- Thank you received: 0
Requiem for Relativity
- Joe Keller
- Offline
- Platinum Member
Less
More
17 years 7 months ago #16627
by Joe Keller
Replied by Joe Keller on topic Reply from
JL Brady determined the mass of "Planet X" (a different "Planet X" revealed by ephemeris discrepancies of Halley's, and other comets) as a multiple of Pluto's mass (Brady, Pub. Astron. Soc. Pacific 84:314+, 1972, Sec. II). Brady's Fourier analysis showed, an influence with the period of Pluto, and another influence 300x bigger with a period corresponding to a circular orbit at 65 AU. When Brady wrote, Pluto was thought to have 0.10 Earth mass (Baker's "Astronomy", 9th ed., 1971). Noting that the influence also should be proportional to the inverse square of the major axis, Brady estimated, in Sec. II, that Planet X should be 1000x heavier than Pluto, i.e., 100 Earth masses according to the presumed mass of Pluto at that time.
Also, the mass should be proportional to the cosecant of the orbital inclination (which Brady eventually estimated at 120 deg, vs. 17 deg for Pluto). This would change Brady's Sec. II estimate, to 300x heavier than Pluto (keeping the same number of significant digits as Brady). The mass of Pluto has been revised drastically downward, to 0.0020 Earth masses (Astronomical Almanac, 2004, p. E88; 2007 ed. is the same). So, Brady's Sec. II estimate becomes 300*0.002 = 0.6 Earth masses, not 100.
In later sections of his article, Brady made computer integrations of gravitational effects, and found 300 Earth masses for Planet X, implying 1 Earth mass for Pluto. The resolution of this paradox, is that the perturbations of Halley's comet, though proportional to the masses of Pluto and Planet X, are not due to gravity "as we know it" (a phrase I borrow from actor DeForest Kelly). The perturbation is 500x that due to gravity, perhaps through the mediation of phenomena at the 52.6 AU barrier, to which Pluto (inside) and Brady's Planet X (outside) make comparably close approaches.
If Brady's Planet X were as heavy as Jupiter, it would have sqrt(65/5.2) = 3.5x Jupiter's angular momentum and would "wag the dog" (the phrase originally appeared in Von Arnum's "Frauelein Schmidt", 1907, was used by F. Scott Fitzgerald in 1935, & was the title of a 1997 movie) of orbital precession in the solar system (Goldreich & Ward, PASP 84:737+). At 1/500 Jupiter's mass, it has 0.007x Jupiter's angular momentum. Not only is its force on the other planets 500x less (which by itself would increase the disruption time from Goldreich & Ward's 1 million yr, to 500 million yr) but it is Planet X which precesses fastest, thereby canceling by averaging, what little disruptive effect Planet X has.
Though Brady predicted magnitude +13 to 14, Klemola & Harlan (PASP 84:736) found nothing in Brady's predicted region, down to mag +15 or, mostly, +17 or 18 (sensitivity varied). The actual 500x smaller mass of Brady's Planet X equates to 4.5 magnitude units, i.e., +17.5 to 18.5. The difference between Pluto's 30% albedo (Astronomical Almanac, 2004) and the canonical Kuiper belt albedo of 4-8%, gives another 1.4 to 2.2 mags, for +18.9 to +20.7. Thus Klemola & Harlan's observational failure also is obviated. (Brady checked that his Planet X was outside Tombaugh's search, i.e. the Lowell Proper Motion Survey, which was down only to +16 or 17 anyway.)
A 500x smaller mass for Brady's Planet X reduces the hypothetical sum of squared residuals due to its effect on the known outer planets, 250,000x. This thoroughly obviates the findings of Seidelmann et al, PASP 84:858.
According to Brady's orbit (Brady, op. cit., Sec. IV last par., & Fig. 5) Planet X now would be near ecliptic longitude 0, ecliptic latitude +52. Using its 0.6 Earth mass, which the modern estimate of Pluto's mass implies, and its "x1" = 65/52.6 = 1.24 (vs. "x2" = 191/52.6 = 3.63 for Barbarossa) Brady's Planet X moves the CMB dipole
0.6/(0.0068*333,000) * f(x1)/f(x2) radian * sin52 = 0.58 degree north,
where f(x)= -2*(1- 3*x^2)/(1-x)^2/(1+x)^2
is the Hermite one-step approximation, for x > 1 (and not too near 1), for an integral arising in my calculation of the CMB dipole caused by a massive body. (The approximation for 0 < x < 1, x not too near 1, is
f(x) = -4*x^3/(1-x)^2/(1+x)^2 .)
Together with the contribution of Neptune ( +0.0243 * 11.8 deg * sin(311-173); conveniently, Neptune was at its descending node) and Uranus ( +0.00323 * 11.8 deg * sin(329-173) ) (the contributions of Jupiter & Saturn are small), this explains 0.58 + 0.21 + 0.03 = 0.82deg of discrepancy between Barbarossa's and the CMB dipole's ecliptic latitudes (observed discrepancy: 0.7deg).
Also, a negligibly larger predicted mass for Barbarossa, results from Brady's Planet X. The rational expression f shows that the induced CMB dipole decreases as inverse distance squared, rendering interstellar effects negligible.
Also, the mass should be proportional to the cosecant of the orbital inclination (which Brady eventually estimated at 120 deg, vs. 17 deg for Pluto). This would change Brady's Sec. II estimate, to 300x heavier than Pluto (keeping the same number of significant digits as Brady). The mass of Pluto has been revised drastically downward, to 0.0020 Earth masses (Astronomical Almanac, 2004, p. E88; 2007 ed. is the same). So, Brady's Sec. II estimate becomes 300*0.002 = 0.6 Earth masses, not 100.
