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Kepler's Third Law as stated on websites I found while searching for "how to convert between a planet's orbital period and semi major axis" is $P^2 = A^3$ where P is the orbital period in years and A the semi major axis in AU.

However, why is mass of the central body not included? If Earth was orbiting at the same distance around a star with more or less mass than our Sun, surely its orbital period would be different?

jwvanderbeck
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    Have you seen this? Also see https://astronomy.stackexchange.com/q/18690/16685 – PM 2Ring Oct 09 '19 at 05:06
  • All systems with a dominant central mass follow Keplers laws - just use the dimensonal one. – AtmosphericPrisonEscape Oct 09 '19 at 10:33
  • Don't worry if your question is closed as a duplicate. It doesn't mean that your question is the same as that question, it means that one or more answers there also answers your question. Here's what I think happened. Your "supposedly boils down to" probably comes from some discussion of the solar system and the orbits of the planets around the Sun. So at about 5.3 AU Jupiter's period is 5.2$${}^{3/2}=$11.86 years etc. Had you included a link to what you'd been reading, or quoted a book or article, it would have been clearer that it only works for things orbiting around the Sun. – uhoh Oct 09 '19 at 11:06
  • Have a look at the answers to the proposed duplicate question, if you have further questions feel free to ask a new one! But this time make sure to explain why you think something is true, because the answer might be found within that reason. – uhoh Oct 09 '19 at 11:07
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    Hi @uhoh I had been searching online for how to convert from orbital period to semi major axis, or back again and I kept running into websites stating Kepler's third law but they all kept stating it as I showed above, which didn't make sense to me given that Mass was not included. Reading that linked question, it isn't exactly what I was asking but it does make me understand my own question. Apparently people shorthand the equation in most places which is what confused me. – jwvanderbeck Oct 09 '19 at 14:01
  • Ah, I'm going to write an answer for you then, thanks for the feedback! – uhoh Oct 09 '19 at 14:04
  • Why would you think mass has to be involved? Have you read the derivation of these laws? – Carl Witthoft Oct 09 '19 at 14:38
  • @jwvanderbeck have a look, I've posted an answer that should help you to calculate Kepler orbits around other bodies. – uhoh Oct 09 '19 at 14:42
  • @CarlWitthoft I have not. I'm not an expert in the field. It just made sense to me that mass would be involved. If I put Jupiter and Earth in the same orbit it feels logical to me that they would have different orbital periods. Or maybe they don't :) That was my reason for askig clarification. As I said all of the websites I had come across when searching for my problem just listed the simple form of he "law". – jwvanderbeck Oct 09 '19 at 15:09
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    By the way IMHO I do' see this as a duplicate. While finding that other question may have helped me understand what I was trying to figure out, the root of the question wasn't about why doesn't Kepler's third law work in X (though honstly I didn't know there were cases where it didn't work) but more "How does it work when dealing with other star systems". The reason for the question as stated being that most websites reduce the equation down with the assumption of Sol, so I didn't realize there were other forms of it. – jwvanderbeck Oct 09 '19 at 15:21

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Though the question may be closed as a duplicate, here's an answer that shows how to calculate periods using Kepler's $\color{blue}{\text{3rd}}$ law but with all the constants and units explained.

Here's the equation from Wikipedia's Orbital_period; Small body orbiting a central body. The idea here is that the the size difference is so big that we can assume the central body doesn't move. In reality there is always at least a tiny motion. The Sun-Earth system rotates around a point about 450 kilometers from the Center of the Sun, and outer planets (especially Jupiter and Neptune) move the Sun so much that the center of mass is sometimes completely outside the Sun!

$$T = 2 \pi \sqrt{\frac{a^3}{GM}}$$

Which is Kepler's $\color{blue}{\text{3rd}}$ law:

$$\frac{T^2}{a^3} = \frac{4 \pi^2 }{GM}$$

What are the units?

While it is convenient to use AU and years, as you've found out that only works for orbits around the Sun. The reason is in that case you're really simplifying the equation to a ratio; period divided by another period around the same body, and semimajor axis divided by the semimajor axis of that same other body.

For general calculations I recommend you use meters, seconds and kilograms. You can use any set of units, but you have to be careful that everything is in the same units.

What is $GM$?

