text:
What does observational evidence say about the number of protoplanets believed to exist in the early solar system?
Currently, the solar system has 8 major planets, a handful of dwarf planets and a plethora of asteroids, comets and natural debris, and of course, a rather locally dominant star (the Sun).  There are many references that describe the early solar system as a chaotic environment with many more protoplanets.
Based on observational evidence, how many protoplanets are believed to have existed in the early solar system?

Earth and moon are thought to be formed by the collision of two planets. So we have at least these two protoplants. The cores probably ended mainly in the core of Earth; that's inferred by the much higher mean density of Earth in relation to Moon.
Iron-nickel meteorites strongly indicate at least one pre-existing planet large enough to form an iron-nickel core.
Comets of the Kuiper belt are hypothized to be a collision product.
As you go further back in time the building blocks of planets get ever smaller and more numerous till the interstellar dust and supernova remnants our solar system is formed of.
Some of the early material will most likely have been ejected from the solar system; this may include protoplanets.
By the chaotic nature of multi-body systems like our solar system it's getting more and more difficult to track single bodies back into the past.
The NICE models try to describe the early solar system.


text:
Periodic error correction in automatic telescopes
I am starting a study on automatic telescopes and the correction of their periodic error.
I found two ways of correcting it : with a CCD camera and manually by recording the error. I was wondering if it could be possible to model the error by a periodic signal in order to add it to the entry signal then correct it with a low-pass filter or an other filter ?

I'm not sure why this was migrated to [astronomy.se] from [electronics.se] since it's relevant to the mount's drive gear and its time to complete revolutions alone. The OP already mentioned correction through the CCD images, and that's pretty much the only "astronomical" way of doing that. There isn't any other way to ask the stars if they know how much some stepper motor errs.

Quoting the most relevant parts from the Richard McDonald's Wiki on Astrophotography Mounts: Periodic Error Correction:

Periodic Error can be reduced to an acceptable level using a variety
  of techniques, only some of which are in the range of a beginner or
  mid-level astrophotographer.

Throw Money at the ProblemVery high-end mounts for astrophotography have very small periodic error because of the time
  and money spent on manufacturing high-precision gears. You can also
  buy higher-precision gear upgrades for some mounts. For the beginner,
  let's call this impractical. Buying a \$1000 replacement worm gear for
  your \$1000 mount is probably not a good balance.
Don't Use GearsThis is really an extreme case of the "throw money" solution, mentioned just for fun. There are some
  very-high-end experimental drive systems showing up on the market that
  don't use worm gears, and that don't exhibit periodic error. Examples
  include direct drive systems and harmonic drives. 'Way out of our
  price range.
AutoguidingA second guide camera and computer can be used to make frequent small corrections to the mount's pointing,
  and this can reduce or eliminate Periodic Error if the onset of the
  error is not too sudden. This is the subject of a separate article.
Periodic Error Correction FeatureTraining PEC with hand-control Most mounts intended for astrophotography include a
  feature called Periodic Error Correction (PEC), which can be used
  alone, or in conjunction with Autoguiding. PEC is used in two phases:

Training. During this phase you use the control panel or menus to say "Hey! Pay attention to this!" to the mount, then you manually
  keep a star perfectly centred for several worm periods. You do this by
  centering a star with a high-magnification eyepiece that includes a
  cross-hair reticle. You then stare at the star for 10 to 15 minutes
  and use the mount's hand controller to manually make the small
  adjustments necessary to keep it perfectly centred on the cross hair.
  The mount records the error corrections you supply, remembering where
  in the worm position each one was needed. By training through more
  than one worm cycle, the mount can record an average correction, in
  case you reacted slowly or over-reacted.
Playback. Once you have trained the mount, you can turn on PEC. The mount will "play back" the recorded error correction information
  by slightly changing the speed of the Right Ascension drive to move
  ahead or back each time your manual error corrections did the same.
  This will cancel the periodic error, resulting in a smoother track.

Training PEC with
  hand-control: Periodic Error Correction (PEC) button on the control
  panel of an equatorial mount.
The manual training phase is quite tedious and a more modern
  alternative is to use a camera and computer - usually the same one you
  will use for auto-guiding - to track a star while the mount records
  the training information.
There is even specialized software available to help collect Periodic
  Error data, analyze and smooth it, and upload it to the mount. You do
  not need such software, as the manual techniques mentioned above will
  work just fine. However, it makes a tedious job simple and pleasant,
  and you may find it a worthwhile investment. I use PEMPRO and am
  very impressed with it. It's not free, but it is inexpensive and works
  very well (and includes another feature to help achieve perfect polar
  alignment).

Quoted text and photograph source, copyright and courtesy of Richard McDonald, no copyright infringement intended. I would also highly recommend reading same author's extensive article on Autoguiding.

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/

sort: votes


text:
What is radiation pressure and how does it prevent a star from forming?
This is a follow up to: Is there a theoretical maximum size limit for a star?
The answer there talks about the radiation pressure preventing a star from forming.

What reaction is causing this radiation pressure and why does this potentially prevent a star from forming?

Radiation pressure is nothing but electromagnetic interaction.
Imagine a hydrogen atom hit by a stream of photons coming from the same direction. Although the atom as a whole is neutral, the electron and the proton are physically displaced, forming a dipole, i.e. a positive-negative charge couple. Some of the photons therefore scatter against the dipole transferring to it some momentum. So the atom start moving in the same direction as the photons. If the photons are in the ultraviolet, the electron can be exited to higher orbitals and possibly stripped from the atom. In this case the scattering is even more efficient.
Now imagine a star surrounded by a layer of hydrogen. Gravity attracts the layer towards the star. The photons emitted by the star try to push the hydrogen atoms away from it, through the electromagnetic force.
Very massive stars are very luminous and hot, which means that they emit a lot of ultraviolet photons. When the pressure transferred from the photons to the layer is larger than the gravitational attraction, then the layer begins expanding, effectively stopping growth of the star.
In the figure posted by the OP there is also dust. I don't know the details of photons-dust-gas interactions (we need a stellar atmosphere expert, I guess), but the basic principle is nonetheless the same.

To add to @Frencesco's answer, if you're talking about why temperature prevents a star from forming in the first place, the same principles apply.
For a star to form you need a cloud of gas to cool off significantly so that it can condense, much further along it can begin fusing hydrogen at its core. For it to cool off and begin collapsing it needs radiates away energy (this is partly why there are thought to be no dark matter stars). This process can take time. However, once the star has formed, ionizing radiation works to act against gravitational collapse, transferring energy/momentum back into the gas.
Summary: You first need to radiate away energy before you can create a star. For the most massive stars, the rate of energy/momentum transferred into the outer layers of the star is simply to great (see above).

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


text:
How can absolute value of azimuth exceed distance from pole?
I'm sure I'm just doing something dumb, but:

Yildun has a declination of approximately 86.5. This puts it 3.5
degrees from the celestial north pole.
Per the snapshot below, Yildun's azimuth can be as high as 4.17
degrees.
Since the celestial north pole always has an azimuth of 0,
wouldn't this put Yildun 4.17 degrees from the pole, contradicting
the distance calculated from declination?

What am I missing?

EDIT: I thought I'd figured this out:

On Earth, one degree of latitude is not the same as one degree of longitude (except at the equator).
Analogously, in the sky, one declination degree is not the same as one right ascension degree.

I thought the situation for right ascension applied to azimuth, but it doesn't.
Both azimuth and declination are measured on "great circles" (declination is measured on a half circle, but same general idea), so degrees in azimuth should equal degrees in declination.

What is the latitude of the observer?

Oh, this is stellarium's default for Albuquerque, NM, which is about 35N and -106.5W. However, I think this question is latitude independent.

It turns out I was right when I thought I was wrong.
Consider the azimuth circle for 85 degrees elevation. It's very small. For example, 85 degrees elevation and 0 azimuth is only 10 degrees away from 85 degrees elevation and 180 azimuth.
Therefore, the azimuth circle for a given elevation is smaller than a great circle. It's only cosine(elevation) the size of a great circle.
At 35 degrees, the azimuth circle is cosine(35 degrees) = 0.819152 the size of a great circle, so 4.17 degrees in azimuth = 4.17 degrees * 0.819152 = right around 3.5 degrees on a great circle.


text:
Is sundial time entirely dependent on solar azimuth?
I've visited several "how sundials work" sites and can't seem to get a
clear answer to this: is "sundial time" just a linear function of
solar azimuth? More specifically:

When the sun is due south (northern hemisphere), it is sundial
noon. All sites I've visited agree on this.
When the sun is due west (azimuth 270 degrees), I say the sundial
time is 6pm, a quarter turn/day from noon. However, I can't find a
site that actually says this, and some sites seem to disagree with
this. Same for it being 6am sundial time when the sun is due east.

I know there are different types of sundials, but had always assumed
they would give the same sundial time. Is that not true?
As a note, the sundial I describe above can be fairly inaccurate at
times, which makes me question whether I'm correct.
EDIT (to clarify question): Forgetting entirely about clock time for a
moment, suppose I build a sundial. When the sun is due west, my
sundial reads 6pm. When the sun is due east, my sundial reads
6am. When the sun is due south, my sundial reads noon. My question:
have I built my sundial correctly? The type of answer I'm hoping to
get:

No. When you build a sundial, it should read (something else) when
the sun is due west.
There are many different types of sundials. Depending on type,
your sundial may or may not read 6pm when the sun is due west.
It depends on your latitude: sundials are
latitude-specific. There's no such thing as a global sundial.
Yes, your understanding of a sundial is correct. Sundials can be
off by as much as (some number/formula) from mean solar time,
depending on your latitude. They can be as much as (some
number/formula) from clock time, depending on your longitude and
time zone.

