# Calculating the force of air entering a vacuum

There is a metal cube that is any size desired. the contents of the cube is a vacuum of space. there is a tube that is also a vacuum of space that also can be any size desired. The third item is a ball that fits air tight in the tube perfectly at the furthest point from the ball. The ball can be hollow or solid with any weight with the weight evenly distributed throughout the ball. There are no obstructions between the ball an the cube. They are in room temperature with normal ground level air density. Then the furthest point from the cube on the tube is opened up letting in the air. How much force would this give the ball as the air rushes in pushing the ball to try and fill the empty void in the cube?

I came up with the idea for this experiment when I was thinking about different ways we could travel into space by using little to no fuel/ thinking of sifi things. If I had an air compressor, chronometer, plenty of scrap metal and a welding kit i could figure it all out on my own.

• WCOLtd

## Answers and Replies

berkeman
Mentor
There is a metal cube that is any size desired. the contents of the cube is a vacuum of space. there is a tube that is also a vacuum of space that also can be any size desired. The third item is a ball that fits air tight in the tube perfectly at the furthest point from the ball. The ball can be hollow or solid with any weight with the weight evenly distributed throughout the ball. There are no obstructions between the ball an the cube. They are in room temperature with normal ground level air density. Then the furthest point from the cube on the tube is opened up letting in the air. How much force would this give the ball as the air rushes in pushing the ball to try and fill the empty void in the cube?

I came up with the idea for this experiment when I was thinking about different ways we could travel into space by using little to no fuel/ thinking of sifi things. If I had an air compressor, chronometer, plenty of scrap metal and a welding kit i could figure it all out on my own.

Welcome to the PF.

The key concept involved is "Conservation of Momentum". Are you familiar with this concept? If not, wikipedia.org is a good place to start learning the basics. :-)

russ_watters
Mentor
There is a metal cube that is any size desired. the contents of the cube is a vacuum of space. there is a tube that is also a vacuum of space that also can be any size desired. The third item is a ball that fits air tight in the tube perfectly at the furthest point from the ball. The ball can be hollow or solid with any weight with the weight evenly distributed throughout the ball. There are no obstructions between the ball an the cube. They are in room temperature with normal ground level air density. Then the furthest point from the cube on the tube is opened up letting in the air. How much force would this give the ball as the air rushes in pushing the ball to try and fill the empty void in the cube?

I came up with the idea for this experiment when I was thinking about different ways we could travel into space by using little to no fuel/ thinking of sifi things. If I had an air compressor, chronometer, plenty of scrap metal and a welding kit i could figure it all out on my own.
Welcome to PF!

Force of pressure is F=P/A

But the device you describe would be a one-shot deal since the air inside the spacehsip pushes the ball into the cube and then the device stops. So all you've done is re-distribute the mass of the spaceship, moving it by the distance the center of mass moved (the center of mass stays in the same place).

Spaceships do use reaction control methods (wheels, mostly) to re-position themselves, but pleae note that if you are looking for reactionless propulsion, you won't find it. It is an explicit violation of conservation of momentum, so it isn't possible. Moreover, perpetual motion machines such as reactionless drives are not allowable subjects for discussion here.

• berkeman
8Hi Mst3!

If I understand you right, you want to know what the force on the ball will be, when it is snugly fitted in a tube that has ambient air pressure on one end and vacuum on the other?
As Russ_watters said above me (although he got the equation a little wrong), the equation for the force is F=P*A where P is pressure in Pascal (the air pressure) and A is the area of which this force acts on. We know the air pressure at the surface of the earth, it's roughly 100 000 Pascal, i.e. 100 kN per square meter. The air rushing into the tube and pushing on the ball will "see" the ball in its way as a flat disc, and it is the area of this "disc" that will be the area A in our equation F=P/A! The area of a disc is given by A=πr2, where r is the radius of the ball in the unit of meters.

So all in all, by substituting the equation for the area of the disc into the equation for the force, we get the equation for the force on your ball:
F=100 000 * πr2

As an example, let's say the ball has a radius of 1 cm. Putting this into the equation we just came up with, gives:
F=100 000*π*(0.01m)2 = 10π Newton

If we imagine the ball is made from iron, which has a density of roughly 8 grams per cubic cm, this gives the ball a mass of 32 grams. The acceleration will thus be a=F/m = 10π N / 0.032kg ≈ 981 m/s. Quite a push!

Last edited:
russ_watters
Mentor
As Russ_watters said above me (although he got the equation a little wrong), the equation for the force is F=P*A where P is pressure in Pascal (the air pressure) and A is the area of which this force acts on.
Whoops. Thanks for the correction -- and welcome to PF!

Whoops. Thanks for the correction -- and welcome to PF!

