Beyond the Book (and What the Greeks Knew About the Earth)

Since the upcoming book is basically done, it’s time for me to launch the next phase of the project — the supplementary material, which will be placed here, on this website.

Any science book has to leave out many details of the subjects it covers, and omit many important topics. While my book has endnotes that help flesh out the main text, I know that some readers will want even more information. That’s what I’ll be building here over the coming months. I’ll continue to develop this material even after the book is published, as additional readers explore it. For a time, then, this will be a living, growing extension to the written text.

As I create this supplementary material, I’ll first post it on this blog, looking for your feedback in terms of its clarity and accuracy, and hoping to get a sense from you as to whether there are other questions that I ought to address. Let’s try this out today with a first example; I look forward to your comments.

In Chapter 2 of the book, I have written

• Over two thousand years ago, Greek thinkers became experts in geometry and found clever tricks for estimating the Earth’s shape and size.

This sentence then refers to an endnote, in which I state

• The shadow that the Earth casts on the Moon during a lunar eclipse is always disk-shaped, no matter the time of day, which can be true only for a spherical planet. Earth’s size is revealed by comparing the lengths of shadows of two identical objects, separated by a known north-south distance, measured at noon on the same day.*

Obviously this is very terse, and I’m sure some readers will want an explanation of the endnote. Here’s the explanation that I’ll post on this website:

Chapter 2, Endnote 1

By the time that ancient Rome’s power was expanding, Greek scholars understood the basics of solar and lunar eclipses. Noting the relation between the phases of the Moon and the positions of the Sun and Moon, they recognized that the Moon’s light is reflected sunlight. They were aware that New Moon occurs when the Sun and Moon are on the same side of the Earth, while Full Moon occurs when they are on the opposite sides. And they knew that a lunar eclipse occurs when the Earth lies between the Sun and Moon, so that the Earth blocks the Sun’s light and casts a shadow on the Moon. This is illustrated (not to scale) in Fig. 1.

From these shadows, they confirmed the Earth was a sphere. Clearly, if the Earth had the shape of an X, it could cast an X-shaped shadow.

If Earth were a flat disk like a coin, then depending on the Moon’s location in the sky, the Earth’s shadow might be circular or might be oval. The important point is the shadow of a circular disk is not always a circular disk. You can confirm this with a coin and a light bulb.

The only shape which always creates a circular, disk-like shadow, from any angle and from any place and at any time, is a sphere, as you can confirm with a ball and a light bulb.

This is consistent with what is actually observed in eclipses, as in Fig. 4. The circular shadow of the Earth is shown especially clearly when sets of photos taken during an eclipse are carefully aligned.

Next, one you are aware that the Earth’s a sphere (and, as the Greeks also knew, that the Sun is far away ), it’s not hard to learn the size of the Earth. Imagine two vertical towers sitting on flat ground in two different cities. To keep things especially simple, let’s imagine one city is due north of the other, and the distance between them — call it “D” — is already known.

In each city, a person observes their tower’s shadow at exactly noon. (No clock is needed, because one can watch the shadow over time, and noon is when the shadow is shortest.) The end of the shadow and the tower’s base and top form the points of a right-angle triangle, as in Fig. 5, whose other angles we can call ⍺ and 𝜃. For a person squatting at the shadow’s end, the angle ⍺ is easily measured: it is the angle formed by the tower’s silhouette against the sky. Because this is a right-angle triangle, the angle 𝜃 is 90 degrees minus ⍺, so each observer can easily determine 𝜃 for their city’s tower. We’ll call their two measurements 𝜃₁ and 𝜃₂. Knowing these angles and their distance D, they know everything we need to determine the size of the Earth.

The key observation is that the difference in these angles is the same as the difference in the latitude between the two cities. To see this, examine Fig. 6 below. The paths of sunlight (the orange dashed lines in Figs. 5 and 6) are parallel to the Earth-Sun line. [This (almost-exact) parallelism is only true because the Sun’s distance from Earth is much larger than the Earth’s size — which the Greeks knew.] Meanwhile the line from the Earth’s center to the first tower (call it L1) is a continuation of the line from that tower’s base to its top. Because (a) L1 forms an angle 𝜃₁ with the sunlight, as shown in Fig. 5, and (b) the sunlight line and Earth-Sun line are parallel, as shown in Fig. 6, the intersection of L1 with the Earth-Sun line is also 𝜃₁! Similarly, the line from the Earth’s center to tower 2 forms the angle 𝜃₂ with the Earth-Sun line. As can be seen in Fig. 6, it follows that the angle between the two lines connecting the Earth’s center to the two towers is 𝜃₂ – 𝜃₁ , the difference in the noon-time sun angles as seen by the two observers!

Now, however, the observers can use the fact that they know D, the distance between the cities. In particular, the distance D is to the Earth’s circumference C just as 𝜃₂ – 𝜃₁ is to 360 degrees (or, in radians, to 2ℼ). In formulas

• D/C = (𝜃₂ – 𝜃₁)/360°

So (dividing and multiplying on both sides) the Earth’s circumference is simply

• C = D [360° / (𝜃₂ – 𝜃₁) ]

and since they know both D and 𝜃₂ – 𝜃₁ , they now know the Earth’s circumference. (Note this correctly says that if the angle were 90 degrees = 2ℼ/4, then D would be C/4.)

Eratosthenes made this measurement (in a slightly different way) around 240 B.C.E. Reports by classical historians do not quite agree on what he found, but in the most optimistic interpretation of the historical record, he was well within 1 percent of the correct answer. And why not? Once you’ve realized what you should do, this is a relatively simple measurement; it’s one that you and a distant friend could carry out yourselves.

Note: You might prefer to see the answer in radians instead of degrees; since 360° = 2ℼ radians, we can write

• C = D [2ℼ / (𝜃₂ – 𝜃₁) ] (in radians)

and since C = 2ℼR, where R is the Earth’s radius, that gives us a particularly simple formula

• R = D / (𝜃₂ – 𝜃₁) ( in radians)

Note: If the two cities are not due north-south of one another, this poses no problem. Measure tower 1’s shadow at the first city’s noon, and tower 2’s shadow at the second city’s noon on the same day; then take D not to be the distance between the two cities but instead the distance between their latitude lines. Practically speaking, we can make a triangle with one side being the distance between the cities and the other two sides aligned north-south and east-west; then D is the length of the north-south line, as in Fig. 7. With this definition of D, the formulas above are still valid.