What kind of equipment, platorm, etc. would be required to project an image on to the surface of a new moon, so that it is visible from Earth?
I’m assuming that you couldn’t project from Earth due to the atmosphere. Should be visible with nothing more advanced than nice binoculars or hobby store telescope.
A new moon is usually only a few degrees from the Sun, meaning that you’d need something pretty bright indeed to show up brighter than the light from the daytime sky. Even if you’re talking about a crescent moon that’s about 3–4 days before/past new — so that it’s fairly far away from the Sun and visible at night — you would need something that’s comparable to the brightness of Earthshine at the Moon’s surface; otherwise it would just get washed out. So you need a light source that, at a distance of a quarter-million miles, is as bright as all the sunlight reflected from the Earth’s surface.
My WAG is that it’s somewhere between “feasible with current technology but ridiculously expensive” and “can’t be done”, but I’m willing to stand corrected if anyone has a better estimate.
So here are some rough calculations concerning how much energy you would need: if you buy this site’s calculations, the intensity of earthshine at the Moon’s surface is about 0.1 W/m[sup]2[/sup]. Moreover, with binoculars you can resolve features on the moon that are tens of arcseconds across. So let’s say that you want to create an image on the moon that is a couple of arcminutes (i.e., 120 arcseconds) across. This would be about 100 km across on the Moon’s surface. The total power of the Earthshine hitting that surface would then be about (0.1 W/m[sup]2[/sup])*(10[sup]5[/sup] m)[sup]2[/sup] ≈ 1 gigawatt. Your Earth-based source would have to provide more light power than that to outshine the Earth on that patch of the Moon.
Beyond this, there are a few more problems I can think of: how efficiently can you turn the electrical power into light? (Most light sources are pretty inefficient). And how do you to focus the light and create an image without melting the lenses/mirrors you’re using to do it? I don’t know if these are insurmountable obstacles, but they’re certainly not trivial.
It would be vastly easier to change the color. I was thinking a while back that you could use fluorescent powder that emitted in a narrow frequency band–you could then filter out everything but that color and see the image (i.e., you don’t have to fight against the entire luminance of the moon). Several hundred kilograms of powder could be enough for something Earth-visible if distributed well.
Would it be absolutely necessary to provide the light source ourselves, instead of using the sun or earth as a giant backlight? Could we just use a lens array made out of microsatellites to filter the incoming light into a certain shape, kind of like a space shadowbox?
Your effort may be in vain and no longer relevant. That aliens who landed on the dark side are already having an image party back there. You just cannot see it. They established their base just yesterday. This was taken during their final approach:
Maybe, like early TVs, we can cheat by scanning the image, thus reducing power needed. If we combine scanning + selective wavelengths + a telescope on the borderline of “hobby store”, I think we might just be able to do it.
Yes, a fun read as usual, but I think a major problem was way at the beginning, when he wrote, “Let’s assume everyone has steady enough aim to hit the Moon…”
Whoa! From this amateur’s experience with telescopes and binoculars, it just ain’t gonna happen. Looking through one of those, I feel like a fool bobbing my aim here and there until I see some random whiteness, and then try to backtrack until I find it again. And that is WITH the excellent feedback provided by the awesome sight in the viewfinder.
But 6 billion noobs aiming their flashlights? They won’t have a clue as to whether their aim is high, low, left, or right. I doubt even 1% will be lucky enough to hit it.
Right, but it’s only lit for 1/Xth of the time, where x is the number of pixels. So, to get the same brightness, it has to be X times brighter than it would if it were static, so the power is the same.