# Difference between the Coriolis effect and the Eötvös effect

The coriolis accelaration is $-2\Omega \times u$, where $\Omega$ is the earths rotation and u is the velocity in a basis following the earth.

When something moves east/west, this results in acceleration upwards/downwards. This is also called the eötvös effect, which wikipedia says corresponds to $2\Omega u cos(\phi) + \frac{u^2 + v^2}{R}$, where u is movement in east/west direction and v is movement in north/south direction.

If the eötvös effect is the vertical component of the coriolis effect, then why are the expressions different? The coriolis effect lacks $\frac{u^2 + v^2}{R}$. Without that term, sufficient velocity west would push an object into the ground, instead of generating a centrifugal effect.

Does the derivation of the coriolis effect assume low velocity, or am I missing something else?

fluff, your problem begins with this assertion: "If the eötvös effect is the vertical component of the coriolis effect..."

In many science disciplines, casual versus formal usages become intermixed, and this is certainly one area. Eötvös is not the vertical component of Coriolis.

The earth is both (a) spherical and (b) spinning. This produces a number of phenomena that affect bodies in motion on or near the surface of the Earth. In casual usage these phenomena tend to be lumped together into all being called "Coriolis," but they are actually discrete physical properties that are not related, except for the fact that they are artifacts of (a), (b), or both.

Coriolis is a conservation of angular momentum consideration when objects move north/south across a spinning sphere. As you move away from the equator latitudinally, the same angular rate of rotation around the Earth's C/G results in a different velocity in the east/west component, and the effects of this difference is the Coriolis Effect. Were the Earth a cylinder instead of a sphere, there'd be no Coriolis Force.

Eötvös on the other hand is a centrifugal force/orbital mechanics problem. Eötvös would still occur on a cylinder, where Coriolis would not.

There IS an angular momentum force that acts east/west based on the height of an object's trajectory or orbit, and thus would affect the vertical component of a projectile's trajectory at long distances involving high trajectories... But this isn't Eötvös at all. If I shoot a projectile perfectly vertically a few miles into the air, conservation of angular momentum dictates the projectile will not land back on me, it will land several feet west of me, opposite the direction of the Earth's spin. It may be more correct to think of THIS motion as the vertical component of Coriolis.

Hope this helps.

Essentially yes

Take a look at the last expression in this link:

https://hepweb.ucsd.edu/ph110b/110b_notes/node15.html

That's the Coriolis force on a ball that is fired from some lattitude $$\lambda$$. The ball is fired with angle $$\theta$$ to the local zenith and $$\phi$$ from the east in the local horizontal plane. So, $$\phi=0°$$ is east and $$\phi=180°$$ is west.

Now take a look at the last expression in that link, the one with the fully expanded Coriolis force for the situation described above.

The Z-component of the acceleration is:

$$\vec{a}=2\omega v\cos\lambda \sin\theta cos\phi\vec{z}$$

Now, take the case in which the cannonball is fired horizontally, meaning, $$\theta=90°$$

The force becomes:

$$\vec{a}=2\omega v\cos\lambda cos\phi\vec{z}$$

For $$\phi = 0°$$, eastward travel is obtained:

$$\vec{a}=2\omega v\cos\lambda \vec{z}$$

For $$\phi = 180°$$, westward travel is obtained:

$$\vec{a}=-2\omega v\cos\lambda \vec{z}$$

Basically, both cases can be subsummed by assuming positive $$v$$ for eastern travel and negative $$v$$ for westward travel.

$$\vec{a}=2\omega v\cos\lambda \vec{z}$$

Which gives you the main component of the Eotvos effect for velocities that are below the local rotational velocity of the earth.

The 2nd term $$(u^2+v^2)/r$$, the correction does not appear unless you assume that the travelling object is trying to keep a circular orbit or in case of a ship, just follow the curvature of the earth.

Another answer mentioned that "the coriolis term wouldn't appear on a cylindrical earth"

But it would. The z-component of it would appear. It even appears on earth for objects fired purely westward or eastward. Bullets for example. And that's for an object that doesn't try to maintain a circular orbit, but one that is just fired freely. If an object is trying to follow the curvature of the earth, the 2nd term (the correction applies) because now it will experience an additional centrifugal force due to the fact that it is moving with velocity $$u$$ relative to earth and is in a circular trajectory relative to already rotating earth.

Whether the whole effect should be called Eotvos effect or just the 2nd term or even just the 1st term, I do not know. But the first term 100% appears just by expanding the Coriolis force. The 2nd term only appears after an orbit is assumed (or in case of a ship, trying to follow the shape of the earth). The 2nd term is nothing but a circular orbit relative to an already rotating system.

For the case of a travelling bullet or a cannonball, the 2nd term does not apply because it does not try to maintain the orbit and the entire Eotvos effect is just due to Coriolis force.