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: $$\vec{T} = \frac{3GM}{r^3} (C − A) \sin(δ) \cos(δ) \begin{pmatrix} \sin(α)\\ −\cos(α)\\ 0\\ \end{pmatrix}$$


  • $GM$ 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.

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    $\begingroup$ This article goes in the other direction, calculating Earth's $C$ knowing its axial precession period. $\endgroup$
    – Spencer
    Jul 18, 2021 at 16:42
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    $\begingroup$ You seem to.know your maths and physics - maybe just calculate it yourself from first principles, for an infinitesimal point at some polar coordinates (easier for 3D spherical symmetry of density and symmetry of size) $\endgroup$
    – Stilez
    Jul 18, 2021 at 19:11
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    $\begingroup$ The issue is that the density distribution is needed as well as the shape to calculate MMOI. This is where the error comes in with the $2/5 M r^2$ estimate. That value is for uniform density. $\endgroup$ Jul 18, 2021 at 22:20

2 Answers 2


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.


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.


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