Would a large, close-by, molten moon be able to support photosynthesis at night?
Setup:
- Earth is the same (somehow, minus the tides -- don't overthink this bit)
- Moon is 2x the mass
- Moon is 2x closer
- Moon contains the derelict fissile fuel-dump of a hyper-advanced civilization, and it is being heated by a long-running natural fission reaction of the meltdown, causing the entire surface to be an ocean of bright-orange magma.
- The same civilization (must've been a civil war, I guess) set up a chain of massive fissile material impactors that hit the "far" side of the moon on a semi-regular basis, adding their fissile mass and kinetic energy and thus helping to keep the moon molten.
How bright would the earth be at night? Would the moon provide enough light for photosynthesis?
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1 answer
Let's assume that the Moon is roughly a black body. Therefore, its luminosity can be approximated by the Stefan-Boltzmann law: $$L=4\pi\sigma R^2T^4$$ As you've said, $T=1100^\circ\text{ C}=1373\text{ K}$. $R$ is related to mass, $M$, by $$\rho=\frac{M}{\frac{4}{3}\pi R^3}\to R\propto M^{1/3}$$ Doubling $M$ multiplies $R$ by $1.26$. Currently, $R=1737\text{ km}=1737000\text{ m}$, so this new moon would have a radius of $R=2189000\text{ m}$. Plugging in the appropriate value for $\sigma$ gives us $L=1.21\times10^{19}\text{ W}\simeq10^{-7}L_{\odot}$. This doesn't seem like a lot.
However, the important thing here is intensity. In this case, the moon would be a distance of $192200\text{ km}$, on average, from Earth. This is 780 times closer than the Sun, and therefore, given that $$I=C\frac{L}{D^2}$$ for some constant factor $C$, we have $$I_{\text{moon}}=C\frac{10^{-7}L_{\odot}}{((1/780)D_{\odot})^2}=0.06I_{\odot}$$ This moonlight would be roughly 1/16th as intense as sunlight. This probably isn't enough for large-scale photosynthesis on its own, although it could certainly provide a little help for any plants that try to gather a little more energy during the night.
Let's also look at the wavelengths this moon would be emitting most strongly at at. Using Wien's law, we find that the wavelength of maximum emission is $$\lambda_{\text{max}}=\frac{b}{T}$$ where $b$ is a constant. Plugging in our $T$ gives me a $\lambda_{\text{max}}$ of $\simeq2.11\mu\text{m}$, which is actually in the infrared band! Yes, there would still be emission of visible light, but this would actually benefit any plants using pigments that have the highest absorption at infrared wavelengths - which is not like most Earth plants.
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