06 Jun 2025
Kepler's Laws and Their Derivation
Derivation of elliptical planetary orbits, equal areas, and the period law from inverse-square gravity.
For a planet of mass $m$ orbiting a much heavier Sun of mass $M$, Newtonian gravity is
\[\mathbf F=-\frac{GMm}{r^2}\hat{\mathbf r}.\]Let
\[h=\frac{L}{m}=r^2\dot\theta\]be the constant angular momentum per unit mass, and put $u=1/r$. A prime will denote differentiation with respect to $\theta$. Since $r=u^{-1}$ and $\dot\theta=hu^2$,
\[\dot r=\frac{dr}{d\theta}\dot\theta =\left(-\frac{u'}{u^2}\right)(hu^2) =-hu'.\]Differentiating with respect to time,
\[\ddot r=-h\frac{du'}{dt} =-hu''\dot\theta =-h^2u^2u''.\]Also,
\[r\dot\theta^2 =\frac1u(hu^2)^2 =h^2u^3.\]Substitution in the radial equation gives
\[m(\ddot r-r\dot\theta^2) =-mh^2u^2(u''+u) =-GMmu^2.\]Cancelling the nonzero factor $-mu^2$ leaves Binet’s equation for gravity:
\[u''+u=\frac{GM}{h^2}.\]Its general solution is
\[u=\frac{GM}{h^2}+C\cos(\theta-\theta_0).\]Choose the direction of closest approach as $\theta=0$ and write $C=eGM/h^2$, where $e\ge0$. Then
\[\boxed{r(\theta)=\frac{p}{1+e\cos\theta}}, \qquad \boxed{p=\frac{h^2}{GM}}.\]First law: the orbit is an ellipse
For a bound planetary orbit, $0\le e<1$. With $x=r\cos\theta$ and $r^2=x^2+y^2$, the polar orbit equation gives
\[r+ex=p, \qquad r=p-ex.\]Squaring the second equation,
\[x^2+y^2=(p-ex)^2,\]and hence
\[(1-e^2)x^2+2pex+y^2=p^2.\]Completing the square in $x$ gives
\[(1-e^2)\left(x+\frac{pe}{1-e^2}\right)^2+y^2 =\frac{p^2}{1-e^2}.\]Define
\[a=\frac{p}{1-e^2}, \qquad b=\frac{p}{\sqrt{1-e^2}}=a\sqrt{1-e^2}.\]Then
\[\boxed{\frac{(x+ae)^2}{a^2}+\frac{y^2}{b^2}=1}.\]This is an ellipse of semi-major axis $a$, semi-minor axis $b$, and focal distance $ae$. One focus is the origin, where the Sun lies. Therefore Kepler’s first law states: each planet moves in an ellipse with the Sun at one focus.
Second law: equal areas in equal times
Gravity is central, so $L$ and $h=L/m$ are constant. The areal velocity is
\[\boxed{\frac{dA}{dt}=\frac{L}{2m}=\frac h2=\text{constant}}.\]Thus the Sun-planet radius vector sweeps equal areas in equal times. This is Kepler’s second law.
Third law: the period relation
In one period $T$, the radius vector sweeps the whole area $\pi ab$ of the ellipse. Since $dA/dt=h/2$,
\[\pi ab=\frac h2T,\]so
\[T=\frac{2\pi ab}{h}.\]From $p=h^2/(GM)$ and $p=a(1-e^2)$,
\[h^2=GMa(1-e^2).\]Using $b^2=a^2(1-e^2)$ and squaring the period,
\[T^2=\frac{4\pi^2a^2b^2}{h^2} =\frac{4\pi^2a^4(1-e^2)}{GMa(1-e^2)}.\]Therefore
\[\boxed{T^2=\frac{4\pi^2}{GM}a^3}, \qquad \boxed{T^2\propto a^3}.\]This is Kepler’s third law: for bodies orbiting the same central mass, the square of the orbital period is proportional to the cube of the semi-major axis.
The Binet equation, ellipse reduction, and third-law cancellation are verified in the Maxima worksheet; every printed residual is zero.
Discussion