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Yes, I can agree with that. The problem for me is fitting it into Munro's argument. He says (in boldface): "You can't use lenses and mirrors to make something
by nkurz 8y ago
Yes, I can agree with that. The problem for me is fitting it into Munro's argument. He says (in boldface): "You can't use lenses and mirrors to make something hotter than the surface of the light source itself".
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)
- ummonk 8y agoAn excellent point! >Now change the moon to be more heat conductive (causing the surface temperature to drop due to more heat loss on the dark side), Yeah, although this isn't too big in the case of the moon (unlike the Earth, it doesn't have an atmosphere and doesn't rotate rapidly, so there isn't too much redistribution of heat across its surface), it is definitely something that would confound the calculations. We'd still be able to just look at the effective temperature of light falling onto the moon and that would limit the temperature that we could light the object up to. But we wouldn't be able to use a direct measurement of the temperature of the surface of the moon. > and more reflective (causing the surface temperature to drop further due to less absorption). To the extent that it's a gray body (and most objects are approximately graybodies), this wouldn't actually lower the temperature. Absorptivity < 1 causes it to absorb less energy from the light, but for a gray body emmisivity equals absorptivity so it also radiates out less light too, and you actually end up reaching the same equillibrium temperature as fully absorptive black body.
- nkurz 8y agoThanks for the kind response. I now see that the final average core temperature should remain the same, but I'm less sure about the surface temperature. There are a lot of complex processes involved, and I'm not sure that one can conclude that everything cancels out to keep the answer constant. I respond late to offer a link (that's hidden on the second page of this thread) that I think presents that argument I was trying to make better than I managed to: https://physics.stackexchange.com/questions/370446/is-randall-munroes-what-if-xkcd-correct-that-magnified-moonlight-cant-get-th/370600 https://physics.stackexchange.com/questions/370446/is-randal.... I thought the comments on the answer were a helpful reframing of the problem.