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Joe Keller
USA
929 Posts |
 Posted - 13 Feb 2007 : 20:07:23
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Three likely theories give coordinates for "Planet X":
I. It is at the (+/-) CMB dipole.
II. Same as (I) but with adjustment for the theoretical gravitational effect of Jupiter, Saturn, Uranus, & mainly Neptune.
III. Same as (II) but with additional adjustment for various other "Planets X".
(I) is least likely. (III) is least useful.
The error bars I'll give are mainly due to the error bars of the 3-yr. WMAP coordinates, because the extrapolation is so small. These are:
0.07 degree of great circle arc either way along the direction of galactic longitude (i.e., 0.1 degree of galactic longitude)
0.04 degree (either way) of galactic latitude
That's an "error ellipse" whose axis will be tilted, in any other coordinate system. It's an acceptable approximation simply to say that the error is 0.07 degree (0.28m or 17s) of Right Ascension and 0.04 degree (2.4') of Declination.
I. Positive CMB dipole, extrapolated coordinates for March 1, 2007:
l = 236.70 b = +48.26
In celestial coordinates:
RA 11h 11m 02s Decl -6deg 48' 45"
(on March 1, 2007)(J2000.0 coordinates)
The monthly change is:
RA -0.38 s/month Decl +4.9"/month
This is ideal observing position: south of the tail of Leo, high in the eastern sky before midnight, in a region with rather few stars.
Ecliptic coordinates (2007.25):
ecliptic longitude 171deg26'41"
ecliptic latitude -11deg6'06"
Per degree of inclination, only half as many asteroids have inclinations of 11 deg as of 0 deg. Furthermore a circular orbit inclined 11 degrees averages only 7 degrees ecliptic latitude. This helps reduce the number of asteroids in view.
II. The adjustment for the four gas giant planets, is +0.226deg ecliptic long. & -0.034deg ecliptic lat.; this gives:
ecliptic longitude (2007.25) 171deg40'15"
ecliptic latitude (2007.25) -11deg08'08"
RA 11h12m08s Decl -6deg55'53"
The monthly change per this theory is:
RA -0.83 s/month Decl +10.7"/month
III. The additional adjustment for an Earth-mass planet at 62 AU, ecliptic long 261, ecliptic lat +63.5 (I chose these coordinates for convenience of estimation, but they correspond very roughly to the "Planet X" inferred above from Davies' Madiera CMB observations) is:
ecliptic longitude (2007.25) 171deg35'36"
ecliptic latitude (2007.25) -11deg21'46"
RA 11h11m23s Decl -7deg10'20"
The adjustment for JL Brady's Jupiter-mass planet (if it were to exist) would be about a radian.
The range of possibilities I-III, plus just over two error bars each way, give a circular search region of radius 17 arcminutes (area 0.25 sq degree) and center:
RA 11h11m35s Decl -6deg59'33"
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Joe Keller
USA
929 Posts |
Posted - 13 Feb 2007 : 20:39:00
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quote:
Originally posted by shando
quote:
Originally posted by Joe Keller
My brown dwarf (or planet) is Lowell's Planet X.
Joe, what would be the period of your dwarf?
How many of our years would it take to circle the sun?
These are just the right questions! See post below. |
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Joe Keller
USA
929 Posts |
Posted - 13 Feb 2007 : 20:45:18
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By the methods above, I estimate 355 AU distance and a 6680 yr period for circular orbit. However, today I discovered that five of the nine proportionally small discrepancies ( < 10% ) in the resonances (m:n for m,n < 11) of the giant planets' orbits (periods per TP Snow, "The Dynamic Universe", 1983) are simple multiples of a discrepancy consistent with a retrograde influence with period about 4430 yr. This would (assuming that the 2:3 ratio is mere chance) correspond to 270 AU distance, 0.0107 solar masses, and a planet with magnitude +16.75, for 5.7% albedo.
This might be a more accurate estimate than that derived from the COBE and WMAP surveys with their nominally large error bars. Alternatively, the orbit might be moderately elliptical ( e > sqr(6680/4430) - 1 = 0.23 ). The other four of the nine discrepancies were related to multiples of Pluto's orbital period; mostly, these were less accurate resonances than the five above.
