Derivation and properties of Clairaut equation

Combining Equations (D.2) and (D.3) with the previous three equations, we deduce that, to first order in $\vert\epsilon\vert$, the total potential (i.e., the sum of the gravitational and centrifugal potentials) can be written

$$U(a,\theta)= U_0(a)+U_2(a)\,P_2(\cos\theta),$$ (D.22)

where

$$U_2(a) =-\frac{8\pi\,G}{3}\left\{\frac{\epsilon(a)}{a}\int_0^a \rho(a')\... ...2}\,da'-\frac{1}{5\,a^{\,3}}\int_0^a \rho(a')\,d[\epsilon(a')\,a'^{\,5}]\right.$$

$$\left.\phantom{=}-\frac{a^{\,2}}{5}\int_a^\infty \rho(a')\,d[\epsilon(a')]\right\} +\frac{{\mit\Omega}^{\,2}\,a^{\,2}}{3}.$$ (D.23)

Here, we have assumed that $\chi \sim {\cal O}(\vert\epsilon\vert)$. However, according to Equation (D.7), if the rotating body is in hydrostatic equilibrium (in the co-rotating frame) then $U$ is a function of $a$ only. In other words, $U_2(a)=0$, which implies that

$$a^{\,2}\,\epsilon(a)\int_0^a \rho(a')\,a'^{\,2}\,da' -\frac{1}{5}... ...^\infty \rho(a')\,d[\epsilon(a')] =\frac{{\mit\Omega}^{\,2}\,a^{\,5}}{8\pi\,G}.$$ (D.24)

Differentiation with respect to $a$ yields

$$\left(a\,\frac{d\epsilon}{da}+2\,\epsilon\right)\bar{\rho}(a) -3\... ...y \rho(a')\,\frac{d\epsilon}{da'}\,da' =\frac{15\,{\mit\Omega}^{\,2}}{8\pi\,G},$$ (D.25)

where

$$\bar{\rho}(a)=\frac{3}{a^{\,3}}\int_0^a \rho(a')\,a'^{\,2}\,da'$$ (D.26)

is the mean density inside the spheroidal surface $r=a\,[1-(2/3)\,\epsilon(a)\,P_2(\cos\theta)]$. Note that

$$\frac{\rho}{\bar{\rho}} =1 +\frac{1}{3}\,\frac{a}{\bar{\rho}}\,\frac{d\bar{\rho}}{da}.$$ (D.27)

Finally, differentiation of Equation (D.25) with respect to $a$ gives

$$a^{\,2}\,\frac{d^{\,2}\epsilon}{da^{\,2}}-6\,\epsilon + 6\,\frac{\rho}{\bar{\rho}}\left(a\,\frac{d\epsilon}{da}+\epsilon\right)= 0.$$ (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 $a=R$, where $R$ is the body's mean radius. [In other words, $\rho(a)=0$ for $a>R$.] It follows that the total mass of the body is

$$M = \frac{4\pi}{3}\,R^{\,3}\,\bar{\rho}(R),$$ (D.29)

The dimensionless parameter $\zeta$, introduced in Section 6.5, is the typical ratio of the centrifugal acceleration to the gravitational acceleration at $a=R$, and takes the form

$$\zeta =\frac{{\mit\Omega}^{\,2}\,R^{\,3}}{3\,G\,M} = \frac{{\mit\Omega}^{\,2}}{4\pi\,G\,\bar{\rho}(R)}.$$ (D.30)

Thus, it follows from Equation (D.25) that

$$\left(a\,\frac{d\epsilon}{da} +2\,\epsilon\right)_{a=R} = \frac{15}{2}\,\zeta.$$ (D.31)

Now, at an extremum of $\epsilon(a)$, we have $d\epsilon/da=0$. At such a point, Equation (D.28) yields

$$a^{\,2}\,\frac{d^{\,2}\epsilon}{da^{\,2}}=6\left(1-\frac{\rho}{\bar{\rho}}\right)\epsilon.$$ (D.32)

However, if $\bar{\rho}(a)$ is a monotonically decreasing function of $a$, as we would generally expect to be the case, then Equation (D.27) reveals that $\rho/\bar{\rho}<1$. Hence, the previous equation implies that, at the extremum, $d^{\,2}\epsilon/da^{\,2}$ has the same sign as $\epsilon$. In other words, the extremum is a minimum of $\vert\epsilon\vert$. This implies that it is impossible to have a maximum of $\vert\epsilon\vert$. Now, a Taylor expansion of Equation (D.27) about $a=0$, assuming that $\rho/\bar{\rho}\rightarrow \beta$, where $0<\beta< 1$, reveals that $\vert\epsilon\vert(a)$ is an increasing function at small $a$. We, thus, deduce that $\vert\epsilon\vert(a)$ is a monotonically increasing function. This implies that $d\epsilon/da$ has the same sign as $\epsilon$. Hence, Equation (D.31) reveals that $\epsilon(a)$ is everywhere positive. In other words, if $\bar{\rho}(a)$ is a monotonically decreasing function then $\epsilon(a)$ is necessarily a positive, monotonically increasing function. Thus, we deduce that all density contours in the body are oblate spheroids.