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RGregoryClark
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Astronomers have calculated that the Russian meteor and asteroid 2012 DA14 have very different orbits:
Astronomers Calculate Orbit of Chelyabinsk Meteorite.
This image shows how greatly the orbit of 2012 DA14 was changed by the close flyby at 17,000 miles away:
Imagine then how greatly the orbit could be altered if a meteor like the Russian one passed by at only 10 miles away.
For calculating how much the Russian meteor could be changed by a close Earth passage, I'm imaging it initially having close to the orbit of 2012 DA14, but being slightly different so that it skims pass the Earth. We can suppose this skim pass occurs at one of the two points where the 2012 DA14 orbit crosses Earth's orbit, as shown here:
La Sagra Observatory discovers very near-Earth asteroid 2012 DA14.
Posted By Jaime Nomen
2012/03/27 05:20 CDT
"The preliminary orbit shows that 2012 DA14 has a very Earth-Like orbit with a period of 366.24 days, just one more day than our terrestrial year. The orbit is nearly circular but just elliptical enough to jump inside and outside of the path of Earth two times per year. Because objects move faster when they are closer to the Sun, the relative motion is similar to some sports races: when the Earth is on the outer track, it is overtaken by 2012 DA14, but when the asteroid crosses Earth's orbit, Earth overtakes it and passes by. It is during the orbit crossings when the closest encounters occur, and when there is potential for a future impact."
http://www.planetary.org/blogs/guest-blogs/3418.html
For this preliminary calculation we don't need to know an actual date of this skim pass but only that the orbit is close to that of 2012 DA14 before the close skim pass and we want to find out how it looks afterwards.
I'm attempting to model it using a patched conic approximation. Anyone familiar with anymore accurate computer simulation programs to calculate this?
Bob Clark
Astronomers Calculate Orbit of Chelyabinsk Meteorite.
Still because of the very unlikely probability of their occurring together purely by chance, I would like to see some simulations of what can happen to an asteroids orbit when it gets as close as the Russian meteor.The Chelyabinsk meteorite is from a family of Earth-crossing rocks called Apollo asteroids and there are 80 million others like it, say astronomers.
http://www.technologyreview.com/view/511691/astronomers-calculate-orbit-of-chelyabinsk-meteorite/
This image shows how greatly the orbit of 2012 DA14 was changed by the close flyby at 17,000 miles away:
Imagine then how greatly the orbit could be altered if a meteor like the Russian one passed by at only 10 miles away.
For calculating how much the Russian meteor could be changed by a close Earth passage, I'm imaging it initially having close to the orbit of 2012 DA14, but being slightly different so that it skims pass the Earth. We can suppose this skim pass occurs at one of the two points where the 2012 DA14 orbit crosses Earth's orbit, as shown here:
La Sagra Observatory discovers very near-Earth asteroid 2012 DA14.
Posted By Jaime Nomen
2012/03/27 05:20 CDT
"The preliminary orbit shows that 2012 DA14 has a very Earth-Like orbit with a period of 366.24 days, just one more day than our terrestrial year. The orbit is nearly circular but just elliptical enough to jump inside and outside of the path of Earth two times per year. Because objects move faster when they are closer to the Sun, the relative motion is similar to some sports races: when the Earth is on the outer track, it is overtaken by 2012 DA14, but when the asteroid crosses Earth's orbit, Earth overtakes it and passes by. It is during the orbit crossings when the closest encounters occur, and when there is potential for a future impact."
http://www.planetary.org/blogs/guest-blogs/3418.html
For this preliminary calculation we don't need to know an actual date of this skim pass but only that the orbit is close to that of 2012 DA14 before the close skim pass and we want to find out how it looks afterwards.
I'm attempting to model it using a patched conic approximation. Anyone familiar with anymore accurate computer simulation programs to calculate this?
Bob Clark