Gravitational Potential of Uniform Spheroid

Let us use Poisson's equation, (1.273), to calculate the gravitational potential generated around a spheroid of uniform mass density $\gamma$ and mean radius $R$. A spheroid is the solid body produced by rotating an ellipse about a major or a minor axis. Let the center of the spheroid be located at the origin, let its axis of rotation coincide with the $z$-axis, and let its outer boundary satisfy

$$r = R_\theta(\theta) = R\left[1-\frac{2}{3}\,\epsilon\,P_2(\cos\theta)\right],$$ (1.328)

where $\epsilon$ is termed the ellipticity. Here, $r$, $\theta$, $\phi$ are conventional spherical coordinates. (See Section A.23.) Moreover,

$$P_2(x) =\frac{1}{2}\,(3\,x^2-1)$$ (1.329)

is a Legendre polynomial of degree 2. It can be seen that the radius of the spheroid at the poles (i.e., along the rotation axis, $\theta=0$) is $R_p=R\,(1-2\,\epsilon/3)$, whereas the radius at the equator (i.e., in the bisecting plane perpendicular to the axis, $\theta=\pi/2$) is $R_e=R\,(1+\epsilon/3)$. Hence,

$$\epsilon = \frac{R_e-R_p}{R}.$$ (1.330)

Let us assume that $\vert\epsilon\vert\ll 1$, so that the spheroid is very close to being a sphere. If $\epsilon>0$ then the spheroid is slightly squashed along its axis of rotation, and is termed oblate. Likewise, if $\epsilon<0$ then the spheroid is slightly elongated along its axis, and is termed prolate. See Figure 1.18. Of course, if $\epsilon=0$ then the spheroid reduces to a sphere. Note that $R$ is the surface-averaged radius of the spheroid, which implies that the volume of the spheroid is equal to that of a sphere of radius $R$. In other words, the slight squashing or elongation of the spheroid along its axis, as $\epsilon$ is varied, does not modify its volume.

Figure 1.18: Prolate and oblate spheroids.

Let ${\mit\Phi}(r,\theta)$ and $\rho(r,\theta)$ be the gravitational potential and the mass density of the spheroid, respectively. Let us write

$${\mit\Phi}(r,\theta) = {\mit\Phi}_0(r) + {\mit\Phi}_2(r)\,P_2(\cos\theta),$$ (1.331)

and

$$\rho(r,\theta) = \rho_0(r) + \rho_2(r)\,P_2(\cos\theta),$$ (1.332)

where

$$\rho_0(r) =\left\{\begin{array}{lll}\gamma&~~~~&r\leq R\\ [0.5ex]0&&r>R\end{array}\right.,$$ (1.333)

and

$$\rho_2(r) = -\frac{2}{3}\,\gamma\,R\,\epsilon\,\delta(r-R).$$ (1.334)

Here, $\delta(x)$ is a Dirac delta function. (See Section 2.1.6.) This function has the unusual property that $\delta(x)=0$ for $x\neq 0$, $\delta(x)=\infty$ at $x=0$, but

$$\int_{-\infty}^\infty \delta(x)\,dx = 1.$$ (1.335)

Thus, a Dirac delta function is an integrable spike function, centered on $x=0$, that has unit area under it. Note that $\rho_0(r)$ is the density distribution of a uniform sphere of density $\gamma$ and radius $R$. On the other hand, $\rho_2(r)\,P_2(\cos\theta)=\gamma\,[R_\theta(\theta)-R]\,\delta(r-R)$ is the density distribution obtained by taking the slight excess or deficit of surface mass, due to the deviation from sphericity of the spheroid, and placing it all at radius $R$. Note that, in writing Equation (1.331), we have assumed that an axisymmetric mass distribution (i.e., a distribution that is independent of the azimuthal angle, $\phi$) gives rise to an axisymmetric gravitational potential.

Now, in spherical coordinates, the Laplacian of ${\mit\Phi}(r,\theta)$ (i.e., a function of spherical coordinates that is independent of the azimuthal angle, $\phi$) takes the form

$$\frac{1}{r^2}\,\frac{\partial}{\partial r}\!\left(r^2\,\frac{\par... ...}{\partial\mu}\!\left[(1-\mu^2)\,\frac{\partial{\mit\Phi}}{\partial\mu}\right],$$ (1.336)

where $\mu=\cos\theta$. (See Section A.23.) Thus, according to Poisson's equation, (1.273), we can write

$$\frac{1}{r^2}\,\frac{d}{dr}\!\left(r^2\,\frac{d{\mit\Phi}_0}{dr}\right) =4\pi\,G\,\rho_0,$$ (1.337)

$$\frac{1}{r^2}\,\frac{d}{dr}\!\left(r^2\,\frac{d{\mit\Phi}_2}{dr}\right) - 6\,{\mit\Phi}_2 = 4\pi\,G\, \rho_2.$$ (1.338)

Here, we have made use of the easily proved result

$$\frac{d}{d\mu}\!\left[(1-\mu^2)\,\frac{dP_2(\mu)}{d\mu}\right]= -6\,P_2(\mu).$$ (1.339)

We have also employed the readily demonstrated result that

$$\int_{-1}^1P_2(\mu)\,d\mu=0,$$ (1.340)

which allows us to separately equate the components of Poisson's equation that are independent of $\theta$, and that vary with $\theta$ as $P_2(\cos\theta)$.

