How much heat is reabsorbed by a high-speed object?

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?

• Do you want temperature vs altitude curves? Apr 23, 2020 at 17:10
• @OrganicMarble I think that I can work with that. Apr 23, 2020 at 18:31
• 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 Apr 23, 2020 at 19:33
• 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.
– Uwe
Apr 23, 2020 at 19:56

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!