How to Figure Out the Size of the Moon Yourself


Having confirmed we live on a spherical, spinning Earth whose circumference, diameter and radius are roughly 25000, 8000, and 4000 miles (40000, 13000, and 6500 km) respectively, it’s time to ask about the properties of the objects that are most obvious in the sky: the Sun and Moon. How big are they, and how far away?

If the Moon were close to Earth, then at any one time it would only be visible over a small part of the Earth, as indicated in light blue. But in fact (except at new moon) about half the Earth can see it at a time.

Historically, many peoples thought they were quite close. With our global society, it’s clear that neither can be, because they can be seen everywhere around the world. Even the highest clouds, up to 10 miles high, can only be seen by those within a couple of hundred miles or so. If the Moon were close, only a small fraction of us could see it at any one time, as shown in the figure at right. But in fact, almost everyone in the nighttime half of the Earth can see the full Moon at the same time, so it must be much further away than a couple of Earth diameters. And since the Moon eclipses the Sun periodically by blocking its light, the Sun must be further than the Moon.

The classical Greeks were expert geometers, and used eclipses, both lunar and solar, to figure out how big the Moon is and how far away. (To do this they needed to know the size of the Earth too, which Eratosthenes figured out to within a few percent.) They achieved this and much more by working carefully with the geometry of right-angle triangles and circles, and using trigonometry (or its precursors.)

The method we’ll use here is similar, but much easier, requiring no trigonometry and barely any geometry. We’ll use eclipses in which the Moon goes in front of a distant star or planet, which are also called “occultations”. I’m not aware of evidence that the Greeks used this method, though I don’t know why they wouldn’t have done so. Perhaps a reader has some insight? It may be that the empires they were a part of weren’t quite extensive enough for a good measurement.

The Method: Concept

Suppose you place a sphere in front of a wall, and you illuminate the sphere with a tiny light bulb that sits on the line that is perpendicular to the wall and passes through the center of the sphere, as in the picture below. The shadow cast by the sphere on the wall will generally be larger than the sphere itself. But the further away you move the bulb, the more the sphere’s shadow will be the same size as the sphere! You can verify this yourself with your hand and the small light on your cell phone.

A sphere lit by a close small bulb will create a shadow on a screen that is larger than the sphere itself. If the bulb moves further away, the shadow shrinks, and eventually has the same size as the sphere.

A shadow, of course, is a place where light from the bulb cannot reach the wall because it is blocked by the sphere. What that means is that an ant on the wall can see the bulb only if it stands outside the shadow; if it stands within the shadow, the sphere has eclipsed — blocked the view of — the bulb. That’s important here, because it could happen that ambient light makes the shadow impossible to see, yet it could remain relatively easy for an ant to tell if the bulb is blocked or not.

So if the bulb were quite far away, and we had a bunch of ants wandering around on the wall, and we asked them, “Can you see the bulb?”, and then we drew a circle around all the ants who replied, “No”, then the circle we drew on the wall would be the same size as the sphere.

Great! Our method is essentially this. The role of the bulb will be played by a distant star or planet. The role of the sphere will be played by the Moon. The role of the wall will be played by the Earth. And the role of the ants will be played by you, me, and all our friends with eyes and binoculars.

Note added: as a reader pointed out, somehow I forgot to state explicitly that in the next section I am assuming that the occulted stars and planets are at least 10 times farther away from the Earth than the Moon is, so that the shadow they cast is no more than 1/10 (i.e. 10%) larger in diameter than the Moon. That assures that the method is accurate to better than 10%. This assumption, of course, has to be tested before the method can be fully trusted; were it false, then the method would give us an over-estimate of the Moon’s diameter. However, as we will see, the second method described in this post, using solar eclipses, gives the same answer as the first. The fact that the two methods agree thus proves the assumption that the planets and stars are far away. Really I should rewrite the post to make this logic less convoluted.

The Method: Details

Now there are a few suspicious things about this method, so let’s deal with them right away.

I started with a tiny bulb, but a distant star or planet is huge! Fortunately, that’s okay, because actually the bulb just has to be so far away that it looks tiny — that it occupies much less sky than the Moon does. A huge but distant star which, unlike the Sun, appears to be a point of light, will do just fine. The planets appear larger than the stars, as you can easily see in binoculars, but they’re still much smaller dots than the Moon’s disk. The Sun will not suffice for this purpose, since it looks as big as the Moon; if you’re interested in how to use the Sun, I have an aside about that at the end of this post.

