Are there any Stars we know don't have planets?

I am beginning to assume that our solar system is not unique and that every star has several planets.

Not quite, but indeed a study published in Nature in 2012 found that, based on our observations so far, roughly 17% of stars host Jupiter-mass planets, 52% host "Cool Neptunes" and 62% host Super-Earths. (Note that these percentages do not add up to 100%, because they are not mutually exclusive possibilities). This was particularly surprising, because half of all visible stars are believe to be in binary systems, which would make planetary systems very unstable, but some binary systems have been found to have planets too.

So indeed it seems the majority of stars have planets, but it's very unlikely that all of them do.

However, the exact answer to your question "do we know of any stars with no planets" has got to be "no", because there remains a possibility that they have planets that we simply haven't been able to detect, because of limitations in our techniques to detect them.


Does this video catch an Eta Aquarid - Earth skimming meteor?

Things entering the atmosphere (whether a meteor or an atmosphere-skimming object) travel too fast to be visible for 10 minutes.  Normally a pass of a satellite (at above atmosphere height) takes 2-3 minutes and most meteors are over in seconds.

Even the famous daylight fireball from the 1970s was over in a very short time.

I would guess what you have there is a plane at high altitude, though it's moving a bit slower than I would expect. (I'm open to suggestions.)


Attach a visible light telescope to the outside of the ISS

A self-sufficient orbiting telescope is basically Hubble mkII and would never get off the ground, literally and metaphorically

Hubble was expensive because it was state-of-the-art, requiring development of many new systems. The systems it needed to function as a standalone satellite (compared to being attached to the ISS) were cheap by comparison (reaction control, power, communications) because they mostly didn't require new development.

So a Hubble successor attached to the ISS would not be much cheaper than a free-flying one.

Any telescope attached to the ISS has to deal with:

high vibration levels which degrade performance
outgassing which degrades performance
severe pointing constraints because it's surrounded by a large structure
bandwidth constraints because it has to share a transmitter with ISS operations
end of mission in 5 years when the station reaches the end of its life.

The advantages of attaching to the ISS:

easier servicing,
you get the option to store lots of data on the ISS and bring it down on a capsule. Downside is a latency of several months.

IMO you'd be better off building a standalone space telescope.


What are the criteria for the classification of galaxies?

There are three main classes of galaxies: Irregulars, Ellipticals, and Spirals. Irregular galaxies, as their name suggests, do not fit into the "normal" classification scheme.

So, how do we distinguish between elliptical and spiral galaxies?

Brighness profile

The radial brightness profile of an elliptical galaxy follows a deVaucouleur law ($r^{1/4}$).

Spiral galaxies have an exponential radial brightness profile, although their central regions ("bulge") also follows a deVaucouleur law.

Star formation

Stars are formed in the spiral arms of spiral galaxies (and can be formed in irregulars), while elliptical galaxies tend to only have old, and consequently low mass, stars.

Components

As far as we can tell, all galaxies consist of a dark matter halo and stars. In addition, spiral galaxies also have clouds of dust and gas. If conditions are right, these can form new stars. (Some ellpticals have a very thin, very hot gas component as well, but there is a lot less of it than in a spiral galaxy).

Kinematics

Spiral galaxies are rotationally supported, while elliptical galaxies are mainly pressure-supported (i.e. they act like an ideal gas, with stars as gas molecules). There are some rotational features present in ellipticals, but they tend to be minor compared to the overall random motion.

A graphical overview of the various galaxy types is usually shown in the Hubble tuning fork diagram. Note that this does not indicate an evolutionary progression from one type to the next.


What was Bradley's observation concerning abevation of light?

You are right that what you must use as a referent is different-- for parallax, your referent will be a much more distant object, because it shows little parallax.  In aberration, it happens to the whole visual field.  But what is shifted is the angle relative to the ground at which the light arrives, so if you set your telescope to see a given star, in principle you would need to correct for both parallax and aberration as the Earth orbits. But aberration is usually much larger than parallax because it scales with v/c rather than D/d (where D is 1 AU and d is the distance to the star).  For the Earth's orbit, v/c is about $10^{-4}$, but stars are much farther than $10^4$ AU.  As for Bradley's observation, a good version is given at http://cseligman.com/text/history/bradley.htm.  He was looking for the parallax effect, and stumbled into the much larger aberration effect.  He could tell it was something different because the sign was opposite and it did not depend on distance to the source.  This was 50 years after Romer, so the speed of light was already known.


I'm certainly not an expert, but from what I understand, the type of observation done in helioseismology might not work with gas giants, hot jupiters, ice giants, etc.

The sun is dense enough and it has enough gravity that, while its a plasma, it's relatively solid most of the way through.   Unlike Red Dwarf stars which are lighter and undergo convection (0.35 solar masses and under are modeled to be fully convective), the sun is layered and doesn't undergo convection.

While oversimplified, the sun behaves like a drum in regards to it's internal soundwaves.  The sun is also very very loud, inside and on the surface.

If you played a drum in a vacuum, you couldn't hear it, but you could observe the motion of the drum's surface and work out what it would sound like by visual observation only.  Same with guitar strings.  The vibration is visible in wavelength and amplitude.

Exaggerated, the pattern on a drum would look something like this.

Source of picture.

That's basically how the Sun's soundwaves are studied.   It's often compared to Sonar but I don't like that comparison because sonar works by sending a soundwave out and seeing what gets reflected back.  It's more like the visual observation of sound vibrations, or, like studying a drum by looking at it's vibrations instead of listening to it.

This is a picture of what those vibrations look like on the sun (again, greatly exaggerated).

Source.

As noted in the article, a drum has one node.  The sun has many nodes, so it's more complicated.  It's the studying of those many nodes that gives insight into the inside of the sun.

The problem with doing that on a gas giant planet is, (and I'm partially guessing here), I don't believe a gas giant planet is solid enough to behave like a drum.   The convection (and gas giant planets are full of convection in their upper layers) would likely dilute any studyable patterns or nodes from it's interior, much like how wind reduces sound waves.   I don't believe this method would work on gas giant planets.


What orbit is the smallest or takes the least time to complete?

In general, the closer you are to the primary, the shorter the orbital period, except of course that you can't orbit inside the primary. So, taking the primary to be a sphere of radius $R$ and density $\rho$ we find that the mass is $$\frac{4}{3}\pi R^3\rho$$ so that the acceleration due to gravity at the surface is $$G\frac{4}{3}\pi R\rho$$ For a surface-grazing orbit, this has to equal $R\omega^2$ where $\omega$ is the angular velocity, so we get $$\omega^2 = \frac{4}{3}\pi G\rho$$ Now the orbital period $P$ is $2\pi/\omega$ so we get  $$P = \sqrt{\frac{3\pi}{G\rho}}$$
So the answer, at least for spherical primaries) is going to be a close orbit around the densest object in the solar system. The densest planet is Earth!
There are denser objects, such as the LAGEOS satellites, but I'm not sure if it is possible to orbit LAGEOS under its own gravity, because of the perturbing effects of Earth's gravity.


What is the "hypothesized lower mass gap" between 2.5 and 5 solar masses?

The "mass gap" is an observed deficit in the number of compact objects with mass between 2.5 and 5 solar masses.
The "mass gap" is/was not understood. Such objects may be rare because they are difficult to detect or because something about the supernova process leads to a bifurcation between the most massive neutron stars and the least massive black holes.
The Ozel et al. references in your question fills in the details. Also see https://astronomy.stackexchange.com/a/8280/2531


What types of star do we know are hostile to life (like ours)?

We so far have a sample size of 1. We know that a largish star, with a surface temperature at about 4.5 billion years of 5800K can be stable enough to support life for long enough for intelligence to evolve. We are fairly sure that very large stars don't last long enough.

With only one data point to extrapolate there is a tendency to think that stars that are like the sun offer the best environment for life. We are fairly sure that very large stars don't last long enough. It seems reasonable to exclude highly variable stars, or ones whose outbursts would strip the atmosphere from any planet. Intuitively these would seem to be less likely sites for life. But there is no strong evidence in the form of multiple discoveries of extra-terrestrial life to support this.

In short we don't know what stars can support life.


What kinds of stars would it be possible to observe in M31 with a small amateur telescope?

LBV stars are very powerful, very large and bright, and even though they may be near the end of their life, still young - since they don't survive for long. These are problematic.

Their powerful solar winds will clear the region around them, meaning that planets are less likely to form close to the star. It is much less likely for a LBV star to be aligned correctly to allow a transit to be visible from Earth. A hot Jupiter close to such a star would rapidly lose mass due to the extreme temperature.

If a transit does occur, because the star is so large, it obscures a small proportion of the light, so a the change in brightness is small. And of course they are intrinsically variable themselves, so detecting the very small signal is harder.

Since they are still young so planets may not have finished forming by the time they go supernova.

We haven't discovered any planets orbiting LBVs in the Milkyway.

Most stars that have known planets are F,K and G spectral type, this is partly sample bias (we look for planets around sun-like stars) But such stars would not be visible in M31 using amateur equipment.

An 8in telescope has a limiting magnitude of about 14, (E. Hubble was observing Cepheids with a mag of 16+) but you are already  close to the limit of the telescope, and distinguishing the very small changes is consequently is harder. Given the difficulties, it is perhaps unsuprising that we haven't found many planets more that 10000 ly away.


Why do aliens have to be carbon based lifeforms?

I want to start with a disclaimer that I'm not a chemist and hopefully I don't screw any of the chemistry up. If I do, please let me know.

I think Bill Oertell hit the nail on the head. Just because you can imagine a species which can drink lead doesn't mean it's actually physically possible. Only specific atoms can form the basis of the (extremely) complex organic molecules needed to form life and Carbon happens to be the atom which can do that. The primary reason for this is that Carbon has four valence electrons and its outer shell can hold eight, allowing it to form strong and varied bonds that most other atoms just can't form. As such, Carbon is the most logical element to form the backbone of organic molecules.

Other ___-based Lifeforms

Moving down the periodic table, you can see that Silicon has the same properties (4 valence elections, 8 possible) and you might be inclined to think that you could form some sort of silicon based life. This topic has been widely considered, but you run into a number of problems that make Silicon-based life unfeasible. The primary reason is that Silicon is just much bigger and more massive than Carbon (having a nucleus with 28 nucleons compared to Carbon's 12). This makes it harder for Silicon to form the necessary bonds that Carbon can readily form and you're out of luck right form the get-go. Similar issues occur with other "Carbon group" elements, except they're even further exacerbated by even higher masses.

