Why were two galaxies named "The Sagittarius Galaxy"? Is this common?

In addition to the Andromeda Galaxy (M31)
and the Triangulum Galaxy (M33),
the Local Group includes several dwarf galaxies named after the constellations in which they appear.
In most cases, nearby dwarf galaxies in the same constellation are distinguished by a letter or Roman numeral suffix.


Can someone explain me this diagram from an article?

The Earth's equator is at an approximate 23.4 degree angle to the plain of its orbit around the Sun. That means that to an Earth-bound observer the Sun appears to move north and south of the celestial equator (which is merely the projection of the Earth's equator on to the sky) during the course of the year.

In fact this is what gives us the seasons - so in the the Sun is at maximum northern declination (the equivalent of latitude when projected on to the sky) at the time of the Summer Solstice (around 21 June - hence summer in the northern hemisphere) and at maximum southern declination at the time of the winter solstice (around 21 December - hence northern winter).

But this movement also means different rising and setting points - in the Northern hemisphere the Sun's rise is between north and ast in the summer and between south and east in the winter (unless, that is, you are within the Artic circle in which case it will not set/rise at all for at least some period in the summer and winter).

There are a variety of online tools that show you this effect - this one: https://www.timeanddate.com/sun/iraq/baghdad?month=12&year=2017 - shows that in Baghdad the sun rises - in 2017 - at 61 degrees East on 21 June and at 118 degrees East on 21 December - a difference of 58 degrees.


What does a pulsar look like if viewed from above?

My understanding is that we "see" pulsars (also called "supernova remnants") by the radio and/or other electromagnetic radiations which they emit, which come to us in regular pulses, like the light-beam on the conning-tower at an airport.

This is because the beam of radiation shoots out along the magnetic-axis, while the object spins around its spin-axis, which might be considerably different from the magnetic-axis.

So if we are directly above, it seems that we might not receive any of the radiation, so we might not be able to "see" the object.


Can Mercury hit Earth or Mars in the next 5 billion years?

Yes, after a close encounter with other planets, almost anything can happen, including a split of Mercury into smaller bodies by tidal forces, or capturing as one or more moons of one or more planets, or a sequence of captures by, and escapes from planets, conversion to a ring of all or some of the fragments, collisions with other moons, fragments falling into the Sun, others ejected from the solar system.


Number of photons in universe? I'll take number from a star in unit time and extrapolate

Extending Zephyrs comment as community wiki

See https://physics.stackexchange.com/questions/196366/number-density-of-cmb-photons

In this response, the number of photons per cubic meter is estimated as $n_\lambda = 10^8$. Since the observable universe has a volume of the order $10^{80}\ m^3$, there are about $10^{88}$ photons in the observable universe.


Would a black hole evaporate via Hawking radiation before you fall past the event horizon?

I've asked this question to a couple of physicists a few days ago. Great minds think alike, huh?

First, bear in mind that Hawking radiation is only hypothetical. It is not theory. If we trust that hypothesis, this is what we can get.

In general relativity, black holes can be described through a number of approximations. For example, the Schwarzschild solution for a black hole describes it as an eternal object — not something that exists for some times, and doesn't exist for others. According to this solution, the event horizon must have always existed, and must remain existing eternally.

Schwarzschild black holes approximate black holes very accurately, but as you can tell, they fail to explain how a black hole can form, and (assuming Hawking radiation is real) they don't explain how one could eventually evaporate.

Of course, that solution won't help us. I've kept looking for one that accurately describes an evaporating, creatable black hole, but I've found nothing. The conclusion I've come to, along with those I've asked, is that our question has a major problem: Hawking radiation is explained via quantum field theory.

Thus, you can't simply use a GR solution for a black hole; you'd need some unholy mixture of quantum field theory and general relativity (keep in mind that both GR and QFT are incompatible in many situations).

In the end, it all comes down to how little we really know about black holes. It's not really possible to determine which solution is the best, and our inability to reconcile QFT with GR poses a big problem. The best answer I could give is "nobody really knows what would happen if you kept approaching a black hole".

We don't know if we would reach the event horizon, we don't know if the black hole will evaporate. We simply don't understand them well enough to know what solution would work, or how we would put QFT into it. If we somehow managed to find an approximation that properly combines GR and QFT, I assume (but don't quote me on this) the situation you described would be possible.

If it is possible, by the way, then we could confidently say that a black hole of any size could rip you apart through tidal forces. Tidal forces become weaker as the black hole's size increases, so one would assume a large enough black hole wouldn't rip you apart.

