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I have been playing around with an app to demonstrate high speed objects entering earth's atmosphere. I am using the drag equation to approximate how much kinetic energy gets converted into heat at each millisecond, which seems to be working OK. However, I also need to figure out approximately what percentage of this heat is dissipated into the atmosphere and how much gets re-absorbed into the object.

I am using a lot of simplifying assumptions in this simulation, like that the earth is a sphere, and (worse) that the atmosphere's temperature doesn't change with altitude, so I am looking for a constant or fairly simple equation to approximate this. I there a reasonable way to do this?

TL;DR: How much aerodynamic heating is left in the atmosphere, vs how much is retained/reabsorbed by the high-speed object?

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  • $\begingroup$ Do you want temperature vs altitude curves? $\endgroup$
    – Organic Marble
    Commented Apr 23, 2020 at 17:10
  • $\begingroup$ @OrganicMarble I think that I can work with that. $\endgroup$
    – RBarryYoung
    Commented Apr 23, 2020 at 18:31
  • $\begingroup$ OK, with your edit, now I'm not going to post an answer. But you could look at pages 10, 11, etc of the US Standard Atmosphere. ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/19770009539.pdf $\endgroup$
    – Organic Marble
    Commented Apr 23, 2020 at 19:33
  • $\begingroup$ It is important that only a little heat energy is reabsorbed by the high-speed object and most of the heat energy is carried away by the atmosphere and the hot gas produced by ablating the heat shield. $\endgroup$
    – Uwe
    Commented Apr 23, 2020 at 19:56

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How much aerodynamic heating is left in the atmosphere, vs how much is retained/reabsorbed

tl;dr: By necessity nearly all of it must be left in the atmosphere or radiated into space or towards Earth, and only a tiny bit could be absorbed.

Using $E = 1/2 m v^2$ and an initial velocity of 7800 m/s, we see that each kilogram of a reentering body starts with 3E+06 joules of energy. Water is a familiar material with one of the higher specific heat capacity materials at about 4200 Joules/kg/K. Using that number and ignoring phase changes it would have to reach 7000 °C to absorb that, and steel at only 500 J/kg/K would have to reach an unphysical temperature of 60,000 °C!

While a heat shield does get toasty, it can not reach anywhere near these temperatures and it represents only a tiny fraction of the mass of the vehicle, so by necessity nearly all of it must be left in the atmosphere or radiated into space or towards Earth, and only a tiny bit could possibly be absorbed.

The atmosphere is heated so high that it glows red hot and since it is somewhat opaque to part of the thermal infrared spectrum becomes an approximation to a black body radiator. So while much of the heat is dumped into the atmosphere a significant fraction is radiated as infrared and visible light, some of which reaches the Earth and some which radiates into space.

See this answer for an overly simplified numerical simulation of a catastrophic reentry.

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