<p>To answer this, let's visit <a href="https://en.wikipedia.org/wiki/Kepler%27s_laws_of_planetary_motion#Third_law" rel="nofollow">Kepler's third law</a> which tells you the orbital period of a body, if you know how far away it is.</p>

<p>$$\frac{P^2}{a^3} = \mathrm{constant}$$</p>

<p>As it turns out, if you use the correct units, for the orbital period, $P$, and the semi-major axis, $a$, namely years and astronomical units (AU) respectively, then that constant is equal to 1! You have our moon orbiting the sun at one light year, so this needs to be converted to AU. Google tells me that $1\:\mathrm{light\:year}$ is $63241.1\:\mathrm{AU}$. For this specific case, you get that</p>

<p>$$P = a^{3/2} = 63241.1^{3/2} = 15,903,734\:\mathrm{years}$$</p>

<p>Notice that this formula didn't depend on the mass of the moon! (Technically the mass of the moon is in there, but it's pretty negligible and so doesn't matter much.)</p>

<hr>

<p><sup><sub>You may be scratching your head over the units $-$ how can a timescale squared equal a distance cubed?. Technically there's some trickery going on because that constant isn't unitless and I've implicitly assumed those units are hanging around, without explicitly saying so. Rest assured, the units do make sense for this equation.</sub></sup></p>

CC BY-SA 3.0
2016-10-21T17:08:50.777Z
zephyr and @joseph.hainline: This way of phrasing Kepler's law is only valid when the central mass is the Sun (and the mass of the orbiting planet can be neglected)! The orbital period definitely depends on this mass (e.g. for infinite mass the object will fall down, and for negligible mass the object won't orbit at all). Using the more general formula yields a period of $P = 2\pi\sqrt{a^3/GM} \simeq 83$ billion years! The gravitational impact of the Moon from the distance of 1 ly is tiny!
pela
CC BY-SA 3.0
2016-10-21T17:11:54.403Z
The above formula also neglects the mass of the object. If its mass is comparable the the mass of the Moon, use $P = 2\pi\sqrt{a^3/G(M_\mathrm{Moon} + M_\mathrm{object})}$.
pela
CC BY-SA 3.0
2016-10-21T17:15:19.843Z
@pela The central body here is the Sun; the orbiting body has a mass of the Moon. I still get an answer more like yours than zephyr's, though.
HDE 226868
CC BY-SA 3.0
2016-10-21T17:30:58.913Z
@pela Yes, that's a good point. I figured it wasn't necessary to go into the full details for such a simple question, but worth pointing out these caveats.
zephyr
CC BY-SA 3.0
2016-10-21T17:31:18.883Z
In AU-M$_\odot$-year units, $G = 4 \pi ^ 2$; if M = M$_\odot$, @pela's formula is equivalent to zephyr's. I concur with 15.9 My.
Mike G
CC BY-SA 3.0
2016-10-21T20:50:36.473Z
You make it sound like an amazing coincidence that the constant is “very nearly 1”, but in fact it is a consequence of the choice of (two closely related) units, the distance Sun–Earth (AU) and the orbital period of the Earth (year). Consider measuring the length of your foot in feet, you will find that it is very nearly 1, too. :-)
chirlu
CC BY-SA 3.0
2016-10-21T22:12:23.693Z
zephyr and @joseph.hainline: Oh, I'm sorry, I misread the question. Apologies!
pela
CC BY-SA 3.0
2016-10-22T10:39:49.070Z
Too many digits! This is a result of treating the period of a massless object orbiting at AU as one year. That "year", however, is the sidereal year, which is slightly longer than a "year" (the tropical year). It would be better to just say 15.9 million years.
David Hammen
CC BY-SA 3.0
2016-10-27T17:49:57.277Z
Comments are not for extended discussion; this conversation has been [moved to chat](http://chat.stackexchange.com/rooms/47525/discussion-on-answer-by-zephyr-orbital-period-for-a-1-light-year-object).
called2voyage
CC BY-SA 3.0
2016-10-21T15:31:43.967Z
2016-10-25T18:04:18.797Z
2016-10-25T18:04:18.797Z
zephyr
<p>How long would the orbital period be for an object the mass of the moon orbiting the sun at 1 light year in a circular orbit?</p>

<orbit><kuiper-belt>
Orbital period for a 1 light year object?
joseph.hainline
# Orbital period for a 1 light year object?

