Potential Outside a Uniform Spheroid

Let us now calculate the gravitational potential generated outside a spheroid of uniform mass density $\gamma$ and mean radius $a$. A spheroid is the solid body produced by rotating an ellipse about a major or minor axis. Let the axis of rotation coincide with the $z$-axis, and let the outer boundary of the spheroid satisfy

$$r = a_\theta(\theta) = a\left[1-\frac{2}{3}\,\epsilon\,P_2(\cos\theta)\right],$$ (132)

where $\epsilon$ is the termed the ellipticity. Here, we are assuming 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 symmetry axis, and is termed oblate. Likewise, if $\epsilon<0$ then the spheroid is slightly elongated along its axis, and is termed prolate--see Figure 5. Of course, if $\epsilon=0$ then the spheroid reduces to a sphere.

Figure 5: Prolate and oblate spheroids.

Now, according to Equation (121) and (122), the gravitational potential generated outside an axially symmetric mass distribution can be written

$$\Phi(r,\theta) = \sum_{n=0,\infty} J_n\,\frac{P_n(\cos\theta)}{r^{n+1}},$$ (133)

where

$$J_n = - 2\pi\,G\,\int\int r^{\,2+n}\,\rho(r,\theta)\,P_n(\cos\theta)\,\sin\theta\,dr\,d\theta.$$ (134)

Here, the integral is taken over the whole cross-section of the distribution in $r$-$\theta$ space.

It follows that for a uniform spheroid

$$J_n = - 2\pi\,G\,\gamma\int_0^\pi P_n(\cos\theta)\,\int_0^{a_\theta(\theta)}r^{\,2+n}\,dr\,\sin\theta\,d\theta$$ (135)

Hence,

$$J_n = -\frac{2\pi\,G\,\gamma}{(3+n)}\int_0^\pi P_n(\cos\theta)\,a_\theta^{3+n}(\theta)\,\sin\theta\,d\theta,$$ (136)

giving

$$J_n \simeq -\frac{2\pi\,G\,\gamma\,a^{3+n}}{(3+n)}\int_0^\pi... ...}\,(3+n)\,\epsilon\,P_2(\cos\theta)\right]\sin\theta\,d\theta,$$ (137)

to first-order in $\epsilon$. It is thus clear, from Equation (118), that, to first-order in $\epsilon$, the only non-zero $J_n$ are

$$J_0 = - \frac{4\pi\,G\,\gamma\,a^3}{3} = - G\,M,$$ (138)

$$J_2 = \frac{8\pi\,G\,\gamma\,a^5\,\epsilon}{15} = \frac{2}{5}\,G\,M\,a^2\,\epsilon,$$ (139)

where $M=(4\pi/3)\,a^3\,\gamma$ is the total mass.

Thus, the gravitational potential outside a uniform spheroid of total mass $M$, mean radius $a$, and ellipticity $\epsilon$, is

$$\Phi(r,\theta) = - \frac{G\,M}{r} +\frac{2}{5}\frac{G\,M\,a^2}{r^3}\,\epsilon\,P_2(\cos\theta) + {\cal O}(\epsilon^2).$$ (140)

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

$$\Phi(a_\theta, \theta) = - \frac{G\,M}{a_\theta} +\frac{2}{5... ...heta^{\,3}}\,\epsilon\,P_2(\cos\theta) + {\cal O}(\epsilon^2),$$ (141)

which yields

$$\Phi(a_\theta,\theta) \simeq - \frac{G\,M}{a} \left[1+\frac{4}{15}\,\epsilon\,P_2(\cos\theta) + {\cal O}(\epsilon^2)\right],$$ (142)

where use has been made of Equation (132).

Consider a self-gravitating spheroid of mass $M$, mean radius $a$, and ellipticity $\epsilon$: e.g., a star, or a planet. Assuming, for the sake of simplicity, that the spheroid is composed of uniform density incompressible fluid, the gravitational potential on its surface is given by Equation (142). However, the condition for an equilibrium state is that the potential be constant over the surface. If this is not the case then there will be gravitational forces acting tangential to the surface. Such forces cannot be balanced by internal pressure, which only acts normal to the surface. Hence, from (142), it is clear that the condition for equilibrium is $\epsilon=0$. In other words, the equilibrium configuration of a self-gravitating mass is a sphere. Deviations from this configuration can only be caused by forces in addition to self-gravity and internal pressure: e.g., internal tensile forces, centrifugal forces due to rotation, or tidal forces due to orbiting masses (see Chapter 6).

We can estimate how small a rocky asteroid or moon needs to be before its internal tensile strength is sufficient to allow it to retain a significantly non-spherical shape. The typical density of rock is $\gamma\sim 3\times 10^3\,{\rm kg}\,{\rm m}^{-3}$. Moreover, the critical compressional stress at which rock ceases to act like a rigid material, and instead deforms and flows like a liquid, is $P_c\sim 10^7\,{\rm N}\,{\rm m}^{-2}$. We must compare this critical stress with the pressure at the center of the asteroid or moon. Assuming, for the sake of simplicity, that the asteroid or moon is spherical, of radius $a$, and of uniform density $\gamma$, the central pressure is

$$P_0 = \int_0^a\,\gamma\,g(r)\,dr,$$ (143)

where $g(r)= (4\pi/3)\,G\,\gamma\,r$ is the gravitational acceleration at radius $r$ [see Equation (129)]. The above result is a simple generalization of the well-known formula $\rho\,g\,h$ for the pressure a depth $h$ below the surface of a fluid. It follows that

$$P_0 = \frac{2\pi}{3}\,G\,\gamma^2\,a^2.$$ (144)

Now, if $P_0\ll P_c$ then the internal pressure in the asteroid or moon is not sufficiently high to cause its constituent rock to deform like a liquid. Such an asteroid or moon can therefore retain a significantly non-spherical shape. On the other hand, if $P_0\gg P_c$ then the internal pressure is large enough to render the asteroid or moon fluid-like. Such a body cannot withstand the tendency of self-gravity to make it adopt a spherical shape. The condition $P_0\ll P_c$ is equivalent to $a\ll a_c$, where

$$a_c = \left(\frac{3}{2\pi}\,\frac{P_c}{G\,\gamma^2}\right)^{1/2}\sim 100\,{\rm km}.$$ (145)

It follows that only a rocky asteroid or moon whose radius is significantly less than about $100\,{\rm km}$ (e.g., the two moons of Mars, Phobos and Deimos) can retain a highly non-spherical shape. On the other hand, an asteroid or moon whose radius is significantly greater than about $100\,{\rm km}$ (e.g., the asteroid Ceres, and the Earth's moon) is forced to be essentially spherical.