Re: Earth, Sun and Vega.
From: Oriel36 (geraldkelleher_at_hotmail.com)
Date: 09/11/04
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Date: 11 Sep 2004 07:03:20 -0700
kencouesbouc@yahoo.fr (kenneth couesbouc) wrote in message news:<1c9636fc.0409100653.60099c0e@posting.google.com>...
> geraldkelleher@hotmail.com (Oriel36) wrote in message news:<273f8e06.0409080651.34e42321@posting.google.com>...
> > kencouesbouc@yahoo.fr (kenneth couesbouc) wrote in message news:<1c9636fc.0409070547.686a72da@posting.google.com>...
> > > geraldkelleher@hotmail.com (Oriel36) wrote in message news:<273f8e06.0409060235.11f46d7a@posting.google.com>...
> > > > kencouesbouc@yahoo.fr (kenneth couesbouc) wrote in message news:<1c9636fc.0409040621.6c33ce2c@posting.google.com>...
> > > > > Hello,
> > > > > I'm trying to calculate something and I wonder if someone can
> > > > > help me.
> > > > > The Earth travels around the Sun and the Sun is travelling towards
> > > > > Vega (not quite, so I'm told?). I'd like to know at what moment (which
> > > > > day in the solar year) the Earth is moving in the same direction as
> > > > > the Sun, towards Vega.
> > > >
> > > > Strictly speaking the Sun does not move in the general direction of
> > > > Vega as both are moving more or less in unison around the Milky Way
> > > > axis.
> > > >
> > > > A more accurate phrasing of your question is when does the orbital
> > > > orientation of the Earth point in the general direction of Vega
> > > > tangential to the Sun Earth line.
> > > >
> > > > http://www.mhhe.com/physsci/astronomy/fix/student/images/04f15.jpg
> > > >
> > > > As constellations are a product of stellar circumpolar motion which in
> > > > itself is an illusion created by the Earth's axial rotation,a common
> > > > error is to determine that Vega is framed within a celestial sphere
> > > > contrary to its true nature as part of stellar rotation about the
> > > > Milky Way axis.
> > > >
> > > >
> > > > http://blueox.uoregon.edu/~courses/BrauImages/Chap01/FG01_008.jpg
> > > >
> > > > As the Earth's orbital orientation does not follow axial rotation
> > > > determined by celestial polar coordinates centered on the star Polaris
> > > > ( as in the above image) it is more productive to drop references to
> > > > constellations altogether in order to calculate which day the Earth's
> > > > orbital orientation points to Vega tangential to the Sun/Earth line.
> > > >
> > > > From the graphic above ,it would be represented by the the position
> > > > 'B' which represents a position sometime early on in the year.
> > > >
> > > Very sorry for framing my question so badly. This is what I wanted
> > > to know. As you put it, it's the Earth's tangential speed which counts
> > > and when it is pointing at Vega. i.e. in the same axial direction as
> > > the Sun.
> >
> > Unfortunately this is incorrect and not at all easy to correct without
> > actually going outside and putting observations into correct
> > observational perspective.
> >
> > The arrows in the following graphic represent orbital orientation to
> > the Sun where the dawn light represents the division between equal
> > amounts of direct sunlight from the Sun as distinguished from the
> > orbital shadow we call 'night'.
> >
> > http://www.mhhe.com/physsci/astronomy/fix/student/images/04f15.jpg
> >
> > Cataloguers would call the following image sunrise while an astronomer
> > would recognise it as the Earth's axial rotation out of its own
> > orbital shadow.
> >
> > http://www.humpa.com/ScenicPlaces/images/las%20vegas%20sunrise%20WEB.jpg
> >
> > The arrows in the previous graphic of Kepler's second law represents
> > the point where orbital orientation changes over the course of an
> > annual orbit,on a given day the orbital orientation will point in the
> > direction of Vega tangential to the Sun/Earth line.
>
> I'm afraid I don't understand how these arrows can point inward,
> towards the Sun. I've been led to believe that a rotating objects
> speed was always tangential.
>
No problem,while the following animation is just a guide,it is best to
go outside yourself and determine that on a given day,Vega will
appear along the axis of orbital motion at the orbital shadow/direct
sunlight division.
http://www.museum.vic.gov.au/scidiscovery/images/mn000783_w150.gif
A common error is that many forget that axial rotation is independent
of orbital motion and that orbital orientation changes over the year.I
am delighted to say you will eventually reach your objective and
perhaps recognise that it is more important than you can imagine at
present.
