I Does the electrical force also respect the Kepler Laws?

[jaumzaum]

I was wondering if the electrical force, which is a radial force that depends on the inverse of the distance squared, also respect the Three Kepler Laws?

If so, what will the constant for the third law be?

 

[vanhees71]

No, because the electromagnetic interaction is very much stronger than the gravitational interaction you cannot neglect the radiation of electromagnetic waves for an electron moving in an ellipse around an atomic nucleus. That's why Rutherford's discovery of the atomic structure lead to a crises of classical physics, because it was all of a sudden no longer understood, why atoms can be stable at all.

The way out was soon after that (apparently) found by Bohr for the hydrogen atom by applying an ad-hoc "quantum condition" allowing discrete stable orbits on which the electrons can move without radiating off em. waves. Of course, this is not very satisfactory, and indeed, Bohr's model even with the more sophisticated generalizations by Sommerfeld didn't work even for the next more complicated Helium atom. This crises lasted, however, not for too long, because in 1925 Heisenberg discovered "modern quantum theory", which is the quantum theory still valid.

Of course the same mechanism also exists for gravity. According to Einsteins General Theory of Relativity also a planet orbiting the Sun radiates off gravitational waves, but these are so weak that it takes much longer for getting visible than our solar system will stay intact (it will end about 5 billion years from now when the Sun becomes a red-giant star, because its hydrogen fuel is used up).

The first indirect evidence for gravitational waves was found by Hulse and Taylor from measuring a double-pulsar star very accurately. Indeed they found that the double-star system looses precisely the energy as expected from radiating off gravitational waves (and indeed pulsar-timing measurements are among the most accurate measurements achievable in physics).

 

[jaumzaum]

Thank you @vanhees71.
I understand we don't see electrons moving in Kepler orbits due to quantum Physics.

I was wondering if bigger particles (for example, a metal ball charged with millions of electrons in the vacuum interacting with another metal ball with millions of electrons, over a distance much greater than their radius).
That is not so practical, but if in theory we create this experiment, would Kepler laws be respected?

 

[PeroK]

I was wondering if bigger particles (for example, a metal ball charged with millions of electrons in the vacuum interacting with another metal ball with millions of electrons, over a distance much greater than their radius).
That is not so practical, but if in theory we create this experiment, would Kepler laws be respected.

Kepler's laws depend to some extent on the gravitational force being proportional to mass. For an electromagnetic force you have the extra parameter of mass to charge ratio of the orbiting body.

 

[jaumzaum]

Kepler's laws depend to some extent on the gravitational force being proportional to mass. For an electromagnetic force you have the extra parameter of mass to charge ratio of the orbiting body.

So they won't?

 

[jaumzaum]

I find it difficult t believe, because if you consider classical Physics, the motion law for two electrical charges would be a constant divided by the distance squared in the radial direction, the same holds for gravitation.

 

[Ibix]

I find it difficult t believe, because if you consider classical Physics, the motion law for two electrical charges would be a constant divided by the distance squared in the radial direction, the same holds for gravitation.

You missed @vanhees71's point. Orbiting charges radiate very strongly, so they rapidly spiral into the "sun". Quantum mechanics stops that happening in atoms, but not in macroscopic charges.

Furthermore, as @PeroK points out, Kepler's laws rely on the fact that mass cancels out because the gravitational force and the centripetal force are both proportional to $m$. The Coulomb force does not depend on $m$, so the orbits would depend on $q/m$ even if radiation were not significant.

 

[jaumzaum]

You missed @vanhees71's point. Orbiting charges radiate very strongly, so they rapidly spiral into the "sun". Quantum mechanics stops that happening in atoms, but not in macroscopic charges.

Furthermore, as @PeroK points out, Kepler's laws rely on the fact that mass cancels out because the gravitational force and the centripetal force are both proportional to $m$. The Coulomb force does not depend on $m$, so the orbits would depend on $q/m$ even if radiation were not significant.

Thanks @Ibix!

I understood small charges would radiate very strongly in very high velocities.
I am just trying to understand if in a scenario where you neglect radiation (i.e. when we don't consider electrons, but big spherical methallic masses charged with low velocities ) the Kepler laws (specially the third) would still work?

I understand @PeroK argument that Coulomb force will not depend on the mass, so the acceleration will depend on q/m. But both charge and mass will be constant, and the kinematics of the motion (considering opposite charges) will still be dictated by the equation:

$\vec {a} = \frac {-k \hat {r}} {r^2}$

Where k is a constant. The same is the case for gravity. So even if the force itself does not depend on mass, will it impact the orbit?

