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For a differentiable central potential, the force is $\mathbf{F}=-V^{\prime }(r)\hat{\mathbf{r}}$. The angular momentum is $\mathbf{J}=m\mathbf{r}\times \dot{\mathbf{r}}$, soThis proves conservation of angular momentum. For $J\ne 0$, the trajectory lies in the fixed plane perpendicular to $\mathbf{J}$. In plane polar coordinates, $J=m{r}^2\dot{\varphi }$. Substituting this into the kinetic energy gives the radial energy equationThe second term is the centrifugal contribution to the effective potential. Its radial derivative reproduces $m\ddot{r}=-V^{\prime }(r)+J^2/(m{r}^3)$. Purely radial motion is the $J=0$ case.

For the satellite, write $\mu =GM$, with $M$ Earth's mass and $m$ the satellite's mass. At apogee and perigee, the radial velocity is zero, so angular momentum conservation and conservation of energy giveEliminating $v_{\mathrm{P}}$ yieldsFor an elliptic orbit, $r_{\mathrm{A}}+r_{\mathrm{P}}=2a$. Therefore the vis-viva equation at apogee isSubstitution into the orbital energy gives $E=-GMm/(2a)$. The circular case follows by continuity, or directly from the radial force balance.

An instantaneous tangential retrograde burn at apogee leaves the position unchanged. For the speed change $v\mapsto (1-ϵ)v$, the orbital energy change isDifferentiating $E=-\mu m/(2a)$ gives $\Delta E=\mu m\Delta a/(2a^2)$ to first order. Since $r_{\mathrm{A}}=a(1+e)$, the vis-viva equation implies $v^2=\mu (1-e)/[a(1+e)]$. HenceThe factor of two distinguishes the major axis from the semimajor axis. As a finite-change check, the new semimajor axis is exactlywithin the instantaneous-burn, unchanged-mass model. Expanding this expression gives the stated result.

The periapsis of an Earth-centred orbit. For an elliptic orbit its distance from the Earth's centre is $a(1-e)$. Both this distance and the apogee distance are measured from the force centre, rather than as heights above Earth's surface.