Analysis

Consider the contribution to the potential at $P$ from the mass contained within a double cone, whose apex is $P$ , and which is terminated in both directions at the body's outer boundary. (See Figure D.1.) If the cone subtends a solid angle $d{\mit\Omega}$ then a volume element is written $dV = r^{\,2}\,dr\,d{\mit\Omega}$ , where $r$ measures displacement from $P$ along the axis of the cone. Thus, from standard classical gravitational theory (Fitzpatrick 2012), the contribution to the potential takes the form

$$d{\mit\Psi} = - \int_0^{r'}\frac{G\,\rho}{r}\,dV - \int_{r''}^0 \frac{G\,\rho}{(-r)}\,dV,$$ (D.2)

where $r'=\vert PQ\vert$ , $r''=-\vert PR\vert$ , and $\rho$ is the constant mass density of the ellipsoid. Hence, we obtain

$$d{\mit\Psi} = -G\,\rho\left(\int_0^{r'} r\,dr + \int_0^{r''} r\,d... ...ght)d{\mit\Omega} =- \frac{1}{2}\,G\,\rho\,(r'^{\,2}+r''^{\,2})\,d{\mit\Omega}.$$ (D.3)

The net potential at $P$ is obtained by integrating over all solid angle, and dividing the result by two to adjust for double counting. This yields

$${\mit\Psi} = - \frac{1}{4}\,G\,\rho\oint \,(r'^{\,2}+r''^{\,2})\,d{\mit\Omega}.$$ (D.4)

Figure D.1: Calculation of ellipsoidal gravitational potential.

From Figure D.1, the position vector of point $Q$ , relative to the origin, $O$ , is

$${\bf x}' = {\bf x} + r'\,{\bf n},$$ (D.5)

where ${\bf x}= (x_1,\,x_2,\,x_3)$ is the position vector of point $P$ , and ${\bf n}$ a unit vector pointing from $P$ to $Q$ . Likewise, the position vector of point $R$ is

$${\bf x}'' = {\bf x} + r''\,{\bf n}.$$ (D.6)

However, $Q$ and $R$ both lie on the body's outer boundary. It follows, from Equation (D.1), that $r'$ and $r''$ are the two roots of

$$\sum_{i=1,3}\left(\frac{x_i+r\,n_i}{a_i}\right)^2=1,$$ (D.7)

which reduces to the quadratic

$$A\,r^{\,2}+ B\,r+C = 0,$$ (D.8)

where

$$A = \sum_{i=1,3}\frac{n_i^{\,2}}{a_i^{\,2}},$$ (D.9)

$$B = 2\sum_{i=1,3}\frac{x_i\,n_i}{a_i^{\,2}},$$ (D.10)

$$C =\sum_{i=1,3}\frac{x_i^{\,2}}{a_i^{\,2}}-1.$$ (D.11)

According to standard polynomial equation theory (Riley 1974), $r'+r''=-B/A$ , and $r'\,r''=C/A$ . Thus,

$$r'^{\,2}+r''^{\,2} = (r'+r'')^2- 2\,r'\,r'' = \frac{B^{\,2}}{A^2} - 2\,\frac{C}{A},$$ (D.12)

and Equation (D.4) becomes

$${\mit\Psi} = -\frac{1}{2}\,G\,\rho\oint\left[ \frac{2\left(\sum_{... ...,3}x_i^{\,2}/a_i^{\,2}}{\sum_{i=1,3}n_i^{\,2}/a_i^{\,2}} \right] d{\mit\Omega}.$$ (D.13)

The previous expression can also be written

$${\mit\Psi} = -\frac{1}{2}\,G\,\rho\oint\left[ \frac{2\sum_{i,j=1,... ...,3}x_i^{\,2}/a_i^{\,2}}{\sum_{i=1,3}n_i^{\,2}/a_i^{\,2}} \right] d{\mit\Omega}.$$ (D.14)

However, the cross terms (i.e., $i\neq j$ ) integrate to zero by symmetry, and we are left with

$${\mit\Psi} = -\frac{1}{2}\,G\,\rho\oint\left[ \frac{2\sum_{i=1,3}... ...,3}x_i^{\,2}/a_i^{\,2}}{\sum_{i=1,3}n_i^{\,2}/a_i^{\,2}} \right] d{\mit\Omega}.$$ (D.15)

Let

$$J = \oint \frac{d{\mit\Omega}}{\sum_{i=1,3}n_i^{\,2}/a_i^{\,2}}.$$ (D.16)

It follows that

$$\frac{1}{a_i}\,\frac{\partial J}{\partial a_i} = \oint\frac{2\,n_... ...,2}/a_i^{\,4}}{\left(\sum_{i=1,3} n_i^{\,2}/a_i^{\,2}\right)^2}\,d{\mit\Omega}.$$ (D.17)

Thus, Equation (D.15) can be written

$${\mit\Psi} = - \frac{1}{2}\,G\,\rho\left(J - \sum_{i=1,3}\,A_i\,x_i^{\,2}\right),$$ (D.18)

where

$$A_i = \frac{J}{a_i^{\,2}}-\frac{1}{a_i}\,\frac{\partial J}{\partial a_i}.$$ (D.19)

