As I understand the cosmos the sun is orbiting the center of the milky
way, "dragging" the earth along with it. I'd like to know in what
direction the sun is headed relative to the calendar year?

A made up answer would be "the sun is headed in the direction of June 13th"

Hello there

The Sun is indeed in motion.  It orbits the center of our galaxy at
the speed of approx 155 miles per second in a counter-clockwise
direction.  And of course we are moving right along with it, being
"dragged" as you put it.

It takes the Sun (and us) about 200 - 250 million years to orbit once
around the center of the Milky Way Galaxy.  So yes, the Sun does have
a "year" as well - about 200 million times or so as long as an Earth
year.

The analogy of the CD spinning around a non-moving center mentioned in
the comment section leaves much out of the equation.  While the CD
disk is spinning around its center, the CD center is moving at
considerable speed as the Earth rotates.  at the equator, that would
be about 1670 km'hr or about 1070 miles per hour.   To find the
rotational speed where you live, take the cosine of your latitude, and
multiply it by the speed at the equator. For example, the speed at a
latitude of 60 degrees, the speed would be cos(60)x1670=835km/hr.

So now we have that CD spinning around a center which itself is moving
at over a thousand miles per hour because it is on the surface of the
Earth.

Now while that disk is spinning around its center and the CD center is
spinning around the Earth, the Earth itself is orbiting the Sun at
about 18½ miles per second - and of course the sun is moving at about
155 miles per second.

So the analogy of the "motionless" center of the CD disk falls far short.

To answer your question - what direction the Sun is headed relative to
"Earth's" calendar year? - - it is headed counter-clockwise around the
center of the galaxy at 155 miles per second and at the end of the
calendar year is approx 1/200 to 1/250 millionth further along the
path than it was at the beginning of the calendar year.

And please don't forget, our galaxy is orbiting within a larger
galactic cluster which in turn is orbiting an even larger cosmic
structure right up to where everything is orbiting something which
might be called the "center of the universe" itself. - - which of
course is me - - and I can prove it by means of concentric circles.
<-- feeble attempt at humor.

I'm getting dizzy.

Search - Google
Terms - orbital speeds

After chasing these web sites several thousand miles through space
(really, as you now know) - I used the following to compose the
answer.

http://searchsmallbizit.techtarget.com/sDefinition/0,,sid44_gci849653,00.html
- "Earth's mean orbital speed" - From searchSmallBizIT.com

http://www.enchantedlearning.com/subjects/astronomy/solarsystem/where.shtml
- information about the Sun's galactic orbit from enchanted
learning.com

http://www.windows.ucar.edu/tour/link=/kids_space/qearth_motion.html -
"Quickie Questions - Extraordinary Earth - Movement of the Earth" -
From University Corporation for Atmospheric Research

If I may clarify anything, please ask.

Cheers
Digsalot

---

Request for Answer Clarification by twbt-ga on 25 Jul 2004 10:27 PDT

Hello,

Unfortunately you didn't answer my question at all :) though of course
I appreciate the detail you went into answering the question you
thought I asked.

I still believe the premise for my question is valid, so I will reword
it in the hopes that its logic is more clear (I could resubmit the
question if desired, i'm easy going..)

Here it is reworded.

PREMISE:
1. The earth has an orbit around the sun. Any position in that orbit
corresponds to a day on the calendar year. (i.e. when the earth is
"left" of the sun its October, I'm guessing on the month, but the
point is that there has to be a corresponding month to the point in
the orbit of the earth where it is located directly "left" of the sun)
2. I don't care where the earth is at all in the equation.
3. The Sun is moving in a straight line (yes it must be moving in a
curve but let's assume to make things simpler that the distances
involved here are small enough that the curve will appear to be a
straight line

Now let's jump to an analogy.

I have a marble that represents the sun. Immediately around it I have
traced a circle in chalk that represents the orbit of the earth.
Clearly the moving sun must CROSS the chalk circle since it is moving.
Since it is moving in one direction only, then it can only cross the
chalk circle at ONE point.

Let's call that POINT X.

Point X is a point on the chalk circle... which is a point on the
orbital path of the earth... which has a calendar date associated with
it.

Therefore The sun crosses the orbital path of the earth at a specific
point, that has an associated calendar date with it.

My question is: What is that date?

---

Clarification of Answer by digsalot-ga on 25 Jul 2004 13:31 PDT

Hi again

AHA! - your restatement of the question creates a new clarity.  Sorry
I misunderstood the first time.

