A Question Concerning Lunar Eclipses and Aristarchus's Work

  • #1
23
1

Main Question or Discussion Point

I have recently been reading To Explain the World: The Discovery of Modern Science by Steven Weinberg. I am currently in the midst of Chapter 7, Measuring the Sun, Moon, and Earth. On Page 66, Weinberg comes to "what in some respects is the most impressive example of the application of mathematics to natural science in the ancient world: the work of Aristarchus of Samos." He discusses four key observations that Aristarchus made in Technical Note 11. However, for this question I will focus on one. Observation 3, to use Weinberg's words, is that "the shadow of the Earth at the position of the Moon during a lunar eclipse is just wide enough to fit a sphere with twice the diameter of the Moon." Before continuing, I want to clarify what the variables mean.

ds is the distance from the Earth to the Sun. dm is the distance from the Earth to the Moon. Ds is the diameter of the Sun. Dm is the diameter of the Moon. De is the diameter of the Earth. do is the distance between point P and the Moon. Point P is the vertex (or apex) of the cone of the shadow cast by the Earth. The shadow ends at point P. Hopefully the picture below is a decent visual of the Earth-Moon-Sun system I am trying to describe. Anyway, there are three similar triangles in this system. The first triangle is formed from the diameter of the Sun and the lines stretching from the disk of the Sun to point P. The second triangle is formed from the diameter of the Earth and the lines stretching from the Earth's disk to point P. Lastly, the third triangle is formed by a line that is twice the diameter of the moon and the lines stretching from the sphere of the aforementioned diameter in the position of the Moon to point P (All the diameters are parallel to each other). From this, we can determine the ratio: (ds+dm+do)/(Ds)=(dm+do)/(De)=(do)/(2Dm). Through some algebra that I can explain later, we find that (do)=((2Dm)(dm))/(De-2Dm). With more algebraic manipulation (that I can also explain if necessary), (De)=((2Dm)(ds)+(dm)(Ds))/(ds+dm). The aforementioned equation is important, because my question involves this and two other equations.

Weinberg states, "if we now use the result of observation 2, that (ds)/(dm)=(Ds)/(Dm), this can be written entirely in terms of diameters": (De)=((3Dm)(Ds))/(Ds+Dm). Observation 2 states that "the Moon just covers the visible disk of the Sun during a solar eclipse". Fairly obvious if you've ever seen a solar eclipse. Anyway, my question is how do I use the result of observation 2, (ds)/(dm)=(Ds)/(Dm), to derive (De)=((3Dm)(Ds))/(Ds+Dm) from (De)=((2Dm)(ds)+(dm)(Ds))/(ds+dm)? I have been struggling to do this and Weinberg doesn't show all of his work, so I have no idea how he was able to do this derivation. Any help would be greatly appreciated. Thanks. Disclaimer: [all of this is the conclusions and methods of Aristarchus re-derived and re-stated in modern terms. Aristarchus, in his treatise On the Sizes and Distances of the Sun and Moon, writes in a Euclid's Elements/deductive style.] Edit: Since the diagram doesn't seem to show up properly, use this link for a decent diagram: https://www.schoolsobservatory.org/learn/astro/esm/orbits/lunar_eclipse
 
Last edited:

Answers and Replies

  • #2
34,053
9,912
(ds)/(dm)=(Ds)/(Dm) implies Dm ds = dm Ds, the two terms in the numerator of the other equation. Therefore, ##(De)=\frac{(2Dm)(ds)+(dm)(Ds)}{ds+dm} = \frac{3(Dm)(ds)}{ds+dm}##. We can also rewrite the first equation as ##dm = \frac{(ds)(Dm)}{Ds}## and plug it in:$$(De) = \frac{3(Dm)(ds)}{ds+\frac{(ds)(Dm)}{Ds}} = \frac{3(Dm)}{1+\frac{(Dm)}{Ds}} = \frac{3(Dm)(Ds)}{(Ds)+(Dm)}$$

An approach without formulas: The Sun is much farther away than the Moon. Every core shadow will reduce its diameter with the same angle (same change in diameter per distance). We know the shadow reduces by nearly exactly the diameter of the Moon over the distance to the Moon. If the core shadow of Earth is still twice the lunar diameter wide at the distance of the Moon behind the Earth, then it has to be one Moon diameter wider at the position of the Earth. 2+1=3 - the Earth has three times the radius of the Moon.
 
  • #3
23
1
(ds)/(dm)=(Ds)/(Dm) implies Dm ds = dm Ds, the two terms in the numerator of the other equation. Therefore, ##(De)=\frac{(2Dm)(ds)+(dm)(Ds)}{ds+dm} = \frac{3(Dm)(ds)}{ds+dm}##. We can also rewrite the first equation as ##dm = \frac{(ds)(Dm)}{Ds}## and plug it in:$$(De) = \frac{3(Dm)(ds)}{ds+\frac{(ds)(Dm)}{Ds}} = \frac{3(Dm)}{1+\frac{(Dm)}{Ds}} = \frac{3(Dm)(Ds)}{(Ds)+(Dm)}$$

An approach without formulas: The Sun is much farther away than the Moon. Every core shadow will reduce its diameter with the same angle (same change in diameter per distance). We know the shadow reduces by nearly exactly the diameter of the Moon over the distance to the Moon. If the core shadow of Earth is still twice the lunar diameter wide at the distance of the Moon behind the Earth, then it has to be one Moon diameter wider at the position of the Earth. 2+1=3 - the Earth has three times the radius of the Moon.
Wow, I can't believe I totally missed that. Something so simple! Anyway, thank you for the feedback. Another thing, how do we actually know that the angle between the lines of sight from the Earth to the Moon and from the Earth to the Sun is 90 degrees when the Moon is half full?
 
  • #4
34,053
9,912
Another thing, how do we actually know that the angle between the lines of sight from the Earth to the Moon and from the Earth to the Sun is 90 degrees when the Moon is half full?
You can just measure that angle in the sky. It is not exactly 90 degrees, but very close.
 

Related Threads on A Question Concerning Lunar Eclipses and Aristarchus's Work

  • Last Post
Replies
5
Views
3K
  • Last Post
Replies
11
Views
3K
  • Last Post
Replies
2
Views
3K
  • Last Post
2
Replies
25
Views
4K
  • Last Post
Replies
17
Views
3K
Replies
8
Views
1K
Replies
5
Views
3K
Replies
11
Views
1K
Replies
4
Views
7K
Replies
2
Views
3K
Top