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Comment by dr_orpheus | Hacker News Reader

Yep, you get about 36% more sun above the atmosphere. 1361 W/m^2 above the atmosphere vs ~1000 W/m^2 on a clear day [0].

Now you just need to bring big enough batteries to survive 14 days in a row with no sun.

[0] https://en.wikipedia.org/wiki/Solar_irradiance#Irradiance_on...

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No need for reliance on batteries. Just use the solar generated to process regolith yielding plenty of hydrogen and oxygen for energy storage, breathing, rocket fuel and additional component elements.

I think you might find places relatively close each other in the moon polar areas that are in aggregate illuminated most if not all the time.

https://space.stackexchange.com/questions/34025/moon-polar-d...

The alternative is that you could also just plan around the lunar cycle by turning machinery off during the lunar night and only having moon landings during the day.

You still need heaters, because a lot of equipment -- including solar panels -- won't survive thermal cycling otherwise.

Wouldn't it be smarter to build on the far side of the moon? Then you don't need to worry about the Earth obscuring the sun at all, right? Or were you referring to some other phenomena?

the moon is tidally locked to the Earth, i.e. it does not rotate about an axis, and takes ~27 days to orbit. therefore around half of those days are in darkness at any given point on its surface, hence the lunar phases

Ah yeah, makes sense... I confused myself.

You still get the Moon obscuring itself. The Earth is not eclipsing the Sun most of the time.

Considering the fact that you have the entire moon as available surface, you'd rather lift a giant rock with that power and store all the potential gravitational energy as a "battery".

A 10 ton rock, lifted up to 50m high, will store up to 810M Joules at moon gravity

>A 10 ton rock, lifted up to 50m high, will store up to 810M Joules at moon gravity

Wrong prefix? I think you meant 810K Joules.

For Earth gravity: 10,000 kg times 9.8 m/s^2 times 50 m = 4.9 megajoules, 1.36 kilowatthours. (Anker sells a 2.04 KWh battery pack for US$1999: https://www.anker.com/products/a1780 The manual says it masses 30.5 kg, so that's a cool 491.88x difference in mass power density.)

Again, that's for Earth gravity. Lunar gravity: 10,000 kg times 1.62 m/s^2 times 50 m = 0.81 megajoules, 0.225 kilowatthours. (2,976x difference in mass power density vs the lithium battery.)

Gravity storage isn't economical on Earth, and it's really not economical on the Moon, where gravity is lighter.

Unfortunately I think you're off by an order of magnitude. I think it would be 810 Kilojoules, which is approximately equivalent to a 1kg lithium-ion battery. Of course, you could move thousands of rocks up and down a big crater, rather than just one, but it would still be a lot of infrastructure for a fairly small amount of energy storage.

Three orders of magnitude*

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