<p>Concerning your first question, a simple estimation can be done assuming the distance Earth-Moon ≅ 4·10⁵km, and the orbit circular. So you can calculate the distance as a circumference (C=2πr) like that:</p>

<p>2π·4·10⁵km
=8π·10⁵km 
≅ <strong>2.4 millions of kilometers</strong></p>

<p>Of course you can do more precise calculations, but sometimes is good to have at first an idea of the orders of magnitude.</p>

CC BY-SA 3.0
2013-11-20T07:54:02.157Z
Ohh really that's an other way of calculating distance thanks would have voted up if i could.. anyways thanks..
Asadullah Ali
CC BY-SA 3.0
2013-11-20T08:00:00.133Z
you're welcome, and don't care about the upvote ahha. I suggest you to use this kind of technique (raw approximations) to have a first clue of what an answer of a problem could be. This will help you furthermore to find errors while using a software or a calculator.
leonard vertighel
CC BY-SA 3.0
2013-11-20T12:26:33.033Z
Amazing, how close this answer is to the other one, although you did quite a few approximations!
Arne
CC BY-SA 3.0
2013-11-20T07:48:36.053Z
2013-11-20T07:48:36.053Z
leonard vertighel
<p>The Moon has an orbital eccentricity of 0.0549, so its path around the Earth is not perfectly circular and the distance between the Earth and the Moon will vary from the Earth's frame of reference (Perigee at 363,295 km and apogee at 405,503 km), see for example <a href="https://astronomy.stackexchange.com/a/22/13">second animation explaining Lunar librations in this answer</a>.</p>

<p>But its orbit can be said, in an oversimplified manner, to be periodic, with no significant <a href="https://en.wikipedia.org/wiki/Precession" rel="nofollow noreferrer">apsidal precession</a> (not really true, but somewhat irrelevant for my following musings here to be still close enough), so we can calculate its orbital length based on its quoted average orbital speed of 1.022 km/s and orbital period of 27.321582 days.</p>

<p>So, plugging our numbers in a calculator, $l = v * t$, we get the Moon's orbital length of 2,412,517.5 km (or 1,499,070 miles). Should be close enough. Source of all orbital elements of the Moon is <a href="https://en.wikipedia.org/wiki/Moon" rel="nofollow noreferrer">Wikipedia on Moon</a>.</p>

