# If a Space Shuttle tile deorbited on its own, what percentage of it's mass if any would remain in one piece until it reached the surface of the earth?

If a Space Shuttle tile deorbited on its own, what percentage of it's mass if any would still be in one piece if and when it reached the surface of the earth?

The question was inspired by one of the comments on this question: Is there a very light material that could deorbit without burning up? specifically: "Trivial example: A properly-shaped piece of the tiles used on the Shuttle. – Loren Pechtel May 3 at 5:22".

Also inspired by this comment: "The question is flawed--nothing can possibly shed heat fast enough in the fire because there's no place to shed it to. Surviving the fire is accomplished by deflecting as much heat as possible (carried away in the shockwave) and slowing the entry of what gets through enough that you don't cook before it's over. – Loren Pechtel May 8 at 0:47" to this question: Is there a very light material that could deorbit without burning up?

• Seems to me this question is rendered practically unanswerable by the fact that said tiles were, I'm almost certain, thermally anisotropic. Can we assume, perhaps, that the "tile" is a sphere, say, with a radius of 8 cm, that is constructed like one of the subject tiles (i.e. with the black coating applied)? – Digger May 12 at 22:37

The question is not flawed - radiating heat away requires no medium.

This NASA Tech Brief (NTRS ID: 19750000042) and many other sources give the maximum temperature of the high-temperature, reusable surface insulation (HRSI) tiles as 2300°F (1543 K). I also found this dated website from Purdue that claims a maximum single use temperature of 2800°F (1810 K). I did not find information on how the tiles handle temperatures beyond these limits.

I ran a simulation of an object entering the atmosphere with the following parameters (blunt 60° body assumed because it's easier to model heating and aerodynamics):

• mass: 150 grams (based on sphere of density $$144 \frac{kg}{m^3}$$)
• diameter: 6" (15 cm)
• drag coefficient: 1.5 [1]
• Ballistic coefficient: $$5.7 \frac{kg}{m^2}$$
• Entry speed: $$7.4 \frac{km}{s}$$
• Entry angle: $$-0.01°$$ (trying to emulate slow orbit decay)
• Thermal emissivity: 0.85 [2]
• Sutton-Graves convective heating (radiative heating ignored)
• Radiative equilibrium to find surface temperature ($$\dot{q}=\epsilon \sigma T^4$$)

Here are the results:

The stagnation (analogous to surface) temperature exceeds both maximums for a period of about one minute. I suspect this is not long enough to ablate a significant amount of material, thus most (if not all) should survive.

• Could you explain what that graph means? – Matthew Christopher Bartsh May 12 at 20:19
• @MatthewChristopherBartsh the convective stagnation point heating (in green, right y-axis) is used as $\dot{q}$ to find the radiative equilibrium temperature (using the Stefan-Boltzmann Law), plotted as the solid black line (left y-axis). The dashed black line is the ambient air temperature (left y-axis). The radiative equilibrium gives a decent approximation of the instantaneous air temperature at the stagnation point (referred to as the wall temperature). – BrendanLuke15 May 13 at 11:22
• @MatthewChristopherBartsh This NASA Aerothermodynamics lecture gives an in depth overview of a lot of these ideas ( I suggest downloading it because the URL routinely changes). – BrendanLuke15 May 13 at 11:23