However, if we take Hawking radiation into account, and if your proposed scenario were indeed correct, the black hole would shrink as it evaporates. Since it would get smaller at a faster rate as we approach the event horizon, it would soon be small enough to rip us apart.


When are the symbols ⊕ and ♁ used to designate Earth?

Yes, I don't know how common the globus cruciger is, but it is used at least here for the electron neutrino pp flux at Earth: NEUTRINO 2006: Proceedings of the XXIIth International Conference on Neutrino Physics and Astrophysics.


Convert MJD date range to number of days elapsed in IDL

Like @James K said, the MJD is simply a count, so you can set some arbitrary date to be 0 and go from there.

For example, take today's MJD date and say there was one event today:

Today = 57657

Add three months and 12 days to the next event:

Today + 103 = 57760

So if you called ten days before the first event 0, then the "dates" of the first two events would simply be the following:

Start = Event_1 - 10 = 0
Event_1 = Start + 10 = 10 <-- Semantics right!
Event_2 = Event_1 + 103 = 113

It really is that simple. Just set a variable to equal whatever starting MJD you want to be zero and count earth days between events.


How to communicate Fine Structure Constant to aliens?

Send two pulses of whatever (light, radio) you are using for communication, one a factor of $\alpha$ longer than the other.


When do radio and tv signals become indistinguishable from background noise of the universe?

If an extraterrestrial civilization has a SETI project similar to our own, could they detect signals from Earth?
  In general, no. Most earthly transmissions are too weak to be found by equipment similar to ours at the distance of even the nearest star. But there are some important exceptions. High-powered radars and the Arecibo broadcast of 1974 (which lasted for only three minutes) could be detected at distances of tens to hundreds of light-years with a setup similar to our best SETI experiments.

SETI FAQ


What is a thermal distribution of eccentricities?

The thermal eccentricity distribution was first calculated by Jeans 1919. The probability density function is indeed $$f(e) = 2e.$$

See this blog post for a nice derivation of this beautiful result, which is independent of the "temperature".

The derivation relies on a small 'swindle' by assuming a population of only binaries and not any single or trinary systems; it also has divergent energy since it allows too many arbitrarily tight or loose binaries - it can only be an approximation to real systems.


Is there a name for the star/supernova hypothesized to have been a precursor to the Solar System?

I couldn't find a name for this hypothetical star/nova after some research.

First off, the existence of this former object is hypothetical. Others working in this field have different hypotheses / conjectures; the science is far from settled. Secondly, if true, this star formed a supernova at the time the solar system was born and hence it no longer exists as a star. Thirdly, what's the point? Naming this hypothetical object would add zero credence to the hypothesis, and arguably would detract as being a bit premature.


Is there anything different about the gravitation around a non-spinning black hole and a neutron star of the same mass?

Birkhoff's theorem is very useful: in general relativity, if you are in vacuum and there is a spherically symmetric gravitational field, then it will be the Schwarzschild solution. This solution only depends on the mass, not on the size of the object. So the neutron star and the black hole will give rise to exactly the same orbits.


The nebula from which the sun and the solar system formed contained mostly hydrogen and helium, but also small amounts of heavier elements, including "dust".

Any dust that ended up in the centre of the accretion disc became part of the sun. The sun contains 0.01% silicon, for example. This is only a small proportion, but is still about $2\times10^{26}kg$, 100 times the mass of the Earth. When you add in the Iron, Nickel, Oxygen, Carbon and other elements that make up the rocks, you find that there is enough matter in the sun to make up thousands of rocky planets. (http://www.space.com/17170-what-is-the-sun-made-of.html)

So it is only that part of the dust that happens not to fall into the sun that goes on to form rocky planets.

The details of the process of planetary formation are complex (and not completely understood) however after the formation of the sun, solar wind cleared out the inner solar system of gas and volatiles, leaving only the rocky planets which were too small to hold on a large hydrogen-helium envelope. Further out, beyond the ice-line, water and other volatiles are solid, and much larger planets can form, that gravitationally attract hydrogen and helium mantles.

see How did the Earth come to be in orbit around the sun? for more on planetary formation.


Code/Package for multiple planet simulation

There are of course serious Python packages to simulate solar systems out there, one of which might be found in The random walk of cars and their collision probabilities with planets but I think that other answers that will be posted will give a more complete answer.

I looked into a "roll your own" script as described in answers to How to calculate the planets and moons beyond Newtons's gravitational force? If you don't need to propagate for millions of years, it should be pretty easy and extremely instructive to just do the numerical simulation yourself.

update: The Python package Poliastro may also do exactly what you need (poliastro in Space SE), and while Skyfield does not do this, there are rumblings in Github issues about adding an orbit propagator to it in the future. PyEphem has been depreciated (same developer as Skyfield) but I believe it also does orbit propagation of user-specified objects, though I'm not sure if it's n-body or just Keplerian orbits.

I think that Universe Sandbox will work for you. See also Is a General Relativity approximation available in Universe Sandbox (1 or 2)?

Though it is a complete program and not in Python, Scott Manley talks about Universe Sandbox in several videos.

old one from 2014 is Universe Sandbox 2 (Alpha) - Simulate The Universe For Fun!
"The Goblin" and an Exomoon The Size Of Neptune - Universe Sandbox
Universe Sandbox² - Planet Nine, Asteroids Orbiting Earth & more


Does near light-speed travel doppler-shift the light from the target into gamma rays?

The equation for Doppler shifting towards the blue spectrum (i.e. when you are traveling towards it) is:

$\lambda_b = \lambda_c/(c + v_b)$

where $\lambda_b$ is the shifted wavelength, $\lambda_c$ is the original wavelength, $c$ is the speed of light and $v_b$ is your velocity towards the source.

BUT! When you get closer to the speed of light you need to go relativistic!  When moving closer to the speed of light you also need to take other effects into account like time dilation.  So we need to move to an equation that looks more like this:

$v_{ob} = v_{source}\times\sqrt{(1+B)/(1-B)}$

Where $v_{ob}$ is the observed wavelength, $v_{source}$ is the emitted source radiation and $B$ is your velocity/speed of light.

So lets say that someone is shining some UV light on us at about $4\times10^{-7}~\rm{m}$ and we want to see at what velocity this light is shifted into gamma i.e. about $1\times10^{-11}~\rm{m}$.  We can re-arange that formula to:

$B = [(\frac{v_o}{v_s})^2-1]/[(\frac{v_o}{v_s})^2+1]$

where $B$ will be a decimal of the speed of light.
The term $(v_o/v_s)^2$ is ridiculously small, we are talking magnitudes of $10^{-10}$.

Then when you add or subtract 1, you really dont move that far from 1. Just under 1/ just over 1 you get the answer that You have to be going ridiculously close to the speed of light for this to happen.  Like so ridiculously close MATLAB wont calculate the difference and just gives me the value of 1, i.e. the speed of light.

IN SUMMARY:

Yes but you have to be going so close to the speed of light it would be physically impossible.


How could we detect planets orbiting black holes?

Gravitational microlensing is a way of finding planets that does not care how luminous or otherwise the hosts of the exoplanets are.

The way it would work is that you stare at a dense background field of stars; then when a foreground black hole passes in front of a background star, the light from the star is magnified by gravitational lensing. Typically, the lensing event takes a few weeks for the magnification to develop and subside.

If the black hole has a planet, it may be possible to "see" its graviational potential, which will manifest itself as an asymmetry in the lens light curve or possibly even a little extra spike in the light curve lasting a few hours.

This technique is well established and is already used to detect planets around unseen objects. It is more sensitive than the transit and Doppler methods to planets orbiting a fair distance from their parent star.

The difficulty here is not finding planets around a black hole, it is proving that the planets you found were actually orbiting a stellar sized black hole. The microlensing event would only be seen once and the black hole system would likely be invisible. There is a possibility I suppose that it may undergo some sort of accretion activity after having been found by lensing, and this accretion might then be picked up by telescopes trying to identify the lensing source. I guess a good place to start would be events where a stellar-mass lens is inferred, but no star can be seen after sufficient time has elapsed that the lens and source star ought to be resolved.


Black Hole Collision & Gravitational Waves

Part of the answer is easy. The strain measured in that event was about $0.25\times 10^{-21}$. That is an object $1m$ long would be squeezed by $0.25\times 10^{-21} m$ in one direction and stretched by the same amount in the orthogonal direction.
The strain drops off linearly with distance from the black hole, so to achieve a distortion of 1mm in something the size of the Earth (ie about $8\times 10^{-11}$ it would need to be about $5\times 10^{10}$ times closer, so about $0.03$ light years away, or about 2000 AU. Two 30 solar mass plus black holes at that distance would have disrupted the solar system quite a bit before they collided.
To achieve a distortion of 1mm in something the size of a human they would need to be another factor of $6\times 10^6$ closer, so about $\mathrm{30000 km}$, closer than geostationary orbit. In this case we would certainly have bigger problems than the gravity waves.
What I don't know how to calculate is the energy absorbed by the Earth, or a human, in one of these scenarios. I suspect it would not be all that much, at least from the wave.
Added later: this answer gives the total energy passing through the Earth (about 34GJ with the black holes at their current distance) but offers no ideas for how much is absorbed. This would increase according to the inverse square law if the black holes were closer.


Your premise is incorrect:

The external observer's view does not imply anything about an observer within the event horizon.

The event horizon itself is just an indicator of the edge of the region where light can no longer escape. Light can head inwards as you would expect.

So an observer inside the event horizon will be able to see light reaching them as normal (well, not exactly as normal, as there will be lensing effects) - there will not be an infinite amount of time visible.


How big a dish do I need for radio astronomy?

I am a member of Astropeiler Stockert e.V., and we are fortunate enough to be able to approach this problem coming from the "large side" :-) We have a 25m, 10m and 3m telescopes as well as an interferometer made from two 1m satellite dishes available. All these dishes can be used to do interesting things, but you'll need to match the instrument to your target (and, in a hobbyist setting, often also the target to your instrument).

First of all, you should think about what frequency band you want to work on. 21cm (1420 MHz) is the classical hydrogen line, which lends itself well for mapping neutral hydrogen in the spiral arms of our galaxy. You can expect comparatively strong signals in a quiet frequency band there. Lower frequencies (and large bandwidths) are interesting for pulsars, higher frequencies give access to more interesting phenomena but will require a lot of work on the high frequency side. So my recommendation would be to start out with 21cm.

Secondly, what to look for? In general, the following areas of observation are easily accessible for amateurs:

continuum emission of the Milky Way, perhaps up to creating maps
spectral measurements of selected, "bright" (as in "intense radio sources") objects
Pulsars

Let's look at them in detail:

Continuum measurements are easily done using dishes of any size, but I'd start from 3m upwards to get more interesting results. Calibration of the backend receiver is not trivial, and you should plan a bit of time for it.

Spectral measurements also benefit from larger collecting areas, but you'll also need to set up a suitable backend. They are, however, possible from 3m upwards if you are happy with a bit of mapping of intra-galactic velocities.

Pulsars require a lot of collecting area and bandwidth, so they are mostly the domain of large instruments. We are currently observing several dozens of the brightest pulsars on our 25m dish. The 10m might be sufficient for some of the very brightest. And while there is a report of pulsar measurements using a 3m dish and an RTLSDR frontend, this is a feat that requires experience and dedication. So, I'd recommend this only for 8m and above dishes.

Interferometry with 2x1m dishes (20 GHz) is rather interesting, but very involved in terms of analysis. Two receiving systems help mitigate local fluctuations, and with that setup, we can observe sources down to 2 Jy (with a looong integration time). This setup will give you access to one or two handful of interesting targets like M1, W51 or Cyg A.

Finally, I'd like to recommend the EUCARA conference series (European Conference on Amateur Radio Astronomy) and the SARA group as great starting points. They have conference presentations available online that show what other amateurs are doing.


Is the expansion accelerating or Decelerating?

The light we see now is not a direct indication of how the galaxy was moving at the time the light was emitted. In the cosmological frame, the galaxies aren't moving (on average, at least); rather, space between them is expanding. The rate at which it is expanding is the Hubble parameter.

If cosmic expansion is homogeneous and isotropic, then all distances on the cosmological scale should be affected in the same manner, proportionally, unless other forces are involved. Thus the distance between some galaxy and us is affected in the same proportion as the wavelength of the light:
$$\frac{D_\text{now}}{D_\text{then}} = \frac{\lambda_\text{now}}{\lambda_\text{then}}\text{.}$$
The light we see now is a direct indication of the scale factor, i.e. by what factor the distance has grown since the time of emission. We can also consider distance $D$ as a function of cosmological time $t$ and define a recession velocity as the rate at which it is changing:
$$v_\text{r} \equiv \frac{\mathrm{d}D}{\mathrm{d}t} = \underbrace{\left[{\dot{D}}/{D}\right]}_{H(t)}D\text{.}$$
Therefore, the recession velocity now is given by the Hubble parameter $H(t)$ now.


Any theories about universal expansion due to action at the universal event horizon?

There are no such theories because there is an immediate problem with both scenarios as stated.