CC BY-SA 4.0
2022-03-18T15:29:15.863Z
Earth orbit around the sun is an ellipse, and so is the moon's orbit (almost). Since you are asking for the distance around the sun _during one orbit around earth_, that distance would be about one 13th of the orbit around the sun, but it would be quite variable depending on where exactly you are on that ellipse around the sun.
gnasher729
CC BY-SA 3.0
2013-11-20T12:47:07.487Z
What if you want to know the motion of the moon around the sun? How would you compute that?
Arne
CC BY-SA 3.0
2013-11-20T12:55:29.997Z
@Arne Heh, I'll take your question as a well meaning brain teaser. :) There is always this question of what's your frame of reference, of course, but one relatively easy way would be calculating the length of one Earth's orbit around the Sun and the Moon's path as a [helix](https://en.wikipedia.org/wiki/Helix) with radius the Moon's Semi-major axis, and height of one rotation 365.25 / 27.321582 days. Should be close enough. ;)
TildalWave
CC BY-SA 3.0
2013-11-20T13:04:21.847Z
Yes, I thought of a similar thing. Wikipedia states that the orbit of the Moon around the Sun is convex however, since the Sun's influence is much greater than the Earth's influence. So I don't know if a helix would be a good approximation... Maybe for the Sun/Jupiter/Ganymede system...?
Arne
CC BY-SA 3.0
2013-11-20T13:08:39.267Z
@Arne Ah yes, [the Moon's path around the Sun is convex](http://www.math.nus.edu.sg/aslaksen/teaching/convex.html), but that is irrelevant for calculating the length of the path it makes. You see, you can bend any spring into a convex curvature, but the wire it's made of will still be of equal length. ;)
TildalWave
CC BY-SA 3.0
2013-11-20T13:14:44.363Z
Good point! Though I still don't quite know how to calculate it. :)
Arne
CC BY-SA 3.0
2013-11-20T13:23:45.293Z
@Arne Well the equation for the length of a circular helix is given by [Wikipedia on Helix](https://en.wikipedia.org/wiki/Helix#Arc_length.2C_curvature_and_torsion), and the remaining required orbital parameters should be also listed in [Wikipedia on Earth](https://en.wikipedia.org/wiki/Earth), but if you're after a fast approximation, then calculate the length of the Earth's orbit the same way I calculated the Moon's around the Earth, and then simply add to it my Moon around the Earth multiplied by 365.25 / 27.321582. It won't be much off from the real helix length (not more than ±5.5% off).
TildalWave
CC BY-SA 3.0
2013-11-20T18:01:25.903Z
@Arne, in ~27.3 days the Moon complete an orbit around the Earth. So in 365 days it completes 13.4 orbits. If one orbit is ≅ **2.4** millions of kilometers (look at my answer), in **13.4** orbits it completes 32.2 millions of kilometers around the Earth. Then, we add the path of the Earth around the Sun: 2π·150 millions of Km (≅ **942** millions of km), and we obtain **974 millions of km around the Sun**.
leonard vertighel
CC BY-SA 3.0
2013-11-21T08:17:06.830Z
@leonardvertighel ok, so orbits are linear and/or additive? I was wondering that. I thought it might be a bit more complicated, since orbits are a function of time, and the orbits are not simple, linear functions.
Arne
CC BY-SA 3.0
2013-11-21T12:10:01.330Z
@Arne No, orbits are nearly as complicated as you want them to be. For example, we didn't even consider perturbations, anomalies, precession, even radiation pressure and space weather. But here's the catch, you _have to_ decide at which point you stop appreciating any effects as meaningful for your needs, otherwise it becomes impossible to calculate, while you're only moving your object a few millimeters, perhaps. Periodic corrections are your best friend with orbits, otherwise it gets insanely complicated, even with relatively simple Keplerian ones in classical mechanics.
TildalWave
CC BY-SA 3.0
2013-11-22T06:03:54.450Z
@Arne, as TildalWave stated, my approach was just a simplification to have a raw estimation. You did not know before if it was 1 million km, 10 millions, 100 millions, or more? Now you have a starting point
leonard vertighel
CC BY-SA 3.0
2013-11-20T07:08:34.320Z
2013-11-20T07:08:34.320Z
2017-04-13T12:59:06.910Z
TildalWave
<p>Also, does it always take the same amount of time, or does it fractionally differ on each revolution?</p>

<the-moon><orbit><earth><distances>
What is the distance that the Moon travels during one orbit around the Earth?
Asadullah Ali
# What is the distance that the Moon travels during one orbit around the Earth?

(Asked by: username_0 on 2013-11-20)

Also, does it always take the same amount of time, or does it fractionally differ on each revolution?

**Tags:** the-moon, orbit, earth, distances

---

## 2 answers

### Answer by username_3 on 2013-11-20. Score: 15 | Accepted answer

The Moon has an orbital eccentricity of 0.0549, so its path around the Earth is not perfectly circular and the distance between the Earth and the Moon will vary from the Earth's frame of reference (Perigee at 363,295 km and apogee at 405,503 km), see for example [second animation explaining Lunar librations in this answer](https://astronomy.stackexchange.com/a/22/13).

But its orbit can be said, in an oversimplified manner, to be periodic, with no significant [apsidal precession](https://en.wikipedia.org/wiki/Precession) (not really true, but somewhat irrelevant for my following musings here to be still close enough), so we can calculate its orbital length based on its quoted average orbital speed of 1.022 km/s and orbital period of 27.321582 days.

So, plugging our numbers in a calculator, $l = v \* t$, we get the Moon's orbital length of 2,412,517.5 km (or 1,499,070 miles). Should be close enough. Source of all orbital elements of the Moon is [Wikipedia on Moon](https://en.wikipedia.org/wiki/Moon).

