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16 years 8 months ago #20717
by Joe Keller
Replied by Joe Keller on topic Reply from
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by Stoat</i>
<br />Hi Joe, I was going to ask TVF what he made of the anomalous findings of helium in the atmosphere of Jupiter. Which suggests that the planets formed very early. I forgot all about it but it would be of possible importance when considering our solar system as something of a "failed" binary system.
<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
Good idea! Any evidence that the sun is borderline, by some criterion, for qualifying to be a member of a binary system, would improve the odds of a small dim companion.
<br />Hi Joe, I was going to ask TVF what he made of the anomalous findings of helium in the atmosphere of Jupiter. Which suggests that the planets formed very early. I forgot all about it but it would be of possible importance when considering our solar system as something of a "failed" binary system.
<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
Good idea! Any evidence that the sun is borderline, by some criterion, for qualifying to be a member of a binary system, would improve the odds of a small dim companion.
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16 years 8 months ago #20718
by Joe Keller
Replied by Joe Keller on topic Reply from
<i>Originally posted by Stoat</i>
<br />Hi Joe, time maybe for another try at a telescope shot.Here's a screen shot of the ones on the Bradford robotic, Pick out the best position one and I'll put it up again as a job. ...
R44042 ("nem4") is good.
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16 years 8 months ago #18174
by Joe Keller
Replied by Joe Keller on topic Reply from
From the data and calculations I have now, I can give the heliocentric celestial coordinates (in degrees) of the center of mass of the Barbarossa-Frey system:
RA = 171.6607 + T * 0.11676 + T^2 * 0.000124 + T^3 * 0.0000015
Decl = -9.0343 - T * 0.06173 - T^2 * 0.000174 - T^3 * 0.0000025
T is the time in years since Joan Genebriera's photograph of Barbarossa, taken 2007.2273 UT (i.e., Greenwich time).
From heliocentric coordinates and the current date, geocentric coordinates (i.e., apparent coordinates, if atmospheric refraction is included) can be calculated. For 00:00 UT March 1, 2008, I calculate the geocentric coordinates (not including atmospheric refraction) of the Barbarossa-Frey center of mass to be
RA 171.8242 deg
Decl -9.1146 deg
At this season, the daily apparent motion of the c.o.m. is
RA -0.0043 deg
Decl +0.00185 deg
The geocentric coordinates I gave above, for March 1, can be corrected by this amount, accurately for the next few weeks while Barbarossa is near opposition.
I think Frey will be about 0.15 deg NE of Barbarossa for the next few months. I think the mass ratio Barbarossa:Frey is about 8771:1229. They have a nearly circular 1.20 AU mutual orbit, with 14.7 yr period, tilted 40 deg to our ecliptic plane, with long apparent axis tilted 14.5 deg clockwise from the celestial equator. Their center of mass is about 197.73 AU from our sun, in a nearly circular orbit around our sun.
The total mass of Barbarossa + Frey is 0.00804 solar masses, and the total Barbarossa system is 0.0103 solar masses.
I base this on data previously discussed:
1. "Disappearing dots" A2 & A (listing the presumed Barbarossa first, then Frey) on the 1954 Palomar sky survey and
dots B3 & B on the 1986 UK-Australian sky survey (online scans).
2. Joan Genebriera's March 25, 2007 photo (Barbarossa) and
Steve Riley's April 1, 2007 photo (Frey). Genebriera used a 16" telescope on Tenerife and Riley an 8" (!) in southern California. Both used electronic detection with stacking.
3. I know of only one prospective (vs. a retrospective finding on a sky survey) photo of Barbarossa (Genebriera's), but I know of as many as four prospective photos of Frey. Genebriera photographed Frey on April 2, 2007. Robert Turner (operating, from England, a 14" robotic telescope also on Tenerife) photographed Frey on April 12, 2007. Riley made his second photograph of Frey on April 24, 2007. All of these photos are near the limit of detection and, like many stars near the limit, are not typical pixel patterns. Genebriera's Frey photo resembles a heart-shaped pixel overdensity which is not seen on another seemingly equally good photo she took only a few minutes later. I recorded Turner's Frey photo in my notes as a "barely significant pixel overdensity". Yet these four detections interpolate perfectly. Furthermore their motion is consistent not only with the presumed distance to the center of mass of Barbarossa-Frey, but also with the presumed orbit of Frey around Barbarossa.
