Can we conclude with reasonable certainty that there is/are no Type 3 Civilizations in our galaxy?

Short Answer: We won't have to worry about the Galactic Empire any time soon.

Long Answer: First we have to consider some of the properties of a Type III civilization - or, to maybe put it better, what kinds of things it would do. Some are shared properties of Type II civilizations, but on a larger scale. You can probably list some off the top of your head:

Advanced interstellar travel
Large-scale engineering projects
The ability to capture energy from stars, black holes, and pretty much any other source you can think of.
Lots and lots and lots of signals
Colonies on most star systems
The intelligence to not contact us

You can extrapolate #1 from just about every science fiction movie ever made. But I don't mean anything like the starship Enterprise, or the Death Star. To become a Type III civilization, a civilization first has to reach Type II status. What does that entail? Well, among a lot of other things, shunning war. I know that might be hard to believe, but think about this: Could a single nation get off this planet on a large scale all by itself?

A good example is the International Space Station. To get the project going, it needed the combined help of the U.S., Russia, Japan, China, and the E.S.A. Admittedly, we're entering a new age of space travel. SpaceX, Blue Origin, Virgin Galactic, Reaction Engines, and other companies are showing that they can get people and things to orbit or suborbital trajectories. Maybe an entire country isn't needed to do something grand; maybe a single company can do it.

So I can conclude that any spacecraft we find from such a civilization would most likely be peaceful - a passenger ship, perhaps, or a freighter. Such a craft would be hard to detect - admittedly, we have trouble finding anything the size of a planet or smaller, once we get outside our Solar System! But that brings us to #2: Large-scale engineering.

Picture the Death Star. "That's no moon!" Yep, and it looks - from afar - a lot like one. Type III civilizations would have the capacity to build such structures - for peace, likely (see #1) - and these would be a lot easier to detect. Wait. Didn't I just say that anything the size of a planet or smaller would be hard to detect? Yep. But a Type III civilization wouldn't be limited to making small faux moons. Planet-sized objects, or bigger (we'll get to that in #3). How could we detect such objects? Well, if they're big enough, we could find them the same way we find (if we're lucky) exoplanets - transits of the primary star, wobbles in the star's orbit, etc. There is another way, though - for specialized creations.

Ah, 3#. Energy-collecting devices. A quick Internet search of "Dyson sphere" should give you a good idea of what I mean (and if that's not enough, you can see Dyson's original paper or see this question). Dyson spheres/rings/any-other-configuration-you-can-think-of would affect the star whose light they were collecting, and we could figure out if the object in question was a Dyson sphere (see the "Dyson sphere" question). But a Type III civilization could transcend that (Dyson spheres are often cited as the trademark of a Type II civilization). Imagine collecting the energy of jets emitted from the accretion disk of a black hole - not an easy job, but still possible.

Up next: signals. Humans have been emitting radio waves since Marconi (okay, he wasn't the only one) turned on his transmitter. Over 100 years later, we're still going at a pretty good rate. A civilization would need millennia to reach Type III status, and should have emitted a bunch of signals along the way. Also, such a civilization would have spread out, and should be emitting them today. Now, the galaxy's a pretty big place. It is tens and tens of thousands of light years in diameter (although there is a debate about where the galaxy ends), so our radio waves won't have reached much of the galaxy. But a Type III civilization should, theoretically, be almost everywhere, and we should have gotten messages (even inadvertent ones) long ago. Unless you think the "Wow" signal was actually a message, there's not a lot of hope. You hit the nail on the head with your SETI remark.

For the next (#5) property, we turn to . . . well, just about any sci-fi show ever made (again). What would Star Wars or Star Trek be without colonies all over the galaxy? Technically, the definition of a Type III civilization is one that merely uses all the galaxy's energy; it doesn't have to be all over. But it's not much of a conceptual leap to realize that such a civilization would actually have to be pretty much everywhere in the galaxy in order to collect all that energy. So we should, in theory, have a decent chance of stumbling upon a colony if we headed to any star. Is that going to happen any time soon? Nope. So there goes that method.

The last property is something of a joke, but also something of a parting message. As Calvin (from the comic strip Calvin and Hobbes) once said,

  Sometimes I think the surest sign that intelligent life exists elsewhere in the universe is that none of it has tried to contact us.

But think about it. We're a pretty violent race. Just turn on the evening news. Remember what I said before: advanced civilizations may not want to contact us. So even if we don't receive any messages, or find any Dyson spheres, of fail to find evidence of any kind, doesn't mean that we're alone on the Milky Way. Unlikely? Yep. But plausible? Yep.

So, to summarize: There are a bunch of different properties that would scream "Advanced civilization" at you. We could detect them by looking at their effects on stars, by searching the sky for radio waves, or by simply hopping over to a few nearby star systems. But being as we really haven't found anything yet: We can conclude, with reasonable certainty, that there is/are no Type III civilizations in our galaxy.


What might a CN filter be in the context of comet watching? Is it showing dust, or gas, or something else?

A CN filter shows abundances of cyanogen, a molecule with the structure $(\mathrm{CN})_2$. Cyanogen emission lines at 387 nm allow us to study gas in the coma of a comet, telling us more about its structure, properties and outgassing. Other gases, like $\mathrm{OH}$, $\mathrm{NH}$ and $\mathrm{C}_2$, also give away a lot of information about a comet, but the often-strong cyanogen lines are a good tool.

The mechanism behind cyanogen production from parent gases isn't entirely clear, although a number of theories have been proposed. For instance, cyanogen could be released by icy grains undergoing sublimation, or by the decomposition of more complex molecules.


Satellite in orbit in front of and behind the Moon

Yes. Trojan orbits (60° before or after the Moon) are even stable in most cases. These points are called the L₄ and L₅ points, which are two of the five Lagrangian points.

These orbits are mostly impractical (nothing is there).

Today the most known is the Chinese Queqiao satellite, which is a communication relay to their probe on the far side of the Moon. Instead of using L₄ or L₅, Queqiao is in a halo orbit around another Lagrangian point; Earth-Moon L₂.


How does the Moon look like from different latitudes of the Earth?

Here are some answers to your various comments and questions:

  you can estimate a direction of south

True, but it would be a very rough approximation at best. In order for it to be accurate, the bright limb of the Moon would need to face either due east or west on the celestial sphere. (In astronomy terminology, the position angle of the bright limb would need to be 90 or 270 degrees.) In reality, the bright limb of the Moon faces the Sun, which can be north or south of due east/west relative to the Moon. Therefore, the cusps are not north/south, so the cusps can point to the "left or right" of due south on the horizon.

The second problem is using two points just 1/2 degree apart (the two cusps) and extending it to where it meets the horizon (along a great circle, too!). Depending on your location and the Moon's declination, that distance is most likely to be much larger than 1/2 degree. At 40 degrees N latitude, the moon can be approximately 27 to 73 degrees from the southern horizon. (These numbers are when the Moon is on the meridian.) The accuracy of extending such a short line over such a long distance of is going to be very low.

  I think you can estimate the latitude of your location

Technically, you can use any astronomical object to determine your latitude. Depending on how you do that, the Moon is probably the worse object to use. In order to estimate your latitude, you need to know the declination of the object (as well as other data or measurements). Because the Moon orbits the Earth, the declination changes "quickly". The Sun or stars are better because their declination moves more slowly or are fixed, respectively. Polaris is the easiest object to use if you are in the northern hemisphere.

  I also think you can use observation of the phase changes throughout a month to find out whether you are at the southern or the northern hemisphere.

I agree.

  An interesting observation made while seeing the Moon during daytime. In the real life at the surface of the Earth this observation would not be possible, but the software allows it. Since the Sun is no longer positioned "under horizon", it illuminates the Moon from different angle, than depicted in the graphics.

The Moon is visible during the day, so I do not understand the second sentence. For the rest of the comment, I suggest that you choose a day when the Moon is closer to the Sun and run the simulations. For example, choose a day when the Moon is within 3 or 4 days of New Moon, and watch the Sun and Moon as they cross the sky. The bright limb of the Moon always faces toward the Sun (along a great circle), so the angle of illumination on the Moon changes throughout the day -- just like Stellarium shows. (The Moon's image for Oct 2017 at 8 hours looks a little off to me, but it may just be the difficulty of estimating the great circle direction from the image. Displaying the lines of right ascension and declination may help.)


Does our current understanding of the mass of black hole only allow for it to be located at the singularity?

As a star or neutron star collapses to form a blackhole there will, of course, be a moment when the matter is distributed everywhere within the event horizon.  So, there may not be a singularity within a full fledged blackhole, at first.  But, the inward forces within the event horizon are irresistible (considering that escape velocity is greater than the speed of light) and so everything within the event horizon must fall into a singularity.  How long this takes is complicated by the issue of which reference frame you chose to time it.


The planets are very, very close to spherical (the Earth has a variaton of about 0.1% between polar and equatorial radii).

The rotational inertial for a solid sphere is

$$I=\frac{2}{5}mr^2$$

If you give us details about why you might be interested in looking at the rotational inertia of planets in greater detail, there might be more motivation for putting effort into finding an answer for you.

Just saying.


The confusing thing is that inclination is not the only thing defining an orbital plane; it also matters where the ascending node is. Only if all ascending nodes would be in the same direction, you could do the trick with the absolute values you mentioned.

I can't draw a picture right now, but imagine you have a table before you now, and use it as a reference plane (the Sun's equator). Put a sheet of paper on the table, and lift the right end a bit so that it inclines 7.155° w.r.t. the table. This is Earth's orbit, and it looks a bit like the picture in the Wikipedia article.

Now take another sheet of paper, and lift the far end a bit so that it inclines 7.155° w.r.t. the table. This is supposed to be the orbital plane of a hypothetical planet with the same inclination to the Sun's equator, but it is also inclined to the ecliptic (i.e. Earth's orbital plane).


I'll address the three sub-questions individually, as a way of fully answering the title question.

