Requiem for Relativity

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17 years 10 months ago #16388 by Joe Keller
Replied by Joe Keller on topic Reply from
If this distant orbit is elliptical, its projection on the celestial sphere is nonetheless a great circle, so the extrapolated sky trajectory is the same except for rate of travel. The extrapolation is so small (2.4 degrees) that for an orbital eccentricity of 0.2, the rate of travel would vary at most 0.8% (causing, at most, an error the size of the rms error from rounding the epoch dates). Furthermore, the orbital period, as implied by the orbital resonances of J, S, U & N, is consistent with the present angular speed (1965.1-1976.1), if the orbit is nearly circular.

The dimmer of the two red magnitudes, for each of the four collinear objects I found in the USNO-B catalog, was +20.61, 20.26, 20.68, and 20.68. I plotted all 61 dim red magnitudes obtained when I searched the USNO-B catalog for all objects in the above seven disks, which had proper motions >80mas/yr in both directions (an indicator of bogus combination) and which also had one red magnitude < +18.99 and one > +19.50. Fifty-nine of the dim magnitudes were roughly evenly distributed between the 19.50 cutoff, and about 20.90. The occurrence of the eight pairs of precisely equal magnitudes conformed everywhere to the Poisson distribution. Two of the magnitudes were > 24 which, according to the documentation accompanying the catalog, usually denotes an invalid reading. One of these "invalid readings" occurred for one of the four collinear stars (p=0.125). One of the equal-magnitude pairs also occurred among them (p=0.09).

Furthermore the chance that four points from a uniform distribution of length 1.40, will span a range of 0.43 or less, is small. The chance that any three of the four will span a range of 0.08 or less, also is small. For four numbers chosen randomly in the unit interval, the mean sum of differences squared is one (by a four-dimensional integral). Scaling the observed magnitudes, above, to a unit interval, gives a sum of 0.245. The only a priori distinction of this group of four, from the other 57 high proper motion, discrepant red magnitude stars, in the searched region, is that they formed the straightest line of any four stars that I could find by eye on my handmade plot on Keuffel&Esser graph paper.

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17 years 10 months ago #16477 by Joe Keller
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I found another object resembling USNO-B catalog objects #2-#4 above:

Object #5. USNO-B 0830-0272239

I found it 4.9" from the extrapolated great circle, searching with overlapping disks, of sometimes 1' but usually 2' radius, so that the band halfwidth always was more than 50" and usually was almost 120", along the chord of the previously unsearched extrapolated (to the present time) portion of the great circle made by objects #2 & #3. This time I searched for either Red1 or Red2 magnitude between 20.54 & 20.75. Roughly 30 objects were found for which the other red magnitude was 19.xx, and comparably many such objects also were found searching randomly a fraction of a degree away.

Only two objects along this arc of the great circle, had the other red magnitude < 18.99. Of these, Object #5, had not only an R2 magnitude of +18.59, but also a large proper motion of the order of that given for Objects #1-#4. Its B2 magnitude was +23, hence "invalid", further accentuating that characteristic of these objects.

The epoch was given as 1974.5. Only one pair (#2 & #3) among the five objects has epochs consistent with other orbital period estimates. Maybe the epoch given, usually is determined by the midrange, not of plates showing the object, but of all plates covering that region of sky. The following post will argue that by considering brightness, even the #2 & #3 epochs must be rejected. It's likelier that the object is, as initially estimated, moving with the angular speed of a circular orbit of 355 AU radius, but has an elliptical orbit whose period is that of a circular orbit of radius 270 AU.


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17 years 10 months ago #16390 by Joe Keller
Replied by Joe Keller on topic Reply from
I discovered and realized the existence of at least one moon, shortly after midnight UT, Friday, Feb. 16, 2007. The distribution of the "moon" magnitudes (not including the fifth "moon" found by searching for such magnitude) is 20.61, 20.26, 20.68, 20.68. This suggests that there is a large moon causing the 20.26 magnitude while a smaller moon causes the others. The closeness of the other three numbers makes an outlying measurement error, or varying albedo, unlikely. Eclipse would be too brief or too rare. My IBM 486 Monte Carlo trial indicated that the chance of any three such close points (sum of squared differences) in a set of four, from a uniform interval from 19.50 to 20.90, is about p=0.02; the closeness of the fourth point (one "Jupiter" & three "Saturn" sightings, but with "Saturn" closer to the "sun" than "Jupiter") reduces this to p=0.01.