4. Dot D (presumably Barbarossa) on the 1997 optical infrared sky survey. Though I've found no Frey on this online scanned plate, when Frey was added to the calculation at its presumed position according to the motion seen on the 1954 & 1986 sky surveys and the Genebriera, Riley, & Turner photos, the center-of-mass position of this infrared object was consistent.
5. Dots C3 & C11 on the 1987 LaSilla sky survey (online scan) have relative position consistent with the Barbarossa-Frey mutual orbit, but their c.o.m. is displaced approx. RA +0.12 deg Decl +0.025 deg, so, they were not used in calculating any orbit. The first dot I ever found, C, is between C3 & C11. Although a very elliptical orbit spends only ~1/6 of its time on the nearer half of the curve, still there probably would be comparably large deviations in the A, B, D, or Genebriera-Riley records if C is another massive Barbarossa moon. Also, C, unlike really any of the other dots, is a streak of length compatible with a Kuiper Belt Object. About half the discrepancy of C3&C11 could be removed if the distance to Barbarossa were much greater, so parallax could be recalculated, but, then Barbarossa would far exceed escape speed from the sun.
RA = 171.6607 + T * 0.11676 + T^2 * 0.000124 + T^3 * 0.0000015
Decl = -9.0343 - T * 0.06173 - T^2 * 0.000174 - T^3 * 0.0000025
T is the time in years since Joan Genebriera's photograph of Barbarossa, taken 2007.2273 UT (i.e., Greenwich time).
From heliocentric coordinates and the current date, geocentric coordinates (i.e., apparent coordinates, if atmospheric refraction is included) can be calculated. For 00:00 UT March 1, 2008, I calculate the geocentric coordinates (not including atmospheric refraction) of the Barbarossa-Frey center of mass to be
RA 171.8242 deg
Decl -9.1146 deg
At this season, the daily apparent motion of the c.o.m. is
RA -0.0043 deg
Decl +0.00185 deg
The geocentric coordinates I gave above, for March 1, can be corrected by this amount, accurately for the next few weeks while Barbarossa is near opposition.
I think Frey will be about 0.15 deg NE of Barbarossa for the next few months. I think the mass ratio Barbarossa:Frey is about 8771:1229. They have a nearly circular 1.20 AU mutual orbit, with 14.7 yr period, tilted 40 deg to our ecliptic plane, with long apparent axis tilted 14.5 deg clockwise from the celestial equator. Their center of mass is about 197.73 AU from our sun, in a nearly circular orbit around our sun.
The total mass of Barbarossa + Frey is 0.00804 solar masses, and the total Barbarossa system is 0.0103 solar masses.
I base this on data previously discussed:
1. "Disappearing dots" A2 & A (listing the presumed Barbarossa first, then Frey) on the 1954 Palomar sky survey and
dots B3 & B on the 1986 UK-Australian sky survey (online scans).
2. Joan Genebriera's March 25, 2007 photo (Barbarossa) and
Steve Riley's April 1, 2007 photo (Frey). Genebriera used a 16" telescope on Tenerife and Riley an 8" (!) in southern California. Both used electronic detection with stacking.
3. I know of only one prospective (vs. a retrospective finding on a sky survey) photo of Barbarossa (Genebriera's), but I know of as many as four prospective photos of Frey. Genebriera photographed Frey on April 2, 2007. Robert Turner (operating, from England, a 14" robotic telescope also on Tenerife) photographed Frey on April 12, 2007. Riley made his second photograph of Frey on April 24, 2007. All of these photos are near the limit of detection and, like many stars near the limit, are not typical pixel patterns. Genebriera's Frey photo resembles a heart-shaped pixel overdensity which is not seen on another seemingly equally good photo she took only a few minutes later. I recorded Turner's Frey photo in my notes as a "barely significant pixel overdensity". Yet these four detections interpolate perfectly. Furthermore their motion is consistent not only with the presumed distance to the center of mass of Barbarossa-Frey, but also with the presumed orbit of Frey around Barbarossa.