Any sundial that gives the same result as this is correct and any other is wrong (but sometimes close enough):
                    _      /############
                    /|    /#############
     skewer        /     /##############
(central) v       /     /###############
          |  north     /################
          | (S in S.  /#################
          | hmsphre) /##################
  ________|________ /###################
hoop      |        /##### LEVEL  #######
(from the | side) /###### GROUND #######
          |      /####### (SOIL) #######
latitude--|-    /#######################
(use      ||   /########################
protractor)|  /#########################
          |V /##########################
          | /###########################
          |/############################
          j#############################
         ,|#############################
        /#|#############################
       /##|#############################
      /###|#############################
     /####|#############################
    /#####|#############################
   /######|#############################
  /#######|#############################
 /########V#############################
/#######################################
########################################

write noon on hoop's inside closest to ground, midnight opposite, 6 pm on the east side, 9 pm midway between the last two, and so on (hours only occurring in darkness optional).
If your sundial reads 6 pm at due west all the time then you're doing it wrong. Let's say I put a vertical stick in the ground, draw a 24 hour clock face around it, put noon poleward and think it's a sundial. In New York City, it could literally be saying 6 am when a genuine sundial says 9 am. That's just middle latitudes. At the equator on the equinox, it would read 6 am all morning and 6pm all afternoon at the equator. If you go 1 mile south of where the next noon, summer solstice and Tropic of Cancer coincide, it would say about 4:30am at sunrise, go forwards at first, then backwards, finally showing the middle 6 hours of the night passing in 7 seconds. Backwards. At noon. Then it will run forwards again until it reaches 7:30pm at sunset. If you place the stick right, you can even make it stay between 4:30a and 6a all morning, stop at 6 am at the instant of noon then instantly become midnight, run infinity years per second backwards for an infinitely short amount of time, go almost 6 hours backwards in seconds, then later forwards again very slowly until it shows about 7:30pm at sunset. This is why sundials cannot be made that way.
(of course, this is theoretical, there are no infinitely thin, vertical, and straight sticks, shadows are fuzzy, they can be too short to see, the Earth wobbles a bit, the speed of light is not infinite, this would only be true if only the Sun and Earth existed, even a flea jumping in Russia moves the Earth etc.)
And yes, the sundial time can be up to 16 minutes away from mean solar time (the equation of time), easily noticeable, but if you wanted correct local clock time instead of correct sundial time then you could put as many dates as needed on another dial and rotate the hour scale until the arrows point at the current time of year. The shadow then shows mean solar time.
That should be close enough to mean solar time that you wouldn't care for a number of centuries, certainly a century if you're real picky. As for clock time, what sundial disagreements are possible is limited only by the whims of man. The sundial is several hours wrong in West China. Cause they use the zone that's good for Shanghai (or Tokyo when daylight savings).
And all sundials without moving parts are latitude specific. Some are adjustible, though. Some designs are more suited for some latitudes or even become impossible in some places, like the kind with a wedge or rod on a level face. They will also not work on days with polar night. Though you could use a moondial if it's also not polar moon's below the horizon for days or weeks. Yes, moondials exist! You need to correct for moon phase and time of year or they're useless.

@MikeG This looks really helpful - is it possible to also add a punch-line or tl;dr somehow? Suppose I believe it unconditionally and don't need proof - but I need a short-cut to the point. Can it be summarized somehow?

Just cleaning up the ASCII art here. I added a separate answer.

The sundial translates the position of the sun to the time of day, so it depends on the path the sun takes across the sky, this is called the ecliptic.
Because the earth's rotation is tilted with respect to its orbit around the sun, the ecliptic shifts across the sky during the year, which is also the cause of seasonal change on earth. Yet the highest altitude of the ecliptic will always be due south, which means noon is well defined every day of the year.
Now to answer the question; if the day length (from sun-rise to sun-down) is exactly 12 hours,  the sun will rise at an azimuth of 90 degrees and set at 270 degrees. However, this happens only twice a year, namely at the vernal equinox (first day of spring) and the autumnal equinox (first day of fall).
At the longest day, the summer solstice, both sun-rise and sun-down azimuths will be at their highest shift northward. On the other hand, during the shortest day, the winter solstice, the azimuth-shift will be maximally southward. This last bit applies to the northern hemisphere, for the southern hemisphere the seasons and therefore day lengths are reversed.
So in short, you have to account for what place on earth your sundial is at and what time of year it is.

The sun is not on the same Azimuth for the same hour (not even at noon!). You need to take the analema figure (http://en.wikipedia.org/wiki/Equation_of_time) into your accounts. If you do not, your clock will be exact only 4 times a year.

The shadow of a vertical stick would show the Sun's azimuth, but that's not really a sundial.
A proper sundial's gnomon is aligned with the celestial poles so that the shadow indicates the Sun's
hour angle.
Then the great circle of hour angle $\pm$90$^\circ$ is projected onto the sundial plane as an east-west line;
the shadow falls on that line at 6am or 6pm local apparent solar time regardless of the Sun's declination.
Unless the sundial is an equatorial type, the other
hour marks
are spaced closer together around noon and farther apart around 6am and 6pm.


text:
USNO moon images look quite different?
I visited:
http://aa.usno.navy.mil/imagery/moon
http://www.usno.navy.mil/USNO/time/moon-phase-images
within seconds of each other, and saw two very different "current moon images":

Is one of these images wrong? The larger image seems to display a much
bigger crescent than the smaller image.
I've noticed this earlier as well.

This is an unofficial explanation, but I doubt you'll get an answer from official sources here, unless you ask them directly (e.g. their Twitter account is @NavyOceans), so I'll give it a shot. The small thumbs appear to be merely approximation, representation icons that aren't resized from the real-time image, but merely prepared in advance images, displayed depending on the current Moon's phase, calculated by the server.
If you try their Moon phase image generator (at the bottom of this page), you'll see why they did it like that. The prepared thumbs are numbered from m180.gif for Full Moon to m360.gif for New Moon, with a 2° step, and apparently the server calculates based on input parameters the Moon phase and displays prepared thumbnail with a minimum approximation precision of 2° (roughly 8 hour step). The same page also offers an explanation for how this thumbnail generation works:

These lunar phase images were created by R. Schmidt from ray-traced
images of the Moon. A Clementine spacecraft mosaic of the lunar
surface was mapped onto a sphere, and scenes were rendered as a
virtual Sun "orbited" the Moon. The depiction of lunar surface
features suffers geometric distortion but the terminator is correct
with respect to the spherical Moon.

The lunar phase calculator offers date selection that goes forward in time or before the time any images of actual observations were made and stored, so it's unreasonable to expect actual images displayed for, say, some date in 22nd or 19th century. Some examples off U.S. Naval Observatory's server:

(source: navy.mil)

(source: navy.mil)

(source: navy.mil)

(source: navy.mil)

(source: navy.mil)
            Full Moon                        Waning Gibbons                     Last Quarter                    Waning Crescent                     New Moon
The larger image of how the Moon looks like now seems to be generated by a lot more precise renderer (image is refreshed every minute by a JavaScript call) that takes into account not merely lunar phase, but also its libration in respect to the surface on the Earth. It would look more like any frame off this animation:

                         Fig. 2: Lunar librations in latitude and longitude over a period of one month (Source: Wikipedia)
And inclination is likely calculated for geolocation of the Astronomical Applications Department of the U.S. Naval Observatory in Washington, DC. The page however doesn't explain if this assumption of mine is correct, but it would certainly appear so.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
How was Io not torn apart by tidal forces during its formation?
Jupiter's moon Io is arguably one of the most volcanically active bodies in the Solar System.  The reason, according to NASA's page Scientists to Io: Your Volcanoes Are in the Wrong Place is believed to be caused by Io being

caught in a tug-of-war between Jupiter's massive gravity and the smaller but precisely timed pulls from two neighboring moons that orbit further from Jupiter – Europa and Ganymede. Io orbits faster than these other moons, completing two orbits every time Europa finishes one, and four orbits for each one Ganymede makes. This regular timing means that Io feels the strongest gravitational pull from its neighboring moons in the same orbital location, which distorts Io's orbit into an oval shape. This in turn causes Io to flex as it moves around Jupiter.

So, how did Io form in the first place, given the tidal stresses acting upon it? Does this suggest (and what evidence is there) that Io 'migrated' into its present orbit?

No, it is not just a matter of migration. You need to take into account two facts.
One is that (as experience shows) Io's own gravity is enough to avoid it breaking by tidal forces. It has been like that through all its history: Io could not have been formed if it started aggregating today, but it was formed at the same time Europa and Ganymede did: they three were growing in parallel.
Other is that of the orbital resonances, which makes precisely that orbit with so simple integer number relations with those of Europa and Ganymede a stable one. Io could not have been formed in another place.

Do you have references/links to add to this answer?

Io's own gravity is self-explanatory. Reference for resonances can be found at http://en.wikipedia.org/wiki/Orbital_resonance

Hmm I was hoping for non-Wikipedia references, like a specific paper about the phenomena

I have not them at hand, just memories from my degree and Google.

I think it would add a lot of extra value to your answer if you found a paper discussing how orbital resonances may constitude evidence that Io could not have formed elsewhere. Wikipedia sources are a good starting point, but sometimes it lacks sufficient detail to answer questions like these.

I think the other answer is correct on migration, but there's a flaw in the way this question is asked, which should be addressed.  It's also worth looking at the formation of Jupiter as well.
One of the rules of planet formation is that angular momentum remains largely constant.   Granted, some angular momentum gets transferred into heat, and some gets lost for any material that escapes from the system and a tiny amount is lost in the escape of thermal radiation (that's a bigger factor with stars than planets).  But these tiny variations aside, we can generally say that angular momentum of the contained material is conserved and not all of that falls into the planet.  Some remains in orbit around the planet, as moons, a ring or a dust cloud.
Stars empty out their closer-in orbital regions fairly quickly.  With planets, that happens much more gradually, so Jupiter likely retained an orbiting nebulous sphere of ice, dust and smaller debris for some time, even after it's moons began forming.
The standard model for Jupiter's moons is that it may have gone through a few generations of moon formation, forming within the orbiting cloud of debris and in time, falling into the planet, while new moons formed and over time, the orbiting disk and gas thinned out.   Based on this model, Io is believed to be part of the latest generation of Jupiter's moons formation.
From Wikipedia link above:

Simulations suggest that, while the disk had a relatively high mass at
  any given moment, over time a substantial fraction (several tenths of
  a percent) of the mass of Jupiter captured from the solar nebula was
  passed through it. However, only 2% the proto-disk mass of Jupiter is
  required to explain the existing satellites.3 Thus there may have
  been several generations of Galilean-mass satellites in Jupiter's
  early history. Each generation of moons might have spiraled into
  Jupiter, because of drag from the disk, with new moons then forming
  from the new debris captured from the solar nebula.3 By the time the
  present (possibly fifth) generation formed, the disk had thinned so
  that it no longer greatly interfered with the moons' orbits.4

Jupiter's fast rotation and tidal forces would suggest that it's moons should move away from it similar to our Moon moving away from Earth, but a cloud of orbiting debris tends to slow moon's orbits and cause them to fall into the planet.  Jupiter's powerful magnetic field and rapidly moving charged particles may also have an effect, the combination is too hard for me to say whether Io is moving inwards our outwards, there's too many moving parts and even an estimate of how those forces combine is above my pay-grade.

But I digress, though I did want to point out that Io is not believed to have formed with Jupiter but formed later.   The question asks how orbiting debris can overcome tidal forces between Jupiter and other larger moons like Ganymede and Callisto.
A cloud of orbiting debris in a disk around a planet can coalesce into a moon provided it's outside the fluid Roche limit.  A solid moon begins to break apart usually closer to the rigid Roche limit, closer to the planet, due to some structural integrity.
For moon formation, all that's required is that there be sufficient debris density and that the debris is outside the fluid Roche limit. It doesn't matter that the density of the orbiting ring is low, what matters is that once the coalescing starts, that the proto-moon is outside the Roche limit, it's the Moon's density, not it's size that determines that Roche limit relative to the planet it orbits.   A moon information might initialy have a lower density, due to being less compact, so it may have a corresponding Roche limit that's more distant from the planet, but the variation is the cube root of the density, so the roche limit boundary wouldn't be all that much further out at the start of formation.
The proto-moon doesn't need to add the ring-debris all at once, it only needs to be able to hold onto what it gets very close to, and that's a product of being outside the fluid Roche limit.   In time, the Moon clears out the region where it orbits, and as pointed out in the other answer, migration likely plays a role in moon formation, but migration isn't the reason Moons form, that's a product of sufficient density of the orbiting disk and gravity.
(hope that makes sense, I'm not sure I explained the last part as well as I could have).

Good answer -- except for one omission. It is incorrect to say that "some angular momentum gets transferred into heat". Heat production implies dissipation of energy. It, however, does not entail, by itself, a loss of the angular momentum.


text:
What is the current accepted theory as to why Titan has retained its atmosphere?
Titan (moon of Saturn) is unique in that it possesses a very thick atmosphere.  However, Titan is certainly is not the largest of the moons - Ganymede being larger.
What is the current accepted theory as to why Titan has retained its thick atmosphere?

It's not really known why only Titan has a thick atmosphere while moons like Ganymede don't.
One thought is that temperatures may have been too high (well above ~40K) in the Jovian subnebula of our solar system due to the greater gravitational potential energy release, mass, and proximity to the Sun, greatly reducing the $NH_{3}$-hydrate inventory accreted by both Callisto and Ganymede. The resulting $N_{2}$ atmospheres might have been too thin to survive the atmospheric erosion effects that Titan has withstood.
EDIT: some links:
http://perso.utinam.cnrs.fr/~mousis/papier3.pdf (though it's from 2002).
http://adsabs.harvard.edu/abs/2012ttt..work...25N (this paper critiques the previous idea of a $CH_{4}$ supported atmosphere)
I would also think the NASA/JPL Cassini pages would have any up-to-date info in regards to Titan.

Very nice answer -  do you have any links to add to your answer?

It has retained it's atmosphere because there is enough water (or ice) to keep it still. That's also how Earth can retain it's atmosphere. If it is a planet or moon like Venus, then the clouds and the greenhouse effect would block the atmosphere, and therefore retain the atmosphere.

I fail to see how that makes sense. I don't believe water is relevant. Venus doesn't have much water, but it retains a very think atmosphere. Titan has water, but it's cold enough that it's essentially a mineral. And the greenhouse effect doesn't "block the atmosphere"; it blocks outward infrared radiation, thus making the atmosphere hotter. I'd think that Venus loses more of its atmosphere to space than it would if it were colder.


text:
What are the current accepted theories of the formation of the Uranian moon Miranda?
Miranda, a moon of Uranus is unique in that it has a very fractured surface

Source: University of Oregon
The surface is said to be jagged and fractured, with comparatively large disjointed cliffs and great faultlines.
What are the current accepted theories for the formation of such a strange world?
Please include authoritive links in any reply

I'd rather say it's too small to be called a "world". Giant rock will be a better term, and those are seldom of regular shape.

As Miranda is a rather small object (only 235km diameter) it may simply have not enough mass to make it matter settle into hydrostatic equilibrium under the influence of its gravity, thus it basically retains the irregular shape most asteroids have. This nice arxiv paper discusses the mass limits necessary for a stellar body to settle into a nice round shape:
http://arxiv.org/ftp/arxiv/papers/1004/1004.1091.pdf

Yes, this is the kind of information I am after.


text:
How many minutes does a sundial change every day?
I'm planning to create a sundial in Ghana (10° 28.852', 0° 26.032') in July 2014.
I will stay there for about 3 weeks.
Do I need to adjust the numbers anywhere in these 3 weeks?
I know the shadow's path changes through the year, but how much does it change?
Is the change significant for a sundial?
(btw, is this the correct site to ask or should I use the astronomy site?)

You can find that information on the Equation of Time page at the USNO.

There are several "Precision Sundial" designs, see "Google Images", that are corrected. Example; http://www.precisionsundials.com/renaissance.htm or from a 1959 article http://www.precisionsundials.com/sunquest%20article/amateur.htm

You need to look up "the equation of time" - this is a correction to the time reported by the sun on any given day - the apparent solar time - (and measured by your sundial) and the standard civil time - the mean solar time - it varies throughout the year and the maximum your sundial will be out is about 15 minutes - but the difference and the rate of change varies. http://en.wikipedia.org/wiki/Equation_of_time
If your sundial is merely for educative/demonstration purposes, this might not matter so much of course - as the sundial will surely demonstrate that the sun's shadow can be used to tell the time.

My sundial will be use to tell the time approximately. A 15 minutes difference doesn't really matter.

According to the wiki entry, the error runs up to roughly 15 minutes.  One can use the
"equation of time" , for which there appear to be published tables, to generate a correction curve. (The main contributors to the error are eccentricity and obliquity of Earth's orbit.

By 15 minutes, they mean 15 in a year, I guess?
That will be equal to just under a minute in 3 weeks.

No, they mean the maximum difference between local time and the time on your sundial will be 15 minutes. Since the sun changes its apparent position over the year, the time difference between the sundial and local time changes too. It's not a cumulative effect, like a clock winding down.


text:
In India, what is the best time to view the geminid meteor shower in 2013
As of the time of writing this question, the meteors should have appeared, but so far I haven't seen any. What time would be the best to view them?

They should be visible any time at night (note that longitude has little to do with observing meteors), but there are many things that could thwart your observing.

Bright lights: In a big city you will see far fewer meteors. In a big enough city, you won't see any. On that note, if you keep checking a mobile device with a backlit screen, you probably won't see anything.
The Moon: The Moon will unfortunately be up until something like 4 AM local time (assuming I plugged numbers in correctly to the US Naval Observatory's table). Just as with cities, this will cover up smaller meteors. Just try not to stare directly at the Moon lest you lose night vision entirely.
Limited view of the sky: The key to watching meteors is to see a large amount of the sky at once. Never use binoculars or a telescope. The idea is to get comfortable by laying down (yes, this is important - people who crane their necks get tired after 20 seconds and then stop looking up!) and just look at as much sky as possible. Your eyes naturally catch and follow movement, so you'll instantly focus on a meteor once it comes into view.

As for where to look, the answer is "everywhere." Meteor showers are named after the radiant - the point in the sky from which the trails seem to originate if you trace them backward far enough. This shower originates in the constellation Gemini, which isn't terribly recognizable, but can be found by locating Orion:

Move from his belt to his right arm (the one indicated by the very red star Betelgeuse) and keep going until you come across two reasonably bright stars. That is Gemini. It is important to keep in mind, though, that meteors streak across other parts of the sky. Just because they seem to originate from Gemini doesn't mean they're actually visible there. Knowing the radiant just helps with knowing what direction to expect the streak to be going.


