Is it possible to start fire using moonlight?

Question

You can start fire by focusing the sunlight using the magnifying glass.

I searched the web whether you can do the same using moonlight. And found this and this - the first two in Google search results.

What I found is the thermodynamics argument: you cannot heat anything to a higher temperature using black body radiation than the black body itself, and Moon isn't hot enough.

It may be true, but my gut feelings protest... The larger your aperture is, the more light you collect, also you have better focus because the airy disk is smaller. So if you have a really huge lens with a really short focus (to keep Moon's picture small), or in the extreme case you build a Dyson-sphere around the Moon (letting a small hole to the let the sunlight enter), and focusing all reflected light into a point it should be more than enough to ingnite a piece of paper isn't it?

I'm confused. So can you start fires using the Moon?

Answer 1

Moonlight has a spectral peak around $650\ \mathrm{nm}$ (the sun peaks at around $550\ \mathrm{nm}$). Ordinary solar cells will work just fine to convert it into electricity. The power of moonlight is about $500\,000$ times less than that of sunlight, which for a solar constant of $1000\ \mathrm{W/m^2}$ leaves us with about $2\ \mathrm{mW/m^2}$. After accounting for optical losses and a typical solar cell efficiency of roughly $20\ \%$, we can probably hope to extract approx. $0.1\ \mathrm{mW}$ with a fairly simple foil mirror of $1\ \mathrm{m^2}$ surface area. Accumulated over the course of a whole night with a full moon, this leaves us with around $6\ \mathrm h\times3600\ \mathrm{s/h}\times0.1\ \mathrm{mW}\approx2\ \mathrm J$ of energy. That's plenty of energy to ignite a fire using the right chemicals and a thin filament as a heater.

Answer 2

At least one point in your favour is that the light we receive from the Moon has barely anything to do with its temperature. Instead it is mostly a secondary light source "reflecting" light from the Sun towards us.

The second point in your favour (I think) is that the thermodynamic argument seems pretty weak. We are not trying to make Earth as hot as the Sun or anything like that. The only thing we want is to gather enough energy in a sufficiently small volume with oxygen and some fuel to light a fire; hence most of the energy for the fire still comes from the enthalpy of the combustion reaction.

Overall, I would think this is not impossible but probably very inefficient because of the minute fraction of power we receive from the Sun's light scattered by the Moon.

Answer 3

If you could fill the whole sky with moons you would not light a fire. It would be the same as looking up and seeing a wide expanse of bright shiny sand on a beach. What you can do with lenses and mirrors is no different than filling the sky with moons, so no: you cannot light a fire that way.

Answer 4

I am bumping this, because there still seems to be no consensus on the matter, even after Randall's What If post and the heated discussions that followed:

What If post

Reddit discussion, where people strongly disagree with Randall

My intuitive take (taking points from the discussions): Obviously, Randall's argument holds for blackbodies. However, a part of moonlight is diffuse reflected sunlight, and so should be able to start a fire.

@Marty Green: If my argument is correct, then "adding moons to the night sky" will certainly increase the temperature beyond the Moons' surface temperature (because all we do is add more mirrors).

Answer 5

Other answers here don't take into account two very important aspects. First, the heated point radiates too. Second, ideal lens with large diameter to focal length ratio don't exist. The latter can be proved with entropy.

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Suppose the "magical" lighting (which might contain the ideal lens) system exists away from Earth. Suppose also we surround this system with a radiation bath of solid angle $4 \pi$ with temperature $T_0$. Then, due to the second law of thermodynamics, the system will be in equilibrium, when the temperature in the "magical" device is uniformly also $T_0$. Then, as to simulate some radiating object (such as Moon), we remove most of the radiation, and leave only a small proportion of the solid angle. Of course, if the "magical" lighting device consists only of lens and mirrors, the radiation towards the heated point can only decrease $\Rightarrow$ it's temperature $T \leq T_0$. Note that the temperature $T$ only depends on the power it is heated, not on the spectrum of radiation.