While there are indeed three planets orbiting PSR B1257+12, note that it's a pulsar, the compact remnant of an energetic event involving the system's progenitor. However, that event would likely have destroyed any planets that originally orbited the star, meaning that the three planets we see today likely formed afterwards (see e.g. Podsiadlowski 1993). By contrast, it seems that WD 1856+534 b may have formed prior to its parent star becoming a white dwarf, after which its orbit was disrupted and it moved into a tighter orbit.

While there are other candidate exoplanets orbiting white dwarfs, some consist of planetesimals, asteroids or other debris (e.g. WD 1145+017), some have only been detected indirectly (e.g. WD J0914+1914), and some are being actively disrupted by the star (e.g. WD 1145+017). On the other hand, WD 1856+534 b seems to be somewhat stable and is quite substantial $(M\sim10\text{-}15M_J$), likely falling into the giant planet regime.

It's certainly true that we don't know that the planet formed before the progenitor became a white dwarf, and it definitely can't be ruled out. The authors (Vanderburg et al. 2020) don't explicitly discuss a capture model, but they do attempt to make a statistical estimate of how close a nearby star is likely to have come within the past ~6 billion years, i.e. since the star became a white dwarf:
$$D\sim\left(\pi vt_{\text{cool}}n\right)^{-1/2}$$
with $v$ the mean local stellar velocity, $t_{\text{cool}}\approx5.85\;\text{Gyr}$ the cooling age of the white dwarf, and $n$ the local stellar number density$^{\dagger}$. Assuming $v\approx60\;\text{km s}^{-1}$, they find that in the last ~5.85 billion years, the closest a nearby star would be likely to come to the system is roughly 600 AU. If we consider the capture model, an encounter at that distance seems unlikely to transfer a planet to a comparatively tight orbit.
That said . . . we obviously can't rule out that sort of interaction. The same holds for a bunch of other proposed scenarios which call for "either
finely-tuned or a priori unlikely initial conditions", as Vandenburg et al. put it. So, sure, capture is possible. It just would presumably require a fairly close approach. The authors also don't (I think) explicitly address the idea of formation after the white dwarf evolved to its present state.

All of that said, yes, this isn't an enormous leap into unexplored territory, but it does seem that we have solid evidence of a massive exoplanet in a stable, close orbit around a white dwarf, and models involving it forming prior to the progenitor's evolution to a white dwarf phase appear to hold water.

$^{\dagger}$ As the system is only $\sim25\;\text{pc}$ away, they make safe assumptions about the values using data from Gaia in the solar neighborhood, i.e. $n\approx0.1\;\text{pc}^{-3}$.


Calculating AZ and ALT from RA and DEC: values are slightly different from Stellarium

Stellarium 0.14 or later has a checkbox under Configuration, Tools, Planetarium options:
Include nutation.
If I uncheck it, the RA and Dec of date for those stars on 2016/1/1
match yours within 0.2 arcsecond.

Some libraries distinguish between astrometric and apparent positions,
aberration
making up to 20 arcseconds difference.


Which moons have cold traps? (i.e. low ecliptic inclination in orbital and rotational axes)

Many of the moons of Jupiter and Saturn are tidally locked and probably move ice from their equatorial regions to their polar regions. Ganymede is the best example where you see bright polar ice caps and a dark equatorial zone.  A migration process occurs on all of these moons in which light from the sun gets absorbed by water ice molecules and sends them on random jumps.  After a long random walk, they eventually end up near the pole where they stay for a long time even if they are not in a perpetual shadow, and they stay there for nearly forever if they are.

But, if the moon has a very thick coating of ice everywhere, this process may not have dug down deep enough, so it is still covered with ice everywhere, in which case it is hard to see the effect happening.


Why does the darkening of the Moon go from left-to-right during an eclipse?

On any given day, you see the moon rise in the east and (apparently) travel across the sky from your left to the right; therefore, you would assume (incorrectly) that the leading edge of the moon's movement must be to the right.

That is not correct.

The apparent motion you are seeing is predominantly from the Earth's ~24 hour rotation. But the orbit of the moon is ACTUALLY from west to east as it revolves around the Earth every 29.5 days. Look a the diagram below.

If you were looking down at the earth-moon-sun system far above the ecliptic over the north pole, you would see the Earth spinning in a counter-clockwise direction… and the moon is also revolving around the Earth in a counter-clockwise direction.

So forget about the Earth's rotation for a moment; it has nothing to do with the moon passing into the Earth's shadow (i.e. the eclipse.) Look at the diagram below, and place yourself on that little ball labeled "Earth" and look into the night sky. Which side of the moon is covered first as it passes behind the shadow cast by the Earth? What do you see?

Left to right.


How did Ptolemy prove in the following excerpt that the Earth is not cylindrical?

If the Earth were a cylinder, and stationary in the centre of the universe, then the apparent daily motion of the stars must be due to the stars' actual motion in the heavens. The stars he thought to be on a sphere that encircles the Earth, and there is a pole to this sphere, at the Northern pole there is a bright star.

If the Earth were a cylinder, and the stars moved in circles around the axis of the cylinder; the axis of the cylinder pointing directly at the poles of the celestial sphere; then for any observer on the curved surface the pole star would be on the Northern horizon. The Southern pole would be on the southern horizon an all the stars would rise and set. That is clearly not the case.

Ptolemy then considers the possibility of a cylindrical Earth with the axis of the cylinder not pointing towards the pole of the sphere. Then some stars would rise and set and others would remain always below the horizon. Ptolemy (in Egypt) would see the Pole Star at about 30 degrees above the horizon. But he knew that the position of the pole star was not the same in the Northern part of the Empire. There were stars that were sometimes visible in Alexandria that remained always below the Horizon in Rome. In Rome the Pole Star was 40 degrees above the Horizon. This is not consistent with a cylindrical Earth.


What else could the Event Horizon Telescope Observe?

How about Betelgeuse?

Betelgeuse is about 640 light years away compared to 54 million light years for M87.  It has an angular diameter of 0.042 to 0.056 arc seconds while the quoted resolution of the EHT was 0.000025 arc seconds so you'd expect some detail on its surface.

Betelgeuse seems to be undergoing a number of rapid changes right now.  It is a young but very massive star and is racing through its evolution.  It is only ten million years old but is expected to explode as a type II supernova within the next million years.

The best image of Betelgeuse is this from ALMA

I guess the problem with this question is that there are many, many researchers and teams who would like EHT time to investigate their stuff, but those individual telescopes are already committed to other projects.  The EHT people managed to pull this off because of the sensational nature of the target, a black hole!  No other subject in astronomy would get so much attention, so many institutes working together, and bring in the funds and time


They are close to circular, although we might quibble about the meaning of "close". Except for Mercury, all planetary orbits have eccentricities below 0.1. That's close enough to a circle to actually pass for a circle at a superficial examination.

Mercury 0.2056
Venus   0.0068
Earth   0.0167
Mars    0.0934
Jupiter 0.0484
Saturn  0.0542
Uranus  0.0472
Neptune 0.0086

Check out these ellipses. The first one has an ecc. of 0.1. Looks like a circle, doesn't it? Even the second one (ecc = 0.2) kind of looks like a circle, too, if you don't examine it too long.

As for why they are not exactly circular, down to the last digit? There will always be perturbations that will squish perfect circles a little. Planets interact with each other, and pull each other off perfect circular orbits. It's very unlikely that in a complex system like ours you'll find highly circular planetary orbits.

But, again, the differences between the current planetary orbits and perfect circles are for the most part quite tiny, and you should think of them as "basically circles", unless you're an astronomer or a NASA engineer.

Comets, OTOH, those do tend to move on narrow elliptic orbits.


Is there a webpage that shows the night sky but can filter out dim stars?

I think your requirements can be met by Stellarium. It is a freely available open-source planetarium software available for PC, and can be used offline.
There is also a web version, which you can try out here.
You can filter stars out by pollution levels, as illustrated in the stellarium wiki.
Here is an Astronomy Stackexchange answer on matching the magnitude of stars to what is visible from specific cities.


One of the most popular ones known as HIPPARCOS lists parallax which is indicative of distance.


When I read that a system is located x light years away, why is there no direction?

The galaxy's catalog name, SDSS J1126+2944, also gives its direction. This means the galaxy is at approximately 11h26m right ascension and 29 degrees 44 minutes north declination.

The "full name" of the galaxy, given here:

http://www.sci-news.com/astronomy/pair-black-holes-distant-galaxy-03546.html

is "SDSS J112659.54+294442.8", which gives its position even more precisely. A finder chart is here:

http://simbad.u-strasbg.fr/simbad/sim-id?Ident=SDSS+J112659.54%2B294442.8


Simulation packages or theory to work with gravitational collapse of massive molecular clouds?

This sounds very ambitious! I think your way in to this topic might be to have a look at the works of two authors - Matthew Bate and Mark Krumholz.

The reason I suggest these two names is that they are representative of the two basic approaches to simulating the collapse of molecular clouds (which has little to do with n-body simulations). These approaches are known as Smoothed-Particle Hydrodynamics (SPH) and Adaptive Mesh Refinement (AMR).

Crudely put, SPH treats the fluid as discrete particles whose properties are smoothed over the size of a kernel - the resolution is controlled by the number of particles in the simulation and the smoothing length. AMR is a more "traditional" grid-based approach where the size of the grid is varied adaptively with time and space to get the requisite resolution.

A quick google reveals a number of public codes -e.g. for SPH there is SPHYSICS or GRLAB; for AMR there is ENZO or PARAMESH.

One of the main activities in model simulations these days is trying to incorporate "feedback" into star formation. i.e. the winds and radiation from newly born high-mass stars and the outflows from protostars. Such things do not form part of standard codes.


Where would an object fly to if it were to leave the gravitational field of Earth? And would it stop somewhere? What would happen to it?

The gravitational field of the Earth has not a border, what you can leave and there is weightlessness. Instead, going more and more far away from the Earth, so will its gravity weaker and weaker.

