The Precession Dialogues--Part Seven

CM: "Before we go any farther, I want to re-do the earlier calculation with better numbers. The earlier calculation was done in the spirit of 'back-of-the-envelope' calculations, using numbers that I remembered, not necessarily the best available. As we shall see this will make a difference.

CM: "I want to perform some calculations on the lunar and solar contributions separately, so I am going to reorganize the calculation somewhat. First I will calculate a common factor for all of the calculations I am considering:"

(3/2)*((C-A)/C)*G*/n

= 1.5 * (1.08263x10^-3 / .3340) * 6.672x10^-11 * 86164/2/pi * 648000/pi

= 9.1760x10^-4.

CM: "The factor of 86164/2/pi is just 1/n (remember that this is the sidereal rate) and the factor of 648000/pi is a conversion factor from radians to arcseconds.

CM: "I have to say a few words about the calculation of ((C-A)/C). Danby's book gives the values C=0.3340 and A=0.3329 (allowing for the misprint). The difference of C and A is 0.0011. What is wrong with this value?"

DB: "Well, it only has two significant digits."

CM: "Exactly! We have fallen victim to 'catastrophic subtraction' and lost two significant digits. We've got to get them back somehow. It turns out that the next contribution to the potential has been measured and you will see the potential written like this:"

U = M/R * (1 - J₂*(r/R)²*P₂(cos(theta))), (remember I use the celestial mechanical convention on gravitional potentials!)

CM: "where J₂ is the value 1.08263x10^-3 used in the formula above, r is the equatorial radius of the Earth, R is the radius vector of the satellite, P₂ is our friend the Legendre polynomial equal to (3/2)x²-1, and theta is the co-latitude, equal to zero at the North Pole, 90 degrees at the equator, and 180 degrees at the South Pole.

CM: "Anyway, J₂ is measured independently of astronomical observations. I wish I had a better value for C that is independent of astronomy."

BH: "What units is this 9.1760x10^-4 quantity?"

CM: "The only dimensional quantities are G, which is m³/kg/s², and n, which is s^-1. Allowing for our use of arcseconds as angular measure, the above quantity is then 9.1760x10^-4 m³*arcsecond/kg/s. Some of this weirdness will go away after the next step.

CM: "Here are the improved values of the masses and mean distances:"

S = 1.9981x10³⁰ kg
M = 7.34834x10²² kg
R_S = 1.496x10¹¹ m
R_M = 3.844x10⁸ m

CM: "From these we obtain:"

Solar part = 9.1760x10^-4 * 1.9981x10³⁰ / (1.496x10¹¹)³ = 5.476x10^-7 arcseconds/s and,

Lunar part = 9.1760x10^-4 * 7.34834x10²² / (3.844x10⁸)³ = 1.1871x10^-6 arcseconds/s.

CM: "Again, note that lunar part is slightly more than twice the solar part, which is typical for tidal effects. Now let's put this all together. Remember that there is a factor of sin(eps)*cos(eps) from the torque which is divided by sin(eps) so that we have an additional factor of cos(eps):"

Precession = (5.476x10^-7 + 1.1871x10^-6) arcseconds/s * cos(23.439 deg) * 86400 s/d * 36525 d/Jcy

= 5022.58 arcseconds/Jcy

CM: "where Jcy is Julian century. This is much closer to the value given in Danby, which is 5029.0966. I'm now off by 0.13 percent. And not a single Sumerian tablet consulted!"

The three of us broke up laughing.

CM: "But before we crow too loudly, let me tell you why I won't be foisting a calculation of the corrections due to the eccentricities and inclinations upon our audience."

DB: "What's wrong?"

CM: "Well, I did the calculations using elements and expressions I developed for cartesian coordinates in the two-body problem, and found the following correction: the individual contributions must be multiplied by (1+3*e²-(3/2)*sin²(I)), where e and I are the appropriate values for the Sun and Moon. It turns out that the solar contribution should be multiplied by 1.0008 and the lunar contribution by 0.997. Only problem is, this actually decreases the calculated precessional rate, making things worse."

BH: "What happened?"

CM: "Well, you remember my reference to the cartoon cliche of a man standing on his front steps, briefcase and keys in hand, wondering to himself 'Did I forget something?' and of course we see that he wears boxers, not briefs. What I was forgetting here is that the motion of the Moon is heavily perturbed by the Sun. The two largest perturbations, which are graced by the names 'Variation' and 'Evection', are intermediate in size between the principal elliptic term and the second elliptic term. I cannot ignore these in any calculation of a correction to the precessional rate.

CM: "One of the keys of pedagogy is to know when to quit. I will give an explanation of the principal terms of the nutation because that only requires the first order in the inclination. I will not give the second order corrections. Of course, if I were teaching a graduate-level course in celestial mechanics, I might present the derivation of the correction I gave above and then assign them the task of deriving the next order correction for the Moon, giving them the necessary terms in the lunar coordinates. After making sure that I can do the calculation myself, of course."

The three of us broke up laughing again.

To be continued...

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