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/
