MacCullagh's Formula

Consider the uniform spheroid discussed in Section 1.10.1. Let us set up a Cartesian coordinate system whose origin coincides with the center of the spheroid, and whose $z$-axis corresponds to the spheroid's rotation axis.

It is clear, by symmetry, that the $x$-, $y$-, and $z$-axes are principal axes of rotation. (See Section 1.7.2.) Hence, given that $x=r\,\sin\theta\,\cos\phi$, $y=r\,\sin\theta\,\sin\,\phi$, $z=r\,\cos\theta$, where $r$, $\theta$, $z$ are conventional spherical coordinates (see Section A.23), the principal moments of inertia of the spheroid are

$$I_{xx} =\int_V (y^2+z^2)\,\rho(x,y,z)\,dV = \int_V\,r^2\,(\sin^2\theta\,\sin^2\phi + \cos^2\theta)\,\rho(r,\theta)\,dV$$

$$=\frac{1}{2}\int_V r^2\,(1+\cos^2\theta)\,\rho(r,\theta)\,dV,$$ (1.374)

$$I_{yy} =\int_V (x^2+z^2)\,\rho(x,y,z)\,dV = \int_V\,r^2\,(\sin^2\theta\,\cos^2\phi + \cos^2\theta)\,\rho(r,\theta)\,dV$$

$$=\frac{1}{2}\int_V r^2\,(1+\cos^2\theta)\,\rho(r,\theta)\,dV,$$ (1.375)

$$I_{zz} =\int_V (x^2+y^2)\,\rho(x,y,z)\,dV =\int_V r^2\,(1-\cos^2\theta)\,dV,$$ (1.376)

where the integral is over the volume, $V$, of the spheroid. In writing the previous formulae, we have made use of the fact that the mass density of the spheroid, $\rho(r,\theta)$, is independent of the azimuthal angle, $\phi$, and that the averages of $\cos^2\phi$ and $\sin^2\phi$ over the volume of a mass distribution that is independent of $\phi$ are both equal to $1/2$.

Let us write $I_{xx}=I_{yy} = I_\perp$ and $I_\parallel = I_{zz}$. (See Section 1.7.2.) Now, according to Equations (1.332)–(1.334), the mass density of the spheroid is

$$\rho(r,\theta)= \rho_0(r)-\frac{2}{3}\,\gamma\,R\,\epsilon\,\delta(r-R)\,P_2(\cos\theta),$$ (1.377)

where

$$\rho_0(r) =\left\{\begin{array}{lll}\gamma&~~~~&r\leq R\\ [0.5ex]0&&r>R\end{array}\right..$$ (1.378)

Here, $R$ is the mean radius of the spheroid, and $\gamma$ is its uniform mass density. Given that $dV = r^2\,\sin\theta\,dr\,d\theta\,d\phi$, we can evaluate the integrals (1.374)–(1.376) to give

$$I_\parallel = \frac{2}{5}\,M\,R^2+\frac{4}{15}\,\epsilon\,M\,R^2,$$ (1.379)

$$I_\perp = \frac{2}{5}\,M\,R^2 - \frac{2}{15}\,\epsilon\,M\,R^2.$$ (1.380)

Here, $M= (4\pi/3)\,R^3\,\gamma$ is the mass of the spheroid.

If we apply the previous two results to the Earth, for which $\epsilon=3.35\times 10^{-3}$ (see Section 1.10.2), then it is clear that the slight rotational flattening of the Earth along its axis of diurnal rotation causes the principal moment of inertia about this axis, $I_\parallel$, to be slightly larger that the principal moment of inertia, $I_\perp$, about an axis that lies in the Earth's equatorial plane. In particular,

$$I_\parallel -I_\perp = \frac{2}{5}\,\epsilon\,M\,R^2.$$ (1.381)

Thus, it follows from Equation (1.349) that the self-generated gravitational potential outside the Earth can be written

$${\mit\Phi}(r,\theta) = -\frac{G\,M}{r} + \frac{G\,(I_\parallel-I_\perp)}{r^3}\,P_2(\cos\theta).$$ (1.382)

This result is known as MacCullagh's formula. It turns out the that the previous formula applies to any axisymmetric mass distribution (i.e., it is not limited to spheroidal mass distributions of uniform mass density). The first term on the right-hand side MacCullagh's formula is the monopole gravitational potential that would be generated if all the Earth's mass were concentrated at its center of mass, whereas the second term is the quadrupole gravitational potential generated by the slight deviation of the Earth's shape from that of a sphere.