# Since the Earth accelerates, does a stationary charge on Earth produce EM waves? [duplicate]

Since the Earth is revolving around the Sun, therefore it is accelerating. This implies that any stationary charge on earth is also accelerating. However, I have never heard anyone saying anything about that. Why is this so? Do charges on Earth not emit electromagnetic radiations?

• I can't do the calculations right now, but I suspect that the acceleration experienced because of the motion of the Earth is much smaller than typical accelerations in e.g. an antenna. – Gremlin Sep 4 '17 at 15:51

Yes... if that's the only charge in the system. However, to a good approximation, the Earth is globally neutral, which means that if you collect a charge $+Q$ in some location, you're creating regions with a total charge $-Q$ elsewhere on Earth, with the same acceleration. In that scenario, while your charge $+Q$ formally radiates, its field will get cancelled with that of the (equal and opposite) radiation from the opposite charge, and the global radiation becomes negligible.
That said, the Earth does have some nonzero charge, however small. The linked question does not have definitive numbers, but this question puts the net charge of the Sun at 77 C, so as a rough estimate, let's put a net charge of 1C on the Earth. How much does this radiate? We can find that via the Larmor radiation formula, which predicts a power dissipation of $$P_\mathrm{yearly} =\frac {q^{2}a^{2}}{6\pi \varepsilon _{0}c^{3}} \approx 8\times10^{-21}\:\rm J.$$ This is absolutely tiny, because it depends on the square of the orbital acceleration, and therefore on the fourth power of the orbital period, $T^4\approx 10^{30}\:\rm s^4$. Moreover, that radiation is at frequencies of the order of 1/year, which is much lower than you can realistically detect.
That said, you can do a good deal better by separating your positive 1C charge on one side of the Earth and putting the negative 1C on the other side, so that you get an electric dipole that oscillates daily, so that it does radiate. If you do that, you get a much higher Larmor power dissipation, of roughly $$P_\mathrm{daily} \approx 10^{-10} \: \rm J,$$ i.e. something that's still negligible. Moreover, once you get into the electric and magnetic dynamics of charge at the scale of the Earth, you get a much more dynamic environment with movement of charges and currents, from thunderstorms to the geomagnetic field to the solar wind to the van Allen belts, all of which produce nontrivial radiation in that frequency range, i.e. radio noise at ultralong wavelengths.