next up previous
Next: Forced precession and nutation Up: Rigid body rotation Previous: Free precession of Earth


MacCullagh's formula

According to Equations (3.59) and (3.64), if the Earth is modeled as spheroid of uniform density $ \gamma$ then its ellipticity is given by

$\displaystyle \epsilon = - \left.\int r^{\,2}\,\gamma\,P_2(\cos\theta)\,d^{\,3}...
...2}\,\gamma\,\left(3\,\cos^2\theta - 1\right)\,d^{\,3}{\bf r}\right/ {\cal I}_0,$ (8.65)

where the integral is over the whole volume of the Earth, and $ {\cal I}_0 = (2/5)\,M\,R^{\,2}$ would be the Earth's moment of inertia were it exactly spherical. The Earth's moment of inertia about its axis of rotation is given by

$\displaystyle {\cal I}_\parallel = \int (x^{\,2}+y^{\,2})\,\gamma\,d^{\,3}{\bf r} = \int r^{\,2}\,\gamma\,(1-\cos^2\theta)\,d^{\,3}{\bf r}.$ (8.66)

Here, use has been made of Equations (3.24)-(3.26). Likewise, the Earth's moment of inertia about an axis perpendicular to its axis of rotation (and passing through the Earth's center) is

$\displaystyle {\cal I}_\perp$ $\displaystyle = \int (y^{\,2}+z^{\,2})\,\gamma\,d^{\,3}{\bf r} = \int r^{\,2}\,\gamma\,(\sin^2\theta\,\sin^2 \phi + \cos^2\theta)\,d^{\,3}{\bf r}$    
  $\displaystyle = \int r^{\,2}\,\gamma\,\left(\frac{1}{2}\,\sin^2\theta+ \cos^2\t...
...3}{\bf r} = \frac{1}{2}\int r^{\,2}\,\gamma\,(1+ \cos^2\theta)\,d^{\,3}{\bf r},$ (8.67)

because the average of $ \sin^2\phi$ is $ 1/2$ for an axisymmetric mass distribution. It follows from the preceding three equations that

$\displaystyle \epsilon = \frac{{\cal I}_\parallel-{\cal I}_\perp}{{\cal I}_0}\simeq \frac{{\cal I}_\parallel-{\cal I}_\perp}{{\cal I}_\parallel}.$ (8.68)

When Equation (8.68) is combined with Equation (3.65), we get

$\displaystyle {\mit\Phi}(r,\theta) \simeq - \frac{G\,M}{r} + \frac{G\,({\cal I}_\parallel - {\cal I}_\perp)}{r^{\,3}}\,P_2(\cos\theta),$ (8.69)

which is the general expression for the gravitational potential generated outside an axially symmetric mass distribution. The first term on the right-hand side is the monopole gravitational potential that would be generated if all of the mass in the distribution were concentrated at its center of mass, whereas the second term is the quadrupole potential generated by any deviation from spherical symmetry in the distribution. Equation (8.69) actually holds for any axially symmetric mass distribution, not just a spheroidal mass distribution of uniform density (this is discussed in the following).

More generally, consider an asymmetric mass distribution consisting of $ N$ mass elements. Suppose that the $ i$ th element has mass $ m_i$ and position vector $ {\bf r}_i$ , where $ i$ runs from $ 1$ to $ N$ . Let us define a Cartesian coordinate system $ x$ , $ y$ , $ z$ such that the origin coincides with the center of mass of the distribution. It follows that

$\displaystyle \sum_{i=1,N} m_i\,x_i = \sum_{i=1,N} m_i\,y_i =\sum_{i=1,N}m_i\,z_i = 0.$ (8.70)

Suppose that the $ x$ -, $ y$ -, and $ z$ -axes coincide with the mass distribution's principal axes of rotation (for rotation about an axis that passes through the origin). It follows from Section 8.5 that

$\displaystyle \sum_{i=1,N}m_i\,y_i\,z_i = \sum_{i=1,N}m_i\,x_i\,z_i= \sum_{i=1,N}m_i\,x_i\,y_i= 0,$ (8.71)

and that the distribution's principal moments of inertia about the $ x$ -, $ y$ -, and $ z$ -axes take the form

    $\displaystyle {\cal I}_{xx}$ $\displaystyle = \sum_{i=1,N}m_i\,(y_i^{\,2}+z_i^{\,2}),$ (8.72)
    $\displaystyle {\cal I}_{yy}$ $\displaystyle = \sum_{i=1,N}m_i\,(x_i^{\,2}+z_i^{\,2}),$ (8.73)
and   $\displaystyle {\cal I}_{zz}$ $\displaystyle = \sum_{i=1,N}m_i\,(x_i^{\,2}+y_i^{\,2}),$ (8.74)

respectively.