There are many other "organic" configurations chemists have considered such as boron-based life or sulfur-based life but they all have some sort of problem that makes it hard for it to exist. Some are more possible than others, but Carbon-based life, as we know it, just seems to be the best solution. If you gave someone all the elements in the periodic table and said, make organic life, they'd probably come back with Carbon-based life because its the easiest. You don't have to overcome as many obstacles.

Observations

Aside from the impracticality of non-Carbon based life, you have to consider that everything we've seen in the universe, as far as organic chemistry goes, points to the fact that the Universe is filled with Carbon-based chemistry. Astronomers have been able to look out at comets, interstellar dust clouds, and all sorts of places and time after time, they see complex, Carbon-based, organic molecules. For example:

We've found Polycyclic Aromatic Hydrocarbons (PAHs) all over the solar system and galaxy,
Astronomers have also found things like i-propyl cyanide in meteorites and in interstellar dust $-$ an organic molecule which is a necessary ingredient in amino acids,
Organic molecules have been found (and believed to have formed) on Ceres.

The point I'm driving at here is that the universe seems ready-made to start forming Carbon-based life everywhere. Everywhere we look, we see organic molecules which have Carbon as the backbone of their structure. Nowhere do we see complex silicon based molecules (or something even crazier). This is highly suggestive that Carbon is the way to go if you want to quickly and easily form life.

There are probably a 100 other reasons to favor Carbon over something else, but that delves into more complicated chemistry than I've touched on here and I'm not as knowledgeable about that. I would think this question would get a more in-depth chemistry answer on the chemistry stack exchange.


Intro for the uninformed: A standard candle is an important concept in astronomy, helping to map out distances in the Universe. Since the observed flux $F$ of a light source decreases with distance $r$ by a known factor ($r^2$), if we know its intrinsic luminosity L, we can calculate the distance. For large distances, where bright sources are needed, we usually use supernovae (SNe). But the luminosity of a SN depends on the mass of its progenitor star which is not known in general. However, for a specific type of SNe — "type Ia" — it is known: This type are SNe that explode when a white dwarf accreting mass from a companion star exceed the mass threshold for explosion of $1.4\,M_\odot$.

In addition to the gravitational waves discussed by Rob Jeffries, I can mention the following candidates for standard candles:
Type II supernovae
Type Ia SNe are so similar in luminosity because they all have (almost) the same mass when they go off. But there is also evidence that type II SNe can act as standard candles. As is also to some extent the case with Ia, their lightcurves (how the luminosity changes with time) are not identical, but can be standardized using the so-called Philips relation (see e.g. Kasen & Woosley 2009).
GRB supernovae
Gamma-ray bursts as a whole are too diverse to be used as standard candles, but when an associated SN is detected, it becomes a standard(izable) candle (Li & Hjorth 2014).
Quasars
My favorite candidate are quasars. This technique doesn't rely on the Philips relation. Quasars are caused by gas accreting onto a supermassive black hole in the center of galaxies, resulting in an "active galactic nuclus" (AGN) with extreme energy outputs (easily over $10^{12}\,L_\odot$, and even up to $10^{14}$–$10^{15}\,L_\odot$; Ibata et al. 1999). It turns out that there is a correlation between the absolute luminosity of the AGN and the size of its broad-line region (BLR), i.e. the region around the quasar where fast-moving gas clouds absorb the continuum$^1$ of the quasar and emit lines$^2$, e.g. H$\alpha$ (Watson et al. 2011). The reason is that the size of the BLR is determined by the depth that the ionizing radiation from the quasar can penetrate into the BLR, which is proportional to the square root of the luminosity. The figure below (from Watson et al. 2011) shows the relation between distances ($D_L$) determined by this technique and distances obtained from type Ia SNe.

Advantages over supernovae
A huge advantage of quasars over SNe is that they don't disappear after a few weeks, meaning that if e.g. we want to refine some measurement, we can go back and observe it again at any time. Another advantage is that quasars, being so luminonous, can be detected out to much larger distances (roughly to $z\simeq4$) than SNe (which are only observed ut to $z\simeq2$).
Reverberation mapping
Since quasars are so far away, the BLR, being less than a parsec in size, cannot be resolved. But luckily, a technique call "reverberation mapping" allows us to determine the size:
The spectrum of the quasar consists of a continuum with spectral lines. Quasars vary in luminosity on rather short timescales. If we measure a quasar's luminosity regularly over some period of time, we get a so-called "lightcurve". But since the lines are created at a distance from the source, a given "bump" in the lightcurve (i.e. a temporary increase in luminosity) does't show up in the continuum and the lines at the same time. Instead, there is a delay, corresponding to the extra distance that the light had to travel from the quasar to the cloud reflecting$^3$ it.
In the figure below, blue shows the continuum, while red shows the lines. The lines are seen to lag behind the continuum, since they first had to travel from the quasar (black) to the clouds (magenta).

Since the clouds lie at a range of distances, they exhibit different time lags, effectively broadening the line:

So, in the figure above, showing observed flux as a function of time, the light in the line increased in luminosity roughly 1.5 days later than the light in the continuum. This means that the BLR is roughly 1.5 lightdays (or ~250 AU) in radius.

$^1$I.e. the continuous and relatively featureless spectrum of many different atomic transitions and physical processes.
$^2$I.e. features in the spectrum resulting from strong atomic transitions. For instance, an eletron falling from the second to the first excited states of a hydrogen atom emits a photon with the wavelength 6563 Å, called "H$\alpha$".
$^3$The light isn't really "reflected". Instead, it is the high-energy photons (UV and X-rays) of the continuum that ionize the atoms in the clouds. When the ions recombine, they emit spectral lines.


Original Hubble diagram units incorrect?

It is in fact supposed to be in $\mathrm{\frac{km}{s}}$ (KILOMETRES per second).

I think that whoever authored the paper, probably took it as a random shortcut and assumed anyone reading it would either have read the paper and understand it is supposed to be $\mathrm{\frac{km}{s}}$, or just assume it from the fact it says velocity.


Can you see the difference with the naked eye whether Venus is at the other side of the Sun or at the Earths' side?

The apparent brightness of Venus does vary.  The distance and phase both affect the brightness, and the light doesn't reflect equally in all directions (though this effect is less significant than in the case of Mercury)

The maximum brightness of Venus occurs at a phase angle of 125 degrees. This is a crescent (0 is full, 90 is half lit and 180 degrees is when Venus transits the sun.)

The brightness curve doesn't vary extremely

In this image generated from NASA data by David Barry you can see the peak brightness is about -4.9.  There is a short central dip. The brightness drops to about -4 as Venus gets further away (even though its phase increases) and note the small rise when Venus is most distant (due to more light being reflected directly than at an angle). These occur when Venus is close to the sun in the sky, and so the brightness can be lost in the dawn or dusk sky.

Judging the difference between -4.9 and -4 is rather difficult, especially as you are not viewing it against a dark background.


UY Scuti is a red supergiant star.

When stars start to run out of hydrogen fuel, their cores start to collapse, causing the core of the star to heat up, and heavier elements start to be used as fuel. This means that the core of the star is hotter and producing more energy.

The effect on the outer layers of the star is to cause them to expand, and as things expand, they cool. The outer visible layer of the star cools (to about 3500K)

So the energy output of the star increases, but the temperature of the outer layer decreases. As you might expect this leaves the star unstable, and UY Scuti undergoes semi-regular variations in brightness.


How come moon's difference in thickness between sides does not create an unbalance on its mass?

As is the case on Earth, the Moon's crust is less dense than the material that comprises the Moon's mantle or core. On the Earth, this means that areas with thick crust rise above sea level. (These are our continents.) On the Moon, this means that the Moon's center of figure (it's center based on surface topography) is offset by a couple of kilometers from Moon's center of mass, with the center of mass closer to the Earth than the center of figure.

This center of mass to center of figure offset does not mean much with regard to the Moon's stability. What matters much more is the distribution of mass top to bottom vs side to side vs front to back. To be stable in a tidally locked orientation, a moon's principal axes deduced from the moment of inertia tensor should be very close to the set of axes as seen from the planet.

Moreover, a very specific ordering of the moments about the principal axes as deduced from the inertia tensor needs to exist. For the moon to be in a stable tidally locked configuration, the axis of rotation needs to be more or less orthogonal to the moon's orbital plane, the moment of inertia about the rotation axis needs to be the largest, and the moment of inertia about the line from the center of the planet to the center of the moon needs to be the smallest, leaving the moment of inertia about the side-to-side as the intermediate axis.

This most definitely is the case with the Earth's Moon. The angular difference between the Moon's geometric axes as seen from the Earth and the Moon's principal axes deduced from its inertia tensor (in technical terms, the angular difference between the Moon-Mean Earth frame and the Moon's principal axis frame) is a few hundredths of a degree.

That the Moon's far side has a much thicker crust than does the near side is a third order effect. The Moon would be nearly as stable as it is now if it was the Moon's far side with its much thicker crust that faced the Earth rather than facing empty space.


How much matter was ejected when the Solar System formed?

It is actually the other way around:
First a massive accretion disc can form, through which material looses angular momentum and accretes onto the star radially, hence being angular momentum poor.
However, during the initial free-fall phase, before the disc forms, infalling material can be 'rejected' at the star, either via high pressure gradients or magnetic interactions. This causes a polar jet to form around the protostar, which emits material back into the interstellar medium. Measured jet mass-loss rates are up to 1/2 the measured accretion rates.
Furthermore, the remaining fraction of material that did not form planets is ejected at the end of the disc lifetime via photoevaporation by the young stars' X and UV photons. What fraction this is in comparison to the mass in the planets is unknown, as the overall planet forming efficiency is unknown.


How do they know the non-uniformities seen on the disk of π¹ Gru are real and not artifacts?

The paragraph you quote continues...

  The images from SQUEEZE and MiRa have very similar characteristics and are shown in Fig. 1 (see also the Image Reconstruction section in Methods)

In that section various tests are described including the reconstruction of simulated data, with similar interferometric coverage and signal to noise, from spotted and spotless (uniform) stars with limb darkening.


