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Assume there is a rocket with 10 kg of fuel in a large empty space without any external forces such as gravity.

Rocket burns 1 kg of fuel and gets a $v_1$ velocity gain. Now it is moving in $v_1$ constant speed.

Then it burns another 1 kg of fuel. If the rocket gains $v_1$ velocity gain this time too, the kinetic energy gain of the rocket is higher this time since the velocity of the rocket is $2v_1$ and kinetic energy is proportional to the square of the velocity.

Since the state of the rocket in both times it burned fuel is the same for an observer inside it (first time it was not moving, and the second time it was moving in constant speed, which is effectively same as not moving, if there is nothing around for a reference), I cannot see any reason why the rocket cannot get the same $v_1$ velocity gain. So it appears that the kinetic energy gain is higher the second time.

So, would the rocket gain a higher kinetic energy the second time?

If the rocket gains a higher kinetic energy at higher speed, let's assume the rocket got accelerated to a very high speed by some external means. Then at this very high constant speed, rocket starts accelerating by its own by burning fuel. The kinetic energy gain may even exceed the total energy contained in the burned fuel. How can this happen?

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marked as duplicate by Ruslan, Dale, Jon Custer, BowlOfRed, PM 2Ring Aug 9 at 16:32

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ Not directly relevant but in general for a rocket each subsequent unit of mass spent in fuel actually will result in a larger increase in speed since the total mass of the rocket is now lower. In the limit where the fuel makes up a small proportion of the rocket mass your question stands though (even though this is no longer a realistic rocket of course). $\endgroup$ – jacob1729 Aug 9 at 10:13
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    $\begingroup$ It might help answer the question if you explain why you think the rocket could not gain more kinetic energy from the second burn. All the analysis above is correct, but you seem to disagree with the conclusion and its hard to work out what exactly needs clarification. The surface answer though is "Yes - each subsequent burn grants more kinetic energy than the last". $\endgroup$ – jacob1729 Aug 9 at 10:46
  • $\begingroup$ @jacob1729 Updated question. Please check. $\endgroup$ – Lahiru Chandima Aug 9 at 11:10
  • $\begingroup$ See the linked 'possible duplicates' - the answer by Knzhou explains the concern about energy conservation. Essentially, the fuel also starts off at some very high speed $V$ and must lose lots of kinetic energy whilst being ejected from the main rocket. $\endgroup$ – jacob1729 Aug 9 at 11:12
  • $\begingroup$ @jacob1729 ok I think I understand what happens now. Thanks! $\endgroup$ – Lahiru Chandima Aug 9 at 11:16
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Use the Tsiolkovsky equation (the rocket equation) to find the speed. Assume $v_0 = 1\ \mathrm{m/s}$ speed of fuel relative to rocket

$$\Delta v = \left|v_0\cdot\ln(m_\mathrm i/m_\mathrm f)\right|$$

Rocket gains this speed $$\Delta v_1 = \left|1\ \mathrm{m/s}\cdot\ln(9/10)\right| = 0.105\ \mathrm{m/s}$$

Then rocket gains this speed $$\Delta v_2 = \left|1\ \mathrm{m/s}\cdot\ln(8/9)\right| = 0.118\ \mathrm{m/s}$$

Rocket gains this energy $$\Delta E_1 = m_1v_1^2/2 =9\ \mathrm{kg}\times(0.105\ \mathrm{m/s})^2/2 =0.0496\ \mathrm J$$

Then rocket gains this energy $$\begin{align} \Delta E_2&=m_2\cdot(v_2+v_1)^2/2 - 0.0496\ \mathrm J\\ &=8\ \mathrm{kg}\times(0.118\ \mathrm{m/s}+0.105\ \mathrm{m/s})^2/2 - 0.0496\ \mathrm J\\ &= 0.149\ \mathrm J \end{align}$$

Every time it burns fuel the gain in energy is bigger.

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  • $\begingroup$ Your equations assume that the dry mass of the rocket is zero, that's not realistic. In the OP's problem, we should assume that the fuel mass is small relative to the dry mass. $\endgroup$ – PM 2Ring Aug 9 at 10:34
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    $\begingroup$ Some typesetting notes: $V_0$ etc look nicer than $V0$. You can do subscripts by using an under_score such as V_0 to produce the above. I personally prefer absolute value bars for $|x|=\text{abs}(x)$ but that may be taste. $\endgroup$ – jacob1729 Aug 9 at 10:36
  • $\begingroup$ So the efficiency of the rocket increases with speed? Also, speed relative to what? If we take the speed relative to a different frame which is moving in the opposite direction at a very high velocity, the energy gain may exceed the total energy contained in the burnt fuel itself, which shouldn't be possible, right? $\endgroup$ – Lahiru Chandima Aug 9 at 10:37
  • $\begingroup$ This clearly answers that the energy increase is larger the second time around (and taking a non-zero dry mass doesn't change that fact). However, I think OP wants an explanation of how this can be the case. $\endgroup$ – jacob1729 Aug 9 at 10:38
  • $\begingroup$ Every time equal quantity of fuel is burned, it has the equal speed (relative to the rocket) and the mass of the rocket decreases. Because of this the efficiency of the rocket increases. When the speed of the rocket = speed of fuel , the efficiency of the rocket starts to decrease. This is theory, the efficiency of the rocket start to decrease before this. $\endgroup$ – user36636 Aug 9 at 14:51
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To get from $v_1$ to $2v_1$ more energy is needed than to reach $v_1$ from zero initial motion.

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  • $\begingroup$ That what it looks from an energy perspective. But, lets assume there is a person who who was awake first, and slept throughout first acceleration, and woke up when traveling at v1 constant speed. The person won't notice any difference in both states, so if he calculates the velocity gain the rocket would get in both situations, he should get the same result. $\endgroup$ – Lahiru Chandima Aug 9 at 10:21
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In addition to the rocket equation, there's some principles from special relativity useful here.

For a particle of mass M, we have $E^2=(Mc^2)^2 + p^2c^2$. Where $M$ is the rest mass, $c$ is the speed of light $p$ is the relativistic momentum. The rest mass is a Lorentz Invariant, the same in all inertial reference frames even though $E$ and $p$ change. So $(Mc^2)^2=E^2-p^2c^2$ has the same value for all observers.

Kinetic Energy is $E-Mc^2=\frac{p^2c^2}{E+Mc^2}$.

But these principles can be generalized.

$$M_{tot}^2c^4=\sum_{k=1}^N (E_k^2-p_k^2c^2)$$ where $k$ is the $kth$ particle of the system, the particles including the expelled propellant as well as the ship proper itself.

The ship loses the mass of propellant as it accelerates, but it gains mass-energy which over time reduces the ability to accelerate.

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