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Space.com writes to say (freeform improved quote):

The LLCD (Lunar Laser Communication Demonstration) space terminal is designed to beam data from the moon at a rate six times faster than the best advanced radio communications systems available today. The system consists of a laser mount on the LADEE (Lunar Atmosphere and Dust Environment Explorer) spacecraft which fires a 0.5-watt laser through a 10-centimeter telescope... The primary ground station, called the Lunar Lasercom Ground Terminal, is located in White Sands, New Mexico, near Las Cruces and features a set of four 40-centimeter telescopes to receive the signal.

Wikipedia indicates 4 db/km attenuation in optical fibres was achieved decades ago. In free space, the attenuation would be higher; which brings me to my questions:

  • How much power does the LLCD lose in transit from Lunar orbit to the ground station?
  • Could the gas in the laser module freeze when in space?
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    $\begingroup$ Why would the loss be higher in free space than optical fibers? $\endgroup$ – PearsonArtPhoto Sep 17 '13 at 12:09
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Pulling some info from the Fact Sheet and this website. The formula, from Wikipedia, is

$\theta \simeq \frac{\lambda}{\pi w_0} \qquad (\theta \mathrm{\ in\ radians}). $

I'm going to assume $W_0$ is the aperture width provided. That gives a beam width, assuming 1.1 um wavelength, of about 3.5 uRads

Okay, at that value at the distance of the Earth will give a spot size of about 2.5 km, or an area of 4,900,000 m^2. The total aperture on Earth is about 0.5 m^2. The difference between them gives a maximum path loss of 70 dB. There will be some atmospheric path loss as well, and no doubt the system isn't quite this ideal, but I would expect the total path loss to be around 70-80 dB.

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