 

[Ibix]

Take Kepler's third law, for example, that radius cubed is proportional to period squared. If you run through the maths you'll find that the law you get for a Coulomb field is $r^3\propto\frac qmT^2$, which is not Kepler's law. The periods of two bodies at the same radius from the same primary are not the same in general.

Assuming that radiation can be neglected and the magnetic field of the orbiting charge can be neglected (I haven't run through the maths but I suspect those two criteria are the same) then you would have more complex versions of Kepler's laws that additionally depended on the q/m values of different orbiting bodies. Further assuming you could insist that q/m was the same for all orbiting bodies then they would reduce to Kepler's laws.

 

[jaumzaum]

Thanks @Ibix !
Do you know what will be the proportionality constant of this equation?

 

[LastScattered1090]

The Coulomb force does not depend on $m$, so the orbits would depend on $q/m$ even if radiation were not significant.

Yeah and another point (that I didn't see being made above, but maybe it was), is that this quantity is not even constrained to be positive, so the direction of the force can change from mutual attraction to mutual repulsion depending on the charges involved. Hard to get Keplerian orbits with a repulsive force!

 

[jbriggs444]

Thanks @Ibix !
Do you know what will be the proportionality constant of this equation?

You should be able to work it out. Fix a value for $\frac{q}{m}$ and for the charge of the "sun" $Q$. Assume a circular orbit. Use variables $r$ for radius and $T$ for the orbital period. Leave the planet's mass as $m$ (it should end up cancelling out nicely). Write down the formula for the electrostatic attraction and for the centripetal force. Equate those two formulae. Solve for $r^3$. If the proportionality holds, it should be some constant multiple of $T^2$.

One would expect the "constant" to depend on the charge of the "sun" ($Q$), the ratio $\frac{q}{m}$ for the planet, Coulombs constant $K$ and perhaps some other small constant factors.

One moment while I try it... Yes. It is straightforward and yields a result along the expected lines.

 

[sophiecentaur]

Furthermore, as @PeroK points out, Kepler's laws rely on the fact that mass cancels out because the gravitational force and the centripetal force are both proportional to m. The Coulomb force does not depend on m, so the orbits would depend on q/m even if radiation were not significant.

Kepler's Laws are not based on anything to do with mass; they are based on observations of bodies of unknown / irrelevant (at the time) masses. Both Coulomb and Gravity follow the inverse square law and that's what accounts for an elliptical orbit.

Newton's laws apply 'in retrospect'. Orbits under Coulomb force would follow all Kepler's laws if the mass were the same - that would 'cancel out' the effect of mass in the experiment. (Hence the references to q/m)

 

[sophiecentaur]

If you run through the maths you'll find that the law you get for a Coulomb field is r3∝qmT2, which is not Kepler's law

Can you give a reference to that? Are you actually implying that the orbit will not be an ellipse?

 

[Ibix]

Can you give a reference to that?

It just comes from equating the Coulomb force with the centripetal force$$\frac{1}{4\pi\epsilon_0}\frac{Qq}{r^2}=m\left(\frac{2\pi}T\right)^2r$$
There's more work to do if you really want to consider ellipses formally, but as you note it's very similar to the gravitational case.

Are you actually implying that the orbit will not be an ellipse?

No (subject to the assumptions about lack of radiation), just pointing out that $r^3\propto\frac qmT^2$ is not the same as $r^3\propto T^2$. Not unless you enforce a charge-to-mass ratio for all orbiting bodies, which doesn't seem to be a part of the OP's problem statement.

 

[jaumzaum]

Thanks @Ibix!

$$\frac{1}{4\pi\epsilon_0}\frac{Qq}{r^2}=m\left(\frac{2\pi}T\right)^2r$$
$$ \frac {T^2} {r^3} = \frac {16{\pi}^3\epsilon_0 m}{Qq} $$

Actually, if you fully derivate Kepler's third law from Newton's equations (considering the big mass also moves), you don't get
$$ \frac {T^2} {a^3} = \frac {4{\pi}^2}{G M} $$
This is only valid when M >>m, the correct formula is
$$ \frac {T^2} {a^3} = \frac {4{\pi}^2}{G (M+m)} $$
And that is also dependent on the mass of the body.

Also, this made me wonder is the m in the numerator is right. If you consider both bodies move, and have comparable masses, would this m be changed to something else (like reduced mass)?