This entry is going to be part of the Fourth Edition of the Carnival of the Physics, whose host is the blog RTFM.es.
What do not you know the Laws of Kepler? I cannot believe it. Anyway, we are going to begin by the simplest thing. The Laws of Kepler are those who govern the movements of the planets and they were discovered by the astronomer and mathematician German Johannes Kepler. But the most curious thing of all this is that the Kepler good one obtained them of the simple observation. In fact, it deduced them after studying meticulously the precise notes of his companion Tycho Brahe, who did it without the help of the telescope, invented with posteriority.
But let’s return to Kepler and his Three Laws (not, those of the Robotics are another Three laws that do not come to story). Kepler (although not in the same order in which today they are known and they are studied), enunción his famous three laws to explain the movement of the planets in his orbits about the Sun:

1. All the planets move about the Sun describing elliptical orbits, being the Sun placed in one of the foci.
2. I remove vector that joins the planet and the Sun it sweeps equal areas in equal times.
3. For any planet, the square of his orbital period (time that is late in giving a return about the Sun) is straight proportional to the bucket of the average distance with the Sun.

In this small article we are going to re-discover the second Kepler law, basing on the Law of Universal Gravitation of Newton:

The force that exercises an object met on mass m1 on other with mass m2 is straight proportional to product of mass, and inversely proportional to the square of the distance that separates them.

For ours intentions, we are going to fix like origin of our system of reference to the Sun, with mass M, and are going to suppose that we have a planet orbiting about him with mass m. And, also, we are going to adopt the system of polar coordinates. This way, if we fix the position of the planet (that we will suppose, as the sun, which is a point of polar coordinates (r, θ)), we are going to call ur to the unitary vector in the direction of the radiovector that joins the Sun with our planet and or θ to the unitary vector perpendicular to the previous one and in the direction in which it increases t.
Whole, which after all this foolish talk, we are going to calculate forces F that the Sun exercises on our planet. Of the Newton’s second law, we know that F =m to, where to es the acceleration of the planet. But if we want to write the acceleration in terms of the polar coordinates, it is necessary to do a few accounts (it avenges, costs, we are going to obviate them, that the stove is not for bollors), after which we will obtain that to = (r · θ "(t) +2r’ (t) · θ ‘(t)) or θ + (r" (t)-r · θ’ (t) 2) ur where t it represents, as almost always, the time.
So, if we decompose the force F in his central component Fr and tangentially F θ, we will obtain that F θ =m (r · θ "(t) +2r’ (t) · θ ‘(t)) and Fr =m (r" (t)-r · θ’ (t) 2)
But skylight, this, in fact, is valid for any type of force, that is to say, that this is the previous formulae there are only the Newton’s Second Law expressed in polar coordinates. Now we are going to introduce the fact that the force that we have is of gravitational type. In our case, only we are going to remain with an aspect of these forces, and the fact is that they are of central type, that is to say, that they do not have tangential component (remember the Law of Universal Gravitation).
Under this new prism, it turns out that the tangential component of our force must be, necessarily, void; which allows us to obtain a Distinguishing Equation r · θ "(t) +2r’ (t) · θ ‘(t) =0 If we multiply this equation for r, is obtained r2 · θ" (t) +2r · r’ (t) · θ ‘(t) =0 or what is the same, (r (t) 2 · θ’ (t)) ‘=0, so that the function between parentheses can be only a constant, that is to say, r (t) 2 · θ’ (t) =h for some constant h.
And now vámonos with the Second Law of Kepler. If To (t) the area is covered for r (t) from a fixed reference position, it is easy to verify (again there are only accounts with which I am not going to overwhelm you) ΔA = (r2 θ ‘(t)/2 · Δt=h/2 · Δt where the symbol Δ represents the increase of the function. And so, between two time moments t1 and t2, it is had that A (t2) – to (t2) =h/2 · (t2 - t1) that word saying is, exactly, what says the Second Law of Kepler:

I remove vector that joins the planet and the Sun it sweeps equal areas in equal times.

In another occasion, we all will make use these of calculations to verify that, as the gravitational force is inversely proportional to the square of the distance, the celestial orbits can be only conical.
I hope not to have bored you very much. Thanks for coming so far.