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, 2 mrad 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.

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.

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, gives a beam width of about 3.5 uRads

Okay, 2 mrad 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.

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, 2 mrad 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.