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?

  • $\begingroup$ Does the derivation of the coriolis effect assume low velocity?: No, it is just the difference between the land speed at that latitude and the a horizontal speed of the air stream. The eötvös effect is due to the increase in centrifugal force when it's speed is added to the earth's spin, forcing it outward towards the equator, or the decrease in centrifugal force when going east to west against the earth's spin, allowing it to fall towards the pole. If it was going east to west at more than double the ground speed, it would be forced towards the equator anyway. $\endgroup$ Commented Apr 30, 2023 at 3:55

2 Answers 2


Essentially yes

Take a look at the last expression in this link:


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.

  • $\begingroup$ Thanks! A question about this: "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" Why not? If you fire a cannonball westward faster than earth's rotation, you are going to see it go upwards, as compared with the earth's surface. Not downwards, as 2𝜔𝑣cos𝜆𝑧 alone would predict. Maybe this is me assuming that it's trying to keep a circular orbit. But if we don't assume that, I don't see why the 2𝜔𝑣cos𝜆𝑧 force would apply (or why it would predict the ball to be pushed downwards!). $\endgroup$
    – fluff
    Commented Apr 17, 2020 at 19:53
  • 1
    $\begingroup$ (1/3) Think about an extreme example. Imagine if you were on a very small planet that is rapidly rotating in the same direction as earth, west-to-east. You throw a ball westwards at t=0 in a straight line. You measure height relative to the place where you are standing. There is a tangential plane that touches the planet at the place where you are standing. Whenever you measure height, you measure height relative to that plane. pasteboard.co/J4iRmJn.png $\endgroup$
    – user238194
    Commented Apr 18, 2020 at 0:56
  • 1
    $\begingroup$ (2/3), left picture is situation at t = 0. red plane is your local coordinate system that rotates with you and that's where you see eotvos acceleration. ball is thrown at t = 0, purely westward. after some time deltaT, right picture appears. ball is still moving in straight line relative to an inertial observer but this time, the planet has rotated and now if you were to measure height of the ball, you will see that it went below your feet because to an inertial observer, your reference frame is rotating alongside earth and so your reference point for height changes direction and origin. $\endgroup$
    – user238194
    Commented Apr 18, 2020 at 0:59
  • 1
    $\begingroup$ (3/3) the extra term would only be necessary if ball was trying to maintain a circular orbit around an already rotating earth. u and v are not absolute speeds but speeds relative to rotating observer. $\endgroup$
    – user238194
    Commented Apr 18, 2020 at 1:06

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.

  • $\begingroup$ Sorry but this is wrong. Coriolis force is a force that's needed, together with centrifugal force, to describe motion in a non-inertial reference frame (both are not needed when describing the same motion in an inertial reference frame). See user238194's answer for the correct description. $\endgroup$ Commented Mar 1, 2020 at 14:43
  • $\begingroup$ I agree that my description above has oversimplified Coriolis. user238194's description of Coriolis is correct, and corrects my oversimplification. However, the vertical component of Coriolis is not Eötvös. Coriolis is a virtual force we calculate to correct the linear tangential momentum imposed on a projectile by the eastward (spin-ward) motion of the "shooter." Agreed, this is required when moving between inertia and non inertial frames. Eötvös is not a correction between reference frames. It is a physical phenomenon, not a virtual one, that can be seen in-situ in either frame. $\endgroup$
    – Max R
    Commented Mar 2, 2020 at 15:54
  • $\begingroup$ There are common ballistic scenarios where Coriolis imposes what appears to be a +Z motion on the projectile in the spherical frame, but that Eötvös is actually -Z. (Think about how that could be.) And think about this... If you change the mass of the earth but leave rotation the same, and use the same velocity and mass of projectile... The net deflection caused by Coriolis remains the same, the net displacement resulting from Eötvös is very different. If they were simply the same phenomena using different names this could not be so. $\endgroup$
    – Max R
    Commented Mar 2, 2020 at 16:17

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