# Moon signal bounce

1. Apr 18, 2013

### Mentallic

I've been reading on how the distance to the moon is calculated by bouncing a laser signal on the retro-reflectors and measuring the time it takes the light to return to Earth, but what I've seen is that the retro-reflectors do a good job of minimizing scattering by returning the signal along the same direction in which it arrived.

Since the light takes approximately 2.5 seconds to reach the moon and back, and since being near the equator of the Earth, the rotational speed is about 1,600km/h on the surface or about 1.2km / 2.5s, so my question is how do the experimenters receive the signal if the retro-reflectors bounce the signal back to where the signal was first emitted which by that point is 1.2km off the mark?

2. Apr 18, 2013

### Mentallic

Actually, it probably has to do with beam divergence.

3. Apr 18, 2013

### Staff: Mentor

Yep. A laser beam spreads to something like a mile or so by the time it reaches the moon if I remember correctly.

4. Apr 18, 2013

### I like Serena

Typical retro-reflectors also give a little divergence.

5. Apr 18, 2013

### Mentallic

And the small portion of reflected light would diverge by the same amount when it hits the Earth again?

By divergence, I'm talking about how laser lights will expand into a cone shape. Do you mean scattering?

6. Apr 19, 2013

### I like Serena

The reflected light would have a diverging angle that is much smaller.
That is because the part of the cone that actually hits the retro-reflectors has a very small diverging angle.

However, the angle between the mirrors in retro-reflector material is intentionally not exactly 90 degrees.
So the returning beam will have a cone shape again.

7. Apr 19, 2013

### willem2

If the retro-reflectors were that accurate, the signal would actually ahead(to the east) of the laser, because the moon moves at about 1 km/s, so the beam has to be sent to a point 1.25 km ahead of the reflector, and will be reflected to a point 2.5 km ahead of the laser, while the earth has only moved 1.2 km in that time.

8. Apr 19, 2013

### CWatters

This article says..

http://www.fesg.bv.tum.de/91872-bD1lbg-~fesg~forschung~llr.html

The pulses are short enough that a very large number can be sent before any come back.

9. Apr 19, 2013

### A.T.

This is just semi related, but might be interesting to some. In the days before human-made communication satellites, we were using a natural one:

http://en.wikipedia.org/wiki/EME_(communications [Broken])

Last edited by a moderator: May 6, 2017
10. Apr 19, 2013

### sophiecentaur

This is not what they achieve - because, when they receive the echo, they are not in the position that they were when the pulse was emitted. All that is necessary is for the reflector to reflect sufficient light in the direction of where they will be 5 seconds later. The angle that this represents is very small, surely? (1.2km in 380Mm = about one second of arc, I think)

@CWatters: If you send the pulses at a rate that is faster than the transit time, you can get a false distance measurement because you can get the phase of the transmitted and received pulse streams wrong. I do agree that you can eliminate this problem by starting with widely spaced pulses and working up, so you can increase the signal to noise ratio and, hence, the accuracy.

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