At some point, the gravity of other bodies (Moon, Sun, other planets) will be stronger than the Earths.

What will happen to a body which leaves the space where not the gravity of the Earth is the strongest, depends on the gravitational force of the other bodies of the solar system. These are also moving (mostly, orbiting the Sun and the moons are orbiting their planets), thus there is no general answer to your question.

For example, in the case of the Starman car (Elon Musk's Tesla shot into the space) will have this orbit in some centuries. Where will it be in millions of years, there are only probabilistic calculations. If the object did not left the Earth too quickly (it went only a little faster than the escape speed), then it it will likely start a roughly "random walk" between the orbits of the Venus and the Mars. This is what also most known asteroids are doing.


There's no back and forth motion on the way towards tidal locking.   There's kind of two ways to look this kind of "settling", there's overshooting and correction, basically a back and forth, and there's a very gradual force that grows smaller, basically moving towards zero but not overshooting.    I realize that's just re-writing your question, but it helps to understand the 2nd option.

I'll use the Earth as an example, because the Earth is in the gradual process of becoming tidally locked to the Moon.  It's the same process, but easier to visualize.

Because the Earth rotates faster than the Moon orbits, the tidal bulge moves ahead of the moon.   This is called the tidal bulge offset.

Source.

It's the size of the bulge and the angle of the offset (and the mass and distance of the Moon) that causes the familiar gravitational effects, the Earth's rotation slowing down and the Moon moving away from the Earth.

Source

The current angle of offset for the Earth's tidal bulge to the Moon is about 3 degrees, often exaggerated in diagrams so it's easier to see.  (I should find a source for that - I've not been able to find a good one).

From the Moon's point of view (and if we assume a circular orbit), the tidal offset stays in the same place.  It's a factor of the relative rotation speed.   As Earth slows down, the angle of offset will move towards zero.    At tidal locking, the angle of offset becomes zero and there's no force to continue in the other direction.    In a simple mathematical model, tidal locking wouldn't ever happen completely, it would just gradually get closer and closer, like an object shot from the Earth at exactly escape velocity.

In reality, because objects aren't perfectly symmetrical and because orbits aren't perfectly circular, the "take forever to get to zero", doesn't happen.  But it's still a gradual approach to zero and no overshooting and returning because the force on the rotation approaches zero as the angle of offset approaches zero and the force needed to change a planet's rotation is enormous and there's essentially zero force to make that happen.

With more eccentric orbits, like Mercury and it's 3:2 rotation to orbit resonance, that might be different.   Not sure how eccentricity might play into this.   That's more complicated.


How far in arc seconds is a planet from its star? The star is 12 parsecs away from earth. The planet is 1 AU from its star

Because a separation of 1 au subtends an angle of 1 arcsecond at 1 parsec. That is the definition of a parsec. As you move further away, the angle gets smaller by a factor of the increase in distance.

Just draw a right-angled triangle where the base length is the distance and the height is the star-planet separation. Now increase the base length and ask yourself what happens to the angle subtended at the Earth? NB All angles are small so $\theta \simeq \tan \theta$.


You can get a position and velocity snapshot easily using the Python package Skyfield. It downloads a JPL Development Ephemeris and then interpolates the positions for whatever time you choose for your snapshot.

You can also use JPL Horizons to produce one or thousands of positions of solar system bodies.

This answer describes how to set up Horizons to save positions to your computer, and includes a script that can read the format of the saved tables. It also happens to be a simulator that does what you are talking about - statrs with a given position and propagates the orbits.


Could a star closely orbit a black hole long enough for the star to have lost 0.5B+ years to time dilation?

Yes, but not very likely.

The closest orbit that does not require constant expenditure of energy to maintain it is the prograde equatorial ISCO. For a Kerr black hole the time dilation factor on this orbit is $$\frac{dt}{d\tau}\approx \frac{2^{4/3}}{\sqrt{3}(1-a/M)^{1/3}},$$ which at the astrophysically likely Thorne limit $a = 0.998M$ gives a dilation of 11.5470. For a Schwarzschild black hole ($a=0$) it is merely 1.22. (Hamilton 2018, p. 654)

So a 13 billion year black hole could have a star orbiting that had just experienced 13/11.5470=1.12583 billion years!

Now, the gravitational wave emission from orbiting like this would no doubt sink the star. After all, stellar remnants starting at least part of an AU apart are merging today, and the emission scales up fast as one gets closer. Ergosphere processes might happen, but they also reduce $a$ and hence the time dilation.


What are these dark spots on Mars' north pole?

They look similar to the "dark dune spots" of the southern hemisphere, associated with accumulation of melting liquid at the bottom of dunes during warming seasons: https://en.wikipedia.org/wiki/Geyser_(Mars)


Why is NEOWISE used for galactic observation?

After WISE ran out of cryogens to cool its detectors it was re-purposed to look for asteroids.  This was possible because the detectors for 3.4 microns and  4.6 microns could work at higher temperatures than the other detectors and passive cooling gets the spacecraft to 74 K.  It had already finished an all sky survey so there were no new deep space targets for it to study and, with only two detectors, the science would not be as significant, but it has been great as a detector of moving objects like asteroids since, for this, just one band-pass is enough.


I'm not sure this fits this site, maybe better for TV & Movies, but it doesn't appear to be a historic or current telescope.

This website advertises that they sell that model Based on the look and it is similar.   It may be a  nice telescope but it's nothing special.  Certainly not worthy of being a family heirloom passed down for generations, so that doesn't really make sense and it's not a perfect match either, though it's similar.

Larger image here.

It not being a perfect match and there being virtually no references to that telescope representing something real, it's likely not a historical telescope, which is fine.   It could be a future telescope and still be in Captain Georgiou's family for generations.  That would work because they used it in the show to look at things the ship couldn't sense due to some kind of obscuring.  A historic telescope probably wouldn't be able to do that.

Ultimately, I think Signourney Weaver as Lt. Madison said it best.


Why did the big bang produce a low entropy universe?

The second law of thermodynamics states:

  the total entropy of an isolated system can only increase over time

If applied to our universe, one can conclude that the entropy of our universe must necessarily increase over time (or, improbably, stay constant), resulting in the final heat-death of our universe. Just as Georges Lemaître saw that the Universe was expanding and surmised that it must have been smaller in the past (implying a Big Bang), we can also surmise that if the entropy of our Universe is increasing to a maximum entropy wherein the Universe experiences a heat death, it must necessarily have been lower in the past. This begs the question, why was it lower? How did the universe come to be in a state with such low entropy such that stars and galaxies formed. Why did it not spring into existence, already in the heat death stage?

The answer to this is, I don't know. And I'm not sure anyone could know or give a solid answer, backed up by facts and observational evidence. At best, there are some guesses and hypotheses which I can go over a bit.

First off, let me point out that the second law of thermodynamics may not in fact apply to our Universe as a whole. The problem with the laws of physics is that they often apply in localized cases, say when we're talking about a science lab or a galaxy. But when you try to apply them to the universe as a whole, there's just too many unknowns and usually the rules change. For example, is our universe an isolated system? Nobody can answer that because nobody can view the entire Universe. We're limited to only observing the observable Universe so already we can't definitively say that the Universe adheres to one of the primary pre-conditions of the second law.

Not to mention, you have the difficulty that it may be impossible for all, absolute laws to apply to the Universe as a whole, when we already know some of them don't. The laws of physics all rely on each other. If you change one, you affect all the others. We know (or at least some believe) that the law of conservation of energy does not necessarily hold for the Universe as a whole. At any local point in space and for any localized event, it does, but not when applied to the entire Universe. So already the laws of physics are messed up and that will have unknown consequences for other laws of physics when applied to the Universe as a whole.

Energy in our universe is not conserved because it is expanding. This expansion is constantly dumping new energy into our universe, effectively making our universe not "isolated" even if it may be based on the dictionary definition of the word. As such, it's entirely possible for the entropy to be driven down by this energy input via expansion. The initial inflationary epoch, which likely violated energy conservation globally, could have resulted in our low, initial entropy state.

But as I said, a lot of this is guessing and hypotheses. I don't think there's a good answer out there to this good question, and if there is, I'd really like to hear it.


There are essentially two reasons.

First, quasars are rare objects, so even though they are homogenously distributed on large scales, the average distance is large. Moreover, the brightest quasars are even rarer, but visible to large distances, so their average distance is even larger.

Second, most quasars were most active at redshift $z\sim2$. This is thought to be because quasar activity is triggered by galaxy mergers, which bring gas into galactic centres so that the supermassive black holes can be fed. Galaxy mergers and star formation peaked at $z\sim2$.

Finally, there is some bias in the sense that astronomers are looking for the most distant (and hence oldest) of objects, because they are more interesting than nearby objects when it comes to learn about the formation history of the universe.


Where do local users of ALMA live and work?

No one travels to ALMA to observe; the observations are all carried out by staff working at the Operations Support Facility. (This is at an altitude of 2900 meters; the actual telescopes are 28 km away at the Array Operations Site, which is at an altitude of 5000 meters and is not permanently staffed.) Any "unusual/non-standard observations" would have to be specified in advance and encoded into the Scheduling Blocks (prepared files specifying the observations you want).

https://almascience.nrao.edu/about-alma/alma-site

I believe the staff at the Operations Support Facility stays in the Residencia. (More pictures here and here; pictures of the OSF, including the telescope control room, can be found here.)

(I wonder if some future movie will film scenes there, as the James Bond movie Quantum of Solace did at the Paranal Observatory's Residencia in 2008...)

Visitors (who cannot stay overnight) are permitted on Saturdays and Sundays, though you have reserve your visit in advance.


What is meant by "short lived" in the duration of a planetary nebula?

In the case of planetary nebulae we are talking of ~10,000 years (follow the link to the main article on Planetary Nebulae).

In general a short lifetime for a phenomena means a time short compared to the total life of the entity it is related to. It is an informal term and does not have a fixed maximum value, but 10,000 years is a short time on most astronomical scales.


