100,000 points of light?

Question

What would happen if you took the sun and divided its light into 100,000 separate points and spread those points uniformily around the earth.

Would we have weather?

What color would the sky be?

Is this the equivalent of blackbox radiation, and if not, why?

Inspired by: http://answers.yahoo.com/question/index;_ylt=AmKHekgjfwvwR5gd3_ONxuLty6IX;_ylv=3?qid=20080510191412AAXkOaO

Answer 1

No matter how many points of lights there are, by definition there will still be some blue sky inbetween, so that the color spectrum will not be exactly the same as the solar spectrum. 100,000 points of light will have an average angular spacing about half the angular size of the sun (or the moon), each point being roughly 5 times brighter than the moon. In other words, you probably can stare at the points, and make out the bluish sky inbetween. A pure blackbody radiation? Nope, the color spectrum wouldn't jibe---wouldn't be a pure Maxwellian. What color would the sky be? On the clearest days we'd be able to discern the bluey space between the points, but even on overcast days the light spectrum will not correspond to that of a pure blackbody, because we'd have mixing of TWO light sources, one the pure solar blackbody, and the other from light scattering of the sky. Finally, will there be weather? If the 100,000 points of light is uniform around the world, ground and atmospheric temperatures would become uniform, so that the only "weather" we'd have would be convection cells between upper and lower atmosphere, much like the convection cells we can see in the solar protosphere. Hence, the world will be divided into a lot of similar micro-climates, kind of like having a McDonalds and Starbucks and WalMart in every village.



Also, if we have points of light, it's moot to say the temperature of the solar blackbody would be changed at all. Even distant pointlike stars can have high blackbody temperatures.



Would the Earth itself eventually reach approximately the same blackbody temperature? No, because the sky would not be 100% covered with the solar blackbody temperature, the inbetween spaces will have a drastically lower apparent temperature (it's not even a blackbody spectrum), so the ultimate ground temperatures would depend on the two, not just one. Probably would be about the same as worldwide average temperature today. To illustrate this point, the night sky has thousands of stars of very high blackbody temperatures, but it can sure be cold out there in the night.



Addendum: remember.kelly, I was sleepy last night when I wrote this answer. But let's divide up the disk of the sun into tiny circles. Then each disk is roughly 1/316 the diameter of the sun. The average spacing is roughly 120Tan((π/2)√(8/100000)) = 1.686 diameter of the sun, or a spacing-to-diameter ratio of over 500. That's pretty point-like, and yet 1) the blackbody temperature and 2) total solar influx is the same as the sun for these tiny disks. The only way the temperatures on the Earth's surface could be close to the solar blackbody temperature is if somehow be trapped by the Earth's atmosphere, so that it itself reaches the same blackbody temperature as to maintain the radiation equilibrium. But it won't---much all of it will pass out into space as it does now.



Addendum 2: How could we have an "heat engine" between the poles and equator? We wouldn't even have nights any more. The only major temperature differences would be in atmospheric altitude, not latitudes. And, remember, the sun spins too, so behavior of its protosphere would be a good guideline of what to expect for Earth in this example. Does the sun have cooler poles?Finally, the size and shape of the convection cells in the Earth's atmosphere would have little or nothing to do with the spacing of the 100,000 points of light, so long there's a lot of them. It would depend more on things like the effective thickness of the atmosphere.



Addendum: Oh I knew that

Answer 2

First of all, the effective temperature (T' ) of the Sun would be greatly reduced, and given by



T' = (r/4R)^(1/2) T,



where T is its present surface temperature r is the actual radius of the Sun and R the radius of the Earth's orbit. The factor of 4 is to take into account the fact that the Sun is a sphere as opposed to a flat disk. The total amount of energy radiated would be the same in both cases, with the difference being that now the Earth intercepts only a very small fraction of it, and in the suggested case it receives all of it.



After some time (how long, I'm not going to estimate now) equilibrium will have been reached and the temperature of the Earth would be the same as the new Sun's, T'. There would not be any weather, winds, etc. on Earth except as possibly produced by any geological energy sources in its interior (e.g. volcanoes). But when true thermodynamic equilibrium arrives, these would no longer exist.



The peak emission wavelength would be that predicted by the black body radiation law and the value of T. However, it probably is a wavelength that falls well outside the visible spectrum. The brightest visible wavelengths would then be in the red, however. To us, the sky would appear a deep red color.



*************

Somebody might want to argue that the temperature of the Earth would actually be much higher than the temperature T' of the "Sun", on the basis that the total flux being the same across the Earth's surface as that leaving the "Sun", and the Earth being so much smaller. Applying the black body radiation law again would give the new temperature of the Earth. The error in this argument is that black body radiation is not directed but is uniform in all directions. Thus of the radiation emitted by a small element of surface of the "Sun" only a very small fraction would intercept the Earth. Carrying out that calculation confirms that the Earth's temperature would equal T', the same as that of the "Sun".



**************************

In concluding that T(Earth)=T' , I assumed the Earth to be a black body radiator. Depending on how high T' is, that may or may not be a good approximation. Thus if the Earth reflects some of the radiation, then T(Earth)
I had overlooked Bekki's point about e.g. mountains creating temperature gradients and thus wind patterns.



***************************

Scythian is right about the grainy source distribution. I chose to change the problem a little, to a uniform distribution. However, I do not blieve that the radiation received by the Earth will be altered significantly due to that fact. Of that which is re-emitted by the Earth a large portion will miss the "Sun" pieces, but under the assumption that these are internally powered, unlike the Earth, they will not be affected significantly by that. Thus I predict the result to be about the same. I haven't read Scythian's argument closely to know if other issues need to be addressed; maybe tomorrow I will have some time.



**************************

OK, one more response to Scythian. The formula I derived applies for a continuous distribution. If you have a granular one, it should be



T' = (r/4a√ N)^(1/2) T,



where a=radius of each piece and N the number of pieces. Note that T' is infinite for point sources, so the question, as worded, makes no sense unless it is interpreted to mean that the pieces are finite but infinitely far away. But if that were the case, the Earth would receive no heat at all. Also, this T' would not be the effective temperature of the sky; that would be close to the value I gave before.



*******************

I agree with Scythian within his model's framework, but the question is too unclear to be able to agree on a model without adding arbitrary but essential assumptions. One is the distance of the "Sun" to the Earth, the other is the size of the pieces. A point source has infinite brightness. If placed at a finite distance, these light sources would be so bright as to be totally blinding to an observer on Earth.



This is a common problem with Yahoo!Answers, that the wording of the question is often so ambiguous that you end up with a variety of answers based on different models of the question.

Answer 3

I guess you could still have weather if certain parts of the earth absorbed more energy than others. You'd have altitude differences and oceans and geographic differences that would drive temperature and pressure gradients. And you'd still have coriolis forces in play.



The sky would be the color of sunlight (white) every way you look with no net impact from scattering. No day/night of course, nor variations of insolation with lattitude.



Yeah, it is pretty much like that.



The link doesn't work for me. It says I can't access the question.



EDIT--oh, I just thought100,000 was shorthand for "peanut-butter spread across the sky". Well I still don't see how that effects things tooooo much. The sky would still be pretty much white and insolation pretty much uniform.

Answer 4

Since the light of the sun is generated because it has a huge gravitating furnace the question is without meaning or consequence.

Answer 5

I think we all would melt so no need for weather and sky.