From: mikkelhaaheim@gmail.com   
      
   Le mardi 7 juin 2016 21:22:15 UTC+2, Christopher Rice a écrit :   
   Here is a little more info to supplement Mark's response. He was being rather   
   generous.   
   Standard pressure is around 1 N/cm^2, or (even more grossly rounded) about 10   
   kg*m/s*cm^2. The larger the hole, the more quickly it will vent out...this   
   means a larger hole will have a greater initial impulse, but that impulse will   
   not last as long. As    
   mark said, this impulse will decrease rapidly, because you are quickly losing   
   the gaseous mass that is supplying the pressure (if some idiot has set the   
   airflow to compensate for pressure loss, however, this impulse could last as   
   long as the air supply    
   does). Now, let's say that the hole is 1 m^2 (about the minimum size for the   
   spinning astronaut to fit through). That means that the initial rate of loss   
   will be about 10 000 kg*m/s. However, since there is only about 300 kg of air   
   in that room, the flow    
   will only last about 3/100 of a second. This means that, generously, the   
   ejected air will have a maximum momentum of about 50 kg*m/s.   
   But that will not be the momentum of the astronaut. As Mark said, although the   
   direction of the ejection out of a cylinder will have a rather insignificant   
   impact on this figure, the orientation of the astronaut's body will have a   
   rather large impact. In    
   effect, the astronaut's body is acting like a sail. The more surface area   
   presented to the outgoing air, the greater the momentum imparted into that   
   body. If the hole is above the body, the astronaut will only present about   
   1/10 m^2 of surface area,    
   while a hole directly in front will present  roughy 1 m^2 (10 times the   
   acceleration). Let's say the latter is the case. Assume (incorrectly) that the   
   body is exposed to the pressure of 1 N/cm^2, all trying to exit the room at   
   once. This means that the    
   force applied will be the same as the outrushof air. However, there is only a   
   maximum volume of about 5 m^3 of air behind her body, or about 6 kg of air.   
   This air will exert a force during the 6/10 000 of a second before it flows   
   around the body and    
   escapes. This will, at most propell the astronaut at 0.1 m/s, assuming an   
   average mass for a fit female astronaut at about 60 kg.   
      
   I am not a professional at this. I could have made any number of mistakes or   
   miscalculations, but I think I have this mostly right. If anything, I think MY   
   figures are probably over generous. I very much doubt the astronaut is going   
   to exceed a few cm/s    
   velocity.   
   Another point to consider: every amount of energy put into spinning the   
   astronaut will be at the expense of velocity. If the (reasonble) unbalanced   
   distribution of force causes her to spin at any substantial rates (difficult   
   to recover from), her    
   velocity will more likely be measured in mm/s rather than a handful of cm/s.   
      
   --- SoupGate-Win32 v1.05   
    * Origin: you cannot sedate... all the things you hate (1:229/2)