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Watsisname
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12 Jun 2017 03:10

I wonder why stellar classes are not proportional nor exponential but their range is like: 3700 - 1500 - 800 - 1500 - 2500 - 20000
I was thinking about putting a brief explanation of that in -- it's an excellent question.  The reason is because the system isn't directly based on temperature or mass or any other obvious physical property of the stars that we often associate the sequence with.  It is instead based on their spectral lines.  For an astronomer, studying things by their spectral properties is the name of the game, because it is much more precise and direct from the observations.  To go by temperature would be to go by a derived property, rather than by a direct measurement.

Much of this is also historical.  Back before astronomers knew much about the astrophysics of stars, they started noticing that different stars had different spectral lines, but they were not simply random.  There were patterns as to what lines were visible, and how strong and wide they were, that roughly corresponded to the color of the star.  So different "spectral" classification systems were developed to describe these.

Some systems used numbers, others letters, and at one point there was even a system using the set A through Q (the "Draper" system).  But eventually much of the letters were dropped, some of them were switched around, and today we have OBAFGKM:

Image

Or in a graphical form:

Image

Of course now we understand that this spectral sequence is related to the surface temperature of the stars, and it's neat to see that much of the order of the letters is maintained (A through M, except for O's and B's being hotter than A's). There is really interesting physics behind it. 

Many people also ask why we don't just fix this to be A through G or something, with A's being the hottest.  We could, but there is a lot of inertia with the system we use, and we just happily stick with it.  It's like electricians not fixing conventional current to correspond to the movement of electrons. :)
 
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12 Jun 2017 11:46

Many people also ask why we don't just fix this to be A through G or something, with A's being the hottest.  We could, but there is a lot of inertia with the system we use, and we just happily stick with it.  It's like electricians not fixing conventional current to correspond to the movement of electrons. :)
It's the same reason why we call the water "water" and not "dihydrogen monoxide" (?)  :lol:
The universe is not required to be in perfect harmony with human ambition.

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13 Jun 2017 16:19

This question is partially directed at Watsisname, but feedback is welcome from anyone if you think you have a good solution.

I want a formula for determining the greenhouse effect on a planet, knowing its albedo and effective temperature and atmospheric pressure and composition. Before going any farther I must point out that I KNOW this can't easily be done with any significant amount of accuracy. Fortunately that's not strictly necessary in this case. What's needed is something that can produce nice, somewhat believable results. For the sake of argument we'll assume that we're talking about the potency of greenhouse gasses for a ~300K planet, like Earth. If we come up with something good, maybe it will be usable in SE to improve its greenhouse generation :)
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13 Jun 2017 17:03

@Salvo:

Dihydrogen monoxide is actually a wrong chemical name perpetuated by the internet when trying to make water seem unfamiliar.  It doesn't fit the chemical naming conventions.  (Consider: is H2S called "dihydrogen sulfide"?  No, it's just called hydrogen sulfide.  Is H2O2 called "dihydrogen dioxide"?  No, it's hydrogen peroxide.)

A more conventional chemical name for water would be oxidane. :)  But we stick with water because everybody already knows the term, including chemists.  No chemist will seriously say "pass me 10mL of that oxidane over there."

With the spectral classifcation system, we keep it the way it is mostly because there is a long history of using it.  If we changed it, then there'd be a lot of problems with referencing anything from the enormous body of literature that used the previous system.  And while a new system could be slightly easier for a new astronomy student to learn, every practicing astronomer would also have to unlearn the old system.
 
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13 Jun 2017 19:39

HarbingerDawn,

As you point out this is extremely difficult to compute precisely.  In fact there is no analytic formula which will precisely determine the greenhouse effect.  Instead it requires computer code, integrating the absorption and transmittance of a spectrum across many thin layers atmosphere and by each small bin of wavelength, accounting for the spectral effects of line broadening, excited molecular states, other mechanisms of heat transfer (convection), and light scattering.  It gets very complex very fast.

