# Coriolis Force in East-West Direction

• person123

#### person123

TL;DR Summary
Why does the Coriolis Force apply on objects travelling east or west?
For objects traveling north or south, the Coriolis force makes sense. The velocity of the Earth's surface changes with latitude, causing an object to drift east or west relative to the Earth's surface.

However, when an object is traveling east or west, the latitude is not changing. How would the Coriolis force still apply?

Summary:: Why does the Coriolis Force apply on objects traveling east or west?

For objects traveling north or south, the Coriolis force makes sense. The velocity of the Earth's surface changes with latitude, causing an object to drift east or west relative to the Earth's surface.

However, when an object is traveling east or west, the latitude is not changing. How would the Coriolis force still apply?
The Coriolis force is more complicated than you imagine. The term is ##-2m(\vec \omega \times \vec v)##, where ##\vec v## is the velocity relative to the rotating reference frame and ##\vec \omega## is the angular velocity - this will be in the north-south direction for the Earth. Note that this means that for an object moving north or south at the equator, the Coriolis force will be zero, but not for an object moving east or west.

See:

https://en.wikipedia.org/wiki/Coriolis_force

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vanhees71
##\omega## is the velocity relative to the rotating reference frame
According to the wikipedia article:

"
is the angular velocity, of the rotating reference frame relative to the inertial frame"

which I think is quite different.

It's hard for me to follow the rest of your post because I'm not sure what was meant by ##\omega##.

At any given point the Coriolus force is perpendicular to the direction of movement. In the northern hemisphere the Coriolus force causes a turn to the right. Free floating buoys tend to move in clockwise circles. This is because the surface of the Earth is moving to the left.

The force is zero at the equator.

According to the wikipedia article:

"
is the angular velocity, of the rotating reference frame relative to the inertial frame"

which I think is quite different.

It's hard for me to follow the rest of your post because I'm not sure what was meant by ##\omega##.
It was a typo. ##\vec v## is the velocity, of course.

person123
At any given point the Coriolus force is perpendicular to the direction of movement. In the northern hemisphere the Coriolus force causes a turn to the right. Free floating buoys tend to move in clockwise circles. This is because the surface of the Earth is moving to the left.

The force is zero at the equator.
This is wrong. See the wikipedia page - or, any other reference on the Coriolis force.

When you travel E-W, your velocity is increasing or decreasing the tangential velocity of your body above or below the tangential velocity of the earth. This results in a change in the centripetal acceleration on you in the N-S direction. This effect only occurs away from the equator. Imagine yourself on a merry-go-round, and you start walking in the same direction as the rotation or in the opposite direction.

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However, when an object is traveling east or west, the latitude is not changing. How would the Coriolis force still apply?
Edit: although the following is true, it is not relevant to the Coriolis force: my apologies for any confusion: Lines of latitude are curved on the surface of the Earth. An object traveling in a straight line is following a great circle route which (apart from at the equator) is not of constant latitude.

Note also the vertical component of the Coriolis force often referred to as the Eötvös effect.

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Lines of latitude are curved on the surface of the Earth. An object traveling in a straight line is following a great circle route which (apart from at the equator) is not of constant latitude.

Note also the vertical component of the Coriolis force often referred to as the Eötvös effect.
A great circle is not a straight line.

A great circle is not a straight line.
On the surface of a sphere it is. If the sphere were stationary it is the path that a moving object will follow that it not accelerating in the reference frame of the surface (or has no component of acceleration perpendicular to its velocity). In the Northern hemisphere of the Earth, the path of such an object drifts to the right with acceleration ## (2 \omega \sin \varphi) v ## perpendicular to its velocity relative to the sphere. The magnitude and direction (relative to velocity) of this force are independent of the direction of the object's velocity.

Edit: note that ## \omega \approx 7.27×10−5 s^{−1} ## so in a stream of 1 m/s (approximately 2 knots which is quite fast but it makes the arithmetic easy) at 30 degrees north (i.e. ## \sin \varphi = 0.5 ##) we have a Coriolis acceleration of ## 2 (7.27×10−5) (0.5) \approx 7.27×10−5 ms^{-2} ## or about Edit: 0.25 m/s per hour. knots per hour (and yes I do mean knots per hour = nautical miles per hour per hour).