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Stoat
United Kingdom
964 Posts |
Posted - 14 Feb 2007 : 04:53:51
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 It doesn't look like they have a plate of that region of the sky. Perhaps that's good news, as if they haven't, then they haven't looked  Maybe the Australian Schmidt would be worth a look, it's been running for about thirty years. |
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Joe Keller
USA
929 Posts |
Posted - 14 Feb 2007 : 16:12:08
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quote:
Originally posted by Stoat
It doesn't look like they have a plate of that region of the sky. Perhaps that's good news, as if they haven't, then they haven't looked Maybe the Australian Schmidt would be worth a look, it's been running for about thirty years.
Above, I've specified a search disk including 0.25 square degrees of sky. Please note that I changed the region slightly to correct a sign error in the increment between (II) & (I). |
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Joe Keller
USA
929 Posts |
Posted - 15 Feb 2007 : 00:56:53
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Planet X will be dimmer than most asteroids but brighter than most Kuiper belt objects. It will move more slowly across the star background, and also retrograde. The average asteroid orbital plane is inclined one degree below the ecliptic near the vernal equinox, so ecliptic longitude -11 in Leo is 10 degrees from the line where asteroids can be seen most often. An extensive plot showed that asteroid inclinations of 10 deg are about 1/2 as common as 0 deg. If the inclinations make a bell curve with half-maximum at 10 degrees, numerical integration (the singularity in the integrand is "removable") shows that asteroids will be seen only 1/10 as often there.
Besides direct observation and studying old photographic plates, I've invented another method of search. The billion-star USNO-B catalog contains bogus objects which arise from two asteroids being accidentally at the same place on plates made at different times (DG Monet et al, Astronomical Journal 125:984+, p. 990). In part because of the unreliability of the magnitudes, the catalog was not filtered for discordant magnitudes. My method is to look for USNO-B catalog objects along the presumed track of Planet X, which have:
1. A difference of 2 or more, between the whole number parts of R1&R2, or B1&B2. E.g., B1=16.98 & B2=18.03.
2. At least one of their published magnitudes ("B1", "B2", "R1", or "R2") between +16.00 and +18.99.
3. Proper motion in either RA or Decl greater than 2 standard deviations from the mean.
In the "VizieR" online catalog, automatic filters can be set to accomplish (2) & (3). Because the typical midrange age of the plates used for a star is 30 years, one should go back 60 years, i.e., 3.2deg.
Three square degrees out of 41,000 in the sky, would have 10^9/14,000 = 70,000 stars, but the automatic filters could remove 90-99% of those depending on the setting. This is a manageable number of stars to check by eye for criterion (1). Asteroids and Kuiper belt objects occasionally would appear in this region of sky, but Planet X might be there all the time. A trail of bogus or dubious objects might be plotted.
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Stoat
United Kingdom
964 Posts |
Posted - 15 Feb 2007 : 02:58:37
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| I was going to set two plates up on top of each other in photoshop, and blink one of them off and on, to check for movement. |
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cosmicsurfer
USA
507 Posts |
Posted - 15 Feb 2007 : 05:11:13
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If 10,800 B.C. is the date for a major holocaust (Younger Dryas-major upliftment/volcanism,possible polar shift) on earth due to the possible gravitational interference from the returning retrograde extreme elliptical orbit of a brown dwarf sister sun; then, a 3600 year period fits perfectly with a year zero (was star of David our sister sun?) and we would not see the so called planet X for another 1593 years.
John |
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MarkVitrone
USA
387 Posts |
Posted - 15 Feb 2007 : 05:53:34
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I wanted to take a brief moment to thank Joe Keller for his use of support in his posts. I like it when a thought is backed by evidence. It is great to see the scientific method unfold before your eyes. Thanks,
Mark Vitrone
Moderator |
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Joe Keller
USA
929 Posts |
Posted - 15 Feb 2007 : 12:01:14
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quote:
Originally posted by Stoat
I was going to set two plates up on top of each other in photoshop, and blink one of them off and on, to check for movement.
That sounds very good, if one has the computer skills to do it. Inaccuracy of the images would cause all the stars to move, so one would be looking for something that moved much more than such background jiggle. Also, it would be good to double check that the image is the actual plate and not modified or reconstructed. |
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Joe Keller
USA
929 Posts |
Posted - 15 Feb 2007 : 12:06:57
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quote:
Originally posted by cosmicsurfer
If 10,800 B.C. is the date for a major holocaust (Younger Dryas-major upliftment/volcanism,possible polar shift) on earth due to the possible gravitational interference from the returning retrograde extreme elliptical orbit of a brown dwarf sister sun; then, a 3600 year period fits perfectly with a year zero (was star of David our sister sun?) and we would not see the so called planet X for another 1593 years.
