# Moving Spacecraft propellent less?

• I
• klwong
In summary, the momentum is conserved and the spacecraft will not have any meaningful motion from this setup.
klwong
TL;DR Summary
I have recently been studying methods of space propulsion. I would like to know if the following approach is feasible for propelling a spacecraft in space.

Assuming there are four robot inside a stationary spacecraft (mass 2 ton) , robot A has an electric gun that can shoot iron balls weighing 1kg at a speed of 8m/s. These balls collide with red balls at a 45-degree angle in a two-dimensional elastic collision. Each ball has a mass of 1kg. The red balls then collide with green and yellow balls in the same manner. Finally, robots B, C, and D catch the balls. After returning the iron balls to their original positions, they repeat the process for 2days (each cycle takes 10 sec). Will the spacecraft move? if yes, what is the speed of it ?

If the entire spacecraft plus balls plus robots end up going in the same direction at the same speed, then no, the assemblage will not have a translational acceleration.

PeroK
This will not work. Momentum is conserved.

A careful analysis of the momentum of each ball, robot, and the spaceship will show you that momentum is conserved and that the spacecraft cannot have any meaningful motion from this setup.

russ_watters
klwong said:
Try this video:

russ_watters
No external force, no movement.

russ_watters
russ_watters
The speed of the balls us considerably below the speed of sound in metal, so the ship will start to move forward when the black ball is fired and stop when the yellow balls are caught, so the hull will move forwards a little bit. In fact, the center of mass of the system has not moved - the balls moved backwards and the hull moved forwards and the center of mass does not move.

Guess what happens during the reset process when the balls are moved forwards. That's right - the hull moves backwards so the center of mass doesn't move. And then we're right back where we started in every sense.

You can't trick your way round momentum conservation by building complicated systems of colliding whatevers. Conservation of momentum applies at every collision, so if you think you have managed to pull momentum out of nowhere you have tricked yourself with the old "can you change a £20?" fraud.

klwong
Ibix said:
The speed of the balls us considerably below the speed of sound in metal, so the ship will start to move forward when the black ball is fired and stop when the yellow balls are caught, so the hull will move forwards a little bit. In fact, the center of mass of the system has not moved - the balls moved backwards and the hull moved forwards and the center of mass does not move.

Guess what happens during the reset process when the balls are moved forwards. That's right - the hull moves backwards so the center of mass doesn't move. And then we're right back where we started in every sense.

You can't trick your way round momentum conservation by building complicated systems of colliding whatevers. Conservation of momentum applies at every collision, so if you think you have managed to pull momentum out of nowhere you have tricked yourself with the old "can you change a £20?" fraud.
What if during the reset process when the balls are moved forwards, we give the balls a speed much slower than 0.1m/s ?

klwong said:
What if during the reset process when the balls are moved forwards, we give the balls a speed much slower than 0.1m/s ?
As I said:
Ibix said:
You can't trick your way round momentum conservation by building complicated systems of colliding whatevers.

russ_watters and PeroK
klwong said:
What if during the reset process when the balls are moved forwards, we give the balls a speed much slower than 0.1m/s ?
Then it will take longer, but you will still end up where you started.

russ_watters
p=mv
since speed is slower， during
the backword process the final momenton of the black ball is smaller?

PeroK
klwong said:
p=mv
since speed is slower， during
the backword process the final momenton of the black ball is smaller?
You must realise that using Newton's laws to disprove Newton's laws is a lost cause. Your spacecraft isn't going anywhere without an external force. Even if all the robots on board do the dance of the fandango.

russ_watters
PeroK said:
Even if all the robots on board do the dance of the fandango.
But what if they do the fandango slowly?

russ_watters
klwong said:
p=mv
since speed is slower， during
the backword process the final momenton of the black ball is smaller?
What you're actually wanting to look at is the momentum transfer between each pair of objects upon acceleration and collision.

Let's simplify things real quick. Let's assume we only have one ball, the black ball.

