Lunar Eclipse
and
How the Ancients used it to determine the size of and distance to the Moon

It is Nov. 8, 2003 at about 9:00pm, cold and clear in Canfield, Ohio.

Click on each of the thumbnails to get the whole picture, taken with the zoom setting on a Sony CDMavica without  any additional lenses. The first pictures are from about 8:20, when the moon was just emerging from the penumbra of the earth. It is difficult for cameras of this sort to do justice to the color at these low light levels BUT one does get a sense of the red-orange of the moon. They are overexposed to near complete circles by the automatic exposure algorithym of the camera. Then there are a few attempts at star fields around M31 which was nearly overhead. The later pictures of a crescent are at about 9:00pm when the moon was about 1/2 out of the penumbra.  These later pictures can be used to determine the ratio of the sizes of the earth and moon and therefore the distance to the moon (see below for details). These are untouched photos.

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From these data it is relatively straightforward to determine the ratio between the earth radius and the moon's radius by fitting iso-intensity contours with circles. Below is an example (blown-up) of an iso-intensity contour (made with GIMP,these are the pixels on which the illumination is "37" or "38".

37_38_filter.jpg

Size of the Moon :

We can most easily determine the radius of the circumscribing circle by, for example, selecting three points on the lower rim (moon)  and three points in the upper rim (earth's shadow). Modern graphics programs (like GIMP) can easily be used to report the x- and y-pixel values of any pixel on the image.  In a plane three points determine a circle. Let the points be (x1, y1), (x2, y2) ,(x3, y3) pixel units.  It is a fun exercise for your students to use a little algebra to write down a general formula, in terms of the points' co-ordinates, for the radius, R,  of the circle through three given points. One finds,

2R = (d12d13d23)/ | (x2- x1)y3 +  (x3- x2)y1 + (x1- x3)y2 |

where d12  is the strightline distance between points '1' and '2' (and d13, d23 are the distances between points '1' and '3' and between  '2' and '3' respectively). That is, for example, d12 = ( (x2- x1)2  + (y2- y1)2 )1/2. The "| ...   |" are absolute value signs.

For example, for the iso-intensity picture above I selected three reasonable points for the moon edge (14,44), (34, 64) and (60,57)
and for the upper edge (earth shadow) I selected (24,19) (40,25), (51,27) , staying well inside the middle-ish lattitudes of the moon (for the illumination always dips as you approach the poles, making the use of iso-intensity contours suspect near the poles. That is what is responsibe for the lima-bean shape!)

Plugging into the above equation, we find that RE~ 80 pixels and RM ~ 27 pixels for the radius of the Earth and Moon respectively. Of course, the ancient greeks knew the earth was round like a ball (as seemingly obvious from the lunar eclipse) and they had a pretty good estimate of its size (lookup erathostenes on the web or in an elementary astronomy text). Using the fact that the radius of the earth is ~6400 Km, we can determine the radius of the moon by ratio and proportion with pixels! For our data this reads,

    R = 27/80*R= 2160 Km.

Which is an overestimate from the accepted value of about 1750 Km, but not too bad for a first try.


Distance to the Moon :

If we know the radius of the moon we can determine the distance to it because we can measure the angular distance across it in the sky.  Things further away take up less of an arc than the same object would closer to the viewer.

figtomoon.jpg

In the above figure we raised a pencil or pen (diameter d) at  some distance (s) from our eyes to just barely block out the full moon. The moon is a distance X away and we already know (an estimate of) its diameter, D,  from the above work. Using ratio and proportion on the similar triangles in the above figure, we have

X/D = s/d

That evening we found that a pen of thickness d=5/16'' just blocked out the moon at arms length of about s=26.5 ''. This leads by the above formula and using our estimate of the moons diameter (2R) to a value of X of 360,000 Km away, not too far off from the accepted value of just over  380,000 Km.