Friday, May 27, 2011

Foreign Skies: Daytime

When I wrote my first post on this blog, I wanted to include some photoshopped images of to-scale Jupiter hanging in the sky as it would above Ganymede and Io. At the time, I didn't have any good photos of the moon to compare with and paste over and I didn't want to steal something from Google Image Search so I put it off. I still don't have any good, cloudless photos of the moon at night with a suitably urban back drop. Instead, I decided to put my (very average) photoshopping skills to use and make a daytime Jupiter in the sky.

A caveat: Ganymede and Io both lack atmospheres that even remotely resemble Earth's. As a result, you wouldn't get a blue sky on either, even under a biodome of some sort (there wouldn't be enough air in the dome to have that effect). Instead the sky would be black outside of the giant orb of Jupiter (which, when full, would definitely be bright enough to make it hard to see stars).

So you shouldn't treat these images as "what the sky would look like on Ganymede/Io" more as an indication of how large Jupiter would be in the Earth's sky if you put it in Ganymede's/Io's place.

First! The original photo with the Earth's natural moon left in (taken with my phone, so it could be awesomer, but it served its purpose):

Unlike Earth, Ganymede and Io are unlikely ever have gum trees. Just saying. (Click to enlarge.)

For the remaining compositions, this APOD image is the shot of Jupiter I used. Credit to NASA, ESA, and E. Karkoschka (U. Arizona).

Next up, Jupiter as seen from Ganymede. I left the moon in for the first one, just so you can visually compare sizes better:

To scale, albeit physically unrealistic. I don't think Jupiter would quite be that colour either (particularly the edges wouldn't be so dark, but I couldn't fix it) and of course the lighting is all wrong. (Click to enlarge.)

Just Jupiter alone in the Ganymedian sky:

Jupiter in Earth's sky if Earth was in Ganymede's orbit around the gas giant. (Click to enlarge.)

And, last one, the size of Jupiter in Io's sky. Large doesn't really cover it.

Jupiter looming as though over the Io skyline. Of course, Io is even less likely than Ganymede to have apartment buildings on it, but shh! (Click to enlarge.)

And there you have it. Once I have a decent night skyline with the moon in it, I will repeat this but at night. It will be significantly awesomer. I just have to get some night photography in first.

EDIT: I have made some similar images for Jupiter from Europa and Callisto and the sun as seen from the Jovian system. See my new post here.

Wednesday, May 25, 2011

Conquering the Horizon

And now for something completely different. The horizon; how far away is it? How different would it look on another planet? We're used to horizons on Earth but if we're writing a story set on an asteroid or on a small moon or planet the horizon will be closer because the planet's surface falls away more quickly.

Distance to the Horizon

When I talk about the distance to the horizon, what I mean is if you're somewhere flat, what's the furthest you can see (including with the aid of binoculars or a telescope etc)? On Earth a good example of this would be how far away the point where the sea meets the sky is, when you're standing on a jetty. Once we have trees and buildings in the way it can get a bit more complicated. Luckily, the sort of extraterrestrial locales where the horizon is going to be most different to Earth's are least likely to have a large abundance of trees. Convenient.

Working out how far away the horizon is takes a little bit of trigonometry. I've drawn a sketch below of all the relevant distances and whatnot.

R is the radius of the planet/moon, h is the height of your person (well, of their eyes) or if they're in a building, it can be how high up they are, d is the straight line distance to the horizon, s is the distance along the surface of the planet/moon and θ is an angle that will be useful in some calculations.
The important thing to that you need to know about your non-Earth planet is how big it is or, more specifically, it's radius which is labelled R in the image above. It's a reasonable assumption that you'll have at least a rough idea of how tall your characters are. If you don't, it doesn't really matter, you can just guess something close since there's not going to be much difference between a tall person's horizon and a short person's (the differences really come into play in non-horizon situations, such as crowds). For the purposes of my calculations later on, I'm going to set h = 1.7 meters. Because I can.