Does Earth's magnetic field become stronger when a solar flare hits?

No, but the magnetic dipole of Earth has a field strength that increases with closer distance. On the direct line between Earth and Sun the magnetic field strength is always $B(r) = M/r^3$ where $M$ is the magnetic moment of the terrestrial dipole and $r$ is the distance to the dipole, aka Earth.
When the solar wind conditions change, the boundary between the wind and the terrestrial magnetosphere changes in location, until it finds a new pressure equilibrium.

The approximate location of the boundary shock can be found by equating the ram pressure of the solar win $p_{\rm ram}= \rho u^2$ with the magnetic dipole pressure $p_{\rm dip}=M^2/(2\mu_0 r^6)$, where $\mu_0$ is the vacuum magnetic permittivity. If you solve this for $r$, you will find a closer location $r$ for the shock for stronger conditions, and then checking at which magnetic field strength this occurs via $B(r) = M/r^3$, you'll find that the magnetic field at the shock has of course increased, but only because the shock has moved closer, not because  $M$ has changed.


What's the terminology of a meteor "ditch" when impact happens at an angle?

I think the answer to this is that it doesn't have a name because such "ditches" don't occur for real meteor impacts. In movies and tv shows, these ditches are often shown as being caused by the meteor first hitting the ground at some angle, then sliding across the ground, digging out a ditch along the way, until they come to their final resting point where the crater is.

When you stop to think about it, this scenario doesn't really make sense. These types of images (e.g., this one or this one) usually show a thin tail or ditch where the meteor first impacts which gets wider and deeper as the meteor scrapes on the ground, resulting in a final, large crater. But, the most energy a meteor imparts to the ground comes from the first instance it hits. If it were to scrape along the ground as portrayed in many movies, the crater would come from the point it hit the ground and the tail would be in the direction the meteor was traveling. The "ditch" or "tail" would get thinner and shallower as the meteor lost energy scraping across the ground. Generally the opposite is shown in movies.

But all that is moot anyway because meteors don't scrape along the ground at impact. That would just violate conservation of momentum and energy. You really have one of two scenarios.

The meteor hits the ground with such force and energy that it mostly destroys itself and you wind up with just a crater. Such an example is the meteor crater in Arizona - in a stroke of brilliance called Meteor Crater. I would hazard a guess that almost all meteor impacts are of this form, be it on Earth or elsewhere.

The other option is that the meteor isn't destroyed when it hits the ground. In this case, rather than dragging or scraping on the ground, your meteor will skip like a rock on water and create another crater on its second impact. In theory, it will keep skipping until it is finally destroyed or runs out of energy. This scenario requires that the impacting meteor is hitting at a very shallow angle with very little force and generally means that it is hitting a smaller body (but larger than itself) such as an asteroid/comet/moon. An example would be the Messier Crater on the Moon, linked by Andy in his comment. These types of craters are unlikely though, as they require pretty specific conditions to occur.


Is the pulsar map on the Pioneer/Voyager probes only meaningful for earthlings?

You aren't missing anything. The pulsar map is fairly useless unless the spacecraft is found within the next million years, when it won't have got too far from the Sun (the probes will have travelled about 30 light years in that time).
The problem is not so much that the set of pulsars won't be visible, because they almost certainly would be if you change your vantage point by a few light years, but that the pulsar activity might cease. Pulsars have a limited lifetime and most likely will have stopped being pulsars in a million years or will not be identifiable from their pulse periods. In addition most pulsars have very high velocities, so they themselves will have moved by $\sim 10$ times as much as the probes.
For something of use beyond timescales of a million years they could have used a network of globular clusters, although again the motions of the globular clusters would need to be accounted for.  It's a tricky problem when everything that's close enough that that their positions would change if you move of order hundreds of light years are themselves in motion. And of course the Sun is moving around the Galaxy too.


Thinking of buying a new telescope. What aspects should be considered?

There are many aspects that should be considered, but a few very important ones are:

New or secondhand? With your budget, buying a new telescope will only get you so far. But if you try to look into secondhand products, you might be able to find something more advanced—well-kept telescopes last years. Of course, you must be careful with whom you are buying it from, and make sure to try it before buying as well as asking about the last time the mirror was cleaned, what eyepieces are included, etc.
Dobson or not? There are many upsides to having a Dobson-type telescope: you get a much larger mirror for the same price, it's easy to set up and maintain... However, it does take more room than other telescopes, and you will have to do more maneuvers to aim and track things with it.
On the same vein, does size matter? If you are looking for the most compact and lightweight telescope possible? If yes, that will have an impact on the kind of telescope you should buy: maybe you will want a Maksutov instead of a Newton, maybe you will want a tripod mount because it can fold and takes the least room.
What do you want to see? As StephenG said earlier, it is important to know your expectations as for what you want to observe with your telescope. If you live in an urban environment and/or are unable to go to spots free of light pollution, that severely limits your possibilities to (mainly) planets and the Moon. To that effect, you might be more interested in a telescope with a bit more focal length. If you are indeed able to go to spots free of light pollution, then you probably want to maximize the diameter of your mirror, because if you get a mirror that is 20% larger, you will get 44% more light.

Some other questions that may influence your choice are: is this a long-term investment or are you looking into making an upgrade in the next few years? Do you want to try astrophotography? How much time are you willing to take to set up your telescope before observations? Do you enjoy using a sky atlas to find the exact object you are looking for? Etc.

To give you an example, I was a similar situation a few years ago. I had a budget of $400, and I wanted a telescope for visual observation only. Because I had access to good observation spots and size was not an issue, buying a Dobson was an easy choice. I joined an astronomy club and a member was going to upgrade his own Dobson (https://www.astroshop.eu/gso-dobson-telescope-n-250-1250-dob/p,6680) to a bigger one, so I was able to test it and buy it secondhand knowing exactly what I was getting.

You should also have in mind the fact that a telescope will often require a few accessories (eyepieces, filters,...) that you might need from day one, so do look at what is included with the telescope you buy, as good eyepieces are things you can keep with your next telescopes, and bad accessories can affect the quality of your observations.

Astronomicably,


How do they measure sub-centimeter wave height on a moon of Saturn?

According to the Wikipedia page on Titan, that data was deduced from Cassini.

  On December 21, 2008, Cassini passed directly over Ontario Lacus and observed specular reflection in radar. The strength of the reflection saturated the probe's receiver, indicating that the lake level did not vary by more than 3 mm (implying either that surface winds were minimal, or the lake's hydrocarbon fluid is viscous).

That claim cites two sources:

newscientist.com (definitely not a primary source)

Harvard ADSABS which does appear to be a primary source analysis of Cassini data. However, I don't have time right now to see what peer-review made of this analysis.

  Cassini RADAR altimetry data collected on the 49th flyby of Titan (2008 December 21) over Ontario Lacus in Titan's south polar region provides strong evidence for an extremely smooth surface, with less than 3 mm rms surface height variation over the 100m-wide Fresnel zone. Histograms of the raw radar echoes imply a mirror-like specular reflection of the transmitted signal. Such an echo is possible only if the surface is extremely flat relative to our 2.2-cm wavelength. The 3 mm upper bound follows from analyzing the strength of the specular return, which declines exponentially with increasing surface height variance. In this experiment, the strength of the echo was larger than expected, severely saturating the receiver. We developed a method to partially correct the echoes for the distortion incurred. While the implied mm-scale smoothness is not proof that the surface is liquid, it is unlikely that a solid surface is so smooth.

From the abstract, it seems they are deriving this conclusion indirectly, but I'm not educated enough in this area to personally vouch for the methods.


The number refers to the number in the catalog compiled by Frank Elmore Ross. From the Wikipedia article on Ross

  At Yerkes Observatory he was the successor to the late E. E. Barnard, inheriting Barnard's collection of photographic plates. Ross decided to repeat the same series of images and compare the results with a blink comparator. In doing so, he discovered 379 new variable stars and over 1000 stars of high proper motion. Some of the high–proper motion stars turned out to be quite nearby, and many of these stars (such as Ross 154) are still widely known by the catalog number he gave them.


Which Saturn satellite passes closest to Saturn's rings and at what distance?

Pan, Daphnis, and various other moonlets, I would argue, are inside the rings.

If you explicitly discount the Encke gap (which Pan orbits in) and the Keeler gap (which Daphnis orbits in) as being part of the ring system, Daphnis would be your answer, as it is a ~8 km object in a 42 km gap. (for comparison, Pan is a ~35 km object in a 325 km gap)

Really, your answer depends on what you want to consider a moon. There are many objects small enough to be classed as moonlets (several hundred metres across) embedded in the rings, and presumably countless more of smaller size.


Such forces do exist.

They are utterly negligible in most circumstances. If you imagine a planet "running into" matter whilst travelling in its orbit. Let's say that on average that matter is at rest with respect to the star the planet is orbiting. Then in order to make a very significant change to the linear momentum of the planet, then it must accrete a significant fraction of its own mass.

Such a thing might be important in the very early history of a planetary system when the system is full of gas and debris (e.g. the formation of the Moon), but later on (and bear in mind that the IAU definition of a planet includes the specification that it has "cleared the neighbourhood around its orbit") this just isn't an issue.

This popular science article suggests that 37,000-78,000 tonnes of material hits the Earth every year. Sounds a lot, but in comparison to the mass of the Earth ($6\times 10^{21}$ tonnes) it is very small. If the material was at rest with respect to the Earth's orbit, then an order of magnitude estimate for how long such impacts would take to significantly affect the Earth's orbit would be $\sim 10^{16}$ years.


How much telescope do I actually need for astrophotography?

Electronic viewing is filling the area between traditional eyepiece viewing and Astro Photography. Electronic images may be viewed live, near live and with digital processing on the fly as the image is viewed or processed after the images have been collected.

