# Coriolis force thought experiment

## Main Question or Discussion Point

Let's say I build a 500 m long circular road around the north pole. Then I drive on the road at speed 20 m/s, to the east.

There will be a noticeable inertial force to the south.

Is that inertial force a Coriolis force?

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Simon Bridge
Homework Helper
No. It is the usual centrifugal effect.

The coriolis force is what you experience if you try to drive due south in a hovercraft ... you have to keep steering to one side to correct.

Let's say I build a 500 m long circular road around the north pole. Then I drive on the road at speed 20 m/s, to the east.

There will be a noticeable inertial force to the south.

Is that inertial force a Coriolis force?
Only one 1729th part of it is the Coriolis Force.

You would complete 3456 laps in a day on this ring, but when you go to the east the earth adds one rotation to this total, so the centrifugal force wich is proportional to the square of the angular speed, will be (1 + 1/3456)^2 ≈ (1+1/1728) times bigger compared to a non-rotating earth. If you would go west it would be smaller by the same amount.

A.T.
Let's say I build a 500 m long circular road around the north pole. Then I drive on the road at speed 20 m/s, to the east.

There will be a noticeable inertial force to the south.

Is that inertial force a Coriolis force?
Inertial forces are frame dependent. You forgot to say which frame you are considering.

- Rotating rest frame of you and north pole : You are at rest so there is no Coriolis force on you, just a centrifugal force due to fast frame rotation.

- Rotating rest frame of the earth surface : There is no "noticeable" inertial force on you in this frame. You are going in circles, so the forces on you are not balanced. Due to the slow frame rotation there are only negligible centrifugal and Coriolis forces. But they are tiny compared to the centripetal force that accelerates you towards the pole.

- Inertial rest frame of the pole: There are no inertial forces at all here. Just the centripetal force that accelerates you towards the pole.

Only one 1729th part of it is the Coriolis Force.

You would complete 3456 laps in a day on this ring, but when you go to the east the earth adds one rotation to this total, so the centrifugal force wich is proportional to the square of the angular speed, will be (1 + 1/3456)^2 ≈ (1+1/1728) times bigger compared to a non-rotating earth. If you would go west it would be smaller by the same amount.
The calculation is tricky and has fooled a number of famous mathematicians, but this seems at least close to correct.

Inertial forces are frame dependent. You forgot to say which frame you are considering.

- Rotating rest frame of you and north pole : You are at rest so there is no Coriolis force on you, just a centrifugal force due to fast frame rotation.

- Rotating rest frame of the earth surface : There is no "noticeable" inertial force on you in this frame. You are going in circles, so the forces on you are not balanced. Due to the slow frame rotation there are only negligible centrifugal and Coriolis forces. But they are tiny compared to the centripetal force that accelerates you towards the pole.

- Inertial rest frame of the pole: There are no inertial forces at all here. Just the centripetal force that accelerates you towards the pole.

People standing on the surface of the earh tend to say that vehicles or winds experience a Coriolis force. So these people's rest frame, which is a rotating frame, is the frame that I meant.

If we think about a person standing on the surface of the earth, he feels a centrifugal force. I think this person's comment about the car driving around the north pole should be: "the car is effected by the same centrifugal force as myself, and then also a centripetal force, the centrifugal force does not depend on the speed of the car"

So I guess we can say: In the rotating rest frame of the Earth surface velocity to the west or velocity to the east does not cause any Coriolis force.

EDIT: actually a person standing on the surface of the Earth should say: "the car is effected by the same centrifugal force as myself, and then also a centripetal force, the centrifugal force does not depend on the speed of the car, and then one more force is needed, to make the car to obey Newton's laws in my frame, the last force is called the Coriolis force."

Moving to west or east does case a Coriolis force, after all.

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A.T.
People standing on the surface of the earh tend to say that vehicles or winds experience a Coriolis force. So these people's rest frame, which is a rotating frame, is the frame that I meant.
Okay. I assumed so. By your "noticeable inertial force" sounded like "directly perceivable by a human". Both inertial forces in the rotating rest frame of the Earth are rather tiny.

If we think about a person standing on the surface of the earth, he feels a centrifugal force.
If he is very sensitive.

I think this person's comment about the car driving around the north pole should be: "the car is effected by the same centrifugal force as myself, and then also a centripetal force, the centrifugal force does not depend on the speed of the car"
This is correct. If you stand on the track the centrifugal force on you and the car are the same, in the rotating frame of the surface.

But there is also a Coriolis force, which acts only on the car, because it moves relative ot the rotating frame of the surface.

Both inertial forces are rather small, because the frame of the surface rotates slowly.

So I guess we can say: In the rotating rest frame of the Earth surface velocity to the west or velocity to the east does not cause any Coriolis force.
No. In a rotating frame there is always a Coriolis force on moving objects, unless they move parallel to the frame rotation axis (here: vertical to the ground).

