When I'm floating in air (no gravity involved), I can imagine that when I blow out air, I will move in the opposite direction to the air I blow out. I will move because of the asymmetric air pressure on my body.
But what will happen if I suck in air? The air I suck in will hit me and send me backward. At the same time, total momentum should be conserved. So if I suck in air, my body just has to move forward, to conserve total momentum. So how does sucking in air produce forward momentum? Is the outside air pressure involved (which is not the case with blowing)?

Now imagine that I magically can inhale air for an indefinite time. Will this be the reverse process of blowing? Clearly not, because the air I blow out will be free, while the air I suck in will end up in my lungs. Only the time-reversed blowing out (which looks like inhaling) will be the same. I give a velocity to the air I suck in, but at the same time (or somewhat later) it stops in my lungs. So in total, I give no momentum to the air. Does this mean that I can't accelerate? Or maybe only at the moment I suck the air in?

This is what I think: At the moment I start to suck the air gets momentum which is directed towards me. As a result, I have to get momentum in the opposite direction. How?? I'm not sure, but I have to acquire it, in accordance with the conservation of total momentum. So initially I get some small velocity (ignoring air friction). If the suck is stationary, there is no net increase of air momentum, so I will keep my initial momentum.

What would happen if I could suck the air in at an increasing rate (or, more realistically, if I stop the suck)? Will I accelerate (or will I have zero velocity after the suck, to be left with a displacement only)?

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    $\begingroup$ Isn't this the same problem as the Feynman sprinkler? $\endgroup$
    – nanoman
    May 22, 2021 at 20:51
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    $\begingroup$ Since the human body is a torus I can easily conceive of a way to get twice the momentum out of any given volume! $\endgroup$ May 23, 2021 at 23:48

3 Answers 3


Yes, while the air you are sucking is moving (say left) into your mouth, due to conservation of momentum you would move slowly to the right. The force would be due to the difference in air pressure in front and behind you.

The trouble is that to maintain a constant force to the right you'd have to maintain the 'suck' which isn't possible for a person to do.

A vacuum cleaner device that propelled the sucked air out of the back could continue to accelerate.

Also what might happen is that the person would rotate, as the sucked air is from the mouth at one end of the body.

Interesting question, perhaps they've tried it in a spaceship. Presumably the effect is tiny, but if a person sucked air, turned their head and blew it out, sucked again etc... maybe they've managed to move themselves.

After the question edit about the suck being continuous. The liberty has been taken of amending the question slightly so we can discuss clearly and avoid rotation problems. Below is a 'robot' sucking device with two 'mouths' in a glass spaceship 'room'. The mass of the glass is small compared to the robot. The air in the room has the same mass as the robot, both $m$ - there is a surprising result.

Let's say there is a vacuum in the robot and the suction occurs when sliding doors at the mouths are suddenly opened. The robot can keep the gas inside with pumps, valves etc... and compress it so that the suction can be maintained.

On each diagram the left part is before the suction and the right part is after all the air has been captured. Movements discussed are relative to the outside observer.

enter image description here

In the first diagram (blue A), the robot is in the middle. As soon as the doors open, there is a reduction of total force on the right side of the robot as moving molecules on that side have less area to hit, so we would expect the robot to be moved right for a short time.

However when the molecules quickly travel the distance of the width of the robot, that effect would stop, perhaps the momentum gained would mean it continued for a short time to the right...

But the main point of the diagrams is to consider the centre of mass of the enclosed system - that position can't change. So it's concluded that when all the air is captured the robot (and captured gas) are back in the middle.

So unless there is a mistake here, it seems as though there would be a slight motion of the robot to the right and then back to the left, ending up stationary in the middle.

In diagram B, the robot starts at the green A and it's centre of mass is at A. The centre of mass of the air is at B and the combined centre of mass is at C. The robot would move to the right when the doors open, then it seems (by the centre of mass reasoning), it must continue to the right but end up stationary at C.