The cameras have lower resolution but higher sensitivity than DSLR's and astro photo cameras. Short exposure time are the norm. Imaging/processing time is generally measured in seconds and minutes instead of hours. Stacking algorithms have been integrated into the viewing software. While tracking is still necessary the software is much more forgiving. People are viewing/imaging DSO's with Dobsonians.

I use Mallincams in my observatory. They are used with a 10" SCT, a 5" SCT and several Canon DSLR lenses.

One has the choice of going big and beautiful or small and good enough is good enough.

Take a look at the Mallincam io forum* and judge for yourself. They have a photo section with thousands of images and answers to the questions those interested might have.

* You'll need to request membership first


What study profiles could land me the job of astronomer?

What you need would be at least the following:

Physics: astronomy is physics in the end, and you'll need strong basis in physics;
Math: as for any good education in physics, you'll need a reasonable amount of math to be able to become an astronomer;
Programming: it's almost impossible nowadays to get away from it. Being an astronomer today implies lots of data processing (to reduce the data, to interpret it etc.), and it involves a lot of coding.

And, as far as I know, in lots of countries/universities, you can find at least few astronomy & astrophysics courses. In some countries, you can even find dedicated master degrees.

With all of that, you'll be ready to start a PhD in astronomy to become a professional astronomer.


Nothing that complex. Habitable zone merely defined a circumstellar volume where a planet (if any) can receive an amount of radiative energy to keep H2O in the liquid phase.

Yes, it has been based on the requirements for life from Earth, lacking any wider samples.

I encourage you to play around with interactive illustration of concept (Flash applet).


What is the future of our universe?

Like StephenG said, nobody knows for sure. But we do have confidence that the universe is expanding, and we also have confidence that the expansion is speeding up. So extrapolating from that, the future looks cold and lonely and bleak. A bit like life for the older generation!

  Is the universe heading towards a Big Freeze, a Big Rip, a Big Crunch or a Big Bounce? Or is it part of an infinitely recurring cyclic model?

I'd say the Big Crunch and the Big Bounce are out. The universe didn't contract when it was small and dense. Instead it expanded, and that expansion is increasing. So it looks like we're in for a Big Freeze. However I wouldn't rule out a Big Rip of sorts. Have a look at page 5 of this paper where Milgrom mentions the strength of space. Then think of the balloon analogy, but make it a bubble-gum balloon, in vacuum.

Image courtesy of the one-minute astronomer.

The skin gets thinner and the balloon expands, then the skin gets even weaker, so the balloon expands even faster, and so on. Bubble-gum bubbles sometimes end badly, and there's something about  this article by Phil Plait that strikes a chord.

  Or is it part of an infinitely recurring cyclic model?

I don't know. I have no evidence to suggest that there's any kind of recycling going on, and I can't think of mechanism by which this might occur. I have of course read about "conformal cyclic cosmology", but I just don't buy it, along with other stuff from Penrose.


To answer your question directly, it is quite unlikely that planets would be habitable without already having life. It would have to be in a near-perfect condition, which would be statistically unlikely. In case of planets already having life, the existence of life provides robustness to small changes and makes habitability possible in a larger range of conditions.

I am assuming 'habitable' here means habitable for man, trees and an ecosystem etc. Remember that there are extremophiles (or certain other hardy species) which completely change the perspective on habitability, and are often the key in making those places habitable for other forms of life.

This is also true for parts of earth which undergo a sudden change and are not 'habitable' for most forms of life till some hardy species takes over and changes the conditions, and consequently allows other species to be able to survive. See, for example, ecological succession (https://en.wikipedia.org/wiki/Ecological_succession), esp. primary succession and pioneer species (follow the links within the article).

Also, this is a long process. In case of the earth, there was also evolution that needed to happen, so it probably won't take as long as a few billion years, but even on earth, it takes a few years to make small disturbed microhabitats habitable. The atmosphere and other factors (soil, moisture etc., for the plants) need to be made suitable for living. Check this interesting article on terraforming: https://microbewiki.kenyon.edu/index.php/Terraforming

Hope this was helpful. I don't really know if a carbon cycle is possible outside of living organisms in some way.


You have it the wrong way around. You have to establish a periodicity before doing the folding.

There are various ways to do this, but the most frequently (pun intended) used is the Lomb-Scargle periodogram.

Start with a datset consisting of a set of points $(t_i, m_i\pm \sigma_i )$, where $t_i$ are times of observation and $m_i$ and $\sigma_i$ are the fluxes and their uncertainties at those times.

You then calculate an array of chi-squared statistics, by fitting a mean level $\bar{m}$ and sinusoidal functions over a range of discrete frequencies $\omega$. The aim is to maximise the statistic
$$p(\omega) = \frac{\chi^2_{0} - \chi^2(\omega)}{\chi^2_{0}},$$
where
$$\chi^{2}_{0} = \sum (m_i - \bar{m})^2/\sigma_i^{2}$$ and
$$\chi^{2}(\omega) = \sum  (y_i - \bar{m})^2/\sigma_i^{2},$$ with $y_i = A\sin(\omega t_i + \phi) + A_0$, where $A$, $A_0$ and $\phi$ are free parameters that are optimised for each frequency (i.e. you find the minimum possible $\chi^2(\omega)$ at each frequency).

Once a period is obtained from a peak in the power spectrum, $p_{best}$, then it must be established whether this peak is "real" or significant.

Theoretically, the probability that $p>p_{best}$ is $(1-p_{best})^{(N-3)/2}$ for $N$ independent measurements (see Cumming, Marcy & Butler 1999). This is the "false alarm probability" - the chance that you would get $p>p_{best}$ from random, aperiodic data with those uncertainties.

However, real data has correlated noise and the points are not independent. So what you do instead (see Collier Cameron et al. 2009) is shuffle all the times of the data points (i.e. randomly reallocate the times of each observation) and go through the process again. Do this a squillion times with different random reallocations and see how many times you end up with $p>p_{best}$. This gives you some idea about how confident you can be that your detected periodic signal is real. This is called a "bootstrap Monte-Carlo method".

There are other ways to do this; my answer is not complete; but it's the one I'm familiar with.


What is the most populated/numerous stellar system in which the orbits of all objects are known?

The answer is none. Other than our own solar system, astronomers don't know if they know all of the large bodies (aka planets) orbiting any given star system. Presumably other star systems have asteroids, comets, and other stuff. The orbits of those small bodies in star systems is unknown, and will remain unknowable for a long time.

We don't even know that for our own solar system. If we did, we wouldn't have to worry about killer asteroids and comets.


As @RobJeffries says, this information is mostly in the relevant wikipedia article.

The answer is really in two parts:

Firstly, they don't have solid surfaces because the elements they are made of don't have solid forms at the temperatures and pressures that arise inside the planets. In particular, Jupiter and Saturn are made mostly of hydrogen and helium which only become solid at rather low temperatures and the planets are still hot due to the heat created by their formation, heating from the Sun and some heat released by internal processes.

This, of course, begs the question of why those two planets are made mostly of hydrogen and helium, when the others are not. I should start with the caveat that a lot of aspects of this are still areas of active research and debate among scientists, and could not be considered settled  or universally accepted.  As more evidence is gathered and computer models (used for making checkable predictions from underlying theories) are refined,views of this may change.

So the solar system started out as a cloud of gas and dust. While it was mostly hydrogen and helium (still around since the big bang) it had significant amounts of other elements present in it which had been produced in older generations of stars. Some chance concentration of mass, or external shock (eg from a supernova) caused it to start to collapse and, under the influence of its own gravity, fluid dynamics and the conservation of angular momentum, a large part of it formed a ball (called a protostar) at the centre which was heated up by the collapse process itself. Around that, gas and dust formed a flat disk (oriented according to the original net angular momentum of the clump). Within that disk, tiny grains of solid matter formed. Far from the protostar, those grains were mostly made of compounds relatively light elements (carbon, oxygen, nitrogen, hydrogen) which we collectively call "ices". Closer to the protostar, its heat stopped ices from solidifying, but still allowed grains of iron, rock-like materials and so on.

When these grains collided, they stuck together, gradually forming bigger and bigger lumps. In the outer system they were lumps of ice, a lot like comets, in the inner system, something more like lumps of rock (like some asteroids). These then started holding together and pulling in more material by their own gravity and eventually grew into planets. At some point during this process fusion started in the Sun, and drove awat most of the remaining gas and dust. After that the planets grew by collisions.

So in the inner system, you got rocky planets like Mercury, Venus, Earth and Mars. A bit further out you got icy planets like Uranus, Neptune and Pluto. A couple of those icy planets  happened to get big enough, quickly enough, to attract a lot of hydrogen and helium gas before the Sun lit up. They are Jupiter and Saturn.


Helium in terrestrial planet atmospheres?

Do you have any reputable estimates as to how much helium has been out
  gassed over the last 4 billion years

Reputable estimates - no.  Bad guesses, maybe.

Short answer:   only trace amounts.  It's impossible to have enough rare radioactive material to produce a real helium atmosphere.  Even if the planet retains all it's helium from radioactive decay it should be measurable in parts per million, not a real atmosphere.

Long answer below:

The number of alpha particles that a radioactive element creates is pretty straight forward, just subtract the Baryons in the final state from the initial state.   Beta decay changes the type of hadron, but not the number, so, from Uranium 238 into Lead 206 - loss of 32 baryons, and all of them through Alpha decay - Alpha particles become helium atoms, so each U-238 decays into 8 helium.

Thorium 232 into Lead 208 - 6 helium.

Uranium 235 into Lead 207 - 7 helium.

Potassium 40, the other common radioactive element only undergoes beta decay into CA40 (89.3%) and AR40 (10.7%), so it's irrelevant in the helium calculation.  There are smaller amounts of other elements that emit alpha particles, like PT-190, but they're rare enough to be ignored.   The bigger unknown is how many elements with shorter half-life were common when the Earth was young but are no longer present.   I have no idea how to calculate that so that will be a bit of an unknown.  One possible approach would be to say that the forming planet would be sufficiently hot to lose much of it's helium, but that's a bit of a cheat.   I think, early on, predictions are rough due to unknown quantities of shorter-half life elements.

