When was the last time all the stars lined up in the sky, and in the
zodiac as they are today?

I.e., what is the cyclicity of the solar system?

Hi there,

Approximately 25,800 years ago was the last time all the stars lined
up in the sky, and in the zodiac as they are today.

A popular name for the cycle is "Precession of the Equinoxes". It is a
cycle of the how the stars appear to us, due to our planet doing a
very slow wobble, caused by the gravitational forces of the sun and
the moon on the equatorial bulge of the rotating earth.

The Columbia Encyclopaedia describes it:

This motion was first noted by Hipparchus c.120 B.C. The precession is
due to the gravitational attraction of the moon and sun on the
equatorial bulge of the earth, which causes the earth's axis to
describe a cone in somewhat the same fashion as a spinning top. As a
result, the celestial equator (see equatorial coordinate system),
which lies in the plane of the earth's equator, moves on the celestial
sphere, while the ecliptic, which lies in the plane of the earth's
orbit around the sun, is not affected by this motion. The equinoxes,
which lie at the intersections of the celestial equator and the
ecliptic, thus move on the celestial sphere. Similarly, the celestial
poles move in circles on the celestial sphere, so that there is a
continual change in the star at or near one of these poles (see
Polaris). After a period of about 26,000 years the equinoxes and poles
lie once again at nearly the same points on the celestial sphere....
The precession of the equinoxes was first explained by Isaac Newton in
1687.
http://www.infoplease.com/ce6/sci/A0840032.html

It is more normally given a time span of 25,800 years, although
popular author Graham Hancock refers to 25,776 years in his book      
 'Fingerprints of the Gods'. Because every source of gravity in the
Universe is going to affect the cycle (mostly at levels so small they
are unmeasurable), it would be impossible to calculate a precise
figure.

It is this cycle that will cause us to move into the Age of Aquarius
sometime soon (nobody knows, because there is no exact spot which is
closer to Aquarius than Pisces).

The stars and constellations are of course not fixed in space, and all
are moving. So technically speaking, the arrangement we see today will
never be duplicated again - unless there is a Big Suck that follows
the Big Bang. But in 25,800 years it should look virtually the same as
today, and the constellations will all still look the same, in the
same place, to the naked eye.

A diagram showing precession is at Astronomy 161:
http://csep10.phys.utk.edu/astr161/lect/time/precession.html

The complex mathematics are explained at Wolfram.com:
http://scienceworld.wolfram.com/physics/PrecessionoftheEquinoxes.html


Google keywords used:

"precession of the equinoxes" 
://www.google.com/search?hl=en&ie=UTF-8&oe=UTF-8&q=%22precession+of+the+equinoxes%22+


I trust this answers your question. Feel free to ask for any
clarifications you may require.


Best wishes,
robertskelton-ga

---

Request for Answer Clarification by nayna-ga on 30 Jul 2002 09:44 PDT

Please see my accidentally double-posted comment -- thanks

---

Clarification of Answer by robertskelton-ga on 30 Jul 2002 18:42 PDT

My answer for a human gazing upwards scenario:

Multiplying the list of synodic periods would be the answer, but there
might be a shorter period than that - if they were all 2 years, for
example, then the answer would not be 2x2x2x2x2x2x2 but rather just
two.

This explanation for just two planets comes from GoMath. If you could
find any common multiples between the planets orbits, then the answer
will be shorter than multiplying them all together. The more accurate
the synodic period figures you use are, the harder it will be to find
a common multiple (so perhaps round them to make it easier)...


Q. Suppose that right now Venus and Earth are lined up on one side of
the sun, as shown in the figure. If Venus circles the sun every 225
days and Earth circles the sun every 365 days, how long will it take
both planets to reach the same position again?

A. you are asked to find the common multiple of 225 and 365
225= 3*3*5*5
365=5*73
so the least common multiple for 225 and 365 is 3*3*5*5*73
http://www.gomath.com/Questions/question.asp?question=1411

==================================

My answer for a mathematically accurate scenario is;

It's impossible to tell exactly - orbits vary slightly over time due
to gravitational interactions, and it is so complex that even NASA
cannot work it out!

The NASA Solar System Simulator only covers 700 years, because:

"We really have very little way of knowing precisely where the planets
and satellites were / will be back then / that far ahead. Each body
acts on each other in small ways, and we can only predict their exact
positions so far. The range in the form is about the longest range
that we can be very accurate."
http://space.jpl.nasa.gov/
http://space.jpl.nasa.gov/faq.html

The problem was first realised by Poincare and his Three Body Problem,
which resulted in the observation that "the Solar System, is not
stable, but chaotic, unpredictable, and doomed."
http://www.musenet.org/utnebury/Poincare.html

=======================================

My answer for a horoscopical scenario:

25,800 or so, for the planets to be in the same places again, relative
to the zodiac.

