Requiem for Relativity

Freya's Orbit

"...Orbits rotate within the plane in the same sense as satellite motion when inclination is less than ~63.4 degrees, and in the opposite sense at higher inclinations. Edit: In exact terms, the cutoff occurs when inclination = arcsin(sqrt(4/5))."

- Grant Hutchison, Senior Member, "Bad Astronomy" forum, Feb. 27, 2008

[This famous fact appears, inter alia, on the first page of the first article published in Celestial Mechanics vol. 1, no. 1, p. 6, June 1969. It is important in placing high-latitude communications satellites. - JK]


Frey's retrograde apsis precession would imply that Freya's orbit is roughly perpendicular to Frey's, thus roughly perpendicular to Earth's ecliptic also. This improves the chance that Freya's orbit is not only perpendicular to Earth's ecliptic, but also seen edge-on. This displacement of Freya, perpendicular to the ecliptic, could explain the displacement of the centers of mass of CC9, E3E4, and D3D.

The maximum displacement of the Barbarossa-Frey c.o.m. was 190" (1997). Because in the previous post, everything seemed consistent with Freya having 1/9 of the total mass, this implies Freya's displacement was about 8 * 190" = 25.3' = 0.4222 deg, which gives a lower bound for the apoapsis of Freya's (highly eccentric, if Frey's varying apsis precession rate is to be explained) orbit about the c.o.m.; the corresponding lower bound for the apoapsis from Freya to Barbarossa (or more accurately to the Barbarossa-Frey c.o.m.), is about 0.4222 * 9/8 = 0.4750 deg. Frey's apoapsis finally was estimated (midrange) 0.243 * 0.435/0.50 = 0.211 deg. If these moons have equal eccentricity, then their period ratio is greater than (0.4750/0.211)^1.5 * sqrt(7/(because of Frey's pull on Freya) = 3.16.

Without contradicting other calculations, the discepancy between 3.16 and 2, is difficult to explain completely, simply by increasing Freya's mass and eccentricity. Recall that I originally calculated the mass ratio Barbarossa:Frey by requiring that the c.o.m. path around the sun, A2A-B3B-JGSR, be straight. If the real mass ratio were known, it would be found that the c.o.m. B3B is deflected laterally also, as are CC9 (until now, said to have -1 quantum of deflection), D3D ( -2 quanta) and E3E4 ( +1 quantum). The true deflections are best estimated as +0.5, -0.5, -1.5, and +1.5 quanta, resp., so they have mean zero. Here D3D's deflection becomes -1.5 quanta instead of -2, so the lower bound of the Freya:Frey period ratio becomes 3.16*(1.5/2)^1.5 = 2.05 ~ 2.

If Freya's eccentricity is 0.65 (i.e., the same as Frey's) then the fast half of its orbit, giving displacement of Freya, equal to the width of its ellipse, i.e. 1.50*a, takes 6.46 * 29.3% = 1.89 yr. The difference in lateral displacement of Barbarossa's orbital path, between B(1986) and C(1987) requires only 0.5/1.5*2 = 2/3 of the maximum possible single displacement of Freya, 1.65*a; this 1.10*a can be achieved in 0.88 yr, if Freya is near periapsis.

On the other hand, 2.03 yr might be enough for Freya's displacement by 1.50*a, but the displacement can't be more than about 2*a (because Freya's orbit doesn't precess greatly in 2 yr); the difference in lateral displacement between E(1995) and D(1997) would seem to require 1.65*a*2 = 3.3*a ("a" denotes the semimajor axis). Furthermore, E4D occurs 9.5 yr ( = 1.5 Freya orbits) later than BC9, so would be near Freya's apoapsis and slowest motion; and, Freya's direction would be wrong (D3D is beneath E3E4; CC9 is also beneath B3B).

A possible amendment is to let most of the displacement be caused by a third, outer, moon, "Lowell", near Freya's orbital plane. Freya could help the displacement E4D by an amount equivalent to 0.9*a, and perhaps make no contribution to the displacement BC9. Then "Lowell" could be in a 3:2 resonant (T=9.5 yr) orbit with the same eccentricity as Freya, near periapsis in both 1986.5 and 1996, with as little as half Freya's mass.

Hsuan & Mardling, Astrophys Space Sci 304:243246, 2006, report their computation, that a 0.00096 solar-mass planet (Jupiter mass), in a near-perpendicular orbit with semimajor axis 0.80 AU, would cause 0.01 deg/yr retrograde apsis precession of the DI Herculis binary star system (semimajor axis 0.50 AU, e = 0.2), yet negligible change in the binary's orbital plane. This binary has 9.67 solar mass total and a mass ratio nearly 1:1.

Barbarossa + Frey have 1/1000 this mass; their mass ratio is 7:1; their eccentricity seems to be 0.65, rather than 0.2; but their semimajor axis was estimated roughly 0.14675deg * pi/180 * 196.8 AU = 0.50 AU, the same as DI Herculis. For 2:1 orbital resonance, Freya's semimajor axis would be 0.50 * ( 2 * sqrt(8/7))^(2/3) = 0.83 AU, close enough to Hsuan & Mardling's hypothetical 0.80. Then I refined my estimate: all distances were divided by 1.232 (which would speed the precession by 1.232^1.5). Freya likely has 1/8 the combined mass of Barbarossa & Frey. If the rate of Hsuan & Mardling's computed apsis precession, is proportional to the mass of the third body, then Freya should cause 0.01 * 9.67/0.00096 / 8 * 1.232^1.5 = 17.2 deg/yr (vs. 16.4 observed).

Deleted; due to idiocy on my part [][][]

Oh! What was I thinking [] If it's on the second one, it won't be on the first one at all.

The first image is the newest one. The planet cannot be on both plates as they are only about 18 seconds across. The only "star" that I've found; just a quick scan, there might be more; that's on the first image but not the second, is the only green dot. Bottom left quadrant.

Stoat, is the green dot the one just left of the two bright stars of about equal brightness in the lower left corner of the picture? Is this possibly Barbarossa then? How could such a massive object be so dim?