U-238 is easiest to calculate with a half-life of about the age of the Earth, so about half of Earth's U-238 has completed its decay change and produced 8 helium per U-238.

Thorium-232, with a half life of about 14 billion years, so only about 20% of the initial thorium-232 has decayed over the age of the Earth.

and U-235, with a half-life of about 700 million years, means that almost 99% of it has decayed over the age of the Earth.  Current estimates say that about 0.72% of Uranium is U-235

I touched on this above, but the big unknown for this scenario is in the range of 100 million year half-life.   That is, elements that are long-lasting enough to survive the Earth's formation but wouldn't remain in high concentrations today.   That's in this range, bottom of the 10^12 seconds, top of the 10^15 seconds.

Curium-247 has a half-life of about 15 million years and it decays into lead-207 releasing 10 alpha particles in the process.   If planet formation happens relatively quickly, that could supply some helium to a planet, but I'm going to ignore it as having slightly too short a half-life.

Uranium-236 has a half life of about 23 million years and it decays into Thorium, emitting just 1 alpha particle.  Presumably much of it would become thorium by the time of planet formation, so I'm going to ignore it as well.

Plutonium-244, mostly synthetic, but it is the rarest primordial element on Earth.   From the Wikipedia article:

  Accurate measurements, beginning in the early 1970s, have detected
  primordial plutonium-244, making it the second-shortest-lived
  primordial nuclide after 146Sm. The amount of 244Pu in the pre-Solar
  nebula (4.57×109 years ago) was estimated as 0.008 of amount of
  238U.

If estimates put it at a bit less than 1% as common as U-238, I'm going to ignore as well, as 1% won't make a big difference.

Finally, Samarium-146 is worth considering with a half-life of about 100 million years (or 68 million depending on which source you use).   It decays into Neodymium-142 releasing a single alpha particle.   Samarium 147 and 148 also undergo alpha-decay but very slowly, half-life of 100 billion and 7 quadrillion respectively.

So these are the 4 primary helium "outgasing" isotopes relevant after planet formation.    U-238 (8 helium), U-235 (7), Th-232 (6) and SM-146 (1).    And as for how much, you have to ask how long the planet has been around.

SM-146 decays fairly quickly, half life 100 million years.
U-235 - 700 million years
U-238 - 4.5 billion years
T-232 - 14 billion years.

If we take a planet about the age of Earth

SM-146 - essentially all of it
U-235 - 99%
U-238 - 50%
T-232 - 20%

Now the trick is, how much of these elements is a terrestrial planet likely to have at formation.

If we go by radioactive decay and radiogenic heat estimates, TH-232 and U-238 combine to generate 20 trillion watts (first article).  That' 6 x 10^20 joules per year or about 3.75 x 10^33 mEV per year, by radioactive decay of those 2 elements.

The Thorium decay chain emits about 42.6 MeV per decay for 7 alpha particles.  (about 7 MeV per helium) and the U-238 decay chain, about 51.7 MeV for 8 alpha particles.  (ballpark 6.5 MeV).   That includes Neutrinos though and that energy needs to be factored out.   I couldn't find precise numbers but about 10 MeV (or 20%) is lost into Neutrinos (technically anti-neutrinos) by this website.   So, very ballpark, lets say 5 MeV heat creates one alpha particle.

current rates of decay of Thorium & U-238 should produce about 7.5 x 10^32 alpha particles, which corresponds to about 1.2-1.3 billion moles of Helium or about 5 million KG, between 5 and 6 tons of helium produced in the Earth's crust and mantle every year - based on radioactive measurements which are probably pretty accurate.

Thorium is currently about (ballpark) 4 times as abundant as U-238 but because it decays 3 times more slowly the helium production and heat generation in the Earth's core is similar (this chart suggests thorium's heat output is higher, but they are close).

The helium production from Thorium has been pretty consistent, as 80% of the Thorium at the formation of the planet is still here.
The production from U-238 was about twice as high when the Earth formed.
The production from U-235 is currently negligible compared to the other two, but was much higher when the Earth was young.   working back from U-235 being about 0.7% of the total Uranium and it having gone through between 6 and 7 half-lifes since the formation of the Earth, (and accounting for U-238 being twice as abundant when the Earth formed), that means between 32 and 64 (lets say 50) x 0.7, so there was some 30-40% as much U-235 as U-238 when the Earth formed.

Bad math again:

Thorium, 2.5-3 tons per year now, 25% more than that when Earth formed.  lets say 3 tons per year, average

U-238, 2.5 tons per year now, twice that when the earth formed.   lets ballpark average that out to 3.5-4 tons per year on average.

I'm not going to average out U-235 because it starts out too high and decays too quickly but if there was about 30-40% as much U-235 as U-238 when the Earth formed I can just multiply U-238's numbers by 35% and by 7/8, working out to about 1 ton per year average.

I'm doing per year/on average just for quick bad math, so 3 + 3-4 + 1 = about 7.5 tons per year x 4.5 billion years - quick estimate for how much Helium was produced by those 3 elements - very bad estimate, 33 billion tons.   Most of this Helium is produced and would be trapped in the Earth's interior until plate tectonics allows it to escape.  The rate of escape is above my pay-grade.

Samarium-146 is the hardest one to estimate.   It's essentially all decayed into Neodymium-142 but both Samarium (it has several stable isotopes) and Neodymium are abundant, and this chain produces a single helium.   It likely doesn't make a huge difference in the total.   Table of elemental abundance in Earth's crust.

Atmospherically speaking, 33 billion tons is nothing.   Even if we adjust to to 40 or 60 or 300 billion tons, that's still essentially too little to be an atmosphere.  Helium is currently about 5.2 ppm in the atmosphere and giving the Earth's atmosphere a mass of about 5,600 trillion tons, 5.2% by molecule, adjusted for the mass ratio of helium (4) to Nitrogen/Oxygen (29), there's about 4 billion tons of Helium in the Earth's atmosphere today.   If we were to say that 400 billion tons of helium was produced on Earth by radioactive decay (about 10 times my estimate), that's still a trace amount, just 100 times the current amount or 520 PPM, which is less than the Argon (currently 9,300 PPM).   Argon-40 which comes from the radioactive decay of Potassium-40 (about 10% of the time), is more common because Potassium-40 is much more common than the Uranium, Thorium and Samarium-146 that decay alpha-particles.

That's a very long answer that basically says there's not enough Alpha-particle emitting elements that last long enough to survive planetary formation, that produce anything more than a trace amount of helium.

Earth also retains some terrestrial helium, about 7% of Earth's helium is terrestrial, 93% formed by radioactive decay.   Most of Earth's terrestrial helium is in it's mantle, very little in it's crust.

Some interesting and loosely related reading.  Yellowstone is able to release more helium than the radioactive material produce because it's crust is unique, very old and it's trapped helium for over 2 billion years, only recently started to release it.

As to your original question.  As I understand it, a planet needs to be very large to acquire significant amounts of helium.     By particle, the universe is about 92% hydrogen and 8% helium (or 75%-25% by mass).

Jupiter's outer atmosphere is slightly helium richer than that, about 10% helium.

Uranus' uppermost atmosphere is about 15% helium, but only it's outer edge.   Uranus, unlike Jupiter and Saturn, isn't primarily a hydrogen-helium planet, it's more of a water/ice planet, sometimes called an ice giant.

It seems that planets need to be very massive to obtain significant amounts of the most common elements, hydrogen and helium, then, as the Wikipedia article implies, they need to get hot to lose their hydrogen, so helium atmosphere planets probably need to be quite large, perhaps more massive than Neptune.   It might be quite a it more difficult for a typical terrestrial planet to do that, unless if it was very massive and a terrestrial planet that massive might resemble a gas giant anyway due to a very thick atmosphere.

I have a tough time thinking of any other way a helium atmosphere is likely to happen, unless you got the planet below 20 degrees K and all the other gases turned to liquid or ice, but I suspect that's not the answer you're looking for.  A planet probably needs to be quite large to retain significant amounts of helium I would think. outside of a very odd circumstance like a helium white dwarf colliding with another large object inside a stellar nursery, even then, helium gas doesn't clump like ices and rocky material, but maybe such a scenario could put enough helium into the planetary formation.  (bit of a crazy what-if), and I'm not sure I buy it.

I tried to reason this one out so, corrections are welcome and encouraged.   I'll try to clean up any grammar a little later.


In a reflecting telescope (Cassegrain), is there a difference between using a concave or convex secondary mirror?

There are many flavors of reflecting telescopes based on the original Cassegrain design which has been attributed by history to Laurent Cassegrain:

  The Cassegrain reflector is named after a published reflecting telescope design that appeared in the April 25, 1672 Journal des sçavans which has been attributed to Laurent Cassegrain. Similar designs using convex secondaries have been found in the Bonaventura Cavalieri's 1632 writings describing burning mirrors and Marin Mersenne's 1636 writings describing telescope designs. James Gregory's 1662 attempts to create a reflecting telescope included a Cassegrain configuration, judging by a convex secondary mirror found among his experiments. (from here; emphasis added)

However, a Gregorian Telescope does indeed use a concave secondary mirror, placed after the primary focus. The diverging beam is then re-imaged behind the primary. According to Wikipedia, examples of Gregorian telescope designs in use include:

The MeerKAT, the Green Bank Telescope, the Arecibo Observatory, and the Allen Telescope Array are all radio telescopes employing off-axis Gregorian optics.
The Vatican Advanced Technology Telescope, the Magellan telescopes, and the Large Binocular Telescope use Gregorian optics.
The Giant Magellan Telescope will also use Gregorian optics.
The Solar Optical Telescope on the Hinode satellite.

Advantages of the Gregorian include the option of putting a field stop at the primary focus, helpful when viewing near bright, interfering sources of light, and providing an upright, non-inverted image.