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Can we calculate the average temperature of the heliosphere?
It is my understanding that one can measure the temperature of the universe by measuring cosmic microwave background radiation. Could we also use this method for calculating the average temperature of the heliosphere?
The average temperature of the universe has supposedly been measured in at 2.735 degrees above absolute zero. Presumably this measurement takes into account the massive amounts of 'empty' space which I imagine has a huge effect on the number.
My question: Can we measure the average temperature of the heliosphere? If so, then what is it?
This might be quite hard to define as how do you determine when our solar system ends? So for the purpose of this question the following assumptions should be made:

The distance from the Sun to the edge of the heliosphere is approximately 100AU (Roughly the size of the heliosphere as measured by Voyager 1).
Assume the heliosphere is a perfect sphere.
Assume we are measuring the black body equivalent temperature (as this is what was measured to calculate the temperature of the universe).

If we don't know how to measure this average temperature (which is likely if we can't differentiate between microwaves from in and outside the heliosphere) then how could we theoretically measure this value assuming a perfect environment?

You need to define what do you want to measure, as there are two things you can see as "Temperature": Black Body equivalent temperature for the radiation, and Statistical Thermodynamic temperature for particles.

@Envite Good point, Black Body equivalent temperature is what i am after, will edit in

The heliosphere is mainly defined by the region dominated by solar wind against the interstellar medium.
"The solar wind is divided into two components, respectively termed the slow solar wind and the fast solar wind. The slow solar wind has a velocity of about 400 km/s, a temperature of $1.4–1.6×10^6 K$ and a composition that is a close match to the corona. By contrast, the fast solar wind has a typical velocity of 750 km/s, a temperature of $8×10^5 K$ and it nearly matches the composition of the Sun's photosphere."
Hence with 1 million Kelvin we get a good estimate of the average temperature of the heliosphere.
The temperature is the freeze-in temperature derived from carbon ion charge states measured by SOHO. Hence it's not the black-body temperature, but a reasonable definition of temperature for solar wind.
"Coronal temperatures are inferred from density ratios of adjacent charge states. The freeze-in temperature derived from a given density ratio is the electron temperature that reproduces this ratio in a static situation. For the analyzed iron charge states we use the ionization and recombination rates of Arnaud and Raymond [1992]". More details about SOHO's temperature measurement here.
If the heliosphere would be in the thermodynamic equilibrium, according to Planck's law or Wien's law, you would get the peak at about 3nm, in the soft x-ray range. But the equilibrium assumption probably doesn't hold.
For the hot interstellar medium extreme ultraviolet radiation measurements have been tried (CHIPSat), some successful (EUVE).
Hence, in an ideal environment we would measure the radiation intensity across the electromagnetic spectrum, look for the peak intensity, apply Wien's law, and retrieve the  effective temperature, which would be identical to the black body temperature in an ideal world.
The plasma of the heliosphere is very rare, therefore almost transparent. That's why we don't feel that heat of about 1 milion degrees. Microwaves play a minor role in that temparature range.
A temperature profile of Sun's corona and heliosphere is shown e.g. on pages 42 and 47 of this survey.

It is too optimistic to expect that the wind which had been travelling for about a year and expanding by at least a factor of $10^6$ in volume would neither cool adiabatically, nor radiate away most of the energy.

@AlexeyBobrick It interacts with the magnetic field of the sun. The strange thing begins with the corona temperature of a million Kelvin, although Sun's surface is below 10,000 K. Where does the heat come from? The plasma cannot be treated as a gas. We have SOHO and Voyager data. The speed of the particles is slowed down at the heliopause by interstellar gas (http://voyager.jpl.nasa.gov/mission/interstellar.html). There seems to be a sharp (not necessarily convex) boundary between heliosphere and interstellar space (http://en.wikipedia.org/wiki/File:Solar_wind_at_Voyager_1.png)

"The rapid and consistent changes in the freeze-in temperature calculated from four pairs of iron charge states confirm the patchy structure of the corona with length scales of some $10^4$ km and reveal the survival of these structures from a few solar radii throughout I AU." of the above referenced SOHO paper. I can't find a hint for adiabatic cooling or net loss of temperature due to radiation, as plausible as it would sound.

It seems to me, that you don't differentiate between the bulk motion of the gas (plasma) and its thermodynamical properties, for example temperature and density. Solar wind surely expands till 100AU, as was pointed out already in the question, and it probably keeps the bulk velocity rather constant until gradually the shock from collision with ISM takes over. However, the temperature will go down almost immediately. You can check Kelvin-Helmholtz timescale of a mole of ideal gas in a reasonable volume. The wind will: 1) Cool down, 2) Possibly stop being in thermodynamical equilibrium.

For Voyager we don't have much more than the bulk motion as rough idea. It's totally different for SOHO: There doesn't seem to be much freeze-in temperature difference between a few solar radii and Earth orbit. The freeze-in temperature is much lower than the temperature-equivalent of the bulk motion kinetic energy, but still $10^6$K. I can only guess, that energy is exchanged by electrons, which are travelling much faster than the ions in the plasma, and that way leading to an electron-based equilibrium, instead of photon-based, feeded by energy of the magnetic field of the sun.

See also Coronal heating problem: http://en.wikipedia.org/wiki/Sun#Coronal_heating_problem

From the SOHO paper you pointed to, it seems likely that due to expansion the relaxation time for the plasma gets rather rapidly long comparable to the timescales it takes to reach the Earth and hence most likely the heliosphere. In this case plasma just gets decoupled from any thermodynamics happening inside heliosphere and any temperatures associated with it, until it starts schocking the ISM.

So given the above said, I don't think this is the correct answer to the question, if there is any. However, there is a great deal of interesting phenomena touched upon in your answer and comments.

Yep, thanks! The question is really difficult to answer, because it contains lots of implicite assumptions which don't match well with reality or usual definitions. So, if you've got an idea for a better answer, feel free to provide one.

I am not entirely happy with the question to answer it. There is no clear specification for the temperature, which is asked: neither location, nor the component is clearly specified. Your answer is much more fun in this sense, it gives eventually a nice picture of what is going on.

Thanks a lot! So it may be better, we leave the answer as it is.


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Do planetary surface temperatures change in unison in a solar system?
Are there any known correlations between the changes in planetary surface temperatures in a solar system?
If so, do the farthest planets have smaller albeit correlated changes?

It could be presumed due to changes in solar/stellar activity.

The activity of stars can change pretty much for some variable stars. Our sun is calm in comparison, but also has an 11- resp. 22-year activity cycle, with varying amplitude. Sometimes the sun produces flares or coronar mass ejections (CMEs), which may affect planets which happen to be located roughly in the same direction from the sun. Over long timescales of billions of years, solar activity and radiation changes. I'm not aware of systematic investigations of short-term surface temperature correlations between planets. But it doesn't look impossible to try it.

Eclipses would rarely lead to a synchronous temperature change for several planets, because constellations with several planets and the sun in exactly one row are very rare.

The simple answer to your question is yes. Taking a simplified equation from Carroll & Ostlie, An Introduction to Modern Astrophysics Second Edition, the temperature of a planet can be estimated as:
$$
T_{p} = T_{\odot}(1-a)^{\frac{1}{4}}\sqrt{\frac{R_\odot}{2D}}
$$
Where $T_p$ is the predicted temperature of a planet in a circular orbit of radius $D$ with an albedo of $a$ around a star with a temperature of $T_\odot$ and a radius of $R_\odot$. If the energy output of the star were to increase, raising $T_\odot$, then there would be a corresponding increase in the temperature of all planets orbiting said star.
In practice there are factors which can make this correlation difficult to measure. The albedo  of a planet during the course of a day can vary greatly and the distance of a planet from the host star changes throughout the year. This equation also assumes that the planet is a perfect black body which most are not which can also change a planets temperature and obscure any changes caused by the host star.

Uh, if I apply this formula to earth and it's distance, I get -24.6°C... is this approximation ok-ish and the difference due to the athmosphere?

The equation assumes that the planet is a blackbody radiator. Anything with an atmosphere is most definitely not a blackbody. I do not offhand know if there is one good equation that can be used to estimate the temperature taking into account atmospheric effects since these would depend greatly on the composition of the atmosphere


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Why does e.g. Hubble's secondary mirror not block part of the picture?
The SOHO observatory has an opaque object in the light path that blocks the direct light from the sun making it possible to observe the corona.
Hubble and many other telescopes have a big primary mirror in the bottom of a tube which bounces light on to a smaller secondary mirror at the opening of the tube. Why doesn't that other mirror cause a circular area of "no data" in the middle of the resulting images?

A particular point in the image is formed by the convergence of all the rays of light that leave from that point in the object, and bounce off the mirror.  As long as there are some rays that can do that, then the only effect of the secondary is to block some of the rays, dimming those parts of the image slightly.

Some references wouldn't go amiss in your answer. For example, you can find a lot of information on these pages (and its links): HubbleSite: The Telescope - Hubble Essentials

The function of a telescope or your eye is to translates angles into distances on a detector. You can see in the figure below (from Hyperphysics' excellent telescope page: http://hyperphysics.phy-astr.gsu.edu/hbase/geoopt/teles.html) that rays of light coming from a single object (such an unresolved point of light from a star in space) enter the telescope from a single direction (angle on the sky), in parallel. They pass through the telescope and an image is formed on the back of your eye (Or in Hubble's case, on an electronic detector, which you can think of being were the little red arrow is). The image is a representation of those angles as positions separated by some distance on a 2-D surface.  On the detector all of the light from that single angle angle is light at a single point.