In later sections of his article, Brady made computer integrations of gravitational effects, and found 300 Earth masses for Planet X, implying 1 Earth mass for Pluto. The resolution of this paradox, is that the perturbations of Halley's comet, though proportional to the masses of Pluto and Planet X, are not due to gravity "as we know it" (a phrase I borrow from actor DeForest Kelly). The perturbation is 500x that due to gravity, perhaps through the mediation of phenomena at the 52.6 AU barrier, to which Pluto (inside) and Brady's Planet X (outside) make comparably close approaches.
If Brady's Planet X were as heavy as Jupiter, it would have sqrt(65/5.2) = 3.5x Jupiter's angular momentum and would "wag the dog" (the phrase originally appeared in Von Arnum's "Frauelein Schmidt", 1907, was used by F. Scott Fitzgerald in 1935, & was the title of a 1997 movie) of orbital precession in the solar system (Goldreich & Ward, PASP 84:737+). At 1/500 Jupiter's mass, it has 0.007x Jupiter's angular momentum. Not only is its force on the other planets 500x less (which by itself would increase the disruption time from Goldreich & Ward's 1 million yr, to 500 million yr) but it is Planet X which precesses fastest, thereby canceling by averaging, what little disruptive effect Planet X has.
Though Brady predicted magnitude +13 to 14, Klemola & Harlan (PASP 84:736) found nothing in Brady's predicted region, down to mag +15 or, mostly, +17 or 18 (sensitivity varied). The actual 500x smaller mass of Brady's Planet X equates to 4.5 magnitude units, i.e., +17.5 to 18.5. The difference between Pluto's 30% albedo (Astronomical Almanac, 2004) and the canonical Kuiper belt albedo of 4-8%, gives another 1.4 to 2.2 mags, for +18.9 to +20.7. Thus Klemola & Harlan's observational failure also is obviated. (Brady checked that his Planet X was outside Tombaugh's search, i.e. the Lowell Proper Motion Survey, which was down only to +16 or 17 anyway.)
A 500x smaller mass for Brady's Planet X reduces the hypothetical sum of squared residuals due to its effect on the known outer planets, 250,000x. This thoroughly obviates the findings of Seidelmann et al, PASP 84:858.
According to Brady's orbit (Brady, op. cit., Sec. IV last par., & Fig. 5) Planet X now would be near ecliptic longitude 0, ecliptic latitude +52. Using its 0.6 Earth mass, which the modern estimate of Pluto's mass implies, and its "x1" = 65/52.6 = 1.24 (vs. "x2" = 191/52.6 = 3.63 for Barbarossa) Brady's Planet X moves the CMB dipole
0.6/(0.0068*333,000) * f(x1)/f(x2) radian * sin52 = 0.58 degree north,
where f(x)= -2*(1- 3*x^2)/(1-x)^2/(1+x)^2
is the Hermite one-step approximation, for x > 1 (and not too near 1), for an integral arising in my calculation of the CMB dipole caused by a massive body. (The approximation for 0 < x < 1, x not too near 1, is
f(x) = -4*x^3/(1-x)^2/(1+x)^2 .)
Together with the contribution of Neptune ( +0.0243 * 11.8 deg * sin(311-173); conveniently, Neptune was at its descending node) and Uranus ( +0.00323 * 11.8 deg * sin(329-173) ) (the contributions of Jupiter & Saturn are small), this explains 0.58 + 0.21 + 0.03 = 0.82deg of discrepancy between Barbarossa's and the CMB dipole's ecliptic latitudes (observed discrepancy: 0.7deg).
Also, a negligibly larger predicted mass for Barbarossa, results from Brady's Planet X. The rational expression f shows that the induced CMB dipole decreases as inverse distance squared, rendering interstellar effects negligible.
Please Log in or Create an account to join the conversation.
17 years 7 months ago #16628
by Stoat
Replied by Stoat on topic Reply from Robert Turner
Oops!
Please Log in or Create an account to join the conversation.
17 years 7 months ago #16629
by Stoat
Replied by Stoat on topic Reply from Robert Turner
Please Log in or Create an account to join the conversation.
- Joe Keller
- Offline
- Platinum Member
Less
More
- Thank you received: 0
17 years 7 months ago #16682
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 />
<br />
<i>Originally posted by Stoat</i>
<br />Here's the nem3 image, which is at RA 11 07 24 Dec -6 38 50
<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
This is the previous version of this coordinate shot, turned upside down. I'm not laughing: this is only the third photo in the world ever made in the search for Barbarossa, and things will go wrong. The important thing is that it's corrected. I still can't correlate it with the Millenium catalog, however. Part of the reason might be that the Millenium catalog typically shows only one or two stars per Bradford frame in this region.
Please Log in or Create an account to join the conversation.
17 years 7 months ago #16633
by Stoat
Replied by Stoat on topic Reply from Robert Turner
It doesn't matter. When we get two shots the same from the Bradford, we can animate them to look for movement.
We look at upside down pictures of the moon without any problems [8D]
We look at upside down pictures of the moon without any problems [8D]
Please Log in or Create an account to join the conversation.
- Joe Keller
- Offline
- Platinum Member
Less
More
- Thank you received: 0
17 years 7 months ago #16634
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 />It doesn't matter. When we get two shots the same from the Bradford, we can animate them to look for movement.
We look at upside down pictures of the moon without any problems [8D]
<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
Good point! Carry on!
<br />It doesn't matter. When we get two shots the same from the Bradford, we can animate them to look for movement.
We look at upside down pictures of the moon without any problems [8D]
<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
Good point! Carry on!
Please Log in or Create an account to join the conversation.
Time to create page: 0.527 seconds