Could our Sun be the product of an ancient stellar collision?

The dynamics of the Solar System and the chemistry of the Solar System bodies don't support a hypothesis of a stellar merger later than formation of the protoplanetary disk which would have mixed-up things considerably and heavily disrupt any circumstellar disk. Thus this basically excludes any collision after the time one can start talking about a protostar, way before it even entered main sequence (as that's already the end of planetesimal formation).


It looks to me that you probably have some misunderstanding of the term meteorite. A meteorite is a remnant body which has reached the surface of a major or minor body (a planet, a moon, an asteroid, etc.). The meteorite's parent body (a meteoroid) would definitely lose its entire water ice (if it had any at all) no later than during its passage through the Earth's atmosphere. That's why no water ice was or will ever be found in meteorites.

So, unless you compare common meteorite impacts with hazardous asteroid impacts, these two body types are incomparable as water ice carriers. And thus, the statement by O'Brien et al still stands firm and will do so ever.


What is the actual temperature of a quasar?

A question/answer over at physics.se  discusses the issue of quasar spectral output.  Part of one answer says,

  The percentage of radiation that is thermal versus non-thermal varies
  from quasar to quasar, and for a single quasar is going to be
  different at different wavelengths, so there isn't one simple answer
  (and also it depends on what angle you are viewing the AGN from
  relative to the jet axis)

  And similarly, there isn't a single value you can quote for the
  accretion disk temperature. Different AGN have different black hole
  masses, different matter infall rates and generally different
  environments. And a single accretion disk doesn't have a single
  temperature either: the temperature is higher towards the center, but
  to the details of what the emission spectrum looks like depends a lot
  on what parameters you put in to the model and what assumptions you
  make about energy transfer within the disk etc.

So it's not really accurate to estimate quasar thermal temperature from the "Brightness temperature," because the latter only applies to objects closely fitting Planck's black body law.


What does a black hole elements consists of?

If your question is "What elements are in a black hole?"
The answer is: The gravity is too great for there to be "elements" (atoms or molecules) inside or near the surface of a black hole. The extreme gravity tears apart everything and compresses it.
Some distance away (especially far away, here on Earth) the black hole doesn't alter matter, but as things get closer they fall at increasing speeds to the surface, becoming a part of the black hole. Everything becomes crushed, becoming elementary particles; what atoms are made of.
NASA's webpage: "NuSTAR Probes Black Hole Jet Mystery" explains that plasma, x-rays, and eventually light can leave a black hole at the poles. See also: "An elevation of 0.1 light-seconds for the optical jet base in an accreting Galactic black hole system" (26 Oct 2017) which says:

"Relativistic plasma jets are observed in many accreting black holes. According to theory, coiled magnetic fields close to the black hole accelerate and collimate the plasma, leading to a jet being launched. Isolating emission from this acceleration and collimation zone is key to measuring its size and understanding jet formation physics. But this is challenging because emission from the jet base cannot be easily disentangled from other accreting components.
Here, we show that rapid optical flux variations from a Galactic black-hole binary are delayed with respect to X-rays radiated from close to the black hole by ~0.1 seconds, and that this delayed signal appears together with a brightening radio jet. The origin of these sub-second optical variations has hitherto been controversial.
Not only does our work strongly support a jet origin for the optical variations, it also sets a characteristic elevation of <~$10^3$ Schwarzschild radii for the main inner optical emission zone above the black hole, constraining both internal shock and magnetohydrodynamic models.
Similarities with blazars suggest that jet structure and launching physics could potentially be unified under mass-invariant models. Two of the best-studied jetted black hole binaries show very similar optical lags, so this size scale may be a defining feature of such systems.".

So, what goes in is not what comes out.
See also this article about AGN: "The largely unconstrained multiphase nature of outflows in AGN host galaxies" (28 Feb 2018).
 Black Hole-Powered Jet of Electrons and Sub-Atomic Particles Streams From Center of Galaxy M87
Astronomical evidence

Relativistic jets may provide evidence for the reality of frame-dragging. Gravitomagnetic forces produced by the Lense–Thirring effect (frame dragging) within the ergosphere of rotating black holes combined with the energy extraction mechanism by Penrose have been used to explain the observed properties of relativistic jets. The gravitomagnetic model developed by Reva Kay Williams predicts the observed high energy particles (~GeV) emitted by quasars and active galactic nuclei; the extraction of X-rays, γ-rays, and relativistic e$^-$—$\,e^+$ pairs; the collimated jets about the polar axis; and the asymmetrical formation of jets (relative to the orbital plane).

See also:

The Penrose process (also called Penrose mechanism) is a process theorised by Roger Penrose wherein energy can be extracted from a rotating black hole. That extraction is made possible because [some of] the rotational energy of the black hole is located not inside the event horizon of the black hole, but on the outside of it in a region of the Kerr metric spacetime called the ergosphere, a region in which a particle is necessarily propelled in locomotive concurrence with the rotating spacetime. In the process, a lump of matter enters into the ergosphere of the black hole, and once it enters the ergosphere, it is forcibly split into two parts.
...
Although momentum is conserved the effect is that more energy can be extracted than was originally provided, the difference being provided by the black hole itself. In summary, the process results in a slight decrease in the angular momentum of the black hole, which corresponds to a transference of energy to the matter. The momentum lost is converted to energy extracted.
The maximum amount of energy gain possible for a single particle via this process is 20.7%. The process obeys the laws of black hole mechanics. A consequence of these laws is that if the process is performed repeatedly, the black hole can eventually lose all of its angular momentum, becoming non-rotating, i.e. a Schwarzschild black hole. In this case the theoretical maximum energy that can be extracted from a black hole is 29% its original mass. Larger efficiencies are possible for charged rotating black holes.

Click to animate.
An unrelated effect is described as Hawking radiation:

Physical insight into the process may be gained by imagining that particle–antiparticle radiation is emitted from just beyond the event horizon. This radiation does not come directly from the black hole itself, but rather is a result of virtual particles being "boosted" by the black hole's gravitation into becoming real particles. As the particle–antiparticle pair was produced by the black hole's gravitational energy, the escape of one of the particles lowers the mass of the black hole.

That shouldn't be confused with the Hawking-Penrose Singularity Theorem either.


When are gravitational waves emitted during a merger of black holes?

The gravitational waves are actually emitted as long as you have a quadrupole moment with a second time derivative that is non-zero. The quadrupole is defined as:
$$ Q_{ij}^{tt}(x) =\int\rho \left(x^i x^j - \frac{1}{3}\delta_{ij}r^2\right)\mathrm{d}^3x.$$
To simplify this, here is a rule of thumb: if you have a system that is not spherically symmetric and that is changing with time (like two black holes), you can expect gravitational waves to be emitted. For example, the system made of the Earth and the Sun is emitting ridiculously small gravitational waves (I think it is emitting a few Joules per year), but it is.

Because of how hard the waves are to measure, you can only expect to see them on Earth for black holes in their last stage, just before they merge, when the emission of gravitational waves is at its peak.

For the LIGO detection, the beginning of the signal as observed by our detectors was in the inspiral phase. This is the phase when the black holes are getting closer one to each other and losing gravitational energy (through the emission of gravitational waves), before their horizons touch. The end of the phase is actually when the black holes get within their so-called ISCO (Innermost Stable Circular Orbit), which is the smallest radius a massless particle could orbit in a stable fashion. Anything inside this orbit will slowly spiral inward and fall into the black hole.


Why have brown dwarf classes been dubbed L, T and Y?

The choice of L and T is explained in Kirkpatrick et al. (1999) 'Dwarfs Cooler than "M": The Definition of Spectral Type "L" Using Discoveries from the 2 Micron All-Sky Survey (2MASS)'. The principles behind the choice are given at the start of section 5.1:

In choosing a letter designation for the new spectral class,
three important points must be considered : (1) The letter
must be unambiguous, having not been used for any currently recognized spectral type. For example, though "N"
follows "M" in the alphabet, it would be a poor choice of
letter since it is used for a class of carbon stars. (2) The letter
must create a taxonomy that is clearly distinguished from
other types of astronomical objects. In this case, the letter
must be accepted by the entire community, both by
researchers involved in low-mass star and brown dwarf
science and by astronomers in general. Although stellar
spectroscopists might find "E0," "E1," "E2," etc., perfectly
acceptable as new spectral subclasses, extragalactic morphologists already recognize these designations as elliptical
galaxy types. (3) The letter must stand the test of time. For
example, choosing "D" to mean that these objects are
"degenerate" brown dwarfs would be flawed reasoning.
Some of these dwarfs certainly are substellar (see § 7), but
such a designation cannot be tied uniquely to any particular
spectroscopic trait. The designation should apply to spectral features alone and be free of physical interpretation.
Our understanding of the underlying physics may change
with time ; our choice of letter should be impervious to such
changes.

Applying these, they find the letters H, L, T and Y to be ok: the rationale for excluding the other letters is given in their Table 5:

Letter
Status
Notes

A
In use
Standard spectral class

B
In use
Standard spectral class

C
In use
Standard carbon-star class

D
Ambiguous
Confusion with white dwarf classes DA, DB, DC, etc.

E
Ambiguous
Confusion with elliptical galaxy morphological types E0–E7

F
In use
Standard spectral class

G
In use
Standard spectral class

H
OK

I
Problematic
Transcription problems with I0 (10, Io) and I1 (11, II, Il)

J
In use
Standard carbon-star class

K
In use
Standard spectral class

L
OK

M
In use
Standard spectral class

N
In use
Standard carbon-star class

O
In use
Standard spectral class

P
Problematic?
Incorrect association with planets?

Q
Problematic?
Incorrect association with QSOs?

R
In use
Standard carbon-star class

S
In use
Standard spectral class for ZrO-rich stars

T
OK

U
Problematic?
Incorrect association with ultraviolet sources?

V
Problematic
Confusion with vanadium oxide (V0 vs VO)

W
Ambiguous
Confusion with Wolf-Rayet WN and WR classes

X
Problematic
Incorrect association with X-ray sources

Y
OK

Z
Problematic?
Incorrect implication that we have reached "the end"?

