Science a GoGo's Home Page Forums General Discussion Sci Fi Forum Orion, Mission to Alpha Centauri

This, coming from the guy promoting a perpetual motion machine in another thread...LOL.



Because that is the problem with your own model. Your space ship is a closed system; matter can neither enter nor leave. You can exert forces within such a system, but the sum of those forces will always be zero at the level of the system itself.



However, none of the above has anything to do with the forces in your closed system.

The propulsive force generated where the air leaves the pipe is due to the momentum of the moving air; not friction, etc, within the pipe (in fact, friction in the pipe reduces thrust by slowing the air). Its all newtons laws - you use pressure to accelerate a mass of air in one direction, and you get a an equal force on the opposite that'll push on your tank of compressed air.

Momentum, BTW, is a specific physical quantity equal to mass * velocity.

Momentum is also conserved in a closed system - you can neither add it, nor take it away.



Sorry, you don't. Missing half the equation does not equal a valid answer.

I'll walk through this slowly; hopefully you'll see where your error is.

The pressurized air in the tank represents a stored (potential) energy. Upon letting air out of the tank, that potential energy generates a force which accelerates the air down the pipe. Upon leaving the pipe the air will have a momentum, defined by p = mass * velocity. The ship will have an equal momentum in the opposite direction, due to newtons 3rd law (every action has an opposite, but equal reaction). For the sake of ease, lets assume that 1kg of air is moving at a velocity of 100m/s. This will give the air a momentum of:

p(air) = mv
p(air) = 1kg*100m/s
p(air) = 100kg*m/s

Newtons 3rd law dictates our tank (and therefore the ship it is attached to) has the same momentum but in the opposite direction, therefore:

p(tank) = -p(air)
p(tank) = -100kg*m/s

Lets assume our ship/tank combo weighs 100kg. In this case this release of air would impart a velocity of:

p(tank) = -mv, therefore v(tank) = -p/m
v = -100kg*m/s / 100kg
v = -1m/s

The negative sign in this case simply means the ship is moving in the opposite direction of the air expelled from the tank.

In an open system this would lead to the tank going one way and the air going the other - exactly what you propose is happening. In a fully open system (space) that air would continue moving forever - newtons first law (an object in motion stays in motion).

But we're not in a open system. In our closed ship that moving air will eventually hit a bulkhead or hull. When this moving column of air meets the bulkhead/hull that impact creates a force which will be proportional to the momentum of the air (in fact, it will be exactly equal to the force which imparted the air its momentum in the first place). As before, newtons 3rd law dictates this will create an equal, but opposite change in momentum on that bulkhead.

So in this case we have the air imparting a momentum to the bulkhead of:

p(air) = p(bulkhead);
since p(air) = 100kg*m/s; p(bulkhead) also = 100kg*m/s

keeping in mind

p(tank) = -p(air) = -100kg*m/s

Now, if the bulkhead and tank were separate entities, you'd have the tank going one way, and the air would blow the bulkhead in the other. However, the bulkhead and tank are connected as they are both part of the ship. Therefore, the total momentum imparted on the ship will be equal to the sum of these two momentums imparted onto these different parts of the ship:

So we have:
p(ship) = p(tank) + p(bulkhead)
p(ship) = -100kg*m/s + 100kg*m/s
p(ship) = 0kg*m/s

Since velocity is a direct product of momentum, (p=mv)
v(ship) = p/m
v(ship) = 0kg*m/s/100kg
v(ship) = 0m/s

Bryan