"A horoscope is simply a chart of the heavens showing a certain
position of the planets and zodiacal signs relative to each other and
the earth. The constellations remain in the same position one to
another, and are therefore called "fixed stars," but the earth and
other planets constantly change. They do not return to the same
relative position until after about twenty-six thousand years."
http://www.rosicrucian.com/ssa/ssaeng01.htm

Oops, my bad -- did a classic mistake in the asking of the question.
Also, probably should have upped the amount frankly, as math is hairy,
but I guess I thought it was a quick answer...  (Still didnt get  that
answer :) but that's okay.) Splendid splendid effort.

As robertskelton mentioned, "all are moving".  Astronomers call this
"Proper Motion".  Most notable fast-moving bright star is Alpha
Centauri (multiple star), which moves about 26 degress in a cycle. 
The fastest (dim) star is Barnard's Star, which moves the diameter of
the moon in only 180 years.
http://astro.estec.esa.nl/Hipparcos/msa-tab8.html
http://astro.estec.esa.nl/Hipparcos/TOUR/tour-fast-stars.html
http://astro.estec.esa.nl/hipparcos_scripts/launchApp.pl?appname=ShowMotion&RA=269.45&Dec=4.66&tol=3

Perhaps you are referring to the planets, instead of the stars?

The following websites might be of interest.

http://www.spirasolaris.ca/times2.html
http://www.thefuture.com.au/acyc.htm
http://arxiv.org/PS_cache/astro-ph/pdf/9501/9501078.pdf
http://www.sweb.cz/vladimir_ladma/english/cycles/reson/synodp.htm
http://scienceworld.wolfram.com/astronomy/SynodicPeriod.html
http://nssdc.gsfc.nasa.gov/planetary/factsheet/index.html

Yikes. Major clarifcation coming:  I am quite familiar with the
precession, and its cyclicity with respect to the "fixed stars".  I am
INSTEAD asking about the cyclicity of the NONFIXED stars of our Solar
System. (As you can see by rereading the question.)

Hence, when was Mars up in the sky as it is now AND Jupiter AND
Mercury AND the Sun AND the Moon AND Venus AND Saturn ? (We can skip
Pluto and Uranus and Neptune and the Asteroid Belt and whatever else
for now.)

(This question has important ramifications in the mythology of
astrology; that is why I am asking.)

Ulu-ga's comment rocks -- it definitely addresses what I'm looking for
-- but I cant figure it out -- but yeah, that's what I'm asking: what
is the syncopation cycle of all the 7 major planets (see my comment)
synodic periods, taken TOGETHER?

I should also note that while ula-ga's links are right-on appropriate
with regard to addressing my question, they do not ANSWER it.  That is
to say, they all discuss the subject that I am asking.  But they do
not anser the particular of when do all 7 of the planets I list repeat
themselves at once.
<P>
My conjecture, for which I am seeking confirmation for my $3 frankly,
is that they would recur only at the multiple of each of their synodic
periods.  So multiplying the list of synodic periods would be the
answer?

Yeah, I don't think there are any submultiples of the planets with
respect to the Earth.  There are some subperiodicities with respect to
eachother, according to  http://www.thefuture.com.au/acyc.htm, but no
real reductions (in the modulos, i.e., no kernels to the set if we
consider this with group theory?)

Plus it also all definitely depends on how lax you are willing to be
on the precision, so that astrologically, Aries spans 30 degrees, etc.
Nonetheless, it seems calculable.
---
Yeah, I love the three-body problem and here we have a nine-body
problem; also nonetheless, I guess I'm boldly assuming Newtonian
approximations on fairly stable cycles...
---

Anyway, thanks, you've been great.

I totally would have upped the ante midstream, had I known the output
that was coming.  I think Google should allow for this possibility!!!!

Sorry Nayna for the delay in responding, but I've been away from the
computer.

I didn't have time to read in detail all of the links I posted, but I
didn't see an answer.

The simple answer is:  It never repeats.

As you said, "definitely depends on how lax you are willing to be on
the precision".  You can get some sense of a period by looking at the
planetary alignments.
http://www.sunspot.noao.edu/PR/alignment.html

There was a popular book, "The Jupiter Effect" by John R. Gribbin.
http://www.amazon.com/exec/obidos/ASIN/0333174186
http://www.earthsky.com/Features/Articles/planets1.html
http://www.atnf.csiro.au/asa_www/info_sheets/alignment.html
http://www.survivingtheapocalypse.com/thefeatures/2000/april/

Orrery (you can check some of dates here)
http://www.fourmilab.ch/solar/solar.html
http://www.cuug.ab.ca/~kmcclary/

Kepler searched for order in the solar system.  Although he didn't
find his "Music of the Spheres", he did uncover some interesting laws
describing the motion of planets.  I recommend the book "The
Watershed: A Biography of Johannes Kepler" by Arthur Koestler.
http://www.amazon.com/exec/obidos/ASIN/0385095767

The least common multiple (LCM) method described above doesn't quite
work.  A problem is the orbit period is not an integer.  Rounding the
value in the beginning will cause problems after multiple orbits.

