How do I calculate the moment of inertia of an idealized Earth's equatorial bulge?

Question

I am working on calculating an idealized Sun-Earth-Moon three-body system. As part of this I want to calculate Earth's axial precession, which requires knowing the torque that is acting on it. Wikipedia gives this equation for the torque caused by a celestial body's gravity acting on the Earth: $$\vec{T} = \frac{3GM}{r^3} (C − A) \sin(δ) \cos(δ) \begin{pmatrix} \sin(α)\\ −\cos(α)\\ 0\\ \end{pmatrix}$$

Where

• $GM$ is the standard gravitational parameter, the product of the gravitational constant $G$ and the mass $M$ of the perturbing body;
• $r$ is the distance between the center of the Earth and the center of the perturbing body;
• $C$ is the moment of inertia around Earth's axis of rotation;
• $A$ is the moment of inertia around any equatorial diameter of Earth;
• $(C − A)$ is the moment of inertia of Earth's equatorial bulge (C > A);
• $δ$ is the declination of the perturbing body (positive for north of the equator, negative for south of the equator); and
• $α$ is the right ascension of the perturbing body (east from vernal equinox)

I have all the numbers needed except for the moments of inertia around Earth's axis of rotation and around "any equatorial diameter of Earth". I don't know how to calculate these, or even what the difference between them is, and Googling hasn't been much help. For the value of $C$ I found one source that says Earth's moment of inertia is $8.04 × 10^{37} \text{ kg m}^2$; and another that says the moment of inertia of an oblate spheroid around its shorter axis is $\frac{2}{5} M r^2$, where $r$ is the major radius, which with my numbers gives me $9.699 × 10^{37} \text{ kg m}^2$. These two numbers are at least the same order of magnitude, which makes me think I'm on the right track, but there's still a large error factor I'd like to eliminate which I'm assuming is related to Earth not being a point mass or of uniform density. That also still leaves me without any idea of what the value of $A$ represents and how it differs from $C$.

In my idealized system, the Earth is an oblate spheroid whose shape is given by the equation $(\frac{x}{6372})^2 + (\frac{y}{6372})^2 + (\frac{z}{6349.875})^2 = 1$, making the equator 22.125 km farther from the center of the Earth than the poles. For density I'm using the Preliminary Reference Earth Model (PREM, en.wikipedia.org/wiki/File:RadialDensityPREM.jpg) and assuming both that the mass is evenly distributed and that the distances in the PREM scale evenly with the radius at any given latitude.

Answer 1

The MMOI tensor of a ellipsoid with semi-radii $(a,b,c)$ is

$$ \mathbf{I} = \begin{bmatrix} \tfrac{2}{5} M (b^2+c^2) & & \\ & \tfrac{2}{5} M (a^2+c^2) & \\ & & \tfrac{2}{5} M (a^2+b^2) \end{bmatrix} $$

and from the shape of the earth $(a= 6732, b = 6732, c = 6349.875)$ and $M$ is the mass.

But the density distribution is not uniform, therefore the factor $\tfrac{2}{5}$ isn't accurate. To find a good estimate use the known value of $f M (a^2+b^2) = 8.04×10^{37} \text{ kg m}^2$ to estimate $f \approx 0.1366 < 0.4$.

Then use this estimate to get

$$ \mathbf{I}_{\rm earth} = \begin{bmatrix} 7.597 & & \\ & 7.597 & \\ & & 8.04 \end{bmatrix} 10^{37} \text{ kg m}^2$$

or the MMOI about any equatorial axis is -5.5% less than that about the polar axis.

Answer 2

You might start with the (I) for the mathematically described spheroid (use circular slices parallel to the equator), and then subtract out the contribution from the inscribed sphere. A problem with the earth would be the variation in density.