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RE: AO40 velocity



I think what he wanted to know is what is the magnitude of the velocity 
vector, and an equations to calculate it at any point in the orbit.


At 09:32 PM 8/29/02 -0400, <w3iwi@toad.net> wrote:
>Margaret is correct in stating that the reference frame needs to be defined.
>The answer that Stacey (W4SM) gave applied to an observer at the center of
>the earth.
>
>Let me expand on this with a couple of simple examples (which also give you
>some fun "magic numbers" to help the next time you have to answer these
>questions in a dingy bar):
>
>                   -----------------------------------------
>
>The earth rotates on its axis every 24 hours = 86400 seconds. The original
>definition of the meter is that 10,000,000 meters is the distance from pole
>to equator along the meridian of Paris. Therefore I can take this number
>(and ignoring the fact that the earth is flattened), I immediately know that
>the circumference of the earth is 40,000 km.
>
>So if you are an observer on the equator, you are spinning at a speed of
>40000/86400 = 463 meters/second. If you are at a higher latitude, then your
>spin velocity is reduced by cosine(latitude), so at 40 degrees (like
>Boulder), you are moving at 355 meters/second. So if the 'how fast "those
>things" go' question is answered for an observer on the earth you need to
>account for this effect, which is ~5% of the speeds that Stacey gave.
>
>                   -----------------------------------------
>
>Now lets move the observer into interstellar space (perhaps at Alpha
>Centauri). The earth's distance from the sun (one AU = Astronomical Unit) is
>about 150,000,000 km and the earth's orbit is nearly circular, so the
>circumference of the orbit is 2*pi*150,000,000 km. The earth goes around the
>sun in one year which just happens to be pi*10,000,000 seconds (to within a
>small fraction of a percent)
>
>So the earth's orbital velocity can add as much 30 km/sec for part of the
>year and subtract 30 km/sec 6 months later when seen from outside. This is
>nearly 4 times the speeds that Stacey answered.
>
>
>
>Define the observer -- It's all relative!
>
>73, de Tom, W3IWI
>
>
>
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