-6
$\begingroup$

In this sketch there is a gun system as a propellant chamber and we are using iron balls as a propellant.

An electromagnet is attached on the other side of this engine .When we fire these Iron balls with the help of this gun system then these balls will move towards forward side and a back force will be create to propel this engine . These iron balls will be piled up on the surface of this electromagnet and we can again reuse these iron balls and send back these iron balls into the propellant chamber or refilling the Gun system again and again with the help of a mechanical system.

In this way this engine will work and the propellant (iron balls) will never be reduced.

I have a question that suppose if the iron balls could reduce its velocity with itself and hit with electromagnet with low velocity then this engine will move or not?

Suppose the iron ball is expelling with 5 meter/second speed from Gun system and is hitting the electromagnet or a wall with 2 meter /second velocity then this device should work or not?

Furthermore I would like to inform you that we will use ferrofluid (magnetic fluid) to reduce the velocity of these Iron balls. These iron balls will be filled up with this magnetic fluid and at the time of expelling this fluid will be in solid state but once expelling from the gun system it will convert into liquid form to reduce the velocity of iron balls. In this way this iron ball will reduce its velocity with itself due to ferro liquid.

Note:

  1. We can use plastic balls also filled up with this magnetic fluid and wall of anything than electromagnet that's not an issue.
  2. The ball will be expelled from the gun system having spin velocity and fluid has very much friction.

Please explain your answer.

Improve this question
$\endgroup$
6
  • 1
    $\begingroup$ "Suppose the iron ball is expelling with 5 meter/second speed from Gun system and is hitting the electromagnet or a wall with 2 meter /second velocity" How is that supposed to work? If there is a magnetic field being generated by some part of the drive, the deceleration force will be necessarily transmitted back through whatever is generating the field. IN short, if the balls are brought to a complete stop within the drive, there will be no net force. IT the balls are not brought to a complete stop, they will be 'reaction mass' that is lost to the drive and cannot be reused. $\endgroup$
    – Andrew Thompson
    Sep 28, 2016 at 7:04
  • $\begingroup$ "..should work or not?" Not. $\endgroup$
    – Andrew Thompson
    Sep 28, 2016 at 7:05
  • 3
    $\begingroup$ As was already explained to you on the NasaSpaceflight forum, this does not work. $\endgroup$
    – Hobbes
    Sep 28, 2016 at 7:28
  • 1
    $\begingroup$ I can't, that thread has been deleted. $\endgroup$
    – Hobbes
    Sep 28, 2016 at 16:08
  • 1
    $\begingroup$ The question shows a lack of understanding of basic physics - voting to close. Vikram - you need to study simple Newtonian physics first, I think... $\endgroup$
    – Rory Alsop ♦
    Sep 29, 2016 at 15:59

3 Answers 3

Reset to default
13
$\begingroup$

NO

As a rule of thumb: if you think you invented either a way to generate infinite energy or infinite propellant: you are wrong.

  • But [...]

  • No. Always.

enter image description here

Your craft will gain a velocity vector of e. To retrieve the bullets you need to apply to them a velocity vector of -e.

If you do this using a liquid, the liquid will accelerate by e. If you do this using magnets, the magnets will accelerate by e.

If the magnet or liquid are attached to the main craft you achieved a final velocity change of 0.

If it is not attached to you craft, the propellant is not the bullets, it's the liquid/magnet. Which is now moving away from your craft.

Improve this answer
$\endgroup$
1
  • $\begingroup$ It's a good rule of thumb. As far as I know, there is no way to prove without question the "always" part, but conservation of momentum just keeps being right whenever it's put to the test, theoretically and experimentally. So it seems to be a darn safe bet. $\endgroup$
    – uhoh
    Nov 7, 2016 at 16:42
5
$\begingroup$

What you are proposing violates the law of conservation of momentum. The force required to pull the iron balls back to re-use them is equal to and opposite of the force required to push them in the first place. If it was this easy we'd have a colony in alpha centauri by now.

To quote TARS in Interstellar: "Newton's third law: the only way humans have ever figured out of getting somewhere is to leave something behind."

Improve this answer
$\endgroup$
5
$\begingroup$

Everyone has this idea at some point, and every time it doesn't work. However, you came to the right place if you want to learn. Antzi explained it very well, but I think there's something deeper to be said: You can never cheat physics.

If you tied a long spool of string to your bullets, firing them quickly and reeling them in slowly might seem like it would work - But it doesn't. Physics doesn't make rounding errors, so the momentum I gain from firing at a high speed is precisely counteracted by slowly stopping them.

"But I'm using magnets, not string." True, but have you ever held a magnet near metal? You'll notice that the magnet is being pulled towards something ferrous just as much as said ferrous thing is being pulled towards it. On their own, the magnet and object gain velocity towards eachother, but the combined magnet-object system (Your gun) has a net velocity change of zero.

"Alright, What if I use gravity to pull the projectiles back in." Again, no. Every object has gravity proportional to its mass, your projectiles included. As your ship pulls them back in, the projectiles are pulling the ship aswell. Long and short, You can't sidestep the issue by changing forces.

However, your idea of using a gun for propulsion isn't entirely without merit. The magnet/self-replenishing-fuel-source idea is, but not the gun. If you ditch the magnet and replace your 5m/s pea shooter with a 40km/s Railgun, you have a truly amazing exhaust velocity, and thus a very efficient propulsion system.

Improve this answer
$\endgroup$
2
  • $\begingroup$ "and thus a very efficient propulsion system" But that would be by virtue of having a 40 km/s exhaust velocity ($v_e$), not by virtue of some magic property of the propulsion system (of which the engine, or rail gun, or whatever, forms one part). $\endgroup$
    – user
    Sep 28, 2016 at 15:20
  • $\begingroup$ @MichaelKjörling I know, just trying to offer something positive. In my experience people don't react well when you take their ideas apart without giving alternate solutions. $\endgroup$
    – UIDAlexD
    Sep 28, 2016 at 15:25

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