EDIT: actually a person standing on the surface of the Earth should say: "the car is effected by the same centrifugal force as myself, and then also a centripetal force, the centrifugal force does not depend on the speed of the car, and then one more force is needed, to make the car to obey Newton's laws in my frame, the last force is called the Coriolis force."

Moving to west or east does case a Coriolis force, after all.
Yes. But note that in the surface frame the inertial forces (centrifugal, Coriolis) do not have to cancel the real centripetal force at the tires. The net force is not zero, because the car still moves in circles in that frame.

In the rotating surface frame the east going car moves in circles slightly slower, than in the non-rotating inertial frame (where there are no inertial forces, just the centripetal force on the tires). So it still needs a centripetal net force, but a slightly smaller one. The slight difference than cancels a small part of the centripetal tire force is composed of both: centrifugal and Coriolis inertial forces.

Cleonis
Gold Member
Let's say I build a 500 m long circular road around the north pole. Then I drive on the road at speed 20 m/s, to the east.

There will be a noticeable inertial force to the south.

Is that inertial force a Coriolis force?

Well: in Meteorology that effect is indeed referred to as 'Coriolis effect'.

Instead of thinking about a road, I suggest to consider a hovercraft, a big one. (The biggest hovercrafts have a mass of several hundreds of tons)

A hovercraft that is hovering in place is co-rotating with the Earth's rotation. A hovering craft doesn't have a tendency to drift towards the equator because the Earth's shape gives rise to the required centripetal force.

The Earth has an equatorial bulge; the distance from the Earth's center to the equator is about 20 kilometers more than the distance from the center to (each of) the poles.

You can calculate that at 45 degrees latitude an incline (towards the nearest pole) of 0.1 degree is needed to provide the required centripetal force. The Earth's bulge provides just that incline.

When that hovercraft has a velocity relative to the Earth then you get dynamic effects.
- When the hovercraft is cruising due east it is circumnavigating the Earth faster than equilibrium speed, so it tends to veer off course, towards the equator
- When the hovercrat is cruising due west it is circumnavigating the Earth slower than equilibrium speed, so it tends slide towards the nearest pole.
- And of course there are also tendencies to veer off course when moving due north and due south.

In meteorology those tendencies in the various directions are grouped under a single name: 'Coriolis effect'. The magnitude of the tendency to veer off is given by the expression: 2mΩv
where:
m = mass of the moving object
Ω = angular velocity of the Earth
v = velocity relative to the Earth

This expression, 2mΩv, is valid for any direction of motion relative to the Earth.

In Meteorology it makes sense to organize things like that. In meteorology you simply say: 'the Coriolis effect is the same in all directions', and you take it from there.

Some people may argue that it's better to restrict the expression 'Coriolis effect' to radial motion only. But for meteorology that is highly unpractical. It would deprive you of a way of thinking about general properties of atmospheric flow efficiently.

Getting back to the hovercraft, cruising due east. What you notice is a need to compensate for a tendency to veer off. When cruising due west the hovercraft's propulsion system must itself provide the required extra centripetal force. It's not necessary to think about that situation in terms of 'an inertial force that tends to deflect you'. You are quite aware of what is going on. When cruising east you need to compensate for insufficient centripetal force. Conversely, when cruising west there is a surplus of centripetal force.

Meteorology is (among other things) the study of fluid motion of the atmosphere. Oceanography is (among other things) the study of fluid motion of the Earth's oceans. Those two fields of fluid dynamics have many parallels, in particular the terrestrial Coriolis effect. The general term is 'Geophysical fluid dynamics'. So the more general statement is that in Geophysical fluid dynamics these rotation-of-Earth-effects are referred to as 'Coriolis effect'.

K^2
jartsa, short answer, if your frame of reference is fixed with the ground, then yes, the radial force due to 20m/s is going to be the Coriolis force. That will be a small addition to the centrifugal force due to Earth's rotation experienced regardless of the speed.

jbriggs444
Homework Helper
2019 Award
jartsa, short answer, if your frame of reference is fixed with the ground, then yes, the radial force due to 20m/s is going to be the Coriolis force. That will be a small addition to the centrifugal force due to Earth's rotation experienced regardless of the speed.
My computations are otherwise.

As Cleonis has pointed out, the centrifugal force due to the Earth's rotation is cancelled by the [normally un-noticed] slope of the ice at the north pole as compared to an ideal sphere.

That centrifugal force is given by m omega^2/r and is independent of the 20 meters/second speed at which you are driving. For a 500 meter circumference track at the rotation rate of the earth, the centrifugal pseudo-force comes to about .000000419 meters/second/second.

The coriolis force is given by 2 m omega cross v. For a velocity of 20 meters/second at approximately right angles to the rotation axis of the earth, the Coriolis pseudo-force comes to about .00290 meters/second/second.