Now, that's a right angle between the line I've labelled d and the left hand radius line. Since we know R and h we can now use Pythagoras's Theorem to work out the distance d. Don't worry if you don't remember any maths, I'm just going to tell you the answer.

Chances are, your planet is significantly larger than your person, so you will usually be able to ignore the h2 but not always (if on a small asteroid, for example). If in doubt, leave it in. It won't make your answer worse.

This is not an unhelpful result. However, I can't help but feel that when people stand in a tall tower and say things like "They're ten miles away but gaining ground!" they don't mean ten miles from their eyes, but ten miles from the bottom of the tower (if nothing else, they'd probably be estimating based on land marks and those are definitely relative to the ground distance).

So how do we find the distance along the ground, s? Unsurprisingly, with more maths. We use the fact that s = Rθ and then work out θ so we can substitute for it and not have to actually calculate it directly. Using the same triangle as before, we can find s in two different ways:

If you're wondering, cos and tan are trigonometric functions all scientific calculators (including the ones hiding in all your computers) can do. The -1 indicates that's it's the inverse of the function which you can usually access by pressing shift/2nd or something like that, depending on the calculator.

For planets/moons which are much, much larger than a person, s and d will be very close; it's the tiny, tricksy moons or asteroids are where it'll really make a difference.

So how far?

Some examples now for a person 170 cm tall and for a ten storey building (30 metres high):
  •  On Earth, ignoring atmospheric effects which actually extend the apparent horizon thanks to bending light, the horizon is 4.7 km away. From a ten storey building it's 19.7 km.
  • On the moon or Io, which are similar in size, a standing horizon is just under 2.5 km and the ten storey building horizon is about 10 km.
  • Ganymede and Titan (moons of Jupiter and Saturn, respectively) are a bit bigger than those two, with standing and ten storey building horizons of 3 km and 12.5 km.
  • Mars is about one and a half times the size of Ganymede and a bit more than half the size of  Earth. It has horizons 3.3 km and 14 km for standing and building respectively.
  • Deimos, Mars's moon, is rounder than Mars's other moon, Phobos, but still not that round. If you stand on a fortuitously round bit, the horizon will be 140 meters away (that's right, metres not kilometres—Deimos is actually an oblong with dimensions only 15⨉12.2⨉10.4 km across. Its average radius is 6.2 km). If you somehow managed to put a 10 storey building on it... well you'd see about 600 meters away.
  • Ceres is a large, round asteroid (or dwarf planet) in the asteroid belt. It was one of the bodies that, when Pluto's planethood was called into question, was up for being classified as a planet if Pluto got to stay. (If you're wondering, it is considerably larger than Deimos, with a radius of 471 km.) A person would see the horizon 1.3 km away (pretty close if you think about how far a kilometre looks when you're driving, for example) and a ten storey building would see the horizon drop off 5.3 km away.
  • And speaking of Pluto, Pluto's largest moon (it also has two tiny ones), Charon, is a bit bigger than Ceres and has a standing horizon of 1.4 km and a building horizon of 6 km.

Seeing things beyond the horizon

The horizons I've talked about above are the limiting distances for seeing things on (or close to) the ground. Things are a little bit different if we want to work out from how far away we can start to see the top of a tall building, for example.

The diagram below shows that although my little stick figure can only see the ground up to d distance away, s/he can see the top of a building which is d+b distance away. Huzzah!

My drawing skillz know no bounds. Close up of previous diagram with a building of height H, that the stick figure can just start to see the top of, added in. The building is b distance further away than the ground point being cut off by the curvature of the planet.
So what if we want to work out how from how far away we start to see the top of a building/monument/spaceport/volcano? Easy. All we have to do is work out the distances to the horizon for both the person and the building/monument/spaceport/volcano and add them together. The distance, D, from which the person starts to see the top of the building/whatever is then approximately given by:

You may have noticed that if we want to work out the distance from which someone can start to see a ten storey building, all we have to do is add the horizons I worked out above together. How convenient! Just quickly, the distances at which the building will start looming out of the ground are:
  • Earth: 25.5 km (assuming you can find an isolated ten storey building in the middle of a 25 km circle of flat ground...)
  • Moon/Io: 12.5 km
  • Ganymede/Titan: 15.5 km
  • Mars: 17 km
  • Deimos: 740 meters if you can manage an ideal situation (I strongly suspect that you can't, but these calculations do give you a good idea of how small Deimos is... the edge of Deimos would look so close!)
  • Ceres: 6.6 km
  • Charon: 7.4 km
There you have it. Note that with Earth being the largest rocky body in the solar system, it by far has the widest plains (or planes, if you prefer to be mathematical about it) around. The larger-but-not-as-bit-as-Earth planets and moons (Mars, the moon, Io, Ganymede and Titan) give us similar results for their horizons, which suggests to me that someone travelling between them wouldn't notice much of a difference. Our small-but-still-roundish bodies (Ceres, Charon) both give results about half that of the larger bodies (and hence a quarter that of Earth). I only included Deimos for fun, but it is important to remember that standing (or floating, as the case may be) on or near the surface of this moon (or any similarly sized asteroid, of course) would, visually, be a very different experience to any other body I discussed.

Monday, May 23, 2011

A couple of cool Io animations

So I was browsing around NASA's photo archive looking for a nice high-res image of Io and I came across this little movie of Jupiter-shine on Io. It's a time-lapse movie in which you can see the Io moving so that the sun is on its far side — it starts off behind and to the left which is why you can see a bright crescent at first. As the sun moves behind Io from the camera's point of view, Jupiter, which is behind the camera becomes more fully illuminated. The sunlight bouncing off Jupiter in turn more brightly illuminates Io's surface which is why the night side of Io gets darker as the crescent of Io gets smaller. Pretty cool, eh?

You can read more about planetshine in this post, which also talks about how much light you'd get from the sun at different distances.

Another cool little movie on the same site is this one, which shows some interesting happenings on Io's dark side. The bright spots away from the edges are volcanoes, whereas the two blue glows on the edges are similar to aurorae on Earth. Except those aren't the poles, but rather at the equator where charged particles are colliding with the tenuous wisps of Io's atmosphere. The thin atmosphere itself is only existent because it's constantly being replenished by volcanic gases. Nifty.

Wednesday, May 18, 2011

Ringing Tides

Saturn has rings. So do all the other gas giants in the solar system. Although we have no ability to confirm whether extra solar gas giants also have rings, chances are some do. Where do these rings come from? Why doesn't Earth have any?

Terrible Tides

The answer to the first question, as you may have guessed from the title of this post, is tides. Last week I talked about tides causing satellites to be locked in synchronous orbits around their primaries (recommended reading if you haven't already). The thing to remember now is that the side of a satellite closest to its primary experiences a stronger gravitational pull than the far side. The difference in forces depends on the mass of the primary, the distance of the satellite from the primary and the size if the satellite.

If you recall from the introduction to gravity post, the force of gravity exerted on a mass, m, a distance, r, from another mass, M, is given by:

If we take M to be the mass of the primary and then consider two smaller masses m1 and m2, one of which is located at r1 on the near side of the satellite, and the other at r2, the far side of the satellite. If we assume our two small masses are equal (you can think of it as considering a kilogram of moon rock in two different locations), then the ratio of the forces experienced by them will be:

That equation might seem a bit abstract, so let's look at it in the context of a few real examples.
  • Io is 4.217⨉108 m from Jupiter, on average, and has a radius of 1.8⨉106 m. The pull of Jupiter's gravity on the far side is just 99.15% that on the near side.
  • Doing a similar calculation for the moon (orbiting the Earth), we find far side gravity 99.10% that of near side.
  • Mercury orbiting the sun has far side gravity 99.99% that of the near side since, even though it's very close to the sun, it's a lot further away than the moons are from their primaries.
  • Let's look at Saturn now. Not one of Saturn's moons, but Saturn's rings. The main rings, according to Wiki, extend between 66 900 km and 480 000 km above the centre of Saturn. The gravitational pull from Saturn on the far edge is 83.6% that of the near edge. Compared with the solid bodies discussed above, that's a much more significant difference.

