Building Vulcan

## The 40 Eridani System

### Introduction

The information about the star system is as accurate and factual as I can manage. Some of it is based on direct observation and measurement. Some of it is guesswork, based on theoretical models of the way stars are thought to work. All of it is subject to possible change or amendment, and the numbers I’ve used may not necessarily be the ones used in other sources. It should only be a matter of fine detail, though. For ease of reference, and to avoid misinterpreting information intended for professionals, I’ve used the Wikipedia page, although I have also cross-checked (as far as I can) with the VizieR astronomical data service to be sure the figures are up-to date.

### The Stars

There are three stars in the 40 Eridani system, usually designated A, B, and C. The system’s distance from Earth has been measured to a fair degree of accuracy, at 16.45 light years. Although that is still provisional. GAIA may measure something different.

#### 40 Eridani B and C

The first important point to note about these two stars is that they follow an orbit separated from the main star of the system by 400 Astronomical Units (A.U.). That’s 400 times further than the Earth is from the Sun, or 77 times further than Jupiter is from the Sun. Put simply, it’s a long way. So far that even though the two stars will be very spectacular when they’re visible from the surface of Vulcan (even during the daytime), they won’t have any significant day to day influence on the planet, any more than Jupiter does on the Earth. Although the presence of other stars in a system may affect the formation of planets, we can ignore that, too. For our purposes, Vulcan’s there, so obviously planets could form.

Starting with the smallest component, 40 Eridani C. The current best guess is that it’s a red dwarf star, spectral class M4.5eV. It has a mass 0.2 that of our Sun, a radius 0.31 of the Sun and has only 0.008 the Sun’s luminosity (that’s basically the brightness). It orbits around 40 Eridani B in a way astronomers call “eccentric.” That just means that in parts of the orbit it’s quite close, and in others it’s much further away, rather than staying at about the same distance all the time (very fortunately, the Earth has an orbit around the Sun that’s almost circular, meaning that astronomers don’t call it eccentric, and we don’t live on a planet with a surface temperature that soars above the boiling point of water, then drops to well below the freezing point every year). Calculating the orbits in multiple star systems is a complicated process, and the life’s work of some astronomers. Again, the current best guess is that C takes 252.1 Earth years to make 1 orbit of 40 Eridani B, and varies in its orbit from being just less that 21 A.U. from B to a little over 49 A.U. The unfortunate thing about 40 Eridani C is that it’s a “flare star.” Just as the Sun has solar flares, so do other stars (naturally). The problem is that the flares of red dwarf stars are extremely spectacular, and emit a lot of X-rays. In real life, even though the star is 400 A.U. from anywhere an inhabited planet might be, that could well not be enough for life to be possible. In “Star Trek”, the Vulcans can obviously cope with it.

40 Eridani B is a white dwarf star, spectral class DA4. It has a mass estimated as being about half that of the Sun, but a radius of only 0.014 the Sun’s (that's about 1½ times the radius of the Earth). The obvious point is that this star is very dense. It is the core of a much larger red star that has lost its outer layers. Although it now has a luminosity only 0.013 of the Sun, at one time it was the largest and most luminous object in the 40 Eridani system, a red giant star. The lifespan of such stars is shorter than our Sun, because they “burn” the hydrogen they’re made of more quickly. Once the hydrogen gets too sparse to sustain the fusion reactions that makes the star shine, it’s effectively a huge explosion that stops exploding. Once that happens, there’s nothing to stop the star collapsing under its own weight. The good news is that reactions start again. The bad news is that there’s a catastrophic explosion, called a “nova” that makes the star incredibly bright as most of its outer layers are blown off into space, leaving an incredibly hot and dense core behind. Very slowly, this will cool down, although “slowly” really does mean slowly in this case. The Universe will be quite a lot older before 40 Eridani B is appreciably cooler. Obviously, the spectacular death of 40 Eridani B didn’t stop life developing on Vulcan, so we can assume (just as we can in real life) that all this happened hundreds of millions of years ago.

For someone standing on the surface of Vulcan and looking in the right direction, 40 Eridani B will have a magnitude of -8 and 40 Eridani C (when it’s not flaring) will have a magnitude of -6. For comparison, the full Moon has a magnitude of -12.6 on Earth, and Venus is -4.6 at it’s brightest. They’ll be bright, but only a bit brighter than the planets are on Earth. You wouldn’t be able to read by their light, and they’d be unusually bright stars in the sky, not “extra suns.”

### 40 Eridani A

Originally a distant companion of the red giant 40 Eridani C was in its youth, A is now the main star of the system, and the only plausible candidate of the three for the planet Vulcan to orbit around. It has a spectral class of K1V, a mass 0.84 of the Sun’s, a radius 0.81 of the Sun’s and a luminosity 0.46 of the Sun’s. Someone standing on the surface of Vulcan wouldn’t notice the difference, but the light from 40 Eridani A is quite a bit more orange than that from the Sun. What it does mean is that Vulcan’s sun is very likely to appear larger in the sky than the Sun does in the sky of Earth. There’s another consequence, too. The upper limit on how long a Vulcan year can be is short. A lot shorter than an Earth year.

## Vulcan

### Introduction

From here, we move into pure speculation, hopefully informed by science, although I don’t claim that my interpretation of the science will be terribly sophisticated. I’ll at least try, though. No planet of any kind has so far been detected orbiting any component of 40 Eridani, although astronomers have been looking. Luckily for “Star Trek”, our ability to spot planets is too limited to rule out planets quite big enough to be Vulcan, at least so far.

### The Vulcan Year

Since we know what star Vulcan is supposed to orbit around, the year is something that can be calculated, rather than guessed at. As a starting point, how long would the years on Vulcan be if 40 Eridani A was exactly as bright in the sky as the Sun is on Earth? That information can be calculated using the luminosity of the star, according to the following formula:

(1)
\begin{align} R=\sqrt L \end{align}

Where R is the radius of Vulcan’s orbit in Astronomical Units, and L is the luminosity of 40 Eridani A divided by the luminosity of the Sun. The answer is 0.678 A.U.

Now there’s a value for R, I can calculate Vulcan’s year, using the formula:

(2)
\begin{align} P=\sqrt { R^3 \over M_{S} } \end{align}

Where P is the year, divided by an Earth year, and MS is the mass of 40 Eridani A divided by the mass of the Sun. Strictly speaking, M should be the combined mass of the star and planet, but the difference between the weights of the two is so great that the planet’s mass can be ignored without noticeably affecting the result. Using the figures given above, Vulcan’s year would be around 0.61 of an Earth year, or 222.6 (Earth) days.

This result raises two problems. The first is that the available evidence (especially Tuvok’s age) suggests very strongly that a Vulcan year is longer than an Earth year. The second is that this represents a maximum, if anything. Vulcan is supposed to be hotter than Earth, with brighter sunlight. It’s shown like that in ENT “Kir’shara,” and strongly implied by TOS “Operation: Annihilate!” If anything, the Vulcan year has to be both longer and shorter than the approximate value I’ve calculated.

As I argued in My Vulcan Calendar, a Vulcan r’tas should be 1.17 Earth years, and the time Vulcan takes to go around its sun will be half that, 0.58 of a year. That’s a bit shorter than the “maximum” length I’ve calculated, implying that Vulcan not only orbits closer to its star than Earth does, but that it gets more sunlight on its surface.

#### A Slight Diversion: Vulcan Has No Moon

In TOS “The Man Trap,” Mister Spock is very certain that Vulcan has no natural satellites. Yet, when TAS “Yesteryear” showed the surface of Vulcan, there was definitely something in the sky. Even if that story is dismissed as “apocryphal,” “Star Trek: The Motion Picture” also showed a large object in Vulcan’s sky. What’s more, there was even a second, much smaller thing that, if it wasn’t a moon, certainly looked very much like one. Again, that particular image didn’t make it into the “Director’s Cut” version of the film. For that reason, I am of the opinion that Vulcan might still be regarded as a single planet. In that case, the length of the Vulcan year and day can remain as I have them, but my speculation about other objects sharing Vulcan’s orbit can be ignored. Basically, if you think there’s nothing in the sky over Vulcan, you can skip over the next bit.

I’ve decided to make Vulcan a little more complicated, not because I regard the evidence as overwhelming (it isn’t, because we never see this mysterious planet in any other of the appearances of Vulcan, and never from space) but because it makes the whole exercise more interesting. So what’s in the sky if it isn’t a moon?

Various suggestions have been made over the years, but if we’re trying to be scientifically accurate, we can reject any planet in a different orbit around 40 Eridani A. Put simply, how large something will appear in the sky depends on how big it is in absolute terms, and how far away it is (I’m guessing you knew that already). What that means in practice is that, ignoring unlikely and exotic forms of matter, if something in the sky is bigger than the sun, then it has to be either a moon of the world you’re on, you’re standing on a moon yourself, or you’re on one half of a twin planet system.

Having decided that Vulcan has a day that lasts about as long as a day on Earth, I’ve ruled out Vulcan being the large moon of an even larger planet. That’s because Vulcan would experience enormous tidal forces from anything that appears so large in the sky. Like our Moon, the end result would be that one side of Vulcan would always face the other planet. That’s something that has to happen, no matter what. Having Vulcan further away from a bigger planet is an option, but only if I accept that the days are going to be considerably longer. If the other planet is a “hot Jupiter” gas giant, then each day is going to last as long as one orbit of Vulcan around the other planet. Our Moon has days a month long. The days on Saturn’s moon Titan last a little under sixteen Earth days.

At this point, I’m pretty much left with one option: Vulcan is one half of a “twin-planet.” They’re both about the same size, and both tidally locked, so the day is the same on each world. Is that a reasonable option? Before going any further, let’s look at a “test” system involving two planet Earths. How far apart would they be, and (more importantly) how big would they appear in each other’s skies? I’ve already decided that I know the orbital period: 1.113 Earth days. From that, we can calculate the distance between the two planets using the formula:

(3)
\begin{align} R= \sqrt[3]{ { {P} \over {0.166} }^{2} \times ( M_{1} + M_{2} ) } \end{align}

Where R is the distance between the two worlds divided by the diameter of the Earth, P is the period divided by Earth days and M1 and M2 are the masses of the two planets divided by the mass of the Earth. The result is that our two planets will be 4.481 Earth diameters apart, which is tremendously close as these things go. So how big would each planet appear in the sky? Using another formula:

(4)
\begin{align} \theta=2sin^{-1} { {r} \over {R} } \end{align}
Here’s the view as we see it:

This is the Moon swapped for an Earth 8.962 Earth radii away:

How does that compare with Vulcan’s sky?

This screencap is from TAS “Yesteryear,” thanks to those diligent people at TrekCore. The rough size of a “twin” seems about right, close enough that it’s worth going on a bit further.

### A Twin

Do we know anything about this twin world, other than it’s absolutely huge in the sky of Vulcan? The answer is maybe. As many of you might have guessed a while ago, the 2009 “Star Trek” film did indeed give us a planet where Vulcan was a large object in the sky. The graphic novel “Nero” was even more explicit. “Delta Vega” orbits very close to Vulcan. Since I’ve already narrowed the options I’m willing to consider down to “twin planet,” it seems that the only available evidence is that Vulcan shares an orbit with an ice planet.

Can that happen? Vulcan is a world hotter than Earth, so how can a world that’s colder be the same distance from a star? Naturally, there are hot deserts and icecaps on Earth, so it’s not impossible for both to be on the same planet, let alone different ones. All the same, the mechanisms that permit both climates on Earth are unlikely to apply to two entire planets. What might? The obvious difference between an ice world and a desert world is the amount of light they reflect. Ice is white and shiny, and the red sands of Vulcan are much darker.

This is where things get complicated. Calculating the surface temperatures of planets is not an exact science. The best that can be done is to calculate an “ideal” temperature, based on the amount of light the planet receives, and the amount of that light it absorbs, ignoring atmospheric effects. This is known as the “black-body” temperature, and it is calculated according to the following formula:

(5)
\begin{align} T_{B} = T_{eff}\sqrt { { r_{S}} \over {2R} } \times \sqrt[4]{ 1 - \alpha } \end{align}

Where TB is the black-body temperature of the planet, Teff is the effective temperature of the star, both in Kelvin; rS is the radius of the star, R is the distance of the planet from the star, and α is the albedo of the planet. You’ll find that there are two figures for albedo, the “visual geometric albedo” that’s used in astronomy, and the “bond albedo” which is a measure of how much total energy is reflected into space, and the one you want. “Bond albedo” is also called “planetary albedo.” Doing these calculations correctly depends on using consistent measurements. For my sums, I’ve taken the radius of the Sun as 695,500,000 metres, and the distance from the Earth to the Sun as 1 Astronomical Unit, or 149,597,870,700 metres. Using a temperature of 5778K as the effective temperature of the Sun, the calculated black-body temperature of Earth is 254.3K, which seems about right.

Now we can apply that to Vulcan and its companion, and that’s where things start getting tricky. Although measurements of the Sun are as accurate as possible, and you can be fairly sure that they’re right, the kind of measurements we need for other stars are much less certain. The figures might be OK as “ballpark” numbers, but there is a margin of uncertainty. Even quite small adjustments to the mass, radius or temperature of 40 Eridani A will have a big impact on the black-body temperature of orbiting planets. What I’m saying is that these are indicative figures, not absolute facts. That said, I’ve already said that the radius of the star is 0.81 of the Sun’s and we can calculate the distance of Vulcan using a simplified version of the equation for getting the distance between Vulcan and its companion, given above:

The letters represent the same things as they did when I earlier calculated the value of P when I knew R. Now when I take P to be 0.5833 years, R will be 0.6587 A.U. That just leaves the effective temperature of the star. I have found a number of different values for that, since there are different ways of working it out, and they give similar, but different, answers. So, with the huge warning that this is only an approximation, I think the effective temperature is 5200K.

To avoid pulling figures out of the air, I’m assigning the albedo of known planets. Vulcan is a reddish desert world, so it gets the bond albedo of Mars, 0.25. The companion has been given an albedo the same as the cloud-covered and reflective world Venus, 0.9. This might seem a bit of a cheat, but the albedo of Venus is comparable of the closest there is to an “ice-world” in the Solar System, Jupiter’s moon Europa. If anything, it might even be a bit low for a true ice-covered planet. Using these figures, I’ve compiled a table:
Planet Bond albedo Black-body temperature Difference from Earth
Earth 0.306 254.3K 0
Vulcan 0.25 258.72K +4.45
Companion 0.9 156.34K -97.93

Stressing once again that any of the figures except for the ones for Earth are rough approximations more than anything else, the results are reassuring. Not only does Vulcan come out having a black-body temperature slightly higher than Earth’s, but the companion ice-planet, with nothing changed but the albedo, comes out very considerably cooler, quite cold enough for a planet where the year-round temperatures are always well below the freezing point of water.

So, I’ve decided that Vulcan is a desert world sharing an orbit with an ice-covered world, and I think that the arrangement is not scientifically impossible, no matter how unlikely it might be. What else might be determined by informed guesswork, rather than just a guess?

Over the years, it has been established that Vulcan is a hot, desert world with intense sunlight, very little water and an unusually thin atmosphere. It has been suggested that the greater strength and stamina of Vulcans is a result of their planet having a higher surface gravity than Earth. One immediate point to consider is that when we see Vulcan, the gravity isn’t noticeably different from Earth’s. In practice, it’s obvious why that is, but it still seems unlikely that the gravity is more than around 110% of ours, since Humans seem to cope with it quite well, and falling objects aren’t moving perceptibly faster than expected.

How do you end up with a thin atmosphere and no water? If the gravity at the surface of Vulcan is more than Earth’s, then air and water molecules are less likely to reach escape velocity, and the planet will retain more, not less, of an atmosphere. There are a number of possibilities. One is a catastrophe that drastically reduces both the air and seas of Vulcan. It might be a natural catastrophe (“Spock’s World” by Diane Duane suggested an unusually fierce solar flare). It might be an artificial disaster. Mister Spock suggests that the Arretians in TOS “Return to Tomorrow” were responsible for some otherwise unexplained aspects of Vulcan prehistory. Since Arret has its atmosphere completely destroyed, it might be linked to the loss of Vulcan’s, too. When the atmosphere was lost is as important as how. Too recently, and how would Vulcan life have become so well adapted? Too long ago, and it becomes difficult to see how life on Vulcan would have had chance to progress as far as it has. I’m intrigued by the possibility that Vulcan has suffered a gradual loss of air and water over time. Long ago, there was quite enough to support the early development of life, but conditions have become less and less hospitable. The process has been slow enough that random mutations have been frequent enough for life to adapt. Where is the atmosphere (and the water) going? My model allows an obvious guess: that ice planet. Making the companion world the larger of the two, even by only a small amount, creates conditions where the loss of some atmosphere from one to the other is not impossible. We end up with a world that has an unusually thick atmosphere, and more water than expected, and one that is more airless and arid than anticipated. I’m not sure that such a process would be enough, even over millions of years, and there may well be other factors involved. Even so, it makes more sense for the companion world to be bigger than Vulcan, not smaller.

Another argument for making Vulcan the smaller of the two worlds is the third object, the little “moonlet” visible in “Star Trek: The Motion Picture”. My own rather rough and ready understanding of gravity and orbits makes me very sceptical that something could orbit either of the two main worlds. The orbit would be unstable, and the moon would either be deflected into the surface of one of the planets, or sent flying off into a separate orbit altogether. Nevertheless, the moonlet’s there, so having it orbit the companion seems the best option. This means that Vulcan has “no moon,” so that Mister Spock wasn’t lying about it. The most stable arrangement I can come up with is that the moonlet orbits the larger of the two planets, again leaving the companion as the larger world.

How much larger, though? Again, we’ve seen the surface of this planet, in the 2009 “Star Trek”. James Kirk found the cold difficult to deal with, but he wasn’t floored by the gravity. More to the point, the enormous ice-beast that tries to eat him doesn’t seem to be the product of evolution on a high-gravity world. I’m left feeling that the figure of about 110% I guessed for Vulcan applies to the companion world, too. At this point, I have to pull some figures out of a hat, because there’s no way I can work this stuff out. Let’s say Vulcan has gravity 1.06 that of Earth, and the companion has 1.13. Perhaps not as high as some would set the figures, but not so high that a healthy human is going to be debilitated by the “extra” weight, or that we viewers can look at the screen and say: “the gravity is obviously not that high, because things aren’t falling visibly more quickly.” (I admit, I’m not sharp enough for that myself, but I know people who are.)

So, having arbitrarily decided on some surface gravities, can we work backwards? The answer is yes and no. Yes, we can assign some value to the density of the planets and work out their radii. No, we can’t do that with any degree of certainty. Although it’s under intense study, there just isn’t anything like enough information available for anyone to say “a star with this composition will have planets of this composition.” Even the “extreme” figures for high and low planet density worked out theoretically might not be extreme enough to accommodate all the planets we already know about, and where planets like Earth are concerned, that’s still very few.

With that important proviso, considerable work has been done on modelling planets with a composition similar to the Earth. Not least because it was simple enough for me to understand, I’ve used an article called “Planetary radii across five orders of magnitude in mass and stellar insolation: application to transits” by J.J. Fortney, M.S. Marley and J.W. Barnes. It was originally published in The Astrophysical Journal on pages 1661 to 1672 of volume 659 in April 2007. Naturally, it’s also available for free on the internet if you look carefully. It is also vitally important to use the corrected formula for planet radii (the correction was published on page 1267 of volume 668 of The Astrophysical Journal in October 2007):

(6)
$$R_{P} = ( 0.0592 rmf + 0.0975 ) (log M_{P})^{2} + ( 0.2337 rmf + 0.4938 ) log M_{P} + ( 0.3102 rmf + 0.7932 )$$
Where RP is the radius of the planet divided by Earth’s radius, rmf is the “rock mass fraction” and MP is the planet’s mass divided by the mass of the Earth. For Earth, the rmf is set at ⅔ (which indicates a planet with about twice as much rock as iron; 1.0 would be pure rock, and 0.0 represents a planet entiely made of iron). Applying this to the rocky planets of the inner Solar System:
Mercury 0.0553 0.383 0.4
Venus 0.815 0.95 0.943
Earth 1 1 1
Moon 0.0123 0.2727 0.259
Mars 0.107 0.532 0.499

As you can see, all of the calculated values, except for Mars, fall within 0.02 of the actual measured values. This is pretty good, especially since there’s no guarantee that the rmf is a constant. Mars certainly seems to have a lot less iron than the calculation anticipates, giving the planet a lower average density, and hence a bigger radius. Certainly the numbers are close enough to be well within the limits of accuracy I’ve been able to establish. For my purposes, it works OK.

You may be wondering what Vulcan’s radius has to do with anything, and how I get to it with the information I’ve worked out and guessed at so far. Well, mass and radius are what determines the density of a world:

(7)
\begin{align} \rho_{P} = {{ M_{P}} \over {r_{ P }} ^3 } \end{align}

where ρP is the average density of the planet divided by Earth’s density, MP is the planet’s mass divided by Earth’s mass and rP is the radius of the planet divided by the Earth’s radius. Radius and density set the surface gravity:

(8)
\begin{align} g_{P} = \rho_{P} \times r_{P} \end{align}

where gP is the surface gravity of the planet, divided by Earth’s. Since I’ve taken a guess at the surface gravities of my two planets, I need to work backwards to get plausible answers for the mass and density.

I could do this by a lot of complicated mathematical analysis. Instead, I’m going to play around with a spreadsheet until I get the numbers to come out how I want. First, I need to make another wild guess. What’s the rmf for planets like Earth orbiting 40 Eridani? I could just pull a number straight out of, er, the air. On the other hand I could try a spurious train of reasoning to get my answer. Stars like our Sun and 40 Eridani A are mainly made of hydrogen. Most of what’s left is helium. A very small proportion of the total is everything else. At least approximately, a measure of all the stuff that isn’t hydrogen, helium or iron, divided by the amount of iron in a star might be vaguely related to the proportions of rock and metal in a planet orbiting it. This quantity is measured, and is known as $\alpha \over Fe$. It is not the “metallicity” of the star, as that is a measure of the amount of iron in relation to hydrogen in a star, compared to the Sun. For 40 Eridani A, the measured $\alpha \over Fe$ is -0.0007 of the Sun’s. That means that there is very, very slightly more iron in the star as a proportion of the other elements that aren’t hydrogen and helium. Although it’s based on an assumption that holds no water at all, I’ve multiplied the rmf for Earth by the changed proportions for 40 Eridani A, and come up with a revised rmf of 0.6662. It’s as good as anything I could make up, reflecting a slightly higher proportion of metals in Vulcan’s make-up, without going overboard on something that is very unlikely to vary significantly across planets very like the Earth orbiting stars not too dissimilar from the Sun.

How does that come out?
 Vulcan Companion Earth Mass 1.141 1.317 1 Surface Gravity 1.06 1.13 1 Radius 1.038 1.079 1 Density 1.022 1.047 1

As you can see, the variations from Earth are quite small, meaning that all my earlier “ballpark” calculations aren’t too far off the mark.

Once again, my arguments are running on, so I’ll start a new page for the next stage, Putting It Together.

by on 30 Dec 2014 14:37, last updated on 04 Mar 2015 08:44

page revision: 41, last edited: 04 Mar 2015 08:44