<p>Well it kind of makes sense, but what is important here is the Earth's velocity relative to the Moon, not the Sun!</p>

<p>Roughly: the Moon-Earth separation is 400,000 km and a complete orbit takes 27 days. Thus the relative orbital speed is more like 1 km/s and so a 20km diameter beam easily covers the expected relative motion of about 2.5 km during the 2.5s round trip. </p>

CC BY-SA 3.0
2016-05-08T20:31:03.033Z
Thanks!  If light follows a "straight line" no matter what, wouldn't it ignore the Earth's orbital velocity around the sun?  If that is the case, wouldn't the emitter be gone by the time light returned? I am a beginner so I am likely missing something conceptually.
Rama Set
CC BY-SA 3.0
2016-05-10T18:39:50.053Z
Would relativity also play a part?
Tanenthor
CC BY-SA 3.0
2016-05-10T19:07:13.690Z
@Tanenthor It might... if any of the velocities were large.
ProfRob
CC BY-SA 3.0
2016-05-08T20:24:55.243Z
2016-05-08T20:24:55.243Z
ProfRob
<p>As I understand it, light that is emitted from a source is not imparted with the motion of the source and so always follows a "straight line". If this is correct, I am having a difficult time conceiving how the Lunar Laser Ranger experiments can detect photons.</p>

<p>In this experiment, a pulse of light is aimed towards a retro-reflector and it is reflected back. It takes about 2.5 seconds round trip and due to diffraction, the returning pulse covers a circle of approximately 20kms in diameter.</p>

<p>The Earth's orbital velocity is approximately 30km/s and in the time the pulse takes to make the round trip, the detector would be 75kms further along the Earth's orbital path. If my first paragraph is accurate, and the light pulse is not imparted with the Earth's orbital velocity, then how is detection of the returning light pulse achieved?</p>

<p>I hope this makes sense and thank you.</p>

CC BY-SA 3.0
2016-05-08T20:33:32.327Z
You have discovered one of the inconsistencies that you need relativistic mechanics to understand. A good frame of reference to choose in this situation is the one where the Earth is stationary. That the whole system has a velocity relative to the Sun is irrelevant. If the Earth has a velocity or not is actually your choice.
SE - stop firing the good guys
CC BY-SA 3.0
2016-05-08T20:43:49.500Z
Thanks for the reply. So then, from the Earth the light pulse goes to the target and returns in an apparently straight line, but from Venus, for example, the light path might appear not straight and perhaps with some Doppler shifting?
Rama Set
CC BY-SA 3.0
2016-05-08T20:46:42.353Z
Sure, from Venus it may seem like it has another wavelength, takes more or less time to return and the like. It just appears different from different frames of reference.
SE - stop firing the good guys
<orbit><the-moon><earth><light><astrophysics>
LLR & Orbital Motion
Rama Set
# LLR & Orbital Motion

(Asked by: username_0 on 2016-05-08)

As I understand it, light that is emitted from a source is not imparted with the motion of the source and so always follows a "straight line". If this is correct, I am having a difficult time conceiving how the Lunar Laser Ranger experiments can detect photons.

In this experiment, a pulse of light is aimed towards a retro-reflector and it is reflected back. It takes about 2.5 seconds round trip and due to diffraction, the returning pulse covers a circle of approximately 20kms in diameter.

The Earth's orbital velocity is approximately 30km/s and in the time the pulse takes to make the round trip, the detector would be 75kms further along the Earth's orbital path. If my first paragraph is accurate, and the light pulse is not imparted with the Earth's orbital velocity, then how is detection of the returning light pulse achieved?

I hope this makes sense and thank you.

**Tags:** orbit, the-moon, earth, light, astrophysics
> **Comments:**
> **username_1**: You have discovered one of the inconsistencies that you need relativistic mechanics to understand. A good frame of reference to choose in this situation is the one where the Earth is stationary. That the whole system has a velocity relative to the Sun is irrelevant. If the Earth has a velocity or not is actually your choice.  
> **username_0**: Thanks for the reply. So then, from the Earth the light pulse goes to the target and returns in an apparently straight line, but from Venus, for example, the light path might appear not straight and perhaps with some Doppler shifting?  
> **username_1**: Sure, from Venus it may seem like it has another wavelength, takes more or less time to return and the like. It just appears different from different frames of reference.  

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## 1 answers

### Answer by username_2 on 2016-05-08. Score: 1

Well it kind of makes sense, but what is important here is the Earth's velocity relative to the Moon, not the Sun!

Roughly: the Moon-Earth separation is 400,000 km and a complete orbit takes 27 days. Thus the relative orbital speed is more like 1 km/s and so a 20km diameter beam easily covers the expected relative motion of about 2.5 km during the 2.5s round trip.

> **Comments:**
> **username_0**: Thanks!  If light follows a "straight line" no matter what, wouldn't it ignore the Earth's orbital velocity around the sun?  If that is the case, wouldn't the emitter be gone by the time light returned? I am a beginner so I am likely missing something conceptually.  
> **username_3**: Would relativity also play a part?  
> **username_2**: @Tanenthor It might... if any of the velocities were large.  

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