# Why does the coriolis effect dissapear at the equator?

I'm studying from the book "Classical Mechanics" by Goldstein and from a coursebook my Professor provided me.

In the coursebook, it says that "the Coriolis effect disappears at the equator (Where the rotation vector $\vec\omega$ of the Earth is horizontal)"

Now here's my reasoning: Take the Coriolis term:

$-2m(\vec\omega\times\vec v_r)$

Nearby the north pole, this would cause a force pointing towards the equator if you were moving from west to east. The closer you get to the equator, the more this force starts pointing "upwards" if you were to describe the vector from the surface of the Earth.
At the equator, this vector is perpendicular to the tangent of the equator and pointing outwards.
Intuitively, this would mean that if you move from west to east on the equator you would be accelerating away from Earth.

Now if my reasoning is correct, this doesn't mean the Coriolis effect disappears, only it turns into some sort of centrifugal force, but this seems weird because it's described by another term.

Where did I go wrong?

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The Coriolis force on the equator indeed does point outwards, if you are moving west to east. This is not the same as the centrifugal force, because the centrifugal force is present always - even if you are not moving. But when you move (west to east), there is an additional force on top of the centrifugal force - the Coriolis force. If you travel east to west on the equator, the Coriolis force is pointing inwards. It has practical effects, as can be seen in this video demonstrating how bullet trajectory is curved upwards or downwards when shooting to the eastern or western direction.

The Coriolis force on the equator disappears if you are traveling in the northern (or southern) direction. In that case your direction is (anti-)parallel to the direction of the $\vec\omega$, so their vector product is zero. Maybe this is what Goldstein had in mind.

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According to Wikipedia:

Because the Earth completes only one rotation per day, the Coriolis force is quite small, and its effects generally become noticeable only for motions occurring over large distances and long periods of time, such as large-scale movement of air in the atmosphere or water in the ocean. Such motions are constrained by the surface of the earth, so only the horizontal component of the Coriolis force is generally important.

Since only the horizontal component is usually considered, then both Goldstein's statement and your description can be correct. Goldstein was only considering the non-vertical effects of the Coriolis force.

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You are right: the coriolis effect in the equator is converted to a kind of centrifugal force. But in many situations only the horizontal Coriolis forces are accountes, because in the vertical one the gravity is so much greater.

The vertical coriolis and the centrifugal force may point parallel, but they are still different forces. Let's think of a stationary object just above the equator:

From a fixed observer there is no force at all (gravity does not count!).

From a rotating observer, it would look like the object is moving at the same speed as the Earth rotates, but in opposite direction (east to west): $-(\vec\omega\times\vec r)$. It will be the subject of three different apparent forces:

1. Coriolis force: $F_c = 2m(\vec\omega\times(\vec\omega\times\vec r))$, directed downwards.
2. Centrifugal force: $F_r=-m(\vec\omega\times\vec(\vec\omega\times\vec r))$, directed outwards.
3. The apparent course of the object, from the rotational frame of reference, follows a curved path following the shape of the Earth, so the Earthlings will assume an additional centrifugal force to be accounted, identical in magnitude to the former: $F_r=-m(\vec\omega\times\vec(\vec\omega\times\vec r))$.

These three forces added together results in a net force of 0, as expected.

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The formula for Coriolis force seems incorrect. The velocity vector is missing and there is one extra $\vec\omega$ and one extra $\vec r$. – mpv Jun 9 '14 at 5:35
@mpv: (You missed negative sign also ;-). I assumed an object over the equator at 0 speed in the fixed frame. That is, from the rotating frame the speed will be the same as the Earth itself but opposite sign. That will be $\vec v = -\vec \omega\times\vec r$. – rodrigo Jun 9 '14 at 8:11
Oh, I see. Thanks for the clarification :-) – mpv Jun 9 '14 at 8:42