Re: Speed and temperature of metorite on impact?

emilio,
If the Earth had no atmosphere, the answer would be pretty simple.  Simply 
consider that the Earth has a potential well of the form 

U=-GMm/r

where G is Newton's gravitational constant, M is the mass of Earth, m is 
the mass of your object, and r is the distance from the Earth's center.  
Assume that the object starts with no net energy.  By conservation of 
energy, the object's kinetic energy (1/2*m*v^2) plus its gravitational 
potential energy must always be zero.
1/2mv^2-GMm/r=0
which through a little bit of algebra yields a velocity of
v=sqrt(2*G*M/r)  (independent of the mass of the object)
Now, plug in the radius fo the Earth for r, and we get a value of 11,173 
m/s as the impact velocity.

However, as you rightly noted, there is air resistance to consider.  
Let's first assume that as the object enters the atmosphere, it is 
moving close to 11,000 m/s.  Let's use the formula for the air friction 
force for fast moving objects:

F=(1/2)*C*rho*A*v^2

Here, rho is the density of air, A is the cross-sectional area of the 
meteorite, v is the velocity, and C is a factor dependent on the geometry 
of the object.  
Spherical objects have a value of C of .5, so I will use .7 due to the 
irregular shape of a typical meteorite.

Terminal velocity occurs when the friction force balances the gravitational 
force downward.  1/2*c*rho*A*v^2 = m*g

Here, we neglect the ablation of the meteorite, that is, the blasting off 
of particles from the surface due to collision with air molecules.  For 
objects of this size range, that is a significant effect.  Let's just 
assume that, to first order, we can do this calculation anyway.

I ran a few calculations for a baseball-sized meteorite (r=3.6 cm). 
I simply used Newton's second law:
 m(dv/dt)=1/2*C*rho*A*v^2-m*g 
(This was done numerically, though it would not be difficult to solve the 
equation exactly if you so desired.)
If you have a baseball-sized meteorite of density 3.2 g/cc, using a value 
of 1.2 kg/m^3 for the density of air, you will find that the meteorite will 
slow from its approach velocity of roughly 11000 meters per second to its 
terminal velocity of 60 m/s in a mere 28 seconds, having traveled only 3 
km.  (By comparison, the speed of sound is roughly 315 m/s.)
It then spends another 100 mins or so falling before it hits the ground, 
giving it ample time to cool down below its original temperature it gained 
during entry into the atmosphere.  
(At 60 m/s, it's moving like a fastball, but not much more.  It'll still 
cause a lot of damage if your car or house is in the way, but it wouldn't 
start a fire or create any appreciable crater.  It would probably be a bit 
warm to the touch.


---Bob Macke
MIT S.B. '96, Physics
 in St. Louis Ph.D. candidate, Physics

References:

Serway, "Physics for Scientists and Engineers," Saunders College Pub., 
Philadelphia, 1990, p. 139. (Explanation of air resistance)