Derivation of Kepler's laws

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 $\tau \propto a^{3/2}$.

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