Say you have two planes flying next to each other at the same speed and one decides to pick up speed by burning a tank of rocket fuel.

If someone on the ground wanted to know that plane's new speed after burning the fuel and knew the plane's mass, initial speed, and energy stored in the fuel, he could calculate it using conservation of energy.

If the pilot of the other plane wanted to calculate the same thing but didn't know his own speed, he could calculate the speed of the first plane relative to himself using conservation of energy--the initial relative speed is 0 and all the work done by the rocket fuel is converted into the plane's new kinetic energy.

He could then relay this relative speed to someone on the ground who knew his plane's speed, but adding the speed of the second plane to the calculated speed of the first relative to the second would give a different value for the first plane's speed than the first guy on the ground would have calculated.

I see why this doesn't work mathematically (a^2+b^2 isn't (a+b)^2), but why can't the second pilot calculate the new speed of the first plane relative to himself by converting the stored fuel energy into kinetic energy? Why when on the ground do we not consider our motion relative to, say, the moon when calculating the first plane's speed relative to us, but the second pilot must consider his speed relative to the ground before calculating the first plane's speed relative to himself? Or maybe to put it better, why is having no movement relative to the Earth instead of relative to the plane's initial speed (the second plane) or anything else the correct way to approach this problem?


marked as duplicate by knzhou, Jon Custer, Kyle Kanos, Qmechanic Sep 28 '18 at 15:04

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    $\begingroup$ When an object gets an increase in kinetic energy and speeds up, the "apparent" increase in kinetic energy is actually different depending on whether you're looking at it in a stationary frame versus a moving inertial frame. This may seem very counterintuitive at first glance, and is likely why you seem to have doubts about the physical reasonableness of the conclusion you've drawn, but it's true. However, one minor point is that when an aircraft burns fuel, it doesn't all get converted into kinetic energy, although that's sort of unrelated to the concept you're asking about. $\endgroup$ – DumpsterDoofus Mar 15 '14 at 1:26
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    $\begingroup$ Possible duplicate of Where does the extra kinetic energy of the rocket come from? $\endgroup$ – knzhou Sep 27 '18 at 9:39

The way the combustion of the fuel increases the kinetic energy of the plane is by doing work on it. Specifically the engine produces a force and the increase in kinetic energy is then the force times distance.

Let's ignore air resistance and start in the rest frame of the two planes. Assume the engine burns for a time $t$ and produces a force $F$ (acceleration = $F/m$). In the rest frame the plane starts at rest and accelerates to a velocity v, and the distance moved is:

$$ s = \frac{1}{2} \frac{F}{m} t^2 $$

so the work done is:

$$ W = F s = \frac{1}{2} \frac{F^2}{m} t^2 $$

This must be equal to the change in kinetic energy so we have:

$$ \frac{1}{2}mv^2 = \frac{1}{2} \frac{F^2}{m} t^2 $$

and a quick rearrangement gives the change in velocity as:

$$ \Delta v = \frac{F}{m}t = at $$

Just as we expect.

Now look at the situation as viewed from the ground. If the initial velocity of the planes is $u$ then the distance moved is:

$$ s = u t + \frac{1}{2} \frac{F}{m} t^2 $$

so the work done is:

$$ W = F s = Fut + \frac{1}{2} \frac{F^2}{m} t^2 $$

The initial KE is $\tfrac{1}{2}mu^2$ and the final energy is $\tfrac{1}{2}mv^2$. As before, we equate the change in kinetic energy to the work done:

$$ \frac{1}{2}mv^2 - \frac{1}{2}mu^2 = Fut + \frac{1}{2} \frac{F^2}{m} t^2 $$

To make progress we need to rearrange this to:

$$ v^2 = u^2 + \frac{2F}{m}ut + \frac{F^2}{m^2} t^2 $$

And the trick is to spot that the right hand side can be rewritten as a square:

$$ v^2 = \left(u + \frac{F}{m}t \right)^2 $$

And square rooting both sides gives:

$$ v = u + \frac{F}{m}t $$

Now just subtract $u$ from both sides to get the velocity change and we get:

$$ \Delta v = v - u = \frac{F}{m}t = at $$

And this is exactly the same velocity change as calculated in the rest frame. So both observers conclude the energy from the fuel is the same as the increase in kinetic energy.

The point to note is that neither work done nor changes in kinetic energy are invariant with respect to Galilean transformations. However they cancel each other out to give a velocity change that is invarient.


The idea presented by DumpsterDoofus in the question comment is correct. The increases in kinetic energy is different depending on which (Galilean) reference frame you're in. The mathematical cause of this is the work-energy theorem $W=\Delta K$; work $W$ depends on the displacement of the object, which is different in different frames.

So, since you can calculate different amounts of work in different frames, one doesn't run into the problem of calculating different changes in velocity.


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