Fortunately, an approximate or believable model can be constructed from some simplifications.  The biggest challenges are from wanting this to be very general -- to apply for any atmospheric composition and any spectrum of incoming sunlight, so it can be applied in Space Engine.

A common first simplification is to treat the incoming and outgoing radiation as being independent (that the atmosphere is mostly transparent to the incoming sunlight but has significant opacity to the outgoing thermal radiation).  This is a reasonable approximation for Earth, but it will fail if the atmosphere is very opaque to the sunlight, or if the star is of cooler temperature (such as around red or brown dwarfs).

If they are independent, then we first compute the surface temperature of the planet by equilibrium (balance the energy in and the energy out).  I can describe this process in more detail in another post.  The resulting formula is

Image

where L* is the luminosity of the star, AB is the planet's albedo, "a" is the orbital distance, and σ is the Stefan-Boltzmann constant.  We may also recognize the 1/4 exponent coming from the Stefan Boltzmann Law.

Then the greenhouse effect arises from the atmosphere absorbing and re-emitting the outgoing radiation.  If we split this up by layers and define each layer to have one "extinction thickness" (absorbing 100% of the outgoing radiation), then we'll arrive at a new surface temperature:

Image

where [tex]\tau[/tex] (tau) is the number of extinction thicknesses, or "optical depth" of the atmosphere to the outgoing thermal spectrum.

In other words, for a planet with a greenhouse atmosphere, its surface temperature grows with the 1/4th power of the optical depth.
 
Now the problem is how to compute the optical depth for general atmospheres.  We'll need to know something about the spectral properties of all the gases Space Engine uses in its atmospheres, and be able to compute effective optical depths for different mixtures at different conditions.  This might be doable, but with a lot of work.  Probably this would involve a community effort.


Finally, we'll have to figure out what to do when the earlier assumption that the planet's atmosphere is transparent to the sunlight but opaque in the infrared is not valid, or if the solar spectrum is comparable to the emitted thermal spectrum.

If the atmosphere is opaque to the sunlight then this reduces the temperature at the surface (an anti-greenhouse effect).  This is important on Titan with the high altitude hazes which absorb much of the sunlight, but let the infrared from below through.  We could model this as an increase in the albedo, although it is not actually making the planet more reflective.

For cooler stars where the incoming spectrum is comparable to the emitted thermal spectrum, I'm not really sure of a good method.  In the literature this is sometimes modeled by computing a cutoff frequency, below which the atmosphere is opaque and above which it is transparent.
 
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14 Jun 2017 00:28

Watsisname, thank you very much for your reply, this is wonderfully useful information. I think I can use it to come up with something workable :)
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14 Jun 2017 18:05

Here's a weather hypothetical, inspired by my dream this morning...

Lets say I have "Elsa" powers, like in the movie Frozen. It's a nice, hot, humid summer day, with a cold front apprpaching creating a line of severe storms. The warm air the cold front is colliding with is, say, 96° F with a dew point of 74° F (arbitrarily chosen).

Now, I use my powers to take the cooler air thats creating the entire cold front, and chill it down all the way to 30° F and a similar dew point, while leaving the warmer air as is, and then just "Let it Go" from there. What happens next? How do the storms react and evolve? How long would it take for the cold front to warm back up to seasonal temps by itself?
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14 Jun 2017 22:43

How is the homunculus nebula so symmetric when eta carinae is a barycenter binary (according to wikipedia's depiction)?
 
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18 Jun 2017 03:14

How is the homunculus nebula so symmetric when eta carinae is a barycenter binary (according to wikipedia's depiction)
I had to do some research to figure out the answer to this question, so apologies for the wait.  It's very interesting though, so I hope it's worthwhile. :)

The binary nature of this system was actually established fairly recently -- just within the last 10 years.  Before then astronomers tried modelling it as an ejection of material from a solitary star.  The bipolar shape was thought to result from a roughly spherical ejection slamming into a torus of older material.  Another model showed it could result from an eruption off of a rapidly rotating star.  There were a number of outstanding issues and open questions with these though, and with the advent of the latest generation of telescopes and computer models we've made a lot of progress on understanding Eta Carinae.

