191–200 of 277 posts
Oh god, can we please get a real physicist in here? This entire thread is a mess of computer programmers “well actually”ing other computer programmers and everyone being wrong.
So you should really be asking for a communications expert, not a physicist.
Earlier quoted context omitted.
>Correct answer - to describe re-radiated energy we need moon's temperature, but to describe scattered we don't. We can ignore re-radiated, since it is visible light and not hot enough. We can use scattered, since it is visible and hot enough. When dealing with a gray-body, the equillibrium temperature of the body will be equal to the effective temperature of the incoming light, which will be equal to the effective t…
Suppose you are right. The effective temperature of the moon is 100 degrees Celsius. It is a grey-body, so its spectrum is dominated by black-body - e.g. it would be similar to spectrum of black-body with temperature lower than 100 degrees Celsius. So your argument is that the light we see from Moon is same as light we would see from a black body heated to less than 100 degrees C? Try putting a charcoal in boiling wa…
Effective temperature is about total power (per the Stefan Boltzmann law), not about the color of the light (per Wien's law).
Compare https://en.wikipedia.org/wiki/Effective_temperature with https://en.wikipedia.org/wiki/Color_temperature
I don't really follow this argument, and I would like to. I think one thing which would help me develop an intuition for it would be to see the calculation of the lens size for heating a one square-centimeter area on the earth to as high a temperature as possible by the light of the moon, and what that optimal temperature is. Anyone reading for whom this is straightforward? Even a description of how to do the calcula…
For a single lens you just want something really big. Make it take up 90+% of the sky from your target.
For temperature, the no-calculation way is to measure a rock on the moon (article says 100C) and use that number for how hot you should be able to get.
The calculation way goes as follows: Near earth you get 1400 watts per square meter of sunlight, so if that bounces perfectly off the center of a full moon and gets through the atmosphere with no losses, your target will get 1400 watts per square meter. That's equal to a black body at 122C. After taking into account the spherical shape and atmospheric losses you might get less than half of that, so ambient heat might drown out your results.
Oh god, can we please get a real physicist in here? This entire thread is a mess of computer programmers “well actually”ing other computer programmers and everyone being wrong.
The linked article is correct and the arguments are well known and accepted in the physics community and have been for a long time (much longer than Randall Munroe has been alive). A lot of the counter-arguments/speculation here in the comments is wrong. Reading this is is equivalent to reading a thread on a physics forum where with people arguing about an article saying that O(n*lgn) is really the best possible runt…
I feel much more frustration when arguing about impractical engineering proposals (e.g. solar roadways, waterseer, hyperloop, or the ocean cleanup project), since people seem to have a much more biased drive to believe in their feasibility, and can't really be reasoned with.
Sun's temperature is 6000 K. Moon's surface is pretty black: it reflects only 12% of the light.
So, effective temperature of the Sun reflected by Moon, considering that thermal radiation is proportional to T^4, is 6000 * .12 ^ (1/4) ~= 3500 K. That's quite enough to light up some fire! Of course, the spectral composition of the light will be not thermal etc, but the estimate should be close enough.
Why doesn't the Moon itself heat up like that? Well, the rocks on Earth don't heat up to 6000 K either... I think, it's partly that they are "not surrounded by sun", partly that the Moon is a giant cold heatsink
Earlier quoted context omitted.
The linked article is correct and the arguments are well known and accepted in the physics community and have been for a long time (much longer than Randall Munroe has been alive). It's interesting what bothers different people. While many of the statements in this thread are probably wrong, not many of them bother me. But I find the lack-of-self-doubt and appeal-to-authority in your message to be genuinely offensive…
>Specifically, I feel certain that one can start a fire with sunlight reflected from a room temperature mirror, and don't understand the difference between a mirror and the moon within Munroe's argument. A mirror does specular reflection and thus conserves the etendue of the sunlight. You're concentrating the image of the sun in the mirror, not light from the mirror itself. The moon in contrast is mostly a diffuse re…
This is true for a black body, but is it always true for an object being illuminated by another? I'm don't know that we can consider a diffuse reflector with a surface temperature of 100C as being equivalent to a black body with temperature of 100C. I think his conclusion is likely true (the moon is too dim to start a fire even with a really big magnifying glass) but I don't think he's right to point to the surface temperature of the moon as being the evidence of this conclusion.
Assume the sun was much brighter, so that ignition on earth is possible with a sufficiently large magnifier. Presumably if the moon was the same, this would mean the surface temperature of the moon was much higher. Now change the moon to be more heat conductive (causing the surface temperature to drop due to more heat loss on the dark side), and more reflective (causing the surface temperature to drop further due to less absorption). I'd guess that if you tweak the parameters sufficiently, you could end up with a surface temperature low enough that Munroe's argument would say that ignition is impossible, even though we've increased the intensity of the moonlight over our baseline.
