Does the Coriolis force act on all objects?

Would, for example, a train traveling from south to north experience this force? If so, why? I would think it does not, since it is in contact with the Earth, so it does not experience the difference in speed when going from one point to the other.

• Mar 4 at 16:35

It does. The fact that it is in contact with the earth (or rather, the rails), just means that it cannot move sideways under the force, because the rail restrict its movement. In the same way, if a train travels down a rail and has wind blowing from one side, there is still a force pushing it, but the rails will exert an opposite force so it can't be moved sideways by that force. Some sources [1–3] even claim the effect of the coriolis force can actually be observed in north-south travelling rails, where one side of the tracks wears off faster than the other.

• Do you have any evidence of the railway wear? Most train tracks run in both directions, so neither side should wear quicker. Even in the unlikely case of of a one directional railway, it would be hard to attribute wear to the Coriolis force rather than a regular wind flow or a slight incline. Mar 5 at 7:59
• @oliversm The wear and tear on the tracks from the coriolis effect is unrelated to the direction the train is traveling in. It is related to the force exerted by the rotation of the earth on the train. Keep in mind that A) train wheels exert force on the inside corner of the train, so the train being pushed to one side by a force would increase the wear on the rail on that side; and B) most train tracks exist in pairs, one for each direction, to allow trains to pass each other more easily, which means that generally a single track on such a pair does have trains going in that direction. Mar 5 at 9:13
• @oliversm That's a good question. Of course that would only appear on tracks with separate sets of rails. I added that remark because I remembered reading about it without giving it too much thought. I was only able to find some anecdotal evidence (e.g. in the New Scientist No. 1435, "The Centrifugal Myth"), but no sources that directly demonstrate it.
– noah
Mar 5 at 9:49
• @oliversm It would probably be easy to distinguish wear form the coriolis force on a pair of tracks from an incline or predominant wind patterns by the fact that the coriolis force acts on both outside or both inside tracks, while an incline or preferred wind direction would push to the same side on each pair of tracks.
– noah
Mar 5 at 9:52
• @Taemyr The coriolis effect is effectively a force exerted by the spin of the Earth. Because the earth always spins in the same direction [citation needed], this means the effect always is applied from the same direction, and thus the rails are always pushed in the same direction. This means that regardless of the direction the train is traveling towards, it's always the same side that gets worn. It might be easier to understand if you don't look at it as "wear and tear from the train driving on the tracks", but rather as "wear and tear from the tracks being pushed into the train wheels". Mar 6 at 21:53

If you recall the Coriolis force given by $$\mathbf{F}=-2m\mathbf{\Omega} \times \mathbf{v}$$ If for a minute, We don't get into the detail of the analysis but just see the magnitude of this. $$|\mathbf{F}|=2m|\mathbf{\Omega}_\text{earth}||\mathbf{v}_\text{train}|$$ Let's take the speed of the train to be $$100$$ Km/Hour which is about $$30$$ m/sec and mass to be $$1000$$ Kg. Then $$|\mathbf{F}|\approx0.3 \ N$$ That doesn't go to do anything to your train. Though the effect can be made observable under high speed.

• Whatever the train speed is, you have to get your East/West speed (in an inertial frame) from 1,000 mph to 0 mph as you go from equator to pole.
– JEB
Mar 4 at 16:33
• I don't understand what do you mean by that? Can you rephrase it? Mar 4 at 16:37
• I got 7.292115x1000x30x2=4N
– Eli
Mar 4 at 17:29
• I'm just trying to give a rough estimate. You can do the real calculation that would not make much difference. Mar 5 at 10:37
• @Tristan The magnitude of $a\times b$ is $\lvert a\rvert\lvert b\rvert\sin\theta$ so if it is not moving perpendicular then the magnitude is even smaller. Mar 5 at 18:43

The fact that there is a force counteracting the Coriolis force just means that the Coriolis force isn't allowed to cause acceleration. It's like sitting on a chair; the chair counteracts gravity, but that doesn't mean that gravity doesn't act on you, it just means that you don't fall down. Just as gravity presses you to the chair, the Coriolis force presses the train to the side of the rail.

One way of thinking about the Coriolis force is that as an object moves towards the axis of rotation, the moment of inertia of the entire system decreases. To preserve angular momentum, the angular velocity must increase.

In the Northern hemisphere, a train traveling North actually causes the moment of inertia of the Earth to decrease slightly, which causes the angular velocity to increase (this, among other factors, means that the length of a day is not constant; it decreases ever so slightly whenever something moves towards the poles). Where does the increase in the angular velocity come from? It comes from the train. The train has to exert a force on the track in an Eastern direction to increase the angular velocity of the Earth. And since forces come in pairs, the Earth must exert a force on the train in a Western direction.

This is somewhat backwards, however, as the wording of the preceding implies that the conservation of angular momentum causes the Coriolis force, which in fact it is the Coriolis force that causes the conservation of angular momentum. It is because an object traveling towards the center of rotation exerts a force on the rest of the body in the direction of rotation that the angular velocity increases to keep the angular momentum constant despite the moment of inertia decreasing.

Yes, this force acts on all objects in motion on a revolving body. You can see this in action even on a merry-go-round. Please note that the force acts tangentially to the pivot/axis of rotation.