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Aristarchus estimated the relative distance of the Sun and Moon by observing the angle between the Sun and the Moon (α in the diagram) when the Moon is exactly half lit. Angle β must be 90° for the Moon to be half lit. By observing the angle α, he could then set the scale of the triangle and thus the relative lengths of the sides. (Sizes and distances are not to scale.)

Does half-lit mean exactly 45 degrees?

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    $\begingroup$ "exactly 45 degrees" of what angle? "Half-lit" means precisely what it says: exactly half the Moon is lit (shining) and the other half is unlit (dark), and the boundary between the two sides is almost exactly a straight line. At other times when we see a crescent Moon, the line is a curve. Clever guy, Aristarchus. $\endgroup$ – Chappo Feb 14 at 3:14
  • $\begingroup$ Considering your activity in Mathematics SE, I don't believe you are having trouble with the geometry. Can you clarify your question? $\endgroup$ – uhoh Feb 14 at 7:17
  • $\begingroup$ @Chappo the Moon itself is always "half-lit"! However, the lit fraction of the part of the Moon visible from Earth certainly does vary ;-) $\endgroup$ – uhoh Feb 14 at 7:19
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    $\begingroup$ @uhoh ah, yes. In which case I guess we’re talking about an exact quarter of the Moon’s surface that satisfies the two conditions “lit” and “visible from Earth” :-) $\endgroup$ – Chappo Feb 14 at 7:52

Half lit means the Moon is at a quarter phase, either First Quarter or Third Quarter.

The original question asks "Does half-lit mean exactly 45 degrees?". I assume this is asking if the angle $\alpha$ is 45 degrees. The answer is no. The angle $\alpha$ is found from trigonometry to be $cos(\alpha)=EM/ES$ where EM is the distance from the Earth to the Moon and ES is the distance from the Earth to the Sun. EM/ES is approximately 1/390 (=238000 miles/93 million miles), so $\alpha$ works out to be 89.85 degrees.

Aristarchus was trying to do the reverse calculation to find the distances. If the angle $\alpha$ could be measured accurately, the relative distances EM/ES could be calculated using the same formula. The problem is the difference between 89.85 degrees (the Sun is 390 times farther than the Moon) and 90 degrees (the Sun is infinitely farther than the Moon) is a very small angle and difficult to measure exactly. If the Sun were a lot closer, then $\alpha$ would be a smaller value, and the error in the measured angle of $\alpha$ would be restrict the value of EM/ES to a more meaningful range than "infinity".


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