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stella

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Everything posted by stella

  1. "On the night of March 1st (2nd UTC) we will have a small asteroid pass by us very close. At this point it looks like .00030 lunar distances." One lunar distance is 384,000 km. So .0003 lunar distances is (.0003x384,000) km =115.2 km. Earth's radius is 6378.2 km at equator. So the asteroid is 6263 km beneath the surface. :blush:
  2. "As the satellites orbiting at those altitudes suddenly encounter thicker air, they slow down." No, they don't, they speed up. Clearly, the writer has no knowledge of orbital dynamics.
  3. "then based on correlations dating back to 1804 when the astronomer Royal, William Herschel, one of the greatest astronomers of all time". Herschel never was Astronomer Royal. In 1804, Maskelyne was the holder of that post.
  4. None of the reports that I've seen quote the closing velocity of the impact, which was 10 km/second (6.2 miles/second). More bang for your bucks (and roubles).
  5. Right first time Magic and Calum. ...and here they are:
  6. How about TWO opportunities to view this eclipse, and guaranteed to be free of cloud? At about 10.13 and again at about 11.30 U.T. about 50% of the solar disc will be covered as seen from the International Space Station. If they do not manoeuvre the station upwards on July 23, as is scheduled, then the crew would get to see a 90% eclipse. Information credit to Herr Holtkamp.
  7. I think you need to check the dates of New and Full Moons in January and February 2009.
  8. ...and what calendar is being used now?
  9. " are those black things stocks? " No, those are pillories.
  10. You won't see any of those, Russell since the location is not specified and they are therefore calculated for a point in the gulf of Guinea at lat=0°, long=0°. You must set your latitude and longitude (50°.6846N, 1°.5092W), and your preferred time zone in order to get predictions for your own observing site.
  11. New Moon is on May 5 at 1218 UT I think.
  12. I used simple binoculars and stopwatch, to obtain the orbit from positions relative to the background stars.
  13. My observations indicate that it was orbiting at a height of 133 nautical miles. The Pentagon states "153 nautical miles". Who do you want to believe?
  14. The middle of the eclipse is also the time that the U.S. SM-3 missile is due to impact the errant secret USA 193 satellite.
  15. This is the way the world ends This is the way the world ends This is the way the world ends Not with a bang but a whimper. T.S. Eliot
  16. That will be Pete Lawrence, probably Britain's best astrophotographer.
  17. An impact that creates a hole 100 ft by 20 ft in dimension, would release enough energy to be detected by seismometers around the world. No such detections have been noted. Any meteor seen in the sky around the time of the explosion, would be coincidental. I would speculate that this was more likely to have been the ignition of a pocket of volatile gas, possibly by lightning.
  18. "There are many things they might be, but no evidence whatsoever that they are 'spaceships' - indeed I would say that logically it's one of the least likely explanations." I find that 98.8% of all the UFOs that I see do turn out to be spacecraft. The last two - on August 2 were TSX5 (00-30A) at 22:51:39.28 U.T and Cosmos 1386 rocket (82-69B) at 23:05:19.40 U.T.
  19. "Do you only post on here to correct mistakes, because it feels like it." No, I would post other events, but I find that I've been beaten to it by you!
  20. I have the partial phase as beginning at 21h 30m U.T.
  21. It's not a rainbow, it's a circumzenithal arc.
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