Suppose I have an object with mass $m$ in vacuum that I propel by applying a constant force, $F$, on it, with a rocket engine that I supply a constant amount of energy, $\frac{\delta E_{supply}}{\delta t}$, to.

Then the object's acceleration is given by $F = ma \implies a = \frac{F}{m}$. Thus, it will have a constant increase in velocity, $\frac{\delta v}{\delta t} = a$.

But, $E = \frac{mv^2}{2}$, and so the objects increase in energy equals $\frac{\delta E_{object}}{\delta t} = \frac{m}{2} \frac{\delta}{\delta t} (v^2) = mv \frac{\delta v}{\delta t} = mva$.

This energy increase, $\frac{\delta E_{object}}{\delta t}$, is not constant, but $\frac{\delta E_{supply}}{\delta t}$ is and so the principle of conservation of energy is violated. Where is my error?


Many of you point out that the problem lies in how I assume a rocket will convert my energy to a constant force (@BMS for example says that $F = \frac{dK}{dx}$ instead of $F = \frac{dK}{dt}$), but there is one thing about that I don't understand:

Imagine that instead of a rocket, I propel my object with an electron accelerator. I accelerate my electrons over an electric potential of $V$ volts and aim the beam opposite to the direction of desired travel. This will consume a constant amount of power since $P = IV$ (power equals current times voltage and the rate of electron ejection (current) can be kept constant).

This should mean that that all my electrons are ejected with the same velocity, $V_e$, relative to my object at the time of ejection, shouldn't it? (or does $F = \frac{dK}{dx}$ make this statement invalid?) And since the electrons should require a known force to be accelerated to $V_e$ and because of Newton's third law, every force has an equal and opposite force, shouldn't this apply a constant force on my object?

  • $\begingroup$ You forget potential energy : your energy is $-Fx + /frac{mv^2}{2}$ $\endgroup$ Commented Sep 11, 2014 at 19:43
  • $\begingroup$ @PierreAlvarez What? $\endgroup$
    – garyp
    Commented Sep 11, 2014 at 19:45
  • 2
    $\begingroup$ What does it mean to supply a constant amount of energy to your rocket? Supplying more fuel? I don't see any reason why adding more energy to your rocket implies that that energy will go to the object that it's pushing. $\endgroup$
    – garyp
    Commented Sep 11, 2014 at 19:47
  • $\begingroup$ His assumption was that was a constant force. From this we can conclude that this force will be conservative and write down the total energy as the sum of kinetics energy and potential energy. $\endgroup$ Commented Sep 11, 2014 at 20:00
  • $\begingroup$ @garyp: I edited the post to provide a reason, so I suppose the mistake now lies somewhere in that reason. $\endgroup$
    – nijoakim
    Commented Sep 11, 2014 at 22:01

3 Answers 3


One mistake you are making is equating a constant power $\frac{\delta E}{\delta t}$ with a constant force. One does not imply the other.

I'm going to switch to a more standard notation and use $K$ for kinetic energy. And let's say $U=\text{const}$ to keep things simple in our system. Finally, let's imagine there's only this one force $F$ on this constant-mass object.

One of the many relations between force and energy is $F=dK/dx$. Note the distinction between that and the quantity $dK/dt$. One outcome of this distinction is that a constant force will supply a non-constant power, and vice versa. Weird, yeah, but oh well. It can be explained by noting for a constant force, the object will move a greater distance in each successive time interval, causing a greater amount of work $F\,dx$ to be done.


"applying a constant force, F with a rocket engine that I supply a constant amount of energy to"

Heere's your problem! Rockets are not as simple as you think. Yes, you could apply a constant force (thrust). And yes, this would consume fuel at a constant rate, but as @garyp points out, this does not mean you are adding to the kinetic energy of the rocket at a constant rate. In other words the energy efficiency of the rocket is not constant, even if the conditions inside the engine itself remain the same.

In fact the energy efficiency can exceed one, as the rocket gains energy by removing kinetic energy from the propellant, as well as its chemical energy.


Conservation of energy only applies to closed (isolated) systems, and your example, with an external source of energy, is definitely not of that kind.

  • $\begingroup$ There are also things as non-conservative forces, such as friction. Although there should be no friction in this example, it is an example of how energy isn't "purely" conserved in earth-life $\endgroup$
    – PipperChip
    Commented Sep 11, 2014 at 20:01
  • 1
    $\begingroup$ Eh? Energy is always conserved. Energy input - energy output = change in the system's internal energy. A closed system is simpler because all of the variables in that equation are 0, but conservation of energy still applies to open systems. (@PipperChip: Friction converts kinetic energy to other forms of energy, such as heat. It's only non-conservative if you're ignoring certain types of energy.) $\endgroup$
    – Brilliand
    Commented Sep 11, 2014 at 23:32
  • $\begingroup$ @Brilliand: If you supply external energy, the energy os the original system is definitely not conserved, but increased (as you have correctly pointed out through your formula). $\endgroup$ Commented Sep 12, 2014 at 12:17
  • $\begingroup$ @Brilliand Yes, the energy goes somewhere; that's why there's the quotes in "purely." It seems most people fail to factor in energy from friction, things bending, and energies which don't obviously affect their experiment. $\endgroup$
    – PipperChip
    Commented Sep 12, 2014 at 16:45

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