Upon firing the gun imparts X momentum to the black ball and -X momentum on itself and the spaceship to which it is attached. Then, when the ball is caught, it imparts X momentum to the spacecraft and the spacecraft imparts -X momentum to the ball. Since ##X + (-X) = 0## and ##-X+X = 0## the net momentum on both the ball and the spaceship at the end of this is zero.

However, the spaceship will have moved to the right in between the ball being fired and being caught. Let's say that distance is ##D##.

Now, the robot carries the ball back to the gun. In order to do so it has to apply a force on the spaceship that points left so that it can accelerate to the right. This force accelerates the spaceship to the left, imparting some amount of momentum to the spaceship. Let's say this is ##X/10##. The robot thus has momentum ##-X/10## by conservation of momentum.

The robot then decelerates when it gets to the gun, which requires a force applied to the spaceship that slows both the ship and the robot to zero momentum. The distance moved by the spacecraft is slightly more than before after the ball was fired, say ##-1.05D##, as we have to include the robot's mass into this.

Then the robot moves back to its starting position, the spacecraft moves ##0.05D##, and everything ends up back in its exact original position. ##D - 1.05D + 0.05D = 0## after all. If you don't believe that the spacecraft will move back to its exact position then you'll have to delve into a more complicated and in-depth analysis that I won't get into here.

Adding six more balls changes nothing. You just have more momentum transfers to take into account, each of which conserves momentum. Just think about it. If every momentum transfer conserves momentum, then there is no setup possible in which the sum of all the momentum transfers can lead to anything but whatever the original momentum was, which in this case was zero. So we don't even need to look at varying setups with many different moving objects or at setups that have complicated, curved paths. By virtue of conservation of momentum we know that there is NO setup that can result in a reactionless drive.

Ibix
klwong said:
p=mv
since speed is slower， during
the backword process the final momenton of the black ball is smaller?
Everything is slower but happens over longer time so the final result is the same.

And as @sophiecentaur just wrote: It's not about the momentum of one object at some time point, but the sum of all momentum transfers over a cycle, which is zero for an isolated system in an inertial frame, due to Newton's 3rd Law.

... and, of course, the centre of mass cannot move.

Drakkith said:
Adding six more balls changes nothing. You just have more momentum transfers to take into account,
...but the increase in complexity is how people start losing track of momentum and start making book-keeping errors and arithmetic slips that let them think they've got momentum out of nowhere.

russ_watters, DaveC426913 and Drakkith
PeroK said:
You must realise that using Newton's laws to disprove Newton's laws is a lost cause. Your spacecraft isn't going anywhere without an external force. Even if all the robots on board do the dance of the fandango.
When we see an illusionist saw a lady in half we KNOW that he is not really doing it. Millions of past experiments are evidence that it can't be done non-fatally. The illusionist is very good and it really looks as though he (and his robots, perhaps) is actually doing it. But we know that, if we looked at it from behind or with a camera in the box etc. that we would see the flaw.

@klwong seems to say that, with enough presentation tricks, the lady could, in fact be cut in half and stuck together again. He's in good company, though. People still come up with designs for Perpetual Motion Machines and however much they try to 'hide' the fatal flaw by making it more and more complex, the machines just won't work (a different 'rule' applies here). This complicated business with the robots and the balls can never get over Conservation of Momentum so why bother trying to add complication when each step in the process will still conserve momentum?

Am I wrong to appreciate the rôte learning of rules that we used to do in my school? Learning that way does not, in fact, damage critical faculties but in fact it puts you half way there in dealing with a new situation. Every morning, we look in the mirror and rehearse our times tables and our Newton's Laws and our Integration rules so that, by the time a serious question comes upon, we are already up to speed. Once in a blue moon, there will be something to upset the paradigm but, if it stands up to Occam's razor, we take the new rule on board.

klwong said:
Will the spacecraft move?
As you have been told, the answer is no because of Conservation of Momentum.

Further, "Reactionless Drives" is a forbidden topic at PF, so this thread is closed.
Forbidden Topics said:
EMDrive and other reactionless drives

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