If your pupils constrict, you don't see less of the world, the angles entering your eye stay the same but the number of rays from each angle decreases and less light from every object (light from every direction) gets dimmer.
The opening at the front of Hubble is like your eye's pupil, if it were smaller, less light would come in, but just as many directions could be seen. The secondary mirror being in the way just means Hubble different shaped pupil from yours, a few rays from the object don't make it to the detector, but the image doesn't change, just the brightness.
If the secondary mirror were not there (as in a harder to manufacturer refracting telescope) Hubble would collect a little bit more light from each direction and could observe things a little quicker, but the view of the sky on the detector would be the same.
(SOHO blocks the sun by effectively placing a block in an image formed inside the telescope, for example where the arrow showing the image is in the simple telescope above, then SOHO's "eye" only sees the rays from around the edges of the sun, by blocking them at an "image plane", this is a "coronagraph", https://en.wikipedia.org/wiki/Coronagraph).

This is a fairly good answer, but the diagram is not the best for explaining what is going on, as it represents a refractor, not a reflector. I could pop up an alternate answer, but maybe you could just replace the diagram with one that shows a secondary obstruction? Also, as the observatories being referred to are not observed with the eye, it may as well show a detector instead of an eyeball...

Why then, if I hold up a mirror at arms length in front of my face, does that "secondary" mirror hide a specific part of my field of view, when on a telescope it doesn't? Because light rays from nearby sources are not parallel, as they are from a distant star? But then what about an apparently large object like the Moon?


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Intrasolar planetary surface temperature change divergence from Earth
Considering the vast amount of insulating gas emitted by human activity on Earth without a coincident release on all other planets in the solar system as well as the phenomenon that planets in a solar system normally exhibit similar coincident temperature changes described here, how far have temperature changes diverged from Earth's?
Please include raw data, graphed if possible.

We've millions of raw temperature datapoints of Mars, a couple of Venus, some of other planets. But we've no detailed long-term climate data of other planets than Earth. For Earth we've e.g. polar ice cores from which long-term temperature fluctuations can be deduced. And we've climate observations over centuries on Earth, of gradually increasing quality. For other planets, this kind of research is still in its infancy. I doubt, that anyone can give you a fully satisfying answer to this question, at the moment.

The formula is based on a simplified theoretical model, which compares the inverse-square law (relative to the distance from the Sun) of incoming radiation with the grey body radiation of the planet (4th power law relative to the surface temperature of the planet, derived from Planck's law), more details, see http://en.wikipedia.org/wiki/Effective_temperature


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How many new galaxies enter the observable universe each day/year/decade?
Each day light has more time to reach our eyes from distant galaxies. In one day, light travels 2.59×10^10 km.
So our observable universe (assuming my simple math skills apply here and there isn't something about relativity to take into account), expands that much every day in all directions. So at what rate are new galaxies being encompassed by the expanding observable universe?

You probably mean how many galaxies? Still, the observable universe horizon isn't limited by the age of the Universe, but its inflation. Number of observable galaxies is actually getting smaller, not larger. And the rate of change would not be linear with time either.

Edited. Someone has suggested closing this question, but it may be better to address the misunderstandings of the questioner in answers.

What we see at the outer reaches of the observable universe is a surface of the cosmic microwave background (CMB), which is at the time when the universe, still quite homogeneous (perturbations were a few parts in $10^5$), cooled enough for the electrons to combine with the protons and make hydrogen atoms.  Each day, we can see a light-day farther and so we slowly see a different (larger) surface of the CMB (but there were no galaxies at this epoch).  Therefore, as we observe the CMB over many years, we will always see the same time epoch, but at different spherical shells.  Will we see new galaxies?  Yes. Galaxies are forming continuously in view at high redshifts (2-20?).  In a few million years, the small perturbations that we see today in the CMB will develop into young galaxies within our view.  One cannot provide numbers for how many we might be seeing because small galaxies merge into bigger ones and those merge again.  At high redshifts there is a lot of merging going, so the rate is particularly ill defined there.
How is this changed by the acceleration from the cosmological constant (aka dark energy)? From the Friedmann equation one sees that the Hubble constant reaches a final constant value of $c\sqrt{\Lambda/3}$ or about 40% higher than its present value.  Eventually, in billions of years, the CMB will shift from $3^oK$ to much lower and then the CMB will fade (redshift) away.  After that, those new distant galaxies will also fade away.  And then more and more of the universe will go dim.


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Can the next possible ice age be projected based upon the Milankovitch cycle?
The Milankovitch cycle

describes the collective effects of changes in the Earth's movements upon its climate[.]

By this theory, the Earth is now experiencing an interglacial.
By the theory, when will the interglacial end and the next ice age begin?

This paper seems to be discussing what you've asked: http://www.sciencemag.org/content/297/5585/1287.full. The next ice age is expected to begin about 50,000 years from now. However, I think it's still debatable how accurate the predictions of Milankovitch cycles are, and how much of an effect humans are likely to have on this earth system.

Your link requires paid membership :(

Try one of the links here: http://scholar.google.co.in/scholar?cluster=6043302048410624765&hl=en&as_sdt=0,5


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Can solar luminosity & activity be predicted?
Can a star's luminosity and its activity, solar flares and sunspots, be predicted?
If so, with what accuracy & precision?

All of these effects are related to the 11 year solar cycle. And while we know there are times when the sun is "active" versus "inactive" we aren't necessarily predicting exactly where sunspots or solar flares will occur, but how many we see in total.

I will be specific to answering the part regarding the prediction of solar flares.
Solar flares are very volatile and dynamic phenomena associated to the breaking and reconnecting of magnetic field lines in the Sun's photosphere.
There is a phenomena associated with solar flares, called quasi-periodic pulsations, which sometimes precede a solar flaring event. As such, monitoring of such pulsations may give rise to a plausible prediction method. However, the accuracy and reliability is still to be debated.
A few good papers on QPPs:
Quasi-Periodic Pulsations in Solar Flares: new clues from the Fermi Gamma-Ray Burst Monitor
Time delays in quasi-periodic pulsations observed during the X2.2 solar flare on 2011 February 15


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Does the universe have an edge?
I've heard the universe is just an empty cavity. Does the universe have any edge or stop or is it just an infinite empty cavity?

Related What is the name of that which exists beyond the Universe?

Of course it has! Where else would it end?

Related question on Phys.SE: http://physics.stackexchange.com/q/128271/2451

Also see https://physics.stackexchange.com/q/136860/123208 & https://physics.stackexchange.com/q/105817/123208

As far as theories I've heard of, we can't see the edge of the universe (if truly exists) because the farther you look out into space through a telescope, the farther back in time you are actually looking. For example, if you are looking at a star 20 light-years away... you're actually currently observing what was happening on that star 20 years ago... the theory is that there is only so far you can look out into the universe because there is only so much history post Big Bang.
Image above: What is the furthest we can see? In 2003, NASA's WMAP satellite took images of the most distant part of the universe observable from Earth. The image shows the furthest we can see using any form of light. The patterns show clumps of matter that eventually formed into galaxies of stars. Credit: NASA/WMAP Science Team

Don't understand why the down vote. [link]http://www.nasa.gov/audience/foreducators/5-8/features/F_How_Big_is_Our_Universe.html

Ya I think a down vote fairy flew by, Good answer by the way, upvoted

That means that as time succeeds our field of view permanently extends doesn't it?

Galaxies extend as far as we can detect... with no sign of diminishing.There is no evidence that the universe has an edge. The part of the universe we can observe from Earth is filled more or less uniformly with galaxies extending in every direction as far as we can see - more than 10 billion light-years, or about 6 billion trillion miles. We know that the galaxies must extend much further than we can see, but we do not know whether the universe is infinite or not. When astronomers sometimes refer (carelessly!) to galaxies "near the edge of the universe," they are referring only to the edge of the OBSERVABLE universe - i.e., the part we can see

If we know the universe is expanding, we are saying the universal boundary is expanding. Before the big bang(s), there was "nothing". Which seems an impossibility, because something happened. So we have the big bang(s) and the boundary begins to move outward, at least at the speed of light (knowing there is debate about variable speeds of light,  I am keeping this open). That means that according to our understanding thus far, there is an edge to the universe.
I try to keep in mind that as a creature with everyday experience in three dimensions, it is next to impossible to really understand all the multidimensional possibilities which may make up "the edge". Just like I cannot imagine a time from my viewpoint where there was a "nothing". I believe that the answer lies in a better understanding of multi-dimensions than we have now.
We are so very, very early in our use of a scientific method of exploration, and it seems that every year we are bowled over by the implications of a new discovery - especially in quantum physics.

I believe the question is about a possible edge of the "whole" Universe, not the observable Universe which is what you're referring to (and which increases its distance to us not only at $c$, but at $4.3c$).

Great comment. I guess I'm extrapolating that as we see the observable universe expand, the entire universe is also expanding. Presumptuous perhaps, but isn't anything we can measure "observable" (directly or indirectly)? In either case, a bang from "nothing" (which I have a hard time with) would seem to create a boundary between that nothing and something...observable or not. And if we are questioning the speed of light, and if it may be variable, how can we convincingly state that the universe expands at  4.3c?

I don't think assuming that the rest of the Universe is expanding is presumptuous. Anyway, a bang from nothing doesn't necessarily create a boundary. The Universe is either finite of infinite, but certainly neither implies that is has an edge. If it's finite, then it "curves back on itself", like the surface of a balloon. In "mainstream physics" we don't question the speed of light, and by Hubble's law galaxies the edge of the observable Universe (which is not a physical edge) recede at $v = H_0 d \simeq 3.3c$.

Because light from increasingly far away will reach us, the distance to the edge increases by another factor of $c$, so the obs. Uni. increases its radius at $v\simeq 4.3c$.