In either the convex/Cassegrain-type or concave/Gregorian type of system, the curved secondary mirror provides several important advantages, including:

Slowing the convergence of the rays, which allows for the diameter to be much smaller than the primary (so as to block less light and be much lighter weight) while still bringing the focus beyond the hole in the primary and out to where an eyepiece and/or other optics can be conveniently located.
The length of the telescope can be quite short while the effective focal length seen at the eyepiece is still quite long, allowing higher magnification for a given eyepiece.

The cartoon illustration shown in the Canada under the stars website shows a meniscus with it's concave side used as a reflector. I think this is just an error by the illustrator. It wouldn't make sense to show a Gregorian telescope in a paragraph about Laurent Cassegrain, and the primary focus does not fall in front of the secondary.

I've tilted the drawing so that the optical axis is horizontal, and highlighted the incoming and reflected rays of light striking the top edge of the secondary. You can see that the reflected ray is more parallel to the axis, which would only happen if the mirror were convex. The very large distance between the lens of the eyepiece, and the actual location of the exit for one's eye confirms that some artistic license has been taken with the optics here.

For reference, look at the realistic drawing of a Cassegrain optical system from Wikipedia shown below.

The drawing from the "A-level Physics Tutor" website is also problematic, showing the primary focus far in front of the primary mirror of an optical telescope, followed by an eyepiece lens with the same diameter as the secondary mirror. Again this is artistic license, but there's no excuse for it in a proper physics tutorial.

There are examples of optical telescopes where the secondary mirror is so strongly convex that the light then becomes divergent, expanding toward the primary before being refocused (made convergent again) by a tertiary mirror, such as the (crazy/beautiful) Large Synoptic Survey Telescope. See also LSST Optical Design.

above: Drawing of a Cassegrain reflecting telescope, from here.

above: Illustration of the optical system of the Large Synoptic Survey Telescope, from here (rotated).


How do we calculate the escape velocity of galaxies?

In short, the escape velocity from the region we find ourselves in the milky way is 544 Km/s +- 10%.

This number, and an extensive explanation of how it was calculated can be found on a paper called The RAVE Survey: Constraining the Local Galactic Escape Speed

Also important to consider is the fact that we already are at aproximately 220 Km/s because of the sun's orbital velocity, so the actual value needed to get there from earth is much lower.


The problem with your approach is that the deformable mirror changes the phase of the light across the mirror, where the light is not focussed.  The light at the sensor array is focussed, and what you get is the intensity which is, roughly, the Fourier Transform of the phase front at the mirror.   By the time you have intensity, the phase information has been lost.

edit for clarity:

The sensor array measures the intensity, at which point the phase info is irretrievable.  If you remove the sensor array and re-measure past the focal plane, yes, you can get phase info -- see "Plenoptic cameras."

Now, there are other techniques -- you might be interested in searching for papers on "Lucky Imaging" , which basically takes as many images as possible and throws away the distorted ones.


The fact that the distance between galaxies is increasing is found by Redshift of light, where, the electromagnetic radiation from galaxies, as they travel large distances, get their wavelengths increased, and turn towards the red side of the spectrum. Upon working on these measurements, Astronomers concluded that the rate of increase of separation between two galaxies is directly proportional to the distance between those two galaxies.

So, consider two galaxies A and B, the distance between them being r

Then,  dr/dt α r
i.e, the speed at which they are separating is proportional to r.

The cause of this is not completely and surely known as of today, so it has been named a mysterious force called dark energy.


In Equation (16) of the paper you link to, $z$ is the observed redshift. In the first paragraph of section 2.2

  The heart of the method is to use a measured redshift, z, to infer a velocity, v(z)


What is the brightest star (by absolute magnitude) that we can see by naked eye?

Mu Sagittarii is a star system, not a single star. If that can be included, then Eta Carinae should be included, and it has an absolute magnitude of -12.0. It's a star system about 7,500 light-years from Earth.

It looks like the brightest (absolute magnitude) single star visible to the unaided eye is WR 24 (in Carina Nebula). Its absolute magnitude is −11.1 and apparent magnitude is 6.48, so just barely visible.

source: Wikipedia - List of most luminous known stars

edit: Rho Cassiopeiae's absolute magnitude is -9.5.


Is there a theoretical maximum size limit for a star?

According to current knowledge, yes. If the gas cloud is too massive, the pressure of the radiation prevents the collapse and the star formation.

The article Stars Have a Size Limit by Michael Schirber, it's about 150 Solar Masses. However, there's the Pistol Star, which is speculated to be 200 SM.

In the article 'Das wechselhafte Leben der Sterne' by Ralf Launhard (Spektrum 8/2013) there's a diagram with information that when the mass is over 100 SM, the star can't form because of radiation pressure. The exact value of the limit is not speculated in the article.


There are multiple categories of Black Hole powered astrophysical jets:

Microquasars: As their name implies, they are somewhat like a highly scaled-downed version of quasars, but unlike quasars, they are powered by accretion from a donor star. They contain stellar-mass black holes.

Quasars: They are huge systems of accretion disk(s) around supermassive black holes.  They are found in the centers of galaxies, and are very powerful, enough to outshine their host galaxies.

The size of jets depends on the type: microquasars have jets only a few light years long, whereas quasars have jets millions of light years long.
The size of the jets depends on the mass of the Black hole and the magnetic field of the black hole and the magnetic field of the accretion disk.


Affect of leap seconds on the prediction of an eclipse

It has been known that changes in earth rotation affect the location of eclipse paths for over a century.  The most active research in this has been Morrison and Stephenson.  See, for example http://adsabs.harvard.edu/abs/2004JHA....35..327M


Are there any studies that prove/disprove the Oscillating Universe Theory?

The main obstacle with the Oscillating Universe model is the pesky second law of thermodynamics — that entropy always increases within an isolated system, and never the reverse. Thus, as time progresses, entropy increases, until it (as far as we know) reaches its maximum — the heat death or Big Rip.

Consider the Big Bang, when entropy was at its minimum, or at least incredibly low. As the Universe evolved, entropy increased, as dictated by the second law. If the Big Crunch were to happen, we'd see a decrease in entropy as conditions in the Universe recede back to how they were at $t = 0$. As we've stated, the Universe's entropy can never decrease.

But say we ignore the whole "Big Crunch problem" and instead focus on the cycles. As stated before, the entropy always increases with time. I'll let Baum & Frampton (2006) explain the problem:

  ...one principal obstacle is the second law of
  thermodynamics which dictates that the entropy increases from cycle to cycle. If the cycles thereby become longer, extrapolation into the past will lead back to an initial singularity again, thus removing the motivation to consider an oscillatory universe in the first place.

As they described, the entropy in the Universe would increase with each cycle. At this point we realize that each cycle in the Universe's history would be different — and if we extrapolated back into the past, we would find that at some point, entropy must have been at its minimum (the Big Bang!) The fact that we would be back to the Big Bang would remove any need for an Oscillating Universe hypothesis — which tries to explain what preceded the Big Bang.

That problem led many astrophysicists to abandon the Oscillating Universe hypothesis. But nonetheless, the idea that the Universe is reborn is quite attractive, and we still know very little about cosmology — including dark energy, which we think has a vital role in the Universe. Some cosmologists continue to research the Oscillating Universe model in hopes that, as we learn more about dark energy and the Universe, we may be able to avoid the problems with the second law of thermodynamics.


Surface brightness of deep-sky objects and telescope specs

Although you have not given any information about your telescope, the Orion nebula is easy to see with almost any telescope, even under urban light polluted skies.

Surface brightness is almost always greater (ie fainter) than magnitude - so for example the total magnitude of the Orion nebula is about 4 (from memory) but you say surface brightness of 13.  The reason is that the area of the Orion nebula is quite large, at least when measured in arc seconds squared, and the eye/brain has some ability to integrate the light.

This works both ways: there are objects, eg M33 and M101 in my experience, which have a magnitude that would make you think they are easy targets for small telescopes, but because the brightness is spread over a large area, they are difficult to see from a typical urban or suburban location.  The important measure in these cases is to compare the surface brightness with that of the background sky.

It's quite a complex topic, and Roger Clark's book "Visual Astronomy of the Deep Sky" has some useful discussion of this, and much more.


What does correcting for LSR mean in galactic velocities?

Proper motions are not velocities. They are angle traversed per unit time. They can be converted to tangential velocities (i.e. tangential to a line joining the star and the Sun) if the distance to the star is known, but you need additional information to get the line of sight (radial) velocity and hence a 3D velocity.

Such velocities are with respect to the Sun (or possibly the solar system barycentre - a tiny difference).

The local standard of rest (LSR) is an estimate of the average motions of stars in the solar neighborhood. The Sun has a peculiar velocity with respect to the LSR. To get a velocity with respect to the LSR, you need to subtract the solar motion with respect to the LSR from your heliocentric velocities.

There are disagreements in the literature (at the level of 1 km/s) as to what the solar motion with respect to the LSR is.


Why type Ia supernova reaches its peak absolute magnitude(~ -19.3mag) at around 15-20 days after the explosion?

The early light curve of a Type Ia SN (i.e., the fact that the brightness increases to a peak, then declines) is the result of the combination of two things, both of which are decreasing over time: energy from the decay of Ni-56 to Co-56 (and Co-56 to Fe-56), and the opacity of the explosion as it expands.

Just after the actual explosion, the expanding fireball is dense and opaque. Only light from the very outermost layer escapes to the outside.[*] Since the Ni-56 created in the thermonuclear burning that caused the explosion is in the interior regions, the initially visible part of the expanding fireball has little or no Ni-56, and thus no energy from its decay.

As the fireball expands, it gets less dense, and thus less opaque, to the point where light from the layers heated by radioactive decay can actually escape. Over time, more and more of the inner part of the fireball becomes visible, and so we see more and more of the regions where Ni-56 (and its daughter product Co-56) is decaying. Roughly speaking, the peak of the light curve is when the opacity is low enough that all the light from the Ni-56 + Co-56 decay can escape.[**]

But since there are fewer and fewer radioactive nuclei as time goes by, the amount of energy produced by the decay decreases with time. This is why the light curve turns over: now we're seeing all the light produced by Ni-56 and Co-56 decay, but that just keeps getting fainter over time.

