= Solution
For a differentiable <central potential>, the <force> is $\mathbf F=-V'(r)\hat{\mathbf r}$. The <angular momentum> is $\mathbf J=m\mathbf r\times\dot{\mathbf r}$, so
$$
\frac{d\mathbf J}{dt}=\mathbf r\times\mathbf F=0.
$$
This proves <conservation of angular momentum>. For $J\ne0$, the trajectory lies in the fixed plane perpendicular to $\mathbf J$. In plane <polar coordinates>, $J=mr^2\dot\phi$. Substituting this into the <kinetic energy> gives the radial <energy> equation
$$
E=\frac12m\dot r^2+V_{\mathrm{eff}}(r),\qquad
\boxed{V_{\mathrm{eff}}(r)=V(r)+\frac{J^2}{2mr^2}.}
$$
The second term is the centrifugal contribution to the <effective potential>. Its radial derivative reproduces $m\ddot r=-V'(r)+J^2/(mr^3)$. 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> give
$$
r_{\mathrm A}v_{\mathrm A}=r_{\mathrm P}v_{\mathrm P},\qquad
\frac12v_{\mathrm A}^2-\frac{\mu}{r_{\mathrm A}}
=\frac12v_{\mathrm P}^2-\frac{\mu}{r_{\mathrm P}}.
$$
Eliminating $v_{\mathrm P}$ yields
$$
v_{\mathrm A}^2=\frac{2\mu r_{\mathrm P}}{r_{\mathrm A}(r_{\mathrm A}+r_{\mathrm P})}.
$$
For an <elliptic orbit>, $r_{\mathrm A}+r_{\mathrm P}=2a$. Therefore the <vis-viva equation> at <apogee> is
$$
\boxed{v_{\mathrm A}^2=GM\left(\frac2{r_{\mathrm A}}-\frac1a\right).}
$$
Substitution 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 $v\mapsto(1-\epsilon)v$, the <orbital energy> change is
$$
\Delta E=\frac12m\left((1-\epsilon)^2-1\right)v^2
=-\epsilon mv^2+O(\epsilon^2).
$$
Differentiating $E=-\mu m/(2a)$ gives $\Delta E=\mu m\,\Delta a/(2a^2)$ to first order. Since $r_{\mathrm A}=a(1+e)$, the <vis-viva equation> implies $v^2=\mu(1-e)/[a(1+e)]$. Hence
$$
\Delta a=-2\epsilon a\frac{1-e}{1+e}+O(\epsilon^2),\qquad
\boxed{\text{reduction of the major axis}= -2\Delta a
=4\epsilon a\frac{1-e}{1+e}+O(\epsilon^2).}
$$
The factor of two distinguishes the major axis from the <semimajor axis>. As a finite-change check, the new <semimajor axis> is exactly
$$
a'=\frac{a}{1+(2\epsilon-\epsilon^2)(1-e)/(1+e)},
$$
within the instantaneous-burn, unchanged-mass model. Expanding this expression gives the stated result.
Back to article page
