It's not. Do you agree that the article claims that the mass and separation of the companion is irrelevant, the precession will always follow the orbital period (they say that clear as day)? Then you must allow that Jupiter provides a very obvious counterexample, as the Sun is in a 12-year orbit due to Jupiter. Thus if the paper authors don't know what they are talking about, why would you accept any of their conclusions?

Even so, you would probably prefer a direct refutation. The problem with their claim is that they are making the same mistake a lot of introductory astronomy students make about the Earth's axis. They imagine that the axis is a rigid rod, connected to the Sun by another rigid rod. If you do that, then the axis precesses every year as the Earth orbits the Sun, and you don't get seasons. At least you get a result where the precession rate follows an orbit, but it's totally wrong. The axis points at Polaris all year long, it does not go around like a rigidly attached rod. The authors are apparently aware of this in the case of the Earth's orbit around the Sun (they know the precession period is 26,000 years, not 1 year), but they still make the same mistake when they imagine the Sun-Earth system orbiting around something else. They imagine that as the Sun orbits something else, the Earth's axis will be dragged around in a similar kind of rigid way. In fact, it's angular momentum will stay the same (if not for the lunisolar torque they ignore) and it will keep pointing at Polaris as it orbits the companion, just as it does as it orbits the Sun. Why should it be different with the companion?

You can also analyze the forces involved and easily show that the companion's gravity is much weaker than the Sun's unless the companion is about a thousand times more massive than the Sun. So it would have to be a very massive black hole to be responsible for the precession, which is a very different kind of claim. At this point, we're writing the paper for the bunglers who came up with the original analysis.

Thanks Ken, for the info.
And let me say that I don't consider BRI's evidence as flawless by any measure.

Additional "evidence:"Transits of Venus vs. NASA Astronomical Data.

__________________
"Where the telescope ends, the microscope begins. Which of the two has the greater view?" - Hugo

"Men occasionally stumble over the truth, but most pick themselves up and hurry off as if nothing had happened." - Churchill

Off topic...the sirius research group (which A.DIM linked to in his post) is anything but serious

Pardon my bad pun...

That's pretty technical stuff, so one must ask onesself if it is worth the effort to work it through and see if they are right. It sounds to me like they are also making the same mistake of thinking that if the Sun orbits something else, the axis of the Earth will magically be twisted around in that orbit, like if you made a rigid model and just rotated it. This is not how physics works, so they would need their own physics to do this. I hope they are ready to build a bridge with it.

I admit most of the info in that article is beyond me, but I'm not prepared to dismiss the possibility of a companion as the cause of precession.

It appears the lunisolar explanation raises more questions than it answers.

__________________
"Where the telescope ends, the microscope begins. Which of the two has the greater view?" - Hugo

"Men occasionally stumble over the truth, but most pick themselves up and hurry off as if nothing had happened." - Churchill

I did wade through some of the Sirius Research Institute stuff, and responded to their article which had been submitted to the online Journal of Theoretics. It's the second letter, and the journal included their reply, but I don't think they really address the points of the letter. O well.

And these questions are?

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How can people who don't know any physics get astronomy funding, perhaps?

You deserve a medal for trying. It seems they are making great hay over a few recent measurements of transits, and are claiming that the tropical year is the same as the sidereal year. I have no idea what that hullaballoo about the 1900 tropical year is, but it does seem pretty clear to me that if the tropical year equals the sidereal year, then the Earth's axis must always point at Polaris, and the zodiacal constellations have to always be in the Sun at the same time of year. So it makes me wonder how they explain the steady march of zodiacal constellations documented over the last several thousand years since the astrological horoscope was developed. Are they trying to unlearn something that has been known since the days of ancient Greece? This is progress?

yes, and I get the impression that it is a backyard setup. The transits are not measured at the meridian, so that makes it harder--maybe some trees were in the way...

The Precession Dialogues--Part One

DB: "What's the dialogue about this time?"

CM: "This time it's about the precession of equinoxes. There was a question earlier about whether or not the equatorial bulge was the cause of the precession of equinoxes. Several answers were given, and then discussion turned to the theory that the precession is caused by a distant as-yet undetected companion of the Sun that orbits every 26,000 years."

BH: "Oh, not that nonsense again!"

