This is one of the well-known paradoxes in physics, which researchers paid attention to for a long time [1,2]. This paradox may be driven by some free energy devices, one of which was presented by the author here. It should be noted that the author of this note does not insist on the unambiguity of the calculations presented below.

The paradox looks very simple at first glance. We have two identical ideal rockets, the engines of which gain speed by discarding a minimum of mass (for example, having ion engines), at the same time, the energy costs for such acceleration, as well as the kinetic energy of the rocket, can be calculated using the classical formula: \[E = W = {m\, v^2 \over 2} \tag{1}\] where: \(m\) is the mass of the rocket, \(v\) is the speed that the rocket has gained. We do not consider the overclocking method here.

We are doing two experiments.
1 experience. We install one rocket on another, and with the help of the engine of the lower rocket we accelerate this structure to the speed \(v\). After that, we disconnect the upper rocket and it continues to gain speed due to its own engine, thus accelerating to \(2 v\). At this moment, we calculate the energy spent on acceleration by the entire structure - \(E_1\), and the kinetic energy of the upper rocket - \(W_1\).

2 experience. We accelerate one rocket to a speed of up to \(2 v\), and at this moment we calculate the energy spent on acceleration - \(E_2\), and its kinetic energy - \(W_2\).

Both missiles were accelerated to the same speed and, it would seem, the costs of their acceleration should be the same. However, let's compare them mathematically.

1 experience. To accelerate two connected rockets to a speed \(v\) energy is required: \[E_{11} = {(m + m)\, v^2 \over 2} = m\, v^2 \tag{2}\] The kinetic energy of this design after acceleration will obviously be the same: \[W_{11} = {(m + m)\, v^2 \over 2} = m\, v^2 \tag{3}\] For further acceleration of one rocket to a speed \(2 v\) energy will be expended: \[E_{12} = {m\, v^2 \over 2} \tag{4}\] Recall that in the second section, the rocket accelerates relative to the already gained speed \(v\). But the kinetic energy will be calculated from the sum of the speeds: \[W_{12} = {m\, (v + v)_1^2 \over 2} = 2 m\, v^2 \tag{5}\] Then the total energy for rocket acceleration will be: \[E_{1} = E_{11} + E_{12} = 1.5\, m\, v^2 \tag{6}\] and the kinetic energy of the upper rocket is (5): \[W_{1} = W_{12} = 2\, m\, v^2 \tag{7}\] It turns out that in this experiment we spent less energy than we received to accelerate the rocket: \[C\!O\!P = {W_{1} \over E_{1}} \approx 1.33 \tag{8}\] And we have not yet considered the kinetic energy of the lower rocket, which continues to fly somewhere in space :)

2 experience. To accelerate one rocket to a speed \(2 v\) energy is required: \[E_{2} = {m\, (2 v)^2 \over 2} = 2\, m\, v^2 \tag{9}\] which is obviously equal to the kinetic energy of our rocket: \[W_{2} = {m\, (2 v)^2 \over 2} = 2\, m\, v^2 \tag{10}\] In this experience \[C\!O\!P = {W_{2} \over E_{2}} = 1 \tag{11}\] we have obtained the classical result expected from the law of conservation of energy.

In this thought experiment, idealized rockets without fuel ejection were considered, which in reality is not yet possible. In our time, we can give an analogue to our experiments - a rocket with two stages, however, there is a difference between a multi-stage rocket and two connected rockets. In the first case, it is necessary to take into account the mass of the discarded stage, because of which, in fact, the steps are made. In addition, in a real rocket today, it is necessary to take into account the energy consumption for the formation of a jet stream [1]. The result (8) of this note, in this case, will be different.

On the other hand, the use of natural static forces can open up completely new sources of energy for humanity, therefore, discarding such a mechanism is impractical. An example of the transformation of the gravitational force into useful kinetic energy, based on the principle presented in this note, is shown here [3].