How does Munroe know that we aren't in that second regime? I don't think there's enough information in his argument to distinguish. Alternatively stated, we know that there is some current temperature to which we can heat an object using concentrated moonlight. We also know that if we can change the shape and composition of the moon, we can reduce the surface temperature without reducing the intensity of moonlight. Unless there is some limit to the effectiveness of the heatsink that we can put on the moon, I think this means there is some possible arrangement that violates the assumption that the surface temperature must always exceed the temperature achievable with a magnifying glass.
(Thanks for helping me to puzzle this out)
Interesting article. Right in many places. Wrong (possibly) in main conclusion. Entropy argument - correct in the sense that using radiation from black body we cannot use lenses to heat another body to the temperature higher than original. Easy to understand why - the first body has a temperature, radiation has the same temperature, if we apply the radiation to another object it will not heat up more than the radiati…
Thermodynamics arguments are always good but you really need to understand that most of them describe closed systems under equilibrium. Real world is messier and if your understanding of thermodynamics is shaky its easy to get to wrong conclusions. Black and grey body discussion is in same realm, so we will avoid it here.
Let's consider piece of wood and try to see what it takes to combust it. We need two things - oxygen and high enough temperature. We are keeping this piece of wood in air, so we have oxygen. The temperature is how fast particles that compose piece of wood are moving. The wood consists of molecules, which are in turn consist of atoms. You can imagine a molecule is a bunch of atoms connected with each other by springs (the springs are created by electromagnetic forces when atoms lend and borrow electrons). The higher the temperature - the larger oscillations of these springs. When the temperature is high enough some of these springs can break and combine with oxygen to release energy - combustion.
Suppose light strikes some piece of wood. What happens? Photons hit molecules and they can interact with an electron or proton - they exchange momentum (and energy), which means that one of the atoms in a molecule gets a bump - springs would start oscillating harder.
Note, low energy photons cannot swing springs a lot. We need photons of high energy to swing spring to high temperature (ok - this part is too oversimplified, I can extend on this if there are questions). The energy of photons depends on their waivelength - the shorter waivelength - the more energetic photons. So we need photons with energy that could break molecular bonds (springs) to be able to heat up to combustion temperature. Visible light definitely has photons of such energy.
So we start sending light down our piece of wood, it starts heating up. But it sits in air and probably fixed by some supporting stand. This piece of wood starts exchanging heat with surrounding materials. To be able to overcome this we need to send lots of photons on this piece of wood. How do we do this? We use lens to collect the photons from broader area and send it down to the wood. Note that it doesn't matter where photons came from, were they scattered or produced by sun, what temperature of scattering surface was, not even what temperature of sun was. All what is matter - can we collect enough high energy photons. So we would need to calculate what is energy flux of visible light photons at the surface earth and see if we could come up with a lens to focus these photons on piece of wood to create sufficient intensity.
We know that we can see Moon in visible light, so it does sends bunch of energetic photons to us. All these photon are coming from moon direction (so we don't care that moon scatters them in other directions as well). So the task is to find what is the energy flux of these photons at earth surface and what size of lens we would need to get sufficient intensity.
I don't really follow this argument, and I would like to. I think one thing which would help me develop an intuition for it would be to see the calculation of the lens size for heating a one square-centimeter area on the earth to as high a temperature as possible by the light of the moon, and what that optimal temperature is. Anyone reading for whom this is straightforward? Even a description of how to do the calcula…
You get the highest temperature by reflecting moonlight from every angle. For a single lens you just want something really big. Make it take up 90+% of the sky from your target. For temperature, the no-calculation way is to measure a rock on the moon (article says 100C) and use that number for how hot you should be able to get. The calculation way goes as follows: Near earth you get 1400 watts per square meter of sun…
I don't really follow this argument, and I would like to. I think one thing which would help me develop an intuition for it would be to see the calculation of the lens size for heating a one square-centimeter area on the earth to as high a temperature as possible by the light of the moon, and what that optimal temperature is. Anyone reading for whom this is straightforward? Even a description of how to do the calcula…
You get the highest temperature by reflecting moonlight from every angle. For a single lens you just want something really big. Make it take up 90+% of the sky from your target. For temperature, the no-calculation way is to measure a rock on the moon (article says 100C) and use that number for how hot you should be able to get. The calculation way goes as follows: Near earth you get 1400 watts per square meter of sun…