John
Thanks for the interesting historical perspective. An ophthalmology journal once reported that because the fibers in the human lens are wound (like the mantle of a baseball) in three directions (for the human lens), a six-pointed star is the most accurate depiction of the scatter that an otherwise-perfect human eye observes. |
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Joe Keller
USA
929 Posts |
Posted - 15 Feb 2007 : 14:43:11
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To go back 60 years, I've specified seven overlapping disks to search. The first disk is centered at the above estimated present position, but I increased the radius from the above 17', to 21' for good measure. The disks grow to allow roughly for up to 5 degree orbital inclination.
Format:
center RA, center Decl, radius in arcminutes
The disks:
11h11m35s, -7deg00', 21'
11h12m59s, -7deg09', 23'
11h14m31s, -7deg19', 25'
11h16m11s, -7deg30', 28'
11h18m03s, -7deg42', 31'
11h20m07s, -7deg55', 34'
11h22m23s, -8deg09', 37'
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Stoat
United Kingdom
964 Posts |
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Joe Keller
USA
929 Posts |
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Joe Keller
USA
929 Posts |
Posted - 16 Feb 2007 : 00:57:56
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This is a small step for man but by the grace of the Gods it is a giant leap for mankind.
The billion-star 2003 United States Naval Observatory B1.0 Catalog suggests that "Planet X" is a double planet with a smaller companion half its radius. The following objects seem to be Planet X and its companion:
Object #1. USNO-B 0827-0286487
Object #2. USNO-B 0824-0279170
Object #3. USNO-B 0820-0274026
Object #4. USNO-B 0813-0233607
A "VizierR" search found that no objects at these positions occur in any other catalogs except NOMAD, wherein essentially the same information is given though slightly less detailed. According to the documentation accompanying the USNO-B catalog, one of the B magnitudes, +24, is probably invalid due to its excessive faintness. Consideration of the other magnitudes shows that there are two bodies accounting for these four observations.
One body had four red magnitudes ranging from 17.41 to 18.57, three blue magnitudes ranging from 18.26 to 19.80, and two infrared magnitudes of 18.36 & 18.78. The aberrantly high blue & small red magnitudes complemented each other energetically consistent with a change in passive reflectance color (weather on the planet).
The other body had four red magnitudes ranging from 20.26 to 20.68, and a blue magnitude of 20.30. At this high galactic latitude most stars with mag > +18 would be red dwarfs. The near equality of red and blue magnitudes for the presumed smaller body, and the near equality (except for the abovementioned aberrant pair of values) of red, blue & infrared magnitudes for the larger body, imply that they are illuminated by the light of our sun.
The pseudo-trajectory was consistent with 27.5 degrees orbital inclination from the ecliptic, too high for most asteroids. Furthermore with only about 10,000 asteroids of this magnitude, asteroidal coincidences on the plates would be far too infrequent. The curvature of the sky path probably is of the correct sign and within a factor of two of what is required.
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Stoat
United Kingdom
964 Posts |
Posted - 16 Feb 2007 : 05:51:38
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Hi Joe, I've been having a read of the robotic telescope. It seems that it takes a couple of days to get one's image back,depending on the seeing. As it's robotic an image can be disappointing if the moon is bright.
Take a look at a the site and see whether it might be an idea to register as a school teacher. That way you can enter all the jobs and make sure that the exposure time is the same for all of them. Then put them all back in a couple of weeks later to check for any movements.
Anyone on this board can then be part of your class. I did put a job up but if you like I can delete it. It was for 11 11' 23" -7 10' 20" |
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Joe Keller
USA
929 Posts |
Posted - 16 Feb 2007 : 10:08:52
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quote:
Originally posted by Stoat
Hi Joe, I've been having a read of the robotic telescope. It seems that it takes a couple of days to get one's image back,depending on the seeing. As it's robotic an image can be disappointing if the moon is bright.
Take a look at a the site and see whether it might be an idea to register as a school teacher. That way you can enter all the jobs and make sure that the exposure time is the same for all of them. Then put them all back in a couple of weeks later to check for any movements.
Anyone on this board can then be part of your class. I did put a job up but if you like I can delete it. It was for 11 11' 23" -7 10' 20"
Thanks, Stoat. From the result of my USNO-B catalog search, I have some narrower coordinates. See below.