[*] At this point, the SN is too faint to be seen. Even the initial thermonuclear explosion is too faint to be seen under most circumstances.

[**] Ni-56 and Co-56 actually produce gamma rays (and energetic positrons) when they decay, so what "the light from Ni-56 + Co-56 decay" means is thermal radiation from the gas heated by the gamma rays. Also, as I understand it, what we're seeing is really the combination of energy from current decay and residual heat from previous radioactive decay.


Can the WISE telescope detect black holes?

Black holes are black. They are only observed directly by telescopes if they are accreting matter. Any radiation observed comes from the matter surrounding the black hole. Generally speaking, the smaller the black hole, the hotter the accreted material becomes. For something of planetary mass, one might expect X-rays and Gamma rays from accreting material.

A non-accreting black hole would produce no signature at any wavelength and there is very little material to be accreted in the outer solar system, except perhaps a small possibility of seeing X-rays if something like an Oort cloud object interacts with the black hole.

It is possible that such a black hole could be detected via its lensing effect on light coming from distant stars, but since the black hole would be very small as seen from Earth, this is unlikely, and WISE would not be the instrument to do that.

EDIT: Just to prevent any confusion. "Naked" black holes are hypothesisd (but not yet observed) to produce Hawking radiation. Planet X, if it exists, is thought to be about ten times the mass of the Earth. Such a black hole would have an event horizon about 20 cm in diameter and would emit Hawking radiation with a luminosity of $10^{-20}$ Watts at a blackbody temperature of around 0.002 K.  This is not going to emit any detectable infrared radiation.


Have we sent radio signals to exoplanets that may be habitable?

Short answer : Yes
Long answer : Yes
We have actually sent radio signals to distant exoplanets that are thought to be capable of hosting (hopefully intelligent) life.
Some examples?

The radio signal directed at a world 12 light-years away included music and math lessons from Earth
Are There Any Habitable Exoplanets That Can Hear Us?

The second link only 'mentions' a planet that had been sent radio planet :

Those waves, which are all electromagnetic like light, would have since traveled about 70 light years away from Earth. The newly confirmed Kepler 452b, at 1,400 light years away, will not be seeing 'I Love Lucy' for more than a millennium.

Hope this answers your question.


Is there any possible connection between Gamma Ray Burst and Fast Radio Bursts?

There is a tentative detection of a GRB simultaneously with an FRB,  DeLaunay et al 2016. If it is true, it would mean that at least some FRBs have the same origin as some GRBs.

Theoretically many possibilities were suggested before and after this tentative, detection. For now they all remain quite speculative, and more observations are required (and planned)
For some selection of models, it could be interesting to look into citations to this paper or its references.


There's a pretty comprehensive wikipedia page on the subject. In summary....

Throughout the 2000's conflicting results were published using the same data. The conflict seems to center on the selection effects of the surveys, and the corrections applied to account for them. I did a quick search for 'redshift quantisation' on the Arxiv and found no papers since 2009, which suggests the supporting data has been comprehensibly discredited, or explained away by natural mechanisms.

There are two explanations for the quantization that do not require exotic cosmologies. The first is the large scale clustering of galaxies that traces the cosmic web. The walls and filaments formed in this web leads to groupings of galaxies around specific redshifts along the line of sight. The second explanation is the quantisation of emission from Hydrogen, known as the Lyman series. Since photons can only be released with certain energies, the measured redshifts are also quantised (this is only true in low pressure and temperature environments, otherwise pressure and thermal broadening occur).


Is there a term for "path of the Sun at a given point of the Earth"

Merriam-Webster defines diurnal arc:

the portion of the diurnal circle of a celestial body that is above the observer's horizon

and diurnal circle:

the apparent circle or parallel of declination described by a celestial body in consequence of the earth's rotation

The NOAA Solar Calculator does not implement the exact feature in question but can show the Sun's azimuth for any given location, date, and time.


In the currently dominant theories, the Universe is basically the same everywhere, if you look on a large enough scale. There may not be a furthest star from Earth at all (the universe may be infinite) or it may be like "the furthest point on Earth from London" which exists, but is not a specially interesting place.  The distinction between those two possibilities is exactly whether the universe "curves in on itself" (in which case it is finite) or doesn't in which case it is infinite. We don't know for sure, but any curvature is definitely pretty slight.

To see the universe as uniform, though, you do have to look on quite a large scale (billions of light years). On smaller scales than this the distribution of galaxy clusters (and so galaxies and so stars) is quite uneven, and there are places where there are no stars close enough to see with the naked eye. Nevertheless, with a big enough telescope in one of those places you would still see stars scattered in all directions.

You mention the big bang, and what you write suggests a common misconception -- that the Big Bang happened in the middle of a big empty space into which matter expands. That is not what the theory suggests. Instead all points in space, when you trace things back far enough, seem to have contained a very hot, very dense, sea of matter, which, as far back as we can extrapolate, was all spreading apart, creating new space uniformly spread throughout the old space in the process.


According to the models of Mordasini et al. 2012 it is about 13 Earth radii (about 1.2 Jupiter radii) for gas giants older than a billion years.

Younger gas giants that are still contracting significantly can be bigger. How much bigger is wrapped up in uncertainties in the formation mechanism and the core mass and composition.

Gas giant mass-radius relationships from Mordasini et al. 2012. The colour coding corresponds to the fraction of heavy elements in the planet. The larger planets have smaller fractions of heavy elements.


How strong is the force between the Sun and the centre of the Milky Way?

Galactic orbits are not Keplerian: There is not a single massive centre, whose gravity attracts the sun, rather the whole disc, and dark matter halo which surrounds the galaxy. We can't use the inverse square law to calculate the force of gravity without knowing the distribution of mass in the galaxy.

Nevertheless, the Sun's orbit is roughly circular, so we can use kinematics to get some idea of the forces involved:  For circular motion $a=\frac{v^2}{r}$. The velocity of the sun is about 225000 m/s, and we are at a radius of about 2.5e20 m from the centre. The formula above gives a very small centripetal acceleration of 2e-10 m/s²

However the sun is quite massive, 2e30 kg, so using $F=ma$, the force on the sun is of the order 4e20 N.  This is about 0.01 of the force acting on the Earth by the Sun. (3.6e22 N)


Eagle Nebula (Pillar of Creation) destroyed?

It's a bit hard to say for sure as the article isn't terribly clear, but what I think they're saying is that there is a dust cloud to the upper right side of the Pillars (as viewed in the photo.) They further say that the dust cloud shows evidence of a supernova shock wave passing through it, presumably from the upper right and presumably from a supernova.  (Note that the supernova was not observed, and they don't report on a remnant being visible.) Then they say that when that shockwave reaches the Pillars, the Pillars will be destroyed.  They estimate that that will be observed from Earth in the next thousand years or so.  (Note that they don't report measuring of the speed of the shockwave -- I'm guessing they assume it's moving as typical speeds.) Since the Pillars are about 7000 light years away, than means that this destruction "already" has happened.

There's nothing impossible about all this, but it looks like speculation based on limited actual facts.  It's also worth mentioning that what you see on the web went through a press office, and press offices are paid for doing publicity, not for doing science.  The quality of what a press office posts (already undermined by being a popularization, which is hard to do well even when you're trying to be accurate) is frequently biased by a publicity-at-any-cost attitude. Always read press releases and the like with your sceptometer turned up to High.

Personally, I'd like to see stronger evidence for the existence of the shockwave, and evidence that the shockwave will actually disrupt the Pillars. (The Pillars are considerably denser than the medium around them, and so may be resistant.)


There is no such unique point, because a key element of the Big Bang model is called the "cosmological principle," which means that every point in the universe is taken as essentially equivalent to any other.  The idea there is that as the universe ages, it changes, but it changes in the same ways everywhere so there is nothing going on that distinguishes one region from another.  This is a simplifying assumption that is supported by the fact that we see no boundary to the universe, and we see no evidence that anywhere else is different from here, the differences we see appear to be due only to the fact that we are seeing farthing into the past as we look farther away.  This assumption means you could pick any point you want, point to it, and say "the Big Bang started there," and you would be no more right nor wrong than if you pointed anywhere else.


What was James Bradley's calculation to calculate the speed of light?

Despite many online sources crediting James Bradley with having calculated the speed of light to a 295.000 km/sec (britannica) or 301.000 km/sec (wikipedia), a small nuance ought to be added. Like Ole Romer before him in 1676, James Bradley calculated the speed of light relative to something else but both never provided a value in Earth-based units.

He wrote,

  This therefore being 20’’,2, AC will be to AB that is the Velocity of Light to the Velocity of the Eye (which in
  this Case may be supposed the same as the Velocity of the Earth's
  annual Motion in its Orbit) as 10210 to One from whence it would follow
  that Light moves or is propagated as far as from the Sun to the Earth
  in 8’ 12’’.

  (source)

The calculation to get to the ratio of 10210 is,

size of a radian (206265’’) / angle of aberration (20’’,2)

This ratio also holds for the speed of light to the speed of the earth. Not sure though exactly how britannica and wikipedia arrive at their respective values.


Has the effect of other planets on Earth's orbit been ever measured?

For decades, geologists have understood that climate changes are due in part to gradual shifts in the Earth’s orbit, which are caused primarily by Venus for its proximity and Jupiter for its mass, and repeat regularly every 405,000 years. In 2018 a team of geologists and Earth scientists unearthed the first evidence of these changes – sediments and rock core samples that provide a geological record of how and when these changes took place.

https://www.pnas.org/content/115/24/6153
"Empirical evidence for stability of the 405-kiloyear Jupiter–Venus eccentricity cycle over hundreds of millions of years"

So given the timescales involved, this is as close to measuring this effect as one could expect to get.