>
> >
> > Then the Earth's cumulated speed would be 30km/s(its speed
> > > around the Sun) + 20km/s(the Sun's speed towards Vega) cos a(the axial
> > > angle between the Earth's mouvement and that of the Sun). Could you
> > > help me to find a more precise date and the angle a?
> >
> > The Sun does not speed towards Vega as both are travelling around the
> > galactic axis at more or less the same speed,in other words the
> > "Sun's speed towards Vega" has no meaningful basis.
>
> I've been sent this link:
> http://www.apnet.com/refer/solar/contents/solarch1.pdf
>
> Apparently the Sun is moving towards Hercules, not Vega and Lyra.
> How is one to know what is what?
>
Most of the difficulties arise from Newton's bulltalk of
geocentric/heliocentric orbital equivalencies.It is only a matter of
going out at dawn and determining the Earth's rotation out of its
orbital shadow to determine axial rotation.To determine the sweep of
the Earth's independent orbital motion is appreceated through seasonal
change and to appreceate Keplerian motion and the Milankovitch
cycles,the ice ages and geological change is the best way to
appreceate the variations in the solar system's galactic orbital
motion.To appreceate the insight of finite light distance(Ole Roemer
in 1676)you move to the next rotation which is the changing
orientation of the Milky Way stars to the other galaxies but that
rotation is challenging.
For your purpose you need only independent axial rotation and orbital
orientation tangential to the Sun/Earth line.I'm afraid you cannot do
this under Newtonian ballistics for he combined axial and orbital
motion into a single sidereal motion to achieve a
geocentric/heliocentric orbital equivalency.In short,it was a
departure from heliocentricity and can be seen here daily in terms of
the common usage of sunrise, sunset and constellations which generate
a fishbowl type universe.
> > What you can do is determine orbital heliocentric alignment with a
> > line drawn through the Sun from orbital perihelion/aphelion
> > coordinates on to Vega or the closest point.Simply use the graphic
> > supplied and calculate on which day the constantly changing orbital
> > orientation of the Earth intersects the line drawn from the Earth's
> > perihelion to aphelion through the Sun and on to Vega.
>
> Can you help me a bit more. The above link gives Hercules: right
> ascension 18h 0m and declination +30°. How do I obtain my vector angle
> a, from that?
>
These coordinates are taken off axial rotational longitude
coordinates.Flamsteed erroneously make axial rotational/stellar
circumpolar coordinates equivalent giving a celestial sphere.In using
orbital orientation,you are required to drop axial rotational
coordinates altogether insofar as orbital orientation does not follow
axial rotation (hence the change in the seasons).
Again,it requires that you go outside and detach orbital orientation
and motion from axial rotation and orientation and treat them as
independent motions for the your purpose.I assure you that your
endeavor has worthwhile implications and the experience is certainly
satisfying.
Apologises if my descriptions are not clear for that is my shortcoming
and not yours.
> > > In another group, someone told me that the Sun's mouvement was
> > > perfectly perpendicular to that of the Earth. In which case cos a = 0.
> > > But that does not seem to be the way things are. Could you confirm
> > > that a is not 90°?
> > > Regards, Ken
> > >
> > > >
> > > >
> > > >
> > > >
> > > >
> > > >
> > > > Also, as the Earth's mouvement is not in the
> > > > > same plane as that of the Sun, what is the angle(a) between these two
> > > > > planes.
> > > > > What I want to find out is the Earth's maximum absolute speed. This
> > > > > would be 30km/s (the Earth's speed around the Sun) plus 20km/s (the
> > > > > Sun's speed towards Vega) multiplied by cos(a). But these two speeds
> > > > > only add up at a precise time of year, which I haven't managed to work
> > > > > out.
> > > > > Hoping to hear from you. Regards, Ken
- Next message: Bob Schmall: "Re: We need to set a minimum moon size"
- Previous message: Jim Miller: "Re: Digital Camera as Sky Meter -- Ongoing Report"
- In reply to: kenneth couesbouc: "Re: Earth, Sun and Vega."
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