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Here it is very simply:

A hypothetical spacecraft has total mass of 1000 kg and a main engine that has an effective exhaust velocity of 3 km/s (or 305.915 seconds if you want Specific Impulse in seconds).

Now burn 1 kg of propellant. So speed = 3000*ln(1000/999) = 3.0015 m/s. Kinetic energy = $E_k = \frac12 mv^2$ = 4,499.997 joules.

Later, burn another 1 kg of propellant. So speed = 3.0015 + 3000*ln(999/998) = 6.0060 m/s. Kinetic energy should now be 17,999.988 joules.

...What in the Universe just happened? It spent the same chemical energy both times (1 kg)...but got a much bigger increase in kinetic energy the second time? Where did this much bigger increase in energy come from?

If burning 1 kg of propellant yields 4.5 kilojoules of kinetic energy, then doing the same thing again should yeild another 4.5 kJ, for a total of 9.0 kJ. But no, we get about 18 kJ instead. Where in the world did this extra energy come from?

Dare I ask, are spacecraft perpetual motion machines? Or free energy machines? I don't see how you can put in the same chemical energy and get out very different kinetic energies. Have I done something wrong in my math?

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  • $\begingroup$ Sorry about the first answer -- I was thinking entirely of the wrong thing. $\endgroup$
    – Russell Borogove
    Commented Jun 27, 2015 at 3:45
  • $\begingroup$ @RussellBorogove Were you thinking of a rocket burn during a gravitational assist? I have that question too and might ask it as a separate question, because I'm pretty sure that any acceleration purely from gravitational attraction will be cancelled out by the same attraction in the other direction once you pass the planet. $\endgroup$
    – DrZ214
    Commented Jun 27, 2015 at 4:04
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    $\begingroup$ Actually you kinda got ripped off. The kinetic energy you put in the exhaust is 4.5 M J twice, but you only got 4.5 kJ + 13.5 kJ back for that in what remained. $\endgroup$
    – Mark Adler
    Commented Jun 27, 2015 at 4:13
  • $\begingroup$ @MarkAdler yep thanks, it was the exhaust's $E_k$ that i forgot about. $\endgroup$
    – DrZ214
    Commented Jun 27, 2015 at 4:21
  • $\begingroup$ Related: Oberth is confusing me. Apparent violation of conservation of energy $\endgroup$
    – David Hammen
    Commented Jun 27, 2015 at 12:26

2 Answers 2

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You forgot about the exhaust kinetic energy.

In the first case, it would be 1kg at speed of nearly 3km/s (3000 m/s for the first molecule exhausted and 2997 m/s for the last one).

After the second burn the speed is slower (approx. by 3 m/s), because the exhaust and the spacecraft move in the opposite directions.

The exhaust energy is about 4.5 MJ in both cases and the difference in its kinetic energy compensates the difference in the spacecraft energy.

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    $\begingroup$ Yep I forgot about the kinetic energy of the exhaust. I knew there was something missing/hidden. I'm trying to redo the math to see if they add up, but run into a problem. As you allude to, only the first molecule of the exhaust will have 3 km/s exhaust velocity. After that, the craft starts accelerating and each molecule will have a slightly different exhaust velocity, down to about 2,997 m/s. ...Am I gonna hafta integrate? Could I maybe just take the geometric average of 3000 and 2997? $\endgroup$
    – DrZ214
    Commented Jun 27, 2015 at 4:20
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    $\begingroup$ Yes, you have to integrate, especially, if you want exact numbers, so you have take in account, that mass of the spaceship decreased during the burns. Similar case is discussed here: quora.com/… $\endgroup$
    – Bohumil Janda
    Commented Aug 30, 2015 at 9:11
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so because of the decreasing mass of the vehicle as you burn fuel the total delta V is given by the rocket equation deltaV=Vexhaust * ln(Minitial/Mfinal). so even if you break up all the burns into multiple events the total deltaV is still governed by the rocket equation

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  • $\begingroup$ The question is about how more energy is transferred later in the burn, this doesn't answer that. $\endgroup$
    – kim holder
    Commented Jul 1, 2016 at 18:39

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