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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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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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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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