Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Here's an proposal on how to get from point A to point B in zero-gravity without using any propellant and the question why it wouldn't work:

A closed tube, filled with water and a round (solid) object. If you need equations, the volume of enclosed water is the same as the round object, but the round object is 10 times lighter. (imagine a glass with water plus a ping-pong ball).

here's a picture:

alt text

On earth the round object will float on the water inside the tube (subject to one G).

In zero-gravity the round object has no preferred position.

If we accelerate the tube in zero-gravity by one G, the situation is the same as on earth, the round object "floats". In this example we are accelerating the tube from the left side to the right side. The round object will consider floating to the right AS LONG AS THERE IS acceleration.

Now consider adding a pipe to the bottom of the tube and connecting it to the top, a loop. Inside the pipe is a small water pump.

If we give this apparatus a push, say one G in a zero-g environment, the round object will move "up", but now we start the water pump and spray the water on the round object, we try to submerge it. It will resist and impart a impulse on the water. like trying to hose down a air balloon floating on the pool. The pump will feel a resistance and hence the whole apparatus will move.

Just running the pump at constant speed, same volume of water per second, will do nothing. but if we run the pump faster and faster the whole apparatus will start moving:

the amount needs to be geometrical. The point is that we need to keep the apparatus feeling an acceleration, since only then will the round-object "float" and resist the incoming water at the top, hence we have something to "push on".

Before you blow the "foul whistle": consider the situation if there were no round floating ball in the apparatus.

(the pump runs on solar power or pre-charged battery)

(disclaimer: i know standing on a sailboat and blowing into the sail will not get me anywhere, action<->reaction)

thanks a bunch Sklivvz for the edit. sometimes the idea just needs to get out, never the mind how it looks like : P

share|cite|improve this question
I'm not going to read this until you make a reasonable effort in presenting it. – Mark Eichenlaub Jan 3 '11 at 17:03
Seconded. From what I can gather it seems like a reaction engine. – Sklivvz Jan 3 '11 at 17:25
I see no question marks. – kennytm Jan 3 '11 at 17:40
Thirded. ThisLooksLikeOneLongStreamOfText... – Noldorin Jan 3 '11 at 17:48
Editing to make at least readable. I won't vouch for content though :-) – Sklivvz Jan 3 '11 at 18:21

Consider the following scenario:

  • A gun is on the left side of a boat, which is very long.
  • The gun shoots, recoil makes the boat moves left.
  • The bullet hits the right side, the boat stops.

The end point of the process is that the boat has moved, but is at rest.

Why does this work? It works because there is a time between the shooting and the impact.

Consider the following scenario:

  • A gun is on the left side of a boat, which is very long.
  • The gun shoots repeatedly with a precise frequency so that at each recoil corresponds an impact at the same time.

The boat will start moving at the first shot (which is unbalanced) and will stop at the last impact (which is also unbalanced).

Why does this work? It works because in the beginning and at the end there is no flying bullet.

Consider the following scenario:

  • Instead of a gun there is a pump shooting a continuous stream of water in an empty tube
  • Initially the recoil makes the boat move left
  • Once the tube is full, the impact of the water on the right side balances the recoil on the left side
  • When you turn off the water, there is a final impact period on the right side that makes the boat stops

The boat will start moving as the tube fills, and it will stop moving as the tube empties

Why does this work? It works because there is no water in the tube initially and at the end.

Finally, consider the following scenario:

  • Same setup as the previous, with an always full tube (e.g. with a reservoir which is simultaneously filled and emptied)

The boat will not move.

In conclusion: I believe that your setup should be equivalent to one of these setups - I am not sure which - but from them you can derive whether your idea would work.

The scenarios are taken from this paper.

share|cite|improve this answer
thank you. initially i had this idea because i was considering the awkward situation an astronaut would be in, if he were stranded two meters from the space station with no rocket and nothing to throw AWAY from space stations direction ... how could he get back? – freeside Jan 3 '11 at 19:50
@freeside: He couldn't get back, if you believe in momentum conservation. Nothing acts on his center of gravity from the outside. Well, he could always wait until he gravitated back to the space station, but...he shouldn't hold his breath for that, so to speak. – Greg P Jan 3 '11 at 21:31
@greg i think i understand "conservation of momentum": every force yields a equivalent counter force? so if the ISS get's a boast from a rocket to higher orbit the passenger and everything inside would move back towards the end of the station that were the boasting rocket is attached, except a ballon with helium inside, it would "rise" and float to the front. if the balloon is big enough, a astronaut could hang on to it? – freeside Jan 3 '11 at 21:48
@freeside: I am referring to your comment about the stranded astronaut. There is nothing he can do if he can't throw some object. If no force is applied to him from the outside, his center-of-mass momentum can't change. So he is stuck. – Greg P Jan 3 '11 at 21:57
@freeside. As for your astronaut hanging onto a balloon inside the spacecraft, I can see what you are saying. But what does this have to do with the stranded astronaut? The buoyancy of the helium only works because of the air in the spacecraft. – Greg P Jan 3 '11 at 21:59

i think it will not work, simply because the round object that floats can only "carry" so much weight before submerging. it's like a life-vest: it can float one person, but not an entire army. i our case of the above mentioned apparatus, the weight of the tube, pipe, pump and water would rest on the small round object, which probably is too much and it would sink, like a overladen life vest, even if the whole apparatus is experiencing an acceleration. -BUT- it could augment the acceleration of a rocket attached to the bottom of the apparatus. so only by itself, it probably cannot accelerate and move itself, but if we have a small rocket attached that would provide 1 G of acceleration, the apparatus could add another x-amount of acceleration when operated in the manner mentioned.

share|cite|improve this answer
@Greg, consider this situation, we are boasting the ISS to a higher orbit. the free floating astronauts will eventually accumulate on the "floor", the side were the rocket is attached. now one astronaut fills a ballon with helium, big enough to carry him, he starts to float away from the floor, to the "ceiling". for an observer on earth, this looks like the ISS is accelerating, but someone inside the ISS is accelerating even MORE! – freeside Jan 4 '11 at 9:35
When a helium balloon floats upward, whatever momentum it gains is lost by whatever it pushed off of. The balloon and astronaut float within the ISS, but they push off the air within the ISS, which pushes off the floor of the ISS. The bottom line is that without propulsion or some outside force, the center of mass of the ISS + everything in it has the same momentum before as after. – Greg P Jan 4 '11 at 15:33

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.