They go with "L" as it is the closest letter to "M" that is still available. They prefer it to "H" because of the occasional use of the term "hydride dwarf" to refer to M subdwarfs which are dominated by CaH and MgH bands.
After that, the sequence is going alphabetically through the remaining available letters, so they put Gl 229B, which has a notably different spectrum to the other (L-class) brown dwarfs known at the time, into spectral class T. Spectral type Y took a few more years to show up in observations, and was chosen by the same principle.
If observations get to the point where it is worth erecting a new spectral class beyond Y then things will get interesting.


Converting from $\mathrm{W \: m^{-2}}$ to $\mathrm{Jy \: km \: s^{-1}}$

You are essentially correct, but here's a more in-depth explanation:
Both units are a measure of the total flux $F$ of a light source in some interval, e.g. a spectral line or a broader region given by a filter. The flux is obtained by integrating a spectrum’s flux density, i.e. the flux per "bin". A spectrum is a distribution of photons as a function of the photons' energy or, equivalently, wavelength or frequency. The bins are then whatever you choose on your $x$ axis.
Typically you will use frequency $\nu$ for low-energy photons (radio to microwaves, with units of e.g. MHz or GHz), wavelength $\lambda$ for mid-range photons (with units of µm for infrared, and Å for optical to UV), and energy $E$ for high-energy photons (keV for X-ray and MeV gamma-rays). The flux density can then be denoted $F_\nu$, $F_\lambda$, and $F_E$, respectively, although the latter notation is not common; these photons are often so infrequent that one simply uses "counts", i.e. the absolute number of individual photons.
Note though that astronomers often are quite slobby and uses the word "flux" when they actually mean flux density (as well as for intensity, radiance, etc.).
Doppler broadening
But there is another very common way to show a spectrum, namely as a function of $\mathrm{km}\,\mathrm{s}^{-1}$. If you consider a spectral line from some electronic transition, it is emitted at a particular frequency (almost a delta function). But because the atoms emitting the photon do not lie still, but have a random distribution of velocities $v$, they will broaden the line. To first order, an atom moving with a velocity $dv$ with respect to the systemic velocity Doppler-shifts its emitted photon by an amount
$$
d\nu = \frac{dv}{c}\nu_0,
$$
where $\nu_0$ is the line center frequency, and $c$ is the speed of light. This means that we can also show a spectral line centered on $0$, with positive and negative velocities on the $x$ axis corresponding to higher and lower frequencies.
This is an informative way to show physical processes such as temperature, turbulence or bulk motion of the gas.
Units
Although astronomers tend to like cgs units, we also dislike huge and tiny numbers with too many powers of tens, plus we're in general not very consistent. Historically, flux in optical and UV light is typically given in cgs ($\mathrm{erg}\,\mathrm{s}^{-1}\,\mathrm{cm}^{-2}$), but wavelengths are measured in Ångström, not cm, so flux densities are measured in $\mathrm{erg}\,\mathrm{s}^{-1}\,\mathrm{cm}^{-2}\,\mathrm{Å}^{-1}$.
On the other hand, radio astronomers tends to use SI units for flux ($\mathrm{W}\,\mathrm{m}^{-2}$) and flux density ($\mathrm{W}\,\mathrm{m}^{-2}\,\mathrm{Hz}^{-1}$). However, since typical astronomical sources are extremely faint, radio astronomers often measure flux densities in jansky (Jy), where
$$
1\,\mathrm{Jy} = 10^{-26}\,\mathrm{W}\,\mathrm{m}^{-2}\,\mathrm{Hz}^{-1}.
$$
Converting from W m–2 to Jy km s–1
Because of the fear of small or large powers of ten, and because as described above spectra can be shown as a function of both frequency and velocity, you can actually integrate you flux density in jansky over velocity, and thus the total flux of radio sources is periodically measured in $\mathrm{Jy}\,\mathrm{km}\,\mathrm{s}^{-1}$. With $F_\nu d\nu = F_v dv$, combining the equations above the relation is (beware of the similarity between $\nu$ and $v$)
$$
\begin{array}{rcl}
\int F_\nu d\nu & = & \int F_\nu \frac{\nu_0}{c} dv\\
\int \frac{F_\nu}{\mathrm{W}\,\mathrm{m}^{-2}\,\mathrm{Hz}^{-1}} \frac{d\nu}{\mathrm{Hz}} & = & 10^{-26} \frac{\nu_0/\mathrm{Hz}}{c/\mathrm{km}\,\mathrm{s}^{-1}} \int \frac{F_\nu}{10^{-26}\mathrm{W}\,\mathrm{m}^{-2}\,\mathrm{Hz}^{-1}} \frac{dv}{\mathrm{km}\,\mathrm{s}^{-1}}\\
\frac{F}{\mathrm{W}\,\mathrm{m}^{-2}} & = & 10^{-26} \frac{10^9\nu_0/\mathrm{GHz}}{3\times10^5} \frac{F}{\mathrm{Jy}\,\mathrm{km}\,\mathrm{s}^{-1}}\\
\frac{F}{\mathrm{W}\,\mathrm{m}^{-2}} & = & \frac{1}{3\times10^{22}}\frac{\nu_0}{\mathrm{GHz}} \frac{F}{\mathrm{Jy}\,\mathrm{km}\,\mathrm{s}^{-1}}
\end{array}
$$
For example, for the CO(1→0) line with a rest frequency of $\nu_0 = 115.3\,\mathrm{GHz}$, to convert from flux in  $\mathrm{W}\,\mathrm{m}^{-2}$ to flux in $\mathrm{Jy}\,\mathrm{km}\,\mathrm{s}^{-1}$ you multiply by $2.6\times10^{20}$.

You can read more about conversions in this document on the low-resolution spectrometer SPIRE on the Herschel telescope. Note though that this conversion differs from mine by a factor of $\pi$. I think this is because in frequency the instrumental resolution of SPIRE broadens a delta line to a sinc function which is normalized to $\pi$, not unity.


Star rising times in a different place given the latitude and time of one place

"Time" in astronomy has many meanings. In essence, let us look at something called Sidereal Time.

On a fixed day, the difference between local and sidereal time is roughly the same. Sidereal time is the sum of the star's right ascension (constant) and its hour angle. The rising hour angle is given by the expression

$$\arccos(-\tan (p) \tan(d)),$$

where p is the latitude and d is the declination.

Since the latitude is "roughly" the same we only need to take longitude into account. This means that the time when an event happens, since it happens at the same sidereal time happens $S_2-S_1$ (S are local sidereal times) later in the second city (the difference is roughly the same since the difference between local and sidereal times is constant for this day).

Now, it happens 80 minutes later since 20 degrees to 360 is 80 min to 24h.


I think that one thing confusing you, is that up/down is a 2 dimensions concept (just like your image by the way), but the Universe is in 3 physical dimensions (let's not talk about time here).

So when you say « going up from Earth is extracting from it's gravity », it can be in many direction. If two persons are in North Pole and South Pole of Earth, and if they are both going up. They are in fact going in opposite directions.

The second point is that you need a frame of reference. You go up/down « to » or « from » something.
For example, you go up (going away) from the North Pole, but from Sun's or Moon's point of view, it's not going up at all.

The up/down/left/right concepts always need a reference. See this very simple example: If we are facing each other, what is on my left, is on your right.
Your question here is exactly like asking : « What is on my left in the universe ? ». There is no real answer to that.

Hope I could help.


You can get into observational astronomy from a range of backgrounds. Some easily, some less so. Physics and data science sounds like a very fine choice as that is what can be a large part of an observational astronomer.

You probably will have to learn a bit about image processing. And possibly about the physics of the objects you are going to observe. Such adaptions are very common when one does not want to become an expert in one thing and one particular thing exclusively.

See also the answers to this question.


What is the evidence that galaxies rotate?

Rotation:

One of the easiest way to show that galaxies rotates are Doppler observations of atomic or molecular lines. For example, if the observed galaxy is seen edge-on and rotates clockwise, the left side of the galaxy is going away from you, and is then Doppler-shifted in the red. The right side, going toward you, is Doppler-shifted in the blue.

This kind of observations was realized already in the '60, in radio.

The spiral patern observed in spiral galaxies is also a clear signature of their rotation. Note that spiral are linked to density waves in the galactic disk (see for example Lin & Shu 1964).

Direction:

As for their directions, you can convinced yourself that there is no tendency, just looking at this Hubble telescope gallery. You can also think that galaxies probably rotate by angular momentum conservation during their formation history, and that it highly dependent on initial conditions, that are clearly random.


This is a very broad topic. You can literally write entire books on it. I'll stick a short answer here for future reference.

You have two main options: refractor or reflector.

https://en.wikipedia.org/wiki/Refracting_telescope

https://en.wikipedia.org/wiki/Reflecting_telescope

Refractor is easier. You'll need a relatively large lens with a long focal length - this will be your primary optics, at the top of your telescope. The length of your scope will be dictated by the focal length of the primary lens.

You can measure the focal length by going outside when the Sun is up and trying to focus the Sun to a tiny pinpoint image - the distance then between lens and image is the focal length.

As an ocular or eyepiece you could use a smaller lens with a short focal length. Much better, you could use an eyepiece from an old microscope or binoculars.

The total distance between the objective (primary lens) and ocular (eyepiece) is the sum of their focal lengths. If the primary lens has focal length F1, and the eyepiece has focal length F2, then when looking at infinity the distance between them is F1 + F2. This will help you calculate the total length of the tube.

However, if the ocular (eyepiece) is a divergent lens (the kind of lens that makes things seem smaller) then the total distance is F1 - F2. By the way, it's hard to measure the focal length of a divergent lens, because it doesn't make an image you can project.

The magnification of the instrument will be:

M = F1 / F2

Do not worry too much about magnification. Make something that works first. More magnification is not always better. In the beginning, aim for something in the 10x ... 50x range. Refine it later.

Make sure the tube is rigid. A plastic pipe like the ones used in water installations should work well.

Make sure all lenses are centered and aligned with precision. You don't want your lenses to be leaning in any direction. Their main axis should coincide with the axis of the scope. This is important for performance.

If you get too many color fringes (little rainbows) around objects, try and reduce the diameter of the primary lens by placing a paper ring in front of it - this makes the image a bit darker but should diminish the pesky rainbows somewhat.

Mount the eyepiece on a smaller tube that's sliding into the main tube. Cut the main tube a little shorter than F1 + F2. This will allow you to move the eyepiece back and forth until the image is in good focus.

