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Say we have an aeroplane cruising along at constant height with velocity $v_0$. It dives by height $h$ and levels out, at a new velocity $v_1$, in a gravitational field of strength $g$.

If all the gravitational potential energy is converted into kinetic energy, $$\frac 12 v_1^2 = \frac 12 v_0^2 + gh$$ So $$v_1 = \sqrt{v_0^2+2gh}$$ And the increase in velocity $$\Delta v = \sqrt{v_0^2+2gh} - v_0$$

So if $h$ = 1 m, $g$ = 10 m/s^2 and $v_0$ = 0 m/s, $\Delta v$ = 4.47 m/s. If $v_0$ = 10 m/s, $\Delta v$ = 0.95 m/s.

But who determines what $v_0$ is? If we measure relative to the earth, it might be 10 m/s, but we could equally take a "moving" frame of reference where $v_0$ = 0. The gravitational potential energy change is the same in both cases. The different frames predict different velocity increases, but from a physical standpoint clearly they must be the same.

How can this be so?

Lodewijk
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2 Answers2

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If we assume that mechanical energy (K+U) is conserved in both the earth frame and the initially co-moving, constant velocity frame, then it's not the differences in velocities which are the same; it's the differences in the squares of the velocities which are the same. $$\frac{1}{2}v_{1e}^2=\frac{1}{2}v_{0e}^2+2gh$$ and $$\frac{1}{2}v_{1S}^2=\frac{1}{2}v_{0S}^2+2gh$$ where S is some other reference frame in which mechanical energy is conserved, say the one moving horizontally at $v_{0e}$ with respect to earth.

Combining these equations we get $$v_{1e}^2-v_{0e}^2=v_{1S}^2-v_{0S}^2=2gh$$

The difference of squares is NOT in general equation the square of the difference. A check of the OP numbers satisfies these factors.

Next, we can factor either of the first two terms. Let's factor the second: $$(v_{1S}-v_{0S})(v_{1S}+v_{0S})=2gh$$ So the actual difference in velocities in the S reference frame must be $$\Delta v_S = \frac{2gh}{v_{1S}+v_{0S}}$$. With $v_{1S}=v_{1e}-v_{Se}$ and $v_{0S}=v_{0e}-v_{0e}$, a quick check of the numbers will show the equation here for $\Delta v$ to be true.

Bill N
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The problem lies in assuming that the plane maintains all its kinetic energy when it turns. How does a plane pull out of a dive? Roughly speaking, through elastic collisions with air molecules!

The hidden actor here is the air. When the plane pulls up out of the dive, it's not true that

$$\frac 12 v_1^2 = \frac 12 v_0^2 + gh$$

Because some of the plane's kinetic energy has been transferred to the air. You can find out how much by considering conservation of momentum.

Once you consider the momentum, and therefore energy, transferred to the air, the expression for $\Delta v$ you find will be equal in all frames.

This is not just a peculiarity of airplane flight - if you used a rocket, the kinetic energy of the fuel would be the hidden energy sink. If you bounced off of a ramp to change direction, the ramp, or perhaps the entire Earth, would take some of the energy. It's conservation of momentum, and you ignore it at your peril!

This is essentially an interesting variation on this question.

The somewhat confusing analogy between these two problems is as follows: The plane in this problem is to the bullet in that problem. The air in this problem is the plane in that problem. The gravitational energy in this problem was the chemical energy in that problem.

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Brionius
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