Pre-parachute air friction equal to post-parachute?

[DocZaius]

I was thinking of the free body diagram of a skydiver and came across something which bothered me.

Fg = Force of gravity
Ff = Force of air friction

1. The skydiver jumps out. Fg (constant) pulls down and Ff increases until terminal velocity is reached. At that point, the skydiver is falling at a constant speed and Ff = -Fg

2. The skydiver opens his parachute. Momentarily, Ff is greater in magnitude than Fg and the skydiver accelerates upwards, losing some of his falling speed until a new terminal velocity is reached.

3. The skydiver is now falling at a constant speed but with a different profile. But since he's falling at a constant speed, Ff = -Fg.

Since Fg is constant, that means step 1 and step 3's Ff's are identical. But why is the force of air friction not higher with the parachute open?

I am sure I am missing some stupid obvious thing.

edited to add: I think I figured it out. After the parachute has opened, even though there are more air molecules hitting the skydiver and parachute, they are doing so at a smaller speed, thus exerting the same total force as pre-parachute. I had erroneously gone through life thinking air friction post-chute-opening was greater. By the way, please let me know if this explanation is correct. Thanks!

 

[arildno]

You've got it.

If we model the frictional force as:
f=c_{f}v^{2} with c_{f} being the coefficient of friction, v the velocity, then in the latter case, its c_f is greater than in the first.

This can be translated into your language that a higher number of molecules hits the open parachute, but at a lower speed.