How can analytically be derived the Kepler's laws?

I found some extremely synthetic equations which from the Newton's laws (in particular $\mathbf{F} = m \mathbf{a}$) tried to obtain the Kepler's laws, but even if it seemed to be a nice procedure they were too much incomplete.

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    $\begingroup$ Read the classics: Classical Mechanics by Goldstein, Mathematical Methods of Classical Mechanics by Arnold, and Mechanics by Landau and Lifshitz. Goldstein is quite detailed, the other two are, in their characteristic style, not quite so detailed. $\endgroup$ May 14, 2014 at 13:31

5 Answers 5


This is the topic of Chapter 8 of Marion & Thornton's Classical Mechanics.

Kepler's second law (equal areas in equal times) is a consequence of angular momentum conservation, $$ \ell = \mu r^2 \dot\theta = \text{constant}, $$ (with reduced mass $\mu$ and coordinates $r$ and $\theta$) because the infinitesimal area swept out per unit time is $$ dA = \frac12 r^2 d\theta = \frac{\ell}{2\mu}dt. $$ This means that the time to sweep out the entire area is $\tau=2\mu A/\ell$, which we'll come back to later.

The first law comes from the equation of motion. The energy of the system is

$$ E = \frac12 \mu\dot r^2 + \frac12 \frac{\ell^2}{\mu r^2} - \frac kr $$

which you can solve for $\dot r$ and integrate to find $r(t)$. (For gravitation, the constant $k=GM\mu$, where $M$ is the total mass of the two interacting bodies.) Ignoring the mathematicians who cry "that's not how differentials work!", we can use the substitution $$ d\theta = \frac{d\theta}{dt} \frac{dt}{dr} dr = \frac{\dot\theta}{\dot r} dr, $$ eliminate $\dot\theta$ using $\ell$, and find $$ \theta(r) = \int \frac{± (\ell/r^2) dr}{\sqrt{2\mu\left( E+\frac kr - \frac{\ell^2}{2\mu r^2} \right)}}. $$

The solution to this integral shows that the orbit is a conic section $$ \begin{align} \frac\alpha r &= 1 + \epsilon\cos\theta & \alpha &= \frac{\ell^2}{\mu k} & \epsilon &= \sqrt{1 + \frac{2E\ell^2}{\mu k^2}} \end{align}. $$ Closed conic sections are ellipses with semi-major and semi-minor axes $a$ and $b$ related by $b=\sqrt{\alpha a}$, and area $\pi ab$. We already learned the time required to sweep out the area of the ellipse $\tau\propto A$, and so we immediately get Kepler's third law $\tau \propto a^{3/2}$.

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    $\begingroup$ Trying to follow your equations: it is up to now quite hard, but thank you so much. Maybe you are referring to Thornton, Marion "Classical Dynamics of Particles and Systems"? $\endgroup$
    – BowPark
    May 19, 2014 at 8:55
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    $\begingroup$ That's the book. Thornton apparently waited two editions and more than twenty years to re-order the names after Marion's death. $\endgroup$
    – rob
    May 19, 2014 at 11:54

If rob's answer is a bit terse for you, see "A self-contained derivation of Kepler's laws from Newton's laws", which assumes less prior knowledge and proceeds in smaller steps. (Yes, I wrote it.)


I am not very familiar with this topic but here is a proof for Kepler's third law in the special case of a circular orbit.

Considering a circular orbit, Kepler's third law states that the square of the orbital period is proportional to the cube of the radius, i.e. $T^2 \propto r^3$.

The period of circular motion is given by: $$T=\frac{2\pi r}{v}$$ Square both sides gives: $$T^2=\frac{4\pi^2 r^2}{v^2}$$ Since the acceleration of circular motion is $a=\frac{v^2}{r}$, we get $v^2=ar$. Substituting this gives: $$T^2=\frac{4\pi^2r^2}{ar}=\frac{4\pi^2r}{a}$$ By Newton's law of gravitation, $F=\frac{GMm}{r^2}$, we get $a=\frac{GM}{r^2}$. Substituting this gives: $$T^2=4\pi^2r\frac{r^2}{GM}$$ $$T^2=\frac{4\pi^2}{GM}r^3$$


Since the derivation of the Keppler's first law given in the other answers involves a non-trivial integration I think it is worth to see an easier way.

Let $\vec p$ and $\vec L$ are the momentum and angular momentum of the planet, respectively, $m$ its mass, $K$ comes from the gravitational force $\vec F=-K\hat r/r^2$, and $\hat r$ is the radial unit vector $\hat r=\vec r/r$.

I will define the so-called Runge-Lez vector, $$\vec A=\frac{\vec p\times\vec L}{mK}-\hat r.$$ The square of this vector is $$A^2=\vec A\cdot\vec A=1+\frac{p^2L^2}{m^2K^2}-\frac{2L^2}{mKr}.$$ To get this result we need to use $\hat r\cdot(\vec p\times\vec L)=\vec L\cdot (\vec r\times\vec p)/r=L^2/r$. The mecanical energy is $E=p^2/2m-K/r$, so $$A=\sqrt{1+\frac{2L^2E}{mK^2}}.$$ Since $E$ and $L$ are constant for any central force, we have that $A$ is also constant. Actually it can be show that the vector $\vec A$ is also constant.

Taking the scalar product of $\vec A$ with $\vec r$, $$\vec A\cdot \vec r=Ar\cos\theta=\frac{L^2}{mK}-r.$$ Solving for $r$ and plugging the value of $A$ we get $$r=\frac{L^2/mK}{1+\sqrt{1+\frac{2L^2E}{mK^2}}\cos\theta}.$$ This is just the polar representation of a conic section of eccentricity $\sqrt{1+\frac{2L^2E}{mK^2}}$.

  • $\begingroup$ Hey, a bit late but it seems to me like you've made a mistake/typo in this answer. I've calculated several times times now and I'm pretty sure $$ A^2 = 1 + \frac{p^2L^2}{m^2K^2} - \frac{2 L^2}{mKr}.$$ This value is also more consistent with the rest of your answer. $\endgroup$
    – Digitalis
    Jan 6, 2019 at 20:38

I have raised a similar question here and finally I ended up answering it. My answer is on the derivation of the $2^{nd}$ law which is the lengthiest of 3. The bonus is that I have done the complete proof using cartesian co-ordinates, so even a high school student with calculus knowledge can understand it.


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