In my previous article on determining the current season, I showed that you needed to know the length of the world’s year in order to track the season. But what if you don’t know the length of the year. Maybe you’ve got a setting book that gives the orbital distance but not the year length, or maybe you don’t have anything at all. In this post I’ll walk you through what you need to know to determine the orbital period of the planet (i.e. its year length) and how exactly to calculate that period.
In order to calculate the orbital period you actually only need two things. You need to know the mass of the central star and the distance from the star of the planet’s orbit. Now, at this point you’re probably thinking “How the heck do I get the mass of the star?” Don’t worry, it’s actually not that hard. But I’m going to walk you through the calculating length of year part first and then we’ll go back and talk about finding the information needed.
How Long is a Year Anyway?
Okay, so we’ve got the mass of our star and the orbital distance, how do we find the length of the year? Before we answer that, a brief history lesson. “It’s a warm summer evening, circa 600 B.C. …” No. Wait. We don’t have to go back that far. In the early 1600′s Johannes Kepler, using the observational data of Tycho Brahe, derived his three laws of planetary motion for planets in our solar system. The third of which is the one were interested in. It says that the square of the orbital period (in years) is equal to the cube of the orbital distance (in astronomical units – the distance from the Earth to the Sun). These laws are true for planets around other stars except the units don’t work for the 3rd law. Those only work for our solar system and you’re probably not doing that calculation.
Kepler’s work was empirical, he simply fit the data that he had. It would not be until about 70 years later, 1687, that it would be possible to generalize this to other systems. 1687 was when Issac Newton published Philosophiæ Naturalis Principia Mathematica which described his theory of gravity. With this law of gravity, it was possible to derive a version of Kepler’s third law that would work for any planetary system. That’s the one were interested in. (If you’re interested in the actual derivation you can read about it on this site.) This form of the law looks like this:
where P is the orbital period in years, a is the orbital distance in meters, G is the gravitational constant (6.67384 × 10-11 m3 kg-1 s-2), and M1 and M2 are the masses of the two objects.
“Hold on!” I hear you shouting. “You said we only needed the distance between the planet and the star and the mass of the star. Now we need the mass of the planet too?” Well, not really. If you’re calculating the orbital period of an Earth-like planet around a star, the truth is that the mass of the planet doesn’t really matter. The mass of the sun is about 2×1030 kg (that’s 2,000,000,000,000,000,000,000,000,000,000) while the mass of the earth is about 6×1024 kg (that’s a six with 24 zeros after it, you can write it out yourself if you really want to). Adding those together is 2.000006×1030 which isn’t much different from just the mass of the sun. If the star is much smaller than the sun or the planet is much larger, there might be a small effect but those combinations don’t typically result in habitable planets so we’re not going to worry about it. If it ever is important you have the full equation (and I may just do a post on determining the mass of your planet someday. Edit: Look I did!).
Okay, so we treat M2 as zero. Now it’s just a matter of plugging in the values and crunching the numbers. Here is an example using the Earth:
The orbital radius of the earth is about 150 million kilometers or 150 billion (US) meters. We have the mass of the sun and G from above so we get:
P = sqrt( 4 * pi2 * (1.5×1011 m)3 / ( 6.67384×10-11 m3 kg-1 s-2 * 2.0×1030kg))
P= sqrt( 1.33×1035 / 1.33 ×1020 ) seconds
P = 3.16×107 seconds
Divide that by 3600 seconds per hour and 24 hours per day and you get 365.68 days Which is a little off since I used approximations for the mass of the sun (it is really 1.989×1030 kg) and the orbital distance (it’s really 1.4959787×1011 m).
You can also use this to calculate the orbital distance if you know the length of the year and the mass of the sun or calculate the mass of the sun if you know the orbital distance and length of a year. These take a little bit of algebraic manipulation of the equation but everyone had algebra in high school so we’re all good. If you don’t want to do the math yourself, there are a number of on-line calculators that will do it for you. This one has a variety of options for the input and output values (although for some reason it doesn’t have kilograms as a option for mass. This one doesn’t look as nice but does have kilograms as an input option.
Now it’s entirely possible that you don’t have all the information you need. I mean how often do game supplements list the mass of the primary star. They rarely even list the orbital distance. Without those it makes it really hard to get the year length. However, all is not lost.
While most system and supplements don’t list out the mass of the star they will often give its spectral type or at least it’s color. And that’s enough for us to estimate the mass. Of course, the spectral type is best, but even the color can give us some clues. The reason this works is that the spectral type is basically just a measure of the star’s color. The color of the star is directly related to the star’s temperature and the temperature is directly related to the mass. So knowing the color or spectral type allows us to estimate the mass.
Now blue stars are hotter and more massive while red stars are cooler and smaller. A yellow-orange star like our sun in kind of in the middle. Now here I’m talking about “normal” or main sequence stars and not things like giants, supergiants, or white dwarfs. While it’s possible for these type of stars to have planets around them, these types represent stars that are dying (or have died) and they probably won’t have habitable planets around them. So for main sequence stars, the color can tell us the approximate mass.
Now given the color or spectral type, we just need to look up the mass. One good site for this is the Atlas of the Universe’s stellar classification page. You want to look at the “Main Sequence (V) table”. The lines are colored the color of the star and the mass is given in multiples of the Sun’s mass (1.989×1030 kg) in the 4th column. This table isn’t very fine grained, however, and only has a few entries.
If you want a super detailed version you can use this Stellar Classification Table. This table has an entry for every single spectral type and stellar luminosity class. I didn’t talk about luminosity class but just know that main sequence stars are luminosity class V. So the Sun is a G2V star. When looking at this table you will be interested in the ones that end in V. This table lists the spectral type in the first column and the mass (again as a multiple of the Sun’s mass) in the second column. A neat thing about this table is that it give RGB colors for the stars in the right most column. The color of the numbers is the color of the stars as they would appear to our eyes.
So if you have the color or the spectral type, you can use those tables to estimate the mass of the star.
This one’s a bit harder to estimate without some prior information. In fact, to do this correctly we need to talk about habitable zones and stellar evolution and that’s a complete post (or two) all by itself that I’ll be writing in the near future. For now we need some way to make a back-of-the-envelope estimate.
First, the Earth is near the inner edge of the habitable zone for the Sun. And that zone extends out nearly to Mars. So if you have a star like the sun, then the habitable planet is going to be about the Earth’s distance. Maybe a shade closer if the climate is generally warmer and a bit further out if it is cooler.
Second, as the star gets redder it puts out less energy. This means that the planet has to be a lot closer in order to get the same amount of total energy from the star. So the habitable zone moves in closer and the orbital distances will be smaller for redder stars. The opposite is true for bluer stars. They put out more energy and so the planet would have to be further away to prevent it from getting fried.
The question is how much closer or further away do we need to go? The energy put out by the star is proportional to the star’s temperature raised to the 4th power. Which means doubling the temperature by a factor of 2 increase the energy output by a factor of 16. Similarly reducing the temperature by a factor of 2 reduces the energy output by a factor of 16.
So that’s how we make our estimate of the distance. Given the star’s spectral type or color, both of the linked tables give us its temperature. Take the ratio of the star’s temperature to that of the Sun’s and square it, then square it again (that gives you the ratio to the 4th power). That value is what you should multiply the Earth’s orbital distance by to get a (roughly) similar temperature range. Then tweak it as desired for a slight climate variation. And now you have the orbital distance and can figure out the length of the year.
And Now You Know
That’s it. Even if we have just the color of the star and the knowledge that the planet is habitable, we can make a reasonable estimate of the length of its year. The more we know about the system, the better our calculation will be, but even just the stellar color can get us something.
Did that all make sense? Or was it all Greek? Are there any tricks you use? Are there any topics mentioned that you’d like me to go into more detail on? Let me know in the comments below.