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I'm taking an introductory course on meteorology and in the lecture notes there is an explanation of the Coriolis effect that confuses me. The components of a wind field are u, v and w: $\vec{U} = u \cdot \vec{i}+v \cdot \vec{j}+w \cdot \vec{k}$. The unit vectors are shown in this figure (the coordinate system is rotating with the earth and $\vec{k}$ is the vector perpendicular to $\vec{i}$ and $\vec{j}$):enter image description here

Then, following a derivation, the change of each of these velocity components is given as a function of the rotational velocity of the earth, the latitude, the longitude and the other velocity components: $\left(\dfrac{du}{dt}\right)_{Coriolis} = 2 \cdot \Omega \cdot v \cdot sin(\phi) - 2 \cdot \Omega \cdot w \cdot cos(\phi)\\ \left(\dfrac{dv}{dt}\right)_{Coriolis} = -2 \cdot \Omega \cdot u \cdot sin(\phi)\\ \left(\dfrac{dw}{dt}\right)_{Coriolis} = 2 \cdot \Omega \cdot u \cdot cos(\phi)$

This first equation I understand just fine. For example, when moving away from the equator, your distance to earth's rotation axis decreases, so conservation of momentum requires your u-component of velocity to increase. This is represented by the first term in the RHS of the first equation. The second term tells us that by going straight up, our distance to earth's rotation axis increases and so our u-component of velocity must decrease. However, I fail to see how having a non-zero u-component of velocity causes your w or v-components of velocity to change (equations 2 and 3). Whether you're moving along the equator or running in a perfect circle around the north pole, to me it seems that angular momentum is conserved. Can anybody shed some light on this?

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  • $\begingroup$ The earth is moving underneath your feet in an arc and not a straight line. That is the origin of Coriolis forces. $\endgroup$ – ja72 Oct 9 at 23:15
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I fail to see how having a non-zero v-component of velocity causes your w or v-components of velocity to change

I’m assuming you meant non-zero u-component.

At the equator, 𝜃 = 0 and since sin(0)= 0, then 𝑑𝑣/𝑑𝑡 = 0. So at the equator the Coriolis effect has no impact on the j-component of the velocity vector. And at the poles 𝜃 = 90 and cos(𝜃) = 0 so there is no impact on the k-component of the velocity vector.

In terms of the conservation of angular momentum, it is of course conserved in the inertial frame but it is important to remember that the coriolis effect happens because of the chosen non-inertial reference frame. The images on the wiki page do a good job of illustrating this: https://en.m.wikipedia.org/wiki/Coriolis_force

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  • $\begingroup$ But what causes, for example, the change of the w-component because of the non-zero u component at the equator? Or the change of both the v-component and w-component due to a non-zero u-component at a 45° latitude? $\endgroup$ – Milan Oct 9 at 21:51
  • $\begingroup$ To your first question: think about if you were standing on the equator of a rotating sphere in space (so we can ignore gravity and air resistance). If you throw a ball straight along the equator you would see the ball rising above you in your non-inertial frame, and it would rise proportionally to how fast your threw it. The ball is of course just traveling straight in the inertial frame, but since you are essentially rotating beneath it as it continues straight, it appears to rise up in the non-inertial frame $\endgroup$ – Tom A Oct 9 at 22:02
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    $\begingroup$ The w-component of the coriolis force is known as the eotvos effect. If you fire a bullet at a target to the west, while that bullet is in flight the ground rotated underneath you—the target rises up for the same reason the sun sets. So the bullet hits low. $\endgroup$ – Ben51 Oct 9 at 22:04
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I have found the answer. Someone in the comments pointed me in the right direction by mentioning the eotvos effect. A non-zero u component causes the centrifugal force to increase. That's why the forces represented by equations 2 and 3 always sum up to a force that is perpendicular to the rotation axis of the earth.

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