OT: Space ship - Printable Version +- HP Forums (https://archived.hpcalc.org/museumforum) +-- Forum: HP Museum Forums (https://archived.hpcalc.org/museumforum/forum-1.html) +--- Forum: Old HP Forum Archives (https://archived.hpcalc.org/museumforum/forum-2.html) +--- Thread: OT: Space ship (/thread-216587.html) |
OT: Space ship - Michael Eckstein - 04-04-2012 For your amusement in free time I present this simple problem: Edited: 4 Apr 2012, 6:59 a.m.
Re: OT: Space ship - Harald - 04-04-2012 The flaw would appear to be that you assume the thrust to be invarient of the velocity.
Re: OT: Space ship - Luiz C. Vieira (Brazil) - 04-04-2012 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 - Ed Wright - 04-04-2012 Wouldn't all the fuel have to be on board initially? And would that be an infinite amount of fuel?
Re: OT: Space ship - Michael Eckstein - 04-04-2012 Yes, all fuel is on board intially, and no, there is not an infinite amount of fuel. Edited: 4 Apr 2012, 7:44 a.m.
Re: OT: Space ship - bill platt - 04-04-2012 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 - Marcus von Cube, Germany - 04-04-2012 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 - Luiz C. Vieira (Brazil) - 04-04-2012 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 - Dave Shaffer (Arizona) - 04-04-2012 Quote:
Bill's got it right. The assumption behind the quote is WRONG!
Re: OT: Space ship - Michael Eckstein - 04-04-2012 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): Re: OT: Space ship - Cristian Arezzini - 04-04-2012 Quote:
Because energy does not go linearly with speed?
Re: OT: Space ship - bill platt - 04-04-2012 It isn't possible.
You can't take the infinitesimal and then use it infinitely. Logical fallacy :-)
Re: OT: Space ship - Dave Shaffer (Arizona) - 04-04-2012 Let me say again: this statement is WRONG:
Quote:
Re: OT: Space ship - Marcus von Cube, Germany - 04-04-2012 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 - Michael Eckstein - 04-04-2012 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 - Jim Horn - 04-04-2012 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 - Michael Eckstein - 04-04-2012 Correct :-)
Re: OT: Space ship - bill platt - 04-04-2012 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 - Marcus von Cube, Germany - 04-04-2012 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.
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