If you find a shop that makes prescription eyeglasses (for people with eye or vision problems), they have so-called "blank" lenses that they use to cut the glasses from. See if you can buy one of those for your primary optics. A focal length of about 1 meter should be fine (opticians call it a 1 dioptre lens, which you can find in pretty weak prescription glasses).

Sometimes you can pull a good lens out of a broken projector or something, but make sure the diameter is big enough (50 mm or bigger should be fine) and the focal length is big enough (I would say anywhere between 0.5 and 2 meters - that's between 2 and 0.5 dioptres - but keep it close to 1 m if possible).

The first telescope I made, many years ago, was very much like this. A blank lens from a prescription glasses maker, a plastic pipe, an eyepiece from a toy microscope. It worked reasonably well, I could see with it craters and mountains on the Moon.

Making a reflector is much harder, but the performance of the instrument is higher. Read the Stellafane guide, it's very comprehensive:

https://stellafane.org/tm/

What's described in there is a dobsonian telescope. That's a newtonian reflector on a simple alt-azimuth mount.


Do most astronomers think that Andromeda will collide with Milky Way?

Relatively high-resolution measurements were made with the Hubble Space Telescope over a 8 year period. From these measurements, astronomers determined that the collision was a certainty.  It is predicted to occur in about 4 billion years. I suppose there may be some astronomers who disagree with the measurements or calculations but I can find no mention of that. The results were reported in an article in "Nature" (http://www.nature.com/news/andromeda-on-collision-course-with-the-milky-way-1.10765#/b1) and published in "Astrophysical Journal"  Both are well-respected publications. The "Astrophysical Journal" reference is below.

Sangmo Tony Sohn; Jay Anderson; Roeland van der Marel (Jul 1, 2012). "The M31 velocity vector. I. Hubble Space Telescope proper-motion measurements". The Astrophysical Journal. doi:10.1088/0004-637X/753/1/7.


Inconsistent online radius value for Proxima Centauri

Wikipedia quotes $0.1542\pm 0.0045$ solar radii which is $107276\,km$ with an error of about $3130\, km$.

They also give a reference to a 2017 paper available on Arxiv Proxima's orbit around Alpha Centauri, P. Kervella, F. Thévenin, C. Lovis that gives this value.

Wolfram's estimate is too far off to be an experimental error, IMO, and I think someone has just inadvertently used the diameter (twice the radius) rather than the radius.


Unlike gravitation, which is entirely a result of space-time curvature, magnetic fields do exist and there is a magnetic field around the Earth and many other planets.

(There is a deeper level of understanding of electromagnetism called Quantum Electrodynamics which explains how fields work in terms of more fundamental concepts, but unlike gravity, the E-M field does exist. (And there is hope -- so far unrealized -- that eventually all of the fields of nature plus gravity will be explained as aspects of one yet deeper thing.  But not today.))

Frame dragging -- the distortion of space-time by a rotating body -- does exist and the link suggested by @PM (en.wikipedia.org/wiki/Frame-dragging) is a good place to start.  Frame dragging has been measured by careful observation of satellite orbits, and possibly by astronomical observations.  The Gravity Probe B satellite measure a closely related effect.  As far as I know, the effect is too small to have been measured in the lab.

Frame dragging does indeed affect the path of a falling body, though for the Earth the effect is really, really small.  It would be large near a millisecond pulsar or a rapidly rotating black hole.  It's also large for merging black holes and is an essential part of the calculations of the gravitational waves emitted during the final rotations before merger.

The effect can be precisely calculated and agrees perfectly with the (still rather crude, alas!) experiments and observations.


Apparent magnitude is measure of how bright an object appears to an observer on Earth, meaning it's a function of both the object's intrinsic luminosity and its distance from us.  The concept of magnitudes dates back to the Ancient Greeks, when stars in the sky were categorized into six magnitudes (the brightest being 1 and the faintest being 6).  Each successively lower magnitude was twice as bright as the one before, meaning the scale was logarithmic.  We still use magnitudes for historical reasons, though the scale was later standardized to use the formula

$m_x - m_{x,0} = -2.5\log_{10}(\frac{F_x}{F_{x,0}})$

where $m_x$ and $F_x$ are the magnitude and flux of the object of interest and $m_{x,0}$ and $F_{x,0}$ are the magnitude of flux of a reference object (where usually Vega is used to define the 0 point in magnitude).  This means any object that appears brighter than Vega has a negative magnitude.  There is no limit to how bright an object can appear, so there is no lower limit to magnitudes.  The sun, for example, being the brightest object in our sky, has a magnitude of roughly -27.


How much does the orbit of Mars deviate when it is in linear conjunction with Jupiter and Saturn?

To properly answer this question, you should apply the 3 body problem, which, gets very long and mathy.  It's technically not solved but it can be approximated with very good accuracy.

Applying basic Newtonian formulas can give you an estimate.

The sun has about 1,048 times the mass of Jupiter, and if we line up Mars at it's Aphelion (1.67 AU) to Jupiter at it's Perihelion, (4.95 AU).  Mars at it's furthest and Jupiter at it's closest, Mars is still about twice as close to the Sun as it is to Jupiter.

For simplicity, I'm going to call Mars at Aphelion an "MAU" - Mars Aphelion Unit, so Mars, furthest from the sun, 1 "MAU", Jupiter closest to Mars, about 2 "MAU" and Jupiter closest to the Sun, 3 "MAU"

So, in perfectly lined up conditions, Jupiter never has more than 1/4,000th the gravitational pull on Mars than the Sun does and we also have to factor in the tidal force.   It gets tricky here, cause all 3 objects exert a force on the other (or stretch space time which tells the objects where to go if you like the relativity version), but I'm going to stick with forces.

Jupiter tugs on Mars and the Sun at the same time so how far it pulls Mars from the sun is more of an an applied tidal force.  At closest point, Jupiter's tug on Mars from 2 "MAU" and it's tug on the Sun from 3 "MAU" means it tugs on Mars 2.25 times stronger.  Take the inverse, the effective gravitational pull - at closest, is the inverse of that, or 44%, so you can only apply 56% of Jupiter's gravitational pull towards Mars relative to the sun.  And that's at peak.  As Mars moves closer to the sun and/or further from Jupiter that number drops.

The effect of Jupiter on Mars orbit relative to the sun is never more than about 1 part in 7200 of that of the sun.

As Mars moved towards Jupiter in it's orbit it speeds up and effectively moves away from the Sun, and as it moves away from Jupiter, this effect is largely cancelled out over time.   In effect it's a slight stretching when moving towards Jupiter and squashing when moving away, of Mars' elliptical orbit every time around the sun.  These are sometimes called orbital perturbations.

1 part in 7,200 works out to about 0.14% and that's the peak Jupiter effect.    Rounding down, a .01% variation on Mars' orbit seems a plausible estimate and that might even be a little high.  Applying that to the diameter of Mars' orbit, that works out to a stretching/squashing of about 20,000 to 25,000 km.  Compared to Mars natural variation based on it's eccentricity of over 26 million miles, that's pretty insignificant.   This stretching and squashing is also largely balanced out as Jupiter circles around the sun.

Now if we apply the acceleration formula, I get a somewhat different number.   g = G*M/r^2.

Gravitational Constant:   6.67E-11
Mass of Jupiter 1.898E27 kg
r = distance between the two.   Closest pass is about 3.28 AU, or 4.91E11 meters

The peak gravitational acceleration Mars feels from Jupiter is about 5.26E-7 m/s^2.   That sounds negligible, but Mars takes 187 days to orbit the sun and it spends about half that time closer to Jupiter than the Sun is to Jupiter.

As a (bad) estimate, because that's the maximum gravitational force at closest pass, if we estimate that over a Mars year, that works out to about 90 days of "falling" towards Jupiter and 90 days falling away.  Obviously the period is longer than 90 days but the force is less when Mars is further away so that's a guestimate.   Apply (d=1/2at^2) to 90 days or about 7.8 million seconds.  The distance using that formula comes to about 16,000 km.

So, both estimates are pretty close, one of 16,000 the other of 20-25,000 km.   This also works well with the 4 or 5 significant figures usually given for Mars' obit.   4 figures on distance works out to nearest 100,000 km, which wouldn't be affected by Jupiter's stretching and squashing of Mars' orbit.

Now as for Saturn, at 30% the mass of Jupiter and roughly twice the distance from Mars and the Sun, you're looking at less than 8% of the gravitational force and with tidal effects figured in, maybe just 4%-6% the orbital perturbations.   Jupiter is the one that really matters.

Orbital stretching and squashing is what lead to the discovery of the planet Neptune, because astronomers noticed that Uranus orbit was effected in a similar way to how Jupiter affects Mars.  Observing those perturbations told them about where to look for Neptune and it was found not long after.

I want to point out that over tens of thousands of orbits, these perturbations can lead to changes in Mars' orbit, primarily it's eccentricity.   It's semi-major axis and period of the orbit remain mostly unchanged, but it's eccentricity operates on two cycles.  Mars' eccentricity varies on a 96,000 year and a two million year cycle.  Like Mars, Earth's eccentricity also has cycles driven by the other planets, 95,000, 125,000 and 413,000 years.   Mars' eccentricity is prone to greater variation, presumably (I think) because it gets closer to Jupiter.

As a rule, these eccentricity oscillations move back and forth unless two planets end up in resonance with each other.   When that happens the variations can continue to grow leading to significant movement of one or both of the planets over millions of years.  This is more likely to happen shortly after the solar-system forms.  Our solar system is thought to be long-term stable with no orbital resonances between planets now, and none expected for at least a couple billion years.   There's a small chance that Jupiter and Mercury will fall into resonance in a few billion years.   Here's a nice detailed article on orbital eccentricity variation and stability of the solar system.

That's a ballpark answer anyway.   Corrections to my math are welcome and if anyone can do the 3 body problem, feel free.  That would be much more accurate, but I think 16,000 - 25,000 km per orbit is in the ballpark.


Use of type-I a supernovae as standard candle

A type 1a supernova forms when a white dwarf grows through accretion to a certain size, at which it becomes unstable. This means that the precursor object is always a white dwarf of mass 1.39 solar masses. As the precursor object is always of the same type and the same size, the supernova is thought to be the same brightness.

