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RE: AO40 velocity

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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