The 20.30 Blue magnitude and 20.66 average Red magnitude of the smaller moon give B - R = -0.36. The USNO-B1.0 catalog gives B - R = -0.365 for Capella (a double star, G6 & G2, vs. the sun's G2). For reflection (generally weak in far blue) to achieve this B-R value using sunlight, it would likely appear bluish to the eye, which is sensitive mainly to near blue (maximum human retinal cone sensitivities are red 0.59 micron, green 0.55 micron, & blue 0.45 micron, according to the CRC Handbook of Chemistry & Physics, c. 1960).

The mass-radius relation for a giant planet or brown dwarf is very flat; due to electron degeneracy at high pressure, 0.019 solar mass would give about 10% less area than Jupiter (Burrows et al, Reviews of Modern Physics, 65:301+, 1993; graph reproduced in the above-cited book, "Protostars..."). An albedo of 5.7% (1/10 of Uranus, but close to average for a common kind of Kuiper belt object or asteroid) gives +18.1 magnitude at 360 AU.

I have five red, two blue and two infrared (0.75-1 micron band) magnitudes for the presumed larger body:

Object 1: R 17.41, B 19.80, I 18.78
2: R 18.17, B 18.95
3: R 18.57
4: R 18.03, I 18.36
5: R 18.59

These nine, average +18.07. The absence of any Red magnitudes between 18.60 & 18.99 is significant at roughly p=0.03, based on the distribution of fellow R values found in the above search of all objects whose smaller R value was near 20.6. The distribution of the magnitudes is most consistent with a large red spot covering almost one hemisphere of the large body, and a rotation period of between a day and a decade.

The spectrum, though red, is unusually flat. This suggests reflection instead of Planck curve emission. I searched USNO-B1.0 within a 20 degree radius centered at RA 11h10m Decl -7, for objects with both RA & Decl proper motions > 80mas (as before), R1 within 0.25 mag of the mean R for the presumed large body, and R2 within 0.25 mag of the mean R for the presumed "first moon" (using only Objects #1-4). Only 150 objects were brighter in both I & B; 1415 were dimmer in both I & B.

This flatness is not due to the averaging of four objects: a calculation based on galactic disk thickness at this galactic latitude, and textbook published main-sequence spectral type counts near the sun, implies that half the stars of the larger body's apparent magnitude, would be spectral class K5-K9, half class M, and all main-sequence. My IBM 486 Monte Carlo trial showed that a group of four equally bright stars with temperatures randomly differing over a +/- 10% range, perceived as one object, would have B-R & I-R, averaged over many groups of four, differing < 0.02 magnitude units from a single object at the population mean temperature. (This seems to grow quadratically with the range.) Furthermore, one side of the curve was raised and the other dropped almost the same amount, so what small effect there was, would mostly cancel, in the statistical test in the preceding paragraph.

In the USNO-B catalog, I checked the photometric values of Procyon (F5 IV-V) and Capella (I called it G6; it's G6 III + G2 III). From these, I interpolated by steps in spectral type, to obtain for the sun (G2 V), B - R = -0.46, I - R = +0.20, B - I = -0.66. For the presumed large body, these figures are +1.34, +0.53, +0.81, resp. This suggests sunlight reflected by a reddish object.

Galactic dimensions and main-sequence spectral type distribution, imply that almost all stars in the magnitude range, were Type K5 through late M (about half late K, & half M). For a mid-TypeK to late-TypeM star, B-R is large positive, but also I-R is large negative. Aldebaran (K5 III, somewhat variable) has B-R=+2.13, I-R=-0.85, B-I=+2.97. Gacrux (Gamma Crucis, in the Southern Cross)(Type M3.5 III, & variable like most Type M stars) has B-R=+2.16, I-R=-0.85, B-I=+3.00. If a dimmer second red magnitude for the object, implies that both B-R and I-R should be corrected downward, then I-R is less discrepant but B-R is even more discrepant.