It is a fact that a spherical shell of material has no net gravitational influence on material anywhere inside it. https://en.wikipedia.org/wiki/Shell_theorem


Not all models of dark energy assume it constant.

However, when this is assumed true, dark energy is usually compared to the energy density of the vacuum. In other words, while the Universe expands, the dark energy increases proportionally, so its ratio with the Universe volume (the density) stays constant.

Or, better, we generally believe those models where the dark energy is a constant because they fit better our observations of the Universe, in particular the Big Bang model, which describes how our Universe evolves.

If you like some math, you can have it here.


How to calculate the position of the sun in long/lat?

By latitude and longitude of the sun, I think you mean latitude and longitude of a point on the earth's surface where the rays of the sun fall perpendicular to the surface.

In astronavigation, this point is called the Geographical Position (GP) of the sun and terminology to describe this point is slightly different:-

Latitude of the GP is called Declination, and it is numerically equal to what one might call "Latitude". It is measured in degrees North or South the equator.

Longitude of the GP is called Greenwich Hour Angle (GHA). Just like longitude, GHA is measured East or West of Greenwich Meridian.

Describing the sun's position in this way is called the Equatorial Coordinate System.

The Wikipedia page you have used gives the additional formulae below to convert from Ecliptic Coordinates (the steps you followed) to Equatorial Coordinates.

Follow these steps to determine Declination $\delta$ (Equivalent to Latitude).

To get the sun's GHA, you need to first calculate $RA$ (Right Ascension) as shown above and then convert this to GHA by these= rules...

$SHA° = 360 – 15 · RA[h] $

$GHA_{Aries}° = 15 · GST[h] $

$GHA = SHA + GHA_{Aries}$

For the second step, you will need $GST$ (Greenwich Sidereal Time) which is you can calculate if you know UT (UT can be obtained from the time shown by a clock). Use this link to convert directly from UT to GST

If you want a simple, concise way to be able to calculate the positions of all manner of celestial bodies, the sun, moon, planets, stars, then I highly recommended the book Astronomical Formulae for Calculators by Jean Meeus. It is the simplest, most comprehensive collection of astronomical formulae you are likely find anywhere on earth.


What are the lines in this Solar Cycle graph?

In the above plot by Hathaway,
the jagged line peaking at 284.5 in June 1989
is the monthly international sunspot number.
The smooth curve starting and ending a few years later is a
prediction
based on recent observations and statistics of past cycles.

The peak of 212.5 in November 1989 is for a
13-month smoothing
of the monthly number.
The Sunspot Index and Long-term Solar Observations (SILSO)
project at the Royal Observatory of Belgium
archives
these data and provides
this plot
of the monthly and smoothed sunspot numbers together.


Back of the envelope? The energy released is about $GM/R$ per unit mass. For a 1.5 solar mass neutron star of radius 10 km, this amounts to $2\times 10^{16}$ J/kg, or approximately 2/9 of the rest mass energy of the accreted material.

Of order half of this would be radiated and half would go into heating the neutron star.

Timescales are more difficult. Starting when? Once the material gets to the innermost stable circular orbit at around 15 km, then the final 5km to the surface takes a tiny fraction of a second. But from where an object is shredded (much further out), much would depend on its angular momentum.


I'll try to add a bit of context to the Wikipedia article, though the major references are all available there.  The article covers things like the potential formation mechanisms but I guess doesn't really put them into scientific perspective.

Consider, for a moment, a red giant of one solar mass, far up the red giant branch. It has a deep convective envelope, at the bottom of which there is a shell of hydrogen fusing into helium, all surrounding an inert, highly degenerate helium core.  The core at this point is much like an isolated white dwarf.  So, back in the late 1970's, Kip Thorne and Anna Żytkow tried to model (Thorne & Żytkow 1977) what would happen if you replaced the white dwarf with a neutron star.  Such hypothetical objects thus became known as Thorne-Żytkow objects (TZOs).

TZOs have mostly remained theoretical exotica, at least in part because it's very difficult to distinguish them from "ordinary" red supergiants. The main giveaway would be the presence of unusual elements in the atmosphere, produced by the much hotter nuclear reactions happening in the fusion shell around the neutron star core. There have been just a handful of papers published about them, almost all by theoretical groups, in the intervening 40 years.

The only candidate that I think anyone takes seriously is HV 2112. As far as I know, Levesque et al. (2014) found the object serendipitously as part of a survey of red supergiants in the SMC.

