Hot bullets and shot cowboys Please excuse the title of the question, there is physics involved here, but I can't think of a way of writing the title without it being more to the point of the question.
A cowboy film is pure drama, and I don't normally expect any correct physics to come from Hollywood.  My question is based on watching old "Gunsmoke" episodes during which, each time an actor is shot, even if he is hit in the body far away from a vital organ, the movie doctor says words to the effect "That bullet has got to be removed now".
Far more important though, is that this may also be the case today, I just don't know.
My question is: if the bullet is powered by a very hot stream of gas to fire it, after which it travels through the air at an average speed of  1,700 mph (2,740 kph), where friction from the air slows it down, (but may also heat it up), then the bullet may be essentially sterile in the short time it takes to hit the "cowboy", and unless it's lodged near the heart say, I wonder am I correct in assuming there is no rush to remove it? 
In fact, the bullet may be so hot, it may cauterize the affected area. 
Over longer distances, say from a rifle shot, this may not occcur and the bullet may well be air cooled enough be a source of infection. This is simply a guess on my part though.
 A: Assuming that the bullet flies at Mach 2.0, the temperature at the front of the bullet is about the same as the surrounding air (since at that relatively low Mach number, there is little thinning of the boundary/shock layer to cause aerodynamic heating). I do not know how much heating there is from the gases in the chamber/barrel (it may be significant, but as you suggested, with a long distance shot, the temperature will settle back to ambient).
One reason suggested for removing bullets like that is that there may be cloth inside the wound, "brought along" by the bullet on entry, which is a great place for pathogens to breed/enter the body.
$$\frac{T_{\text{bullet}}}{T_\infty} = \frac{T_{M=2.0 (\text{pre-shock})}}{T_{M = 0}}\cdot \frac{T_{\text{post-shock}}}{T_{\text{pre-shock}}}\cdot \frac{T_{\text{bullet}}}{T_{\text{post-shock}}}$$
$$\frac{T_{\text{bullet}}}{T_\infty} = \left(\frac{5}{9}\right)\cdot 1.6875\cdot \frac{1}{0.9375} = 1$$
A: A basic question is - how hot does a bullet actually get? Here is an answer. Note that the biggest effect is frictional heating in the barrel lands, which produces peak temperatures of ~320 C. Also note, though, that the affected areas are quite small in proportion to the overall bullet size, so the average bullet temperature rise is also quite small. Although the propellant gases are quite hot, they only contact the base of the bullet for a few milliseconds, which is not time for the heat to penetrate very far into the projectile. 
As has been remarked, the big problem with bullet wounds which are not immediately fatal is the introduction of dirt and cloth, and getting debris out of the wound is important to avoid infection. But it's not urgent if you can't go in and actually repair organ damage.