The observed centripetal acceleration calls for a net force of m v^2 / r and is independent of the rotation rate of the earth (*). For a velocity of 20 meters/second on a circular track with a circumference of 500 meters, the centripetal acceleration comes to about 5.03 meters/second/second.

The real force of the ice on the tires is going to be the vector sum of these three forces with appropriate sign conventions depending on the direction of travel.

(*) The value for v may vary depending on your choice of reference frame, but the computed acceleration will be m v^2 / r regardless.

A.T.
My computations are otherwise.

As Cleonis has pointed out, the centrifugal force due to the Earth's rotation is cancelled by the [normally un-noticed] slope of the ice at the north pole as compared to an ideal sphere.

That centrifugal force is given by m omega^2/r and is independent of the 20 meters/second speed at which you are driving. For a 500 meter circumference track at the rotation rate of the earth, the centrifugal pseudo-force comes to about .000000419 meters/second/second.

The coriolis force is given by 2 m omega cross v. For a velocity of 20 meters/second at approximately right angles to the rotation axis of the earth, the Coriolis pseudo-force comes to about .00290 meters/second/second.
Didn't check the numbers, but the approach seems reasonable so far.

The observed centripetal acceleration calls for a net force of m v^2 / r and is independent of the rotation rate of the earth (*). For a velocity of 20 meters/second on a circular track with a circumference of 500 meters, the centripetal acceleration comes to about 5.03 meters/second/second.

The real force of the ice on the tires is going to be the vector sum of these three forces with appropriate sign conventions depending on the direction of travel.

(*) The value for v may vary depending on your choice of reference frame, but the computed acceleration will be m v^2 / r regardless.
Here I disagree. What is frame independent here is the real force of the ice on the tires. The centripetal acceleration and thus the observed net force is very much frame dependent. For example:

- In the rotating restframe of the car the net force is zero, because the centrifugal force cancels the real centripetal force on the tires.

- In the rotating restframe of the Earth the net force is centripetal, but slightly lower than in the inertial frame, because the sum of centrifugal and Coriolis forces cancels a part of the real centripetal force on the tires. This is consistent with the slightly slower rotation and thus lower centripetal acceleration in that frame, compared to the inertial frame.

jbriggs444
Homework Helper
2019 Award
Here I disagree. What is frame independent here is the real force of the ice on the tires. The centripetal acceleration and thus the observed net force is very much frame dependent. For example:
I think that we actually agree.

Yes, the real force of the ice on the tires is frame independent.
Yes, the centripetal acceleration and thus the computed net force is frame dependent.

[You can't directly observe the net force -- you can only observe the acceleration that it produces]

The formula for the centripetal acceleration will be v^2/r regardless of reference frame, as long as it is a reference frame in which uniform circular motion is observed. As I wrote opreviously, the value of v is not invariant.

- In the rotating restframe of the car the net force is zero, because the centrifugal force cancels the real centripetal force on the tires.
In this frame, v = 0.

- In the rotating restframe of the Earth the net force is centripetal, but slightly lower than in the inertial frame, because the sum of centrifugal and Coriolis forces cancels a part of the real centripetal force on the tires. This is consistent with the slightly slower rotation and thus lower centripetal acceleration in that frame, compared to the inertial frame.
As I interpret the problem, it is in this frame that v = 20 meters/sec.

Cleonis
Gold Member
I'd like to reexamine the context of the original question.

The original question was, intentional or not, about the Earth. It was not, for example, about a disk-shaped spacestation that is rotating, with an astronaut on a space-walk moving hand over hand along the outside structure.

The special situation for the Earth is that it's not only rotating, but also that on every latitude the required centripetal force is provided by the equatorial bulge. With that bulge being created by the rotation, of course)

So, the context of the original question is a rotating system where at every distance to the center of rotation the required centripetal force is provided by gravity.

In that context the noticable rotation-of-the-system-effect is the same for any direction of motion, and it's a natural move to refer to that as 'the Coriolis effect'.

By contrast, if required centripetal force is not provided, then the noticable rotation-of-the-system-effect is dominated by centrifugal effect. Then there is no separately noticable Coriolis dynamics; the dynamics is overwhelmingly centrifugal effect.

I emphasize the noticable. On Earth required centripetal force is provided by gravity, hence there is no centrifugal effect. That sets the stage for noticing a Coriolis effect.

jbriggs444
Homework Helper
2019 Award
By contrast, if required centripetal force is not provided, then the noticable rotation-of-the-system-effect is dominated by centrifugal effect. Then there is no separately noticable Coriolis dynamics; the dynamics is overwhelmingly centrifugal effect.
The required velocity to make the Coriolis force dominate the centrifugal force is roughly equal to the rotational velocity of the Earth.