The theoretical predicted surface temperature for this brown dwarf is 430K (interpolation or slight extrapolation, for 0.02 solar masses & 5 billion yr, in Table I, p. 316, Burrows, op. cit.; there was a log-log relation to age and a linear relation to mass). This might make it look more like Jupiter or Venus than like Pluto: its albedo would be too high. Maybe it cooled faster or burned less deuterium than theorized. Maybe it is less massive or accreted gradually.

Theoretically, "late M" dwarfs progressively show a blueward shift in their color temperature (Burrows, op. cit., Fig. 5, p. 309; quoting Allard, 1991). That is, according to Kirchhoff's law (Condon et al, Handbook of Physics, 2nd ed., p. 5-37, 1st partial par.) they emit & absorb blue better, and reflect red better. This trend might continue into brown dwarfs and giant planets. Also, common kinds of Kuiper belt bodies are reddish.

At 0.08 solar masses, one has a red dwarf (Burrows, op. cit.); not only does it begin to emit its own light, but the radius considerably increases. Even before that, one has a "hot" brown dwarf with self-produced IR magnitude. So, the distance can't be much more than 360 AU or the mass needed would be too great. A 440 AU distance likewise makes it too bright, even with the lowest likely albedo.

According to the 2003 AJ article giving information on the USNO-B catalog, it used almost 8000 plates; a plate sometimes included 5 million stars. For a billion stars, this implies that each small region of sky is included in < 40 plates.

Let a moon orbit Planet X; the moon's orbit might lie near the plane of Planet X's orbit around the sun. In a 6600 yr circular orbit, Planet X moves 196"/yr. That's 0.34 AU, at Planet X's orbital radius. Because Planet X has roughly the geometric mean of the sun's and Jupiter's masses, its main satellites might orbit at roughly the geometric mean of the distances, of the main satellites of the sun and Jupiter. A quarter of the time, a moon in a circular orbit, will be within 22.5 deg of its greatest elongation. For a 0.34 AU orbit, that is within 15". A 5deg orbital inclination for the moon, would cause typically only another 12" displacement perpendicular to the first. A somewhat larger orbit would be somewhat less efficient at placing the moon near 0.34 AU before or behind Planet X.

Generally, objects on the plates might be combined, if no more than 30" apart. Maybe half the time, the Red1 & Red2 plates were taken one year apart. Then, a quarter of the time, Planet X's moon would be positioned to impersonate Planet X on the plates. So, such a mysterious object would appear, in one of the 40 plates, 40/2/4 = 5 times. I found five objects in a thorough search of the USNO-B catalog.

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17 years 10 months ago #16393 by Joe Keller
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This new giant planet (or small star as the case may be) shall be named Barbarossa. Its largest moon shall be named Frey and its next-largest moon, Freya. The estimated diameter of Barbarossa is 83,000 miles; of Freya (sighted four times among the above five USNO-B catalog objects) and Frey (sighted once), 18,000 miles and 23,000 miles, respectively. The radius of Freya's orbit is about 1/2 AU and of Frey's orbit, 1 AU. Barbarossa rotates with a period of many days; it is mostly red on one side, dark gray on the other, with a visual albedo of about 6% on both sides. Frey and Freya resemble Neptune: blue, with albedos of 30%.

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17 years 10 months ago #16479 by Joe Keller
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If the proper motion were computed without regard to the dates of the plates on which an object actually appears, but rather the dates of all plates covering that region of sky, it still could be accurate for real stars. The planet and moon, appearing in one position on only one pair of plates a year apart, will have their proper motion underestimated by a factor of roughly (2000-1940)/2=30; that is, the typical 15"/3=5" separation will be reported as 5000/30=170mas/yr proper motion, typically 120 pmRA and 120 pmDec.