On the other hand, type II supernovae form from the collapse of giant stars. The precursor object can have a range of masses from about 10 Solar masses, to a maximum of over 100. The brightness of the supernova depends on the mass of the precursor (and other factors such as the metallicity) and so type II supernovae are not all equally bright.

The other types also have precursors which vary in size or other factors, only type 1a supernovae are thought to always have a white dwarf of mass 1.39 as a standard precursor, so only type 1a are suitable as a standard candle.


What (actually) does "peculiar periodic spectral modulations" refer to in this preprint?

The idea of Trottier & Borra is to analyze signals that might be hidden IN the stellar spectrum. I know this sounds weird; it did sound weird to me when I first heard of it too.

You know how you can decompose light in diferent colors, right? Well, we astronomers like to think of colors in terms of wavelength (with $~4500$ Angstroms being blue and $~7000$ angstroms being red) or in frequency ($1.5\times 10^6$ GHz and $2.3\times 10^6$ GHz respectively. To get these numbers, just divide the wavelengths by $c=3\times 10^8$ m/s, the speed of light). In the range of wavelengths in which the Sloan Digital Sky Survey (SDSS) works, as you can see, it is much more clear to work in wavelength, but people, for different reasons (like in this paper), can also work in frequency-space rather than in wavelength-space.

The thing is that the SDSS has spectra for millions of objects, including stars. That is, they have measurements of the flux as a function of wavelength (color) for these objects. Well, the idea in this paper is actually an idea Ermanno Borra published years ago (2012; here is the paper: http://adsabs.harvard.edu/abs/2012AJ....144..181B). In this paper, he proposes that maybe extraterrestials would like to communicate by injecting extra signals in that spectra, i.e., that they would add extra light in different colors to the stellar spectra we observe at Earth. However, he proposes that maybe they would put colors in frequency-space rather than in wavelength-space (because it is easier to modulate that with, e.g., lasers), and that these color modulations will be periodic in this frequency-space spectra. So, basically, this is what they search for in that spectra: periodic modulations in the colors, but in frequency-space rather than in wavelength. This is what they show, for example, in their Figure 4:

The top panel has an added signal with a periodic modulation that is $10^4$ times larger than the one they detected on the real star (lower panel).

As you can see, they don't analyze a time series: they analyze the stellar flux in frequency space. Now, I think I have quite a bunch of experience with stellar spectra, and quite frankly I think this result might just be spurious signals due to instrumental problems and/or data reduction issues. I've seen similar modulations before reducing both high-resolution, mid-resolution and even low-resolution spectra (the process of going from the image to the final spectra easily introduces things like this); I wouldn't be surprised if this is the case too.


Why are the magnetic poles of a pulsar so far off the rotational axis, yet stable?

The magnetic field of a star is not entirely a result of the global spin of the star. The global spin is part of it, but there are other mechanisms as well. Within the star, there are convection zones, meridional flow, etc.

http://solarscience.msfc.nasa.gov/dynamo.shtml

All these flows generate their own field components. The overall field is simply the sum of all little fields. Its general orientation might be close to the global spin, if the strongest components are aligned to it, but there are many smaller components with different orientations. Therefore, the total field of the star can be somewhat slanted.

And then the star collapses into a neutron star, and its field is compressed. The collapse itself may be slightly asymmetrical, and may further deviate the magnetic field axis.

As a result of all of the above, it's by no means unusual that the magnetic field of the neutron star is not aligned with the spin.


How many galaxies are there in the Hercules–Corona Borealis Great Wall?

I'm sorry for the long answer. The result is in the bottom. :)
The number of galaxies in the "wall", as in any volume of space, is the number density $n_{\mathrm{gal}}$ times its volume $V_{\mathrm{wall}}$. However, the number density of galaxies depends on their masses/sizes: Small galaxies are much more common than large galaxies. Usually we describe this by the halo mass function (HMF), which gives the (differential) number density dN/dM of dark matter halos hosting galaxies as a function of halo mass M. Depending on your favorite cosmology, it looks something like this (created with this tool):

The solid line shows the HMF today in the local Universe, i.e. at redshift zero, while the dashed and dotted line shows how it was at redshift 1.6 and 2.1, respectively, roughly 10 billion years ago, which mark the beginning and end of the wall (according to Horváth et al. 2014, A&A, 561, L12). Notice how the number of large halos grow with time, at the expense of small halos. This is due to the smaller ones accreting onto larger ones.
The HMFs are characterized by a power law at low masses with an exponential cut-off at high masses. To get the total number density we need to integrate over the entire range of masses, and here comes the problem: Whereas the upper limit is set by the maximum size a galaxy can have (roughly $3\times10^{13}M_\odot$, but the exact number isn't very important, as these galaxies are so rare), the lower limit depends on your definition of a galaxy. Large galaxies have smaller galaxies orbiting as satellites, and there isn't a well-defined threshold for when a clump of gas, stars, and dark matter is called a galaxy. These guys report on the detection of a $\sim10^3M_\odot$ clump which they call a galaxy. That's probably not what you think of as a galaxy, so for the sake of this exercise, let's use Small Magellanic Cloud-sized galaxies as our lower minimum, i.e $\sim 6.5\times10^9 M_\odot$.
The total number density of galaxies is thus found by integrating the HMF. Luckily, the online HMF tool can do that as well. The result, i.e. the number density of halos above a given mass, looks like this:

Note that the $x$ axis is given in terms of $h^{-1}M_\odot$. Since $h \simeq 0.7$, the mass of the SMC is $M_\mathrm{SMC} \simeq 5\times10^9 h^{-1}M_\odot$. From the graph, the number density of these objects is $\sim0.5\,h^3\mathrm{Mpc}^{-3} = 0.17\,\mathrm{Mpc}^{-3}$ (1 Mpc, or megaparsec, equals 3.26 million lightyears). In principle we should subtract the number density of halos above $3\times10^{13}M_\odot$, but since these are four orders of magnitude smaller, they can be neglected.
The wall starts at redshift ~1.6 and ends at redshift ~2.1. This translates into starting at a distance of 4.65 Gpc (billion parsec) and ending at 5.47 Gpc, i.e. a depth of ~800 Mpc. These numbers are in comoving coordinates, i.e. what it has expanded to today, as opposed to physical coordinates, i.e. what it was at the time it emitted the light that we see today. But the HMF is given in terms of comoving volume, so that's cool.
To get the volume, we also need the area. The finders report that it's roughly 2 Gpc times 3 Gpc, but they don't seem to state whether this is in physical or comoving coordinates (which differ by a factor (1+z) ~ 3). I suspect it's comoving, though.
Thus, the total volume is $V_{\mathrm{wall}} = 0.8 \times 2 \times 3\,\mathrm{Gpc}^3 = 4.8\,\mathrm{Gpc}^3 = 4.8\times10^9\,\mathrm{Mpc}^3$, or 166 cubic gigalightyears. Here, I have ignored the fact that the physical area spanned by the angle of observation is a bit different at the front and the back of the wall, but since these numbers are already quite uncertain, it's of lesser importance.
The considerations above are all for the average Universe. But the wall represents an overdensity in space, so the number will be larger. I can't find any estimates of the overdensity of this particular supercluster, but in general galaxies in superclusters are more numerous by a factor of $\delta \sim 5$–$10$. Let's use 5.
That is, the total number of galaxies the size of the Small Magellanic Cloud or above is
$$N_\mathrm{gal,tot} = n_\mathrm{gal} \times V_\mathrm{wall} \times \delta$$
$$= 0.17\,\mathrm{Mpc}^{-3} \times 4.8\times10^9\,\mathrm{Mpc}^3 \times 5$$
$$\sim 4 \mathrm{\,billion\,galaxies}.$$
But if you go through the same exercise for halos of masses above, say, $10^3\,M_\odot$, you'll find a number which is a million times larger.
In reality, the number density of luminous galaxies is smaller. At the high-mass end, the gas has trouble cooling sufficiently to condense and create stars, and active galactic nuclei in massive galaxies heat the gas. At the low-mass end, gas is more easily blown out of the potential well by stellar feedback from supernovae.


Nothing comes out from the BH (well, with a single known, very small exception).

The accretion disk of the Black Holes most likely has various streams, pressure differences and so in it. Calculating it or measuring it is surely a very hard task, if you have luck, a professional explains the results in a better answer.

The accretion disc of the typical, star-sized black holes is much smaller as 1AU. And sound doesn't propagates in vacuum. Thus, from 1AU distance from the Black Hole we most likely wouldn't hear anything.

Other radiations (mainly gamma from the disc) may be deadly even from 1AU, but not its sound.


lunar calendar accuracy for predicting new moons

The answer depends on what you mean with "precisely".  The ELP theory seems to have an accuracy of ~10-6° in longitude over a century, which comes down to about a millisecond for phases.


We have no practical means to travel to either planet and set up even a base, let alone a colony.  As such neither is better.

Also note that we have no reason to do it and we would need an extremely good reason because it would almost certainly involved the expenditure of vast resources to even set up a base, let alone a viable colony.

In a practical sense neither have any real human habitability options other that entirely sealed bases.

However current knowledge suggests that the geology on Titan would be really challenging.  We'd be operating a base at significantly higher temperatures than the local norms and as such we'd always be subject to problems in a atmosphere with active weather.  The only attraction on Titan is studying the locale, but we'd have done that in enormous depth before we could even consider a colony, so it's a mute point.

Mars is closer, less hostile and more familiar.  We'd go there before Titan for the simple reasons it's easier, cheaper and more likely to succeed.  But again, why ?

But we'd be unlikely to do either before we'd establish a viable self-sustaining Moon base of at least "colony" status.  Again that's not even remotely possible and would not be possible without a unified global effort.  Well this particular globe can't even agree the wording of one sentence in a document about it's own report on climate change.  They can't even agree about one lousy word !  So no one is going to even start talking about the possibility of setting up a committee to discuss the likelihood of setting  up another committee to start planing for a colonization effort research program.

In short, even if we solved the technical problems to make it worth deciding which place to colonize, we have no political chance in Hell of actually doing it.  Humanity will die squabbling in a committee room.  No colonists required until we replace humans with something that works.  Skynet anyone ? :-)


Where can I find a catalog for stars within 450 light-years of Earth?

