I know this sounds basic but I've wondered about this for a long time and I'm still confused.

In aviation, air speed is ground speed minus wind speed. If wind speed is zero, air speed equals ground speed.

But the earth is rotating 360 deg/day. At sea level this rotation imparts to you an eastward linear velocity of just over 400 m/s. The atmosphere is generally close to static at the surface, so the air moves at this speed also with the rest of us.

But what happens at 25 km above ground? If all winds magically disappeared, would you find the air move eastward at >400 m/s (with a tiny correction for the increased elevation of 6360 + 25 km instead of just 6360 km at sea level)?

This seems to be a requirement if air speed is to be calculated as ground speed + wind speed, since both speeds need to be relative to the same reference (the ground), so I want to say yes... but can someone confirm or dispute?

Thanks! And apologies for littering with basic questions like this. I should know this, but I don't.


1 Answer 1


I think you’re mixing up frames of reference a bit here. That ~400 m/s speed of the surface is relative to an absolute, non-rotating frame of reference in which an observer sees the Earth as spinning on its axis. In meteorology we typically use a rotating frame of reference that spins with the Earth, in which an observer sees the surface as moving at 0 m/s. When we talk about wind, we talk about air motion relative to this rotating frame. You essentially say this in different words:

… since both speeds need to be relative to the same reference (the ground)...

So, if all winds disappear (i.e., wind speed = 0 m/s) then all the air is moving with the rotating frame by definition. There’s no meteorology or fluid dynamics invoked here, this just falls out from our chosen coordinate system and definition of “wind”, so it would be the same at 2 m or 25 km altitude.

NB Your definition of air speed changes from "minus" in the first para to "plus" in the last para; I've assumed that this was a typo.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy