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The main drawback in Rutherford's model of the atom as pointed out by Niels Bohr was that according to Maxwell's equations, a revolving charge experiences a centripetal acceleration hence it must radiate energy continuously as a result of which its kinetic energy is reduced and it eventually should fall into the nucleus. Bohr then proposed his own model in which he explained this through quantum theory stating that the general laws of electrodynamics and classical mechanincs do not apply in the case of electrons. Does this imply that if a body is revolving around another, then it will tend to spiral in towards the centre? If so, then how is the stability of the solar system explained? This question has been already asked but no satisfactory answers were recieved. Hence , I am reasking the same question

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  • $\begingroup$ I have modified the title to, I hope, more closely represent what the question is actually about (it is now too long unfortunately), & added some tags. Feel free to revert this if you think it's wrong: I think the original title was misleading however (although I see why you chose it!). $\endgroup$ – tfb Jul 12 '17 at 9:59
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    $\begingroup$ > "as a result of which its kinetic energy is reduced" actually due to radiation total energy should decrease. Since total energy = kinetic + potential = - kinetic, kinetic energy would increase (the electron orbits faster on lower orbits). $\endgroup$ – Ján Lalinský Jul 12 '17 at 11:04
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In classical electromagnetism, accelerated charges radiate. But the solar system is not held together by electromagnetic forces: it's held together by gravitation. In Newtonian gravity, accelerated objects do not radiate: indeed there is no wavelike solution in Newtonian gravity at all. So to the extent that Newtonian gravitation provides a good solution for the solar system, there is no radiation and the system is stable, at least in that sense (it may still lose energy by occasionally evicting objects due to close interactions).

But this is not the complete story: Newtonian gravitation is only an approximation to a more correct theory of gravitation, General Relativity. And in GR orbiting bodies do radiate power, in the form of gravitational waves, and therefore they do spiral in. But for the solar system the effect is extremely small: the power radiated by two orbiting bodies is

$$P = \frac{32 G^4}{5 c^5} \frac{\left(m_1 m_2\right)^2 \left(m_1 + m_2\right)}{r^5}$$

where $m_i$ is the mass of the $i$th body, $r$ is their separation (the orbit is assumed to be circular).

Well, you can compute this for the Earth-Sun system and $P \approx 200\,\mathrm{W}$, which is tiny. In particular compare it with the total energy in the Earth-Sun system according to Newtonian gravitation, which is given by:

$$E = -\frac{Gm_1m_2}{2r}$$

with assumptions that $m_2\ll m_1$. Note that $E$ is negative because the system is bound.

So, for Earth-Sun, $m_1 \approx 2\times 10^{30}\,\mathrm{kg}$ (mass of Sun), $m_2 \approx 6\times 10^{24}\,\mathrm{kg}$ (mass of Earth), $r\approx 1.5\times 10^{11}\,\mathrm{m}$ (semimajor axis of Earth's orbit), giving

$$E\approx - 2.6\times 10^{33}\,\mathrm{J}$$

Now I will be lazy and not do the integral I should do, but simply ask how long, assuming a constant $200\,\mathrm{W}$ of lost power (this is the lazy bit!), it would take to reduce $r$ to $0.9\times r$. Well, the change in energy is $\Delta E \approx 2.9\times 10^{32}\,\mathrm{J}$ (the lower orbit has a more negative energy, so energy is being extracted from the system as we would expect) and

$$\frac{\Delta E}{200\,\mathrm{J/s}} \approx 1.5\times 10^{30}\,\mathrm{s}$$

Well, the age of the universe is $4.3\times 10^{17}\,\mathrm{s}$ so this is about $3.4\times 10^{12}$ times the age of the universe, to radiate enough energy to lower Earth's orbit by a tenth. The real answer would be different as I've assumed constant power and been generally lazy, but not enough different to matter: this process takes stupidly long periods of time and the Earth's orbit around the Sun is, to all intents and purposes, stable over relatively short periods like the age of the universe: Newtonian gravitation is indeed a rather good approximation for the Earth-Sun system.


Of course, Newtonian gravitation fails to be a good approximation for very massive bodies in very close orbits, such as orbiting black holes, if their orbits are close enough (which eventually they will become). Two 10-solar-mass black holes orbiting each other at the radius of Earth's orbit (which is not close) radiate about $2\times 10^{16}$ times as much power as the Earth-Sun system, for instance. And very famously these systems do radiate a lot of power in their last moments as they spiral in: on the 14th of September 2015 the first such event was detected.

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  • $\begingroup$ Tfb, where can I find a derivation of your P formula? (textbook if possible). $\endgroup$ – magma Jul 18 '17 at 10:53
  • $\begingroup$ @magma: I always steal it from Wikipedia (note I quietly assumed that eccentricity is small), which does not give a derivation but has a reference to the original paper (which is behind a paywall). But I think it is in Misner Thorne & Wheeler: I will check this evening if I remember. There probably are more recent textbooks which have it, although I don't know them. $\endgroup$ – tfb Jul 18 '17 at 11:01
  • $\begingroup$ Tfb, found in MTW eq. 36.16a page 988 $\endgroup$ – magma Jul 18 '17 at 14:12
  • $\begingroup$ @magma: cool, I don't have to dig it out from under the terrifying pile of books on top of it! $\endgroup$ – tfb Jul 18 '17 at 19:49
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As you said, charged objects produce electromagnetic radiation as they accelerate, and electrons can't have classical-style orbits for this reason. In contrast, the bodies in the solar system don't have very much net charge, so they don't generate very much electromagnetic radiation as they orbit.

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