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The moon is smaller than the Earth, but how do we know that (without the use of modern technology)? To be more specific, how can we show that the moon is smaller than the Earth (smaller diameter) with technology before 1800s?

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  • $\begingroup$ See: physics.stackexchange.com/questions/136160/… $\endgroup$ Dec 26, 2022 at 7:08
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    $\begingroup$ Because we went there and looked. Not, admittedly in the 1800s $\endgroup$
    – Valorum
    Dec 26, 2022 at 10:01
  • $\begingroup$ @Valorum well did the Apollo mission actually take a measurement of the moon’s diameter? $\endgroup$
    – Obama2020
    Dec 26, 2022 at 16:25
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    $\begingroup$ @Obama2020 - Actually yes. en.wikipedia.org/wiki/Lunar_Laser_Ranging_experiment Because we have the precise distance of the moon at various points on the moon we can work out the moon's approximate curvature by firing ranging pulses at the same time. $\endgroup$
    – Valorum
    Dec 26, 2022 at 16:29
  • $\begingroup$ How could 'technology before 1800s' matter, please? Why not ask about Babylon or Egypt or ancient Arabia? $\endgroup$ Dec 27, 2022 at 21:56

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Because when the shadow of the moon hits the earth for eclipse, it's only a small shadow that covers a little zone of the earth and lasts a brief moment.

When the earth shadow passes on the moon, it lasts a lot longer and it's a bigger shadow. They measured the time 2250 years ago and found the earth is 3.5 times bigger than the moon:

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  • $\begingroup$ A good application of Occam's razor $\endgroup$
    – user47732
    Dec 29, 2022 at 11:13
  • $\begingroup$ Can you give a reference to (a source detailing) the measurement 2250 years ago? $\endgroup$
    – Walter
    Jan 1, 2023 at 13:33
  • $\begingroup$ Ok, which of these references is best? google.com/search?q=aristarchus+moon+shadow $\endgroup$ Jan 1, 2023 at 23:16
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Another way is to have two people on opposite sides of the Earth measure the position of the Moon at the same moment. Knowing the size of the Earth, it’s easy to triangulate the Moon’s distance—imagine you have a triangle with you and your friends as the base, and the Moon as the summit: you know the size of the base and two angles, so you can easily calculate the length of the other two sides. From there, it’s then easy to figure out its physical diameter from its apparent diameter:

$ \delta = 2 \arctan \left(\frac{d}{2D}\right) $

where $ \delta $ is the angular diameter in degrees, d the actual diameter, and D the distance, both measured in the same units.

EDIT (Addition):

In the image below, the Earth is the circle on the left and the Moon is the grey one on the right. (Diagram is NOT to scale!) You are located on the top of the Earth circle (which doesn’t mean it’s the pole…), and your friend is at the bottom. You both measure the position of the Moon with respect to the same star (for example, the fourth one from the top; the one just below the line that goes from your friend to past the Moon), and you compare your measurements. The stars are presumed to be at an infinite distance, so you can easily figure out the angle difference.

enter image description here

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  • $\begingroup$ It would have been feasible even without modern technology, as our distant ancestors did have hourglasses, clepsydras, and other ways to know the time… $\endgroup$ Dec 24, 2022 at 22:24
  • $\begingroup$ I’m not sure if I’m picturing your example correctly. So the angles would be the angle that the distance to the moon forms with the ground, right? How do you calculate that? Could you link a picture? $\endgroup$
    – Obama2020
    Dec 24, 2022 at 22:28
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    $\begingroup$ @PierrePaquette: Even so, a timepiece that is simultaneously accurate and portable is quite a bit more recent than you might expect. $\endgroup$
    – Kevin
    Dec 25, 2022 at 4:04
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    $\begingroup$ @Kevin A lunar eclipse is a timepiece that is perfectly accurate and also completely portable, being present everywhere on Earth at the same instant. Ancient measurements of lunar parallax used lunar eclipses as a source of simultaneity. $\endgroup$ Dec 25, 2022 at 6:37
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    $\begingroup$ precision time pieces are not needed, just have observers at different lattitudes record the Moon's passage through a known constellation on the same night. the size difference is so exetreme that high precision is not needed to find that there is a difference. $\endgroup$
    – Jasen
    Dec 26, 2022 at 0:11
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There are two parts of the problem of proving that the Moon is smaller than the Earth.

One) Proving the Size of the Earth.

Two) Proving the size of the Moon.