CM: "'Fraid so! I don't see what the attraction (pardon the pun) of this theory is, given that the cause of the precession is nearby, easily observed, understood and calculated in its broadest outlines. Of course these various Nemesis-like theories are better (though not much better) than the theory that holds that the Solar System is orbiting around the star Sirius every 26,000 years."

DB: "Now that's just silly. We couldn't even get to Sirius in that much time, much less orbit it in 26 millennia!"

CM: "Well, actually we could if we could travel at sufficient velocity. Eight light-years divided by 26,000 years is about 1/3,000 of the speed of light, which would be 100 km/s. It would take a lot of fuel, but we might be able to do it. Of course the time is another matter. As for orbiting Sirius, that would imply a velocity 2 times pi times this velocity, or 630 km/s. Could the Solar System really be orbiting Sirius at that velocity? If it were, we would observe Sirius making a complete circuit of the sky every 26,000 years, or moving at about 50 seconds of arc annually. We simply do not observe Sirius moving at such a rate against the background stars. Sorry."

BH: "So what is moving at 50 arc-seconds per year then?"

CM: "The axis of the Earth's rotation precesses clockwise, causing the plane of the equator to shift and its intersection with the ecliptic to shift 50 arc-seconds per year west along the ecliptic. Notice that this is defined only in terms of the ecliptic and the equator; we need to look at the Earth since that is where the equator is defined, after all, and to the nearby Solar System since the nearest and most massive objects are found in the near Solar System.

CM: "But first, let me draw your attention to this BAUTForum thread from October, Explaining precession simply but accurately. The thread was started by grant hutchison and contributed to by Ken G. It is a true dialog, not some cheesy literary device like this one!"

CM: "In fact, why don't we take a break while the readers have a look at that thread. Let's refill our coffee cups!"

To be continued...

__________________
Microsoft is over if you want it.

The bar has been lowered for the promotion of ATM ideas; the bar for the acceptance of ATM ideas must remain high.

The Precession Dialogues--Part Two

Our coffee cups refilled, we resumed the dialogue.

CM: "I think it is a wonderful thread and a very good example of what can be done with a qualitative argument. We will now try to use this type of argument to deduce the direction of the precession of the equinoxes.

CM: "We start by noting that only the equatorial bulge matters. A perfectly spherical Earth with a spherically symmetric mass distribution would behave as if all the mass were concentrated at its center, that is, like a point-particle. So we will imagine the Earth as having a 'spare tire' of mass concentrated on the equator and treat it like a gyroscope.

CM: "As a further simplification we will consider how the Earth appears from the Sun, but the principles are the same for the Moon. It's just that with the Sun we can associate the various positions that I mention with the seasons and these can be pictured easily.

CM: "We consider ourselves to be in the plane of the ecliptic, looking out towards the Earth. Since this is being written just after the December solstice, we would see the North Pole turned away from us, the South Pole towards us, and the 'spare tire' above us on the side of Earth facing us and below us on the side of Earth facing away from us. Since the Earth rotates counter-clockwise in this coordinate system, points on the 'spare tire' are passing from left to right relative to us on the facing side. If we pull down on the facing side, this causes the motion to go from being left-to-right to being (slightly) left-up to right-down, and that causes the North Pole to shift to the right and the South Pole to the left."

DB: "But aren't we also pulling on the far side?"

CM: "Yes, we are, but remember on the far side the 'spare tire' is moving from right to left and we are pulling up, so the motion goes from being right-to-left to right-down to left-up and that also gives a rightward nudge to the North Pole and a leftward nudge to the South Pole. Now consider the June Solstice. Here the North Pole is turned towards us, the South Pole away from us, the 'spare tire' is now below us, so we are pulling up on it. This causes the North Pole to be nudged to the left and the South Pole to the right, and by the same argument as earlier, pulling down on the far side compounds this motion."

BH: "If the North Pole is first being pushed to the right and then to the left, how do we get a steady precession?"

CM: "Consider where the North Pole is during the solstices. During the December Solstice the North Pole is facing away from us and a rightward nudge causes it to move slightly in the clockwise sense about the vertical, which points to the ecliptic north pole. During the June Solstice the North Pole is facing towards us and a leftward nudge also causes it to move in a clockwise sense.