We now know that the system is indeed a binary, with a highly eccentric orbit and interacting stellar winds:

[youtube]HGqXMEWiRvs[/youtube]
Based on the timing of the orbits, the previous eruptions are thought to have been caused by the influence of the companion near periastron, while the primary happened to be unstable.  These eruptions are very energetic, and the material is sent out at several times the escape speed from the system.  So the shape of the resulting nebula does not care a whole lot about the presence of the companion star.  It ends up being fairly axisymmetric.  You might think the companion and its winds should at least affect it a little bit though, and recent observations do show deviations from axisymmetry that may be related to it.

But as the above video shows, where the companion really matters is in the structure and behavior of the inner region, where the stellar winds are colliding.  This is a big part of how we were able to determine the companion's existence, because we cannot resolve it directly.

Very good question!


PlutonianEmpire, I'm hoping other people would like to put thought to your scenario, but I'll take a stab at it afterwhile if no one else does.  It's hard to give a precise answer, but shouldn't be too hard to describe qualitatively.
 
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18 Jun 2017 21:38

How is the homunculus nebula so symmetric when eta carinae is a barycenter binary (according to wikipedia's depiction)
I had to do some research to figure out the answer to this question, so apologies for the wait.  It's very interesting though, so I hope it's worthwhile. :)

The binary nature of this system was actually established fairly recently -- just within the last 10 years.  Before then astronomers tried modelling it as an ejection of material from a solitary star.  The bipolar shape was thought to result from a roughly spherical ejection slamming into a torus of older material.  Another model showed it could result from an eruption off of a rapidly rotating star.  There were a number of outstanding issues and open questions with these though, and with the advent of the latest generation of telescopes and computer models we've made a lot of progress on understanding Eta Carinae.

We now know that the system is indeed a binary, with a highly eccentric orbit and interacting stellar winds:

[youtube]HGqXMEWiRvs[/youtube]
Based on the timing of the orbits, the previous eruptions are thought to have been caused by the influence of the companion near periastron, while the primary happened to be unstable.  These eruptions are very energetic, and the material is sent out at several times the escape speed from the system.  So the shape of the resulting nebula does not care a whole lot about the presence of the companion star.  It ends up being fairly axisymmetric.  You might think the companion and its winds should at least affect it a little bit though, and recent observations do show deviations from axisymmetry that may be related to it.

But as the above video shows, where the companion really matters is in the structure and behavior of the inner region, where the stellar winds are colliding.  This is a big part of how we were able to determine the companion's existence, because we cannot resolve it directly.

Very good question!


PlutonianEmpire, I'm hoping other people would like to put thought to your scenario, but I'll take a stab at it afterwhile if no one else does.  It's hard to give a precise answer, but shouldn't be too hard to describe qualitatively.
Thank you so much for researching in your time, I really appreciate it! What'd I'd like to know though is that the orbit of the primary star doesn't affect the symmetry of the homunculus nebula as well? Considering it's bipolar in nature.
The guy in the video says that the primaries stellar wind is extremely dense, thus making observations hard. So is the 'wind' from the star visibly dense enough to be semi-transparent or somewhat opaque, or is that only in certain wavelengths?
One final question, do some really massive stars (don't know what mass range) never transition into a red giant phase and supernovae from really hot blue giant phases (such as in a wolf-rayet star, though I'm not surebecause I've heard they briefly enter a red giant phase before blowing out their outer layers and becoming wolf-rayet again.)?
 