But, if you're floating in the inside of a balloon, and travel to the balloon itself, that is an edge to whatever is beyond it. I totally get the "edge" idea. But I don't think we can answer it any more than we can answer "what was there before the big bang?", and if the speed of light is indeed variable (or slowing down), that has major implications in our understanding of all kinds of things...age of universe, how relativity would work, where the edge of the universe is, and how long that light would indeed take to reach us.

The balloon analogy is a 2D analogy of a 3D space. You can't be "inside" the ballon, since then you leave the 2D space. But just as you can travel the surface of the balloon and never reach an edge, you can travel 3D space and never reach an edge. If space is homogeneous and isotropic (the "mainstream" assumption) we can answer the question: Space is either infinite, in which case there's no edge and you can travel forever and meet new galaxies, or it is finite, in which case there is also no edge, but you can (in principle) travel and get back to the beginning (like on the balloon).

As for the variable speed of light hypothesis, that's not mainstream physics and, to my knowledge, has no experimental verification whatsoever. As for "how long that light would take to reach us", that's quite easily calculated from the Friedman equation. I will leave it as an exercise :)

Thanks Pela, very informative!  I caught on that OP was referring to the observable universe, to which there obviously is an "edge" of sorts. Wasn't getting that the balloon was a 2D analogy, sorry. Makes much more sense now. Maybe the question is better thought of as a membrane on the universe. What could lie beyond it. I wish I could come back in 1,000 years to see what we've discovered, if we haven't completely annihilated the earth.

Oh, and the question of how long light would take to reach us wasn't something I mentioned. Obviously we can calculate that if we accept the older assumption of a constant speed of light. I personally have major doubt about that, and like Magueijo and Afshordi's theory because the uniformity of temperature and matter in the universe are puzzling under our current assumptions. Hawking's expansionist theory is hard to accept without knowing the mechanism which triggered that expansion. It really feels like the theory was massaged to fit the observation of uniformity. Digressed, sorry.

Dark energy is driving the accelerated expansion of our universe.
Is it accelerating into itself?
Or is it accelerating into nothing, An outer void?
If we could see the furthest away accelerating object that would be the edge of the Universe.  Thus there is an edge to the Universe.
Remember, just because we can't look and see today doesn't mean it doesn't exist.

Your last two statements appear contradictory. Your first states that the observable universe is the entire universe, while your last statement seems to indicate the exact opposite.


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Do we expect life to develop only in certain regions of the galaxy?
If I understand correctly, the Earth lies in an area of the galaxy - the "local bubble" - that is relatively free of dust and gases and where no new stars are being born.
What is the current theory on the relation between life on Earth and this location? Or in other words: Would we expect life to exist anywhere in the galaxy, e.g. also inside some nebula, or only in specific areas, and how are they characterised?

There's a short answer here and a long answer.
Short Answer: It isn't special.
Long Answer: Yes and No.
The Local Bubble really isn't so special with regards to where it's placed in the galaxy. Location-wise, it's perched on the inner edge of the Orion Arm, one of the Milky Way's spiral arms. The Orion Arm, too, isn't incredibly special. It's relatively small - it's often referred to as the "Orion Spur" - and near 1/4 of the way out from the galactic center. Not too far away, but not too near. The donut-shaped area 4,000-10,000 parsecs out from the galactic center is, however, sometimes called the Galactic Habitable Zone. In theory. life as we know it could exist quite nicely any where in this area (I should note that the scientific community is divided on whether the Zone is important, with some saying that it is arbitrary).
But let's dive in deeper. Actually, we're in a decent spot in the "Arm." We're on the edge of it - not too near large groups of stars that could effect our system via gravity, or near any supernova candidates. And the "Arm" itself isn't really a spiral arm. It is, as I said earlier, more of a spur, or bulge. There aren't as many stars to watch out for. Also, we're near the corotation circle - i.e., we orbit the galaxy at such a distance that we move at a similar speed to the arms, reducing further the likelihood of crossing them. Is this necessary for life? Nope. But does it help? Oh, yes.
There are, of course, other factors that make this area a nice spot: No black holes nearby, gamma-ray bursts, etc. But could life survive elsewhere? Yep. These same characteristics can be found in many other places in the galaxy - and life can probably survive in even more dangerous spots.
By the way, I'd avoid using the term "nebula". Technically, we're not in one, or anywhere near one. The term can refer to a multitude of objects - planet-forming areas, "clouds" surrounding white dwarfs, and other rather photogenic places - none of which we inhabit.
I hope this helps.

It only seems reasonable you could restate "there's nothing around us that kills life" to "the life that spawned around here isn't killed by the various things around here", no?

I found your mention of the Galactic Habitable Zone (GHZ), the corotation and being outside the galactic arms most helpful hints in your answer. Googling around and reading some scientific articles on the GHZ, those that favour such a theory put forth similar arguments. So in sum it seems that maybe life in our galaxy is most probable in (prerequisite 1) a circular area not too far and not too close to the galactic centre and (prerequisite 2) between the galactic arms (or, more specifically, outside dense and high radiation areas), [contd.]

[contd.] resulting in a space very roughly akin to the three green sections in this image from the final episode of the Fox tv series Cosmos: http://www.evolutionnews.org/ghab.png

@CorsiKa, where did I use that phrase?

You didn't use the phrase specifically. But you implied it by saying we're in a nice spot followed by a small list of things that would potentially kill us.

@CorsiKa - True. How would you suggest I edit it?

As my username indicates, I'm eager to point out that we are currently inside the "local fluff", which is "stuff" which has been ejected long ago by some disturbingly active young stars in that evil Scorpius-Centaurus association through the local bubble. The waves of their stellar winds just happen to pass us by in our current million year era.
However, our Sun does circle the entire Milky Way every 250 000 000 years. Dinosaurs appeared about one galactic orbit ago, but died out a quarter of an orbit ago. Earth has had quite advanced life during at least one galactic orbit. That's why I argue that there is little dependence of a solar systems habitability on its orbital position in a galactic scale. (Or that life on Earth has been incredibly lucky, and finally someone will, right? So what do we know from observing no one but ourself?)

Thanks. No relation to distance from the galactic core either? I understand: the farther from the core the older. And since life takes some time, maybe it cannot exist in "young areas"? Or on stars outside of galaxies? And "finally someone will" what?

There's no relationship between distance from the core and age.

@what I don't think the Sun has changed its distance to the galactic core by more than a few or ten percentages per orbit. But even half way towards the center, no hostile environment has been observed. Mostly just empty space there too. "We've" survived several orbits around the Milky Way! That's a huge trip! --- By "Someone will eventually get lucky", I mean just the anthropocentric idea that given whatever anyone can observe, they can only exist in conditions which made them able to observe that. If a thousand men run across a minefield, someone will survive. We might be him.

This may not be a very scientific answer. I think it gives some incredible insight, so I thought I'd share it...
What "Vlad the Astrophysicist" said:

If we condense the entire time that has passed since the Big Bang into 5 min.
And we consider the vastness of space and the time each civilization lasts.
Then life and civilizations that develop from it rise and fall in less than a split second. Likely never coming into contact with any others.

We can only assume life is quite possible anywhere where the conditions happen to right. But it's very hard to actually quantitatively put a number to this since the universe is so HUGE. All we can say is basically, life (as we define it) is more likely develop in the "Goldilock" zones found in the galaxy, which are bound to exist. Notice this may vary depending our definition or understanding of "life".


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Name of area close to Local Bubble?
The following image is a map of the area surrounding the Local Bubble. One side of the image is 1700 lightyears. From the Sun to the Hyades (immediately below the sun) it is 150 lightyears.
The local bubble, with the sun in its center, has three immediate neighboring bubbles, numbered in roman numerals, Loop I, II, and III. At the lower right edge we can see a section of the Orion-Eridanus-Superbubble.
What is the name of the area marked in green?

Image source: Henbest/Couper, The Guide to the Galaxy. Found online at http://www.astronomy.ohio-state.edu/~pogge/TeachRes/Ast162/ISM/Local500pcAnn.gif

A guess (not enough for a definitive answer) is that it is the Orion Molecular Cloud Complex.

As far as I know the Orion Molecular Cloud Complex centers at the lower right end (as this image is presented) of the Orion-Eridanus-Superbubble (here "O-E-Loop"), outside the image, approximately where the line of your comment ends (if viewed not in the mobile version). That's where the Orion Nebula, Barnard Loop, Orion OB1 Association and so on lie. I'm not sure if the O-E-Superbubble is part of the Orion Molecular Cloud Complex, but I believe so. In anyway, the region I seek the name for lies outside the Orion Molecular Cloud, and (at least partly) between it an the Local Bubble.

Okay, so much for that idea. I haven't been able to find much about the region you're talking about.

The situation kind of stinks, really. I love your question, but I can't seem to find anything on it.

