Blue Origin manufactured solar cell prototype from lunar regolith simulants

Re:Blue Origin manufactured solar cell prototype from lunar ... by leidenfrost on Tuesday February 14, 2023 @03:05PM (#34795166)

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

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Re:Blue Origin manufactured solar cell prototype from lunar ... by sacred_numbers on Tuesday February 14, 2023 @03:28PM (#34795482)

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.

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Re:Blue Origin manufactured solar cell prototype from lunar ... by sbierwagen on Tuesday February 14, 2023 @03:31PM (#34795512)

Three orders of magnitude*

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Re:Blue Origin manufactured solar cell prototype from lunar ... by sbierwagen on Tuesday February 14, 2023 @03:23PM (#34795425)

>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)

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.

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

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