# If the ISS let out all it's air in retrograde, would the resulting force be enough to deorbit it? [duplicate]

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?

• 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. – Mark Adler Mar 6 '17 at 22:33
• I know this, I was just wondering how much DV can be generated with the escaping air from the ISS. – Jake Blocker Mar 7 '17 at 0:39
• 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. – Mark Adler Mar 7 '17 at 1:44
• I'm sorry I should I have been more clear – Jake Blocker Mar 7 '17 at 3:13

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 $v_{max} = 766\,\textrm{m/s}$.