# Direction of deflection of Coriolis force on Earth

I'm a bit confused about Coriolis force direction on Earth. The observed deflection is to the right in the northern emisphere and to the left in the southern one.

But what about an observer in the northern hemisphere throwing a ball southwards?

In what direction will the ball be deflected? Will it be deflected in two different ways if observed from the northern or the Southern hemisphere, or will the deflection be independent from where the observer is? (northern or southern hemisphere)

This confuses me because if I use the right hand rule with $\vec{\omega}$ directed upwards (as it is), then I get that the deflection should be to the right (from the perspective of the person throwing the ball). Nevertheless in the Southern hemisphere objects are deflected to the left and $\vec{\omega}$ is considered directed downward. Again, does the considered direction of $\vec{\omega}$ depends on where the observer is?

The direction of deflection of the ball should be independent where the observer is. Left or right deflection is with respect to an observer facing in the direction the ball is moving.

Are you asking about a ball being thrown from the northern to the southern hemisphere? In that case the deflection would change from deflecting right to deflecting left as it crosses the equator, though near the equator the amount of deflection is minimal.

Let's just say, for the sake of argument, that the equator has a linear eastern velocity of 1,000 mph (actually 1,039). If you release a ball (presumably fired from a cannon) southward from a latitude of 70° north, your ball has an eastern velocity of about 342 mph (cosine of 70° = 0.342). As your ball travels over 45° north, the ground beneath it will have an eastern velocity of about 707 mph (sine or cosine of 45° = 0.707), but your ball will still have its original eastern velocity of 342 mph. The ground beneath your ball will, therefore, be traveling 365 mph eastward faster than the ball. This will make the ball appear to be deflecting to the right, or westward, relative to you.

If your projectile still had enough velocity to cross the equator, the ball would still have that same eastward velocity of 342 mph that it had at launch (neglecting air friction, outside forces, etc.). As it made its way over 45° south latitude, the ground would be moving eastward at 707 mph and the ball eastward still at 342 mph, so the ground would be moving eastward faster than the ball. Relative to you, an observer in the north, the ball would still be deflecting to the right, though that deflection would decrease as it neared the opposite southern latitude. Once the ball made it to the opposite southern latitude, it would then start deflecting to the left (eastward). This is true for any motion crossing the equator. The ball will not start to deflect in the other hemisphere's direction until the ball has passed the latitude opposite from its originating latitude. For instance, if a ball is launched from, say, 35° north latitude towards the equator, it will deflect to the right, cross the equator, continue deflecting to the right (while lessening its deflection as it approached 35° south latitude), and then start deflecting to the left once past 35° south latitude. To a stationary observer in space the ball would appear to travel in a straight line while the earth rotated under it.

If the ball/projectile was launched from the North Pole, it would have no eastward velocity, thus the ball would appear to be travelling to the right (west) equal to the velocity of the ground at that latitude (i.e. 707 mph at 45° North). As it crossed the equator, it would immediately appear to start deflecting to the left (east) and would eventually land at the South Pole, assuming sufficient initial velocity. To a stationary observer in space the projectile would have travelled in a straight line from pole to pole.

Just for clarification: if your ball was launched south from the equator, the ball would have an eastern velocity of 1,000 mph, and as it crossed 45° south latitude, the ground would be moving eastward at 707 mph, while the ball was moving at 1,000 mph, thus appearing to deflect to the left.

• This does not answer the question, and you seem to mix latitude and colatitude... Commented Apr 20, 2016 at 13:10
• @L.Levrel, what specifically to you find unacceptable about my answer? It is accurate. Commented Apr 20, 2016 at 19:39
• The OP wants to know if the position of the observer matters, and the figure shows a body crossing the equator. Also, if you throw a ball southwards from latitude 20°, there's no way it reaches 45°, which is northwards; additionally, tangential velocity is given by the cosine of latitude. That's why I think you wrote "latitude" (+90° at north pole, 0° at equator, -90° at south pole) where you were thinking "colatitude" (0° at north pole, 180° at south pole). I hope you didn't find my comment to be rude. Commented Apr 20, 2016 at 20:14
• @L.Levrel, sorry, my bad. I meant 70°, not 20°. I will correct my answer. My apologies. And no, I didn't find your comment rude, just inadequate. If I'm wrong, I'd like to know why instead of a blanket statement stating so. I think that goes for just about anybody. Commented Apr 21, 2016 at 0:41
• I was thinking 20° from the North Pole, and for some reason 20° stuck in my head. I have no idea why I didn't catch it; I do know the difference. Apologies to all for the confusion. Commented Apr 21, 2016 at 5:34

The above answers do not take into account the vector nature of $\omega$. I venture to provide an answer that hopefully should satisfy the questioner. We Consider a point at colatitude $\lambda$ on the Earth's surface in the Northern hemisphere. We draw a local coordinate system at this point such that the x-axis points south, the y-axis points east, and the z-axis points in the vertical direction. Then $\vec{\omega}=-\omega\sin\lambda\hat{i}+\omega\cos\lambda\hat{k}$ (because $\vec{\omega}$ points towards the north-pole). The general formula for the Coriolis force is $\vec{F_{c}}=-2m(\vec{\omega}\times\vec{v})$, where $\vec{v}$ is the velocity of a moving body as measured by the observer on the Earth. If we put $\vec{v}=v_{x}\hat{i}+v_{y}\hat{j}+v_{z}\hat{k}$ and the value of $\omega$ above into the Coriolis force formula we get $\vec{F_{c}}=(-\omega\cos\lambda v_{y})\hat{i}+(\omega\cos\lambda v_{x}+\omega\sin\lambda v_{z})\hat{j}-(\omega\sin\lambda v_{y})\hat{k}$. Now, let us imagine that the observer at colatitude $\lambda$ throws a body towards the south. Then $\vec{v}$ for such a body according to the local coordinate system at that point is $v\hat{i}$, where v is the speed of the body.The y-, and z-components of velocity of the body is zero. The formula for $\vec{F_{c}}$,then, is equal to $-\omega v\cos\lambda\hat{j}$ confirming that the body deflects to the right of the observer in the Northern hemisphere. In the southern hemisphere, an equivalent location has colatitude equal to $(180^{\circ}-\lambda)$. If we repeat the above calculation for a body thrown southwards in the southern hemisphere, we find that the body deflects to the left of the observer.

• RE: "body deflects to the right of the observer in the Northern hemisphere" and "southern hemisphere...body deflects to the left of the observer." No arguments there. But one of my main points (I hope) was that for an object originating in one hemisphere and crossing the equator, it does not deflect in that hemisphere's direction until it passes the same but opposite latitude. Commented Apr 29, 2016 at 23:39