Please correct me if I’m wrong with Facts. All we need is a corner reflector on the moon. 3 mirrors glued together so they are perpendicular. 3 sides of a cube. The other 3 are open. Like a radar reflector on a ship. It is not perfect but nearly so, such that a perfect powerful red laser focused at that distance reflects back to the earth with a dot a few feet away from where I’m standing. You would calibrate this error by measuring the x,y distance, with respect to the reflector, at the north pole, or when the moon is near the horizon. Your target would be perpendicular to the moon. It’s not so hard as it seems because the cube would not rotate, as the same side of the moon is always facing us. You would expect this x,y offset to remain for awhile. So instead let’s just make it a perfect cube always coming back to exactly the laser source. This is easier to imagine.
But would it come back on the same line exactly? Please excuse any math errors. It takes 2.5s for light to travel to the moon and back. During that time the equator rotates 1160m. So the dot would be 1160m to the West of where it started. Am I correct so far? There’s more.
Sending any light source to the Moon’s retroreflectors is going to be dispersed in a cone shape - and it’s going to come back via another cone shape that covers the ground lost due to Earth’s rotation.
At the Moon’s surface, the beam is about 6.5 kilometers (4.0 mi) wide
By the time it gets back to Earth, it would be even wider, so movement of about a kilometer won’t matter so much. But there’s also this:
The reflected light is too weak to see with the human eye. Out of a pulse of 3×1017 photons[25] aimed at the reflector, only about 1–5 are received back on Earth, even under good conditions.[26]
This isn’t something you’d whip up in your backyard. You’d need really good telescopes and detectors to accurately detect 5 photons.
Well, then, yes, it would be a factor, but just one amongst many. From that page:
To compute the lunar distance precisely, many factors must be considered in addition to the round-trip time of about 2.5 seconds. These factors include the location of the Moon in the sky, the relative motion of Earth and the Moon, Earth’s rotation, lunar libration, polar motion, weather, speed of light in various parts of air, propagation delay through Earth’s atmosphere, the location of the observing station and its motion due to crustal motion and tides, and relativistic effects.[20][21]
Most of these won’t matter. We are measuring the location of the dot, not the precise time. If we could do this experiment it would be close to the results we predicted. ±1%
It’s totally unclear to me what the OP is actually asking.
If you assume a perfect laser beam with zero spread or loss, and you assume a perfect retro-reflector with zero spread or loss, and you assume no atmosphere to add spread or loss or diffraction or whatever, then …
What are you asking about?
Does the beam return to Earth? Yes. Does it return to the same spot on Earth? No. For the big reason you mentioned, Earth’s rotation during the transit time, and then for another dozen minor reasons because you haven’t magically assumed them out of existence with your perfect this, that, and the other thing, but not magic perfect everything.
I probably don’t understand the question but does the OP know that the reflected beam is offset from the original beam even if you’re directly in front of the retroreflector?
It’s a thought experiment. I am only asking about the thing, much bigger than everything else, causing the dot to move from it’s origin. Thank you for your answer LSLguy.
You’ll be happy to know I’m coming back to reality in the next part. Had to overcome some theory first to understand the next bit.
Probably also depends on where you are on Earth. At the equator, the return beam will be offset a lot more than it would be at the poles. If you were at the rotational north pole, could you send the perfect beam and have it come back to you?
Perhaps you don’t realize this, but even a “perfect laser” spreads out. Once you get beyond a distance called the confocal parameter (or half the Rayleigh Range) from the laser’s smallest point – its beam waist the beam spreads out at a constant angle, making that “cone shape” people have brought up. Lasers don’t produce narrow, line-like beams forever. Get far enough away and it’s a cone. A pretty wide one by the time you get to the moon. So any offset is going to be lost in the width of any real laser beam.
A Gaussian Beam is as close to a “perfect laser” as you’re going to get
Maybe you’re right. I don’t want to debate this. The exact possible size of the dot is not relevant. Even if the cone is 100 miles across, it is brighter in the center, since it is a gaussian beam. Measurably so. This is not my point.