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19 Jun 2017 10:24

PlutonianEmpire, I'm hoping other people would like to put thought to your scenario, but I'll take a stab at it afterwhile if no one else does.  It's hard to give a precise answer, but shouldn't be too hard to describe qualitatively.
Understandable. :)

I posted here because people on Reddit are downvote-aholics and Yahoo Answers sucks. :lol:
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20 Jun 2017 15:57

So, I have a worldbuilding question related to the habitable planet of my custom system.

So, the planet is a super-earth with 1.2 Earth radii and 1.84 Earth masses. It is placed in the Delta Pavonis system, which is at the end of the of the main sequence, therefore the planet is finding itself straddling the inner boundary of the habitable zone. The planet's atmospheric pressure is 3.239 atmospheres, and 88.8% the surface is covered by water.

Assuming the remaining 11.2% of land is all hot sandy desert, is there enough dust to make the atmosphere red like Mars'?
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26 Jun 2017 16:26

I have a question regarding Betelgeuse distance: In Wikipedia it is reported to be ~ 643±146 ly away from earth. But in space engine it is reported to be ~425 lys. Which one is correct?
Also I watched a documentary regarding Betelgeuse in YT and whether it will affect us when it goes nova. It said in this documentary that Betelgeuse's gamma ray burst from its poles will miss us by 20 degrees. But how are they so certain? I mean all stars are moving inside the galaxy. So maybe it won't miss us. OK the chances are probably practically zero but still..

Also they say that the Japanese have developed a system underneath the ground that will detect neutrinos coming from the dying star a few minutes before it explodes. So that when they detect an increase flux of neutrinos, they will notify all astronomers and the scientific community around the globe to turn the telescopes towards Betelgeuse before it starts to shine brighter. Does this mean neutrinos travel faster then light? Probably not..

Aslo if Betelguese goes nova now, we will see it 6 centuries later, I mean our grand grand grand grand grand grand..etc.. children will. Or it has exploded about 6 centuries ago and the light information will come any moment now..
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26 Jun 2017 17:26

Also they say that the Japanese have developed a system underneath the ground that will detect neutrinos coming from the dying star a few minutes before it explodes. So that when they detect an increase flux of neutrinos, they will notify all astronomers and the scientific community around the globe to turn the telescopes towards Betelgeuse before it starts to shine brighter. Does this mean neutrinos travel faster then light? Probably not..
Neutrinos travel at lightspeed, but they are emitted a few moments before the star actually goes supernova, that's why they arrive earlier.
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26 Jun 2017 21:56

Neutrinos travel at lightspeed, but they are emitted a few moments before the star actually goes supernova, that's why they arrive earlier.
Yes.  The core collapse that triggers the supernova is very fast (fraction of a second!) and emits the neutrinos, which due to their ghostly nature escape from the star easily and at (or at least *very* close to) the speed of light.  But the shockwave from the core takes hours to reach the surface and become visible as a brightening of the star.
I have a question regarding Betelgeuse distance: In Wikipedia it is reported to be ~ 643±146 ly away from earth. But in space engine it is reported to be ~425 lys. Which one is correct?
Space Engine's figure is based on HIPPARCOS parallax data, which were less precise and prone to error. Newer data which combine VLA measurements with HIPPARCOS yield the figure given by Wikipedia.
 It said in this documentary that Betelgeuse's gamma ray burst from its poles will miss us by 20 degrees. But how are they so certain? I mean all stars are moving inside the galaxy.
They probably mean if it went off in the foreseeable future (hundreds of thousands of years).  Relative motions of the stars won't affect the orientation very much on timescales we care about.  (I don't think the proper motion will bring the pole into alignment with us anyway, but I'd have to check to be sure.)
The way we know the direction the jet would be relative to us is by measurement of the star's rotation axis:
Image
Aslo if Betelguese goes nova now, we will see it 6 centuries later, I mean our grand grand grand grand grand grand..etc.. children will. Or it has exploded about 6 centuries ago and the light information will come any moment now..
I hope it exploded its light travel time ago minus about half a year. ^_^

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