On Dec 28, 2009, at 2:09 PM, Rick Monteverde wrote:

Saw an orange fire colored UFO last night just after nightfall. The path was that of an object flying in a curved path at high altitude (a u-turn, basically), definitely not a satellite, and a bit brighter than a good space station sighting. Even through 8x binoculars it appeared as a point source. The time of day, the sighting angle, and the variable characteristics of the light over the two minutes or so the object was visible suggests it was reflected light from the setting sun on the lower surface of a solid object, but given that the sky was almost completely dark at my location at that point, my guess is that its altitude could have been above the atmosphere. I’ve seen conventional aircraft reflecting sunset’s light after local sunset, and though the appearance was similar, in those cases it was much closer to sunset and sky was still quite light. Once it gets dark, I think such reflections tend to be in satellite territory.

Anyway, since I can find the sight angle because of its passage near identifiable stars, the time, and my location on that date, it should be possible to calculate the earth’s shadow line from the setting sun and see where my sight line crosses it. That would tell me if it was just an airplane at very high altitude, or something maneuvering up a bit higher than conventional aircraft can reach. Anybody have an idea how I would go about that (umbra?) line?

Thanks,

-          Rick

Hi Rick,

Coincidentally, I saw something similar yesterday (Dec 28, 2009) around noon AKST, (about 11 orbits later) west of Palmer AK, but heading SW. It was one small finger width at arms length above the horizon. It had a periodic (about 10 second) flash to it, so I assumed it might be a booster, but strange it was heading SW, not SE or NE, or just S. Of course a U-turn is not a typical satellite maneuver, nor did I see that!

The altitude h to the directly overhead sun midline is given by:

   h = r_earth * ( SQRT(1 + sin^2 theta) -1)

Given time after sunset t we have:

theta = (t/(8.64x10^4 s))*(2*Pi) radians = (t/(1440 min))*(2*Pi) radians

Earth radius, r_earth, at Hawaii is about 3951 mi. Here are some numbers:

t (min) theta (radians) h (miles)
                
1       0.00436331944   0.03760073165
5       0.02181659722   0.93976780755
10      0.04363319444   3.75594358
20      0.08726638889   14.973936498
30      0.13089958333   33.506081478
60      0.26179916667   130.1553394
90      0.39269875      279.3533269


Since the above is time after total sunset, you don't have to correct for the angular width of the sun. However, even total sunset is not good enough to black out an object though, due to light diffraction. Clearly not enough time, i.e. "shortly after sunset", passed to rule out an airplane.

Best regards,

Horace Heffner
http://www.mtaonline.net/~hheffner/




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