Updated 23 May 26
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| Geometry of a Solar Eclipse |
The diagram shows a simplified representation of a particular type of solar eclipse in which the Sun is obscured for an instant only. If the Moon were closer, the eclipse would be total for a period of time and if it were further away, the eclipse would be annular.
At right is a red vertical bar representing the Sun, radius r₃; in the middle is a green vertical bar representing the Moon, radius r₂; at left is a black vertical bar representing the extent of the observed partial eclipse (penumbra), radius r₁.
Point 1 represents the location of an observer (on the Earth) of this momentary total eclipse.
The distance between point 1 and (the centre of) the Moon is d₁ (known); the distance between (the centre of) the Moon and point 2 is d₂ (unknown) and the distance between point 2 and (the centre of) the Sun is d₃ (also unknown). The total distance between the point 1 and (the centre of) the Sun is therefore d₁ + d₂ + d₃.
and:
Hence, the ratio of the Earth-Moon distance to the Astronomical Unit is (minus the radius of the Earth):
6. d₁/ (d₁ + d₂ + d₃) = ((r₁ - r₂)/r₁).(d₁ + d₂)/ (d₁ + d₂ + d₃)
= (r₁ - 2r₂)/r₁.
So, all we need to do is measure r₁, plug it into the above equation along with the known value for r₂ and we are done! However, there is a problem. Let's say the ratio in the above expression is 1/n, then:
7. r₁ = 2r₂.(n - 1)/n.
This means we must measure r₁ with an accuracy (much) greater than 1/n - an estimate will not do.
I had hoped that this could form the basis of a simple method to estimate the AU, but that does not seem to be the case. And I've not even considered yet just how to estimate r₁ from the timings of a solar eclipse on the spinning Earth. Not simple then and I can only think that this was never a practical method either for finding the size of the AU. However, for high n, r₁ to all intents and purpose equals 2r₂, so this is instead one way that the diameter of the Moon could be measured.

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