Talk:Tsiolkovsky rocket equation
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[edit] Derivations
Perhaps we could have more mathematically rigorous derivations for the various formulae? -- The Anome 10:51, 10 Oct 2004 (UTC)
- I would welcome that. But is this a request or an offer?--Patrick 21:32, 2004 Oct 10 (UTC)
- The derivation is simple. Force is assumed constant, and a = F/m. The mass decreases as fuel is consumed, so you are integrating 1/m, which is ln(m). Tsiolkovsky did this calculation in 1897, according to his notebooks. DonPMitchell 07:42, 26 August 2006 (UTC)
[edit] historical note
The original form of the rocket equation, and the form which is still used in practice, corresponds to the original definition of specific impulse (given as an alternative definition in Wikipedia): m1/m0 = exp(-dv/g/Isp). This has the distinct advantage that Isp is the same in both English and metric units; furthermore, delta-V expressed in various units (e.g., m/s or km/s) can be accommodated by converting g to the same units.
[edit] historical note (correction)
On further reflection I am almost certain that Tsiolkovsky's work predates the development of the concept of specific impulse (Isp) by many decades, so the original form of the rocket equation must have used exhaust velocity. But I can state pretty authoritatively that the form generally used today includes g*Isp to replace exhaust velocity, since propellants are almost universally characterized by Isp in seconds (i.e., the alternative definition of specific impulse given in Wikipedia). [note and correction by Ted Sweetser, tsweetser@aol.com]
[edit] More concerning Isp
Specific impulse is indeed in seconds. Specific impulse is not equivalent to exaust velocity, so describing Ve as Isp is false. [correction by T.Cooper, the.centipede[at]gmail[dot]com
[edit] More about Isp
It's a measure of impulse per mass of fuel consumed. Impulse equals momentum, incidently. The units are often given as "seconds", but it is more accurate to give them as "kgf-sec/kg", kilogram-force seconds per kilogram, or pound seconds per mass-pound. The number is the same in both cases. DonPMitchell 07:40, 26 August 2006 (UTC)
[edit] energy gain?
I'm not an expert, so forgive me if I am making a silly question: If at some point in the future we get to make a rocket propulsion system in which the propellant mass is negligible and its potential energy goes entirely to the rocket acceleration itself (so that the percentage of that energy taken by the propellant being expelled is not inversally proportional to its mass), wouldn't we get a kinetic energy gain in the rocket that would increase for the same amount of fuel spent in a given time frame as velocity got higher? If the rocket was at 20.000 m/s, for instance, and we used a given energy amount to accelerate it to 20.001 m/s, wouldn't we get a greater increase in energy than if we were just accelerating from 100m/s to 101m/s, using the same quantity of fuel? If it is so, couldn't it mean a kinetic energy gain to the rocket greater than the energy contained in the fuel that was spent to obtain it, if we got our rocket at a sufficiently high speed?
- It's not a silly question, it's a good question, but if you do the maths it turns out that energy is conserved (of course). The reason is that the kinetic energy of the propellant in a rocket going 20,000 is much higher than one going at 100 m/s. So, when emitted by the rocket, the propellant loses much more energy at high speed, than at low speed, and that's why the high speed rocket gains more energy than at low speed. Rockets are actually optimally efficient when the exhaust speed and the speed of the rocket match- in that case 100% of the energy ends up in the rocket and the exhaust just stops dead. Above and below that speed, efficiency drops. Hope this helps.WolfKeeper 22:03, 17 September 2006 (UTC)
- Yep, as Wolfkeeper says, you have to include the rocket plus all propellant (including spent propellant) to get this right. You'll find, of course, that energy and momentum are conserved, since these are the principles from which the rocket equation itself was derived. As the mass of propellant gets smaller, its velocity becomes correspondingly greater, and so you can never neglect the spent propellant no matter how quickly it exits the rockets. (Of course there are also relativistic effects to consider.) --P3d0 02:13, 18 September 2006 (UTC)