Near the pole on the circumference of a 500 meter track, that velocity is 500 meters in 24 hours. For a cart moving at a velocity of 20 meters/second, Coriolis dominates.

In the middle latitudes, the rotational velocity is more like 300 meters per second. For a cart moving at a velocity of 20 meters/second, centrifugal force would indeed dominate, were it not for the slope of the geoid.

The required velocity to make the Coriolis force dominate the centrifugal force is roughly equal to the rotational velocity of the Earth.

Near the pole on the circumference of a 500 meter track, that velocity is 500 meters in 24 hours. For a cart moving at a velocity of 20 meters/second, Coriolis dominates.
What dominates is the centrifugal force you get because of the acceleration in the frame that's rotating with the earth. The coriolis force is what you would get if you moved in a great circle with a circumference of 40000 km, wich also passes a point 500m/2 pi from the south pole. This force is very big compared with the centrifugal force you get rotating around the pole in 24 hours along with the earth, but very small compared to the centrifugal force going round the track in 25 seconds.

K^2
As Cleonis has pointed out, the centrifugal force due to the Earth's rotation is cancelled by the [normally un-noticed] slope of the ice at the north pole as compared to an ideal sphere.
Yes, idealized surface is going to be an equipotential, so the centrifugal force is balanced by gravity and normal force. I'm not sure how that changes the question, however.

As for the ratio between the two, I read 500m as the radius. For track whose circumference is 500m, yes, Coriolis force dominates at given speed.

Well: in Meteorology that effect is indeed referred to as 'Coriolis effect'.

Instead of thinking about a road, I suggest to consider a hovercraft, a big one. (The biggest hovercrafts have a mass of several hundreds of tons)

A hovercraft that is hovering in place is co-rotating with the Earth's rotation. A hovering craft doesn't have a tendency to drift towards the equator because the Earth's shape gives rise to the required centripetal force.

The Earth has an equatorial bulge; the distance from the Earth's center to the equator is about 20 kilometers more than the distance from the center to (each of) the poles.

You can calculate that at 45 degrees latitude an incline (towards the nearest pole) of 0.1 degree is needed to provide the required centripetal force. The Earth's bulge provides just that incline.

When that hovercraft has a velocity relative to the Earth then you get dynamic effects.
- When the hovercraft is cruising due east it is circumnavigating the Earth faster than equilibrium speed, so it tends to veer off course, towards the equator
- When the hovercrat is cruising due west it is circumnavigating the Earth slower than equilibrium speed, so it tends slide towards the nearest pole.
- And of course there are also tendencies to veer off course when moving due north and due south.

In meteorology those tendencies in the various directions are grouped under a single name: 'Coriolis effect'. The magnitude of the tendency to veer off is given by the expression: 2mΩv
where:
m = mass of the moving object
Ω = angular velocity of the Earth
v = velocity relative to the Earth

This expression, 2mΩv, is valid for any direction of motion relative to the Earth.

In Meteorology it makes sense to organize things like that. In meteorology you simply say: 'the Coriolis effect is the same in all directions', and you take it from there.

Some people may argue that it's better to restrict the expression 'Coriolis effect' to radial motion only. But for meteorology that is highly unpractical. It would deprive you of a way of thinking about general properties of atmospheric flow efficiently.

Getting back to the hovercraft, cruising due east. What you notice is a need to compensate for a tendency to veer off. When cruising due west the hovercraft's propulsion system must itself provide the required extra centripetal force. It's not necessary to think about that situation in terms of 'an inertial force that tends to deflect you'. You are quite aware of what is going on. When cruising east you need to compensate for insufficient centripetal force. Conversely, when cruising west there is a surplus of centripetal force.

Meteorology is (among other things) the study of fluid motion of the atmosphere. Oceanography is (among other things) the study of fluid motion of the Earth's oceans. Those two fields of fluid dynamics have many parallels, in particular the terrestrial Coriolis effect. The general term is 'Geophysical fluid dynamics'. So the more general statement is that in Geophysical fluid dynamics these rotation-of-Earth-effects are referred to as 'Coriolis effect'.

Wikipedia says that in a carousel a Coriolis force does push objects that move relative to the carousel surface, and the magnitude of the force does not depend on the direction of the motion. (This is in the rotating frame of the rotating carousel surface)

Adding gravitating chunks of mass onto the carousel would not change the Coriolis force at all.

Now, I thought Wikipedia must be very confused, but actually I was slightly confused with all the frames, as was pointed out by A.T.

http://en.wikipedia.org/wiki/Coriolis_effect

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Cleonis
Gold Member
Now, I thought Wikipedia must be very confused,

http://en.wikipedia.org/wiki/Coriolis_effect

Yes indeed.

Over the years inconsistent edits by various contributors have been accumulating. You can get ideas from that Wikipedia article, but you'll have to think it through for yourself. Any paragraph may be either good or just rambling.