High altitude, high velocity wind chill?

Question

I know that the air temperature at 35000ft is between -50 and -70 degrees celcius but what i want to know is that if an object is falling at 200 mph (89.40826m/s), what is the wind chill factor and how cold does that make it.

Answer 1

v=200mph
t1=-60C=-76F (use minus 76 deg F), temp reading
w=wind chill factor
(heat loss per unit area due to wind velocity)
w=(10.45-0.447*v+6.686*v^0.5)*(91.4-t1)/1.8
w=1451
t2=wind chill temp in deg F
t2=-0.04544*1.8*(w+91.4)
t2=-126F (minus 126F)
or
t2=minus 88C

Answer 2

The object would approach ambient temperature.

Wind chill is only the amount of cooling on a relatively warm body (such as skin). All those calculations for wind chills make assumptions. Assumptions like skin temperature is equal to 30ºC and a specified amount of clothing is worn.
It comes up with a wind chill like minus 20ºC, for example, which would be equivalent to the amount of heat lost from a warm person walking at 5 km/h in an actual temperature of minus 20ºC with no wind.

Other factors to consider:
The heat capacity of the object will affect how quickly it approaches the ambient temperature.

The air is thinner at higher altitudes and therefore has less thermal conductance.

For very fast objects, friction between the air and the object becomes significant enough to warm the object to a significantly higher than ambient temperature.

Answer 3

Wind chill does not affect objects . It would affect u because your skin has moisture on it and there will be a chill affect on how much moisture evaporates.

Answer 4

It will depend on the ambient temperature at the time you fall. However, I doubt it would get any colder than -70C

Answer 5

well you have to take into acount that the air is thinner so therefore you will have less wind chill

Answer 6

http://www.nws.noaa.gov/om/windchill/index.shtml


here has a calcultor on the bottom