I know that if an object moves in the air, it can experience two types of drag, laminar and turbulent. For instance, I have a meteor of ideal spherical shape falling from edge of space, say, 100 km up at the Earth surface with initial velocity of 1 km per second. I would consider turbulent drag, but is it still applicable at supersonic velocities? How do I estimate meteors velocity when it hits the ground?
[Physics] Air friction at supersonic velocities
airdragfrictionmeteors
Related Solutions
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.)
Imagine a blunt object like a space-capsule entering the atmosphere. It experiences a decelerating force, right? If you divide this force by the surface area of the blunt front facing surface, we get an effective pressure. The atmosphere has to create this pressure in front of the object, otherwise there would be no force (a molecule streaming by the sides of the object without hitting it can't create such a force). This compression also heats the gas in front of the object.
The hot, dense gas now streams along the sides of the capsule. If we want to keep the capsule cool, then we certainly don't want this hot gas to touch the body again, which is why capsules are entering with the broad side and are not flying like planes with a sharp nose cone. The body angle has to be small enough that the gas can pass the entire body before it expands enough to reach the walls.
You can see these effects nicely in old NASA images showing the supersonic bow shock around models of their capsules, the dark areas are dense gas under high pressure, the light ones are less dense, low pressure regions:
The majority of the kinetic energy in the capsule will be converted to heating of gas in this bow shock, only a fraction of it will be absorbed by heat shield and an ever smaller amount will heat the backside walls. Without this phenomenon re-reentry would be an even harder thermal problem than it already is.
Best Answer
The drag force, $F$, on the meteor is given by:
$$ F = C_d \frac{1}{2}\rho v^2 A$$
where $\rho$ is the fluid mass density, $v$ is the meteor speed, $A$ is the meteor cross-sectional area, and $C_d$ is the drag coefficient (discussed below).
The fluid mechanics of meteor reentry is quite complicated. Over the important ranges of altitude, the meteor is supersonic and a bow shock forms in front of the meteor. Consequently, the flow adjacent to the meteor is subsonic. The flow field is therefore non-trivial.
For a quick answer, though, meteor trajectories can be calculated by assuming $C_d=0.7$. If you want to be more accurate, a plot of $C_d$ vs far-field mach number can be found in this paper: "ESTIMATING THE DRAG COEFFICIENTS OF METEORITES FOR ALL MACH NUMBER REGIMES" by R. T. Carter, P. S. Jandir, and M. E. Kress:
From the plot, you can see that the rule-of-thumb value of 0.7 underestimates the drag at high speed and overestimates at low speed.
You will also need to know atmospheric density vs altitude. A NASA report from 1976 defines the standard values for this and is available online here. There are also online web calculators, such as this one, that are based on the same report. (For data above 86 km, see this report.)