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A follow-up to After what speed air friction starts to heat up an object?

I understand there may be technological limitations at present ... but, is it theoretically possible for a body to travel through the atmosphere so fast that the shock wave transforms to plasma? What velocity would a body need to achieve for this to happen?

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I considered two different approaches to this problem.

An obvious solution method would be to use the same method found in the linked air friction question (see "internal references" at the end). That is, consider the leading edge to cause an adiabatic compression of the air. The pressure can be found from the velocity, which then implies a temperature since the properties are known. You know what temperature air will ionize itself, so working backwards you can find a velocity. Here is what I obtained in miles per hour:

$$ T_2 = \frac{1086.4 kJ/mol}{N_A \frac{3}{2} k} \approx 87,000 K$$

$$\frac{T_2}{T_1} = \left( 1+\frac{1}{2} \rho v^2 / ( 1 atm) \right)^{2/7}$$

$$v = \sqrt{ \left( \left( \frac{87,000}{293} \right)^{7/2} -1 \right) \frac{2 (1 atm)}{1.3 kg/m^3} }$$

$$1.9 \times 10^7 mph$$

(note: this was revised down from 4e7 because I previously neglected the 3/2 factor on k)

FYI, this can also be written as $0.025 c$, which sounds ridiculous. It seems doubtful that the physical mechanisms would continue to work in the same way as you go so many times faster than the speed of sound, and the velocity of the air molecules themselves. We continue in search of alternatives...

Here is another answer, using the approach suggested in the comments. That approach is using this compressible flow equation. This is an alternative to the above solution.

$$ \frac{v^2}{2}+\frac{\gamma P }{(\gamma-1) \rho} = \text{constant}$$

We have to make a substitution from the ideal gas law, $P/\rho = R_{specific} T$. The specific gas constant for Air is 287.04 J/(kg*K). That yields this equation.

$$ \frac{v_1^2}{2}+\frac{\gamma R_{specific} T_1 }{(\gamma-1) } = \frac{v_2^2}{2}+\frac{\gamma R_{specific} T_2 }{(\gamma-1) } $$

Everything above is known except for $T_2$. Solve for that temperature, which comes from my first equation in this answer, or whatever number you want to use for the onset of ionization of air. The answer:

$$ 29,354 mph$$

Dare I say, this looks like the right way to answer the problem. I should qualify, I omitted a $\sqrt{-1}$ factor. As you can see, the $T$ and $v$ in the equation appear to be on the wrong sides. I haven't figured out a way to argue this away, so I felt I should report it.

A problem that struck me as similar was the relativistic baseball (see internal references) question. In this question a baseball magically starts moving at $0.9 c$ and it then follows that the air particles have sufficient energy to cause nuclear fusion on the surface of the baseball. This can be said without using the air thermodynamic properties of air because for the purposes of the problem, air molecules are basically stationary. The current problem could basically reduce to the same thing, with the exception that instead of nuclear fusion, we are talking about electron ionization. This is an interesting dichotomy because one way treats the problem with gas dynamics, and the other treats it as a transport problem. Qualitatively, it could be difficult to argue for one method over the other. After all, the high pressure region on the leading edge of the object is basically a cushion to the next row of gas molecules.

The counter-argument to that air is mostly empty space. It could be that the majority of gas molecules don't even "see" the high pressure region in front, so they can't be affected by it. The path length of air molecules is $ \lambda_{mfp} = \frac{kT}{P \sigma} \approx 282 nm$. Combined with a speed in atmosphere of about $500 m/s$, that means the time between collisions is about $0.5 ns$ (other sources give 0.2). Let's compare to, say, a space capsule reentry vehicle of $3 m$ diameter, if it travels at $0.05 c$ that means that by the time a Nitrogen molecule collides once, it has traveled about $0.75 cm$. It seems that even at that speed, we can't credit direct collisions between the air molecules and the craft (unless it's really tiny). Nonetheless, some N2 molecules would probably be hitting the surface, and at this energy they would ionize the living daylights out of the surface.

Let's continue with this calc anyway. Just take the ionization potential of Carbon, which may be used in the structural materials (the material doesn't matter all that much), and you have $1086.4 kJ/mol$, which translates to an energy of about $11 eV$ per atom. Use the molecular mass of N2 and you can get a velocity. Ignore the thermal motion within the gas (the object will be flying much faster), and that's the speed the craft will be traveling at in order for an N2 molecule impinging it directly to ionize some of the surface. With this method I get:

$$ 16,500 mph$$

Incidentally, this is almost exactly orbital velocity. However, I should qualify that this has limited convincing power. The high pressure region on the front of an object moving in atmosphere will only extend outward to a length on the scale of its diameter, and at this speed molecules have plenty of time to collide with each other in that distance. This is still a better answer than my first number, so I can say I've narrowed it down to 3 orders of magnitude.

What does NASA say about this? Well, they think that ionization happens in reentry.


But now, NASA, is it really a dense plasma, or is it one of those lame sparse plasmas? I would be inclined to think the latter. Not all molecules have the same speed, and in the chaotic mix near of gas near the thermal shields of a reentry craft I don't doubt that there is some ionization taking place.

Are there some other physics about supersonic flow that could further narrow this down? There could be, but I don't think it's likely. Even if you're supersonic, the drag still roughly follows $1/2 \rho v^2$, which dictates the basic energy balance. Sparse ionization should happen well before the wavefront reaches temperatures that we typically think of as the transition to plasma. That is more-or-less what I get from the NASA reference.

For the naive calculation to clearly say the air becomes plasma, the craft itself has to travel on the order of the desired plasma temperature, which is around $100,000 K$, which leads to my first number.

Internal References:

Air friction question: After what speed air friction starts to heat up an object?

Relativistic baseball: “Relativistic Baseball”

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Given that the Space Shuttle create a plasma during reentry, I doubt your calculation of 4E7 mph was correct... I'm looking for a chart that shows Cp and $\gamma$ of air vs Mach number and altitude but I can't find it... –  FrenchKheldar Nov 12 '12 at 11:57
Here it is, sort of. I can't link to the chart I was thinking about from Anderson, High Temperature Gas Dynamics, but this should do Check Fig. 1.1, they put the ionization limit at around 10 km/s which seems higher than what the Space Shuttle reaches, and yet I'm pretty sure there is a plasma cloud around the SS. I will research this further... –  FrenchKheldar Nov 12 '12 at 12:07
I don't understand your first calculation. Ignoring real gas effects, the stagnation temperature is 10 eV at a Mach number of $\sqrt{(400-1)\frac{2}{1.4-1}}\approx 44$, a speed of about 13 km/s. Though, for communication purposes, ionization starts being significant at Mach 10. –  mmc Nov 12 '12 at 12:16
@mmc What you're referring to is the approach for the 2nd calculation. The first calculation assumes the gas is compressed until the temperature reaches the ionization temperature. That's why it's so high, and that's why I didn't just ignore it when I got the other answer. The difference between the two as I understand it is basically between some ionization versus near 100% ionization. –  AlanSE Nov 12 '12 at 13:16
@AlanSE You are using the wrong Bernoulli Equation :) The compressible version must be used for transonic and supersonic flows but it's easier to get the temperature directly by using conservation of enthalpy: $c_p\,T+\frac{v^2}{2} = c_p\,T_s$. –  mmc Nov 12 '12 at 23:00
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