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#1
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The recent lunar eclipse has come and gone. Like many other
observers, I went out to see what I could see from my location. Unfortunately, my location on the Earth's surface placed me at a distinct disadvantage. I was able to see only the last nine minutes of the umbral phase of the eclipse. It's very rare that I observe celestial objects as they are rising or setting; but after doing so I've never regretted it. Under such circumstances it's possible to see phenomena that are not otherwise visible. On this occasion I was treated to two phenomena of interest. First was the observation of a full Moon, in partial eclipse, distinctly flattened along its vertical axis by atmospheric refraction. Even without an eclipse, this was worth observing. The second phenomenon involved color. Fortunately I had chosen to use a small, 102mm SCT that was relatively free from chromatic aberration. (My 80mm, f/5 achromat was left inside!) A thin, distinct ring of color hugged the lunar limb. The bottom portion of the ring was red. On both sides of the Moon, as I observed the ring progressively higher up, I was treated to a familiar sequence of colors. In order, the colors we red, orange, yellow, green, and blue. For me, the highlight of this eclipse were the above mentioned colors, not artifacts from a telescope, but the result of atmospheric refraction. As it turned out, my location on planet Earth wasn't all that bad after all! -- Bill Celestial Journeys http://cejour.blogspot.com |
#2
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This reminds me of one night years ago when I was casually sitting out
on my lawn looking out over the open fields to my East with a pair of 11x80 binoculars. I saw a glow over the hill which forms my horizon, and homed in on it to watch the moonrise. I wasn't expecting much, but when the limb of the moon rose over the hill, it was outright shocking! Even with the atmospheric distortion, the just past full moon looked like a giant, three dimensional golf ball rising up behind the hill! Hardly a scientific observation, but I've never forgotten it! ![]() Marty |
#4
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but the Earth wasn't rotating fast enough
to satisfy me. Gerald oughta be satisfied with this... ![]() Now I'm reminded of another truly serendipitous observation: I was observing Uranus (or was it Neptune?) one evening when I noticed a 'star' that showed some very obvious motion relative to the planet. It didn't take long for me to realize that the 'moving star' was in reality neither moving, nor a star. I had stumbled upon a geosynchronous satellite! Cool! I've never (knowingly) seen one of those... Marty |
#5
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Marty wrote (about geosync sats):
Cool! I've never (knowingly) seen one of those... The geostatationary satellites (a subclass of geosyncs that stay more or less above the same spot on the Earth's equator) cluster around the celestial equator, for what I hope is a reasonably obvious reason, so looking just south of the equator if you're in the northern hemisphere, or just north of it if you're in the southern hemisphere, will allow you to pick one up eventually. I've seen a couple of them while observing the Trapezium. Puzzle: Estimate my latitude. ![]() -- Brian Tung The Astronomy Corner at http://astro.isi.edu/ Unofficial C5+ Home Page at http://astro.isi.edu/c5plus/ The PleiadAtlas Home Page at http://astro.isi.edu/pleiadatlas/ My Own Personal FAQ (SAA) at http://astro.isi.edu/reference/faq.html |
#6
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Brian Tung wrote:
Marty wrote (about geosync sats): Cool! I've never (knowingly) seen one of those... The geostatationary satellites (a subclass of geosyncs that stay more or less above the same spot on the Earth's equator) cluster around the celestial equator, for what I hope is a reasonably obvious reason, so looking just south of the equator if you're in the northern hemisphere, or just north of it if you're in the southern hemisphere, will allow you to pick one up eventually. I've seen a couple of them while observing the Trapezium. Puzzle: Estimate my latitude. ![]() Answer #1: google isi.edu, see that it's in Marina del Rey, CA = latitude is about 33 degrees. ('Tis a shame Google Maps lacks lat/lon grids....) Answer #2: Geosynchronous radius is R = sqrt (GM/n^2), with GM = 398600 km^3/s^2, n = 2pi/86164 radians/sec, so R = 42164 km. Declination of the Trapezium is -5.4 degrees. I'm not going to try an ASCII plot, but it's obvious that tan Dec = -r sin lat / (R - r cos lat) with r = Earth's radius (assuming a sphere) = 6378.14 km. The solution is lat = Dec - arcsin (R sin Dec / r) = +33.07 degrees. Thanks for giving me a little bit of fun this evening. -- Bill Owen P.S. I've seen a geostationary satellite show up in my pictures too. Nice straight line running e-w, and if you take the same field one sidereal day later, it'll still be there. Perplexed the heck out of me until we figured out what it was. |
#7
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Puzzle: Estimate my latitude.
![]() Brian, you overestimate my intelligence by several magnitudes. However, playing with a globe, a ruler, the Scalnate Pleso Atlas, and a trig site on the internet, (I don't know trig...) and hopefully remembering the distance of the geostationary satellites as 22,000 miles, I'd crudely guesstimate around 35 degrees...?? Marty |
#8
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The solution is
lat = Dec - arcsin (R sin Dec / r) = +33.07 degrees. Bill, I get that but only after dropping the sign in sin Dec. Isn't that a trigonometric sin? Ben |
#9
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On Mar 6, 9:03 pm, (Marty) wrote:
Puzzle: Estimate my latitude. ![]() Brian, you overestimate my intelligence by several magnitudes. However, playing with a globe, a ruler, the Scalnate Pleso Atlas, .... You got a Skalnate Pleso Atlas? !!! Where on Earth did you get it? I looked for one for the longest time and came up zilch. Ben |
#10
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You got a Skalnate Pleso Atlas? !!! *
Where on Earth did you get it? I looked for one for the longest time and came up zilch. Ben I bought it when it was new. ![]() Marty |
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