120,000 feet...

For a normal person, it represents the distance travelled during a fairly long commute to work. For Felix Baumgartner, the Austrian daredevil, it represents the altitude from which he plans on free-falling towards the Earth this coming Monday, October 8, 2012.

For someone like myself,

*any* height is too high to jump from with nothing but a parachute to save me from death. However, even sky divers, who are themselves barely sane, see Baumgartner's jump as nothing short of lunacy.

You see, 120,000 feet is 36,576 m - that's more than 36 kilometers! To put this into perspective, his descent will begin at an altitude that is three times that at which typical commercial airplanes fly. It is above the troposphere, in the middle of the stratosphere. So, "How will he get there?" you ask. Why, he will wait inside a man-sized pod that is lifted by a large balloon, of course. When the balloon reaches the correct altitude, the pod will open, and down he will fall.

There are literally countless risks associated with this particular sky dive that aims to crush the previous altitude record of 102,000 feet.

To begin with, the air way up there is extremely cold. Should Baumgartner's special suit fail even a little, the convection associated with the high speed sub-zero air flowing by him will freeze him almost instantly.

Not only is it much colder up there, but the air pressure is just 1% of that on the surface. For this reason, the daredevil will have an oxygen tank strapped to him. And, with this pressure change, comes a change to the most important environmental factor when it comes to aerodynamics: fluid density.

The density of air on the surface of the Earth is about 1.2

kg/m^{3}. In the middle of the stratosphere, it is more like 0.01

kg/m^{3}. A quick application of Newton's second law shows that this has a dramatic effect on the terminal velocity of the dive...

**Terminal velocity**
A sky diver reaches his or her terminal velocity when his or her body ceases to accelerate. When this inertial term vanishes, we are left with a simple force balance: Drag force = Gravitational force. The force of gravity can be approximated as

*mg* (even at an altitude of 36 km, using the gravitational acceleration one experiences on the surface of the Earth, 9.8

m/s^{2}, introduces very little error). The drag force is a bit more

complex, and is given by:
Drag force = (1/2)*ρC*_{D}Av^{2}

In this expression, *ρ *is the density (kg/m^{3}) of the fluid, *C*_{D }is the drag coefficient (unitless) of the falling body, which is essentially a measure of how aerodynamic it is (it is greater for objects that are not streamlined), *A* is the projected surface area (m^{2}) of the body, and *v* is the relative velocity (m/s) of the body with respect to the fluid. It is clear that drag is largest when large objects move within dense fluids at high speeds. This is why we can often ignore drag for, say, a ball that is tossed through the air by a child.
Substituting these parameters into the force balance and solving for

*v*, we get the terminal velocity equation as follows: