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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?

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To answer this, let's visit Kepler's 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.

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  • $\begingroup$ 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! $\endgroup$
    – pela
    Oct 21, 2016 at 17:08
  • $\begingroup$ The above formula also neglects the mass of the object. If its mass is comparable the the mass of the Moon, use $P = 2\pi\sqrt{a^3/G(M_\mathrm{Moon} + M_\mathrm{object})}$. $\endgroup$
    – pela
    Oct 21, 2016 at 17:11
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    $\begingroup$ @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. $\endgroup$
    – HDE 226868 ♦
    Oct 21, 2016 at 17:15
  • $\begingroup$ @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. $\endgroup$
    – zephyr
    Oct 21, 2016 at 17:30
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    $\begingroup$ 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. :-) $\endgroup$
    – chirlu
    Oct 21, 2016 at 20:50

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