So I was watching the Martian and in order to get the rendezvous speed down they let out cabin air to provide a force to slow down. I was wondering if somehow all the air escaped the iss in a retrograde direction, and assume it didn't spin out of control, would the iss dip low enough into the atmosphere to completely deorbit?

  • $\begingroup$ All ISS has to do to deorbit is nothing at all. Just wait six months to a year, not doing any orbit raising maneuvers over that time, and it will deorbit. $\endgroup$ – Mark Adler Mar 6 '17 at 22:33
  • $\begingroup$ I know this, I was just wondering how much DV can be generated with the escaping air from the ISS. $\endgroup$ – Jake Blocker Mar 7 '17 at 0:39
  • $\begingroup$ That was answered already in another question, and it is not what you asked. The answer to the question you actually asked is: yes, since the $\Delta V$ required to deorbit ISS is exactly zero. $\endgroup$ – Mark Adler Mar 7 '17 at 1:44
  • $\begingroup$ I'm sorry I should I have been more clear $\endgroup$ – Jake Blocker Mar 7 '17 at 3:13

ISS has approximately 915 cubic meters of pressurized volume. If we, for the sake of this exercise, ignore the volume of pressurized gases stored in reserve tanks and only focus on what the effects of venting the pressurized volume would be, that gives us a total available propellant mass of roughly 1160 kg of air.

To deorbit ISS more or less immediately, its perigee would need to be lowered below 100 km altitude. At a circular orbit at 400 km, that would require a roughly 90 m/s retrograde change in velocity.

By conservation of momentum, we can explore how fast we much accelerate the ISS atmosphere to effect this change. The ISS has a mass of approximately 500,000 kg. Thus, the average velocity of the escaping air must be given by $0 = 1160\,\textrm{kg} \times v_{air}\,\textrm{m/s} + 498840\,\textrm{kg} \times 90\, \textrm{m/s}$, which gives $v_{air} \approx 38700\,\textrm{m/s}$.

That is, the escaping air on average would have to self-accelerate to more than three times earth escape velocity. This is a problem, because the maximum possible exhaust velocity for a supersonic flow nozzle is given by $$v_{max} = \sqrt{2\left(\frac{kR}{k-1}\right)T_0}$$ or $$v_{max} = \sqrt{\frac{2}{k-1}}v_{sound}$$ which for $k=1.4$ and $v_{sound} = 343\,\textrm{m/s}$ gives $v_{max} = 766\,\textrm{m/s}$.

TL;DR: No.


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