What equations/constants were used to calculate the Kármán Line for Earth?

Question

I am interested in how the original value of ~100 km was calculated for the Kármán line of earth. What equations and constants would need to be used to reproduce this value?

Note: By constants, I'm not referring to things like the gravitational constant but am referring for instance to the wing area in the equation for lift coefficient.

Answer 1

In principle calculating the latitude of the Kármán line for any planet is simple. For some planet of mass $M$ the orbital speed $v_o(r)$ at a distance $r$ from the centre of the planet is just:

$$ v_o(r) = \sqrt{\frac{GM}{r}} $$

where $G$ is the gravitational constant. So far so good, though from here it gets a bit messier. If you're flying a plane with some mass $m$, then the gravitational force pulling you down is (this is just Newton's equation):

$$ F_g = \frac{GMm}{r^2} $$

and using the expression for the lift from the Wikipedia article you mention, the lift opposing the gravitation force is:

$$ F_{lift} = \frac{1}{2} \rho v^2 A C_L$$

To work out what speed you need to fly you just set $ F_g = F_{lift} $ and you get (after a bit of rearrangement):

$$ v_{fly}(r) = \sqrt{\frac{2GMm}{\rho A C_Lr^2}} $$

The Kármán line is the height at which $v_o(r) = v_{fly}(r)$. Setting these equal and removing the common factors of $G$, $M$ and $r$ gives:

$$ \frac{2m}{\rho(r) A C_Lr} = 1 $$

This is a simple equation and solving for $r$ gives you the height of the Kármán line. The problem is that the atmosphere density $\rho(r)$ is a function of height, and it varies in a non-trivial way because it depends on the temperature and the temperature varies in a non-trivial way with height. To work out the Kármán line on an extraterrestrial planet I suspect you'd need to measure the temperature as I doubt it would be easy to predict it.

I note that the mass of the plane $m$ appears in the formula. I'd guess Kármán took some representative value of the ratio of the plane's mass to the wing area, $m/A$, and used that in his calculation.