Relative to the Earth? When your upwards acceleration equals the acceleration of the Earth.
But, I've said this already.
Made this using LaTex and I forgot to resize the page so sorry
Your 'FE' equation is wrong. Since you let us all have a good laugh, I'll go ahead and post the derivation for you.
The drag force on a body in a fluid is given by
F = .5*C
d*v
2*A*rho (1)
where C
d is the drag coefficient, v is the velocity of the object, A is the cross sectional area of the object and rho is the density of the fluid.
We know from Newton's Second Law of Motion that
F = m*a
o (2)
where F is force, m is the mass of the object and a
o is the acceleration of the object. Rearranging (2) gives us
a
o = F/m (3)
Combining (3) and (1) gives
a
o = C
d*v
2*A*rho/(2*m) (4)
Now, an object that has reached terminal velocity, v
t, has no relative acceleration to the Earth. Applying this to the FE, that means that the object must have an upwards acceleration equal to that of the Earth's. It follows then, that
a
e - a
o = 0 (5)
where a
e is the acceleration of the Earth and a
o is the acceleration of the object.
Rearranging (5) gives
a
e = a
o (6)
which, by examination, is correct for all objects whose height is not changing relative to the Earth (aircraft for example).
Combining (6) and (4) leaves us with the equation
a
e = C
d*v
t2*A*rho/(2*m) (7)
Performing simple algebra on (7) yields the equation
v
t = sqrt(2*a
e*m/(Cd*A*rho) (8
)
which can easily been seen to be the exact same equation as the one for the RE, with the exception that a
e refers to the acceleration of the Earth and a in the RE equation refers to the acceleration due to gravity.
Using your numbers, we see that the terminal velocity on the FE is 22.1426 m/s.
p.s. Your RE equation is wrong.