If someone (or some robot) on the moon were to point a laser at the Earth, how many watts would the laser need, to be easily seen with the un-aided eye from Earth?

Obviously this depends on a number of factors. Assume the beam is made divergent enough to cover latitudes 60S to 60N, wavelength 532nm, and the beam originates from a fully lit area of the moon, and the observing is done in full darkness.

I can work out watts/meter-square from the laser, but don't know how to work out the watts/meter-square from the moon, and don't have a clue about how the eye would perceive the contrast.

  • $\begingroup$ I think the atmospheric conditions and pollution would play a very important role here, right? $\endgroup$ – fortran Feb 16 '11 at 11:44
  • $\begingroup$ @fortran - yeah, no doubt. So assume nice clear skies w/little pollution. $\endgroup$ – JustJeff Feb 16 '11 at 11:49

There are two things that make it difficult to see the laser: 1) the large area illuminated (as BarsMonster pointed out) and 2) the resolving power of the eye.

The eye has an angular resolution of roughly 10^-4 radians (from the diffraction limit), so your eye will "blur" the light from the laser with all the light from the moon's surface within an angle of 10^-4. Using the earth-moon distance of 4e8 m, that means you're competing with the light from a 10^9 m^2 area. Since you say it's a fully lit area of the moon, I'll take an intensity of 1 kW/m^2 for the solar intensity and assume the moon is white (a bad assumption) for a total power of 10^12 W.

The light from the moon's surface is scattered into something like 2-pi of solid angle (as opposed to the laser). So the fraction of light that will be competing with the laser (assuming the laser diameter at the earth is 4000 km) is roughly ((4000 km)/(400000 km))^2= 10^-4.

So if we want the intensity from the laser to equal the intensity of the unresolved scattered light from the moon (which, as a ballpark figure, should make it noticeably green and thus "easily seen"), we need a laser that's 10^8 Watts.

Good luck building that.

  • $\begingroup$ Beautifully done, a very complete answer. +1 when I get votes again tomorrow. $\endgroup$ – Carl Brannen Feb 16 '11 at 2:00
  • $\begingroup$ i think you've convinced me. conclusion - you'd need ridiculously stupendous amounts of power, and even then you'd just be changing the tint of one spot, which could easily be missed. so, if you wanted to undeniably prove to people 'hey, someone is mucking about on the moon', you'd probably have better luck as BarsMonster suggests, using a large Xe-flash. $\endgroup$ – JustJeff Feb 16 '11 at 3:04
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    $\begingroup$ There are ways to "be seen" from the moon that would require much less power: illuminate a smaller area on the earth with your laser, and wait until the location of your laser on the moon is dark. $\endgroup$ – Anonymous Coward Feb 16 '11 at 21:12
  • $\begingroup$ Given that a 1.25 petawatt (1.25 * 10^{15} watt) laser was built in 1996 ( llnl.gov/str/Petawatt.html ), building a 10^8 watt laser is not as difficult as you might think. $\endgroup$ – las3rjock Feb 16 '11 at 22:27

I afraid if you going to light entire earth with a "laser" it will not be visible from earth. With the same success you can take 10kW xenon lamp which is easily focusable to the required beam for way lower price, weight and better efficiency.

Rough estimation: You are lighting 4000km x 4000km area, human eye has surface 7x7mm, so you will 'see' 0.32x10^-18 of the power you emit.

1 photon (555nm) has energy 1.54x10-12 J, so if you will have 1000W laser, your eye will get 1 photon each 5000 seconds. Which is obviously very very far away from being visible.

  • $\begingroup$ Nice answer. +1 when I get my vote back tomorrow. $\endgroup$ – Carl Brannen Feb 16 '11 at 2:00
  • $\begingroup$ yeah, it's how to think about how the laser (or Xe lamp) competes with the diffuse scattering that stumps me. fwiw, i only said laser b/c i figured the inherent directionality would deliver light on target better, but if you start from input power, you're probably right about the efficiency. $\endgroup$ – JustJeff Feb 16 '11 at 2:55

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