If arbitrarily large diameter to focal length lens existed, these could be used to focus light of a radiating body (such as Sun or Moon) to arbitrary intensities, giving rise to arbitrarily high temperatures (contradicting the proof).

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Thus if the heated point is black, the maximum intensity of the black-body radiation is the intensity of the reflected light. Thus the maximum temperature might indeed be in the range of $0^{\circ} C$.

If the heated point is not black, but radiates only very high-frequency spectrum, the achievable temperatures would be higher, probably temperatures up to the temperature of the sun.

Answer 6

It appears the question is specifically concerned with using a lens.

In the original question: "So if you have a really huge lens...and focusing all reflected light into a point it should be more than enough to ingnite a piece of paper isn't it?"

In which case using solar cells doesn't answer the question.

The answer is no. No matter the lens you cannot make the surface brighter than the surface of the moon. That's thermodynamics. see: second law of thermodynamics

The second law of thermodynamics states that the total entropy of an isolated system always increases over time

In other words energy cannot flow from a colder area to a hotter one.

It can also be explained using the optical calculations point of view. See: CuriousOne's comment about a passive optical system and conservation of etendue. You should see this post. Especially the answer by CountIbis which will explain the limits using optic calculations.

If the object radiates as a black body, has a radius of R, a temperature of TT and is a distance d away, then the flux of radiation reaching the lens is: $$F = \sigma T^4 \left(\frac{R}{d}\right)^2 = \sigma T^4 \alpha^2$$ The total power of the radiation entering the lens PP is the area of the lens opening times the flux: $$P = \pi r^2 F = \pi \sigma T^4\alpha^2r^2$$ This power ends up heating the area of the image in the focal plane. The flux of radiation there is: $$F_{\text{im}} = \frac{P}{\pi\alpha^2f^2} = \sigma T^4\frac{r^2}{f^2}$$ Suppose then that you put a black body in the image plane, then the temperature there would be $T_{\text{im}}$ where $\sigma T_{\text{im}}^4 = F_{\text{im}}$ therefore: $$T_{\text{im}} = \sqrt{\frac{r}{f}}T$$ The ratio of the focal length f and the lens diameter is called the F-number and this is always larger than 1. So, the factor multiplying TT in the above equation will always be smaller than 1, therefore you can never reach a higher temperature than the temperature of the object in this way.

You should also see @zubo 's links http://what-if.xkcd.com/145/

Answer 7

• Suppose we have a blackbody cooking utensil, with surface area $A_0$ , and we are allowed to use any geometrical-optical device to focus as much moonlight as possible upon the utensil, then what is the maximal moonlight power we can focus on the utensil, and what is the maximal temperature reachable?
• Under a full moon, the surface of earth receives moonlight power per land area of $2 \mathrm{mW/m}^2$ .
• The full moon, viewed on earth, has visual diameter of roughly 0.5°. This means its solid angle is about $$\pi \left(\frac{0.25}{180}\pi\right)^2 \mathrm{sr} = 6\times 10^{-5}\mathrm{sr}$$- Since the moon is roughly Lambertian, and we are cooking in open air, the etendue is conserved, and the maximal area of moonlight we can focus on the body is $A$ , such that $$A_0 \times (2\pi)\mathrm{sr} = A \times(6\times 10^{-5}) \mathrm{sr}$$ since only $2\pi$ steradians (half of a sphere) of focussed moonlight can shine upon any point of the utensil's surface.
• We have $\frac{A}{A_0} = 10^5$ , and the radiative power on the utensil surface is at most $200 \mathrm {W/m}^2$ . By Stefan-Boltzmann law, the maximal temperature reachable by the utensil is $$\left(\frac{200}{5.7 \times 10^{-8}}\right)^{1/4}K = 243 K$$

Answer 8

Since the question is not constrained in any way, the answer is yes, it is possible. Is it "realistically" possible, no.
To make it possible, all you have to do is concentrate the moonlight (optically,or electrically) until you have the energy required to reach the ignition point of the material. what makes it not realistic is the size, cost, and/or time involved in creating a "concentrator."