4. Dot D (presumably Barbarossa) on the 1997 optical infrared sky survey. Though I've found no Frey on this online scanned plate, when Frey was added to the calculation at its presumed position according to the motion seen on the 1954 & 1986 sky surveys and the Genebriera, Riley, & Turner photos, the center-of-mass position of this infrared object was consistent.
5. Dots C3 & C11 on the 1987 LaSilla sky survey (online scan) have relative position consistent with the Barbarossa-Frey mutual orbit, but their c.o.m. is displaced approx. RA +0.12 deg Decl +0.025 deg, so, they were not used in calculating any orbit. The first dot I ever found, C, is between C3 & C11. Although a very elliptical orbit spends only ~1/6 of its time on the nearer half of the curve, still there probably would be comparably large deviations in the A, B, D, or Genebriera-Riley records if C is another massive Barbarossa moon. Also, C, unlike really any of the other dots, is a streak of length compatible with a Kuiper Belt Object. About half the discrepancy of C3&C11 could be removed if the distance to Barbarossa were much greater, so parallax could be recalculated, but, then Barbarossa would far exceed escape speed from the sun.
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16 years 8 months ago #14303
by Stoat
Replied by Stoat on topic Reply from Robert Turner
Here's an article on the helium in Jupiter's atmosphere.
www.space.com/scienceastronomy/solarsyst...elements_991117.html
I've posted the job on the Bradford. Might take a while as the weather isn't that good at the moment I might be an idea to see if you can get another from your two contacts, they would expect to do another plate after this time, to do a blink comparison.
I've posted the job on the Bradford. Might take a while as the weather isn't that good at the moment I might be an idea to see if you can get another from your two contacts, they would expect to do another plate after this time, to do a blink comparison.
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16 years 8 months ago #14529
by Joe Keller
Replied by Joe Keller on topic Reply from
Last year, I adjusted the mass ratio Barbarossa:Frey so that the three centers of mass A2&A, B3&B, and JG&SR (Genebriera's & Riley's objects) would lie on a great circle (which then happened also to be of constant angular speed, to about 1 arcsec accuracy). The quadratic interpolant through A2&A, B3&B, and JG&SR, for the heliocentric celestial coordinates of the c.o.m, in degrees, is
RA = 171.6607 + T * 0.115124 + T^2 * 0.0000159
Decl = -9.0343 - T * 0.058890 + T^2 * 0.0000148
(Again, T is the time in years since Genebriera's photo of Barbarossa, taken 2007.2273 UT.) This is practically equal to the exact constant-speed great circle, from which, the deviation of C3&C11, RA +0.1258, Decl +0.0339, is about the same as its deviation from the cubic interpolant (see previous post).
The cubic interpolant given in the previous post, was chosen to pass through an additional point, the presumed c.o.m. when Barbarossa is the 1997 optical infrared sky survey point D, and Frey is at the estimated position based on the three observed Frey positions of 1954, 1986 & 2007. The 4-pt. cubic and the great-circle interpolants necessarily agree perfectly in 1954, 1986 & 2007. In 1997, D (that is, the estimated c.o.m. with D as Barbarossa) deviates from the constant-speed great circle through A2&A, B3&B, & JG&SR, by RA -0.0068 deg, Decl +0.0123 deg.: a distance of 0.014 deg = 50 arcsec = 0.048 AU = 4.5 million miles, at Barbarossa's distance from the sun.
RA = 171.6607 + T * 0.115124 + T^2 * 0.0000159
Decl = -9.0343 - T * 0.058890 + T^2 * 0.0000148
(Again, T is the time in years since Genebriera's photo of Barbarossa, taken 2007.2273 UT.) This is practically equal to the exact constant-speed great circle, from which, the deviation of C3&C11, RA +0.1258, Decl +0.0339, is about the same as its deviation from the cubic interpolant (see previous post).