(Asked by: username_0 on 2016-10-21)

How long would the orbital period be for an object the mass of the moon orbiting the sun at 1 light year in a circular orbit?

**Tags:** orbit, kuiper-belt


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## 1 answers

### Answer by username_1 on 2016-10-21. Score: 6 | Accepted answer

To answer this, let's visit [Kepler's third law](https://en.wikipedia.org/wiki/Kepler%27s_laws_of_planetary_motion#Third_law) which tells you the orbital period of a body, if you know how far away it is.

$$\frac{P^2}{a^3} = \mathrm{constant}$$

As it turns out, if you use the correct units, for the orbital period, $P$, and the semi-major axis, $a$, namely years and astronomical units (AU) respectively, then that constant is equal to 1! You have our moon orbiting the sun at one light year, so this needs to be converted to AU. Google tells me that $1\:\mathrm{light\:year}$ is $63241.1\:\mathrm{AU}$. For this specific case, you get that

$$P = a^{3/2} = 63241.1^{3/2} = 15,903,734\:\mathrm{years}$$

Notice that this formula didn't depend on the mass of the moon! (Technically the mass of the moon is in there, but it's pretty negligible and so doesn't matter much.)

---

You may be scratching your head over the units $-$ how can a timescale squared equal a distance cubed?. Technically there's some trickery going on because that constant isn't unitless and I've implicitly assumed those units are hanging around, without explicitly saying so. Rest assured, the units do make sense for this equation.

> **Comments:**
> **username_2**: username_1 and @username_0: This way of phrasing Kepler's law is only valid when the central mass is the Sun (and the mass of the orbiting planet can be neglected)! The orbital period definitely depends on this mass (e.g. for infinite mass the object will fall down, and for negligible mass the object won't orbit at all). Using the more general formula yields a period of $P = 2\pi\sqrt{a^3/GM} \simeq 83$ billion years! The gravitational impact of the Moon from the distance of 1 ly is tiny!  
> **username_2**: The above formula also neglects the mass of the object. If its mass is comparable the the mass of the Moon, use $P = 2\pi\sqrt{a^3/G(M_\mathrm{Moon} + M_\mathrm{object})}$.  
> **username_3**: @username_2 The central body here is the Sun; the orbiting body has a mass of the Moon. I still get an answer more like yours than username_1's, though.  
> **username_1**: @username_2 Yes, that's a good point. I figured it wasn't necessary to go into the full details for such a simple question, but worth pointing out these caveats.  
> **username_4**: In AU-M$_\odot$-year units, $G = 4 \pi ^ 2$; if M = M$_\odot$, @username_2's formula is equivalent to username_1's. I concur with 15.9 My.  
> **username_5**: You make it sound like an amazing coincidence that the constant is “very nearly 1”, but in fact it is a consequence of the choice of (two closely related) units, the distance Sun–Earth (AU) and the orbital period of the Earth (year). Consider measuring the length of your foot in feet, you will find that it is very nearly 1, too. :-)  
> **username_2**: username_1 and @username_0: Oh, I'm sorry, I misread the question. Apologies!  
> **username_6**: Too many digits! This is a result of treating the period of a massless object orbiting at AU as one year. That "year", however, is the sidereal year, which is slightly longer than a "year" (the tropical year). It would be better to just say 15.9 million years.  
> **username_7**: Comments are not for extended discussion; this conversation has been [moved to chat](http://chat.stackexchange.com/rooms/47525/discussion-on-answer-by-username_1-orbital-period-for-a-1-light-year-object).  

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