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We know that the sun uses nuclear fusion to generate sunlight and heat energy. If we are using solar panels to harvest solar energy, aren't we putting some electrical load(resistance) on the sun? If yes, does it have any effects on it?

Edit: Does the presence of a receiver affect the working of the broadcasting sender? It is more of a philosophical question.

  • The Sun is 99.8% of the mass of the solar system. Earth is puny. 5.9e+24 kg is nothing. – doug65536 Nov 14 '22 at 06:31
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    The sun is opto-isolated from the solar panels, so... – Michael Nov 15 '22 at 15:07
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    Purely philosophically spoken: I am surprised and excited about the "waves", an apparently naive question is producing into an avalanche of very interesting answers - down to the smallest problems. – kampmannpeine Nov 15 '22 at 18:42
  • @doug65536 The question here is not about mass. It is the resistive effect of any observer on the sun-an energy source. – user18398875 Nov 16 '22 at 12:35
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    Is it more difficult to throw a baseball at catcher which absorbs the energy versus a small water-wheel which harvests the energy via rotation and transfers it to a battery? – MonkeyZeus Nov 16 '22 at 15:06
  • The most recent edit "Does the presence of a receiver affect the working of the broadcasting sender? It is more of a philosophical question." should be asked as a separate new question, not added several months later to the original question. You'd need to explain a bit about what kinds of "sender" and "receiver" you are asking about. You can refer back to this question, since the new question is sort of a generalization of this one. – David Bailey Feb 14 '23 at 16:11

3 Answers3

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There is no electrical connection between our solar panels and the Sun. The Sun radiates electromagnetic energy (photons) out into space. If we catch some of those photons, the Sun neither knows nor cares about it.

In fact, we'll catch those photons regardless of whether we set up a solar panel or not. If you took away the solar panel from somebody's roof top, then the same photons that would have been absorbed by the panel, will be absorbed by the rooftop instead. Whereas some of their energy would have been converted into electrical energy by the panel, All of their energy will be converted to heat in the roof when the panel is taken away.*


* Except, these days, people who live in warm sunny climates are learning to paint their roofs white. A white roof reflects a lot of the photons back out into space instead of absorbing them. But even then, the Sun still neither knows nor cares. As far as the Sun is concerned, once the photons leave it's surface, they are gone forever.

Solomon Slow
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    Solomon, this is an excellent answer, but isn't there a tiny, tiny, tiny feedback from the earth to the sun? Some of the photons reflected/radiated from the earth do reach the sun. If you think of the earth as $5\times 10^{-10}$ of a Dyson sphere, the discussion about Could a Dyson sphere destroy a star? may be infinitesimally relevant. – David Bailey Nov 13 '22 at 18:32
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    @DavidBailey, Dang! I just +knew+ somebody would say that. Tell you what, You go chase after the Parker Solar Probe, and when you catch it, look back toward Earth. If you'll send me a link where I can watch the video of your liftoff, then I'll start painting my roof white, and by the time you get there, it'll be ready for you to count the photons that make it from my roof, back to the Sun. – Solomon Slow Nov 13 '22 at 19:05
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    @DavidBailey I imagine that the displacement of the sun due to the radiation pressure exerted on it by the tiny number of photons reflected from the tiny amount of sunlight that falls upon the tiny solar panel that is tiny compared to the Earth which, itself, captures only a tiny portion of photons from the sun which then needs to travel back through the immense distance between the earth and the sun without being absorbed by any stray matter that may be in the way or ejected matter near the sun, is smaller than the Plank Length. Certainly smaller than the diameter of an atom. – DKNguyen Nov 13 '22 at 19:29
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    @DKNguyen does "the position of the sun, to within a millimeter" even have a well-defined meaning? – Lawnmower Man Nov 13 '22 at 20:50
  • @LawnmowerMan I mean I suppose you would have to model the movement and position with and without the presence of the reflected photons. But good look with that with enough accuracy to track down the deviations you're interested in. lol Unless you want to use the stars to define the position of the sun but good luck with that for the same reason. – DKNguyen Nov 13 '22 at 20:56
  • Oh yeah, let's also not forget that the reflected light is in the shape of a cone and the sun fills only very small solid angle in that cone, further reducing the number of reflected photons that even head in the sun's direction to begin with. – DKNguyen Nov 14 '22 at 00:09
  • Hmm question has me pondering a funny situation: SUN: as big as Mnt Everest EARTH: as big as a tennis ball Hmm Sun wins. #micdrop – Fandango68 Nov 14 '22 at 01:22
  • @LawnmowerMan Good point. Modern astronomy uses the International Celestial Reference Frame, which is based on extragalactic sources. It's quite difficult to precisely locate Solar System bodies, as I mention here. – PM 2Ring Nov 14 '22 at 07:54
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    (cont) The Sun is extra tricky to locate precisely because we can't bounce radar off it, and because it substantially warps spacetime. The Sun wobbles around a fair bit, relative to the nominal Solar System barycentre (SSB), as I discuss here. – PM 2Ring Nov 14 '22 at 07:54
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    If you want to provide an analogy, if someone throws a ball at you, it doesn't matter to the thrower whether you catch it or not. – Barmar Nov 14 '22 at 07:55
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    @DKNguyen When modelling binary stars we do take into account the energy of the light that each star receives from the other, since that has a noticeable effect on the stars' temperatures. I'm fairly sure the momentum of the intercepted light is ignored, but I guess we need to ask an actual astrophysicist about that. – PM 2Ring Nov 14 '22 at 08:02
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    @Fandango68 - Sun is ~333000 * heavier than the earth, while the diameter is only about ~116 times bigger. I think your tennisball would have to be exceptionally big compared to Mt Everest. Going by ~8850 m which would make your tennisball about 76.3 m in diameter, a tad too big to actually play tennis. – eagle275 Nov 14 '22 at 08:49
  • @Barmar Unless you play Baseball, or you have an entangled identical ball on your side. – Peter - Reinstate Monica Nov 15 '22 at 08:43
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    @Peter-ReinstateMonica Even adding quantum mechanics wouldn't be enough to get nerds to enjoy baseball. – Barmar Nov 15 '22 at 14:39
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    @Barmar "sabermetrics" – user3067860 Nov 18 '22 at 14:24
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No, but your intuition that there should be some effect isn't entirely incorrect. The term for the type of phenomenon you're describing is "near field effects," and it comes up in antenna engineering. There are two types of near field effect - radiative and reactive. Radiative effects are caused by EM waves bouncing back off loads, interfering with outgoing waves as they travel back towards the source. Reactive effects are caused by energy stored in the electric and magnetic fields around a source.

