# Derivation of Kepler's laws

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.

Is there a book or a link where I can find an exhaustive treatise about this? I am looking for a (as complete as possible) set of hypothesis and the steps which lead to the laws.

Could you suggest me some resources?

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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. –  Robin Ekman May 14 at 13:31

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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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"? –  BowPark May 19 at 8:55
That's the book. Thornton apparently waited two editions and more than twenty years to re-order the names after Marion's death. –  rob May 19 at 11:54

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: $$T^2=4\pi^2r\frac{r^2}{GM}$$ $$T^2=\frac{4\pi^2}{GM}r^3$$

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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.)

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