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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

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


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

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

Differentiation with respect to $ a$ yields

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


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

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

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

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

$\displaystyle a^{\,2}\,\frac{d^{\,2}\epsilon}{da^{\,2}}-6\,\epsilon + 6\,\frac{\rho}{\bar{\rho}}\left(a\,\frac{d\epsilon}{da}+\epsilon\right)= 0.$ (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

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

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

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

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

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

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

next up previous
Next: Derivation of Radau equation Up: Darwin-Radau equation Previous: Calculation of gravitational potential
Richard Fitzpatrick 2016-03-31