It is now possible to come up with a scenario where the pull of the primary on the near side of the satellite is bigger than the pull of its own gravity. Let's look at one of Saturn's tiny moonlets. Pan orbits inside Saturn's A ring (towards the outer edge of the ring system). It's radius is only 14.2 km and it weighs 5⨉1015 kg, making it's surface acceleration due to gravity 0.0016 m/s2 (less than a ten thousandth of a percent that of Earth's). By comparison, the acceleration due to gravity from Saturn at that distance is 2.12 m/s2, more than 1300 times greater. Clearly, Pan could not have formed where it now orbits since it's very much held together by chemical forces, not gravitational.

EDIT: Correction made to the above paragraph. Previously I had stated that if you stood on the Saturn-side of Pan you would fall up into Saturn. This is not true. The more accurate statement I should've made was that if you were floating around in the vicinity of Pan's orbit and Pan came past you, its gravity would not be strong enough to pull you in over Saturn's gravity. No matter how close to it you were (even if you could touch the surface), if you were not already moving along with it (and hence had enough centripetal acceleration to balance Saturn's gravitational acceleration), then Saturn's gravity would win out and you would fall towards Saturn, rather than towards Pan.

Making rings

Even further out than the point at which the primary's gravity becomes stronger than the satellite's gravity, the primary's gravity will start to deform the satellite. This effect is not dissimilar to the tidal bulge the moon causes on Earth. It is also part of the reason the Galilean moons of Jupiter are tidally locked.

(Interesting fact: over time, the tidal bulge of the Earth is causing the Earth to slow down its period of rotation since the change in shape (which isn't constant, remember, as the moon's orbit is much slower than the Earth's day) alters the way it rotates (catch phrase: conservation of angular momentum). The drag of the water in the tidal bulge is also pushing the moon back, slightly, in its orbit. Eventually (and we're talking a pretty long eventually) the Earth-moon system will settle into a mutually tidally locked rotation with the moon significantly further away than it is now. Here is an interesting article about it from

In the case of a satellite which is reasonably fluid and only being held together by its own gravitational pull (called self-gravity), then there is no reason for it to stay together as one lump. It will disintegrate because the part closer to the primary wants to orbit faster than the part further away. A disintegrated satellite will turn into a system of rings around the planet. The point at which this happens is called the Roche limit and the equation which tells us the distance from the primary of the Roche limit is:

d is the distance of the Roche limit from the centre of the primary,  R is the radius of the primary and M and m are the densities of the primary and the satellite respecively.

You'll notice that there are densities in the above formula. The density of the satellite is relevant because it's a measure of both mass and gravity (since we're talking satellites that are only held together by their self gravity and not chemical bonds). The density of the primary comes into it because we need to know the mass (which is proportional to radius cubed times density and R will becomes cubed if you move it inside the brackets) but the radius is also relevant because if the Roche limit is inside the primary, we can pretty much ignore it.

Of course, most satellite aren't balls of dust but are held together by other chemical forces (like Pan is). For example a rock on Earth isn't held together by gravity, it's held together by the chemical bonds between the different atoms and molecules inside (slightly different bonds depending on it's composition). Similarly, once a satellite has formed (outside of the Roche limit), then it probably goes through other experiences (such as tidal heating) which fuse it into a more solid lump. If it then wanders inside the Roche limit, it's not going to dissolve just because it couldn't've formed there. Pan and a handful of other moons in Saturn's rings prove this point. So what's the Roche limit for satellites held together by more than just gravitational forces? It sort of depends on the forces, but Roche himself derived an approximation for fluid satellites which deform a bit before they break up due to the tidal forces:

If you're wondering whether rock counts as fluid, it does. Everything will deform a bit under sufficiently strong forces.

So this last equation is the point at which a satellite will start to break up if it spirals in too close to its primary. For Earth-moon system, the moon will disintegrate if it wanders within 11 000 km. Luckily for the moon, this isn't likely to happen until the sun end's it's main sequence life.

Wednesday, May 11, 2011

Tides and their locks

Gravity causes tides. On Earth, a planet with a whole bunch of wet stuff sloshing around on the surface, this leads to the sort of sea and ocean tides that most of you have probably encountered at some point.

Tidal interactions

Tides on Earth are mainly caused by the gravitational force of the moon pulling on the Earth. Water, unlike the rock making up most of the surface of the Earth, is able to move a little bit towards the moon in response. Obviously, it's not a massive effect—we're not talking about losing chunks of ocean into space—but it's significant for the sea level to rise a few meters in certain places. The sun also has a similar effect on the Earth so if we didn't have a moon, we'd still have some tides, just on a smaller scale and rather more regularly. As it is, the reason tides are so irregular is because moon and sun aren't in sync (this is also why lunar calendars and solar calendars are so different).

I should also add that while the water on the side of the Earth closest to the moon becomes deeper due to the gravitational pull of the moon, it is conservation of angular momentum which causes the water on the side of the Earth furthest from the moon to also bulge out. I won't go into the specifics unless someone asks in the comments, but the short version is that that opposite bulge of water is require to "balance out" the bulge formed by the gravitational pull of the moon.

OK, so the only thing that really sloshes around on the Earth is water, but what would happen if the moon was bigger or the Earth was closer to the sun and had no water? Or what if we had a rocky moon orbiting close to a gas giant? Well, instead of water sloshing about, it's possible that the gravitational pull of the large planet would pull on the moon strongly enough to deform rock. This is exactly what happens with Io and Europa, Jupiter's two innermost (Galilean) moons (although Europa is more ice than rock). The tidal tug of gravity on Io is what keeps its core molten and causes so many volcanoes on its surface. It's what keeps the interior of Europa liquid (or at least, what keeps some water in liquid form below the surface) and it's also what keeps Earth's core molten (because of our moon's tidal forces). Without these tidal interactions, there would have been more than enough time for these planets and moons to cool enough for their cores to solidify. In the case of Io, I believe the smaller gravitational tugs of the other Galilean satellites, particularly Europa and Ganymede, may also play a small part in its tidal heating.

The take home message is: tidal forces cause volcanoes. It's an important point to remember if you're situating your planet/moon close to its star/primary.

Locked with tides

Let's move away from the Earth and the Jovian satellites for a moment and think about a miscellaneous rocky planet, close to it's star. Like Mercury, for example. For a long time, it was thought that Mercury was tidally locked, meaning that the same side always faces towards the sun. It turns out this isn't quite true thanks to gravitational tugs from some of the other, bigger planets. So let's ignore the other planets. We have a star and a planet forms in place around it. I've mentioned before that conservation of momentum dictates which direction planets and moons will initially rotate and orbit. Any deviations from this will be a result of later collisions. So the planet will form orbiting in the same direction that its star rotates and also rotating in this same direction. (If you're a bit confused about how an orbit and a rotation can be in the same direction, stick your thumb out and curl your fingers around. Your thumb is pointing in the direction of angular velocity of something rotating in the direction your fingers are curling. You could make a looser curl with your fingers to represent an orbit and, so long as your thumb continued pointing in the same direction, that orbit would be in the same direction as the previous rotation.)

The angular momentum of the star-planet system has to be conserved. (Angular) Momentum is mass multiplied by (angular) velocity for each body and then summed. Since the mass of the system isn't going to change much (ignore comets and spare dust/gas that might accrete), then for the planet's rotation to change (that is, get faster or slower) one of the other angular velocities has to change to compensate. This is exactly what happens when a planet becomes tidally locked around its star or when a moon becomes tidally locked around its primary (example: the Galilean moons of Jupiter); rotational angular momentum is slowly converted into orbital angular momentum resulting in a slightly faster orbit but a slower period of rotation. Given enough time, the planet will become locked in a synchronous orbit (the same side always facing its sun). This is where the "tidal" part of "tidally locked" comes from.

Now, it's possible to work out how long this process takes or, for a given time frame, what the "tidal lock radius" is; that is, the distance from the star inside of which planets will be tidally locked after that time period has elapsed. The general equation for this, as given by von Bloh et al. (2007) (and Kastings (1993), pdf sorry) is:

This isn't the most helpful equation ever. And the units are confusing
Where P0 is the original period of the planet, Q is a factor to do with the rotational properties of the planet and M* is the mass of the star. As the caption says, this isn't super useful and they use somewhat baffling units. Subbing in the values they assume for P0 and Q and assuming the same time they assume which is 4.5 Gyr (the G stands for giga—yes, just like in your computer—and means 4.5 billion years or 4 500 000 000 years), I get something close to:

See, isn't that nicer to deal with? And now we have rT in AU and M* in solar masses—yay!
So if you put in the mass of your star in units of solar masses (or the mass of your gas giant, but you still have to use solar masses) then you will learn the tidal lock radius in AU for systems which have had 4.5 Gyr in which to evolve. I think 4.5 Gyr was chosen because that's roughly the age of the Earth/solar system (actually, it's more like 4.6 Gyr, but close enough).

If you want to read more about orbital mechanics, I found a review written by the guy who originally worked this stuff out in 1977: Peale (1999) (pdf again, sorry). I haven't had the chance to read through all of it yet, and it's a bit heavy on the maths, but it looks interesting.

There is more that I want to say about tides, but I feel this post has gotten long enough so I'll leave it for a future blog. Coming up soon: the Roche limit and why Saturn has rings (and all the other gas giants too). Stay tuned!

Wednesday, May 4, 2011

Stars in their Skies

My intention had been to write another gravity post, this time about tides, tidal locking and Saturn's rings, but unfortunately I've been too busy to fully cover the scope I wanted to. Instead of a half-hearted post on the above, I bring you: stars! Stay tuned for tidal forces next week.

It's full of stars!

There are many different types of stars, as you may recall from the HR diagrams I've discussed previously (I've also drawn a rather crude, annotated HR diagram below). Stars come in a whole spectrum of colours, based on their (surface) temperatures, and these are loosely correlated with their masses. In general (for the main sequence), small stars are red and hence cool (only 3700 Kelvin) and large stars are blue and hence very hot (more than 33 000 Kelvin).

These large blue stars are sometimes called blue giants and small red stars are often called red dwarfs. In terms of mass, these stars can range from over 100 times the mass of the sun for blue giants down to about a tenth of the mass of the sun for red dwarfs. The sun, by the way, is on the cooler and smaller end of the middle of the main sequence.

Aside from descriptive words, stars are also separated into lettered classes based on their temperature. The hottest stars are designated O type, then it goes B, A, F, G (the sun is a G-type star), K and the coolest red dwarfs are M type. I'm not terribly fond of the mnemonic I use to remember them, so if you can think of a good one, please leave it in the comments. ;-)

Main sequence stars generate energy by fusing hydrogen and helium in their cores. They need to generate enough energy to over come the pressure of gravity trying to pull them into their central point. This means that bigger stars burn their hydrogen faster because they need to produce more energy to prevent gravitational collapse. Well, technically it's increased gravitational pressure that drives faster fusion in the core. Smaller stars don't have as much gravitational pressure acting on them, so the hydrogen nuclei in their cores are pushed as close together, meaning that their fusion reactions happen much more slowly and it takes longer for them to burn through all their fuel. So big stars burn bright and fast and die young. Small stars burn more conservatively and lead much longer lives. For reference, the lifetime for the hottest blue stars is around a million years whereas red dwarfs can expect to live a lengthy fifty billion years or so (the universe is currently only 13.4 billion years old). Our sun has been around for four and a half billion years and can expect to keep going for another five billion or so.

A very crude representation of where different types of stars fall on the HR diagram.

Stellar life

Talking about how long stars last is all well and good, but where do they come from and where do they go?

Nebulae (singular: nebula) are giant clouds of dust and gas in space. Even though the particles in a nebula are fairly spread out, over enough time gravity pulls them together into clumps. When these clumps get big enough that the gravitational pressure holding them together is great enough for fusion to start in the core, they "turn on" and switch from warm balls of gas into blazing young stars. The rest of the gas and dust that didn't make it into the protostar before it turned into a star proper either get blasted away by the new stellar wind unless they already clumped together enough to form planets.

After that, the star lands on the main sequence, based on its mass. What happens when it uses up all its fuel and reaches the end of its main sequence life varies depending on its mass. Smaller stars, like our sun, will throw off their outer layers and swell up into a red giant. The star will eventually eject all its outer layers and all that will be left is the star's exposed core; a white dwarf. A white dwarf no longer fuses hydrogen or helium or anything else. Instead it just slowly radiates away all of its heat until it eventually (over trillions of years) cools. White dwarfs are basically just spheres of carbon or oxygen or a mix of the two, depending on the initial star. Those news stories you might have seen about the "largest diamond in the universe"? Those are talking about carbon white dwarfs.

If we start with a larger star—big enough that after the star has ejected all its outer layers the core that's left behind is more than 1.4 times the mass of the sun—then the core will be too massive to remain merely a white dwarf. Instead it will collapse in a giant explosion known as a supernova. In the immense pressures exerted in the supernova, the protons in the old stellar core combine with the electrons to form neutrons (atoms are usually made of protons, neutrons and electrons). This type of star, composed entirely of neutrons and not found on an HR diagram, is called a neutron star. A really massive star can, after a supernova, collapse into a black hole, an even denser object (previously mentioned here).

Life elsewhere

Since the sun is a G type star and has a life-bearing planet, of course G type stars elsewhere could have life-bearing planets too. There is also a reasonable chance that F and K type stars, which are fairly similar to our sun, could also harbour life. The biggest problem is when we look at very massive stars. When we're talking about stellar lifetimes in the millions of years, then there probably isn't enough time for life to form. It took four or so billion years for life on Earth to get to the stage it is now. If the sun had only lasted for ten million years before exploding, then we wouldn't be here.

What about red dwarf stars then? They live for a very, very long time, so there's certainly ample time for life to develop. However, because the habitable zone is so close in to the star (because the star is so dim and cool), there are two issues:
  1. The planet will probably be tidally locked, meaning that one side is always facing its sun while the other never gets any direct heat or light. It's likely in this case that both the day side and the night side will be permanently too hot or cold. The ring of the terminator (the boundary between the day side and the night side) would probably be the best bet for life developing, temperature-wise.
  2. Small stars seem to have more flares than larger stars. Flares aren't terribly conducive to life, particularly at such close proximity. This is still an active area of research, however.
But, theoretically, in optimal conditions life could develop on a planet orbiting a red dwarf. Also, both of those issues are things I will probably blog about in the future.

As I mentioned last week, just because life can't develop there, doesn't mean humans can't try to colonise planets around different and interesting stars. Of course, colonies with a five or so million year time-limit might be a bit sort sighted, but if there are planets in suitable places, it could be good for a laugh.


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