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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?

Lahiru Chandima
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    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). – jacob1729 Aug 09 '19 at 10:13
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    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". – jacob1729 Aug 09 '19 at 10:46
  • @jacob1729 Updated question. Please check. – Lahiru Chandima Aug 09 '19 at 11:10
  • 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. – jacob1729 Aug 09 '19 at 11:12
  • @jacob1729 ok I think I understand what happens now. Thanks! – Lahiru Chandima Aug 09 '19 at 11:16
  • Re, "...for an observer inside [the rocket]..." The observer inside the rocket is accelerated when the rocket accelerates. After each burn the observer is in a different reference frame than the one they were in before. An outside observer in any inertial reference frame, who adds up the potential energy stored in the un-burned fuel, the kinetic energy of the rocket, and the kinetic energy of the exhaust particles should get the same total energy before and after each burn. – Solomon Slow Aug 09 '19 at 13:24

2 Answers2

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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.

user36636
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  • 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. – PM 2Ring Aug 09 '19 at 10:34
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    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. – jacob1729 Aug 09 '19 at 10:36
  • 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? – Lahiru Chandima Aug 09 '19 at 10:37
  • 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. – jacob1729 Aug 09 '19 at 10:38
  • 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. – user36636 Aug 09 '19 at 14:51
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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.

R. Romero
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