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.
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: $$T^2=4\pi^2r\frac{r^2}{GM}$$ $$T^2=\frac{4\pi^2}{GM}r^3$$
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 $\sqrt{1+\frac{2L^2E}{mK^2}}$.
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.