I wonder if the area I am looking for is GSH 238+00+09.
In the color image quoted in my question, the galactic core lies beyond the top center. This would be 0° galactic longitude. Going counterclockwise, 238° is somewhere around Beteigeuze.
Here is a "[p]rojection of the Local Cavity onto the Galactic plane":

Figure 9 from Lallement et al. (2003)
But if you try to superimpose this image on the other one (I matched the positions of the Sun, the Ophiuchus clouds and the Pleiades bubble, the Lupus cloud is a bit off), it seems as if the area marked in green was actually part of our Local Bubble! And GSH 238+00+09 then appears to be identical to the Orion-Eridanus-Superbubble:

I'm sure I didn't match this correctly. Especially Loop III would be in totally the wrong place. But I found an article that used the same black and white image to illustrate possible locations of Loop III, and the second alternative to the one shown here was in the upper right hand corner, where it would then overlap with "Loop II" of the color image. But probably we should not worry too much over Loops II and III. The creator of the 3D Galaxy Map writes: "I haven't added Loop II, III or IV because I couldn't find any accurate information about their size and location." I'm sure he has been looking more thoroughly than I have, so I feel the mismatch of Loop III is not a problem.
But then the color image is from 1994, and understanding and cartography of our Galactic neighborhood has advanced quite a bit in the ten years between the two publications, so maybe the placement of the Local Bubble in the color map is only a rough equivalent and the mismatch between the two images is not my fault but a result of scientific progress.
Here is another image (rotated 180° in relation to the color image from my question) of local loops, bubbles, and superbubbles, where it becomes clear that GSH 238+00+09 (termed "New Superbubble" (NSB) in this image) is not identical to the Orion-Eridanus-Superbubble:

Figure 8 from Heiles et al., (1998)
If you compare the scales of this last and my first color image, you will note that the color image falls completely inside the circle denoting the Local Bubble, supporting the thought that the green area lies within it (and not within GSH 238+00+09 or the Orion-Eridanus-Bubble). But this second image also supports the insight that the color image might not be that accurate, and that maybe the area whose name I am looking for does not exist in that shape at all.
In the following image (which shows Loop I as the Scorpius-Centraurus Shells and the Orion-Eridanus-Loop as the Orion Shell) the red blot immediately rimward (lower right) of the sun is the Taurus Dark Cloud. This is not from a scientific but a popular publication (just as my color image), so we should not expect it to be totally accurate, but it still suports the notion that the green area might be part of the Local Bubble.

Unnumbered 3rd figure from Frisch (2000)
Here is a final image, apparently a different rendering of the image from Lallement et al. (2003), the first image given in this answer. In this image the location of some important stars is marked. Most notably ε CM, which is the scientific name for the Arab name Adhara, with which the same star is marked in the color image from my question and there lies far out in the green area, **lies still within the Local Bubble in this image, close to the edge.

Figure 3 from Sfreir et al. (1999)
Here I overlayed the previous image and the image from my question. I matched the sun an Adhara. Other objects are still a bit off, but not much, and we get a good idea of where the Local Bubble ends.

So it seems the local bubble extends to a bit beyond the star Adhara, halfway from the Sun to the right edge of the image in my question and does indeed include the closer part of the green area.
This thought is supported in the abstract of Heiles (1998), where it says: "[GSH 238+00+09] is at least partially responsible for the apparent extreme elongation of the Local Bubble toward l ~ 230° ..." (my emphasis).
I'll leave my first thoughts and the badly matched image in this answer, to document how I came to the final conclusion.
I'd welcome any thoughts and corrections of this attempt at an answer.

Sources:

Frisch, P. (2000). The Galactic Environment of the Sun The heliosphere appears to protect the inner solar system from the vagaries of the interstellar medium. American Scientist, 52-59. Available online at https://www.americanscientist.org/issues/issue.aspx?id=862&y=0&no=&content=true&page=5&css=print
Heiles, C. (1998). Whence the Local Bubble, Gum, Orion? GSH 238+ 00+ 09, a nearby major superbubble toward Galactic longitude 238. The Astrophysical Journal, 498(2), 689. Available online at http://iopscience.iop.org/0004-637X/498/2/689/pdf/0004-637X_498_2_689.pdf
Lallement, R., Welsh, B. Y., Vergely, J. L., Crifo, F., & Sfeir, D. (2003). 3D mapping of the dense interstellar gas around the Local Bubble. Astronomy and Astrophysics, 411(3), 447-464. Available online at http://www.aanda.org/articles/aa/full/2003/46/aa3635/aa3635.right.html (under "Article" in the sidebar)
Sfeir, D. M., Lallement, R., Crifo, F., & Welsh, B. Y. (1999). Mapping the contours of the Local bubble: preliminary results. Astronomy and Astrophysics, 346, 785-797. Available online at http://adsabs.harvard.edu/full/1999A%26A...346..785S

This paper (http://iopscience.iop.org/0004-637X/498/2/689/pdf/0004-637X_498_2_689.pdf) seems to suggest it lies somewhere nearer to the Linblad Ring and the Eridanus Loop. See the picture on page 698. Also, what is your scale in the original picture?

Part of the problem here probably lies in the fact that we are looking at 2D projections of 3D space.

@called2voyage Since all images are projected onto the Galactic plane, that should not be a problem (if done correctly).

@HDE226868 You mean in the color image in my question? From the Sun to the Hyades it is 150 lightyears (according to the scale that's below that image in my book). Note that scale in some of the other images is in parsecs.

@called2voyage brings up an interesting point. I agree that it might not be a huge issue, but it could influence some distance measurements.

Also, what looks like the boundary of a bubble from a 2D perspective may not actually be the boundary. In your original image, the orange-yellow strip between Hyades and Betelgeuse where the "Local Bubble" label rests may not be a boundary. There may be a cavity above and/or below that line of density which connects the cavities on either side of it. Notice in this 3D representation of the Local Bubble that it is has an amorphous boundary.

@called2voyage Yes, but the idea that the local bubble does not have spherical form (but rather looks like an hourglass open at the bottom and top) is more recent that the book that contained that color illustration. I actually read the article that proposed the non-sphericality during my research. That is why I emphasized the age of the book. In it, the loops are round. It is only recently that we have come to understand the true form of the Local Bubble, which, in more recent texts, is being called the "Local Cavity" or "Local Tube" (since it is open at both vertical ends).


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Are there stars that don't emit visible light?
Are there stars that do not emit any light in the visible part of the EM spectrum?

do you count black holes as stars?

@JanDvorak - Black holes don't emit any EM radiation.

@HDE226868 See Hawking Radiation

@Yashbhatt That's a completely different process.

I think you could start by outlining your definition of a star, to help the answers.

@HDE226868 Why is it different?

Stars emit light as the end result of nuclear processes. Hawking radiation is (and I will quote your link) "due to quantum effects near the event horizon."

@Harogaston I agree, and I think JanDvorak does, too.

There are two possible reasons why a star would not be detectable in the visible part of the spectrum (even with the most powerful telescopes not yet invented), apart from the trivial (too far away, hidden behind screens of dust).

It has too low a temperature. To not emit much in the visible part of the spectrum, a star has to be very cold indeed, at most a few 100K, when radiation is mostly in the infrared (which is still detectable). By definition, a star is an object that undergoes (or has undergone) hydrogen burning (H$\to$He fusion). This sets an lower mass limit of 0.08M$_\odot$. All objects of that mass (or more) ever formed in the universe are still much hotter than 100K. However, in the very distant future, some 'dead' stars (without remaining fusion energy source) will cool down to such temperatures, this includes white dwarves.
It has too high a gravitational redshift for any light to appear in the visible. In fact, the so-called stellar-mass black holes (remnants of supernova explosions of massive stars) may in fact be such objects: strange stars denser than a nucleus (consisting of a quark-gluon plasma as dense as a neutron) but with such a small size and high mass that any radiation emitted from their surface is redshifted by a factor 1000 or larger.

So if you consider those (still hypothetical) strange stars, then these are possible candidates.

I've seen in the past some sources suggesting Jupiter may be or become one of your first category stars. It emits a lot of radiation, but no visible light.

@jwenting Jupiter is not a star, but a planet. Stars are defined to undergo (or have undergone) hydrogen burning (H$\to$He fusion). This requires a minimum mass of about 0.08M$\odot$. Objects smaller than that, but massive enough to have some fusion (Li) early on are brown dwarves. The mass limit here is about 10M${\mathrm{Jupiter}}$.

Maybe a very old Neutron Star?
A black dwarf would not emit any visible light, but the universe is not old enough for that.
Even the oldest and coolest white dwarfs still have a temperature between 2500-4000K (sorry for not remembering the reference for this).
Brown dwarfs (or planemos/sub-brown dwarfs) like WISE 0855–0714 could be as cool as ice.
but they don't count as stars anyway.

This could be right, if the OP ever gets back to the definition of a "star".

This article http://arxiv.org/abs/1501.07083 deals with the hypothesis that Fomalhaut b is an old neutron star located in the background of Fomalhaut, but still very close (11Pc).
If confirmed, one could imagine how intrinsically faint an old NS can be, given the apparent faintness of Fomalhaut b.


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During an eclipse, does the size of the moon and sun match perfectly?
During a lunar or solar eclipse, from our perspective when we look towards the moon and the sun is behind the moon, does it fit exactly?
Has anyone calculated the sizes of each in a relative context?  Obviously the sun is thousands of times larger, but I am talking about the size of the sun/moon when looking from our perspective.
I hope my question makes sense.

Short comment due to lack of time. Since both distances, moon-earth and sun-earth, vary periodically, so do the apparent sizes of moon and sun. The moon can be smaller or larger than the sun; see 'annular eclipse' for example. How much larger the moon appears determines (among others) the length of the totality phase of the eclipse.

To expand on what taupunkt commented, on average, according to the article Solar Eclipse: What is a Total Solar Eclipse & When is the Next One? (Rao, 2014), it is a unique quirk of nature that

The sun's 864,000-mile diameter is fully 400 times greater than that of our puny moon, which measures just 2,160 miles.  But the moon also happens to be about 400 times closer to the sun than the Earth

This is often stated in middle and high school science textbooks.
However, the orbits of the Earth around the sun and the moon around the Earth are elliptical, as illustrated in the diagram below:

Image source: NOAA
This results in the exact ratio mentioned above not occurring resulting in an annular eclipse, where the moon is too 'small' to fully cover the solar disc.
A comparison between a total and annular eclipse is shown below:

Image source: ScienceJedi
NASA's Eclipse data page, in particular, their Eclipses and moon orbit information page discuss in detail (with considerable data), the

interaction and harmonics of the synodic, anomalistic, and draconic months not only determine how frequently eclipses occur, but they also control the geometric characteristics and classification of each eclipse.