The answer is the day of the northern Winter Solstice.

The northern summer takes place when the Earth would be positioned to
the "right" of the Sun as you use in your analogy..  That is when the
northern hemisphere is tilted most directly toward it.  Remember, the
Earth orbits in a counter-clockwise direction so it should be easy to
visualize.  As the northern hemisphere moves into autumn, the Earth
has moved approx one quarter of its orbit and is now directly in
"front" of the sun as the Sun moves on its orbit around the galaxy. 
So we now know that the 'figurative' crossing of the Earth's orbit by
the Sun's orbit takes place sometime in the northern autumn.  Actually
in the very last instance of the northern autumn or the first instance
of the northern winter, whichever way you want to look at it.

When we consider that both the Earth and the Sun have
counter-clockwise orbits, the earth is directly in 'front' of the
Sun's direction of travel on the day of the Winter Solstice. - the
21st or the 22nd of December, depending on the year.

I have posted a diagram which might make the explanation a little
easier.  The direction of the solar galactic orbit is from bottom to
top and the positioning of the Earth's orbit is marked by season. 
Both orbits are counter-clockwise.  The solar orbit is based on the
presumption it is a straight line, though we know it is actually a
curve.  This file is a temporary posting which will vanish in a few
days so if you want a copy of it, do it soon.
Link for your file is : 
http://68.15.21.151/uploads/researchers/Earthorb.gif

Still dizzy
Digs

---

Clarification of Answer by digsalot-ga on 25 Jul 2004 18:40 PDT

A further narrowing down of the date thanks to efn-ga.  Since we are
dealing with a hypothetical fixed point where the planet earth is in
its orbit on the Winter Solstice, it takes a fraction over 7 days for
the Sun to reach that point.  In 7 days, the sun will have moved
93,744,000 miles along its path.  So if you wish we can make the date
for the Sun crossing that hypothetical point on Dec. 28.  You can
accept that or winter soltice, whichever one you want.

Now I can blame efn-ga for making me dizzy - though I'm glad he did.

Digs

---

Clarification of Answer by digsalot-ga on 30 Jul 2004 11:02 PDT

Thank you for the kind words and the extra.  Please keep us in mind if
we can of service in the future.

Digs

Thanks for answering my question perfectly. The diagram was helpful too.

Since the earth orbits the sun eliptically (nearly a circle), the sun
is almost always in the center of the earth's orbit. The year is
determined by the earth's position in earth's orbit about the sun. A
complete circuit is one earth year.

So, unless you feel that the sun is orbiting the earth, the sun's
position is not relative to the year. It's pretty much asking the
question of where is the center of a CD heading relative to the
spinning of the outside of the CD? It isn't moving. It could be
moving, but the spin is also moving with the center.

We'd actually be in trouble if the sun started to move very much with
respect to the earth's orbit. It would mean that the planets, starting
with Mercury, would be more affected by the Sun's gravitational pull
at one part of the orbit, and possibly spin into the sun or out of the
Sun's orbit. Each loss of an interior planet would affect the pull on
subsequent planets.

:) That thing about the earth orbiting the sun?
http://roland.lerc.nasa.gov/~dglover/dictionary/y.html Year: A period
of one revolution of the earth around the sun. :)

Digsalot :) Hey, thanks for that information. Is the motion of the sun
relative to the year significant? How much does  1/200 millionth mean
in actual movement (distance, KM)?

Also, I considered the CD center to be the motion of the sun, not the
rotation of the earth... The CD's outside edge would have been the
motion of the earth relative to the sun.

The question assumes that the earth's orbit around the sun and the
sun's orbit around the center of the galaxy are in the same plane, but
they are not.

http://www.physicsforums.com/showthread.php?t=9666

http://amateurastronomy.org/Events/EH451.html

We could project the earth's orbit perpendicular to the ecliptic plane
to define a cylindrical surface and ask where radially on that surface
the sun's galactic orbit would intersect it, to get a time of year
answer.  To make this meaningful, we would have to pick a particular
time at which to project, since the earth's orbit is moving with the
sun.

--efn

Efn, based upon your comment, there may not ever be a possibility that
the sun crosses the orbital path of the earth, right?

I really would be concerned that the sun would cross the orbital path
of the earth. How close is the orbit of the earth in the nearest to
the sun? How long would it take the sun to travel that distance?

I'm willing to guess that it's not even close to one earth year. Even
as fast as the sun might move, I think we'd really be in trouble if
the sun ever came close to the orbit of the earth.