Equations (1.333) and (1.337) must be solved subject to the physical boundary conditions that ${\mit\Phi}_0(r)$ be finite at both $r=0$ and $r=\infty$, and that ${\mit\Phi}_0(r)$ and its first derivative both be continuous at $r=R$. The latter constraint ensures that the gravitational acceleration is both finite and continuous at $r=R$. It is easily seen, by inspection, that the appropriate solution is

$${\mit\Phi}_0(r) = \frac{G\,M}{2\,R}\left[\left(\frac{r}{R}\right)^2-3\right]$$ (1.341)

for $r<R$, and

$${\mit\Phi}_0(r) = -\frac{G\,M}{r}$$ (1.342)

for $r>R$. Here, $M= (4\pi/3)\,\gamma\,R^3$ is the mass of the spheroid. If we calculate the associated gravitational acceleration, ${\bf g}=-\nabla{\mit\Phi}_0$, then we can see that this solution is the same as that found for a uniform sphere in Section 1.8.3 using Gauss's law.

Equation (1.334) and (1.338) yield

$$\frac{1}{r^2}\,\frac{d}{dr}\!\left(r^2\,\frac{d{\mit\Phi}_2}{dr}\right) - 6\,{\mit\Phi}_2 =- \frac{8\pi}{3}\,G\,\gamma\,R\,\epsilon\,\delta(r-R).$$ (1.343)

Now, ${\mit\Phi}_2(r)$ must be continuous across $r=R$ otherwise the gravitational acceleration would be infinite. Hence, integrating the previous equation across $r=R$, and making use of Equation (1.335), we obtain

$$\left[\frac{d{\mit\Phi}_2}{dr}\right]_{r=R-}^{r=R+} = - \frac{8\pi}{3}\,G\,\gamma\,R\,\epsilon.$$ (1.344)

Note that the discontinuity in the gradient of ${\mit\Phi}_2$ at $r=R$ is just an artifact of the fact that we have placed all of the excess or deficit of surface mass at this radius, and is not a real phenomenon (i.e., the discontinuity would be resolved if we were to spread the mass out slightly). Now, for $r\neq R$, the Dirac delta function, $\delta(r-R)$, is zero, and Equation (1.343) reduces to

$$\frac{1}{r^2}\,\frac{d}{dr}\!\left(r^2\,\frac{d{\mit\Phi}_2}{dr}\right) - 6\,{\mit\Phi}_2 =0.$$ (1.345)

This equation must be solved subject to the physical boundary conditions that ${\mit\Phi}_2(r)$ be finite at both $r=0$ and $r=\infty$, that ${\mit\Phi}_2(r)$ be continuous at $r=R$, and that ${\mit\Phi}_2(r)$ satisfy Equation (1.344). It can be seen, by inspection, that the appropriate solution is

$${\mit\Phi}_2(r) = \frac{2}{5}\,\epsilon\,\frac{G\,M\,r^2}{R^3}$$ (1.346)

for $r<R$, and

$${\mit\Phi}_2(r) = \frac{2}{5}\,\epsilon\,\frac{G\,M\,R^2}{r^3}$$ (1.347)

for $r>R$.

Hence, we deduce that the net gravitational potential is

$${\mit\Phi}(r,\theta)= \frac{G\,M}{2\,R}\left[\left(\frac{r}{R}\right)^2-3\right] +\frac{2}{5}\,\epsilon\,\frac{G\,M\,r^2}{R^3}\,P_2(\cos\theta)$$ (1.348)

inside the spheroid, and

$${\mit\Phi}(r,\theta)=-\frac{G\,M}{r} +\frac{2}{5}\,\epsilon\,\frac{G\,M\,R^2}{r^3}\,P_2(\cos\theta)$$ (1.349)

outside the spheroid. In particular, the gravitational potential on the surface of the spheroid is

$${\mit\Phi}(R_\theta,\theta)= -\frac{G\,M}{R}\left[1+\frac{4}{15}\,\epsilon\,P_2(\cos\theta)+{\cal O}(\epsilon^2)\right].$$ (1.350)