> **Comments:**
> **username_2**: What if you want to know the motion of the moon around the sun? How would you compute that?  
> **username_3**: @Arne Heh, I'll take your question as a well meaning brain teaser. :) There is always this question of what's your frame of reference, of course, but one relatively easy way would be calculating the length of one Earth's orbit around the Sun and the Moon's path as a [helix](https://en.wikipedia.org/wiki/Helix) with radius the Moon's Semi-major axis, and height of one rotation 365.25 / 27.321582 days. Should be close enough. ;)  
> **username_2**: Yes, I thought of a similar thing. Wikipedia states that the orbit of the Moon around the Sun is convex however, since the Sun's influence is much greater than the Earth's influence. So I don't know if a helix would be a good approximation... Maybe for the Sun/Jupiter/Ganymede system...?  
> **username_3**: @Arne Ah yes, [the Moon's path around the Sun is convex](http://www.math.nus.edu.sg/aslaksen/teaching/convex.html), but that is irrelevant for calculating the length of the path it makes. You see, you can bend any spring into a convex curvature, but the wire it's made of will still be of equal length. ;)  
> **username_2**: Good point! Though I still don't quite know how to calculate it. :)  
> **username_3**: @Arne Well the equation for the length of a circular helix is given by [Wikipedia on Helix](https://en.wikipedia.org/wiki/Helix#Arc_length.2C_curvature_and_torsion), and the remaining required orbital parameters should be also listed in [Wikipedia on Earth](https://en.wikipedia.org/wiki/Earth), but if you're after a fast approximation, then calculate the length of the Earth's orbit the same way I calculated the Moon's around the Earth, and then simply add to it my Moon around the Earth multiplied by 365.25 / 27.321582. It won't be much off from the real helix length (not more than ±5.5% off).  
> **username_1**: @Arne, in ~27.3 days the Moon complete an orbit around the Earth. So in 365 days it completes 13.4 orbits. If one orbit is ≅ **2.4** millions of kilometers (look at my answer), in **13.4** orbits it completes 32.2 millions of kilometers around the Earth. Then, we add the path of the Earth around the Sun: 2π·150 millions of Km (≅ **942** millions of km), and we obtain **974 millions of km around the Sun**.  
> **username_2**: @leonardvertighel ok, so orbits are linear and/or additive? I was wondering that. I thought it might be a bit more complicated, since orbits are a function of time, and the orbits are not simple, linear functions.  
> **username_3**: @Arne No, orbits are nearly as complicated as you want them to be. For example, we didn't even consider perturbations, anomalies, precession, even radiation pressure and space weather. But here's the catch, you _have to_ decide at which point you stop appreciating any effects as meaningful for your needs, otherwise it becomes impossible to calculate, while you're only moving your object a few millimeters, perhaps. Periodic corrections are your best friend with orbits, otherwise it gets insanely complicated, even with relatively simple Keplerian ones in classical mechanics.  
> **username_1**: @Arne, as TildalWave stated, my approach was just a simplification to have a raw estimation. You did not know before if it was 1 million km, 10 millions, 100 millions, or more? Now you have a starting point  
> **username_4**: Earth orbit around the sun is an ellipse, and so is the moon's orbit (almost). Since you are asking for the distance around the sun _during one orbit around earth_, that distance would be about one 13th of the orbit around the sun, but it would be quite variable depending on where exactly you are on that ellipse around the sun.  

---

### Answer by username_1 on 2013-11-20. Score: 10

Concerning your first question, a simple estimation can be done assuming the distance Earth-Moon ≅ 4·10⁵km, and the orbit circular. So you can calculate the distance as a circumference (C=2πr) like that:

2π·4·10⁵km
=8π·10⁵km
≅ **2.4 millions of kilometers**

Of course you can do more precise calculations, but sometimes is good to have at first an idea of the orders of magnitude.

> **Comments:**
> **username_0**: Ohh really that's an other way of calculating distance thanks would have voted up if i could.. anyways thanks..  
> **username_1**: you're welcome, and don't care about the upvote ahha. I suggest you to use this kind of technique (raw approximations) to have a first clue of what an answer of a problem could be. This will help you furthermore to find errors while using a software or a calculator.  
> **username_2**: Amazing, how close this answer is to the other one, although you did quite a few approximations!  

---

