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The Coriolis Force, Part 2

Part 2 of 2: A Quantitative Explanation

In my last article, I provided a qualitative explanation of the Coriolis force, which is a fictitious force that deflects large-scale motion on any rotating planet (including earth). But at the end of the article, I posed three questions about the Coriolis force which I had not yet answered:

1. Why does the Coriolis force deflect things to the right in the northern hemisphere and to the left in the southern hemisphere?

2. Why is it stronger when you’re farther from the equator?

3. Why is it stronger when the motion is faster?

In this article, I will use math and physics to answer these questions. In order to understand these answers, you need to have a familiarity with Newton’s Laws of Motion and with the concept of a cross-product.

Newton’s Second Law, one of the most famous and important equations ever discovered, holds that force equals mass times acceleration. The net force on a particle is equal to the product of its mass and its acceleration. It is actually a vector equation: F = ma, where a bold letter represents a vector.

In fluid dynamics, it makes more sense to divide both sides by m: a = F/m. The acceleration of some little parcel of air in the atmosphere is equal to the net force per mass that acts on it. Thus, if we want to figure out the path that the parcel will travel, we need to identify all the forces (expressed as force per mass) that act on it and add them up.

There are three main forces that act on a parcel in the atmosphere: the pressure gradient force, the Coriolis force, and friction. The pressure gradient force turns out to be (-1/ρ)grad(p) and friction is usually just written as F. But an expression for the Coriolis force does not come so easily.

(“grad” is supposed to be the upside-down triangle symbol, which means “gradient”. I am not tech-savvy enough to know how to write this symbol in a Medium article. Also, p is pressure and ρ is density.)

Recall that the Coriolis force is a fictitious force. It doesn’t actually exist; it just appears to exist because we are living in a rotating reference frame (the earth is rotating). Thus, if we want to derive an expression for the Coriolis force, we would start by defining a, the acceleration, to be the relative acceleration (also called the apparent acceleration, because it is the acceleration that we would observe from our rotating perspective), and we would then have to derive an expression for a fictitious force (the Coriolis force) that accounts for the fact that our reference frame is actually rotating.

The derivation of this expression is long and complicated. I went through the whole derivation once, and it took me several hours. It also involves several approximations. It is quite a difficult derivation. I will not waste your time by going through the entire derivation, but I will tell you the final result: — f x v. The Coriolis force is approximated as — f x v (notice the negative sign). The vector f is equal to (0, 0, 2Ωsinθ), where Ω is the rate of rotation of the earth (7.292 x 10^(-5) radians per second) and θ is latitude. Meanwhile, v is the parcel’s velocity (its relative velocity, to be exact), and the x refers to a cross product. (Again, in order to understand this article, you need to know what cross products are and how they work.)

Thus, we have our version of Newton’s Second Law for the earth’s atmosphere:

a = (-1/ρ)grad(p) — f x vF

Notice that the vector f will always point either straight up or straight down. In fact, it points straight up in the northern hemisphere and straight down in the southern hemisphere. The reason for this is that the sine of a number between 0 and 90 is positive, while the sine of a number between -90 and 0 is negative.

And that is the key to answering the first of the three questions. Recall the right-hand rule for cross products and visualize — f x v in your head. You can see that in the northern hemisphere, where f points upwards, the Coriolis force points 90° to the right of v, while in the southern hemisphere, where f points down, it points 90° to the left of v. Since v is the (apparent) velocity, this tells you how motion will be deflected by the Coriolis force in each hemisphere.

And that’s the answer to the first question: due to the right-hand rule, the Coriolis force will deflect a moving air parcel to the right in the northern hemisphere and to the left in the southern hemisphere.

The right hand rule explains why the Coriolis force deflects things to the right in the northern hemisphere and to the left in the southern hemisphere.

The answers to the second and third questions both lie in the magnitude of — f x v. Recall that the magnitude of a cross product a x b is given by absinφ, where a = |a|, b = |b|, and φ is the angle between a and b. Thus, if we increase either the magnitude of f or the magnitude of v, we will increase the magnitude of the Coriolis force.

I’ll start with answering the third question, because that’s easier. If you increase the magnitude of v, you will therefore increase the magnitude of the Coriolis force. So when an air parcel is moving faster, the Coriolis force that acts on it has a greater magnitude than when the parcel is moving more slowly. And if an air parcel is stationary (v = 0), the Coriolis force has no effect on it at all.

The answer to the second question lies in the formula for f: f = (0, 0, 2Ωsinθ), where Ω = 7.292 x 10^(-5) and θ is latitude, which ranges from -90° (south pole) to 0° (equator) to 90° (north pole). Recall that sinθ increases continuously as θ ranges from -90° to 90°. sin(-90) = -1, sin(0) = 0, and sin(90) = 1. Thus, when θ is farther away from 0, in either direction, the magnitude of sinθ increases. That means when an air parcel is farther away from the equator, the magnitude of f will be larger.

And that’s the answer to the second question. When an air parcel is farther away from the equator, in either hemisphere, the Coriolis force will have a greater magnitude. That’s why the deflection due to the Coriolis force is greater for an air mass near one of the poles than for an air mass near the equator. In fact, if an air mass is exactly at the equator, there is no Coriolis force acting on it at all, because sin(0) = 0.

The fact that the Coriolis force disappears at the equator presents a challenge for some approximations about the atmosphere. It would be easier for us if the Coriolis force were never 0. But unfortunately for us, the Coriolis force has a magnitude of 0 at the equator.

I have now answered all three questions, by using math and physics. That shows you the explanatory power of mathematics. Mathematics can explain things that a qualitative explanation just can’t.


The Coriolis Force, Part 2 was originally published in ILLUMINATION on Medium, where people are continuing the conversation by highlighting and responding to this story.

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