There is a rocket $R_1$ that is traveling with velocity $V$ with respect to another rocket $R_0$ of the same mass. $R_1$ has $J$ joules of kinetic energy with respect to $R_0$. $R_0$ has fuel onboard with $J$ joules of chemical energy. Hence, the system has $2J$ joules of total energy. We light the fuel on $R_0$ and burn it all with $100\%$ efficiency, so that $R_0$ is traveling with the same velocity as $R_1$. $R_0$ and $R_1$ now have $0\ \mathrm{J}$ of kinetic energy with respect to each other, and no fuel remains. The system now has $0$ joules of total energy. The system had $2J$ joules of energy and now has 0 joules of energy. What mistake was made to make it seem that this scenario has violated the conservation of energy?
2 Answers
What mistake was made to make it seem that this scenario has violated the conservation of energy?
You neglected the energy of the rocket exhaust.
The energy is conserved in an inertial reference frame. Once you light the fuel of $R_0$, it begins to accelerate, so its reference frame is no longer inertial, so you can no longer apply conservation of energy in its reference frame. If you want to apply conservation of energy here, you need to look at the inertial reference frame in which the initial velocity of $R_0$ is zero. In this frame, the velocity of $R_1$ does not change, so its kinetic energy does not change, and the final kinetic energy of $R_0$ + its exhaust will be equal to the initial energy of the fuel.
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$\begingroup$ Thinking about this a little more - since the final velocity of R0 is equal to R1, is the final kinetic energy of R0 in R1’s inertial reference frame equal to 0? $\endgroup$– plafrattCommented Jan 27, 2023 at 15:18
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$\begingroup$ Thank you. So, initially, R0 has J joules of kinetic energy in that inertial reference frame. And the fuel has J joules of chemical energy. Finally, R0 has 0 joules of kinetic energy in that inertial reference frame, and the fuel has 0 joules of chemical energy. So, all 2J joules of energy were transferred to the exhaust? Or, am I still missing something somewhere? $\endgroup$– plafrattCommented Jan 28, 2023 at 6:57
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1$\begingroup$ @plafratt Yes, that is also correct. $\endgroup$– SandejoCommented Jan 28, 2023 at 21:47