The Coriolis force is given by the vector equation $-2m \Omega \times v'$.

Where $v'$ is the velocity with respect to the rotating frame.

For the case of a projectile on the Earth, $\Omega$ is along the axis of rotation. It seems from this definition of the Coriolis force that a projectile fired in a northwards direction will have a $v'$ parallel to $\Omega$, so the right-hand rule would give a force of zero magnitude. Likewise, for a projectile fired in an eastwards direction, the right-hand rule would give a force directed towards the center of the Earth.

However, I know that the true result is that the northward projectile experiences an eastward force and the eastward projectile experiences a southward force. How is this compatible with the right-hand rule?

(This is NOT a duplicate of Is there an intuitive explanation for the Southward force caused by the Coriolis Effect on rotating spheres?, because my question is specifically about the vector expression of the force, which is not addressed in that post.)

  • $\begingroup$ I'm going to delete my answer since the actual content is wrong, but your issue with movement to the North is that you're thinking of the velocity vector pointing to an absolute $\hat{z}$ direction; this is only the case at the equator. The velocity needs to be projected into a plane parallel to the surface of the Earth if we're talking about something moving at roughly constant height, which mixes up the components and creates a non-zero Coriolis force. Note that the Coriolis force is still zero at the equator. $\endgroup$ Mar 1, 2018 at 3:22

2 Answers 2


The horizontal component of the Coriolis force for non-equatorial latitudes arises because the axes of the rotating frame are not parallel to the axis of rotation, except for when the rotating frame is on the equator.

The statement in the question, that $\Omega$ is parallel to $v'$, is incorrect.
diagram of rotating axes
You can see that if an object is traveling North, there is a component of the angular velocity that is perpendicular to the object's velocity. The right hand rule gives $\Omega \times v'$ pointing in the +y' direction, but there is a negative sign on the expression for the Coriolis force, so the horizontal component of the Coriolis force points in the -y' direction. This corresponds to East, which is the correct direction.

The magnitude of the Coriolis force can be found using the trigonometric breakdown shown below.
diagram of trigonometry components

  • $\begingroup$ Is it possible to restore the pictures previously associated with this post? I would like to learn from them. Thank you. $\endgroup$
    – Kirk A
    Aug 16, 2020 at 10:03

At the equator, the northward project has no Coriolis force, while the eastward projectile has no horizontal forces. You are correct. This is why hurricanes don't form at the equator.

Regarding the northward projectiles, the horizontal component of the Coriolis force points East (West) in the Northern (Southern) hemisphere--that change in sign comes with a zero crossing at the equator.

The vertical component of the Coriolis force contributes to the Eotvos effect, which also includes a centrifugal term. For velocities less than the true rotation speed the Coriolis term dominates leading to an apparent decrease (increase) in the local gravity for objects moving East (West).

  • $\begingroup$ Could you please clarify how the horizontal component arises at non-equator latitudes? $\endgroup$
    – dfghbvcx
    Mar 1, 2018 at 2:43
  • $\begingroup$ @dfghbvcx The force is perpendicular to both the angular velocity and the protective velocity. Considering horizontal velocities, this means there's going to be a horizontal component when away from the equator--moreover it's completely horizontal for any velocity at the poles. $\endgroup$
    – JEB
    Mar 1, 2018 at 4:16

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