# I'm having trouble wrapping my head around force, and thus acceleration, diminishing with a higher velocity when using a constant power

Here is an hypothetical situation I thought about that I don't understand.

Let's say I have 2 identical rockets with a fixed power. These rockets start to race in the same direction using the full power of their engines.

So, by the equation $P=Fv$ we get that with the increase of speed of the rockets, the force exerted by each of them, diminishes. And we also know that $F=ma$, so with increased velocity, the acceleration decreases.

At any given time, the rockets are travelling at the same speed and at the same acceleration one from another, so it looks from their perspective that they are in the same position.

But let's say rocket B breaks down. And now rocket A starts accelerating away from it. At that moment, the acceleration of rocket A is $a_0$

$$a_A=a_0$$ $$a_B=0$$ $$a_{AB}=a_A - a_B = a_0$$

But A is travelling at $v_0$, while it's travelling at $0$ relative to B.

$$v_A=v_0$$ $$v_B=v_0$$ $$v_{AB}=v_A - v_B=0$$

But then, by this logic...

If $P=Fv$, and $F=ma$, then $a=F/m=P/mv$

$$a_{AB} = P/mv_{AB}$$ $$a_A = P/mv_A$$

Which are clearly different.

So, what am I doing wrong? Where is the flaw?

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Your mistake is that you are using two reference frames at once. Rearrange, and you can see that the power is clearly not the same in both frames: $P = m a_0 (v_a - v_b) \neq m a_0 v_a$. This is because, as you said in your title, with constant power, acceleration decreases as velocity increases. Conversely, with constant acceleration, power increases as velocity increases. Ship A clearly has the same acceleration in both frames, and so must be using more power in the one where it is going faster.

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But I don't understand, how can an engine use 2 different amounts of power at the same time? How is that physically possible?? I never said acceleration to be constant, I meant the power to be constant, so accelerations diminishes with velocity. – Zequez Oct 13 '13 at 23:33
Should you not consider the energy gained by both the rocket and the reaction mass being ejected from the exhaust? – DJohnM Oct 14 '13 at 1:00

So, by the equation P=Fv we get that with the increase of speed of the rockets, the force exerted by each of them, diminishes.

This can't be correct since you're completely ignoring the KE of the exhaust products.

Think about it. According to your analysis above, in the instant the rocket engine is first turned on, the speed of the rocket is zero which implies the force on the rocket is infinite!

In fact, the rocket engine is converting the chemical energy of the propellents into the KE of both the exhaust products and the rocket.

In the center of mass reference frame, the frame in which the rocket is initially at rest, we see that, as time increases, the speed of the exhaust products decreases while the speed of the rocket increases and the force on the rocket due to the engine is actually constant.

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But that is completely contingent on the type of fuel used. He could in principle be getting this energy from anywhere, in a way that does not create exhaust products, e.g. from solar panels, from nuclear fission (bringing the fission products along with him), or from some hypothetical process which turns mass completely into energy. – ZachMcDargh Oct 14 '13 at 3:31
@ZachMcDargh, in rocket propulsion, some of the rocket's mass must be expelled in the opposite direction and this mass is the exhaust product. – Alfred Centauri Oct 14 '13 at 10:34
sure, but the same questions could be asked about two cars driving down the road. I'm not clear on why there has to be some kind of exhaust product. – ZachMcDargh Oct 14 '13 at 14:56
@ZachMcDargh, the rocket has nothing to "push against" other than the mass it expels. en.wikipedia.org/wiki/Tsiolkovsky_rocket_equation#Derivation – Alfred Centauri Oct 14 '13 at 15:06
Ok, so, what if it's an electric rocket? – Zequez Oct 15 '13 at 10:26

You correctly quote the two equations P = Fv and F = ma, but then apply them improperly. You also haven't explained what reference frame each of these forces and velocities are relative to.

For a inertial observer watching the rockets fly away, the power does go up as the rockets get faster, assuming the thrust stays constant. Most of the rest of your confusion seems to follow from this error.

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Everything but $a_{AB}$, and $v_{AB}$ is relative to an stationary frame of reference. Also, I meant that the power is constant, it stays the same. – Zequez Oct 13 '13 at 22:07