Let’s say the roofs are all red, how big does it have to be to be visible as a little red dot?
http://www.waloszek.de/astro_mond_0_e.php
tl;dr: between 120 and 350 kilometres depending on how good your eyes are.
Paris is around 15km Tokyo is 90km
Whew. Don’t have to worry about them ruining the view in my lifetime!
That’s if they don’t put a giant ad on the moon. Just need a little bit of paint!
Oh god, that’s the real future.
“we’re trying to reach you about your car’s extended warranty!”
Why would a huge neon sign be so cool yet so dystopian.
I’m all for giant neon sign, but it has to spell fart or be dick shaped.
Since the Moon is tidally locked, being on the opposite side of the Moon would mean you could never see it.
Right now there might be a massive base manufacturing… Astronaut ice cream and we would never know
I was about to reply that op didn’t ask about the other side of the moon… Then you went to making astronaut ice cream!
Well played…
Or a nazi base
What movie is this from?
Thanks!
The Inhumans would never allow it.
Or Jewish space lasers
Yeah, but we already know about that one.
So that’s what China was doing back there! Their “sample return” must’ve just been a shipment of astronaut ice cream.
Oh yeah, you gotta do periodic quality control… You gotta, don’t Cha know.
Phineas and Ferb reference? They made moon ice cream…
Depends on how bright the lights are, and the phase.
Cool idea for an SOS beacon: Just shoot an incredibly bright red light at Earth and they’ll know something bad is happening :D
If you can shoot an incredibly bright red light, chances are you can shoot an incredibly bright radio light too.
Someone’s already given an answer for a non-illuminated structure, but the necessary brightness of a light to be visible is also an interesting question.
We’ll assume the light is located on the dark portion of the Moon. From experience, the dimmest stars clearly visible with the naked eye when right next to the Moon are around magnitude 1, which is about 3.6x10^9 photons/sec/m^2.
If we focus the light on the near hemisphere of the Earth (which has an area of 2.5x10^14 m^2) we need to produce 9x10^23 photons/sec. A green photon has an energy of around 3.7x10^-19 joules, so the total power output is 9x10^23 x 3.7x10^-19 = 333 kW.
For reference, this is roughly comparable to the wattage of the fastest electric car chargers. It’s a lot of power, but well within the capability of a small lunar solar farm.
Geographically big, if you need to see it as more than a point. 20/20 human visual acuity is around an arcminute, or 1/60 of a degree. The whole moon is about a 1/2 a degree across as seen from Earth. Below that, it’s a matter at how good you are at picking reddish grey out from normal grey. Red is also scattered and absorbed pretty strongly by the atmosphere IIRC.
If it’s allowed to emit light, you can go a lot smaller. I’d guess an LED array with an emitting area of 10 m2 would be visible from Earth as a sort of star when in shadow, without doing any actual math on it. When competing with sunlight it becomes a visual processing problem again.
If it’s allowed to focus at you, too, a 1 cm aperture should be able to resolve a single kilometer on Earth, and at a kilometer of distance you can see a candle flame if it’s dark enough. It’s just a beefy laser pointer at that rate.
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