Combining Equations (D.2) and (D.3) with the previous three equations, we deduce that, to first order in
, the total potential (i.e., the sum of the gravitational
and centrifugal potentials) can be written
![$\displaystyle U(a,\theta)= U_0(a)+U_2(a)\,P_2(\cos\theta),$](img4352.png) |
(D.22) |
where
Here, we have assumed that
. However, according to Equation (D.7), if the rotating body is in hydrostatic equilibrium (in the
co-rotating frame) then
is a function of
only. In other words,
, which implies that
![$\displaystyle a^{\,2}\,\epsilon(a)\int_0^a \rho(a')\,a'^{\,2}\,da' -\frac{1}{5}...
...^\infty \rho(a')\,d[\epsilon(a')]
=\frac{{\mit\Omega}^{\,2}\,a^{\,5}}{8\pi\,G}.$](img4357.png) |
(D.24) |
Differentiation with respect to
yields
![$\displaystyle \left(a\,\frac{d\epsilon}{da}+2\,\epsilon\right)\bar{\rho}(a) -3\...
...y \rho(a')\,\frac{d\epsilon}{da'}\,da' =\frac{15\,{\mit\Omega}^{\,2}}{8\pi\,G},$](img4358.png) |
(D.25) |
where
![$\displaystyle \bar{\rho}(a)=\frac{3}{a^{\,3}}\int_0^a \rho(a')\,a'^{\,2}\,da'$](img4359.png) |
(D.26) |
is the mean density inside the spheroidal surface
. Note that
![$\displaystyle \frac{\rho}{\bar{\rho}} =1 +\frac{1}{3}\,\frac{a}{\bar{\rho}}\,\frac{d\bar{\rho}}{da}.$](img4361.png) |
(D.27) |
Finally, differentiation of Equation (D.25) with respect to
gives
![$\displaystyle a^{\,2}\,\frac{d^{\,2}\epsilon}{da^{\,2}}-6\,\epsilon + 6\,\frac{\rho}{\bar{\rho}}\left(a\,\frac{d\epsilon}{da}+\epsilon\right)= 0.$](img4362.png) |
(D.28) |
This differential equation was first obtained by Clairaut in 1743 (Cook 1980).
Suppose that the outer boundary of the rotating body corresponds to
, where
is the body's mean radius. [In other words,
for
.] It follows that the
total mass of the body is
![$\displaystyle M = \frac{4\pi}{3}\,R^{\,3}\,\bar{\rho}(R),$](img4366.png) |
(D.29) |
The dimensionless parameter
, introduced in Section 6.5, is the typical ratio of the centrifugal acceleration to the gravitational
acceleration at
, and takes the form
![$\displaystyle \zeta =\frac{{\mit\Omega}^{\,2}\,R^{\,3}}{3\,G\,M} = \frac{{\mit\Omega}^{\,2}}{4\pi\,G\,\bar{\rho}(R)}.$](img4367.png) |
(D.30) |
Thus, it follows from Equation (D.25) that
![$\displaystyle \left(a\,\frac{d\epsilon}{da} +2\,\epsilon\right)_{a=R} = \frac{15}{2}\,\zeta.$](img4368.png) |
(D.31) |
Now, at an extremum of
, we have
. At such a point, Equation (D.28) yields
![$\displaystyle a^{\,2}\,\frac{d^{\,2}\epsilon}{da^{\,2}}=6\left(1-\frac{\rho}{\bar{\rho}}\right)\epsilon.$](img4370.png) |
(D.32) |
However, if
is a monotonically decreasing function of
, as we would generally expect to be the case, then Equation (D.27)
reveals that
. Hence, the previous equation implies that, at the extremum,
has the same sign as
. In other words, the extremum is a minimum of
. This implies that it is impossible to have a maximum of
. Now, a Taylor expansion of Equation (D.27) about
, assuming that
, where
, reveals that
is an increasing function at small
. We, thus, deduce that
is
a monotonically increasing function. This implies that
has the same sign as
. Hence, Equation (D.31) reveals that
is everywhere positive. In other words, if
is a monotonically decreasing function then
is necessarily a positive, monotonically
increasing function. Thus, we deduce that all density contours in the body are oblate spheroids.