At this stage, it is convenient to adopt the spherical angular coordinates, $\theta$ and $\phi$ (see Section C.4), in terms of which

$${\bf n} = (\sin\theta\,\cos\phi,\,\sin\theta\,\sin\phi,\,\cos\theta),$$ (D.20)

and $d{\mit\Omega}=\sin\theta\,d\theta\,d\phi$ . We find, from Equation (D.16), that

$$J= 8\int_0^{\pi/2}\sin\theta\,d\theta\int_0^{\pi/2}\left(\frac{\s... ...theta\,\sin^2\phi}{a_2^{\,2}}+ \frac{\cos^2\theta}{a_3^{\,2}}\right)^{-1}d\phi.$$ (D.21)

Let $t=\tan\phi$ . It follows that

$$J = 8\int_0^{\pi/2}\sin\theta\,d\theta \int_0^\infty \frac{dt}{a+b\,t^2} = 4\pi\int_0^{\pi/2}\frac{\sin\theta\,d\theta}{(a\,b)^{1/2}},$$ (D.22)

where

$$a = \frac{\sin^2\theta}{a_1^{\,2}} + \frac{\cos^2\theta}{a_3^{\,2}},$$ (D.23)

$$b =\frac{\sin^2\theta}{a_2^{\,2}}+\frac{\cos^2\theta}{a_3^{\,2}}.$$ (D.24)

Hence, we obtain

$$J = 4\pi\,a_1\,a_2\,a_3^{\,2}\int_0^{\pi/2}\frac{\sin\theta\,\sec... ...2}+a_3^{\,2}\,\tan^2\theta)^{1/2}\, (a_2^{\,2}+a_3^{\,2}\,\tan^2\theta)^{1/2}}.$$ (D.25)

Let $u=a_3^{\,2}\,\tan^2\theta$ . It follows that

$$J = 2\pi\,a_1\,a_2\,a_3\int_0^\infty \frac{du}{\mit\Delta},$$ (D.26)

where

$${\mit\Delta} = (a_1^{\,2}+u)^{1/2}\,(a_2^{\,2}+u)^{1/2}\,(a_3^{\,2}+u)^{1/2}.$$ (D.27)

Now, from Equations (D.19), (D.26), and (D.27),

$$A_i = \frac{2\pi\,a_1\,a_2\,a_3}{a_i^{\,2}} \int_0^\infty\frac{du}{\m... ...tial a_i}\!\left( 2\pi\,a_1\,a_2\,a_3\int_0^\infty \frac{du}{\mit\Delta}\right)$$

$$=-\frac{2\pi\,a_1\,a_2\,a_3}{a_i}\int_0^\infty \frac{\partial}{\p... ...ght)du =2\pi\,a_1\,a_2\,a_3\int_0^\infty\frac{du}{(a_i^{\,2}+u)\,{\mit\Delta}}.$$ (D.28)

Thus, Equations (D.18), (D.26), and (D.28) yield

$${\mit\Psi} = - \frac{3}{4}\,G\,M\left(\alpha_0-\sum_{i=1,3}\alpha_i\,x_i^{\,2}\right),$$ (D.29)

where

$$\alpha_0 =\int_0^\infty \frac{du}{\mit\Delta},$$ (D.30)

$$\alpha_i =\int_0^\infty \frac{du}{(a_i^{\,2}+u)\,{\mit\Delta}}.$$ (D.31)

Here, $M= V\,\rho$ and $V=(4/3)\,\pi\,a_1\,a_2\,a_3$ are the body's mass and volume, respectively.

The total gravitational potential energy of the body is written (Fitzpatrick 2012)

$$U = \frac{1}{2}\,\int{\mit\Psi}\,\rho\,dV,$$ (D.32)

where the integral is taken over all interior points. It follows from Equation (D.29) that

$$U = -\frac{3}{8}\,G\,M^{\,2}\left(\alpha_0 - \frac{1}{5}\,\sum_{i=1,3}\alpha_i\,a_i^{\,2}\right).$$ (D.33)

In writing the previous expression, use has been made of the easily demonstrated result $\int x_i^{\,2}\,dV = (1/5)\,a_i^{\,2}\,V$ . Now,

$$2\,\frac{d}{du}\!\left(\frac{u}{\mit\Delta}\right) = -\frac{1}{\mit\Delta}+ \sum_{i=1,3}\frac{a_i^{\,2}}{(a_i^{\,2}+u)\,{\mit\Delta}},$$ (D.34)

so

$$\sum_{i=1,3}\alpha_i\,a_i^{\,2} = \int_0^\infty \sum_{i=1,3}\frac... ...u}\!\left(\frac{u}{\mit\Delta}\right)+\frac{1}{\mit\Delta}\right]du = \alpha_0.$$ (D.35)

Hence, we obtain

$$U =-\frac{3}{10}\,G\,M^{\,2}\,\alpha_0.$$ (D.36)