OT: Space ship



#20

For your amusement in free time I present this simple problem:

Let's assume a space ship with a mass M and zero velocity, which is in free space with no outer gravity field. Then a negligible (for simplicity) amount of fuel with mass m << M was burned and the ship achieved a velocity v. Then a same amount m<<M was burned and thus the ship achieved a velocity approximately 2v.


From the other point of view, with sufficent accuracy we can say that after the first fuel burn the space ship had kinetic energy Mv2/2. After the second fuel burn the ship had kinetic energy 2Mv2.


So the second burn of the fuel added three times more energy, than the first burn. We can thus continue several steps and deduce, that after N-th such step the ship can gain arbitrary big increase in energy with just a very small amount m of fuel.


Hence we have a perpetuum mobile of the first kind ;-)

Where is the flaw in this consideration?

Edited: 4 Apr 2012, 6:59 a.m.


#21

The flaw would appear to be that you assume the thrust to be invarient of the velocity.

#22

To achieve a relatively big increase of kinetic energy (although hypothetical, we are talking about a conservative system, right?) we need infinite 'portions' of m fuel, hence an infinite fuel tank and an infinite time span (you did not consider time in your equations).

Edited: 4 Apr 2012, 7:26 a.m.

#23

Wouldn't all the fuel have to be on board initially? And would that be an infinite amount of fuel?


#24

Yes, all fuel is on board intially, and no, there is not an infinite amount of fuel.
The question is how it's possible to gain ever end ever increasing amount in energy burning the same amount of fuel.

Edited: 4 Apr 2012, 7:44 a.m.


#25

E=0.5*M*(V0+dV)2 = 0.5*M*(V02+2*V0*dV+dV2) = E0 + M*V0*dV+0.5*M*dV2

Thus dE = M*V0*dV+0.5*M*dV2

The energy is not rising with the square of V0 but with the square of dV and proportional to V0. It still looks a bit tricky to me but not as intimidating as it was.

#26

That's obvious. You cant make the speed double just because you want it so. The kinetic energy should simply be equated to the fuel energy.


#27

Quote:
Then a same amount m<<M was burned and thus the ship achieved a velocity approximately 2v

Bill's got it right. The assumption behind the quote is WRONG!

#28

Is it just an impression of mine or there is a missing post in this thread? I cannot find a post I read some minutes ago...

????

#29

OK, now is the right time for the answer to the problem. Here it is (don't read if you'd like to find the solution yourself):


The only flaw was this statement:


Hence we have a perpetuum mobile of the first kind ;-)



All other statements are correct.

But there is still one unanswered question remaining - how it's possible to gain ever end ever increasing amount in energy burning the same amount of fuel?


#30

Quote:
how it's possible to gain ever end ever increasing amount in energy burning the same amount of fuel?

Because energy does not go linearly with speed?

#31

It isn't possible.

You can't take the infinitesimal and then use it infinitely. Logical fallacy :-)

#32

Let me say again:

this statement is WRONG:

Quote:
Then a same amount m<<M was burned and thus the ship achieved a velocity approximately 2v.


As Bill first pointed out, you have to consider the energy content of m amount of fuel and equate that to the change in kinetic energy of the rocket, which goes as v^2.


#33

Are you sure? In the reference system of the space ship after the first burn, it has zero kinetic energy. In other words, setting relativistic effects aside, each burn starts from square one.


#34

It doesn't matter which reference system. If you change reference systems, you can't "start fresh". If you are in the rocket's reference system, it doesn't absolve you of what Feynman termed "accounting".


#35

I think the tip with the amount of fuel left in the rocket gaining energy through the speed increase hits the nail on the head. In my opinion it's equivalent to my view of starting fresh with a changed reference system, just a different description of the same effect.

#36

In fact, after the second burst the velocity would be even (very slightly) greater than 2v, So with sufficent accuracy 2v is valid.

#37

The flaw is that the first incremental unit of propellant supplies energy to the ship *and all the rest of the fuel*. Successive units of propellant release the energy contained in themselves including the energy given to them by all that was expended earlier. So yes, propellant used at the end of thrusting provides much more energy than propellant used at liftoff.

For small delta propellants, the linear model does indeed apply as proposed but it breaks down as the quantities become non-insignificant.

Reaction dynamics has all sorts of non-intuitive characteristics. Yes, it really *is* rocket science!


#38

Correct :-)


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