The Hipparcos catalogue (use the revision by van Leeuwen 2007) contains trigonometric parallaxes and is complete down to about 9th magnitude.

What this means is that you will have a complete set of measurements for F-type main sequence stars and hotter, and all giant stars within 450 light years. However, the sample will not be complete for G dwarfs or cooler, since these are fainter than 9th magnitude at 450 light years.

The catalogue does not contain mass, since mass is not a measurable quantity for single stars. If you want masses you will have to use relationships between absolute magnitude and mass for main sequence stars (e.g. Kroupa 2007), but note that these depend on metallicity and a little bit on stellar age. Masses are extremely difficult to estimate for giants.

If you can wait a couple of years, the Gaia mission will report parallax distances for all stars to magnitudes of 19 or 20. This will be almost complete at 450 light years, even for M dwarfs.


How is information preserved on the surface of a Black Hole?

How is information preserved on the surface of a black hole?

We have no evidence that it is. We have some claims of this nature, associated with the holographic principle. But I'd say the physics concerned is hypothetical and speculative. It isn't in the same league as general relativity, which is one of the best tested theories we've got. See Clifford M Will's paper on the confrontation between general relativity and experiment for details of that.

  Is there an compression algorithm for storing the information?

Not that I know of.

  What if there is a uniformly complex object dropped to the black hole? Then there will be no compression of information.

Instead there will be a gamma ray burst. Friedwardt Winterberg wrote about this. See the 2013 AMPS paper an apologia for firewalls. Tucked away in the conclusion is footnote 31, containing a reference 87 to Winterberg’s 2001 paper gamma ray bursters and Lorentzian relativity. I'm confident that Winterberg is essentially correct.

  How is there enough space on the surface (2D) to encode something which is in. The planck distance is the same in the black hole as in the universe. So how can the black hole encode information about an object in 3D that is both uniformly complex and as dense as the black hole?

I don't think it can.

  Or is there a law that says 3D objects cannot be more dense then some level - which is sufficiently sparse so that when collapsed to a black hole all its information can be encoded on the surface of a sphere? Has this limit been found - given that a sphere has the lowest surface to volume of any object: https://en.wikipedia.org/wiki/Surface-area-to-volume_ratio?

As far as I know there is no law or limit. And as far as I know there is no evidence to support the holographic principle. I for one do not take it seriously. But please advise if you receive any useful answers about this subject on any other website.


How galactic density falls using Oort constants

From the definitions of the Oort constants we know that
$$A + B = -\frac{dv}{dr} $$
at the solar radius.
But $A + B$ is small, implying that $dv/dr \simeq 0^{\dagger}$ and we are dealing with a flat(ish) rotation curve.
Assuming a spherically symmetric mass distribution with a density that is proportional to $r^\alpha$, we can write the centripetal acceleration at a radius $R$
$$\frac{v^2}{R} = \int^R_0 \frac{4\pi G r^2 \rho(r)}{R^2} \ dr\ . $$
Then, if $v$ is constant, we can just deal in proportionalities:
$$ R^{-1} \propto R^{-2}\int^R_0 r^2 r^\alpha \ dr\ , $$
$$ R^{-1} \propto R^{-2} R^{3+\alpha}\ ,$$
and so $\alpha =-2$.
Another variant is to tell you that $A=0$ in some galaxy. This means $dv/dr = v_0/r_0$, which means $v \propto r$ (i.e. solid body rotation). This in turn means $\alpha =0$.
$\dagger$ by small, what we mean is that the velocity gradient is small compared with $v_0/R_0$ where $v_0$ is the rotation speed where the Oort constants are defined. Since $A-B = v_0/R_0$ this means that
$$ |A+B| \ll |A-B| $$
which in this case is 2.5 vs 26.5 km/s per kpc.


What is the average temperature of all planets at 1 bar?

At 1 bar (0.9869 atm / 14.5 psi) pressure level the planets have the following average temperatures:

Venus: 30 degrees Celsius (1 bar level is about 33 mi or 53 km high)
The Earth: 15 deg C
Jupiter: minus 145 deg C
Saturn: minus 178 deg C
Uranus: minus 197 deg C
Neptune: minus 200 deg C
Titan's temperature ranges from minus 180 deg Celsius on the surface to minus 93 deg Celsius in the upper stratosphere. Titan's surface air pressure is 1.45 atm (1.47 bar).

Mercury has no atmosphere while Mars' atmosphere doesn't reach 1 bar anywhere on the planet.


When did the first annular eclipse happen?

A quick and dirty estimate is as follows.

An angular eclipse occurs first if the Moon is at apogee and Earth at perihelion. Assuming that these two events coincide (which happens every ~8.85 years owing to lunar apsidal precession), an angular eclipse requires that ($a$ denoting orbital semi-major axis and $R$ radius)

$$
      \frac{R_\supset}{(1+e_\supset)a_\supset} < \frac{R_\odot}{(1-e_\oplus)a_\oplus}
$$

or

$$
      a_\supset > \frac{R_\supset}{R_\odot} \frac{1-e_\oplus}{1+e_\supset} a_\oplus\approx 348300\,\mathsf{km}
$$

where I assumed that the orbital eccentricities, Lunar and Solar radii, and Earth's semi-major axis didn't change (over the period considered here) and are given by
$e_\supset=0.0549$, $e_\oplus=0.0167$, $R_\supset=1737.1\,$km, $R_\odot=695510\,$km, and $a_\oplus=149.6\times10^6\,$km.

At the present we have $a_\supset=384400\,$km. If we assume an average annual change by 22 mm, we find a time of

$$1.6\times10^9\,\mathsf{years},$$

though it could have been more recent if the annual change of $a_\supset$ was larger some hundred million years ago (which may well have been the case).

In a similar way, one may also estimate when the last full Solar eclipse will occur. This time, the Moon is at perigee and Earth at apohelion, when

$$
      a_\supset < \frac{R_\supset}{R_\odot} \frac{1+e_\supset}{1-e_\oplus} a_\oplus\approx 400800\,\mathsf{km}
$$

which last occurs in $430\times10^6$ or $745\times10^6$ years, depending on whether the current lunar recession rate of 38mm/yr or the long-term average of 22mm/yr is used.


Jupiter-Venus conjunction is vertically aligned. I'm at 49° N. How is this possible?

A combination of several movements can lead two objects in the sky to appear vertically aligned. The planets are constantly moving relative to each other, they are orbiting on slightly different planes, the Earth is orbiting and rotating, so such configurations between planets can occur frequently.

An observer does not need to be in the tropics to see two planets being vertically aligned at some point. The planets orbit close to the ecliptic, but their orbital planes are slightly inclined relative to ours. Jupiter's orbit is about 1° from the ecliptic, and Venus is about 3°. This means that both planets can move above or below the ecliptic by a few degrees. When they pass each other as seen from the Earth in a conjunction, they are rarely on top of each other, most of the time, they are separated by a few degrees, and they are not exactly on the ecliptic line. (Such occurences of planets occulting or transiting each other do happen, but are rather uncommon: the next time Venus will transit in front of Jupiter will be on 22 November 2065)

I'll add a few pictures to help visualizing the positions, movements and alignments involved here.

Here is a view of the sky on 20 January at 49° North latitude:

As you can see, Jupiter and Venus appear pretty much vertically aligned in the morning sky. The yellow line is the ecliptic. Note that the ecliptic does not pass through the planets here. Jupiter is pretty close, but Venus is a few degrees north of it. As both planets pass each other in the next days, at some point, their alignment will be perpendicular to the ecliptic (called the ecliptic conjunction), as we can see from this January 22 simulated image:

But they won't appear vertical anymore from your vantage point. (In fact, they would apprear vertical again later during that day, because of Earth's rotation, but the Sun will be up, so it wouldn't be observable directly)

A few days later, on January 26, they will appear horizontally aligned!

(Images generated in Stellarium)


What is the maximum number of planets in the habitable zone?

This question not only makes you think about the temperature, pressure but also the impact of gravity. Not moving away from topic, I must say that the region you are saying about is quite large in the view of metrics but quite short again if we really sense gravity.

The distance of a planet from the sun is maintained by gravity, but we must also not forget that distance between two planets is also controlled by the same. You must appreciate gravity because most of the features on earth are because of the same thing.

Now, if you want to calculate it more precisely you should know each and every postulates, theorems and properties shown by the space in nourishing our Earth for habitability.

Perturbation may give you an idea of adjustments, that how our earth is adjusting day by day for its decreasing distance from sun or increasing size of the sun. Talking according to theory of perturbation there may be more than 2-3 planets in the habitable zone. But still you cannot annoy the word "more" in that.

It shall depend mainly on the following factors : -

Size of Planets
Distance between each of them
Distance from sun accordingly
Their mass (priorly)
Speed of revolution around sun

I know this information is not much for your question but this is enough according to recent knowledge about space. We even don't know the drastic changes happening in the space....the dimensions of space or how vast it is....whether it is measurable or not.....and if yes then are the measurements precise. Each answer tackles you more to have questions....But don't stop, just keep asking!


Theoretically what happens to a black hole that evaporates?

Utterly unknown.

So Hawking proposed radiation, that reduces the mass of the black hole (BH) by black body radiation.  The effective temperature is antiproportional to the mass of the BH.  In the linked paper, Hawking concludes that BHs end in "fairly small explosions by astronomical standards."

There are different solutions to the equations of general relativity that produce black holes, not only the common Schwarzschild metric.  Whether the curvature singularity or the event horizon vanish before the BH completely evaporates, depends on the internal structure of the BH.  As any mass, even very small ones, can become BHs if you compress them enough, there might be a singularity until the very end (not taking into account quantum mechanics).  On the other hand a "vanishing event horizon" is not something we should be strictly speaking about, because then it wasn't an event horizon in the first place.  You may see the looming problems with black hole remnants.

Also: according to the no-hair theorem, BHs are completely characterized by mass, electric charge and angular momentum.  If you can see a remnant, e.g. a clump of matter, could you derive its complete properties from the previous three only?

In research, a remnant seems to be generally undesired
 but possible.  This fresh and unreviewed paper jumps on the bandwagon and proposes white holes as remnants. So the debate about the end of BHs and their information theoretical implications is still a hot topic.

Disclaimer: I'm not a BH specialist.  (And I'm waiting for the time where "black hole specialist" is a craft job designation.)


