# How to describe the inner curve of a crescent?

What is the equation which describes the inner circle of the crescent that a celestial body displays when view at an angle from its light source, as a function of the crescent-cycle period? For instance, the cycle of Earth's moon?

I know that on a new moon the inner curve is coincident with the circumference, so the inner curve's radius is the moon's radius (let's call that 1 unit) and it's displacement from the moon's center is 0 units. After one week (one fourth of the cycle period) the inner curve's radius is infinite, and it's center is infinite distance from the center of the moon. The in-between values are not so clear, though!

Does r=1/cos(2x·π) describe the inner curve's radius (x would be the period)? It at least fits the end-values. How about determining the distance of the inner curve's center from the center of the moon?

I figure that this is a physics question and not a mathematics question as I am looking to describe a phenomenon that occurs in nature, which is only one of the infinite curves which could be described. If this question is better placed in mathematics.SE or astronomy.SE then it can be moved.

Thanks.

-

This is only barely physics, but I'll answer it any way.

First of all I will assume that you are at a great distance from the object so that we don't have to deal with parallax issues. Similarly, I will assume the light source is at a great distance and that the celestial body is a perfect sphere.

In this case, the terminator (the boundary between the light and unlighted portions of the sphere will always be a circle with the radius of the sphere and centered at the center of the sphere.

You are thus viewing the terminator circle from an angle and a circle viewed from an angle is an ellipse. The semi major axis of the ellipse is the radius of the sphere and the semi minor axis will be the radius of the circle times the cosine of the angle of rotation of the object (with 0 angle assumed at the "new" moon time).

I'll leave the writing of the equation up to you. Hope that is helpful.

-
Thank you Frank, it seems to me that you are correct. Once it is recognised that the terminator is described by an ellipse the pieces fall into place. Have a great day! – dotancohen Feb 1 '12 at 14:48
Note that you have "eclipse" in place of "ellipse". I don't have edit privileges, but that might confuse some other readers. – dotancohen Feb 1 '12 at 14:52
You are welcome and thanks for noticing the eclipse typo... – FrankH Feb 1 '12 at 15:55