Derivation of Radau equation


$\displaystyle \eta = \frac{a}{\epsilon}\,\frac{d\epsilon}{da}.$ (D.33)

It follows that

$\displaystyle \frac{d\epsilon}{d\eta}$ $\displaystyle =\frac{\epsilon\,\eta}{a},$ (D.34)
$\displaystyle \frac{d^{\,2}\epsilon}{d a^{\,2}}$ $\displaystyle =\left[\frac{1}{a}\,\frac{d\eta}{da}+\frac{\eta\,(\eta-1)}{a^{\,2}}\right].$ (D.35)

Equation (D.28) transforms into

$\displaystyle a\,\frac{d\eta}{da} + \eta^{\,2} -\eta -6 + 6\,\frac{\rho}{\bar{\rho}}\,(1+\eta)=0.$ (D.36)

Making use of Equation (D.27), this reduces to

$\displaystyle a\,\frac{d\eta}{da} + \eta^{\,2}+5\,\eta + 2\,\frac{a}{\bar{\rho}}\,\frac{d\bar{\rho}}{da}\,(1+\eta)= 0.$ (D.37)

However, it is easily demonstrated that

$\displaystyle \frac{\left.d\left[\bar{\rho}\,a^{\,5}\,(1+\eta)^{1/2}\right]\rig...
+\frac{1}{2\,(1+\eta)}\,\frac{d\eta}{da}.$ (D.38)

Hence, Equation (D.37) yields Radau's equation (Radau 1885),

$\displaystyle \frac{d}{da}\left[\bar{\rho}\,a^{\,5}\,(1+\eta)^{1/2}\right] = 5\,\bar{\rho}\,a^{\,4}\,\psi(\eta),$ (D.39)


$\displaystyle \psi(\eta) =\frac{1+\eta/2-\eta^{\,2}/10}{(1+\eta)^{1/2}}.$ (D.40)

Figure: D.1 The function $\psi (\eta )$.

Figure D.1 shows the function $\psi (\eta )$. It should be noted that $\eta(a)$ is necessarily positive, because $\epsilon(a)$ is a monotonically increasing function. Moreover, $\eta=0.57$ on the surface of the Earth, and the largest value of $\eta$ for any planet in the solar system is about $1.4$ (Cook 1980). It is thus clear, from the figure, that it is an excellent approximation to take $\psi(\eta)\simeq 1$ for such bodies. This leads to the so-called reduced Radau equation,

$\displaystyle \frac{d}{da}\left[\bar{\rho}\,a^{\,5}\,(1+\eta)^{1/2}\right] \simeq 5\,\bar{\rho}\,a^{\,4},$ (D.41)

which can immediately be integrated to give

$\displaystyle (1+\eta)^{1/2} =\frac{5}{\bar{\rho}\,a^{\,5}}\int_0^a\bar{\rho}(a')\,a'^{\,4}\,da'.$ (D.42)