CM: "Finally, consider the equinoxes. At the equinoxes one pole is facing left and the other right, but both are at equal distance. Every point on the 'spare tire' that is above the plane is exactly matched by one below the plane at the exact same distance from us, so there is no turning of the axis. Thus the Earth's rotational axis is turned most strongly in the clockwise direction at the solstices and not at all at the equinoxes. But it all adds up and turns out to be about 50 seconds of arc per year."

BH: "Can you calculate this? I see that Nereid often asks the hard questions of many of the ATM proponents. Not all of them answer, some of them just blow right on by on their way to banishment. But I never see a mainstream proponent being asked to 'put up'."

CM: "Well, I could be arrogant and say we don't have to, that the mainstream calculations are there in the literature, from elementary treatments in the standard textbooks of celestial mechanics on up to the definitive treatments that our ephemerides are based on. But I think I could give a good elementary treatment here."

DB: "You've presented more difficult calculations on paper napkins at Couch Cafeteria!"

CM: "Yes, let me get a few paper napkins. It'll be just like old times!"

To be continued...

__________________
Microsoft is over if you want it.

The bar has been lowered for the promotion of ATM ideas; the bar for the acceptance of ATM ideas must remain high.

The Precession Dialogues--Part Three

I returned with a handful of paper napkins and set to work.

CM: "Our goal is to calculate the torque on the Earth caused by the gravitational pull of the Sun and Moon. To perform this calculation we will approximate the effect as being due to a ring of additional mass at the equator of an otherwise spherically symmetric Earth. Once we have calculated the torque and its direction we will compare it to the total angular momentum of the Earth and derive the period of the precession of the equinoxes.

CM: "Ironically, the part that gave me the most trouble was the beginning of the computation, trying to estimate how much of the Earth's mass is contained in the equatorial bulge. I kept coming up with a torque that was half of what was in Danby's book, so I knew something was wrong with the assumption I was making, but what was it? Do we have to know the details of the mass distribution of the Earth in order to compute this? Fortunately we do not. The key is that the Earth has two different values for the moment of inertia, one called 'C' that is evaluated for the polar axis, and the other called 'A' that is evaluated for any equatorial axis (they're all equal). My proposal is this: imagine that the moments of inertia are due to the spherical distribution of the bulk of the Earth, yielding a value of 'X', plus the contribution due to the equatorial bulge which is imagined as a ring of mass right at the equator.

CM: "Now in some xyz coordinate system with z pointing in the polar direction, the moment of inertia C is the integral of rho*(x²+y²) over the volume of the Earth, with rho being the density (and a function of x, y, and z) and x and y being coordinates. The contribution from the bulk of the Earth will be X but the contribution from the ring will be m*r². Likewise A will be the integral of rho*(y²+z²), and again the bulk of the Earth will contribute X to this but this time the ring at the equator only contributes half as much because only the y² term contributes since the ring is at z=0. We thus have two equations involving m:

A = X + (1/2)*m*r²,

C = X + m*r²,

CM: "which may be solved for m to obtain 2*(C-A)/r². This was the critical first step for me. Now we should also divide this mass by 2*pi in order to obtain the density of the equatorial bulge per radian of longitude of the Earth, because we are going to be performing integrals over the longitude. Fortunately, we are only going to have sines and cosines of the longitude, and only the integrals of sin² and cos² will have non-zero values. These values would be pi, so what we'll do is omit the division by 2*pi of the equatorial mass and just replace all sine-squareds and cosine-squareds with 1/2 when we integrate over the longitude.

CM: "Now the coordinate system. We are going to use ecliptic coordinates but place the Earth at the origin. We will ignore the eccentricity of the Earth and Moon and we will also ignore the inclination of the Moon's orbit. The coordinates of the Sun or the Moon will thus be R=(R*cos(L), R*sin(L), 0), with R as the appropriate distance and L as the appropriate ecliptic longitude.

CM: "Likewise we would have r=(r*cos(phi), r*sin(phi), 0) where r is the radius of the Earth and phi is the longitude measured in ecliptic coordinates for that instant (not the geographical longitude!), but there's just one thing: the plane of the equator is not the plane of the ecliptic. The Earth's axis is tilted through an angle of 23.47 degrees with respect to the ecliptic's z-axis, an angle that we'll call 'eps' (short for epsilon) in what follows. We must rotate the coordinates of the equatorial bulge through eps in the yz-plane, thus arriving at the following coordinates for points in the bulge: r=(r*cos(phi), r*sin(phi)*cos(eps), -r*sin(phi)*sin(eps)).