Meanwhile, the screen is flat, but the Earth is curved. That’s ok, as long as we use the part of the Earth closest to the Moon. That’s because any small part of a huge sphere is pretty close to flat, and the part closest to the Moon acts like a nearly-flat screen. (This only works because the Moon’s much smaller than the Earth, so we only need to see the shadow on a small part of the Earth.) The Earth’s curvature will make the shadow a bit larger than it would be on a flat wall. And if the shadow gets too close to the edges (say, to the north or south pole) it will be greatly distorted and the method I’m outlining won’t work. But if we can find the Moon’s shadow when it’s more or less centered on the Earth behind it, then the curved surface of the Earth is close enough to a flat wall for our purposes. Fortunately, this turns out to be easy.

Another issue is that a star or even a planet is far too dim to cast an obvious shadow on the Earth. This is mainly because, except at new Moon, the Moon’s own light (i.e., the sunlight that it reflects) is much brighter than the light from the star or planet, so it washes out its own shadow. That’s why we’ll instead use the “ask the ants” method. I’ll still refer to the blocked region as “the shadow”, for lack of better terminology, but remember it’s really the region where the star or planet cannot be seen because the Moon is in the way.

And what about all of us ants? What are we supposed to do to carry out this method?

One option is to wait for the next occultation of a planet or bright star by the Moon. In advance of this event, we can organize a big international star party where we’re all watching, and at a particular chosen moment, we all report whether we can or cannot see the star. Not only is that a good idea, astronomers (professional and amateur) do this all the time. And we won’t have to wait long, because occultations of naked-eye stars happen several times a month. Occultations of planets are less common, but not rare either; I’ve seen five myself, without trying very hard.

For 2022, here’s a long list of occultations. Note particularly that Venus and Mars are both occulted this year; all the rest are stars. For each entry on the list, click on the link under “NAME” to see a map that shows where it will and won’t be visible.

If you don’t trust the maps, you can tune in to the next star party and participate to see for yourself; here’s a nice introduction (pdf format) for those interested in watching occultations. But if you’re too impatient for that, and if you do trust the maps (which are used constantly without anyone complaining) then we can read off the Moon’s size just from them.

You can see this in the predictions posted to the right; there are three of them, one some years ago, one this week, and one later this year. Each prediction is not a circle, but is instead a strip. That’s because the Moon is moving and the Earth is spinning, so the location of the circle of obstructed ants moves around over time. At one moment the circle might be over Africa, and a few hours later it might be over Asia. So even though at any one moment it’s a circle, over time it forms a strip — whose width is the diameter of the circle!

Predictions for three occultations of fairly bright stars by the Moon, visible from near the equator, from the past, the present and the future. At any one moment, the region where the occultation can be seen has the shape of a disk the size of the Moon; over several hours, the disk moves, forming the strip shown in each map.

I should point out that I’ve selected examples where the Moon’s shadow crosses the equator, or nearly so. As I mentioned earlier, the shadow gets distorted when it lies near the edges of the Earth as seen from the Moon; there the Earth’s curvature becomes a bigger effect. On top of that, these predictions show a flat map to represent the Earth, and since no flat map can ever represent a sphere accurately, the resulting distortion makes the shadow all wonky, much more than it actually would if you were looking at a globe. So the occultations that can be seen from near the equator — about half of them — are the best ones for measuring the Moon’s size.

In fact, in the figure’s three examples, the strip where the occultation can be seen is always about the same size. And that, my friends, is the size of the Moon. If we measure the width of the predicted (or measured) strip where at some time the occultation will be (or was) visible, and we do so near to the equator — and even better, in the middle of the strip, which will be on the part of the Earth closest to the Moon — then we are estimating the diameter of the Moon.

The Size of the Moon

Already you can tell from the maps that the Moon’s diameter must be less than half that of the Earth, but clearly more than one tenth. Let’s see what we get using a rough measurement.

Below is a (successful) prediction for the occultation of the planet Mars by the Moon on April 17th of last year. The shadow’s motion across the Earth’s surface switched from northward to southward, and I’ve put a dark disk at that point, showing the shadow as it would have been at that moment. This is a particularly easy location in which to interpret the map. You can see the shadow’s northward edge is halfway across Afghanistan, and its southward edge is just north of the equator.

