Substitute this first approximation in the small term~$3mu^{2}$, then \Eq{(40.1)} becomes \[ \frac{d^{2}u}{d\phi^{2}} + u = \frac{m}{h^{2}} + 3\frac{m^{3}}{h^{4}} + 6\frac{m^{3}}{h^{4}} e\cos(\phi - \varpi) + \frac{3}{2}\, \frac{m^{3}}{h^{4}} e^{2} \bigl(1 + 2\cos(\phi - \varpi)\bigr). %[** TN: Handwritten correction \cos 2(\phi - \varpi); unable to verify] \Tag{(40.3)} \] Of the additional terms the only one which can produce an effect within the range of observation is the term in $\cos(\phi - \varpi)$; this is of the right period to produce a continually increasing effect by resonance. Remembering that the particular integral of \[ \frac{d^{2} u}{d\phi^{2}} + u = A\cos\phi \] is \[ u = \tfrac{1}{2}A \phi \sin\phi, \] this term gives a part of~$u$ \[ u_{1} = 3\frac{m^{3}}{h^{4}} e\phi \sin(\phi - \varpi), \Tag{(40.4)} \] which must be added to the complementary integral~\Eq{(40.2)}. Thus the second approximation is \begin{align*} u &= \frac{m}{h^{2}} \Bigl(1 + e\cos(\phi - \varpi) + 3 \frac{m^{2}}{h^{2}} e\phi \sin(\phi - \varpi)\Bigr) \\ &= \frac{m}{h^{2}} \bigl(1 + e\cos(\phi - \varpi - \delta\varpi)\bigr), \end{align*} where \[ \delta\varpi = 3\frac{m^{2}}{h^{2}} \phi, \Tag{(40.5)} \] and $(\delta\varpi)^{2}$~is neglected. Whilst the planet moves through $1$~revolution, the perihelion~$\varpi$ advances a fraction of a revolution equal to \[ \frac{\delta\varpi}{\phi} = \frac{3m^{2}}{h^{2}} = \frac{3m}{a(1 - e^{2})}, \Tag{(40.6)} \] using the well-known equation of areas $h^{2} = ml = ma(1 - e^{2})$.% \PageSep{89} Another form is obtained by using Kepler's third law, \index{Kepler's third law}% \[ m = \left(\frac{2\pi}{T}\right)^{2} a^{3}, \] giving \[ \frac{\delta\varpi}{\phi} = \frac{12\pi^{2} a^{2}}{c^{2} T^{2}(1 - e^{2})}, \Tag{(40.7)} \] where $T$~is the period, and the velocity of light~$c$ has been reinstated. This advance of the perihelion is appreciable in the case of the planet Mercury, and the predicted value is confirmed by observation. For a circular orbit we put $dr/ds$, $d^{2}r/ds^{2} = 0$, so that \Eq{(39.31)}~becomes \[ -re^{-\lambda} \left(\frac{d\phi}{ds}\right)^{2} + \tfrac{1}{2} e^{\nu-\lambda} \nu' \left(\frac{dt}{ds}\right)^{2} = 0. \] Whence \begin{align*} \left(\frac{d\phi}{dt}\right)^{2} &= \tfrac{1}{2} e^{\nu} \nu'/r = \tfrac{1}{2} \gamma'/r \\ &= m/r^{3}, \end{align*} so that Kepler's third law is \emph{accurately} fulfilled. This result has no observational significance, being merely a property of the particular definition of~$r$ here adopted. Slightly different coordinate-systems exist which might with equal right claim to correspond to polar coordinates in flat space-time; and for these Kepler's third law would no longer be exact.