Suppose Barbarossa's moons, like Jupiter's moons, orbit in Barbarossa's equatorial plane. Let this plane be tilted 5 degrees from Barbarossa's orbital plane. Typically then we would see the moons describe ellipses tilted 3 degrees, implying 7" vertical error; the horizontal error was estimated at 5" in the previous paragraph. In absolute value, the mean putative PM in Decl then would be 1.4x the mean putative PM in RA; for the five objects, the observed ratio is 1.5.

It's likelier that the moon's orbit is slightly larger than optimal for mutual impersonation, than slightly smaller. So, usually the putative proper motion will be rearward, i.e., positive in RA (because there is reason to believe Barbarossa's orbit is retrograde, like the motion of the CMB dipole). Barbarossa's orbit line is inclined -26 degrees to the RA & Decl axes. The putative PMs would tend to be 5 units down and right along the orbit line, then perpendicularly 7 units up & right (if Barbarossa's equator is so inclined); indeed 4 of the 5 PM's (128,426; 274,394; 308,234; 302,142; but not -120,-502) lie in the first quadrant clustered about this point. The aberrant PM (Object #5) could be accidental; the moon's orbit isn't edge-on. Likewise a second moon accidentally could cause the PM of Object #2.

The earliest epoch of any object is given as 1965. This implies that the plates used for this region span about (2003-1965) x 2 = 76 yrs. The distance between Objects #5 & #4, corresponds theoretically to 69 years of orbit.

When I searched for Object #5, I extrapolated (westward only) one degree from Object #1 (using the line through Objects #2 & #3, because of Object #1's suspiciously different magnitudes). I narrowed the dim red search window 7-fold. There had been 61 candidate objects found in 4 sq degrees, so I expected only 61/4/7 = 2 per sq deg. Yet I found Object #5 within 5" of the predicted orbital great circle (p=0.006).

The RAs of the objects show periodicity analogous to the Saros eclipse cycle. Objects #5, #1 and #3 (all thought to represent the moon Freya) are on even minutes of RA. Object #2 (thought to represent the moon Frey) is near an odd minute of RA. Object #4 (representing Freya, with exactly the same magnitude as one of the other Freya sightings) is on a half minute of RA.

It's mere chance (1 in 3) that Frey's sighting occurred ten seconds away from a minute of RA (as measured from the Freya sightings to either side). The other four Object sightings (Freya) occur at intervals which are approximately multiples (1x, 3x, & 3x) of 115 seconds of RA; this corresponds to 9.8 years, considering the 26 degree slope of Barbarossa's orbital line. Freya displays "beats", because plates of the constellation Leo tend to be made at the same season, if not the same date.

If Barbarossa has period 4400 yr but present angular speed 2/3 of average, that's approximately 270*sqrt(1.5) = 330 AU. To give the gravitational effect I theorize on the CMB dipole, Barbarossa would be 0.016 solar masses. If Freya were in a circular orbit of 0.31 AU radius (just big enough to impersonate Barbarossa, next year or last year) Freya's year would be 1.36 Earth years. Really, Freya's orbit probably is slightly bigger. The period depends strongly (i.e., ^1.5) on radius. Freya might travel 0.5 *(1 + 1/9.8) orbits in one Earth yr., an orbit of 1.82 yrs., which, if circular, would have 0.37 AU radius, but which could have a radius only slightly more than 0.31 AU at greatest elongation on one side or the other, if moderately elliptical. Approximately every 9.8 yr, Freya would be roughly synchronized with Earth so that Freya is near maximum elongation when Leo is high in the night sky.

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17 years 10 months ago #16395 by Stoat
Replied by Stoat on topic Reply from Robert Turner
And I thought i was being a bit controversial with the "planet stoat." [:)] Barbarossa, was a crusader, so that might raise a frown from Islam. Barbarossa was the code name of the Nazi invasion of Russsia, so I think you might be stepping on a few toes there as well.

What happens is that you would get the credit and the first suggestion as to what to call the thing but it will be decided by an international astronomy body. I think it will end up with a name from a mythology other than the Eurocentric.

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