# Coriolis force: why is pole-to-equator air flow eastly?

I have little knowledge in fluid dynamics, so this may be naive. But I have a question while reading a textbook about the Coriolis force, by which the rotation of the earth from west to east changes the air circulation pattern.

The text states that

As a ring of air about the earth atmosphere moves toward the poles, its radius decreases. In order to maintain angular momentum, the velocity of air increases with respect to the land surface, thus producing a westerly air flow

The converse is true for a ring of air moving towards the equator -- it forms an easterly air flow.

I can't understand why an easterly flow should be generated in the "converse" part. To me, suppose we have a satellite staying put (w.r.t to earth coordinate system) above the earth. Since the earth is rotating west-to-east, the satellite should move east-to-west (westerly) relative to the earth surface, right?

How does the Earth rotation cause a easterly movement?

-- EDIT --

I was probably confused about the word "easterly", and thought it meant "to the east". As @rob pointed out, it means "from the east". In that case, the question still remains, how does the Earth rotation cause a "westerly" (i.e. west-to-east) movement as described in the first half of the quoted text?

(I really don't see the symmetry in the "conversely" part of the reasoning in terms of physics).

An "east wind" blows from the east; your intuition is correct.

As far as why, in the northern hemisphere, the Coriolis force makes a south wind bend into a west wind, and a north wind bend into a east wind: it's conservation of momentum. An observer on the ice at the North Pole doesn't have any east-west momentum — he can't, because "east" isn't a well-defined direction at the poles. But an observer stationary on the ground at the equator is circling the entire Earth once per day, which corresponds to an eastward speed of about a thousand miles and hour relative to the Earth's center. When an equatorial observer goes north she doesn't lose that momentum; as constant-latitude path around the poles get shorter, her "extra" eastward momentum will bend her path to the east. Likewise an observer heading south from the pole has less eastward momentum than the ground under his feet, so his path bends to the west.

Your satellite analogy is good for the observer leaving the pole. The satellite moves "due south" in some reference frame where the Earth isn't rotating, while the Earth moves eastward underneath it; an observer on the ground would see that satellite coming from the north-east.

For an air mass moving away from the equator, your polar-satellite analogy doesn't work as well — you have to account for the fact that, at the equator, your air mass was stationary with respect to the ground.

• +1: Thanks for the great answer. Your argument using momentum/inertia is better than the satellite analogy I had. Commented Apr 6, 2015 at 23:46

$R = r(\cos A\cos B,\,\cos A\sin B,\,\sin A)$ where $A$=latitude, $B$=longitude and $r$=Earth radius

Let \begin{align}\frac{dB}{dt}&=\frac{2\pi}{24\,{\rm hours}} \\ \frac{dA}{dt}&=v/r\end{align} mass $m$ moving due N with speed $v$.

Differentiate the vector $R$ twice with respect to time to find the acceleration that is experienced by mass $m$. There are 3 component accelerations. The $r\frac{dA}{dt}\frac{dB}{dt}$ acceleration component is the Coriolis's acceleration vector.

• Might be worth clicking on the help link (icon top-right of Answer pane) and reading up on how maths mark-up works.. To start you off, try quoting things in dollar signs: $F=ma$. Commented Apr 25, 2018 at 10:57