Why is metallicity important in the death of stars?

I can't give a detailed answer; the details are buried in the depths of numerical stellar evolution models.

The thing that changes most with the metallicity of a newborn star is the radiative opacity of the gas. Higher metallicity leads to more opacity.

This has two immediate effects - it makes energy harder to get out of the stellar interior and makes it more likely that convection will take over.

Convection has the property of mixing up all the material within the convective zone. This can have knock on effects as to how long each nuclear burning phase lasts and how much material is consumed. It also mixes synthesised material from the interior outwards.

A further important effect is that mass loss from massive stars is very extensive and is due to radiatively accelerated winds. For a given luminosity, high metallicity gas is more opaque and easier to accelerate. Hence mass loss is very sensitive to metallicity and determines how massive the star is as it reaches the end of its life. This in turn has a large bearing on what the remnant will be.

There is a further feedback in that the wind metallicity is the metallicity at the surface, but this in turn can be affected by interior mixing that is in turn metallicity-dependent.

If that sounds complicated, that's because it is, and detailed numerical models are required to see how these things play out.


Hawking radiation isn't some special kind of radiation. It's perfectly ordinary "thermal" radiation, mainly photons, with a few neutrinos and electrons and positrons if the temperature is high enough. Black holes shrink when they emit it, not when the absorb it. So aiming simulated Hawking radiation (or any other kind of radiation) into a black hole just makes it bigger.


Is the diamond ring effect visible in a 40mm hydrogen alpha telescope?

Experimental answer, at least with a 40mm PST: No.

Don't worry, I didn't miss much of it :)


Did I see a meteor(oid) or space debris or something else?

The OP's clarifying comment under the question offers an opportunity to examine further:

Do meteoroids really get that slow? I estimate its speed at about 1/6 that of a typical shower meteor.

The apparent or angular speed of an object in the upper atmosphere depends on several things, including

the actual linear speed in say km/sec
the distance from observer to the object
the cosine of the angle between the direction of motion and direction to object (i.e. the dot product of the normals of those vectors); if it's moving straight towards you the angular speed is zero.

The linear speed of meteors with respect to Earth is a function of the orbital speeds of both objects. See the (currently unanswered) question How to calculate the position of a meteor shower's radiant point based on its associated comet's orbit? and links therein. Just for example if the meteor shower is associated with a comet in a circular orbit at 1 AU with an inclination of 90 degrees (a "polar orbit") then the relative speed is
$$\sqrt{2} \sqrt{G M_{Sun}/1 \text{AU}} \sim  29.7 \sqrt{2} \text{ km/s} \approx  42 \text{ km/s.}$$
Speeds of meteors relative to Earth can theoretically be somewhat higher or much lower than this. So despite 42 being an aesthetic number let's just us 30 km/s for a typical value.
The orbital speed of a spacecraft at 100 km above the Earth before drag stars slowing it down is
$$ \sqrt{G M_{Earth}/(6378+100) \text{km}} \approx  7.8 \text{ km/s.}$$
Since both are roughy at the same distance, a burning reentry might appear to move (in angular velocity) four to six times faster than a small bit of space junk burning up.
That ratio could be reduced if the meteor was sighted near the radiant point, where it will appear to move much more slowly by a factor of the cosine of the angle mentioned above.
So in some cases a meteor might appear to move the same speed as a bit of reentering junk, but in that case the trail will be very short, it will not cross the sky. From the OP's question it sounds like the motion was much larger than a few degrees so we can rule this out.
But what if the comet associated with the meteor was moving in an orbit nearly tangent to Earth's with a similar to velocity to that of Earth at the intersection point (i.e. low inclination)?
There can't be absurdly slow slow meteors like that because of Earth's gravity. An object moving nearly the same as Earth will accelerate as it gets closer to Earth. $v_{escape} = \sqrt{2} v_{orbit}$ means even a co-orbiting object will reach almost 10 km/s by the time it enters the atmosphere. And I don't know of any meteor showers associated with comments who's orbits fit this description.
Conclusion
Most likely it was space junk. Bits and pieces reenter regularly so while it's not common to see, from time to time people see it. And soon some people may pay to see it on demand!

Why would a box full of 1cm balls released into LEO be so scary to an engineer supporting the ISS?


Is there a single, most prominent helium line?

I am unaware of there being a single Helium line that is as prominent or useful as the 21cm line is for Hydrogen.

The 21 cm line is due to a hyperfine transition in the 1s ground state of hydrogen. It is notable due to its strength, and also its usefulness for mapping neutral hydrogen in the universe, which is impossible in the optical since it is absorbed by neutral Hydrogen. It was in fact the first such spectral line observed in the radio part of the spectrum, and is so fundamental that the Pioneer and Voyager probes used the length of the transition as a time and distance unit on the plaques they carry, so that potential extraterrestrials could decode the messages.

There are many more spectral lines in the radio from heavier elements, and a whole field dedicated to observing and understanding them: radio frequency spectroscopy.

For Helium in particular, hyperfine lines can also be observed; the hyperfine transition of the $^3$He isotope has a wavelength ~4mm. The same isotope also emits a recombination line, where an electron recombines with an ion and falls to lower energy levels, emitting a photon in the process. The wavelengths of such recombination lines for Hydrogen like atoms (singly ionised in the case of Helium, so that it has a single electron) can be estimated using the Rydberg Formula:

$$\nu = R_{m}c\left[\frac{1}{n^2}-\frac{1}{(n+\Delta n)^2}\right]$$

where $R_m$ is the reduced mass Rydberg constant, $n$ is the final state and $\Delta n$ is the change in state.

Estimating the strengths of such lines is a little more complicated, but there are details here.


This assumes that when a pair particle-antiparticle is created, a bias towards antiparticles being captured and particles escaping must exist.

No such assumption or bias exists.

As far as we know the universe treats anti-particles just like particles with different properties (e.g. charge is opposite compared to the corresponding particle).

  Otherwise, the mass of the black hole might fluctuate, but in average for big numbers of pairs, equal percentage of particles and antiparticles would be captured/escape.

You seem to be implying that anti-matter has negative mass.  All particles (including anti-matter) have either zero mass (like photons) or positive mass (like electrons).

So regardless of whether the anti-particle or particle escapes, the black hole looses mass (more precisely it looses energy).

  So, is there really a bias needed for the black hole to loose mass, or is it something wrong about my understanding of Hawking radiation?

Your understanding of the mass of anti-particles is likely the problem.  They have positive mass, just like particles, because they are particles.


How do we define "up and down" in space?

For humans up and down is related to gravity. So you would need to choose a reference first. If you are in outer space and your feet point to the sun and your head to Jupiter, you'd be rightside up (if the sun is the center) and upside down (if Jupiter is the reference).

North and south do not have any physical relationship with up and down. Neither in our planet, nor in space.

In space you have to choose a coordinate system. In those systems you have a reference plane, latitude and longitude. For latitude (+90 to -90) the positives numbers are those that would correspond to the north of the equatorial plane.


What made the recent supermoon so super?

The thing you have to really understand is that not all supermoons are equal. The official definition of a supermoon, as you state, is that the full moon must coincide with the moon being at perigee.

However, this never actually happens! The moon being at perigee happens for only a single instance of time - less than a second. The moon being officially "full", again only happens for a single instance of time (i.e., when the Sun, Earth, and Moon all form a single line). The chances of these two instances of time happening simultaneously are almost non-existent.

So, when someone calls a full moon a supermoon, they really just mean that the exact full moon instance is "pretty close" to the exact moment of reaching perigee. The closer the better and the more super your supermoon will be.

If you take a look at the Supermoon which occured on Nov. 14, widely proclaimed to be the "superest" supermoon for a long time, you can see that perigee was reached precisely at 11:23 UTC, while the full moon occurred precisely at 13:52 UTC, two and a half hours later. For pretty much every other supermoon, these two events are close enough for the moon to be called a supermoon, but not two-and-a-half-hours close. The closeness in time between reaching perigee and being truly full made this moon especially big and more important, especially bright.

There is also another thing which made this supermoon the superest and that was the actual distance of the Moon's perigee. As with anything in our solar system, the Moon's orbit is not perfectly elliptical, but instead has small variations and perturbations over time. This means that the perigee of the Moon's orbit is slightly changing over time. The November supermoon occured during a particularly close perigee that won't be matched until 2034 supermoon which you reference.


Is it fair to describe a solar flare as a big lightning strike on the sun?

ProfRob gives a more detailed explanation of what a solar flare is in his answer to "What triggers a solar flare?" It helps a lot in distinguishing between CME's, eruptions, prominences and solar flares. There is a lot of confusion, even among space enthusiasts, between these phenomena. This is largely because images of prominences and CMEs always seem to be labeled as solar flares. This has been confusing to me, too. A solar flare might be associated with a large eruption of matter, but that is secondary, and the primary feature of a solar flare is plasma being suddenly moved from one location to another by magnetic reconnection, causing it to collide at relativistic speeds with the chromosphere and photosphere, producing the intense light and x-rays seen from earth. Unlike lightning, a solar flare is not an electrostatic discharge. A solar flare is like lightning in the sense that there is a near light-speed flow of charged particles that result in the emission of bright electromagnetic energy, spanning the entire spectrum from radio to gamma waves.


What is the gaseous-to-rocky ratio of exoplanets?

>  Do other stars have similar gaseous-to-rocky ratios among their planets? For any given stellar system, are there typically as many gaseous planets as there are rocky planets?

With the current instruments and methods we have, we can only access certain population of exoplanets. Terrestrial-mass exoplanets are quite hard to find hence we have more discoveries of Jupiter-mass exoplanets, which does not mean that in reality, we have more chance to have a formation of Jupiter-mass, it just means that they are easier to detect so we detect more. Therefore it is hard to conclude and make statistics right now.

Or do most stars have either mostly gas giants or mostly rocky planets, making our own Solar system an exception?

This is more linked to planetary formation. I invite you to read about the 'core accretion model'. It explains very well the formation of terrestrial and gaseous planets in the proto-planetary disk. But in general, both can form and our Solar System is not an exception, we just don't have the instruments to find a system like it for now.
For example, an alien with the same instruments as us that would study the Sun with the radial velocity method would only detect Jupiter, the other planet would be invisible to it.