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Free floating buoys tend to move in clockwise circles. This is because the surface of the Earth is moving to the left.
This is wrong.
Not entirely wrong, just incomplete. Completing it:

In the Northern hemisphere objects floating in large bodies of open water tend to move in clockwise circles ellipses relative to the solid surface of the Earth. This is because the surface of the Earth is moving to the left whilst the object drifts with the tidal stream.

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In the Northern hemisphere objects floating in large bodies of open water tend to move in clockwise circles relative to the solid surface of the Earth. This is because the surface of the Earth is moving to the left whilst the object drifts with the tidal stream.
This Coriolis force is related to the angular velocity in a rotating reference frame. It's not about "motion to the right or left", or caused by ocean tides.

This Coriolis force is related to the angular velocity in a rotating reference frame. It's not about "motion to the right or left", or caused by ocean tides.
I think we must be at cross-purposes: the Coriolis force is given by ## − 2 m (\boldsymbol \omega \times \boldsymbol v) ## where ## \boldsymbol v ## is the velocity in the rotating reference frame. The "motion to the right or left" or tide is that ## \boldsymbol v ##.

On the surface of a sphere it is. If the sphere were stationary it is the path that a moving object will follow that it not accelerating in the reference frame of the surface (or has no component of acceleration perpendicular to its velocity). In the Northern hemisphere of the Earth, the path of such an object drifts to the right with acceleration ## (2 \omega \sin \varphi) v ## perpendicular to its velocity relative to the sphere. The magnitude and direction (relative to velocity) of this force are independent of the direction of the object's velocity.

Edit: note that ## \omega \approx 7.27×10−5 s^{−1} ## so in a stream of 1 m/s (approximately 2 knots which is quite fast but it makes the arithmetic easy) at 30 degrees north (i.e. ## \sin \varphi = 0.5 ##) we have a Coriolis acceleration of ## 2 (7.27×10−5) (0.5) \approx 7.27×10−5 ms^{-2} ## or about 0.25 m/s per hour.
The surface of a sphere is a curved subspace in 3D space, and a great circle is definitely curved in 3D space.

Do you realize how irresponsible it is to bring this concept up to a student at the apparent level of the OP?

When an object travels on a circle, the centrifugal force is mv2/r, which equals mωr, because ω=v/r.
Now, on rotating Earth, the centrifugal force is always present... but it causes and is balanced by the bulge of Earth equatorial flattening, showing in a decrease of gravity at equator.

ΔBut when an object moves east-west anywhere east and west exist (anywhere away from poles) then the direction of the movement matters. Because the total velocity is now (v+Δv) where v is the rotation speed of ground beneath and Δv the speed relative to ground.
Total centrifugal force then becomes m(v+Δv)2/r=m(v2+2vΔv+Δ2v)/r
The component mv2/r is the part already included in ambient gravity and curvature of the horizontal. The part mΔ2v/r is the small centrifugal force due to moving on round though big earth. But the part m*2vΔv/r is the Coriolis force.
At equator, Coriolis force is directly vertical - but it is actually there.

person123
At equator, Coriolis force is directly vertical - but it is actually there.
The vertical force is called the Eotvos force.

berkeman
When an object travels on a circle, the centrifugal force is mv2/r, which equals mωr, because ω=v/r.
Now, on rotating Earth, the centrifugal force is always present... but it causes and is balanced by the bulge of Earth equatorial flattening, showing in a decrease of gravity at equator.

ΔBut when an object moves east-west anywhere east and west exist (anywhere away from poles) then the direction of the movement matters. Because the total velocity is now (v+Δv) where v is the rotation speed of ground beneath and Δv the speed relative to ground.
Total centrifugal force then becomes m(v+Δv)2/r=m(v2+2vΔv+Δ2v)/r
The component mv2/r is the part already included in ambient gravity and curvature of the horizontal. The part mΔ2v/r is the small centrifugal force due to moving on round though big earth. But the part m*2vΔv/r is the Coriolis force.
At equator, Coriolis force is directly vertical - but it is actually there.
Thank you, that clarifies it a lot for me. I missed that there would be a cross term when adding the velocities from Earth's rotation and an object's motion.