I think your question is valid. There should be a date (and time to
the second) each year on which the Earth is circling the galaxy faster
than the sun, by a maximum amount, and another date about 6 months
later when Earth's speed is less than the sun by a maximum amount. The
date will shift slightly due to leap year, and drift a few parts per
billion each year. The tilt of the sun's path and the tilt of the
Earth's path complicate the calculation, but IMHO there is an answer,
and it is of some significance reguarding the speed (relative to
Earth) at which particals pass though our solar system. Sorry, I don't
know the dates.   Neil

Digsalot, you have the earth crossing the sun's orbit. When does the
sun cross the earth's orbit, according to the clarification of the
asker?

Quote from twbt-ga: 
"Therefore The sun crosses the orbital path of the earth at a specific
point, that has an associated calendar date with it.

My question is: What is that date?"

Hi crythias-ga 

Take a look at the diagram again.  As the sun moves forward, it
crosses Earth's orbit on the winter solstice.  While the diagram is
stationary, the sun and earth are not.  The Sun is moving forward at
155 miles per second.  It will 'figuratively' cross Earth's orbit at
the point of winter solstice a very short time after Earth has passed
that point.  Short enough that the event will take place within a few
hours of the instant of solstice.

We need to imagine that point when Earth reaches the solstice as being
stationary while the Sun moves.  Not a reality but necessary for the
question.

In reality, the Sun will never cross the Earth orbit because the earth
orbit is moving forward at the same speed as the Sun since it is
centered on the sun.  It would, in diagram form, be a forward moving
spiral rather than a circle.

The hypothetical crossing of the Earth orbit by the Sun is as stated,
winter solstice.

Digsalot,

What's the evidence that the sun's galactic orbit moves it toward
where the earth is on the winter solstice?

I accept that if the equinoxes are to the right and left, the
solstices are in front and behind in orientation.  But this is just an
arbitrary orientation, and I don't see what it has to to with the
sun's orbit through the galaxy.  Why couldn't we just as well say the
solstices are to the left and right and the vernal equinox is in
front, so the sun is going toward the vernal equinox?

By the way, for those who are interested, it takes the sun about seven
days to move in its orbit a distance equivalent to its average
distance from the earth.

--efn

Hi efn-ga  - - You say the orientation is arbitrary?  Not really. 
Since the orbital directions are counter-clockwise for both the Sun
and the Earth, there is only one way we can orient it as a "map."  The
orbit of the Earth must move around the dot representing the Sun in
that counter-clockwise direction and in this case, the orbit of the
sun itself must be from the bottom of the page to the top,
counter-clockwise around the galaxy. Nothing arbitrary about it.

We must also take into account the Earth's 'tilt.'  For the northern
hemisphere to experience summer, that tilt must have the northern
hemisphere pointed toward the Sun.  On a map aligned with the
counter-clockwise orbit, the earth must be to the "right" of the Sun
on the map for that to happen.  So once again, positioning is not
arbitrary.

But - and thank you - I did the calculations and it does take just
over 7 days for the Sun to move that distance.  I will correct my
answer.

LOL - If we want to take it to the ultimate - the Sun is 'always'
crossing Earth's orbital path.  Since the Sun takes 225 million years
to orbit the Galaxy, it has done that many times in the 4.5 billion
years of Earth history.  The path is pretty well covered by now.

Digs

http://hypertextbook.com/facts/2002/StacyLeong.shtml
The Sun has orbited the galaxy, more than 20 times during its 5
billion year lifetime.

http://www.physicalgeography.net/fundamentals/6h.html
(More info supporting the above comments)...

But... what's keeping earth (or mercury, for that matter) from
crashing into or careening off the Sun's orbit? It would seem that you
have a Sun moving fast enough to be in the way of the earth's orbit if
it were moving in the same plane, and then be so far out of the orbit
that the earth couldn't orbit again.

How does this get rectified? What are the angular momentums and ...
Sorry, I will ask this in a real question...

I am perplexed by the answer.  Does the orbital path of Earth rotate
as Sun orbits the galactic center?  Would not this have to be the case
for the Winter Solstice to always occupy the point of the
intersection?  i.e. If the sun is located on the "left" side of its
orbit, moving counter-clockwise "top to bottom", is Earth's orbital
path rotated so that the Winter Solstice occupies the exact "bottom"
of Earth's orbit?  Or, does Sun in fact intersect at the Summer
Solstice when it is located thus?