[FiatLuxDave]

Hmm... the paper plane has an initial kinetic energy of (0.004 kg) * (7800 m/s)^2 / 2 = 121.6 kJ. But it is supposed to experience 10^5 W of heating for several minutes? The atmospheric heating comes from converting the kinetic energy to thermal energy. What am I missing?

[mjw1007]

It's supposed to be 10^5 W heating per square metre, and the aeroplane's cross-section is much smaller than a square metre.

[FiatLuxDave]

This is the best answer. The area of the plane is 0.0196 m^2, so the heating of 10^5 W/m^2 is actually around 2 x 10^3 W, which seems much more reasonable.

Also, I noticed that I missed that the plane is made from 4 sheets of A4 paper. Table 1 lists the mass as 4 grams, but 4 grams is typical for a single sheet of A4 paper, so the listed mass is probably for a single sheet. The actual plane mass is likely 16 grams. This means that the kinetic energy is likely closer to 480 kJ.

Thank you to afeuerstein for pointing out that I was missing the potential energy! However, that is not enough to make a huge difference. A quick estimate because I'm lazy is 9.8 m/s^2 * 480 km * 0.016 kg = 75.2 kJ. Yes, gravity decreases slightly as you get farther out, so this is an over-estimate.

So a total of around 550 kJ, and a power around 2 x 10^3 W gives a duration of 275 seconds or... a couple of minutes. I feel much better about the numbers now.

[chris_va]

Not that the paper was clear, but I assume that is the maximum heat rate across the entire airplane. If only 1/4 of the plane sees the heating (say wings), that adds up about right vs the kinetic energy.

[afeuerstein]

You are missing the potential energy.

[SAI_Peregrinus]

That adds roughly (6.674e-11(m^3)(kg^-1)(s^-2) * 5.972e24kg * 0.004kg) / (6380km + 420km) = 234kJ. So 355kJ total. That's 5.9kW over 1 minute, or 590W over 10 minutes, etc. It's 10^5W over 3.55s, not several minutes. So either the power is incorrect or the time is incorrect.