Figure 8.3: A general mass distribution.
\begin{figure}
\epsfysize =2.25in
\centerline{\epsffile{Chapter07/fig7.03.eps}}
\end{figure}

Consider the gravitational potential, $ {\mit\Phi}$ , generated by the mass distribution at some external point $ P$ whose position vector is $ {\bf r}\equiv (x$ , $ y$ , $ z)$ . According to Section 3.2,

$\displaystyle {\mit\Phi}({\bf r}) = -G\sum_{i=1,N}\frac{m_i}{\vert{\bf r}_i-{\bf r}\vert},$ (8.75)

which can also be written

$\displaystyle {\mit\Phi}({\bf r})=-\frac{G}{r}\sum_{i=1,N}m_i\left(1-2\,\frac{r_i}{r}\,\cos\psi + \frac{r_i^{\,2}}{r^{\,2}}\right)^{-1/2},$ (8.76)

where $ \psi$ is the angle subtended between the vectors $ {\bf r}$ and $ {\bf r}_i$ . (See Figure 8.3.) Suppose that the distance $ OP\equiv r$ is much larger than the characteristic radius of the mass distribution, which implies that $ r_i/r\ll 1$ for all $ i$ . Expanding up to second order in $ r_i/r$ , we obtain

$\displaystyle {\mit\Phi}({\bf r})\simeq -\frac{G}{r}\sum_{i=1,N}m_i\left[1+\fra...
...si+ \frac{1}{2}\,\frac{r_i^{\,2}}{r^{\,2}} \left(3\,\cos^2\psi-1\right)\right].$ (8.77)

However,

$\displaystyle \sum_{i=1,N} m_i\,\frac{r_i}{r}\,\cos\psi= \sum_{i=1,N} m_i\,\fra...
...dot{\bf r}}{r^{\,2}}= \sum_{i=1,N} m_i\,\frac{x_i\,x+y_i\,y+z_i\,z}{r^{\,2}}=0,$ (8.78)

where use has been made of Equations (8.70). Hence, we are left with

$\displaystyle {\mit\Phi}({\bf r}) \simeq - \frac{G\,M}{r} - \frac{G}{2\,r^{\,3}}\sum_{i=1,N} m_i\,r_i^{\,2}\,(2-3\,\sin^2\psi).$ (8.79)

Here, $ M = \sum_{i=1,N} m_i$ is the total mass of the distribution. Now,

$\displaystyle \sum_{i=1,N}2\,m_i\,r_i^{\,2}$ $\displaystyle =\sum_{i=1,N} m_i\,(y_i^{\,2}+z_i^{\,2})+\sum_{i=1,N}m_i\,(x_i^{\,2}+z_i^{\,2}) + \sum_{i=1,N}m_i\,(x_i^{\,2}+y_i^{\,2})$    
  $\displaystyle = {\cal I}_{xx} + {\cal I}_{yy} + {\cal I}_{zz},$ (8.80)

where use has been made of Equations (8.72)-(8.74). Furthermore,

$\displaystyle {\mit I} \equiv \sum_{i=1,N} m_i\,r_i^{\,2}\,\sin^2\psi$ (8.81)

is the distribution's moment of inertia about the axis $ OP$ . Thus, we deduce that

$\displaystyle {\mit\Phi}({\bf r}) \simeq -\frac{G\,M}{r} - \frac{G\,({\cal I}_{xx}+{\cal I}_{yy}+{\cal I}_{zz}-3\,{\cal I})}{2\,r^{\,3}}.$ (8.82)

This famous result is known as MacCullagh's formula, after its discoverer, the Irish mathematician James MacCullagh (1809-1847). Actually,

$\displaystyle \sum_{i=1,N} m_i\,r_i^{\,2}\,\sin^2\psi$ $\displaystyle =\sum_{i=1,N}m_i\, \frac{ (y_i^{\,2}+z_i^{\,2})\,x^{\,2} + (x_i^{\,2}+z_i^{\,2})\,y^{\,2} +(x_i^{\,2}+y_i^{\,2})\,z^{\,2} }{ r^{\,2} }$    
  $\displaystyle = \frac{ {\cal I}_{xx}\,x^{\,2}+ {\cal I}_{yy}\,y^{\,2}+{\cal I}_{zz}\,z^{\,2} }{ r^{\,2} }.$ (8.83)

Hence, MacCullagh's formula can also be written in the alternative form

$\displaystyle {\mit\Phi}({\bf r}) \simeq -\frac{G\,M}{r} -\frac{G\,({\cal I}_{x...
... I}_{xx}\, x^{\,2}+{\cal I}_{yy}\,y^{\,2}+{\cal I}_{zz}\,z^{\,2})}{2\,r^{\,5}}.$ (8.84)

Finally, for an axisymmetric distribution, such that $ {\cal I}_{xx}={\cal I}_{yy}={\cal I}_\perp$ and $ {\cal I}_{zz}= {\cal I}_\parallel$ , MacCullagh's formula reduces to

$\displaystyle {\mit\Phi}({\bf r}) \simeq - \frac{G\,M}{r} + \frac{G\,({\cal I}_\parallel - {\cal I}_\perp)}{r^{\,3}}\,P_2(\cos\theta),$ (8.85)

where $ \cos\theta=z/r$ . Of course, this expression is the same as Equation (8.69), which justifies our earlier assertion that this equation is is valid for a general axisymmetric mass distribution. Incidentally, a comparison of the preceding expression with Equation (3.66) reveals that

$\displaystyle J_2= \frac{{\cal I}_\parallel-{\cal I}_\perp}{M\,R^{\,2}},$ (8.86)

where $ R$ is the mean radius of the distribution, and the dimensionless parameter $ J_2$ characterizes the quadrupole gravitational field external to the distribution.


next up previous
Next: Forced precession and nutation Up: Rigid body rotation Previous: Free precession of Earth
Richard Fitzpatrick 2016-03-31