The cubic interpolant given in the previous post, was chosen to pass through an additional point, the presumed c.o.m. when Barbarossa is the 1997 optical infrared sky survey point D, and Frey is at the estimated position based on the three observed Frey positions of 1954, 1986 & 2007. The 4-pt. cubic and the great-circle interpolants necessarily agree perfectly in 1954, 1986 & 2007. In 1997, D (that is, the estimated c.o.m. with D as Barbarossa) deviates from the constant-speed great circle through A2&A, B3&B, & JG&SR, by RA -0.0068 deg, Decl +0.0123 deg.: a distance of 0.014 deg = 50 arcsec = 0.048 AU = 4.5 million miles, at Barbarossa's distance from the sun.
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16 years 8 months ago #18195
by Joe Keller
Replied by Joe Keller on topic Reply from
Barbarossa and Frey Now Identified on 1954, 1986, 1987, & 1997 (Optical IR) Sky Surveys, and 2007 Photos of Joan Genebriera, Steve Riley & Robert Turner
radius of Barbarossa's circular solar orbit: 197.7283 AU
mass ratio Barbarossa:Frey = 0.8771:0.1229
(The object designations A2, etc., are those used in my notes.)
observed geocentric celestial coordinates in Julian epoch 1954.1516 Palomar sky survey, online scan of plate:
Barbarossa A2 11 2 25.26
-5 56 11.3
Frey A 11 3 12.4
-5 58 9
heliocentric celestial coordinates
of center of mass A2/A 165.5836
-5.86275
geocentric, Julian epoch 1986.2015 UK/Australia sky survey
Barbarossa B3 11 16 51.55
-7 49 41.1
Frey B 11 16 56.07
-7 55 14.3
heliocentric, c.o.m. 169.2360
-7.7851
geocentric, Julian epoch 1987.08215 LaSilla sky survey
Barbarossa C 11 18 3.18
-7 58 46.1
Frey C9 11 17 29.15
-7 57 28
heliocentric, c.o.m. 169.3289
-7.8601
geocentric, Julian epoch 1997.1693 optical infrared sky survey
Barbarossa D3 11 22 1.33
-8 34 36.9
Frey D 11 22 16.77
-8 29 30.9
heliocentric, c.o.m. 170.4712
-8.48695
geocentric, first successful photos taken prospectively at coordinates given by me, Julian epoch 2007.226648 (JG, Tenerife) & Julian epoch 2007.246595 (SR, southern California) (I used weighted ave. of time)
Barbarossa JG 11 26 22.2
-9 4 59
Frey SR 11 26 25.0
-8 57 48.5
heliocentric, c.o.m. 171.6471
-9.0281
Above, I use epochs I calculate, from the (mid-exposure) times of the plates, by the formula and tables in the 1989 Astronomical Almanac. For the 1954 and 1997 plates, these epochs differ appreciably (more than mere rounding error) from the epochs given by some online sources; I think it is better to use my calculated epochs, which are based on the times reported on the DSS online "plate finder".
Last year I determined the mass ratio Barbarossa:Frey by requiring that the c.o.m. heliocentric path A2/A-B3/B-JG/SR be a great circle. Recalculating from scratch this week, with a first-order correction for Earth's orbital eccentricity, confirms my statement of last year, that this mass ratio also renders the c.o.m. angular speed constant. The discrepancy in location at B3/B is +11.5 arcsec along the path.
This morning I discovered that the same mass ratio applies to C/C9 and to D3/D. Using the same mass ratio as above, the heliocentric paths A2/A-C/C9-JG/SR and A2/A-D3/D-JG/SR are both straight (i.e., great circles), possibly within 1 arcsec, if adjusted for an unexplained quantized deviation of approx. +/- n * 90", perpendicular to the path. These c.o.m. angular speeds also are constant. The location discrepancy at C/C9 is +19.5 arcsec along the path, and at D3/D, +25 arcsec, assuming constant angular speed.
