# How do chemical propulsion systems not break the law of conservation of energy?

Lets say that hypothetically we have a rocket with constant mass, and an engine providing a constant force. I understand that it is unrealistic to have a rocket of constant mass with a rocket engine that ejects mass to accelerate, however this simplifies the problem.

As the rocket accelerates the force from the engine will be travelling at a larger speed, therefore increasing the power of the rocket engine, since $$W=Fm$$, and this makes sense due to the constant acceleration that would occur with a constant force, since $$E=\frac{1}{2}mv^2$$.

The rate of fuel flow into the engine will remain constant, so the chemical energy that is going into the engine will remain constant. This seems to contradict the increasing power from the force of the engine.

I have answered a similar question on NasaSpaceFlight.com years ago. The answer is that, we do not only calculate the chemical energy of the fuel, we need to include the kinetic energy the fuel carries also. Fuel on the rocket is moving at certain speed thus has certain kinetic energy. When it is ejected after burnt, the speed of the exhaust ejected (seen by static observer) is different. The difference contributes to the energy you see in your question.

The higher speed the rocket is, the higher this contribution is.

As you have noted, the effect that goes into accelerating the rocket increases with time. However, when the fuel is accelerated in the direction opposite direction to give the rocket speed, the fuel has a change in specific kinetic energy $$k$$ (kinetic energy per mass unit) according to

$$\Delta k = \frac{1}{2}v_2^2 - \frac{1}{2}v_1^2,$$

where $$v_1$$ is the velocity of the fuel right before it has been used and $$v_2$$ is the velocity of the fuel right after it has been used. Since the velocity difference is always the same, we can write

$$v_2 = v_1 - \Delta v,$$

where $$\Delta v$$ is a constant. We therefore get

$$\Delta k = \frac{1}{2}(v_1-\Delta v)^2 - \frac{1}{2}v_1^2 = \frac{1}{2}\Delta v -v_1,$$

which means that the fuel used to accelerate the rocket looses more and more energy. This explains the increase in effect in accelerating the rocket.

There seems to be a confusion about the energy and the power: "the chemical energy that is going into the engine" is the rate at which energy is supplied, i.e., the power of the engine, $$P$$, whereas the energyof the rocket is its full energy, accumulated over time since the engine started. In other words: $$\frac{dE(t)}{dt} = P(t) \Leftrightarrow E = \int_{0}^td\tau P(\tau).$$

• Wouldn't a constant rate of energy going into a rocket's velocity reduce the acceleration over time, since its a $v^2$ relationship? Mar 9, 2021 at 14:59

What you are describing is what is known as the Oberth Effect, where a rocket's fuel efficiency increases the faster it is going. It is possible that the kinetic energy gained exceeds the total amount of chemical energy in the rocket to start, which seems to violate conservation of energy principles.

However, the reason this is not the case is that the energy contained in the propellant is the sum of its chemical energy and its kinetic energy, such that propellant moving faster has more energy stored in it than propellant moving slower. When the propellant burns, it transfers both its chemical and kinetic energy to the exhaust (and thus to the rocket), giving what appears to be extra energy.

The force provided by the rocket, however, is the same regardless of its kinetic energy. You can use this to show that the change in energy is proportional to its acceleration times its velocity, and if the force (and mass) are constant then the acceleration is constant -- and thus a faster rocket gains more energy than a slower one. The linked Wikipedia page has the derivation.