PS, following comment:
The consensus view is that a single star couldn't form a TZO because if the core collapses into a neutron star, models predict that the envelope will be ejected.  That is, single star's only evolve into "naked" neutron stars.  Under this assumption, the only way to form a TZO is to have a neutron star collide with or accrete enough material from another star. This could happen if two wandering stars collide. Alternatively, it could happen if one star in a binary collapses into a neutron star.  In particular, it's known that the collapse process is probably not symmetric, so the neutron star gets a "kick" in some direction. If the neutron star is kicked into the companion, it will presumably undergo some viscous drag and settle into the centre.


How to attach binoculars on starwatcher tracker?

I own such a mount. You can turn it to wherever you want by loosening the right ascension (the black “knob” that’s touching the white part just below right of the greenish dovetail plate. You can also loosen the declination with the “knob” that’s around the mounting bolt.


How can I observe Sgr A* with itelescope.net

I expect that all the itelescope.net instruments work at visible wavelengths. Therefore you have no chance at all to image the stars around Sgr A*, since it is behind about 25-30 magnitudes of optical extinction, caused by dust between us and the Galactic centre.

The published images you have seen were taken by large telescopes working with adaptive optics in the near-infrared (wavelengths longer than 1.5 $\mu$m). At these longer wavelengths the extinction caused by dust is only 2-3 magnitudes.


Primordial supermassive black holes. How old are they?

The first supermassive black holes must have been formed earlier than about 750 million years after the big bang, since there is evidence for luminous quasars at redshifts of up to 7 and with likely black holes of a billion solar masses or more at their centres (e.g. Momjian et al. (2013).

The candidate mechanisms to produce such objects in a short period of time are hyper-Eddington accretion onto stellar black hole seeds; the direct collapse of large, primordial gas clouds into black holes; or the merger of stellar-sized black holes in dense clusters, followed by gas accretion. These possibilities are accessibly reviewed by Smith, Bromm & Loeb (2017).


Are Barlow lenses stackable for bright objects?

Yes, stacking Barlow lenses is a common practice to effectively increase focal length by multiplying their individual focal lengths. When I say common, most advanced eyepieces actually have many glass elements and are a type of a Barlow lens themselves, so just by using a single Barlow lens in front of your eyepiece you'd already be, technically, stacking Barlows. It is a technique that is particularly useful for bright object like our Solar system's planets, with Jupiter that you mention perhaps even the best example.

Your image quality might vary substantially tho with the Barlow and eyepiece quality of glass elements and the manufacturing precision. You might also want to play with the order at which you stack them and find an optimal arrangement of all of the glass elements they add to the system. The more precisely they're made, the less the order will matter though. And you'd technically want the less precise ones at the end of the stack, so their deficiencies don't magnify with each added element. And yes, with any additional glass element you'll be losing a bit of the illumination so this is, like mentioned, most suitable for bright objects.

Do not however, under any circumstances, do this when observing the Sun! You shouldn't need additional glass elements anyway, but should you try, you'll most certainly cause them to overheat, expand and permanently deform quite fast, regardless what material their tube and thread are made of, metal, alloy or plastic.


What are the dark spots in this Hubble image of the Crab Nebula?

They are most likely dense clumps of dust absorbing whatever wavelength of light this image was captured in. It's hard to say just how big they are as I can't find any scale for this image, but generally nebulae are lightyears massive, with the Crab Nebula according to wiki being 11 light years. So whatever they are, they're still pretty big.

Edit: Just look at the horizontal band of darker lumps, you can tell that they are between us and the brighter region. The band would be of the same material as the bright green cloud, just less energised as it is further away from the star.


Can we see the Big Bang happen if we look far enough?

No. The furthest we can see is the cosmic microwave background radiation (CMB). Early on (after the big bang), matter was fully ionised and the electrons frequently interacted with the photons. That has two consequences. First, the radition was that of a blackbody at the same temperature as the matter. Second, the universe was opaque, i.e. photons couldn't travel very far.
Due to the explosive expansion of the universe, the temperature decreased all the time until eventually atoms formed. This is called the epoch of re-combination, though the "re" makes little sense. At that point the universe became suddenly transparent and we can see most radiation emitted ever since. In particular the radiation field from the epoch of recombination has been redshifted and appears to us as the CMB.

edit
To answer the speculation in the comments what if the universe was not opaque. Well, all we would be able to so is some (nearly) isotropic radiation field emitted very early on (and redshifted by the cosmic expansion since the big bang). The CMB is exactly that, but emitted at redshift $z\sim$1000 (the epoch of re-combination) rather than larger $z$.


In what form was the space at singularity?