Perhaps the surprising result is the last diagram, C. Here the robot starts at E and the centre of mass of the air is at D, the combined centre is at F. When the doors open, , even though the doors are still on the right and even if it moves right initially, by the centre of mass argument it seems that it must then move left and end up stationary at point F.

For the robot in an infinite 'sea' of air, perhaps others would like to comment on whether diagram A represents that case best or whether it would be different again?!

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    $\begingroup$ Haha! I can already see the astronauts doing this. :-) Won't the air I suck in hit me, to send me in the same direction (as that air coming in)? $\endgroup$ May 22, 2021 at 10:59
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    $\begingroup$ Yes, the acceleration that first occurred would then be cancelled out, but you would be left with a tiny movement of the person to the right, and the tiny rotation. In principle the first suck would give you the tiny movement and the air (on average) in the 'room' would be moving the other way. $\endgroup$ May 22, 2021 at 11:02
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    $\begingroup$ If you're in free fall and vacuum, and you reach out and pull an object to you, the end result will be to shift yourself toward the object sightly, but you won't end up with any change in velocity if you and the object were not moving relative to each other to begin with. If you could suck in a blob of air the same way the same thing would happen. Now in an ambient atmosphere both of these would go slightly differently, but to a first approximation I expect the principle is the same. $\endgroup$ May 22, 2021 at 16:01
  • $\begingroup$ @DescheleSchilder I'm guessing that the way you're thinking about it is something like this - you start sucking in the air, and nothing significant occurs until the air particles hit you, so your expectation is that they'll send you flying back. However, that's not what happens - remember, the air is already pushing on you on all sides, before you do anything. You're interacting with the air as soon as the suction starts. $\endgroup$ May 22, 2021 at 19:09
  • $\begingroup$ Not sure now if, after the suck stops, the motion of the person will stop completely or whether you'll be left with a tiny velocity to the right, it might even depend on whether you are in an enclosed room or a theoretically infinite space of air. So answer has been edited slightly. What do others think on this detail? $\endgroup$ May 22, 2021 at 21:55

Here's the device you need. A suck-and-blow pipe™.

enter image description here

By rotating the pipe you can travel in different directions and impart different spins.

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    $\begingroup$ I wanna buy one! $\endgroup$ May 23, 2021 at 15:59

A Feynman sprinkler, also referred to as a Feynman inverse sprinkler or as a reverse sprinkler, is a sprinkler-like device which is submerged in a tank and made to suck in the surrounding fluid. The question of how such a device would turn was the subject of an intense and remarkably long-lived debate.
The behavior of the reverse sprinkler is qualitatively quite distinct from that of the ordinary sprinkler, and one does not behave like the other "played backwards." Most of the published theoretical treatments of this problem have concluded that the ideal reverse sprinkler will not experience any torque in its steady state. This may be understood in terms of conservation of angular momentum: in its steady state, the amount of angular momentum carried by the incoming fluid is constant, which implies that there is no torque on the sprinkler itself. There are two counterbalancing forces: the pressure differential pushing on the back of the nozzle, and the inflowing water impacting on the opposite side.[8]
Many experiments, going back to Mach, find no rotation of the reverse sprinkler. However, in setups with sufficiently low friction and high rate of inflow, the reverse sprinkler has been seen to turn weakly in the opposite sense to the conventional sprinkler, even in its steady state. Such behavior could be explained by the diffusion of momentum in a non-ideal (i.e., viscous) flow.[9]
However, careful observations of the actual behavior of experimental setups show that this turning is associated with the formation of a vortex inside the body of the sprinkler.[10]


  • $\begingroup$ This is indeed the same as I have in mind! The difference being that linear momentum is involved in my question. Instead of angular momentum. Won't the sprinkler get some initial angular momentum (after which it stops fast due to friction)? $\endgroup$ May 22, 2021 at 22:56

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