electromagnetism - 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?

Answer

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.