A singularity in the strict sense is not assumed in the beginning of the big bang, but instead a Planck epoch governed by a yet to confirm quantum gravity. General relativity and quantum theory, are not consistent in this phase. This means, that space and time, if they existed at all in some sense, didn't follow the laws of physics we inferred from experiments.

This epoch is thought to have been followed by a grand unification epoch, where gravity is already separated from a unified strong, weak and electromagnetic interaction.

Then inflation is thought to have followed, "driven by a negative-pressure vacuum energy density". That's when spacetime expansion is thought to have accelerated rapidly, weak and electromagnetic interaction still unified.

All this is thought to have happened within the first $10^{-32}$ seconds.

Most of the physics in those early phases is not (yet) accessible to experiments. A few cosmic ray experiments may try to look into the energy ranges, thought to have been present in the early epochs of the universe.

More on the chronology of the universe.


Cubic spline interpolation in light curve of pulsating stars

The main thing a cubic spine lets you do is have a continuous second derivative.  If all you need is a continuous curve, connecting the dots with straight lines suffices, with no overshoot, but a very discontinuous derivative.  So if you want to think about the derivative of the light curve, you'd have to go at least to a quadratic spline, which allows a continuous first derivative but introduces the potential for overshooting (a curve that goes outside the range of the points themselves).  If you actually want to think about the second derivative also, then you'll need that to be continuous, so that will require a cubic spline.  But the potential for overshoot keeps getting worse when you add more continuous derivatives.  Personally, I find cubic splines to be very unreliable, the overshoot can be terrible unless you have a very nicely resolved curve with not much noise, but that may be just what they do have.


How can space expand in accelerating speed when dark energy is constant?

This is a very good question and it comes down to what is meant by "accelerating expansion".

In cosmology we define something called the scale factor $a$. By definition at the present time $a=1$, billions of years ago when linear distances on a cosmic scale (e.g. the distance to a faraway galaxy) were half what they are now $a$ was equal to $0.5$ and many years in the future when cosmic linear distances have increased to twice their current size the $a$ will equal $2$.

$a$ is a function of cosmological time and we also define its 1st derivative and 2nd derivatives, with respect to cosmological time, $\dot{a}$ and $\ddot{a}$.

$\dot{a}$ is a measure of the Universe's rate of expansion and so when $\dot{a}$ is increasing with time we say that the expansion of the Universe is accelerating. $\ddot{a}$ is a measure of the change of the rate of the Universe's expansion so equivalently we say if  $\ddot{a}$ is positive then the expansion of the Universe is accelerating.

However note we have defined $a$ at the present time to be $1$, but the present time is entirely arbitrary. This means that $a$, $\dot{a}$ and $\ddot{a}$ depend on when we define $a$ to be equal to $1$. To get round this can instead think of the Hubble Parameter $H = \frac{\dot{a}}{a}$ as a better measure of the rate of expansion as it does not depend on when $a=1$.

In a de Sitter Universe (spatially flat, empty, positive cosmological constant) the Friedmann equations reduce to:

$$ \frac{\ddot{a}}{a} = H^2 = \frac{\Lambda c^2}{3}$$

From this we can see that $H$ and $\ddot{a}$ are both positive constants. So in one sense (the sense of $H$) the rate of expansion of the Universe is constant, but in another sense (in the sense of $\dot{a}$) the rate of expansion is increasing. So we can see a link between constant energy desnity (the energy density of the cosmological constant is constant) and a constant rate of expansion as long as we define the rate of expansion in a certain way.

In the currently-favoured cosmological model dark energy is a cosmological constant and we are currently in an era when dark energy dominates the dynamics of the Universe. (Looking at the full Friedmann equations) the contribution of matter reduces $\ddot{a}$, but as the cosmological constant dominates it is still positive and we say that the Universe's acceleration is expanding. Looking at $H$ however we say that, because the energy density of matter decreases with time asymptotically to zero, $H$ is decreasing asymptotically to the value given above for the de Sitter Universe.


Will all of the gas in the universe be converted into red dwarf stars?

No. The reason is that gas recycling only recycles only about 40-50% of the gas in a sun-like star, leaving the rest as a white dwarf that slowly cools off. Heavier stars are even less effective in converting their gas into a condensed core. So the process will tend to produce a fair fraction of heavier white dwarfs now in the early stages of the stelliferous era, followed by a slow emergence of lighter white dwarfs from red dwarfs in the rest of the era.

