When we see a crescent Moon that means it should be seen a half Moon from another position in the sky, because the terminator line (i.e. the line that separates the bright and dark areas of the Moon) is in fact a meridian. And a Meridian is a straight line which is seen curved when it's looked from an angle.

So when we see a regular narrow crescent Moon on the Earth, how far should we be from the planet in its orbit to be in front of that Meridian?

  • $\begingroup$ Quick answer is 384,400 km, or the distance of the Moon to the Earth. You’d then be over the terminator and (except for horizon effect) you would see the moon exactly half-lit. Moving away from the Moon to see it whole and cancelling the horizon effect brings you further, but your question implied the minimum distance… $\endgroup$ Jun 28, 2023 at 0:33
  • $\begingroup$ What about on the Earth's orbit? $\endgroup$ Jun 28, 2023 at 0:37
  • $\begingroup$ Roughly 5.4 million km. Demonstration coming up. $\endgroup$ Jun 28, 2023 at 0:42
  • $\begingroup$ The title asks about a full Moon, but the body of the question implies a quarter Moon. Which do you want? $\endgroup$
    – JohnHoltz
    Jun 28, 2023 at 21:45
  • $\begingroup$ I will edit the title. However, I don't recall I wrote Full Moon (also with a capital F!). Update: A user called eshaya has edited my post. $\endgroup$ Jun 28, 2023 at 22:17

1 Answer 1


In the image below, the blue curve is the Earth’s orbit, and the green circle is the Moon’s orbit around the Earth. The corner of the red angle is the position of the New Moon. A crescent would be slightly to the left or to the right of this point, but it doesn’t change much if we restrict ourselves to a “narrow crescent” as you mention in your question.

In order to see the Moon half-lit (as a “quarter”), while staying along the Earth’s orbit, we would need to be at the leftmost point of the red angle, at the far left of the drawing.

Given that the Moon’s orbit is roughly 384,400 km in radius, a rough measurement on the drawing gives the distance to the intersection of the blue curve (Earth’s orbit) and red line (line of sight for half-moon) as approximately 5.3–5.4 million kilometres EDIT: I had forgotten to multiply by 2 (as my circle for the Moon’s orbit was 384,400 in DIAMETER), so the answer is ≈10 million kilometres.

Your actual mileage may vary. Slightly.

Zoomed-in portion of Earth’s orbit

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    $\begingroup$ Can use Pythagorian Theorem: $d =\sqrt{(R^2 - (R-r)^2)}$. Where R is orbital radius of Earth and R is orbital radius of Moon. $\endgroup$
    – eshaya
    Jun 28, 2023 at 17:03
  • $\begingroup$ @eshaya The orbital radius of Earth is equal to the distance of Earth to Sun. Also, the orbital period of Moon is equal to the distance of Moon to Earth. So is this formula still true? $\endgroup$ Jun 28, 2023 at 22:24
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    $\begingroup$ @eshaya: Indeed. Your formula gives about 10 million kilometres. It seems that I measured wrong (forgot to multiply my drawing’s length by 2) yesterday; apologies. Snack Exchange: Yes, the formula is valid. The hypotenuse is the line from the Sun to the left part of my drawing, so $ R $, while the long side of the triangle is from the Sun to the bottom right of my drawing, so $ R - r $ indeed. $\endgroup$ Jun 28, 2023 at 22:29

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