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Next: Exercises Up: Ellipsoidal Potential Theory Previous: Introduction


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

$\displaystyle d{\mit\Psi} = - \int_0^{r'}\frac{G\,\rho}{r}\,dV - \int_{r''}^0 \frac{G\,\rho}{(-r)}\,dV,$ (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 $ P$ 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.
\epsfysize =2.25in

From Figure D.1, the position vector of point $ Q$ , relative to the origin, $ O$ , 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 $ P$ , and $ {\bf n}$ a unit vector pointing from $ P$ to $ Q$ . Likewise, the position vector of point $ R$ is

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

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

which reduces to the quadratic

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


$\displaystyle A$ $\displaystyle = \sum_{i=1,3}\frac{n_i^{\,2}}{a_i^{\,2}},$ (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), $ r'+r''=-B/A$ , 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)


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


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


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


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


$\displaystyle \alpha_0$ $\displaystyle =\int_0^\infty \frac{du}{\mit\Delta},$ (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 up previous
Next: Exercises Up: Ellipsoidal Potential Theory Previous: Introduction
Richard Fitzpatrick 2016-03-31