Once the sizes of the Earth and the Moon are known, their relative sizes will be obvious.

Eratostenes of Cyrene (c.276 BC to 195/94 BC) was an ancient Hellenistic philosopher who calculated the circumference of the Earth (and thus its diameter). Knowing that Alexandria was about 5,000 stadia north of Syene (Aswan), Eratostenes found the difference in the angles of the noonday Sun on the same day at Alexandria and Syene. Thus Eratostenes could find how many stadia were in what fraction of the Earth's circumference, and calculated the circumference of the Earth fairly accurately.

https://en.wikipedia.org/wiki/Eratosthenes

Thus the approximate size of the Earth has been known for about 1,900 years before 1800s levels of technology.

Ancient people could roughly measure the angular diameter of the Moon at about half a degree of arc. So once the distance to the moon was calculated, it would be a simple matter to calculate the Moon's actual diameter from its angular diameter at that distance.

And estimates of the relative sizes of the Earth and the Moon were made at about the time of Eratostenes or even before.

Aristarchus of Samos (c. 310-230 BC), another Hellenistic philosopher, wrote On the Sizes and Distances, discussing the sizes and distances of the Sun and the Moon, expressed in relative terms, in radii of the Earth.

Aristarchus estimated and/or calculated that the radius of the Earth was 2.85 times the radius of the Moon (it is actually 3.5 times, and the distance to the Moon was 20 Earth radii (it is actually 60.32 Earth radii). Thus his figures for the Moon were underestimates. Aristarchus calculated the radius of the Sun as 6.7 Earth radii and the distance as 380 Earth radii (actually they are 109 Earth radii and 23,500 Earth radii). Clearly his methods were much less accurate with the Sun than the Moon.

https://en.wikipedia.org/wiki/On_the_Sizes_and_Distances_(Aristarchus)

On Sizes and Distances (of the Sun and Moon) (Περὶ μεγεθῶν καὶ ἀποστημάτων [ἡλίου καὶ σελήνης], Peri megethon kai apostematon) is a text by the ancient Greek astronomer Hipparchus (c. 190 – c. 120 BC) in which approximations are made for the radii of the Sun and the Moon as well as their distances from the Earth. It is not extant, but some of its contents have been preserved in the works of Ptolemy and his commentator Pappus of Alexandria. Several modern historians have attempted to reconstruct the methods of Hipparchus using the available texts.

https://en.wikipedia.org/wiki/On_Sizes_and_Distances_(Hipparchus)

Hipparchus also found that the Earth was larger than the Moon.

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    $\begingroup$ Incoherent reasoning because (a) the Moon can only measured relative to the Earth and the size of the Earth does not need to be known for this and (b) the distance to the Moon can only be measured by deducing it from its size. $\endgroup$ Dec 25, 2022 at 6:40
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    $\begingroup$ @MartinKochanski I thought we could calculate distances to celestial bodies using parallax. Does that not work for the moon specifically for some reason? $\endgroup$ Dec 25, 2022 at 10:52
  • $\begingroup$ @MartinKochanski With modern techniques using parallax (measuring the angles to the Moon as seen from 2 places on Earth with known distance from each other) works fine to determne the distance of the Moon. Cassini measured the parallax (and thus distance) of Mars ,many times farther than the Moon, in 1672. And for 50 years laserbeams bounced off of reflectors left on the Moon have been used to measure the moon's distance including its minute yearly increase. Thus the distance to the Moon is known independently of its size, and can prove its size. $\endgroup$ Dec 25, 2022 at 20:17
  • $\begingroup$ @MartinKochanski Keper's Somnium, written in 1608 and published in 1634, is sometimes called the first Science Fiction story. In it a daemon describes Levania, a name for the Moon": "The island of Levania is located fifty thousand German miles high up in the air"". frostydrew.org/papers.dc/papers/paper-somnium And obviously the Moon would be much smaller than Earth to have an angular diameter of about half a degree at the distance of 50,000 German miles. $\endgroup$ Dec 25, 2022 at 20:29
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The necessary technology is the eyes and the mind. Nothing else.

  1. Solar eclipses show that the shadow of the Moon gets smaller by one Moon-width as it goes from the Moon to the Earth. That is, the width of the shadow when it lands here is either a few hundred miles (total eclipses) or less than zero (annular eclipses).

  2. Therefore the shadow of the earth when it hits the moon in a lunar eclipse is one Earth width minus one Earth width.

  3. The Moon travels its own width each hour. A lunar eclipse lasts up to 90 minutes. So the Moon is totally immersed over a distance of 1½ Moon widths.

