# Is there an intuitive explanation for the Southward force caused by the Coriolis Effect on rotating spheres?

From the Coriolis Effect article on Wikipedia, the following with regard to the Coriolis Effect on a rotating sphere:

By setting vn = 0, it can be seen immediately that (for positive φ and ω) a movement due east results in an acceleration due south. Similarly, setting ve = 0, it is seen that a movement due north results in an acceleration due east. In general, observed horizontally, looking along the direction of the movement causing the acceleration, the acceleration always is turned 90° to the right and of the same size regardless of the horizontal orientation.

My intuitive (but possibly incorrect) understanding is that if there are two points, Point A and Point B, at different latitudes in the Northern hemisphere, the Eastward velocities of these points are different because they are at different distances from the Earth's axis of rotation, and this causes the Coriolis effect for a rotating sphere.

If a projectile is fired due North from Point A near the Equator towards Point B near the North Pole, the projectile will start off with the higher Eastward velocity of Point A, and will land to the East of Point B, which is moving East at a slower velocity than the projectile.

Firstly, is that correct?

If that IS correct, that brings me to the quote off Wikipedia. The quote implies that, if a projectile is fired due East, it will experience a Coriolis force to the South. My intuitive explanation based on differences in velocities between origin and destination does not account for a Southwards movement at all, since the Point A and Point B velocities are identical if the points are on the same latitude.

What am I missing? Would a projectile fired due East or due West experience any North or South drift caused by the Coriolis effect (or anything else for that matter)?

Why?

• I'm puzzled by it too, although there's an obvious effect. If a projectile starts at 1 meter away from the north pole and moves due east or west, very quickly it will be moving south. – Mike Dunlavey Oct 31 '12 at 14:34
• @MikeDunlavey True, and tricky to conceptualize! However, that effect is just because it's a sphere, it has nothing to do with the fact that the sphere is rotating. It will be the same for a stationary sphere, and thus has nothing to do with coriolis. – Pieter Müller Oct 31 '12 at 14:41
• There's something strange. For example, if $a=-2\Omega\times v$, shouldn't it be 0 when you throw something in north direction? in that case $v$ is parallel to $\Omega$. But this is obviously nonsense... – Bzazz Jan 29 '13 at 23:58
• Thanks for asking! I've been meaning to ask for about a year – aPhysicist Aug 21 '14 at 2:27
• @Bzazz at the equator a $v$ of northerly velocity is parallel to the axis of rotation... and so there Coriolis deflection (except of course as it heads further north of the Equator, it will start transitioning its own motion into increasing component perpendicular to Ω). But near the North Pole, north (or any other horizontal direction for that matter) is just about entirely perpendicular to Ω) – JeopardyTempest Jun 29 '18 at 14:29

## 2 Answers

Firstly, is that correct?

Yes your intuitive understanding for this part of the Coriolis effect is correct.

The second part, that is, why wind in the East direction is deflected South, is a bit trickier, and involves the use of centripetal force. this is given by the equation:

$F = \frac{mv^2}{r}$

If we re-arrange the above equation, we can find $r$ in terms of $v$, and we arrive at:

$r = \frac{mv^2}{F}$

This tells us that as velocity increases, the radius required to maintain the orbit also increases.

Now let's apply this concept to winds on the Earth. If we feel no wind on the Earth, then the air in the atmosphere is travelling at the same velocity as the Earth. The Earth is naturally spinning towards the East.

In the case of an additional Eastward wind felt on the Earth, this wind has effectively increased its velocity, and therefore the above equation tells us that the radius of orbit must increase as well. Radius in this case is the distance, measured perpendicularly of the Earth's axis, between the axis and the wind.

In order for the radius to increase, the wind moves southwards, where the radius is larger.

Similarly, wind moving in the West direction, is moving in the direction opposite of that to the Earth, and therefore its velocity is decreased. Consequently this wind moves towards the North, where the radius is less.

The above image shows what happens. The wind moving East begins to expand its radius, thus moving outwards. Gravity pulls it back, and the wind moves South, in order to maintain the larger radius required for its increased velocity.

• Thanks Chris, that makes a lot of sense and is very well presented! – Pieter Müller Jan 30 '13 at 14:29
• If an object moved eastward at a low speed, then by your explanation, wouldn't it try to move North. But if an object moved eastward at a great speed wouldn't it try to go south? – aPhysicist Aug 21 '14 at 3:07
• @aPhysicist, no because even if the Eastward velocity is small, it is nevetheless still an increase in radial velocity and thus will require an increase in radial orbit to sustain. Remember, even when wind is 0, there is still a radial velocity due to Earth's rotation. – Kenshin Aug 21 '14 at 4:21
• @Kenshin but why it doesn't go up in order to increase the radius? In the case of winds perhaps density differences would make it easier to go south instead of up, but in the case of a projectile that density difference wouldn't apply. However, the effect is the same. Why? – Camilo Rada Feb 12 '18 at 18:45
• Camilo appears to have known the answer well based upon his post in How to deduce the direction of the Coriolis effect: the thing is, going "up" doesn't increase the radius to the axis at the Pole, but fully does at the Equator. Full Coriolis has a vertical term, and applies everywhere on Earth. Moving outwards from the rotation axis is fully "horizontal" at the Poles, fully "vertical" at the Equator, and the proportion transitions from one to the other in between. – JeopardyTempest Jul 2 '18 at 12:46

It's a geometric effect. Consider a point stationary in a non-rotating frame initially coincident with a point on the sphere north of the equator. In the rotating frame, vE is negative (magnitude equal to the local rotational velocity). The coordinate of the point in the rotating frame acquires an instantaneous acceleration directly towards the axis of rotation. The projection of the apparent acceleration onto the local N/E coordinate components has a nonzero N component.

This is the argument at negative vE equal to the local rotational speed. Other values require tweaking the geometric picture, but the idea is the same.