Lift due curvature of the earth?

[ATCast]

The ratio of the Schuler time to the time that it takes to travel around the world is not the percentage of lift that you get for free.

The centripetal force is given by F = m * V^2 / r
Where:
m - aircraft mass
V - aircraft speed
r - radius of the earth

Lift is given by L = m *g
Where:
g - gravitational acceleration

The percentage of Lift "generated by curvature of the eart"
V^2 / (g*r) * 100

Substituting:
r =~ 6400000 m
V =~ 280 m/s
g =~ 9.81 m/s

Gives 0.12%

My guess is that is within the error margin of most measurements and calculations and therefore it is probably not used.


For validation of the above you can calculate the "Schuler speed" and substitute that:

V =~ 40000000 / (84*60) =~ 7936 m/s

Gives 100.31% ~ 100% Q.E.D.