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...
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