The vertical force is called the Eotvos force.
Eötvös force is just a component of the Coriolis force as defined in physics. It's only in Earth science, that Coriolis force sometimes refers to the components tangential to the Earth's surface, while the normal component has a special name.

Thank you, that clarifies it a lot for me. I missed that there would be a cross term when adding the velocities from Earth's rotation and an object's motion.
One example to see why you need the radial Coriolis component for tangential motion: An object that at rest in an inertial frame is moving in circles when seen from a rotating frame (when it has some constant offset to the frame axis). To account for the circular motion in the rotating frame you need a net centripetal force, towards the frame axis. This cannot be the centrifugal force, which points always away from the axis. In this case the Coriolis force must cancel the centrifugal force and provide the centripetal acceleration, hence the factor 2, in the Coriolis formula.

East-West makes sense only in the context of Earth science, so presumably this is what the original poster had in mind.

pbuk
East-West makes sense only in the context of Earth science, so presumably this is what the original poster had in mind.
I don't think the decomposition of the Coriolis force based on the Earth surface (like done in Earth science) is relevant to the OP's question. The question can be answered using the basic definition of the Coriolis force in a rotating reference frame, which is independent of any specific surface. East-West translates to tangential movement with or against the reference frame rotation.

person123 and PeroK
I don't think the decomposition of the Coriolis force based on the Earth surface (like done in Earth science) is relevant to the OP's question.
I do. Look at the highlighted words in the OP:
Summary:: Why does the Coriolis Force apply on objects traveling east or west?

For objects traveling north or south, the Coriolis force makes sense. The velocity of the earth's surface changes with latitude, causing an object to drift east or west relative to the Earth's surface.

However, when an object is traveling east or west, the latitude is not changing. How would the Coriolis force still apply?
All of those words refer to the 2D spherical geometry of the Earth's surface so I cannot understand how an answer in Euclidian 3-space is relevant.

PeroK
I don't think the decomposition of the Coriolis force based on the Earth surface (like done in Earth science) is relevant to the OP's question. The question can be answered using the basic definition of the Coriolis force in a rotating reference frame, which is independent of any specific surface. East-West translates to tangential movement with or against the reference frame rotation.
The surface of a sphere is a curved subspace in 3D space, and a great circle is definitely curved in 3D space.

Do you realize how irresponsible it is to bring this concept up to a student at the apparent level of the OP?
Both these comments assume that the coordinates of a 3D rotating reference frame are more 'basic' to everyone than rotating spherical coordinates, however for anyone who has navigated an aircraft or a ship on a long passage the opposite is true. If, as in this case, the OP indicates that they are familiar with spherical geometry then with respect to @Chestermiller I think it is irresponsible to insist with apparent authority that a great circle is not a straight line in that geometry.

I started a thread in the SA forum to address this apparent confusion among us.

I don't think the decomposition of the Coriolis force based on the Earth surface (like done in Earth science) is relevant to the OP's question.
I do. Look at the highlighted words in the OP:
Just because the OP framed the question in terms of the Earth surface, doesn't mean the Earth surface is actually relevant to understanding the core issue the OP is having, which is much more basic and general.

I cannot understand how an answer in Euclidian 3-space is relevant.
You don't need much 3D, because the Coriolis force always lies in the plane perpendicular to the frame rotation axis. So you are just dealing with a simple plane instead of a sphere. Once you fully understand the Coriolis force in that plane, including for tangential motion, you can then still decompose it based on the local normal of some sphere.

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PeroK
Why does the Coriolis Force apply on objects traveling east or west?
Recall the formula: ##\mathbf{F} = -2m \boldsymbol{\Omega} \times \mathbf{v}##. Here the angular velocity ##\boldsymbol{\Omega}## is aligned with the axis of the Earth.

For motion on the Earth's surface, i.e. for ##\mathbf{v}## parallel to the Earth's surface, one usually recasts this in the form* ##f = 2m\Omega v \sin{\lambda}##, parallel to the direction ##(v_y, -v_x, 0)##, where ##\lambda## is the latitude angle. Both components of the velocity (E-W and N-S) contribute to the Coriolis force!

