How Fast Is The Earth Spinning?


To get to my next point about Arthur Christmas I needed to know how fast an arbitrary point on the Earth is moving, as the Earth rotates. This required me getting out a sheet of paper and doing some sketches, so, I figured it’s worth a side article to explain what I was doing.

The first thing was that I simplified stuff. In particular, I decided the Earth is near enough a sphere that I’m not bothering with the fact that it isn’t. The difference between an actual sphere and the geoid is not worth bothering with unless you’re timing the retrofire for a ballistically-reentering space capsule. That’s … actually fairly close to the problem I want, about how long it might take the reindeer and sleigh to get back to Arthur Christmas and Grand-Santa, but that’s also too much work for the improvement in the answer I’d get.

The next thing was deciding how long the day is. The first thought is, of course, 24 hours, or 86,400 seconds, but that’s not a touch off. It’s 24 hours to go from noon to noon, or midnight to midnight, if you ignore stuff like the tilt of the Earth changing the length of the day, but that’s because over the course of the day the Earth moves in its orbit around the Sun, and the Sun needs to take a bit longer than it might otherwise to get back to the same position in the sky. For the Earth to spin around once completely — so that something which isn’t in any particular motion around the Sun or the Earth gets back to the same position — is closer to 23 hours 56 minutes 4 seconds, a time called the sidereal day. So, 86,164 seconds. To my eye that’s a more irritating number than 86,400, but I have a computer to deal with these numbers anyway. I can just use some letter like T to represent how long the day is.

If Arthur and Grand-Santa are on the equator, then, we have a pretty easy job figuring out how fast they’re moving around the center of the Earth: they travel the circumference of the Earth in one day. The circumference of the Earth is … ah … well, it’s going to be pretty near 40,000 kilometers. The meter was defined as the length it is, in part, so that the circumference of the Earth would be a nice round number, and 40 million meters is a very nice round number. So, on the equator, Arthur and Grand-Santa would travel 40,000,000 meters in 86,164 seconds, or about 464 meters per second.

Now, this isn’t quite right; Wikipedia says the speed for something on the equator is 465.1 meters per second. But the Earth’s circumference isn’t exactly 40,000,000 meters; the meter was originally defined using the longitudinal circumference of the Earth, and the equatorial circumference is different. I chose to ignore that when I pretended the Earth was a perfect sphere. And the meter’s original length was slightly miscalculated, so the longitudinal circumference wasn’t quite even, for reasons complicated enough at this point I can just point to Ken Alder’s The Measure Of All Things: The Seven-Year Odyssey That Transformed The World and promise it’s explained in there and touchingly human.

And for that matter, the sidereal day isn’t 86,164 seconds either. It’s a touch longer than that, and it’s not perfectly uniform. The Earth’s spinning is a startlingly complicated thing, when you really look at it. There’s the one big, obvious motion, with the subtle point that changes the rotation time from 24 hours to 23 hours 56 minutes and a touch. There’s smaller motions, that reflect the complex inner structure of the planet, like how the molten interior of the planet drags the rocky crust; or — this one made me stop and revel in awe — how small fluctuations in the salinity of sea water alter the ocean’s circulation patterns, and those in turn affect the weight pressing on the sea bed, and that produces a measurable influence on the rotation of the Earth. There are tidal influences from the Moon and the Sun, and subtler tugs from all the other planets of the solar system. You can find these influences, and identify them, and characterize them, and yet, the real thing will always be a little more complicated and always have another little bit to discover. It can be hard to quite understand how any person can ever be bored, when you realize that a fact like “the Earth spins around in 24 hours” is capable of infinite discovery.

Of course, most of these corrections are tiny things. Millisecond and microsecond modifications that are important to the National Institute of Standards and Technology, and to the graduate students turning measurement anomalies into presentable theses, but way beyond the level of precision needed for my little magic reindeer problem. Just as I’m pretending the Earth is a sphere 40,000,000 meters around, I’m pretending the sidereal day is 86,164 seconds, and the speed of points on the equator 464 meters per second and I don’t care to get more exact than that.

What’s really important is what happens if you aren’t on the equator. For two points — the North and South poles — the answer is easy: if you’re on either of those, you aren’t moving, relative to the center of the Earth at least. So that just leaves the rest of the Earth’s surface unanswered-for.

But the basic approach used to find the speed at the equator still works here. Imagine a spot at an arbitrary latitude — call it \theta since that’s the natural variable to use for angles, in the same way x is the default variable to use for an unknown quantity — and find how far that point travels in a day, then divide it by 86,164 seconds. That distance is the circumference of the little circle made by all the points at that same latitude. At this point the reader suspects there’s trigonometry involved and goes off to hide under the mattress. But it’s not that bad.

Why? Well, the radius of the little circle, at whatever line of latitude you want, is easy to find when you know the radius of the Earth. It’s how far some point on that line of latitude is from the axis of the Earth’s rotation, and that, neatly, is the radius of the Earth — you knew that had to be in there somewhere — times the cosine of the latitude. If you’re really comfortable with what cosine means, that’s obvious. If you’re not, drawing a little diagram with the situation might help you get comfortable with cosines.

(The elements of your drawing: the Earth, drawn in cross-section, so it looks like a circle; a line across the middle, showing the equator; a point on the edge of the circle, showing where you are; a line parallel the equator and going through that point, showing the line of latitude; a line from the center of the Earth to your point, marked ‘R’ or maybe ‘Re‘ showing the radius of the Earth; and a \theta symbol which happens to be your latitude. If you get stuck how this all fits together sleep on it, or at least walk away and think about other things a while, although if you’re still stuck I don’t mind showing the diagram and how to interpret it.)

The radius of the Earth I don’t know offhand either, but since I know the circumference of the Earth I know 2 \pi R and, nicely enough, the total distance travelled around the Earth’s axis of rotation in a day at the latitude \theta is going to be 2 \pi R \cos\left(\theta\right) . So I can skip working out the radius of the Earth times anything and replace it with the circumference of the earth times the cosine of the latitude, and the circumference is that nice round number. The distance travelled in a day is, then, 40,000,000 meters times the cosine of the latitude. A day is 86,164 seconds, so dividing the two — well, we can do the 40,000,000 divided by 86,164 first; we already did that when working out how fast points on the equator move. Then we can multiply that result by the cosine.

So to conclude: if you’re at the latitude \theta , then the linear speed, how fast you’re moving around the center of the Earth, is roughly 464 \cdot \cos\left(\theta\right) meters per second.

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