It's the gravitational constant $G$ times the mass $M$ of the central body. You can look them up separately, but it turns out that you can find $G$ times $M$ listed as the standard gravitational parameter. These are more accurate than multiplying $G$ times $M$ because each one alone is known to much less accuracy. That's because we can't (easily) measure the mass of a planet by comparing to a standard kilogram. Instead we look at periods and distances of orbits and determine the product $GM$ together.

Here are a few of the numbers from the Wikipedia article, I've rounded them to three decimal places. You can see more decimal places and discussion in the question Where to find the best values for standard gravitational parameters of solar system bodies?.

body      GM (m^3/s^2)
-------   ------------
Sun       1.327E+20
Earth     3.986E+14
Moon      4.905E+12
Jupiter   1.267E+17

So to get the period of the Earth around the Sun:

$$T = 2 \pi \sqrt{\frac{(\text{1.496E+11})^3}{\text{1.327E+20}}} = \text{3.156E+07 sec} = \text{1 years.}$$

To get the period of Europa around Jupiter:

$$T = 2 \pi \sqrt{\frac{(\text{6.709E+08})^3}{\text{1.267E+17}}} = \text{3.067E+05 sec} = \text{3.550 days.}$$

Wikipedia gives 3.551 days, which is the same considering we're only using three decimal places here. If you take advantage of using $GM$ instead of multiplying $G$ times $M$ and use many more decimal places, you should get agreement to the known periods to many more decimal places as well.

But why don't I get the right period for the Moon?

The Moon's mass is more than 1% that of Earth, so the approximation treating the central body as not moving doesn't work.

uhoh
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  • oops too late :-) :-) – Carl Witthoft Oct 09 '19 at 14:38
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    @CarlWitthoft nope! As long as a post exists and isn't deleted, it can always be edited/updated. – uhoh Oct 09 '19 at 14:40
  • Thanks for the answer. It really helps me out, but maybe edit it to be a bit more applicable to the actual asked question for posterity? After all the original question was "How does Kepler's third law work for exoplanets", but this answer is very Sun centric. I assume I just need to substitute the mass of my (in my case fictional) star for M. – jwvanderbeck Oct 09 '19 at 15:17
  • @jwvanderbeck well I gave two examples of orbits around two different objects, one happened to be the Sun, one happened to be a planet, but you can use the equation for any object around any larger object. However, now I do see better what your question is or should be: How can I look up the masses of stars in order to calculate periods of exoplanets? You use the same equation above, but you need to get either the mass $M$ and multiply by $G$ or the standard gravitational parameter $GM$ for each star. Why not post a new question about that? – uhoh Oct 09 '19 at 15:21
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    Thanks for including the info about the standard gravitational parameter. People interested in orbit calculations need to know about that! – PM 2Ring Oct 09 '19 at 15:26
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    @uhoh No I don't need to know how to look up the mass of stars, I already have that information. It was how does kepler's third law apply in other star systems where the mass of the star is different from the sun's mass given that mass wasn't a part of the equation. The answer really was that mass IS part of the FULL equation, just not part of the simplified equation as given on most websites. – jwvanderbeck Oct 09 '19 at 15:54
  • Okay that's great! In that case I'm not sure how to further modify this answer for you. If you like you are welcome to edit the answer yourself to improve it. It's late here now, I'll check back tomorrow. Thanks! – uhoh Oct 09 '19 at 15:55
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    I'm going to edit the question to be more clear, though it is ironic that I can only do that because the answers helped me better understand what to ask. – jwvanderbeck Oct 09 '19 at 15:56
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    @uhohThe edit I made was very minor and just clarifies the question more precisely. – jwvanderbeck Oct 09 '19 at 16:12
  • @jwvanderbeck looks great! – uhoh Oct 09 '19 at 16:14
  • It's Kepler's 3rd law, not Kepler's law. Kepler's 3rd law (as you have written it) works fine for whatever units you use (with an appropriate choice of $G$). You mean that Kepler's 3rd as written in the OP only works for the Sun, au and years. – ProfRob Oct 09 '19 at 19:49
  • @RobJeffries ya I know. I thought that "the Kepler law who's enumeration I can't remember at the moment" was too long to write each time. I'll add the "3rd" now. – uhoh Oct 10 '19 at 03:40