Elementary Calculation of the Nodal Precession (Part 1--Vocal)

As promised, here is an explanation of the nodal precession in terms of the Solar torque on the Earth-Moon subsystem. Enjoy!

As in the torque example in an earlier post, we ignore general relativistic corrections, other planets, and the distribution of mass in the component bodies, reducing the problem to that of three point-masses. We also ignore the eccentricities, but NOT the inclination.

We define the following longitudes and arguments:

• L, mean longitude of the Moon. Since we are ignoring the eccentricities, this is also the true longitude.
• L', mean longitude of the Sun, also its true longitude for our purposes here.
• Omega, the longitude of the node.
• D, mean elongation, equal to L-L'. This is 0° near new moon, 90° at first quarter, etc.
• F, mean argument of latitude, equal to L-Omega . It is equal to 0° when crossing the ascending node, etc.

The Earth-Moon barycenter is at the origin and the x-y plane is the ecliptic. The coordinates of the Sun are therefore ( R * cos(L') , R * sin(L'), 0 ) = R * R_u, where R_u is a unit vector directed towards the Sun. The coordinates of the Earth and Moon are as follows:

• Earth: -(M / (E+M)) * r * r_u
• Moon: +(E / (E+M)) * r * r_u

The unit vector for the Earth-Moon separation is a little tricky, but we can start with the unit vector in the x-direction (1, 0, 0) and apply three rotations to bring it into position.

1. Start: ( 1 , 0 , 0 )
2. Rotate about z-axis through angle F: ( cos(F) , sin(F) , 0)
3. Rotate about x-axis through angle I: ( cos(F) , cos(I) * sin(F) , sin(I) * sin(F) ). Here I is the inclination of the Earth-Moon orbital plane to the ecliptic. For convenience define c = cos(I) and s = sin(I) so that the above unit vector becomes ( cos(F) , c * sin(F), s * sin(F) ).
4. Finally, rotate about z-axis through angle Omega: r_u = ( ((1+c)/2) * cos(L) + ((1-c)/2) * cos(L-2*F) , ((1+c)/2) * sin(L) + ((1-c)/2) * sin(L-2*F) , s * sin(F) ).

Now the angular momentum vector and the line of nodes vector:

• Orbital angular momentum of Earth-Moon subsystem: h = (E*M/(E+M)) * n * r^2 * h_u. Here n is the mean motion of the Moon.
• Unit vector h_u = ( s * sin(Omega) , -s * cos(Omega) , c)
• Unit vector in direction of line of nodes: N_u = ( cos(Omega) , sin(Omega) , 0 )
• Unit vector perpendicular to angular momentum and line of nodes: ( - c * sin(Omega) , c * cos(Omega) , s ).

The three unit vectors in the above list are mutually perpendicular ("verification left as an exercise for the student!") and form a basis. For reasons that will become apparent later, we are mainly interested in the line of nodes vector.

Now the distances:

• Sun-Earth: D_E^2 = (R * R_u + (M / (E+M)) * r * r_u)^2.
• Sun-Moon: D_M^2 = (R * R_u - (E / (E+M)) * r * r_u)^2.

Once again we will only be concerned with the first order correction in (r/R), so by a process similar to my previous example:

• D_E^n = R^n * (1 + n * (M / (E+M)) * R_u · r_u (r/R) ).
• D_M^n = R^n * (1 - n * (E / (E+M)) * R_u · r_u (r/R) ).

Notice the n this time instead of (n/2) and the presence of a dot product. This dot product is:

R_u · r_u = ((1+c)/2) * cos(D) + ((1-c)/2) * cos(D-2*F) ("verification left as an exercise for the student!")

We can now write:

• D_E^n = R^n * (1 + n * (M / (E+M)) * (((1+c)/2) * cos(D) + ((1-c)/2) * cos(D-2*F)) (r/R) ).
• D_M^n = R^n * (1 - n * (E / (E+M)) * (((1+c)/2) * cos(D) + ((1-c)/2) * cos(D-2*F)) (r/R) ).

The forces are then:

• Sun on Earth: (G*S*E) * ( R * R_u + (M / (E+M)) * r * r_u ) / D_E^3
• Sun on Moon: (G*S*M) * ( R * R_u - (E / (E+M)) * r * r_u ) / D_M^3

Next: The torques.

_________________
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<font size=-1>[ This Message was edited by: Celestial Mechanic on 2002-07-08 00:51 ]</font>

<font size=-1>[ This Message was edited by: Celestial Mechanic on 2002-07-08 00:53 ]</font>