Three of the five known images of Barbarossa, C (the first sky survey object I found), D3, and JG, are elongated several arcsec along a direction angle 60 or 70 deg (clockwise from north), similar to Barbarossa's orbital inclination. This might indicate moons, rings, an accretion disk, dust, or a corona. These elongations make the true location uncertain enough to explain some of the location discrepancy.
From these five pairs of points, a consistent (apparent counterclockwise) elliptical orbit of Frey around Barbarossa likely can be calculated. The results of my very approximate calculation, assuming a circular but tilted orbit and using only the three pairs A2/A, B3/B, & JG/SR, were reported to this messageboard earlier this week.
radius of Barbarossa's circular solar orbit: 197.7283 AU
mass ratio Barbarossa:Frey = 0.8771:0.1229
(The object designations A2, etc., are those used in my notes.)
observed geocentric celestial coordinates in Julian epoch 1954.1516 Palomar sky survey, online scan of plate:
Barbarossa A2 11 2 25.26
-5 56 11.3
Frey A 11 3 12.4
-5 58 9
heliocentric celestial coordinates
of center of mass A2/A 165.5836
-5.86275
geocentric, Julian epoch 1986.2015 UK/Australia sky survey
Barbarossa B3 11 16 51.55
-7 49 41.1
Frey B 11 16 56.07
-7 55 14.3
heliocentric, c.o.m. 169.2360
-7.7851
geocentric, Julian epoch 1987.08215 LaSilla sky survey
Barbarossa C 11 18 3.18
-7 58 46.1
Frey C9 11 17 29.15
-7 57 28
heliocentric, c.o.m. 169.3289
-7.8601
geocentric, Julian epoch 1997.1693 optical infrared sky survey
Barbarossa D3 11 22 1.33
-8 34 36.9
Frey D 11 22 16.77
-8 29 30.9
heliocentric, c.o.m. 170.4712
-8.48695
geocentric, first successful photos taken prospectively at coordinates given by me, Julian epoch 2007.226648 (JG, Tenerife) & Julian epoch 2007.246595 (SR, southern California) (I used weighted ave. of time)
Barbarossa JG 11 26 22.2
-9 4 59
Frey SR 11 26 25.0
-8 57 48.5
heliocentric, c.o.m. 171.6471
-9.0281
Above, I use epochs I calculate, from the (mid-exposure) times of the plates, by the formula and tables in the 1989 Astronomical Almanac. For the 1954 and 1997 plates, these epochs differ appreciably (more than mere rounding error) from the epochs given by some online sources; I think it is better to use my calculated epochs, which are based on the times reported on the DSS online "plate finder".
Last year I determined the mass ratio Barbarossa:Frey by requiring that the c.o.m. heliocentric path A2/A-B3/B-JG/SR be a great circle. Recalculating from scratch this week, with a first-order correction for Earth's orbital eccentricity, confirms my statement of last year, that this mass ratio also renders the c.o.m. angular speed constant. The discrepancy in location at B3/B is +11.5 arcsec along the path.
This morning I discovered that the same mass ratio applies to C/C9 and to D3/D. Using the same mass ratio as above, the heliocentric paths A2/A-C/C9-JG/SR and A2/A-D3/D-JG/SR are both straight (i.e., great circles), possibly within 1 arcsec, if adjusted for an unexplained quantized deviation of approx. +/- n * 90", perpendicular to the path. These c.o.m. angular speeds also are constant. The location discrepancy at C/C9 is +19.5 arcsec along the path, and at D3/D, +25 arcsec, assuming constant angular speed.
Three of the five known images of Barbarossa, C (the first sky survey object I found), D3, and JG, are elongated several arcsec along a direction angle 60 or 70 deg (clockwise from north), similar to Barbarossa's orbital inclination. This might indicate moons, rings, an accretion disk, dust, or a corona. These elongations make the true location uncertain enough to explain some of the location discrepancy.
From these five pairs of points, a consistent (apparent counterclockwise) elliptical orbit of Frey around Barbarossa likely can be calculated. The results of my very approximate calculation, assuming a circular but tilted orbit and using only the three pairs A2/A, B3/B, & JG/SR, were reported to this messageboard earlier this week.
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