At right is a calculation, using Google Maps, of the distance between these two points on the Earth. I just eyeballed it, and remember the Earth’s curvature will extend it just a bit, so I really don’t have a right to expect it to be more accurate than 5%. (You’re invited to do it more carefully if you like.) I ended up with about 2200 miles (3600 km.)

Left: the region where an occultation of Mars by the Moon was visible in April 2021. At the moment the path reached its most northerly extent, the occultation would have been visible within the dark circle. Right: the diameter of the dark circle is estimated using Google Maps to be about 2200 miles (3600 km).

According to precision studies using modern science, the Moon’s average diameter is 2,159.1 mi (3,474.8 km). So I did indeed come within 5%.

Now you may complain that I used someone’s predictions to come to this estimate, rather than an actual measurement. But remember, this is a prediction for something that happened in the past; the occultation was widely observed, so if the prediction had been wrong, you would certainly have heard about it. And you always have the option of waiting for future similar events, when at a given moment, astronomers will report whether they have seen the star or planet disappear behind the Moon. If, for an occultation not far from the equator, you compare the northernmost observer to the southernmost observer at that particular moment, you will find the distance between them is quite close to the Moon’s diameter.

No triangles, no trigonometry. Just timing and location reports from some star-gazers on the right night; that’s all we need.

Extra: Doing the Same Thing with a Solar Eclipse

There’s another slightly more complicated way to do this same thing using a solar eclipse, when the Moon blocks the view of the Sun. It’s similar, but interestingly different, because unlike a distant star, which looks like a point from our perspective, the Sun occupies the same amount of sky as does the Moon. Of course the Sun has to be further away, or the eclipse couldn’t occur, so its true size has to be bigger than the Moon. (Whether it’s bigger than the Earth is something we’ll address later.)

When you look at the shadow of an object cast by a large source of light — for instance, the shadow of a tennis ball cast by a standard-sized light bulb — you’ll see it can have two components if you move the ball close enough. There’s a central shadow, the “umbra”, which is dark. There, the bulb’s light is completely blocked by the ball. Outside that is a region where there’s a faint shadow, the “penumbra”; there the ball only blocks part of the bulb’s light. Try it.

The trick for an eclipse is the following. Because the Moon and Sun appear similar in size in the sky, the umbra of the Moon’s shadow is a very narrow region, usually only a couple of hundred miles (km) wide. There the Sun is completely blocked and the eclipse is total: the sky goes pretty dark and the Sun’s enormous outer atmosphere (“corona”) can be seen. The penumbra, where the eclipse is partial, is much larger, thousands of miles (km) wide. There the Moon incompletely blocks the Sun, and so the sky gets a bit dimmer but the world remains in daylight.

Now here’s a remarkable fact: the diameter of the Moon is approximately one half of the difference between the diameter of the penumbra and the diameter of the umbra. Since the umbra is much smaller than the penumbra, the diameter of the Moon is approximately one-half of the diameter of the penumbra.

Top: from a distant star, the Moon’s image is directly projected onto the screen. Bottom: Since the Sun and Moon appear similar size in the sky, a nearby Sun (drawn much closer and smaller than it actually is) will be just barely blocked by the Moon (blue lines), creating a small dark umbra where the sky goes dark: a total eclipse. In the penumbra, part of the Sun is blocked; the red lines which indicate the edge of the penumbra mark locations where the Moon will appear to just barely touch the Sun. The black dashed lines indicate the Moon’s image projected as in the top figure. One can see by eye, or by the use of triangle geometry, that the Moon’s diameter is about half that of the penumbra (because the umbra is very small.).

The geometry that makes this true is shown in the figure above (which is not to scale; it is drawn to make the conceptual ideas easy to understand.) While a distant star’s light rays are essentially parallel by the time they reach Earth, the Sun’s rays can arrive at the Earth from various directions from different parts of the Sun. Only in the umbra — very small because the Sun and Moon appear almost the same size in the sky — can no rays from the Sun reach the Earth; there the eclipse is total. This is indicated by the blue lines, which show that neither the top nor the bottom (as the picture shows them) of the Sun can be seen. Right at the edge of the penumbra, as shown by the red lines, the Sun and Moon would seem barely to touch in the sky. Inside the penumbra, some but not all of the Sun’s rays are blocked by the Moon, and the sky remains bright; there the eclipse is partial, with a crescent Sun that becomes narrower as one approaches the umbra.

