I Optimizing Lunar Power: North vs. South Poles for Mining and Moonbase Sites

[dmac257]

Opposite sides of a hill. On a perfect sphere with no seasons you could stick the panel on a vertical mast and rotate it.

Places that are receiving sunlight are not the same places that are most likely to have ³He. At the middle equator you could always see clearly with Earthshine. The terracycle would not include a new Earth at night. The new Earth would be above close to the noon sun. At sunset you have a quarter Earth waxing and at sunset you have a quarter Earth waning. A full Earth on Luna is brighter than a full moon on Earth.

So how much energy would be produced by a panel from "a full earth" on a lunar night compared to "lunar daylight". As for the vertical mast wouldn't it be fairly easy to erect a mast at a high point near one of the poles to keep a panel in sunlight all the time?

dmac257

 

[Al_]

Very little power from a panel in Earthshine. The albedo of the Earth is 0.37 (de Pater and Lissauer), and the Earth receives 1000W/m^2 from the Sun and appears as a disc covering 3 degrees seen from Moon. So, ballpark less than 1 Watt per square metre.
The mast would have to at least 100m high I estimate. Lunar gravity makes it a bit easier, but transport and construction... waaaaaay hard.

 

[stefan r]

So how much energy would be produced by a panel from "a full earth" on a lunar night compared to "lunar daylight".

The flux is lower by more than 10⁴. Not much energy but the view is impressive. NASA astronauts are effectively soldiers so the view will not matter. For a tourism industry a poor view can kill the project.

...As for the vertical mast wouldn't it be fairly easy to erect a mast at a high point near one of the poles to keep a panel in sunlight all the time?

dmac257

If you build a thriving steel industry you could build sky scrapers.

...
The mast would have to at least 100m high I estimate. Lunar gravity makes it a bit easier, but transport and construction... waaaaaay hard.

NASA has this article with nice topography pictures of Shackleton crater. The top of the rim on any side is less than a kilometer wide. Is much easier to lay a conductor line 1 km than to build a 100m tower.

The slope on all sides gets over 30% grade. You could use a cable or build roads with a lot of switchbacks. I suspect the cable is less effort unless you assume there is already an extensive road building infrastructure. The tow cable needs to haul loaded vehicles 7 to 8 kilometers. Presumably the solar panels are also providing power for vehicles working inside the crater which means we need to run 10+ km of conductive cable. You can use double the panels (place multiple locations) and have continuous coverage. A crew could also haul the panels from one side of the slope to the other if there was a demand for extra power. It is possible that the lower section of rim would be a better location because it cuts the the vertical ascent by half a kilometer.

You could also place moveable mirrors around the rim. Pillage the water in the basin and launch it out. If the primary goal is hydrogen fuel for inter-planetary missions that should work. Would be interesting to see how much fuel exhaust resettles back into the crater.

 

[Al_]

Thanks, nice article in the link.
Yes, a power cable running out to two panel sites makes more sense.
I think a vehicle can be built that can handle 30% grade, and initially it's much less mass than a long cable.
How about, thinking about a minimum mass solution, a steerable mirror on the rim that directs sunlight down to solar panels on the rover?