Attention: anyone else on this board! Please look! I'm busy with these calculations and I don't mind a bit if someone else sees it first! |
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Joe Keller
USA
929 Posts |
Posted - 16 Feb 2007 : 12:03:43
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The second and third of the four USNO-B candidates above, seem to have accurate epochs, because their great circle angular separation, divided by the difference in epoch (1976.1 vs. 1965.1) implies a period, for retrograde circular orbit, of 4535 yr, in close agreement with the 4430 yr derived above from the outer planet resonances. I used these epochs to extrapolate an accurate location for 2007.25 (i.e., 07:30 UT, April 2, 2007):
RA 11h 06m 0.50s = 166.5021deg
Decl -6deg 28' 20.5" = -6.47236deg
daily correction from April 2, 2007:
RA -0.0468 sec/day
Decl +0.354"/day
The epochs are given to 0.1 yr, implying a possible track of locations, equivalent to up to a 103-day correction (-4.84s RA and 36.6" Decl) either way.
The fourth USNO-B object also seems confidently from its magnitudes, to be Planet X. Using the 3rd & 4th objects (instead of the 2nd & 3rd objects), the slope d(Decl)/d(RA) is 1.35% less steep. Such a true slope would result in a correction of -54.2" Decl.
Based on these USNO-B objects, then, the narrowed search region is a parallelogram with area 2.2 sq arcminute (0.0006 sq deg) and corners:
RA 11h 06m 5.34s
Decl -6deg 28' 57.1"
RA 11h 05m 55.66s
Decl -6deg 27' 43.9"
RA 11h 06m 5.34s
Decl -6deg 29' 51.3"
RA 11h 05m 55.66s
Decl -6deg 28' 38.1" |
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cosmicsurfer
USA
507 Posts |
Posted - 16 Feb 2007 : 13:58:22
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Hi Joe,
Would your extrapolations for time period of a circular orbit of 4535 yr change with an elliptical orbit? Orbital speeds would accelerate towards and away from the sun, plus Planet X may be surrounded by planetoids, debris fields and trailing asteroids and comets (former oceans) from previous collisions, e.g., destruction of Fifth Planet forming asteroid belt. Thus the former retrograde circular orbit of a former binary star is now in a extreme elliptical orbit from being thrown out of our solar system from such a collision or gravitational interference near miss causing EPH.
An elliptical orbit would result in a constantly shifting tidal effect from Planet X that would be felt by all planets in our solar system during its flight around sun.
Regarding historical perspective, the myths of a world wide deluge and even biblical revelations depictions of stars falling from heaven and skies turning red might account for a six sided refraction or “Star of David” during close encounters. Also, we could search for periods of higher intensities of meteorite bombardments as a possible indication of passage of Planet X. I did manage to find one large impact around the 10800 B.C. period near the Andes leaving a large crater.
Most star systems are binary and it may be that planetoids are the result of solar plasmas separating into two halves causing the initial spreading of mass and following orbital planetoids around the larger gravitational body (all in concert with overall structure and motion of Universe).
John
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Joe Keller
USA
929 Posts |
Posted - 16 Feb 2007 : 15:07:14
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quote:
Originally posted by cosmicsurfer
Hi Joe,
Would your extrapolations for time period of a circular orbit of 4535 yr change with an elliptical orbit? Orbital speeds would accelerate towards and away from the sun, plus Planet X may be surrounded by planetoids, debris fields and trailing asteroids and comets (former oceans) from previous collisions, e.g., destruction of Fifth Planet forming asteroid belt. Thus the former retrograde circular orbit of a former binary star is now in a extreme elliptical orbit from being thrown out of our solar system from such a collision or gravitational interference near miss causing EPH.
An elliptical orbit would result in a constantly shifting tidal effect from Planet X that would be felt by all planets in our solar system during its flight around sun.
Regarding historical perspective, the myths of a world wide deluge and even biblical revelations depictions of stars falling from heaven and skies turning red might account for a six sided refraction or “Star of David” during close encounters. Also, we could search for periods of higher intensities of meteorite bombardments as a possible indication of passage of Planet X. I did manage to find one large impact around the 10800 B.C. period near the Andes leaving a large crater.
Most star systems are binary and it may be that planetoids are the result of solar plasmas separating into two halves causing the initial spreading of mass and following orbital planetoids around the larger gravitational body (all in concert with overall structure and motion of Universe).
John
Thanks for passing along this wealth of interesting ideas. I think I can address some of them. |
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Requiem for Relativity