  4. So the Earth’s shadow “out there” is 2½ Moon widths wide.

  5. So the Earth’s shadow “down here” is 3½ Moon widths wide.

  6. So the Earth is 3½ times the diameter of the Moon. A

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    $\begingroup$ look at the graphic posted: Your point 2. should end with "one Earth width minus one Moon width" besides this mistake your post is a nice textual animation of that graphic $\endgroup$
    – Bobby J
    Dec 25, 2022 at 9:59
  • $\begingroup$ *The Moon travels its own width each hour. A lunar eclipse lasts up to 90 minutes. So the Moon is totally immersed over a distance of 1½ Moon widths.* Are you implicitly assuming the earth's shadow doesn't move during a lunar eclipse? (it travels at the speed of the sun, which is much smaller than the speed of the moon, but I think it should be mentioned) $\endgroup$
    – JiK
    Dec 25, 2022 at 22:00
  • $\begingroup$ Also to avoid confusion, I would mention that the speeds you mention are relative to the stars in the sky, not affected by Earth's rotation. $\endgroup$
    – JiK
    Dec 25, 2022 at 22:03
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“Gravity really does exist,” Newton stated in 1687. “[It] acts according to the laws which we have explained, and abundantly serves to account for all the motions of the celestial bodies.”1

In particular, it establishes that the mass of the Moon is much smaller than the mass of the Earth, because the barycenter of the Moon-Earth system is much closer to the center of the Earth than of the Moon.

Unless the Moon is much less dense than the Earth it must be much smaller; if the density is comparable one can directly compute the size from the orbits.


1https://www.amnh.org/explore/videos/space/gravity-making-waves/newton-einstein-gravity

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    $\begingroup$ But like you mentioned, Newton’s gravitational theory tells us nothing about the diameter of the moon relative to that of Earth’s. Though this leads to an interesting question: how do we know that the Moon orbits the Earth (circles around it) without knowing their masses? Is a model where the Earth orbits the moon more complex than one where the moon orbits Earth? I guess I can ask a question about that next time. $\endgroup$
    – Obama2020
    Dec 26, 2022 at 16:22
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I wanted this to be a comment, but it looks like that needs 50 reputation which, I don't possess.

@LifeInTheTrees got it right.

The shadow of the Earth is not the same size of the Earth at the Moon's distance. The cross section of the umbra is always smaller than the Earth's cross-section , and the cross-section of the penumbra is always larger. The size differences are significant at the Moon's distance, however, the reduced-size umbra of the Earth is still larger than the diameter of the Moon at the Moon's distance. ~@notovny [corrigendum]

Next, we wait for an lunar eclipse. Possibilities:

  1. The eclipse is partial i.e. a thick ring of moon is visible around earth's shadow. The moon is larger than the earth.

  2. The eclipse is complete but like total solar eclipses, lasts only for a short while. The moon is nearly as big as the earth.

  3. The eclipse lasts for a relatively long time during which no moon is visible at all. This is what happens. Ergo, the moon is smaller than the earth.

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  • $\begingroup$ The shadow of the Earth is not the same size of the Earth at the Moon's distance. The cross section of the umbra is always smaller than the Earth's cross-section , and the cross-section of the penumbra is always larger. The size differences are significant at the Moon's distance, however, the reduced-size umbra of the Earth is still larger than the diameter of the Moon at the Moon's distance. $\endgroup$
    – notovny
    Dec 26, 2022 at 18:17
  • $\begingroup$ Answer edited. Muchas gracias. $\endgroup$ Dec 26, 2022 at 18:23
  • $\begingroup$ Just a quick thought: Does this assume that the sun is at a greater distance from Earth than the moon? Would ancient people have been able to prove that? $\endgroup$
    – Jim
    Dec 26, 2022 at 19:49
  • $\begingroup$ @Jim, my guesstimate, yep! The earth's shadow has to be either the same size or smaller than the earth itself for the argument to work. Size (of the sun) matters too if I'm correct. $\endgroup$ Dec 26, 2022 at 20:01
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    $\begingroup$ Strictly speaking, the Moon is very much visible (and reddened) during a total lunar eclipse. It comes from the effects of light refraction and scattering from the convex lens that is our atmosphere, and is a popular setting for pictures. $\endgroup$ Feb 13, 2023 at 19:43

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