Elementary Calculation of the Nodal Precession (Part 2--Instrumental)
The forces that we just computed are the sums of multiples of two unit vectors,
R_u and
r_u. Since the cross-product of a vector with itself is zero and our moment arm vector only contains
r_u, the torques are simply:
- Earth: -(G*S*E*M/(E+M)) * ( r * R / D_E^3) * r_u × R_u.
- Moon: +(G*S*E*M/(E+M)) * ( r * R / D_M^3) * r_u × R_u.
So the total torque is:
(G*S*E*M/(E+M)) * ( r * R ) * (D_M^-3 - D_E^-3) *
r_u ×
R_u.
The cross-product (also left as an exercise!) is:
( - s * sin(F) * sin(L') , +s * sin(F) * cos(L') , -((1+c)/2) * sin(D) - ((1-c)/2) * sin(D-2*F) ).
Now, if we were developing the full lunar theory we would be considering much more than these few terms, but here we are only interested in a constant part, not the periodic perturbations. We note the following properties of sine and cosine series:
- The product of a sine series with a sine series is a cosine series.
- The product of a cosine series with a cosine series is a cosine series.
- The product of a sine series with a cosine series is a sine series.
- A constant part can only appear in a cosine series.
The cross-product calculated above has three components that are cosine, sine, and sine series respectively. The series for D_E^n and D_M^n are cosine series. Of the three unit basis vectors considered, only the line of nodes unit vector is also of type cosine, sine, sine (0 may be considered as either), so only that component of the torque that projects onto the line of nodes can have a constant part.
This dot product (also left as an exercise!) is:
-s * sin(F) * sin(L'-Omega) = -(s/2) * cos (D) + (s/2) * cos(D-2*F).
The projection of the total torque onto the line of nodes is then:
(G*S*E*M/(E+M)) * ( r * R ) * (D_M^-3 - D_E^-3) * (-(s/2) * cos (D) + (s/2) * cos(D-2*F)) *
N_u.
As in the previous example, D_M^-3 = D_E^-3 = R^-3 at the lowest order so that it cancels at that order. To the next order, however we find:
3 * (G*S*E*M/(E+M)) * ( r * R ) * (((1+c)/2) * cos(D) + ((1-c)/2) * cos(D-2*F)) * (r / R^4) * (-(s/2) * cos (D) + (s/2) * cos(D-2*F)) *
N_u.
We are only interested in the constant parts, which arise from cos(D) * cos(D) = 1/2 + 1/2 * cos(2*D) and similarly for cos(D-2*F). Therefore, throwing away the periodic terms we are left with:
3 * (G*S*E*M/(E+M)) * ( r^2 / R^3 ) * (-s*(1+c)/8 + s*(1-c)/8) *
N_u
= -3/4 * (G*S*E*M/(E+M)) * s * c * ( r^2 / R^3 ) *
N_u.
As claimed, there is a constant part of the torque directed along the line of nodes. Because it is perpendicular to the angular momentum it does not change the magnitude of the angular momentum, only the direction.
Now let's calculate the torque another way. The angular momentum of the Earth-Moon subsystem is (E*M/(E+M)) * n * r^2 *
h_u, where n is the mean motion of the Moon. In our model we can consider n and r to be constant, so remembering that the torque is also defined as the time derivative of the angular momentum, we have:
(E*M/(E+M)) * n * r^2 * (d/dt)
h_u
as the torque. But (d/dt)
h_u = (d/dt) ( +s * sin(Omega) , -s * cos(Omega) , c )
= dOmega/dt * ( s * cos(Omega) , +s * sin(Omega) , 0 ) = dOmega/dt * s *
N_u. Equating the two expressions for the torque projected along the line of nodes gives:
(E*M/(E+M)) * s * n * r^2 * dOmega/dt = -3/4 * (G*S*E*M/(E+M)) * s * c * ( r^2 / R^3 ) and some cancellation yields:
dOmega/dt = -3/4 * (G*S) * c / n / R^3.
We can massage this a bit more: because the motion of the Sun about the Earth-Moon barycenter is a Keplerian ellipse, we can write Kepler's Third Law for it as N^2 * R^3 = G*(S+E+M), where N is the mean motion of the Earth-Moon about the Sun, and finally obtain:
dOmega/dt = -3/4 * (S/(S+E+M)) * c * N^2 / n = -3/4 * N^2 / n (approximately),
where I'm ignoring (S/(S+E+M)) and c since they are both nearly one.
Next: Number crunching time!