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Originally posted by LightSource
second, it shows the distance the asteroid will be in relations to the sun and to the earth. On the 9th the asteroid will be .991 AU to the sun and .0022 AU to the earth. If you add those 2 numbers together you get .9932 AU. We are missing .0068 AU (that is a big number) why?
second, it shows the distance the asteroid will be in relations to the sun and to the earth. On the 9th the asteroid will be .991 AU to the sun and .0022 AU to the earth. If you add those 2 numbers together you get .9932 AU. We are missing .0068 AU (that is a big number) why?
Originally posted by LightSource
When I look at it on November 8th it will be .0079 AU from the earth or just over 3 LD from earth however on the 9th it will be .0022 AU or .859375 LD.
second, it shows the distance the asteroid will be in relations to the sun and to the earth. On the 9th the asteroid will be .991 AU to the sun and .0022 AU to the earth. If you add those 2 numbers together you get .9932 AU. We are missing .0068 AU (that is a big number) why?
Originally posted by TheGrandWazoo
Originally posted by LightSource
second, it shows the distance the asteroid will be in relations to the sun and to the earth. On the 9th the asteroid will be .991 AU to the sun and .0022 AU to the earth. If you add those 2 numbers together you get .9932 AU. We are missing .0068 AU (that is a big number) why?
This is pretty weird. The two numbers should be at the very least 1 AU.
Wait a minute. The Astronomical Unit is an average, because the earths rotation is elliptic. Sometimes the actual distance is shorter, sometimes it's longer, depending on where we are in the orbit
This is because the distance between the Earth and the Sun is not fixed (it varies between 0.983 289 8912 AU and 1.016 710 3335 AU
Originally posted by ngchunter
reply to post by trekwebmaster
Once again, our average distance from the sun is 1au, it is not a fixed distance, only an average. Our orbit is slightly elliptical.
Originally posted by ngchunter
reply to post by spikey
I don't see any evidence it jumped but using a 2 body java applet for this is wrong to start with. It even says, don't use for planetary encounters. I've already analyzed the orbit myself with much better software.
Originally posted by LightSource
second, it shows the distance the asteroid will be in relations to the sun and to the earth. On the 9th the asteroid will be .991 AU to the sun and .0022 AU to the earth. If you add those 2 numbers together you get .9932 AU. We are missing .0068 AU (that is a big number) why?