<p>Originally <a href="https://space.stackexchange.com/questions/65632/is-the-light-time-correction-for-the-moons-position-in-meeus-astronomical-al?noredirect=1#comment219992_65632">posted on Space Exploration SE</a>, I was encouraged to post it here.</p>
<p>I'm looking at the second edition of Meeus's &quot;Astronomical Algorithms,&quot; chapter 47, &quot;Position of the Moon.&quot; At the bottom of page 337, right before getting into the actual formulae of the algorithm, Meeus states</p>
<blockquote>
<p>Moon's mean longitude, referred to the mean equinox of date, and including the constant term of the effect of the light-time (-0&quot;.70):</p>
</blockquote>
<p>The size of the correction corresponds with the light travel time between Earth and the Moon and the Moon's orbital angular velocity, so it's clear to me that it's correcting for the light travel time between these two bodies. My question is, <strong>which direction is the correction?</strong> Is this converting from an inertial calculation (what someone on the Moon would say is the Moon's position, not subject to any light travel time effects) to the apparent Moon position (what an observer on Earth would say is the Moon's observed position, subject to light travel time effects)? Or is this correcting from a calculation of the apparent location to the true/inertial location?</p>
<p>Phrased another way, <strong>is the Moon location algorithm in Meeus providing a true/inertial location (what the Moon would say its location is), or is it providing the apparent position as viewed from Earth, corrected &quot;backwards in time&quot; to account for the light travel time between the two bodies?</strong></p>

CC BY-SA 4.0
2024-03-12T23:26:29.677Z
The way I understood it, it’s to compensate for the fact that we see it as it was about 1 second ago, so that would be for the apparent position.
Pierre Paquette
CC BY-SA 4.0
2024-03-13T02:43:01.363Z
See https://en.wikipedia.org/wiki/Light-time_correction FWIW, [Horizons](https://ssd.jpl.nasa.gov/horizons/manual.html) uses the term "geometric" to refer to the uncorrected state vector.
PM 2Ring
CC BY-SA 4.0
2024-03-13T03:12:14.843Z
You may enjoy these posts on the speed of gravity: https://physics.stackexchange.com/q/27845/123208 & https://physics.stackexchange.com/a/75124/123208
PM 2Ring
CC BY-SA 4.0
2024-03-13T03:51:31.637Z
@PM2Ring I got [taken to school by DavidHammen on that topic](https://space.stackexchange.com/a/21472/12102)
uhoh
CC BY-SA 4.0
2024-03-13T18:21:20.060Z
@PierrePaquette To make sure I understand, you're saying that according to your understanding, Meeus is correcting the true position into the apparent position by including that term?
NeutronStar
CC BY-SA 4.0
2024-03-13T18:22:49.097Z
@PM2Ring Thanks for the link. It doesn't seem to answer my question though, just provides an article on correcting for light travel time. Or am I missing something, does it answer my question?
NeutronStar
CC BY-SA 4.0
2024-03-13T18:26:01.157Z
@NeutronStar: Yes. Position is calculated for where the Moon was when light left it before coming into our eyes. In the meantime, the “real” Moon has moved a little bit, but the position is calculated for the apparent moon.
Pierre Paquette
CC BY-SA 4.0
2024-03-13T20:38:03.657Z
For the importance of this effect, see for example [How does a laser from Earth manage to hit the Moon with precision?](https://physics.stackexchange.com/q/800394/83380) and to a lesser extent, [When they shoot lasers at the Moon for ranging, what is the shape of the beam?](https://space.stackexchange.com/q/50389/12102)
uhoh
<the-moon><algorithm>
Is the "light-time" correction for the Moon's position in Meeus "Astronomical Algorithms" putting the Moon in inertial position or apparent position?
NeutronStar
# Is the "light-time" correction for the Moon's position in Meeus "Astronomical Algorithms" putting the Moon in inertial position or apparent position?

(Asked by: username_0 on 2024-03-12)

Originally [posted on Space Exploration SE](https://space.stackexchange.com/questions/65632/is-the-light-time-correction-for-the-moons-position-in-meeus-astronomical-al?noredirect=1#comment219992_65632), I was encouraged to post it here.

I'm looking at the second edition of Meeus's "Astronomical Algorithms," chapter 47, "Position of the Moon." At the bottom of page 337, right before getting into the actual formulae of the algorithm, Meeus states

> Moon's mean longitude, referred to the mean equinox of date, and including the constant term of the effect of the light-time (-0".70):

The size of the correction corresponds with the light travel time between Earth and the Moon and the Moon's orbital angular velocity, so it's clear to me that it's correcting for the light travel time between these two bodies. My question is, **which direction is the correction?** Is this converting from an inertial calculation (what someone on the Moon would say is the Moon's position, not subject to any light travel time effects) to the apparent Moon position (what an observer on Earth would say is the Moon's observed position, subject to light travel time effects)? Or is this correcting from a calculation of the apparent location to the true/inertial location?

Phrased another way, **is the Moon location algorithm in Meeus providing a true/inertial location (what the Moon would say its location is), or is it providing the apparent position as viewed from Earth, corrected "backwards in time" to account for the light travel time between the two bodies?**

**Tags:** the-moon, algorithm

> **Comments:**
> **username_1**: The way I understood it, it’s to compensate for the fact that we see it as it was about 1 second ago, so that would be for the apparent position.  
> **username_2**: See https://en.wikipedia.org/wiki/Light-time_correction FWIW, [Horizons](https://ssd.jpl.nasa.gov/horizons/manual.html) uses the term "geometric" to refer to the uncorrected state vector.  
> **username_2**: You may enjoy these posts on the speed of gravity: https://physics.stackexchange.com/q/27845/123208 & https://physics.stackexchange.com/a/75124/123208  
> **username_3**: @username_2 I got [taken to school by DavidHammen on that topic](https://space.stackexchange.com/a/21472/12102)  
> **username_0**: @username_1 To make sure I understand, you're saying that according to your understanding, Meeus is correcting the true position into the apparent position by including that term?  
> **username_0**: @username_2 Thanks for the link. It doesn't seem to answer my question though, just provides an article on correcting for light travel time. Or am I missing something, does it answer my question?  
> **username_1**: @username_0: Yes. Position is calculated for where the Moon was when light left it before coming into our eyes. In the meantime, the “real” Moon has moved a little bit, but the position is calculated for the apparent moon.  
> **username_3**: For the importance of this effect, see for example [How does a laser from Earth manage to hit the Moon with precision?](https://physics.stackexchange.com/q/800394/83380) and to a lesser extent, [When they shoot lasers at the Moon for ranging, what is the shape of the beam?](https://space.stackexchange.com/q/50389/12102)  

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## 0 answers

