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I know that capsules typically require heat shields to survive reentry from orbit. I'm wondering how an object's size, density and aerodynamic profile affects this.

For more specificity:

  • The feather would be the tail feather of an adult pigeon
  • The orbit would be that of the ISS at +-7.6 km/s
  • By "drop", I mean throw downwards toward earth at 5 m/s so that it de-orbits in a reasonable timeframe
neelsg
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    An African or a European pigeon? – Mark Adler Jun 15 '15 at 13:34
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    @MarkAdler Not sure if you're actually asking in earnest, but let's say a Common wood pigeon with tail feather of 14 cm – neelsg Jun 15 '15 at 13:46
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    A wooden pigeon would certainly burn up. – LocalFluff Jun 15 '15 at 13:48
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    Your premise that throwing it downward at 5 m/s is flawed. In fact, it would be rather difficult to make it re-enter in a reasonable time., although it would happen eventually... – PearsonArtPhoto Jun 15 '15 at 13:48
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    @MarkAdler The reference is actually to an African or a European swallow, which i had to say just to nerd it up. – kim holder Jun 15 '15 at 14:15
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    I'm glad that at least someone got the reference. – Mark Adler Jun 15 '15 at 14:21
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    See also: https://en.wikipedia.org/wiki/Paper_plane_launched_from_space – szulat Jun 15 '15 at 14:31
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    If you were to try "throwing it downwards", you would simply increase the eccentricity of the orbit. Since the "impulse" of 5m/s is downwards (and hence perpendicular to the direction of motion), the impulse does not change the amount of energy in the feather. – Aron Jun 15 '15 at 16:30
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    How many pigeon feathers would you have to eject at 5 meters per second from the space station every minute in order to make up for its drag in the atmosphere? Would this count as "flying"? – Adam Davis Jun 15 '15 at 19:46
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    Note to @Aron: A perpendicular $\Delta V$ does change the energy, but by much less than a parallel $\Delta V$. The latter changes the energy by a factor of $1\pm{\Delta V\over V}$, whereas the former changes it by a factor of $1+\left(\Delta V\over V\right)^2$. – Mark Adler Jun 15 '15 at 20:42
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    @MarkAdler second order effects where, dV << V. – Aron Jun 15 '15 at 23:57
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    We know dE = F . dx. How then does the impulse change E? It first needs to change the direction of dx. Second order effects – Aron Jun 16 '15 at 00:04
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    unladen swallow – qneill Jun 16 '15 at 00:52
  • @LocalFluff: Perhaps a clay pigeon would do better? – Vikki Jun 23 '18 at 01:08
  • I truly and genuinely propose that NASA conduct this experiment for educational purposes. It would be fascinating. The cost of launching the mass to orbit would not be exorbitant, and an astronaut could simply release it as gently as possible on their next spacewalk. Certainly someone here has a NASA contact that can make this happen?? :) – pbristow Mar 07 '22 at 23:25

1 Answers1

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Throwing it down at 5 m/s will do basically nothing. That will simply cause it to advance in its orbit a bit. To deorbit, you need to throw it backwards, not down. However in this case, since the feather has a such a low ballistic coefficient, it will promptly deorbit from ISS altitude on its own, without you having to do anything at all. Just wait a bit.

Given a 10 cm feather with a mass of 0.05 g, I looked at two cases intended to bound the possibilities. The first case is that the feather trims to a face-on attitude, with the lowest possible ballistic coefficient. It reenters from the ISS altitude in less than three hours. The maximum deceleration is about 10 G's at 100 km altitude. My heating estimate is highly suspect (I am applying a blunt body formula with a nose radius based on feather feature sizes), but my rough estimate is $30\,\mathrm{W/cm^2}$, which is rather a lot for organic material. The feather would not look like a feather anymore.

However it is not clear how it would maintain that attitude. More likely would be a trim with the heavier part of the root of the feather forward. For that I used about 1/40th of the previous $C_D A$, measured from a Vulture feather. (Albeit at completely the wrong Reynolds numbers, but hey, this is just for fun.) Then I get that it decays from the ISS orbit in less than four days, with a maximum deceleration of 8 G's at about 80 km altitude. The heating is much worse, at $200\,\mathrm{W/cm^2}$. That is a typical Mars entry heat rate. The feather would be gone.

If anything, I suspect that my choice of approach and parameters underestimates the heating. So my conclusion is that, alas, the feather will burn up. And I had such high hopes for the feather.

Mark Adler
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