Within the NASA link, data about how much of the sun's disk that would be obscured by the moon has been calculated (from observations and predicted). For example, using the data for solar eclipses from 2011 to 2020, The average fraction of the sun's disk that is and is expected to be obscured (this is referred to as 'Eclipse magnitude') in an annular eclipse varies from about 0.95-0.99 (95% to 99%) for that time period.
A catalog of eclipses up to the year 3000 is provide on this NASA page.
The greatest difference would occur if the moon is at its furthest from the Earth (apogee) and the Earth closest to the sun (perihelion).

It would be good to include some information about the magnitude of variability discussed here.

Variation in moon size at APOD http://apod.nasa.gov/apod/ap140121.html


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Is it conceivable that the presence of chlorophyll (or similar) could be distinguishable from other chemicals on an exoplanet?
If an Earth-like exoplanet had a sufficiently extensive plant biosphere (i.e large forests, aquatic algal mats etc), is it at all conceivable that scientists could spectroscopically detect the presence of biochemicals, such as chlorophyll, anthocyanins etc?
If possible, how would the spectra of these biochemicals be distinguished from other chemicals?

Take a look at this article: http://www.nature.com/nature/journal/v513/n7519/full/nature13785.html
Every more or less stable compound has its identifying absorption lines, based, well, on what the compound is made of. The question is, what is the minimum amount for a compound in an given atmosphere at a given distance with a certain sun to be still detectable. In deed, this also depends on how "empty" the light's way to us is. But guessing, you want to know, if we propably could detect an exo-planet with vegetation on it - yes, this is possible under circumstances. Most of all, the area covered by the possible organisms plays an important role.

Finding Soret bands http://en.wikipedia.org/wiki/Soret_peak from porphyrins like chlorophyll, hemoglobin or cytochromes would be a particulalrly interesting discovery.


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How did scientists determine an estimate of the number of planets greater than Sedna's size to exist in the Inner Oort Cloud?
In the article, New Dwarf Planet Found at Solar System's Edge, Hints at Possible Faraway 'Planet X' (Wall, 2014) where they report on research and observations performed by astronomers that discovered the dwarf planet 2012 VP113 which is smaller than, but has an orbit severely elongated like that observed for Sedna.
The article reports the claim from the astronomers who discovered 2012 VP113 that

Based on the amount of sky the scientists searched, Trujillo and Sheppard estimate that about 900 bodies larger than Sedna may exist in this faraway realm, which the astronomers dub the inner Oort Cloud.

How was the estimate, in bold above, calculated?

Wild guess: they surveyed approximately 1/900-th of the sky, found one such object, and concluded they can expect 900 total such things. If so, expect a large number of statisticians to collapse into black holes of anger as a result.

@zibadawatimmy best. comment. ever! The first sentence makes good sense, but of course I am looking for any papers etc about it. Your second sentence is just awesome!

Unfortunately, the paper is not available on ArXiv (oh, what hardships we must overcome!), but I have found it here. In it, where the "900" figure is mentioned (2nd page, I believe), the authors (Trujillo and Sheppard) say that they ran simulations with the data already found and their additional findings, and found that 900$^{+800}_{-500}$ bodies larger than 1,000 km in diameter could exist. They also found that 430$^{+400}_{-240}$ relatively "bright" bodies could exist, as well. Notice that there is quite a lot of uncertainty there; they are rather cautious (and rightfully so) when announcing the results of their simulations.
Interestingly enough, they also mention a paper in that section, written partly by none other than Mike Brown (!) that also used simulation methods. The authors, Schwamb et al, concluded (in 5.4, as well as the conclusion) that 393$^{+1286}_{-264}$ and 74$^{+249}_{-47}$ objects greater than or equal to Sedna in brightness, not size, could exist (393 is for the "hot" population, while 74 is for the "cold" population). The reason Trujillo and Sheppard cited this paper is that their simulation results fit the results of the simulations by Schwamb et al., which is encouraging.
Summary: Trujillo and Sheppard arrived at 900 via simulations based on prior observations. There is a large amount of uncertainty in this number, but it does agree with prior simulations by Schwamb et al. While zibadawatimmy 's comment is perhaps the funniest I have seen, it is, fortunately, inaccurate. Those statisticians must be relieved.
I hope this helps.

I hope you don't mind that I've edited in a reference. I had no idea who Mike Brown is and it is hard to Google for as there is a current news item about an unrelated man with the same name.

@dotancohen That's a good idea.

Noteworthy as Mike Brown is, Chad Trujillo was also on the Palomar team that discovered Eris.


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Is Zeta Reticuli within the Orion Arm?
My research has told me astronomers have labeled the galaxy arm that our system is located the Orion Arm but I have yet to find out if that also applies to Zeta Reticuli or not (not asking about aliens or anything, just trying to get my locations straight).

Let's take a look!
While I can't quite track down the source of this image, I've found it throughout the web, so hopefully it's roughly accurate. I'm pulling it from the Wikipedia article on the Orion Arm.

In this image you can also see the Orion Nebula sitting (relatively) cozily to the Sun.  The distance between the two is believed to be around 1350 light years.  Zeta Reticuli, on the other hand, is believed to be a mere 39 light years from Earth. Given the orientation of these two locations and the relative distances, it would be hard to place Zeta Reticuli anywhere but firmly within the Orion Arm.
Not exactly rock solid evidence, but sometimes some back of the envelope deduction will do.

I'm not sure how accurate the image is, either, but you're clearly right, even taking just logic into account.

The Orion Arm is somewhat loosely defined (and many astronomers do not regard it as a proper spiral arm of the galaxy, merely a spur or some other minor feature). But at the very least this neighbourhood is hundreds of light-years across.
The Zeta Reticuli system is about 39.9 light-year away. So it is in the arm.


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When do Mercury/Venus reach greatest elevation at sunset/twilight for a given location?
On what day does Mercury reach its greatest elevation (in degrees from
the horizon) at sunset a given location?
The obvious answer is the day of Mercury's greatest elongation from
the Sun, but, since the ecliptic is slanted with respect to the
horizon, I'm not convinced this is correct.
In other words, on the day after greatest elongation, Mercury's total
angular distance from the Sun will be smaller, but it's vertical
distance in elevation (for a given location) at sunset might be
higher.
Same question for Venus, and for when the sun is 6 degrees below the
horizon (ie, civil twilight), and for sunrise/dawn.
I'm guessing the date might vary based on position (mostly latitude) since the ecliptic's slant varies at different locations.
I googled and found nothing. My (preliminary and possibly wrong)
expierments with stellarium show that Mercury's elevation at sunset IS
higher 1-3 days after its greatest elongation, but by less than 1/2
degree.
So, it's possible that the date of greatest elongation is a close
enough approximation.

You can take a look at this table - it's in Dutch, but the dates and numbers in the table shouldn't be too hard to understand.  The first date is that of the greatest elongation (GE), the second (halfway the columns) that of the best visibility (BV).  You should ignore the lines with red dates - they're unfavourable appearances, and the code may find a date far from that of the GE.  The table shows that the difference between the moment of GE and that of the BV is typically a few days, although it can be about a week. (You can click on the first date for a page which includes a sky map of the appearance at constant Sun altitude (-6° I believe)).
Venus is a different story, because it is usually visible for months.  As you may know, in the northern hemisphere morning appearences of the inner planets are unfavourable in spring and favourable in autumn, and for evening appearances this is the other way around.  The table shows that Venus had its GE (west/morning) on 8 June 2001, but was best visible on 22 August.  In the mean time, the planet had moved somewhat closer to the Sun, but the effect of moving from spring towards autumn was clearly stronger (click the date link for the weird map that goes with this).  Hence, for Venus the difference can be several months.
Of course, these are just examples for one location (52°N).  I think the process may be too complex to be described by a single, simple equation.


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Observed sun/moon rise/set times to nearest second?
Where can I find a list of observed (not computed) sun/moon rise/set
times to the nearest second, along with the location (latitude,
longitude, altitude [or enough information to find these three
values]) and, ideally, weather conditions (temperature/pressure)?
Since it's hard to tell where the "middle" of the sun/moon is when
it's setting, these would be the setting times for the lower or upper
limb of the sun/moon.
Reason: I'm trying to compute sun/moon rise/set to the nearest second
using high-precision data, and I'm curious to see if I can get
anywhere near the actual observed time. My technique:

Use NASA's Chebyshev approximations to find the positions of the
Sun and center of the Earth.
Use an elliptical (not spherical) model of the Earth to find the
ICRF xyz coordinates of a given location at a given time. This
includes compensating for altitude and precession.
Model refraction based on weather conditions, including adjusting pressure for altitude (weather stations report air pressure normalized to sea level, which isn't the same as actual air pressure, especially at higher altitudes).

I realize down-to-the-second computations of sun/moon rise/set are
probably impossible, in part because refraction depends on more than
just local conditions, but I'm curious to try.

That would depend on what is on the horizon of your observation site. If you are in the mountains, the observed sunset time will be earlier than if you are in a desert plain.

As far as I know, there is no such database set. However, you might be interested in reading (if you haven’t already) this article: https://skyandtelescope.org/astronomy-news/we-dont-really-know-when-the-sun-rises/