reply to post by Phage
I have doing been some celestial thinking of how the moon is illuminated by the sun and what angle the sun will be seen in the sky if standing on the
moon.
This is how i see it. The moon is sphere shape that completes one orbit of the earth every 28 days. Its not spinning like the earth so therefore for
a person seen standing on the centre Latitiude 0.00°, and Longitude 0.00° on the lunar map; will observe the rise and fall of the sun along the
longitude line.
At this 0.00 coordinate an observer will should see the sun in the horzion is tilted on the horizion in Latitiude about 20 degrees°. This is due to
the Moon orbiting closer with the eliptical equator and not the celestial equator.
The celestial equator differs by about 23 degrees from its eliptic plane as shown in the picture. The moon orbits facing directly the earth at an
angle close to the eliptic plane and therefore I would expect the moons Longitude 0.00 coordinare on the LRO lunar map is aligned closely with the
eliptic plane (or very close to within 5 degrees).
Another variable to consider is the Moon takes 28 days to go once around the earth. There are 360 degrees angle spread in a circle circumference.
Therefore in 1 day (24 hour period) there is 360/28 = 12.85 degrees. A person standing at the lunar coordinate of longtitude and lattitude 00, 00
will therefore see the sun move in the horizion about 0.5 dgrees per hour. That can explain the variations in sun angle position as observed by
Astronauts while walking on lunar surface. The link below takes you to a web bage which documents sun elevation angles observed on the lunar landing
sites for each Apoolo mission.
www.hq.nasa.gov...
I will work with Latitiude 20.00° , and Longitude 0.00 as a reference position on the lunar surface or considering rise and fall of the sun along the
longitude line as observed by an observed at this coordinate. The highest position/height of the sun in the lunar sky as observed from this cooridnate
position will allow determination of the maximum sun elevation angle possible at Anaxagoras crater 73.4°N 10.1°W.
Since the moon is sphere that goes around the earth every 28 days, in a 360 degrees cycle. Then at one particular quarter of this cycle (i will
arbitrarily set a reference for this as 0 degrees) a person standing at Latitiude 20.00° , and Longitude 0.00 on the lunar suface will see the sun
directly over head. At +/-90 degree movements of the moon moving around earth from this reference; the lunar observer will see the sun falling behind
the east or west horizions of moon.
If it had been a +/-45 degree moon/earth position around the reference then the observer standing at the lunar surface at Latitiude 20.00° and
Longitude 0.00 will see the moon at 45 degrees over the west or eastern horizions.
For example on Apollo 15 26.13222° N latitude, 3.63386° E longitude (which are close to the coordinates i using as a reference) the sun elevation
was recorded being 44.3 deg at completion of the astronauts lunar walk around.
The working out the "maximum possible" sun elevation possible at Anaxagoras crater 73.4°N 10.1°W is actually very simple. 90 - 73.4 = 14 degrees.
Well actually for the more accurate calculation >>
The sun in its highest altitude for year at Latitiude 20.00° , and Longitude 0.00 on the lunar suface, requires rescaling the lattitude coordinate
line to compensate for the Latitiude 20.00°. Therefoe 90-20 = 70 degrees arc between Latitiude 20.00° and Longitude 0.00 position and where it goes
into darkness at North pole Lat 90, Long 0 degrees.
Therefore 70/90 x 14 = 10.888 degrees or rounding off 11 degrees. My calculation lookings surprising familiar.
Since the photo of Anaxagoras crater rim was taken directly overhead of the crater rim I have ignored as negligible the 10 degrees West coordinate the
sun illumination has on objects located around the rim of the crater.
Therefore depending on where the moon was in its orbit with relation to earth means that a sun elevation angle of 11 degrees at sunrise is possible
for the 73.4°N 10.1°W location.
To put into a practical context what the above all means, is that when Nasa planned to send their satellite to take photos of Anaxagoras crater, they
picked the right time of year where the sun has its maximum elevation around the Central latitude 0.00 and longitude 0.00 coordinates. And it makes
sense for Nasa to do that for obtaining best sunlight angle on the crater which allows for the clearest photo picture of the crater.
edit on 17-2-2013 by AthlonSavage because: (no reason given)