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.
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".
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,
RM
= 27/80*
RE =
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.
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 (2
R) to a value of
X of 360,000 Km away, not too far
off from the accepted value of just over 380,000 Km.