CM: "Each piece of the ring contributes rxF to the torque, where r is the moment arm and F is the force that we shall evaluate shortly. Since the Sun or the Moon is located at R, the attraction for a point at r is G*M*m*(R-r)/D³, where M is the mass of the Sun or Moon and m is the mass of a piece of the ring. D is the distance between the two, defined as D²= (R-r)²=R²-2R*r+r². But there is also the force on the Earth as a whole to be considered, because this is really a tidal effect. So the formula for the torque is:

N=G*M*m*rx((R-r)/D³-R/R³).

CM: "Since rxr is zero, we can omit the r inside the parentheses and rearrange a little to obtain:

N=G*M*m*rxR*(1/D³-1/R³).

To be continued...

__________________
Microsoft is over if you want it.

The bar has been lowered for the promotion of ATM ideas; the bar for the acceptance of ATM ideas must remain high.

The Precession Dialogues--Part Four

CM: "Our goal is to calculate the torque on the Earth caused by the gravitational pull of the Sun and Moon. To perform this calculation we will approximate the effect as being due to a ring of additional mass at the equator of an otherwise spherically symmetric Earth. Once we have calculated the torque and its direction we will compare it to the total angular momentum of the Earth and derive the period of the precession of the equinoxes. So far we have made several simplifications to the distribution of the Earth's mass, defined our coordinates, simplified the motion of the Sun and Moon, and arrived at the following expression for the torque:

N=G*M*m*rxR*(1/D³-1/R³).

CM: "Now we've got to evalute the cross-product rxR and the term in parentheses. As we found before, the components of R and r are:

r = (r*cos(phi), r*sin(phi)*cos(eps), - r*sin(phi)*sin(eps))

R = (R*cos(L), R*sin(L), 0)

CM: "So that it is straightforward (exercise left to student! ) to calculate:

rxR = (R*r*sin(L)*sin(phi)*sin(eps), -R*r*cos(L)*sin(phi)*sin(eps), R*r*(sin(L)*cos(phi)-cos(L)*sin(phi)*cos(eps)))

CM: "Whew!! Don't worry, it will start simplifying real soon. Also consider the formula for D:

D² = R² - 2*R*r + r²

= R² - 2*R*r*(cos(L)*cos(phi)+sin(L)*sin(phi)*cos(eps) + r².

CM: "Since the ratio r/R is about 1/60 for the Moon and much less than that for the Sun, we can write:

D^-3 = R^-3 + 3*(r/R⁴)*(cos(L)*cos(phi)+sin(L)*sin(phi)*cos(eps) + higher order terms

CM: "And therefore the parenthesized term is just 3*(r/R⁴)*(cos(L)*cos(phi)+sin(L)*sin(phi)*cos(eps)). Now we are ready to combine our results together, remembering that we will only consider terms with the products sin²(phi) or cos² in them, which we will promptly replace with 1/2:

N = G*M*m*3*r²/R³*((1/2)*sin²(L)*sin(eps)*cos(eps), -(1/2)*sin(L)*cos(L)*sin(eps)*cos(eps), (1/2)*sin(L)*cos(L)*sin²(eps)).

CM: "As stated earlier, the torque is zero when sin(L) = 0, that is at the equinoxes. At the solstices, the torque has its maximum value and is equal to (3/2)*G*M*m*r²/R³*sin(eps)*cos(eps)*(1, 0, 0) in value, confirming our earlier argument.

CM: "We are only interested in the non-periodic part, so we will also replace the functions of the longitude by their averages, obtaining:

N_avg = (3/4)*G*M*m*r²/R³*sin(eps)*cos(eps)*(1,0,0).

CM: "Now if we remember our formula for m, m=2*(C-A)/r², and replace M with S for the Sun and M for the Moon, we have:

N_avg = (3/2)*G*(C-A)*(S/R_S³+M/R_M³)*sin(eps)*cos(eps)*(1,0,0).