But these effects vanish within a few wavelengths of the source. Radiative near field effects decline with the inverse fourth power of number of wavelengths of distance, and reactive near field effects decline with the inverse sixth power.

Given that solar panels are on the order of meters large, they're only going to meaningfully interact with EM waves with wavelengths on the order of meters or smaller. The earth is more than a few meters from the sun, therefore solar panels can't impact the sun's near field.

Now, if you sat a cell phone using 4G LTE (which has a minimum carrier wavelength of ~5cm) within 5cm of a solar panel, you could reasonably say that the conductive solar panel may be acting as a load on the cell phone due to near field effects. You mention resistance, but this would depend on capacitance and inductance as well.

Willa
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    The sun, en bulk, isn't radiating light via EM coupling. The radiator, being the exotic particles in the plasma of the sun, near field in this case is going to be well within the photosphere, where any antenna will quickly lose structural integrity. – Aron Nov 15 '22 at 05:56
  • As a point of clarity, are you referring to situations in which an object could attract and harvest more energy than a "perfect energy sink" of the same size? – supercat Nov 15 '22 at 22:51
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While there may not be an electrical effect, per se, there will be an effect, however tiny. Many of the responses seem to focus on the tininess of the effect and conflate "really really small" with "doesn't exist".

Specifically (and maybe there are others that I'm not thinking of), the temperature of the sun is a function of the radiation leaving it minus the radiation it receives from around it. This fact is encapsulated in the radiative heat transfer law where heat exchange is equal to the Stefan-Boltzmann constant times the difference of the 4th powers of the temperatures. Because the earth is absorbing radiation from the sun, its temperature increases, increasing the amount of thermal radiation it, in turn, radiates outward. Some very tiny fraction of that light will hit the sun, increasing its temperature by such a tiny bit. To be sure, the effect is compounded by the fact that the earth also reflects some light to the sun, which has the same effect, it just doesn't follow the 4th power law.

In other words, the earth is a tiny piece of insulation, blocking the heat of the sun from escaping and warming it up ever so slightly.

So when you throw a solar panel in the mix, you're actually (and usually very temporarily) converting some of that radiation into ordered work instead of pure heat, So the temperature of the earth increases just slightly less, causing just slightly less heat radiation back to the sun.

The are other effects of course; since solar panels absorb most light and reflect very little, they may actually increase the absorption of heat from the sun, although the albedo of black asphalt single is also pretty low (which is why the white roofs are mentioned elsewhere). So if the albedo of the solar panels plus their efficiency in converting light to electricity is less than the albedo of the background (the roof), then the temperature of the earth actually goes up, but the reflection goes down. Either way, there's an effect.

I think part of the confusion lies in the fact that many (maybe all? ) of our physical laws, especially those dealing with wavelike phenomena are pragmatic approximations. So things like the near-field effect don't suddenly vanish after a few wavelengths, they just become so small that they make no difference in the predictive power of those models for practical use. Which is to say, they become smaller than other sources of error in the prediction of how the system will behave. Negligible does not mean nonexistent.

  • That's a great comprehensive answer! So the near field effect of a solar panel on the sun isn't literally zero? – M. Y. Zuo Dec 13 '22 at 02:20