk

As we know, risk can be computed as = 1-(1-Pc)^N

where Pc = each sat's collision probability and N = # of sats
Now, you don't have to be a math expert to know which variable is most important in that equation. Yep, the exponent!

Say you've been offered X^Y dollars, where each variable is between 1-9 and you get to choose one of them. You'd choose Y every time! Who cares what X is? 💵💰💵
Surprisingly, the FCC has asked if they should regulate risk by only limiting Pc, or if they should also take into consideration the N part of the equation.

Isn't the math obvious? Why would anyone ignore the most important part of the equation?

Well, it makes it hard to comply
Indeed, some operators who plan on launching 1000s of sats told the FCC that N should not be considered 🤔

Both @Amazon and @SpaceX advocated against a complete calc of risk or a limit of that risk, and instead say to focus on Pc alone. Read it here: fcc.gov/ecfs/search/fi…
Conversely, those in favor of proper math include @SES_Satellites, @Boeing, @Maxar, @ViasatInc, @MyriotaGlobal, Pico/Nanosat Developer Group, Lynk, @AstrodynamicMIT, @AerospaceCorp, @OneWeb, CSSMA, @Eutelsat_SA, @_CONFERS, the Association of Space Explorers, & @astroscale_HQ 😎👍
Limiting the negative impact we have on our environment requires quantified understanding before we can take direct and effective action.

If we can't get the math right in the first step though, well, then that's just disappointing.

#spacesustainability #spacedebris

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More from @mikeclindsay

3 Jul
A thread about satellite collision probability 🛰️💥🛰️

The FCC & others consider "satellites equipped with propulsion capability to have a collision risk of zero or near zero."

So let's ask ourselves: is that a reasonable assumption?

docs.fcc.gov/public/attachm… (see paragraph 20)
Well, if met with a collision risk (Pc) of 1/1000, an operator might avoid it with a maneuver (they don't have to), thus buying the risk down to 1/1,000,000.

But if Pc is initially 1/100,000, the operator likely won't maneuver at all! That's not zero risk, but is it "near zero?"
Depends on the # of times you take that risk!

Math tells us that the probability of having NO collisions = (1 - Pc)^N, where N is the # of times you "roll the dice."

So if Pc = 1/100,000 and you roll the 🎲 100 times, the probability of no collisions = 99.9%.

Not bad, right?
Read 5 tweets

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