Why do electrons have to fall on the nucleus in the Rutherford atomic model? As I read on Wikipedia, the Rutherford atomic model is not correct according to classical electrodynamics, as it states that electron must radiate electromagnetic waves, lose energy and fall onto the nucleus.
I don't understand this explanation. 
It is clear to me that with given acceleration directed to nucleus and proper speed, electron can move around the nucleus. 
I don't understand explanation about energy, but I understand that there must be some force directed to nucleus. Also this force must not be constant because if it is, a larger speed could keep electron moving around the nucleus.
So what is that force? Why does this explanation on Wikipedia and on other resources operate with energy, not with force?
 A: Well, I dont see a problem in any of those answers here, but, since you want in force terms... lets go.
The Lorentz force is:
$$
\mathbf F = q(\mathbf E + \mathbf v\times\mathbf B)
$$
Lets assume the nice and simple atom of hydrogen. A single electron is classically orbiting it. Lets say there is no magnetic field. Only electric. The electric field is a central field, meaning it is pointing only radially, meaning it will result in an orbit. And more: Its a kepler orbit (same of the planets).
$$
\mathbf F = q\mathbf E
$$
But then, the electron when accelerated irradiates electromagnetic energy. Conservation of energy must apply, such that the irradiation takes away the energy of the electron. The electron loses then its energy. Energy is proportional to the momentum (kinetic energy). Thus, electron loses momentum. Changing in momentum is force. If we take Larmor Formula and make this process, we will arrive at Abraham-Lorentz force.
Now the complete force of this is:
$$
\mathbf F = 
\frac{d\mathbf p}{dt} = 
m\frac{d^2\mathbf r}{dt} = 
q\mathbf E(\mathbf r) + \frac{\mu_0 q^2}{6\pi c}\frac{d^3\mathbf r}{dt^3}
$$
Note that, for a circular orbit in xy-plane: $\mathbf r = r(\cos\omega t, \sin\omega t, 0)$, and thus:
$$
\omega^2\mathbf r = -\frac{d^2\mathbf r}{dt^2} \quad\Longrightarrow\quad
\omega^2\frac{d\mathbf r}{dt} = -\frac{d^3\mathbf r}{dt^3} \quad\Longrightarrow\quad
\mathbf F = q\mathbf E - \frac{\mu_0 q^2}{6\pi c}\omega^2\mathbf v
$$
Meaning, the third order derivative has a relationship with the speed. And not only that: Has a minus sign over there, indicating a drag force: A force always opposite to the velocity, and thus will tend to stop the motion. So, an electron orbiting a proton with no magnetic field present, will drag because this force, spiral in, and collapse into the proton.
A: Excellent question, which is rarely addressed in introductions to quantum mechanics. Maxwell's equations clearly show that the electron in a classical Rutherford atom would radiate EM fields with a total power given by the Larmor formula. But it does not immediately follow that this radiation would cause the electron to spiral in toward the nucleus.
For that, you need to separately postulate a radiation reaction force (often, but not always, assumed to take the form of the Abraham-Lorentz-Dirac force) that supplements the usual Lorentz force. There is no universally agreed-upon way to do this, or even a consensus among physicists as to whether it's necessary to do at all. There are different versions of the theory of classical EM that handle the radiation reaction in different ways, each of which has (somewhat subjective) pros and cons. There is no single "correct" way to do it, because classical EM is just an approximate theory that can't fully capture the (ultimately quantum) nature of physical reality at tiny length scales.
(It's often claimed that conservation of energy logically requires that the Lorentz force law be supplemented with a radiation reaction force for internal consistency, but this isn't actually true. Since the self-energy of a point particle is formally infinite, it's actually mathematically self-consistent - although perhaps subjectively distasteful - for a charged particle to radiate forever without ever changing its trajectory, because strictly speaking, it has a bottomless reserve of potential energy to borrow from.)
So it's actually an oversimplification to say that the radation emitted by a classical Rutherford atom's electron necessarily causes it to fall inward towards the nucleus. Whether or not that's true depends on your choice of postulated radiation reaction mechanism, for which there is no single canonical form.
