[Laser] heliograph vs. laser indeed!

[F1AVYopto at aol.com]

I did rough calculations with your initial conditions.
These  calculations are simplified to get only orders of range.
A 1 m² mirror  gives to 200 Km a (400pi / 360) x 0,5 = 1,7 Km in diameter 
spot with a 0,5° beam  divergence.
In vacuum, the spot power density will be 1000 / 850².pi =  0,00044 W/m² for 
1 KW reflected power.
With a 5 mW laser the same laser  density will be in a 0,005 / 0,00044 = 11 
m² spot  (3,8 m in diameter  !)
The laser beam angle must be 2pi / (400pi / 0,0038) = 2.10-5 radians = 20  
micro radians !
A 20 micro radians beam is only possible with a very big  telescope as 
collimator.
On the moon the spot will be smaller than 7 Km  !
A so small spot needs a more than 1 m aperture telescope !
For the  heliograph or the laser, in a very clear atmosphere, the loss should 
be 0,2 dB /  Km.
The laser wavelength must very well chosen for this minimal loss...
At  200 Km the loss will be 40 dB.
The real power density becomes 44.10^-9 W/m².
For conclusion, for the  same optical brightness, the heliograph is very 
better than the laser for  simplicity …
In the same principle I did another calculation.
I imagine a  simple 1 m² LEDs matrix (for example LUXEONs) on the moon with 
10 W total  optical power.
Each of the modulated LEDs is fitted with a lens that gives an  angular 
dispersion to just cover the earth diameter (about 2°).
The power  density on the earth will be near 8.10^-8 W/Km².
In a square meter 8.10^-8 /  10^6 = 8.10^-14 W.
The atmospheric losses are rather small for a nearly vertical  aiming.  
In a 30 cm telescope with a femto watt capability photo  receiver, a FFTDSP 
extracting, this signal can be detected.
Of course because the light background, the detection will be only possible  
during the moon phases that give darkness to the beacon location.
73 Yves  F1AVY