This is NOT a homework question, there are few informations that I found on the Internet about this specific case of the Coriolis Force

We currently are dealing with the Coriolis Force, and I struggle to understand how to calculate the force when moving east or west ALONG the Equator. I know that the force is zero when moving north or south at the Equator, because the Latitude Angle used to calculate the force is zero.

I know that generally $$F_c = -2m (\Omega \times \overrightarrow{v})$$ where Omega is the Angular Velocity, in the case of Earth it is $\frac{2\pi}{24\cdot 60 \cdot60}$ because we divide the period by the total time in seconds it takes for Earth to go around, so one day. And $\overrightarrow{v}$ is the velocity vector. The resulting force $F_c$ points perpendicular to both velocity and angular velocity

I also know that if moving tangetially along the Surface of earth, the formula is $$F_c = 2\cdot m\cdot v \cdot \omega \cdot \sin{\phi}$$ where $\phi$ is the latitude at which the object lies. So at the Equator, $\phi = 0$ so obvisouly $\sin\phi$ is also zero.

But there is a Coriolis Force when moving East or West along the Equator, I know it is directed inside the Earth when moving West, and directed toward outer space when moving East. But how I am supposed to calculate it in this case ? The latitude angle is zero, so I can't use the formula above even tough we are also moving tangetially along the surfcae of Earth

I would really appreciate any support, because either I didn't find it, or there are only few informations about this specific case of the Coriolis Force on the Internet

Thanks a loot for your help, feel free to comment if anything is unclear

  • $\begingroup$ Feel free to ask if anything is unclear $\endgroup$
    – wengen
    Commented Oct 4, 2022 at 9:05

1 Answer 1


You are supposed to calculate the Coriolis force using the vector equation you provided, not the latitude dependent resolution tangential to the Earth's surface.

See https://en.wikipedia.org/wiki/Eötvös_effect e.g.,:

enter image description here

At the equator the vector Coriolis effect is entirely vertical, so the portion resolved parallel to the surface is zero. This is all in the Earth-fixed frame.

In the frame of the moving object, there is no Coriolis effect (as $v=0$ in that frame), and the pseudo-force is called the Eötvös effect, which is a change in the apparent local gravity strength caused by a centrifugal force.

  • $\begingroup$ So I can just take the Eotvos Formula with velocity in latitude direction = 0 and velocity in longitudinal direction equal to the velocity of the object because it moves east-west along the equator ? The phi angle in this case is also zero because it's the latitude angle, so $\cos\phi = 1$ ? The angular velocity is the same as above, that is 2pi/(24*60*60), and the radius of the earth is 6.378 * 10^6 meters ? This formula gives the acceleration, to get the force can I just mutiply by the mass of the object, so F = m*a ? This seems a bit too easy, am I missing something? Thanks for your support $\endgroup$
    – wengen
    Commented Oct 4, 2022 at 10:35
  • $\begingroup$ By the way, I would like to upvote your answer but I need at least 15 reputation to do so xD. As soon as I get them, I will upvote your answer $\endgroup$
    – wengen
    Commented Oct 4, 2022 at 10:42

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