I was reading "Russia threatens criminal charges against a NASA astronaut" in arstechnica.com. Somewhere in the article it says: " At the time, a 2 mm breach was discovered in the orbital module of the Soyuz MS-09 vehicle docked with the International Space Station" and later it says: "Left unchecked, the small hole would have depressurized the station in about two weeks."

I'm thinking there must be either enough air to leak out so slow or something can compensate for the leak or both. The question is what size of a small hole would cause a catastrophic failure? Meaning that astronauts would need to evacuate International Space Station (ISS). I'm talking about a small hole big enough to cause the evacuation ... I'm sure a foot size hole would be catastrophic but I'm talking small hole or holes. Thanks.


2 Answers 2


There are some numbers in The International Space Station - Operating an Outpost in the New Frontier page 348:

A hole that measures 0.6 cm (0.25 in.) in diameter will cause the ISS to depressurize to the minimal atmospheric level for supporting human life (490 mm Hg, 9.5 psi) in about 14 hours, whereas a 20 cm (8 in.) hole will reach that level in about 50 seconds.

However, the individual modules making up the ISS have hatches separating them, so time permitting (as described in the linked document) a sort of binary search would be performed by closing hatches to split the volume into halves until the leaking module was isolated. So unless the leak is massive, the entire ISS is unlikely to be depressurized.

Therefore, if "catastrophic" == "evacuate the ISS", a small hole is unlikely to cause that. If the leak is one of the crew vehicles (Soyuz or Dragon), the crew of that vehicle might have to return early, if the leak couldn't be fixed.

Also relevant: Can they isolate individual modules on the ISS?

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    $\begingroup$ A back of the envelope calculation suggests that the 6mm and 20cm holes in that doc depressurise at about the same rate per mm^2, but the "two weeks" for the 2mm one would have it about half as fast. Wonder if it's a different measure of depressurisation - total rather than "down to safe level"? $\endgroup$
    – Andrew
    Nov 30, 2021 at 23:12
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    $\begingroup$ @CuteKItty_pleaseStopBArking I'm not so sure: I ran the numbers properly just now, and the leak rate implied by the larger holes (~0.3psi/day/mm2) is very close to the leak rate for a 2mm hole to reduce to zero over 14 days (~0.33psi/day/mm2) $\endgroup$
    – Andrew
    Dec 1, 2021 at 14:49

If the station develops an unexpected rapid loss of pressure, the initial process is to go to a safe location when the alarm is sounded, then begin trying to locate the leak if possible. There is a detailed summary in section 3.1 of this paper.

The threshold for evacuating rather than staying to locate the problem is when the time left before dropping to the minimum safe pressure level (9.5 psi) is less than ten minutes. This seems to be the estimated time required to safely allow everyone to get to their spacecraft seats.

So the threshold for "big enough to cause a catastrophic loss" is probably "a hole that would depressurise the station in ten minutes or less" - anything that causes that big a hole will probably also have caused other problems, but let's leave that aside. Can we work out how large a hole that would be?


To expand my comment on OM's answer, we have two quoted values for estimated depressurisation rates to the 9.5psi threshold:

  • 20cm hole = depressurise from 14.7psi to 9.5psi in 50 seconds (not enough time to evacuate)
  • 6mm hole = ditto in approximately 14 hours (definitely enough time to evacuate)

If we model this linearly (but see comments on the other answer as to whether this makes sense), then it suggests that a hole of approximately 4-5cm diameter would give a time to minimum safe pressure of 10-20 minutes. This would probably be on the limit of what is survivable for the crew but not the station - the crew would be able to evacuate to a safe location before the pressure dropped too low, but would have to then abandon the station.

A larger hole than this, and the crew might not be able to safely evacuate at all.

A smaller hole than this, and there is a good chance they would have enough time to isolate the leak and seal off the damaged module(s), leaving at least part of the station safely pressurised and operational. I cannot work out how long it would take to go through the full process of finding the leak - probably a couple of hours, but hopefully this question will have an answer.

(Whether you would still want to keep the crew up having had that level of damage is another question, of course.)

I cannot find a clear statement as to whether they would continue putting air into the station to try and maintain a safe level while a rapid depressurisation is ongoing, but I think the answer is probably not. This paper discusses the nitrogen reserves as providing "capability to isolate modules and recover cabin pressure in the remaining volume during a potential “rapid” depress event", but "remaining volume" suggests it would be used to bring the safe area back up to pressure after the leaking section was sealed off, not while it's still losing air.

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    $\begingroup$ At least on shuttle, they explicitly turned off the repressurization systems at the beginning of the leak procedure, to try to better characterize the leak, and to conserve the gas for later. I haven't seen an updated ISS "rapid depress" procedure in many years. Nice answer. $\endgroup$ Dec 1, 2021 at 18:45
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    $\begingroup$ @OrganicMarble aha, it looks like it's the same here - from the 2020 paper. "Any gas introductions that can be terminated automatically are stopped...". $\endgroup$
    – Andrew
    Dec 1, 2021 at 18:49
  • $\begingroup$ +1 There's a bit of related calculations in answers to How could the 2018-08-30 Soyuz MS-09 / ISS leak be so slow? $\endgroup$
    – uhoh
    Dec 1, 2021 at 22:14

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