# Proof/explanation that Sun is bigger and more distant from Earth than Moon is, Middle Age technology allowed only

If I was living in the Middle Ages, how can someone prove or at least explain to me in a simple way that the Sun is bigger and more far from Earth than the Moon?

Can a parallax be used for it?

$$\text{ratio}_1=\frac{\text{diameter}_\text{sun}}{\text{diameter}_\text{moon}}=400.8$$

$$\text{ratio}_2=\frac{\text{distance}_\text{earth-sun}}{\text{distance}_\text{earth-moon}}=385.2$$

$$\text{ratio}_2$$ is only $$3.9$$ % lower than $$\text{ratio}_1$$.

I know that there were four astronomers, that were important in developing the right opinions about the mutual positions and movements: Nicolas Copernicus, Tycho Brahe, Johannes Kepler, Galileo Galilei.

• Eclipses, man. Eclipses. How can the moon shadow the sun, if it aint in front of it. Commented Nov 12, 2021 at 9:29
• wow a lot of inadmissible theorems. very nice question. academia.stackexchange.com/questions/116019/…
– BCLC
Commented Nov 12, 2021 at 10:45

## 3 Answers

As the answer by Jonas pointed out, a solar eclipse is an obvious way to realise that the Sun must be further away than the Moon and hence larger, since their angular size is almost the same.

However, this does not tell you how much larger and further away the Sun is. The most basic way I have heard of, but which is hard to measure with ancient technology, is to realise that the Moon is lit by the Sun, and that seeing a half-moon means we have a right triangle, with the right angle between the axis Moon-Earth and Moon-Sun. You can then try to estimate the angle between the axis Moon-Earth and Earth-Sun, which basically appears to be close to $$90^\circ$$, which would render the Sun infinitely far way. Aristarchus, an ancient Greek, measured an angle of $$87^\circ$$, rendering the ratio of distance to be around $$1/\cos(87^\circ)\approx20$$, see here.

• Related question, echoing the end of your answer: Why does the light side of the moon appear not to line up correctly with the evening sun? Commented Nov 12, 2021 at 8:43
• @Ruslan I understand that this a projection phenomenon... straight lines in space are not necessarily projected onto 'straight lines' at the sky. However, I never thought about how this may make the measurement of the angles harder. Commented Nov 12, 2021 at 10:58

Easy to measure since ancient times are the angular diameters of sun ($$0.53°$$) and moon ($$0.52°$$), which happen to be nearly equal to each other.

This gives you $$\frac{\text{diameter}_\text{sun}}{\text{distance}_\text{earth-sun}} =0.53°\frac{2\pi}{360°}=0.0093 \tag{1}$$ $$\frac{\text{diameter}_\text{moon}}{\text{distance}_\text{earth-moon}} =0.52°\frac{2\pi}{360°}=0.0090 \tag{2}$$

Historically, the next thing determined was the distance of the moon. This was first done by measuring the parallax of the moon (i.e. the apparent position difference of the moon in the sky when viewed from two different locations on earth at the same time).

(image from Lunar Parallax)

The difficult part here was to figure out how to do these two observations at the same time even before choronometers were invented. When observing the moon from different continents (i.e. separated by a distance $$s$$ of several $$1000$$ km) you measure a parallax $$\delta$$ between $$1°$$ and $$2°$$, which is easily detectable even without a telescope. From this you can calculate the distance of the moon by $$\text{distance}=\frac{s}{\delta}\frac{360°}{2\pi}$$ and get $$\text{distance}_\text{earth-moon}=380{,}000\text{ km}.$$

The parallax of the sun can be measured in the same way as above. But it is much more difficult because the measured parallax $$\delta$$ of the sun turns out to be much smaller (between $$0.002°$$ and $$0.005°$$ when measured from different continents). From this result you already see without any calculation that the sun must be much further away than the moon. When calculating the distance of the sun by $$\text{distance}=\frac{s}{\delta}\frac{360°}{2\pi}$$ you get $$\text{distance}_\text{earth-sun}=150{,}000{,}000\text{ km}.$$

From these distances of moon and sun you can, by using (1) and (2), calculate their diameters and find $$\text{diameter}_\text{moon}=3{,}400\text{ km}$$ $$\text{diameter}_\text{sun}=1{,}400{,}000\text{ km}.$$

• You seem to be mixing conventions for your thousands separators and decimal points. All your distances use . for the thousands separator, but everything else seems to use . for a decimal point. You might want to stick to one convention! Commented Nov 12, 2021 at 3:56
• Indeed, there is a different notation for that in different countries: docs.oracle.com/cd/E19455-01/806-0169/overview-9/index.html Commented Nov 12, 2021 at 5:27
• @Hearth Yes, sorry. Now using the English thousand separator convention. Commented Nov 12, 2021 at 6:21
• @JanN. Yes, I'm aware; it's not really a problem as long as you're consistent with it, I brought it up because both the English and the European convention were being mixed in this same answer, which runs the risk of being confusing. Commented Nov 12, 2021 at 6:31
• Whatever the use of comma inside a number, it should be typeset without spaces around it. I.e. $1{,}400{,}000$ instead of $1,400,000$. Otherwise it's just a sequence of numbers. (Use Show Math AsTeX Commands to learn the difference in code.) Commented Nov 12, 2021 at 8:45

On a solar eclipse, the moon moves in front of the sun. From that it should be obvious that it must be closer to earth than the sun. From everyday experience it should also be clear that if two objects appear roughly the same size (which could perhaps even be observed with the naked eye, but don't do that), but one of them is farther away, it must be larger.

• "These cows are small. These cows are far awaaay." (Father Ted) Commented Nov 12, 2021 at 8:19
• Do we (approximately) know since when people were aware that the sun is actively shining whereas the moon is just reflecting light? Otherwise it would be impossible to distinguish the mechanism behind the lunar eclipse and the solar eclipse IMHO. Commented Nov 12, 2021 at 10:01