I've run out of time again, but if you (or some other
commenter/researcher) can program this up, it will come close to your
answer.

Get very precise sidereal orbit periods (you can start with each
planet from the earlier NASA link)
Earth =? 365.25636042 
(but they seem to disagree with the values here???)
http://www.blueskymining.com/saddleback/mse/astro/astro_equations.html

Consider each orbit to be a circle with 0 to 1 mapped onto it.
You will add an earth's year worth of orbit to each path and see if it
is within range.  (This will place the sun and other planets in the
same place relative to the stars)

For each planet, Divide Earth's period by the other planet's period
("years" worth) and store in a table.

Initialize a table of planet's positions to be .5 (easier range
checking)

year = 0
do( forever )
  year = year + 1
  for each planet, add a year's worth to the planet position.
  for each planet, while position is greater than or equal to 1.0,
subtract 1.0
  if all positions are near .5 (within .01 is 3 days) then print year
and positions

There may be a mathematical way of solving this, but I can't think of
it now.

If you keep track of a precision variable, starting with .5, you can
decrease it each time you find a year where the orbit difference is
less than the current precision.

This will print out several "periods" that will get longer as it gets
more precise.  (Don't forget to print out the precision variable at
the same time you print out the rest)

Dudes! How can I get you some more money?

I asked google to install a tip jar on these things -- don't you think
that's a good idea?

Good idea on the tips.  I think that would work better than having a
person repost another question.  I've seen other questioners suggest
something like that too, especially when they want more information
than they originally thought or they thought the researcher did an
exceptional job.

Sorry for the non-answers, but thought you still might be interested
in the material and someone else could carry the ball.

I don't think you can take the "multiple of each of their synodic
periods" to find an answer.  Synodic periods occur at different
positions in the orbit.  Earth/Mars would occur at a different
position than Mars/Jupiter.  This also would not work if you wanted
them to appear in the same constellations of the zodiac.  For outer
planets, they would meet at opposition.  These approximations also
don't take into account elliptical orbits (yet another complication).

Continued fractions might be a way to approximate the orbit periods
and look for common denominators.
http://mathworld.wolfram.com/ContinuedFraction.html

Precision is very important.  The difference in the orbital position
in 50 years with the obital period being 365.25 vs 365.26 is the other
side of the sun!  Those little numbers add up, especially when dealing
with such a long time.  Some web sites find interesting patterns,
because they just drop a few digits of precision.

Good luck!  (Be sure to watch the upcoming meteor showers in Aug. and
Nov.)

Sorry about my mistake on precision.  That error would only be one day
in position different.  If it's over 25,800 years, you could be on the
other side of the sun.

If you find an answer, be sure to post it.  It's been an interesting
question.

"(This question has important ramifications in the mythology of
astrology; that is why I am asking.) "

One of the big ones in astrology is due to precession. For example,
plug your birthday into one of the many free planetarium software
packages out there.  Look for the Sun.  For most people, the Sun is in
the constellation _next_ to their "sign" because of precession.  Over
the thousands of years since the signs were invented, precession has
shifted things around.

If you exclude our sun, planets, moons, asteroids and comets from the
definition of stars, the line up is approximately the same at about
one year intervals, but changes of several parts per million are
occurring annually for most of the stars, so they never lined up
exactly as they are now, nor will they in the future. Over a period of
several thousand years, this "proper motion" reaches a degree or more
for several important stars. Could all the effects cancel and return
to about the same pattern in some very long time?Perhaps, but by then
some of the present stars will be cold white dwarfs, nuetron stars or
black holes or collision products. Similar stars could move in to
replace the caualties, but only a simple pattern of approximately the
same shape is likely even in a google years. Do you want to consider
the position of stars in other galaxies? If so we are talking about
200 billion times 200 billion stars repeating the same pattern.   Neil

I read your clarification after I posted. Approximate line-ups of
today's seven planet positions in the zodiac occured at several almost
random dates in the past 100,000 years. The dates will be radically
different (I think) for last month or a year from now. If 700 years is
state of the art for good accuracy, we probably can find one of the
past or future dates that match approximately on several days next
year or this year. Other reference dates will yeild no dates with
reasonable confidence. I have no idea how to do this and I may be dead
wrong. If you find a common denominator by multiplying the nearest
whole number of period in days you will get somewhat different dates
than if you use numbers in hours. Better accuracy would be period in
seconds. I think. I suspect the eccentricity and the tilt of the
orbits with respect to the plane of the ecliptic make this approach
fatally inaccurate.  Neil