CM: "Notice, by the way, that the radius of the Earth has dropped out. Of course the distribution of the Earth's mass is still reflected in the quantity C-A. The expression above represents the change per unit time of the rotational angular momentum of the Earth. The total angular momentum of the Earth is C*n, where n is the sidereal rotation rate, 2*pi/86164 s^-1. It is a vector with components (0, C*n*cos(eps), C*n*sin(eps)) in the ecliptic coordinate system that we are using. It projects down onto the ecliptic plane as a circle of radius C*n*sin(eps), and combined with our torque calculated above this results in a precessional rate of N_avg/C/n/sin(eps) in radians per second. This rate is then:

(3/2)*((C-A)/C)*G*(S/R_S³+M/R_M³)*cos(eps)/n.

CM: "Now let's put in some numbers."

To be continued...

__________________
Microsoft is over if you want it.

The bar has been lowered for the promotion of ATM ideas; the bar for the acceptance of ATM ideas must remain high.

The Precession Dialogues--Part Five

CM: "This represents the change per unit time of the rotational angular momentum of the Earth. The total angular momentum of the Earth is C*n, where n is the sidereal rotation rate, 2*pi/86164 s^-1. It is a vector with components (0, C*n*cos(eps), C*n*sin(eps)) in the ecliptic coordinate system that we are using. It projects down onto the ecliptic plane as a circle of radius C*n*sin(eps), and combined with our torque calculated above this results in a precessional rate of N_avg/C/n/sin(eps) in radians per second. This rate is then:

(3/2)*((C-A)/C)*G*(S/R_S³+M/R_M³)*cos(eps)/n.

CM: "Now let's put in some numbers. For the mass of the Sun, 1.98x10³⁰ kg and the distance, 1.5x10¹¹ m, for a contribution of 0.000587 kg/m³ from the Sun. For the Moon, we have a mass of 7.35x10²² kg and a mean distance of 3.84x10⁸ m, for a contribution of 0.001298 kg/m³. As is usually the case with tidal effects, the Moon contributes about twice as much as the Sun.

CM: "As for the values of A and C, it appears that Danby's book Fundamentals of Celestial Mechanics has a typo in Appendix C.4 on page 434 where he gives 0.3340*M*r² (our notation) for C but 0.0029*M*r² for A. I believe the correct value for A should be 0.3329, which is consistent with the value of 1/304 that he gives for ((C-A)/C) in the text. Notice that this implies that 0.0022 of the total mass of the Earth is contained in the equatorial bulge, which sounds plausible. We will use 1/304 for the expression ((C-A)/C). We thus have:

1.5*(1/304)*6.672x10^-11*(0.000587+0.001298)*cos(23.47)*(86164/2/pi) <-- remember, 86,164 seconds in a sidereal day!

= 7.806x10^-12 s^-1.

CM: "This is in radians per second, so we calculate:

2*pi/7.806x10^-12 = 8.0492x10¹¹ s.

CM: "Divide this by 86400 seconds per day:

8.0492x10¹¹/86400 = 9.3162x10⁶ days,

CM: "and finally divide by 36525 days per Julian century:

9.3162x10⁶ / 36525 = 255.06 Julian centuries,

CM: "or 25,506 years. And if we divide 1,296,000 seconds of arc by 25,506 years we obtain 50.81 arc-seconds per year."

BH: "Excellent! But isn't 50.81 a little high?"

CM: "Yes, the adopted value for the epoch 2000 is 50.29 from Danby's book, page 435, so I'm off by, oh, one percent. Let's see someone get that much accuracy from some moldy old carpet! (Just say no to rugs!)"

CM: "So there you have it, our crude model shows that the gravitational pull on the Earth's equatorial bulge by the Sun and Moon together account for practically all of the observed precession of the equinoxes. A more detailed and rigorous calculation including the various factors we ignored (such as orbital eccentricities and the inclination of the Moon's orbit) would no doubt take care of our one percent error.

CM: "No ancient mythologies needed to be consulted, nor distant undetected companions invoked. The ball is in your court, A.DIM. What multiplicity of questions about the precession of the equinoxes does Newtonian gravity as applied in this exposition fail to answer?"

DB: "Wow, that's a tough question!"

BH: "Almost like one that Nereid might ask."

We all broke up laughing.

__________________
Microsoft is over if you want it.

The bar has been lowered for the promotion of ATM ideas; the bar for the acceptance of ATM ideas must remain high.

I'm suitably honoured to find myself featuring at the starting point of this lovely series of posts.



Grant Hutchison

Yes, that is a level of depth in a calculation that is suitable for a textbook, let alone a forum thread. Well done, CM.