*derivation: let ##(x,y,z)## be local Cartesian coordinates in a small region of the Earth's surface, with ##\mathbf{e}_z## the local vertical. Put ##\mathbf{v} = (v_x, v_y, 0)##. In this basis one can also express ##\boldsymbol{\Omega} = (0, \Omega \cos{\lambda}, \Omega \sin{\lambda})##, therefore \begin{align*}
-2m\boldsymbol{\Omega} \times \mathbf{v} &= 2m\Omega (v_y \sin{\lambda}, -v_x \sin{\lambda}, v_x \cos{\lambda})
\end{align*}The projection onto the Earth's surface is ##\mathbf{f} = 2m \Omega (v_y \sin{\lambda}, -v_x \sin{\lambda}, 0)##, having a size ##f = |\mathbf{f}| = 2m \Omega v##.

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Yes with a bit of editing:
the Coriolis force always lies [in] the plane perpendicular to the [axis of rotation] ... Once you fully understand the Coriolis force in that plane [you can decompose it into components tangential and normal to the surface].
this is now heading towards an explanation which I think works better for those without a background in 3D vector mechanics.

Yes with a bit of editing: ... this is now heading towards an explanation which I think works better for those without a background in 3D vector mechanics.

Yes, and the 3D stuff only comes in, when you try to relate it to the Earth's surface. The Coriolis force itself can be understood in the 2D plane of rotation:

PeroK
having a size ##f = |\mathbf{f}| = 2m \Omega v##.[/I]
I think you mean ## f = |\mathbf{f}| = 2m \Omega v \sin \lambda ##. Also note that ## \omega ## is normally used for scalar angular velocity and ## \varphi ## for latitude, and of course we have east and north instead of x and y, hence the expression of the surface Coriolis acceleration relevant to navigators of ## 2 \omega \sin \varphi \begin{pmatrix} v_n \\ -v_e \end{pmatrix} ##.

This obviously only works with a background in vector mechanics, but put this quantitative solution together with a qualitative one and I think we have hit the target (ballistics pun intended).

Klystron
Yes, and the 3D stuff only comes in, when you try to relate it to the Earth's surface.
Ah, but if my starting point is the Earth's surface then this understanding:
The Coriolis force itself can be understood in the 2D plane of rotation:
is no help to me: I cannot fire a cannonball into the surface of the Earth, and an object on a rotating disc cannot move north (up).

I think you mean ## f = |\mathbf{f}| = 2m \Omega v \sin \lambda##.
Thanks - typo!

Also note that ##\omega## is normally used for scalar angular velocity and ##\varphi## for latitude.
Well, notation is the prerogative of the author. I used ##\Omega## and ##\lambda## sub-consciously because that's what my lecturer used last year.

Ah, but if my starting point is the Earth's surface then this understanding:

is no help to me: I cannot fire a cannonball into the surface of the Earth, and an object on a rotating disc cannot move north (up).
Then start with the Earth's surface near the North Pole. It's almost like a rotating disc.

I don't think the decomposition of the Coriolis force based on the Earth surface (like done in Earth science) is relevant to the OP's question. The question can be answered using the basic definition of the Coriolis force in a rotating reference frame, which is independent of any specific surface. East-West translates to tangential movement with or against the reference frame rotation.
Yes, I do actually regret framing the question specifically about the Earth since I would have the same exact confusion with respect to an object traveling tangentially on a rotating disc.

Thank you very much to everyone for all the posts; I'm still not that clear when it comes to relating the derivations with a conceptual understanding but it sort of makes sense to me.

pbuk, PeroK and A.T.
Thank you very much to everyone for all the posts; I'm still not that clear when it comes to relating the derivations with a conceptual understanding but it sort of makes sense to me.
You're welcome. The radial Coriolis force component is a common source of confusion, because the definitions of the inertial forces in a rotating frame are based on mathematical simplicity, rather than ease of intuitive conceptualization. Here my explanation from a previous thread:
One intuitive way to think about the radial Coriolis force, is as a velocity dependent modification of the centrifugal force. In fact you could order the inertial force terms in a rotating frame by direction, and lump the radial Coriolis force with radial centrifugal force. But for mathematical and historical reasons we separate them as position dependent term (centrifugal) and a velocity dependent term (Coriolis).

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