The scenario is as follows. In a hypothetical future drive I spend energy E to accelerate a mass of propellant m to velocity v (e.g., plasma state propellant in the very powerful circular accelerator), so $v = \sqrt {2E/m} $.

Then I use my accelerated propellant as exhaust for my jet engine, using all mass in time t, getting thrust:

$$T = vm/t $$

Then my spaceship of mass M will be accelerated to speed (assuming m<<M):

$$V_1 = Tt/M = vm/M = \sqrt {2E/m}*m/M = \sqrt {2Em/M^2} $$

Now, if I spend the same energy E directly to accelerate my ship (by some unknown mechanism but transforming all energy E into kinetic energy of the ship), I get: $ E = MV^2 / 2 $ and $V_2 = \sqrt {2E/M} $

So, $V_2/V_1 = \sqrt{M/m}$ and my ship's velocity in the 2nd case is much bigger than in the first, while I seem to have spent the same amount of Energy in both excercises.

I am sure I am missing something very obvious but for the life of me can't figure it out, my physics became a bit rusty I guess :( Any pointers appreciated - why this difference? where does the energy "disappear" in the 1st case?

  • $\begingroup$ What's the energy in the mass of the "exhaust"? en.wikipedia.org/wiki/Tsiolkovsky_rocket_equation $\endgroup$
    – John Doty
    Oct 3 at 11:21
  • $\begingroup$ This sounds like an illustration of the Oberth effect. $\endgroup$ Oct 3 at 11:26
  • $\begingroup$ BTW, your second method is unphysical because it doesn't conserve momentum. $\endgroup$
    – PM 2Ring
    Oct 3 at 12:33
  • $\begingroup$ @PM2Ring how about I spend energy into curving the geometry of space-time and in effect the ship is accelerated by this artificial gravity field? Or even more realistically, the whole impulse-nuclear-explosion design, where we detonate small nuclear devices and the resulting energy is caught by a "sail" of sorts as here wikiwand.com/en/Nuclear_pulse_propulsion $\endgroup$
    – J Ho
    Oct 3 at 13:18
  • $\begingroup$ @JHo Spacetime curvature is fully determined by the stress-energy-momentum tensor. There are various ways to modify that tensor, but you always have to conserve momentum & energy locally (global energy conservation isn't well-defined in GR). You can shuffle energy around, but you can't make it magically appear or disappear. Rest mass is a very concentrated form of energy, and if you convert mass to a less concentrated form of energy you'll reduce the curvature. $\endgroup$
    – PM 2Ring
    Oct 3 at 16:54

1 Answer 1


The missing energy goes into increasing the KE of the exhaust.

Note that at high rocket speeds there is actually “extra” energy rather than missing energy. This comes from decreasing the KE of the exhaust


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