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

John's answer is a good one, I just wanted to add some equations and addition thought. Let me start here:

Heating is really only significant when you get a shock wave i.e. above the speed of sound.

The question asks specifically about a $200^{\circ} C$ increase in temperature in the atmosphere. This qualifies as "significant" heating, and the hypothesis that this would only happen at supersonic speeds is valid, which I'll show here.

When something moves through a fluid, heating happens of both the object and the air. Trivially, the total net heating is $F d$, the drag force times the distance traveled. The problem is that we don't know what the breakdown is between the object and the air is. This dichotomy is rather odd, because consider that in steady-state movement all of the heating goes to the air. The object will heat up, and if it continues to move at the same speed (falling at terminal velocity for instance), it is cooled by the air the exact same amount it is heated by the air.

When considering the exact heating mechanisms, there is heating from boundary layer friction on the surface of the object and there are forms losses from eddies that ultimately are dissipated by viscous heating. After thinking about it, I must admit I think John's suggestion is the most compelling - that the compression of the air itself is what matters most. Since a $1 m$ ball in air is specified, this should be a fairly high Reynolds number, and the skin friction shouldn't matter quite as much as the heating due to stagnation on the leading edge.

Now, the exact amount of pressure increase at the stagnation point may not be exactly $1/2 \rho v^2$, but it's close to that. Detailed calculations for drag should give an accurate number, but I don't have those, so I'll use that expression. We have air, at $1 atm$, with the prior assumption the size of the sphere doesn't matter, I'll say that air ambient is at $293 K$, and the density is $1.3 kg/m^3$. We'll have to look at this as an adiabatic compression of a diatomic gas, giving:

$$\frac{T_2}{T_1} = \left( \frac{P_2}{P_1} \right)^{\frac{\gamma-1}{\gamma}}$$

Diatomic gases have:

$$\gamma=\frac{7}{5}$$

Employ the stagnation pressure expression to get:

$$\frac{P_2}{P_1} = \frac{P1+\frac{1}{2} \rho v^2}{P1} = 1+\frac{1}{2} \rho v^2 / P1 $$

Put these together to get:

$$\frac{T_2}{T_1} = \left( 1+\frac{1}{2} \rho v^2 / P1 \right)^{2/7}$$

Now, our requirement is that $T2/T1\approx (293+200)/293 \approx 1.7$. I get this in the above expression by plugging in a velocity of about $2000 mph$. At that point, however, there might be more complicated physics due to the supersonic flow. To elaborate, the compression process at supersonic speeds might dissipate more energy than an ideal adiabatic compression. I'm not an expert in supersonic flow, and you can say the calculations here assumed subsonic flow, and the result illustrates that this is not a reasonable assumption.


addition:

The Concorde could fly at about Mach 2. The ambient temperature is much lower than room temperature, but the heatup compared to ambient was about $182 K$ for the skin and $153 K$ for the nose. This is interesting because it points to boundary layer skin friction playing a bigger role than I suspected, but that is also wrapped up in the physics of the sonic wavefront which I haven't particularly studied.

You have to ask yourself, what pressure is the nose at and what pressure is the skin at. The flow separates (going under or above the craft) at some point, and that should be the highest pressure, but maybe it's not the highest temperature, and I can't really explain why. We've pretty much reached the limit of the back-of-the-envelope calculations.

(note: I messed up the $\gamma$ value at first and then changed it after a comment. This caused the value to go from 1000 mph to 2000 mph. This is actually much more consistent with the Concorde example since it gets <200 K heating at Mach 2.)


I suspect this is a formidably difficult question to answer from first principles.

The cooling depends on the details of the air flow around your cylinder, and also on the temperature difference. An aeroplane flies at an altitude where the temperature is -50C, so the cooling is obviously going to be very different to say a baseball travelling at sea level. Calculating the cooling would be an exercise in computational fluid dynamics.

Heating is really only significant when you get a shock wave i.e. above the speed of sound. That's because above the speed of sound the air can't get out of the way fast enough so you get a shock wave at the front of the object. This shock wave increases the pressure and therefore compresses the gas and increases its temperature. However, as with the cooling, calculating this from first principles is complex.

If you Google you'll find lots of articles describing such calculations, but they're complex and not very non-physicist friendly. If you just want a simple answer this article reports the breakeven speed for an aeroplane as Mach 1.2.


Another issue with the object getting heated up, apart from the concerns already raised by AlanSE, is that even if you can put the air at the front of the object at 500 K, that doesn't bring the object heat to 500 K, not even after a long time, and certainly over its whole volume !

The main issues the size of the object and the fact that there will heat transfer from the object to the air over a large area. The air is at 500 K only in a small region at the aft of your sphere. So only a small area will experience significant heat transfer. That heat transfer is also pretty slow since it is only due to molecular diffusion (there is no convection since at the stagnation point, where your air is at 500 K, there is by definition no velocity). But alright, let's say we can maintain a constant speed forever. Well you have to heat up the whole sphere which given its mass and heat capacity is going to take a long time. Given that the heat conduction in your object is likely quite high (given its mass), the temperature distribution inside is going to be fairly uniform.

Which brings to my last point. As the rest of the object heats up, the downwind part of it is in contact with air that is much cooler than the 500 K at the front. Not quite 300 K but still, but probably less than 350 K. So the heat transfer there is going to be over a much larger surface and might make your sphere cool down much faster than it heats up. I've just realized, you didn't specify how your object was at 400 K initially with air being at 300 K. Because if anything, based on what have been said, it is likely your object will be cooled by the flow of air... Heating up the object might actually be much easier with a small (dare I say bullet-size?) object than your 1 m sphere.