No problem! And thanks, might stick around :)

The way I understood this experiment, the ball would experience 0 additional force because the tube is already a vacuum, making the vacuum space bigger by opening up a cube of vacuum on the other end of the tube is not going to cause any additional force to be exerted. What is pushing the ball in, is not the vacuums suction but the air pressure on the outside, if you don't increase that it doesn't matter how big you make a vacuum the fact that it is a vacuum means it have pressure exerted on it in this case the surface area of the tube times the outside pressure per unit square area. In other words cross sectional surface of tube times 1atm of pressure.

Welcome to PF!

Force of pressure is F=P/A

But the device you describe would be a one-shot deal since the air inside the spacehsip pushes the ball into the cube and then the device stops. So all you've done is re-distribute the mass of the spaceship, moving it by the distance the center of mass moved (the center of mass stays in the same place).

Spaceships do use reaction control methods (wheels, mostly) to re-position themselves, but pleae note that if you are looking for reactionless propulsion, you won't find it. It is an explicit violation of conservation of momentum, so it isn't possible. Moreover, perpetual motion machines such as reactionless drives are not allowable subjects for discussion here.

the device i described is not a device at all but rather a representation of space, earths atmospheric pressure, launch tube, and a vessel. I should have been more clear on what was what.
the cube represents space.
the tube is a elevator like structure that reaches into space from earth(outside earths atmosphere)
the ball is the vessel that is propelled into space.

Obviously as the vessel gets further and further from earth the air pressure drops giving the vessel less thrust. The idea is just meant to give the vessel a quick boost and then powerful magnets along the elevator like structure would continue to proficient sufficient force to push the vessel into space. The force of the magnets, however, are for another question.

i am going to do some research on reactionless drives as they seem interesting.

The way I understood this experiment, the ball would experience 0 additional force because the tube is already a vacuum, making the vacuum space bigger by opening up a cube of vacuum on the other end of the tube is not going to cause any additional force to be exerted. What is pushing the ball in, is not the vacuums suction but the air pressure on the outside, if you don't increase that it doesn't matter how big you make a vacuum the fact that it is a vacuum means it have pressure exerted on it in this case the surface area of the tube times the outside pressure per unit square area. In other words cross sectional surface of tube times 1atm of pressure.

thank you
your answer hit my question square on.
I'd like to ad another question if i may.
If the tube is pointing straight up, is a vacuum just like before, would the change in atmospheric pressure effect the driving force behind the ball as it reaches further and further up? In my mind it would stop having any force behind it once the outside air pressure lessens enough. Just like putting a straw in water. The water in the straw will not rise over the level of the water.

Welcome to the PF.

The key concept involved is "Conservation of Momentum". Are you familiar with this concept? If not, wikipedia.org is a good place to start learning the basics. :)

Thanks for the reply. I'm new to physics so any help is much appreciated. Do you know of any sites that i can learn about the Conservation of Momentum" other than Wikipedia? Perhaps a more trust worthy web page?

berkeman
Mentor
Wikipedia is fine for the basics of CoM. :-)

8Hi Mst3!

If I understand you right, you want to know what the force on the ball will be, when it is snugly fitted in a tube that has ambient air pressure on one end and vacuum on the other?
As Russ_watters said above me (although he got the equation a little wrong), the equation for the force is F=P*A where P is pressure in Pascal (the air pressure) and A is the area of which this force acts on. We know the air pressure at the surface of the earth, it's roughly 100 000 Pascal, i.e. 100 kN per square meter. The air rushing into the tube and pushing on the ball will "see" the ball in its way as a flat disc, and it is the area of this "disc" that will be the area A in our equation F=P/A! The area of a disc is given by A=πr2, where r is the radius of the ball in the unit of meters.

So all in all, by substituting the equation for the area of the disc into the equation for the force, we get the equation for the force on your ball:
F=100 000 * πr2

As an example, let's say the ball has a radius of 1 cm. Putting this into the equation we just came up with, gives:
F=100 000*π*(0.01m)2 = 10π Newton

If we imagine the ball is made from iron, which has a density of roughly 8 grams per cubic cm, this gives the ball a mass of 32 grams. The acceleration will thus be a=F/m = 10π N / 0.032kg ≈ 981 m/s. Quite a push!

wow thanks for the detailed reply.
it is quite a push. 981 meters per second is a bit over 2100 miles per hour or 3300 km per hour. Is there an equation to calculate the resistance from the friction of the materials used?

ok thanks, I'll be sure to check it out then.

thank you
your answer hit my question square on.
I'd like to ad another question if i may.
If the tube is pointing straight up, is a vacuum just like before, would the change in atmospheric pressure effect the driving force behind the ball as it reaches further and further up? In my mind it would stop having any force behind it once the outside air pressure lessens enough. Just like putting a straw in water. The water in the straw will not rise over the level of the water.

