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...

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