Next: Exercises Up: Ellipsoidal Potential Theory Previous: Introduction

Analysis

Consider the contribution to the potential at from the mass contained within a double cone, whose apex is , 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 measures displacement from along the axis of the cone. Thus, from standard classical gravitational theory (Fitzpatrick 2012), the contribution to the potential takes the form

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

(D.2)

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

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

(D.3)

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

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

(D.4)

**Figure D.1:** Calculation of ellipsoidal gravitational potential.
$\begin{figure} \epsfysize =2.25in \centerline{\epsffile{AppendixD/fig5.06.eps}} \end{figure}$

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

$\displaystyle {\bf x}' = {\bf x} + r'\,{\bf n},$

(D.5)

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

$\displaystyle {\bf x}'' = {\bf x} + r''\,{\bf n}.$

(D.6)

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

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

(D.7)

which reduces to the quadratic

$\displaystyle A\,r^{\,2}+ B\,r+C = 0,$

(D.8)

where

$\displaystyle A$	$\displaystyle = \sum_{i=1,3}\frac{n_i^{\,2}}{a_i^{\,2}},$	(D.9)
$\displaystyle B$	$\displaystyle = 2\sum_{i=1,3}\frac{x_i\,n_i}{a_i^{\,2}},$	(D.10)
$\displaystyle C$	$\displaystyle =\sum_{i=1,3}\frac{x_i^{\,2}}{a_i^{\,2}}-1.$	(D.11)

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

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

(D.12)

and Equation (D.4) becomes

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

(D.13)

The previous expression can also be written

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

(D.14)

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

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

(D.15)

Let

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

(D.16)

It follows that

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

(D.17)

Thus, Equation (D.15) can be written

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

(D.18)

where

$\displaystyle A_i = \frac{J}{a_i^{\,2}}-\frac{1}{a_i}\,\frac{\partial J}{\partial a_i}.$

(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

$\displaystyle {\bf n} = (\sin\theta\,\cos\phi,\,\sin\theta\,\sin\phi,\,\cos\theta),$

(D.20)

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

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

(D.21)

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

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

(D.22)

where

$\displaystyle a$	$\displaystyle = \frac{\sin^2\theta}{a_1^{\,2}} + \frac{\cos^2\theta}{a_3^{\,2}},$	(D.23)
$\displaystyle b$	$\displaystyle =\frac{\sin^2\theta}{a_2^{\,2}}+\frac{\cos^2\theta}{a_3^{\,2}}.$	(D.24)

Hence, we obtain

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

(D.25)

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

$\displaystyle J = 2\pi\,a_1\,a_2\,a_3\int_0^\infty \frac{du}{\mit\Delta},$

(D.26)

where

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

(D.27)

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

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

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

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

(D.29)

where

$\displaystyle \alpha_0$	$\displaystyle =\int_0^\infty \frac{du}{\mit\Delta},$	(D.30)
$\displaystyle \alpha_i$	$\displaystyle =\int_0^\infty \frac{du}{(a_i^{\,2}+u)\,{\mit\Delta}}.$	(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)

$\displaystyle U = \frac{1}{2}\,\int{\mit\Psi}\,\rho\,dV,$

(D.32)

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

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

(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,

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

(D.34)

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

(D.35)

Hence, we obtain

$\displaystyle U =-\frac{3}{10}\,G\,M^{\,2}\,\alpha_0.$

(D.36)

Next: Exercises Up: Ellipsoidal Potential Theory Previous: Introduction

Richard Fitzpatrick 2016-01-22