Supposing that this tube goes up into the stratosphere yes I suppose the pressure would be less, but in that case you would have gravity pushing the ball in as well, an additional force. Perhaps if you made the tube long enough there would be relatively no pressure at all outside the tube. The reason it works over less distance with a straw is because of the weight of the water vs the weight of the air. in a straw it is true that the pressure of the air is less than the pressure than the bottom of the straw's water, but the water doesn't go rushing up to the top of the straw,. Why is that? because what is preventing that is the weight of that water it has to overcome gravity. One of my first 'inventions' as a high school student was this 'brilliant' idea for a tube that would go up a mountain, and it would have turbines in it, the tube would operate in the following way, because the air pressure at the bottom of the mountain was more, and the top it was less, the air would go up the tube and generate wind energy. Lol, of course it doesn't take much thinking to point out the error in the thought, for instance why doesn't air just constantly float upwards, or "what is it about a tube that allows air to behave in a manner it wouldn't otherwise do?" I was so fixated on this brilliant invention for a time that I did little else but fantasize about the millions I'd be raking in with my patented idea.

russ_watters
Mentor
the device i described is not a device at all but rather a representation of space, earths atmospheric pressure, launch tube, and a vessel. I should have been more clear on what was what.
the cube represents space.
the tube is a elevator like structure that reaches into space from earth(outside earths atmosphere)
the ball is the vessel that is propelled into space.
[separate post]
If the tube is pointing straight up, is a vacuum just like before, would the change in atmospheric pressure effect the driving force behind the ball as it reaches further and further up? In my mind it would stop having any force behind it once the outside air pressure lessens enough. Just like putting a straw in water. The water in the straw will not rise over the level of the water.
Yes: f the ball weighed nothing, then it would be propelled to the "top" of the atmosphere and stop, as the tube stopped filling-up. Of course, if the ball weighed nothing, you wouldn't need the tube: it would be a balloon. Sure, balloons work.
The idea is just meant to give the vessel a quick boost and then powerful magnets along the elevator like structure would continue to proficient sufficient force to push the vessel into space. The force of the magnets, however, are for another question.
Well, a railgun is a differerent issue. Sure, a big railgun could fire an object into space.
i am going to do some research on reactionless drives as they seem interesting.
Don't. They are a hoax, you will be wasting your time and we won't allow you to discuss them here because we don't discuss nonsense physics here, only real physics. I only brought it up because I didn't understand what you were after with the OP.

Supposing that this tube goes up into the stratosphere yes I suppose the pressure would be less, but in that case you would have gravity pushing the ball in as well, an additional force. Perhaps if you made the tube long enough there would be relatively no pressure at all outside the tube. The reason it works over less distance with a straw is because of the weight of the water vs the weight of the air. in a straw it is true that the pressure of the air is less than the pressure than the bottom of the straw's water, but the water doesn't go rushing up to the top of the straw,. Why is that? because what is preventing that is the weight of that water it has to overcome gravity. One of my first 'inventions' as a high school student was this 'brilliant' idea for a tube that would go up a mountain, and it would have turbines in it, the tube would operate in the following way, because the air pressure at the bottom of the mountain was more, and the top it was less, the air would go up the tube and generate wind energy. Lol, of course it doesn't take much thinking to point out the error in the thought, for instance why doesn't air just constantly float upwards, or "what is it about a tube that allows air to behave in a manner it wouldn't otherwise do?" I was so fixated on this brilliant invention for a time that I did little else but fantasize about the millions I'd be raking in with my patented idea.[/QUOTE

That's an interesting idea, has some flaws, but if you can move it with the wind like wind mills it may work better. Ie predict the direction of the wind.

I always thought that the cohesive force of the water sticking together was what caused the water to not rush up the small tube that is the water. Like when you put your finger on the top of the straw and pull it out then flip it then put it back in the cup of water. Now there is water on the top and air on the bottom of the straw. Though you are right it is gravity that makes it rise slower but fall down the tube faster.

Yes: f the ball weighed nothing, then it would be propelled to the "top" of the atmosphere and stop, as the tube stopped filling-up. Of course, if the ball weighed nothing, you wouldn't need the tube: it would be a balloon. Sure, balloons work.

Well, a railgun is a differerent issue. Sure, a big railgun could fire an object into space.

Don't. They are a hoax, you will be wasting your time and we won't allow you to discuss them here because we don't discuss nonsense physics here, only real physics. I only brought it up because I didn't understand what you were after with the OP.

It won't be a waste of time if i learn something new. Like a failed experiment isn't a waste of time because you learn a new way NOT to do something. I also view things like an unsolved equation. If one way to solve an equation doesn't work try another way till it does. Thanks for the advice though. I'll be sure to google "why reaction-less drives don't work" instead of just "reaction-less drives."