[Laser] Re: another lunar experiment

[Tom Becker]

 > ... on the outbound leg the path loss is proportional to 
(1/distance**2)...

Is it not true that a collimated beam will deliver the transmit power - 
reduced only by absorption, and not by inverse distance^2 - to a "spot" 
on the lunar surface?  When the spot then reradiates that energy, 
essentially omnidirectionally, inverse-distance applies as the power is 
disbursed over an ever-increasing-by square spherical surface area.  
Isn't the best-case round-trip attenuation then 1/distance^2, and a more 
practical path loss somewhere between 1/distance^2 and 1/distance^4?

If an Earth-borne mirror can reflect a solar beam that's less than the 
lunar disk at that distance, why would not essentially all of that 
energy be delivered to the lunar surface on the outbound path?


Tom
Cape Coral