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Say spaceship $\alpha$ burns a portion of its fuel to leave planet A and is cruising through space at 10 m/s relative to the surface from which it launched. Spaceship $\alpha$ is being observed by spaceship $\beta$, which launched from planet B. The relative velocities of the two planets are such that spaceship $\alpha$ is moving at 50 m/s relative to planet B.

Spaceship $\alpha$ plans to increase its speed by burning some amount of fuel (which is converted completely to kinetic energy), and it communicates its intentions to spaceship $\beta$. Astronauts on each ship use conservation of energy to predict the change in velocity of the ship after the fuel is burned, with $$\frac{1}{2}m v_2^2 = \frac{1}{2}m v_1^2 + E_{burn}$$ and $$\Delta v = v_2 - v_1$$ The chemical potential of the fuel burned shouldn't be affected by the observer's reference frame. However, astronauts on spaceship $\alpha$ consider their initial velocity to be 10 m/s, but from the perspective of astronauts on spaceship $\beta$ who launched from planet B, it is 50 m/s. This will cause the astronauts to calculate different values for $\Delta v$.

Clearly, $\Delta v$ won't depend on reference frame. What, then, is the criteria for selecting $v_1$ such that the correct $\Delta v$ is calculated? With the spaceship flying through space, choosing planet A to have zero velocity seems as arbitrary as choosing planet B.

Dan
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When you burn the fuel it releases an energy $E_\text{burn}$, but your mistake is to assume that this all goes into the kinetic energy of the ship.

Rocket fuel works by accelerating itsef relative to the ship and leaving the ship at high velocity as exhaust gases. Momentum is conserved so when the fuel goes one way the ship goes the other way. The energy $E_\text{burn}$ is divided between the kinetic energy of the ship and the kinetic energy of the exhaust gas:

$$ E_\text{burn} = \Delta E_\text{ship} + \Delta E_\text{exhaust} $$

As you change reference frames you change the velocities, and hence the kinetic energies, of both the ship and the exhaust gas. The individual values of $\Delta E_\text{ship}$ and $\Delta E_\text{exhaust}$ will be different for different observers, however when you add together the change in the two kinetic energies it always adds up to the energy released by the fuel regardless of what observer is doing the calculation.

John Rennie
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