Shown below at left is the path of a famous solar eclipse from 1991; I remember it as a partial eclipse from my graduate years in California, and still regret that I did not find a way down to Baja California to see it as a total eclipse. The umbra’s path across the world is marked in dark blue, the edge of the penumbra by the outermost light blue lines. The red dot at center marks mid-eclipse, which occurs when the Earth’s surface is closest to the Moon; conveniently, the distance from the equator is not large either. This means the Earth’s surface is relatively flat along the 1900UT line (a line of fixed time), and an estimate of its length from midpoint to edge is obtained using Google Maps in the picture at right.

Left: the path of the 1991 solar eclipse, showing the umbra (dark blue) and penumbra (light blue). At the time 1900 UT, the umbra was centered just off Baja California (Mexico) while the penumbra extended into Ontario (Canada). This distance, half the diameter of the penumbra, should be approximately the diameter of the Moon. Right: The distance is estimated using Google Maps.

This method gives an answer very similar to the last one, and again accurate to better than 5%, which is as much as I’d trust my rather haphazard approach. A dedicated observer could do considerably better. But for a rough estimate, it’s already pretty good!

11 responses to “How to Figure Out the Size of the Moon Yourself

  1. Wouldn’t we need to know that a certain planet or star is far away first? Also, how far would it have to be to for the size of its shadow to accurately match the moon’s size? This method seems to require that we know the distance to the planet or star first, otherwise we could easily get a shadow that is too big or too small, and thus inaccurately calculate the size of the moon.

    • Fair question; there is an issue here of which assumptions to make in which order while doing this, and you are correct that I did not state my assumptions in a logical order. It’s not that the logic is circular but that it is iterative.

      The planets and stars look like points to our eyes. If you look at the first picture, we can’t get a shadow that’s smaller than the Moon from a bulb that appears to our eyes to be a point. We can only get one that’s too large. And if you think about the triangles involved, if the object’s distance is ten times the Moon’s then your error on the shadow can only be 10%. So really one only needs to know that planets and stars are more than 10 times further than the Moon in order to get a good estimate.

      It’s not that hard to show that the Sun and the outer planets, which orbit it, are more than 10 times further away. These are things that the ancient Greeks knew, in ways that I’ll soon discuss. And these are things that we, too, can even more easily know, using modern technology.

      But it is true I have not *yet* shown this. So really I ought to say this is an assumption that we will confirm later.

      As for the stars, the question of how to put an easy lower bound on their distances may require first confirming that the Earth goes round the Sun; or maybe there’s a simpler method that I haven’t thought of yet.

      The need for iterative logic — where we make an assumption, derive a consequence, then use that consequence to learn new things (still subject to that assumption) which then in turn allow us to confirm the assumption — is a common and effective practice in science. Perhaps a little discussion of that is merited, because indeed it’s important to track all of our unconfirmed assumptions.

      Thanks for the question.

  2. How precise does the timing need to be? Synchronized clocks weren’t a thing until the advent of railroads. Noon was just whenever the sun was at its highest at your exact location.

    • I think it depends on the technique you use, which in turn might depend on how bright the object is that’s being occulted. If you really do exactly as I described here, then the timing has to be pretty decent — probably within half an hour if you want to get a decent estimate. But it’s not so hard to do that if you can state which bright stars are directly overhead on your north-south line at the time you made the observation; that’s latitude-independent, and easily corrected for longitude. An alternative, If you can see the object disappear and reappear — easy only for planets, not stars, because the Moon’s so bright even when it’s a crescent — then you can draw a line across the Moon from disappearance point to reappearance point. Compare your line with the line drawn by someone else’s who’s significantly further north or south, and just by looking at how the two lines are different relative to the Moon’s apparent position, you already have all the information to predict how wide the shadow was and how large the Moon is — no timing necessary. And there are other methods you could try. So I suspect that the Greeks didn’t use this method only because they couldn’t predict lunar occultations of even planets with confidence (or, perhaps, couldn’t do so until after they already had other methods to measure precisely the Moon’s size.)

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