[Laser] pictures from the Moon

[Art]

At 12:07 PM 2/22/04, you wrote:
>On Sun, 22 Feb 2004 11:30:24 EST, you wrote:
>
> >The Moon is tide locked
> >to the Earth so that once aligned, neither the camera nor the downlink
> >antennas would need to be adjusted.
>
>If I remember right, there's some wobble to it. I wonder how the
>wobble compares to the spread of the laser beam?
>
>Of course, the trick would be to make the laser beam wide enough that
>the variation in the moon's orientation w.r.t. the earth wouldn't
>matter.

The wobble is called libration. It happens because the Moon wants to turn, 
but the Earth's gravity yanks it back-there is a proverbial tug of war.

The Earth has libration as well, which is influenced by the Sun, other 
planets and the Moon itself.

I doubt that a laser beam that is spread out over 470 miles diameter would 
be readable on earth for 2 reasons....

#1      The Moon itself has libration, which causes the ambient light level 
from the Moons reflection to vary in phase and amplitude. This variation is 
much stronger than the laser is and would be interference.

#2      The field strength of a laser spread over a 470 mile diameter would 
be nil, you just wouldn't be able to have a light collector large enough to 
recover any real amount of laser power. Let's say the laser was 1 watt and 
that there was no loss in the receiver optics or the collimation optics on 
the transmitter. Let's also say there is no attenuation or scattering from 
the Earth's atmosphere. Let's also assume our receiver has a 10 femtowatt 
NEP, which is about what a PM tube has. Let's also assume there was no 
interference from the ambient light reflected from the Moon (which is a 
combination of Sunshine and Earthshine).

The area of the footprint on the Earth is Pi X R X R, which is 3.14 X 235 X 
235 (in miles), which is about 173,000 square miles, which is also about 8 
X 10e17 square feet.

The total area of the footprint on the earth is 8 X 10e17 square feet.

10 femtowatts divided by the output power of the laser is 1watt/10 X 
10e-15= .1 X 10e15 or 1 X 10e14.

The area of our lens (in square feet) needs to be 1 X 10e14/8 X 10e17=125 
square feet.

Thus, we would need a 125 square foot lens to recover 10 femtowatts of the 
original signal.

My math is rusty, perhaps I made an error?

It seems to me that a simple fixed laser pointed towards the Earth is not 
going to be useful for any sort of real data transfer.

I think the laser must be collimated to a very narrow beam and that it has 
to be a tracking laser in order to keep it pointed at the same spot on Earth.

I would think that a radio link would be much more practical.

Regards,

Art