The Museum of HP Calculators

HP Forum Archive 21

 OT: Space ship Message #1 Posted by Michael Eckstein on 4 Apr 2012, 6:53 a.m. 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<

 Re: OT: Space ship Message #2 Posted by Harald on 4 Apr 2012, 7:21 a.m.,in response to message #1 by Michael Eckstein The flaw would appear to be that you assume the thrust to be invarient of the velocity.

 Re: OT: Space ship Message #3 Posted by Luiz C. Vieira (Brazil) on 4 Apr 2012, 7:25 a.m.,in response to message #1 by Michael Eckstein 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.

 Re: OT: Space ship Message #4 Posted by Ed Wright on 4 Apr 2012, 7:34 a.m.,in response to message #1 by Michael Eckstein Wouldn't all the fuel have to be on board initially? And would that be an infinite amount of fuel?

 Re: OT: Space ship Message #5 Posted by Michael Eckstein on 4 Apr 2012, 7:40 a.m.,in response to message #4 by Ed Wright 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.

 Re: OT: Space ship Message #6 Posted by Marcus von Cube, Germany on 4 Apr 2012, 9:27 a.m.,in response to message #5 by Michael Eckstein 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.

 Re: OT: Space ship Message #7 Posted by bill platt on 4 Apr 2012, 7:50 a.m.,in response to message #1 by Michael Eckstein 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.

 Re: OT: Space ship Message #8 Posted by Dave Shaffer (Arizona) on 4 Apr 2012, 11:59 a.m.,in response to message #7 by bill platt Quote:Then a same amount m<

 Re: OT: Space ship Message #9 Posted by Luiz C. Vieira (Brazil) on 4 Apr 2012, 9:44 a.m.,in response to message #1 by Michael Eckstein 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... ????

 Re: OT: Space ship Message #10 Posted by Michael Eckstein on 4 Apr 2012, 12:07 p.m.,in response to message #1 by Michael Eckstein 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?

 Re: OT: Space ship Message #11 Posted by Cristian Arezzini on 4 Apr 2012, 12:11 p.m.,in response to message #10 by Michael Eckstein 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?

 Re: OT: Space ship Message #12 Posted by bill platt on 4 Apr 2012, 12:12 p.m.,in response to message #10 by Michael Eckstein It isn't possible. You can't take the infinitesimal and then use it infinitely. Logical fallacy :-)

 Re: OT: Space ship Message #13 Posted by Dave Shaffer (Arizona) on 4 Apr 2012, 12:16 p.m.,in response to message #10 by Michael Eckstein Let me say again: this statement is WRONG: Quote:Then a same amount m<

 Re: OT: Space ship Message #14 Posted by Marcus von Cube, Germany on 4 Apr 2012, 12:25 p.m.,in response to message #13 by Dave Shaffer (Arizona) 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.

 Re: OT: Space ship Message #15 Posted by bill platt on 4 Apr 2012, 12:53 p.m.,in response to message #14 by Marcus von Cube, Germany 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".

 Re: OT: Space ship Message #16 Posted by Marcus von Cube, Germany on 4 Apr 2012, 1:51 p.m.,in response to message #15 by bill platt 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.

 Re: OT: Space ship Message #17 Posted by Michael Eckstein on 4 Apr 2012, 12:25 p.m.,in response to message #13 by Dave Shaffer (Arizona) In fact, after the second burst the velocity would be even (very slightly) greater than 2v, So with sufficent accuracy 2v is valid.

 Re: OT: Space ship Message #18 Posted by Jim Horn on 4 Apr 2012, 12:29 p.m.,in response to message #1 by Michael Eckstein 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!

 Re: OT: Space ship Message #19 Posted by Michael Eckstein on 4 Apr 2012, 12:31 p.m.,in response to message #18 by Jim Horn Correct :-)

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