Substitute from \Eq{(38.5)} and \Eq{(38.32)} in these, and collect the terms. The equations to be satisfied become\label{eqn:(38.6)} \begin{align*} G_{11} &= \tfrac{1}{2}\nu'' - \tfrac{1}{4} \lambda' \nu' + \tfrac{1}{4} \nu'^{2} - \lambda'/r = 0, \Tag{(38.61)}\displaybreak[0] \\ G_{22} &= e^{-\lambda} \bigl(1 + \tfrac{1}{2}r(\nu' - \lambda')\bigr) - 1 = 0, \Tag{(38.62)}\displaybreak[0] \\ G_{33} &= \sin^{2}\theta · e^{-\lambda} \bigl(1 + \tfrac{1}{2}r(\nu' - \lambda')\bigr) - \sin^{2}\theta = 0, \Tag{(38.63)}\displaybreak[0] \\ G_{44} &= e^{\nu-\lambda} (-\tfrac{1}{2} \nu'' + \tfrac{1}{4} \lambda' \nu' - \tfrac{1}{4} \nu'^{2} - \nu'/r) = 0, \Tag{(38.64)} \\ %[** TN: Omitted extra "= 0"] G_{12} &= 0. \Tag{(38.65)} \end{align*} We may leave aside \Eq{(38.63)} which is a mere repetition of~\Eq{(38.62)}; then there are left three equations to be satisfied by $\lambda$ and~$\nu$. From \Eq{(38.61)} and \Eq{(38.64)} we have $\lambda' = -\nu'$. Since $\lambda$ and~$\nu$ are to vanish together at $r = \infty$, this requires that \[ \lambda = -\nu. \] Then \Eq{(38.62)} becomes \[ e^{\nu} (1 + r\nu') = 1. \] Set $e^{\nu} = \gamma$, then \[ \gamma + r\gamma' = 1. \] Hence, integrating, \[ \gamma = 1 - \frac{2m}{r}, \Tag{(38.7)} \] where $2m$~is a constant of integration. It will be found that all three equations are satisfied by this solution. Accordingly, substituting $e^{-\lambda} = e^{\nu} = \gamma$ in~\Eq{(38.2)}, \[ ds^{2} = -\gamma^{-1}\, dr^{2} - r^{2}\, d\theta^{2} - r^{2} \sin^{2}\theta\, d\phi^{2} + \gamma\, dt^{2}, \Tag{(38.8)} \] where $\gamma = 1 - 2m/r$, is a particular solution of Einstein's gravitational equations $G_{\mu\nu} = 0$. The solution in this form was first obtained by Schwarzschild. \Section{39.}{Planetary orbits} \index{Orbits of planets}% \index{Planetary orbits}% According to~\Eq{(15.7)} the track of a particle moving freely in the space-time given by~\Eq{(38.8)} is determined by the equations of a geodesic~\Eq{(28.5)}, viz.\ \[ \frac{d^{2}x_{\alpha}}{ds^{2}} + \{\mu\nu, \alpha\}\, \frac{dx_{\mu}}{ds}\, \frac{dx_{\nu}}{ds} = 0. \Tag{(39.1)} \]