Adams and Laughlin estimate that the final mass fractions will be brown dwarfs (never fused) 9.7%, white dwarfs 88%, and 2.4% neutron stars and black holes. In the post-stelliferous era some of the brown dwarfs will merge and make a few extra red dwarfs .


What are the "lines" in a constellation or asterism called?

My understanding is that the stick-figure lines (in terms of applying them to every constellation) are a fairly modern invention - before that the individual stars were not (generally speaking) linked together directly but were instead highlighting important areas of each constellation figure. The constellation figures were fully 2D and potentially quite detailed, and there was not always a one to one correspondence between stars and details. In this sense one is free to modify the existing stick figure patterns to suit one's needs. Many of the stars visible to the naked eye have individual names (sometimes more than one name in fact) and these names usually mean/meant something about the constellation figure they are/were in. It's not always possible to follow these names when redrawing the figures (the constellations we use today are not all ancient and star names remain from figures we no longer use), but usually you can accommodate most of them. This is one of my main problems with H.A.Rey's versions - they often pay no attention at all to what the star names mean and you end up with a star named for a tail now marking the figure's mouth. :)


Why do zodiac constellations not look like their names?

Who says they don't? Surely it just depends on how you play "join the dots"? The human brain will make out patterns where it wishes and what things look like is largely in the eye of the beholder.
The second explanation isn't true. Although the relative positions of the stars do change, they have not changed very significantly over a period of 3000 years, certainly not enough to change the basic shape of the constellations.


If you know how to reduce raw spectrum data, you may look at the official archives of big observatories (ESO, MAST, CADC). But a quicker and much easier way could be to use the following portal:

http://archive.eso.org/scienceportal/home

Typing the name of some Wolf Rayet stars will lead you to reduced data and thus spectra.


Any book on `How do they know it (in Astronomy)`?

There are books that treat the general history of astronomy, such as those mentioned in the previous post.

The History of Astronomy: A Very Short Introduction by Michael Hoskin.
History of Astronomy by George Forbes

Some books focus on discoveries from a certain period of history

The Cosmic Century
by Malcolm S. Longair
details discoveries that were made in the twentieth century.

Other books retrace the "how we know it" of specific phenomena

Flash
by Govert Schilling
traces the story of how we discovered what we know about gamma ray bursts.
Cosmic noise : a history of early radio astronomy,
by Sullivan, Woodruff Turner III
focuses on what we learned from radio astronomy.


Some classical Cepheids pulsate simultaneously in two or even three modes.
Their lightcurves can be explained as a overposition of fundamental plus overtone modes.

The terminology you used is also found in this paper (where you have also other references):
"the discovery of many double-mode Cepheids (DMCs) pulsating in both the fundamental and first-overtone modes (F/10) and the first and second-overtone modes (10/20)"

Other reference: Classical cepheids


It's an observational fact that many disk galaxies have a color gradient with redder (or as you say 'yellow') stars in the central regions and bluer stars in the disk.
There are two things that can cause this central redness.

Age, older stars being redder than younger stars
Metallicity, with red stars being more metal rich than blue stars

Whether it's age or metallicity or both causing these gradients is topic of current research. It's a popular theory that galaxies grow inside-out, which causes a red center.


Do celestial objects need to be big to have liquid water on their surfaces?

Liquid water can't exist in a vacuum. If there is no pressure, then the boiling point will drop to the freezing point and so there will either be ice or water vapour.
And if the world is "small" then its gravity won't hold on to any water vapour, and it will be lost to space.  The Earth can have liquid water because its gravity is strong enough to hold onto water vapour, and provide vapour pressure to raise the boiling point to 100 degrees, which is hotter than the temperature due to the sun.
A small world can have a sub-surface water layer, as the ice above it will stop the water from boiling off into space. Enceladus has such a layer, but if Enceladus was a moon of the Earth, the sun would have melted the ice and the water would have boiled off long ago.


Why does the Tycho-2 catalog not include Hipparcos stars?

Could you give a source or example: The Tycho-II catalogue paper by Hog et al. (2000) suggests there are about 19,000 stars in Hipparcos or Tycho-I that are not in Tycho-II. This can be because they are too bright (but that is not very many stars) or for other reasons that are not clearly specified (binarity is an issue I think).

Looking at this website, it suggests 4% of stars did not have proper motions and could not be assigned epoch 2000 positions in Tycho II but I don't think these are missing.


Why are the spherical and cartesian galactic coordinates in the ATNF Pulsar Catalogue different?

The ATNF Pulsar Catalogue's galactic longitude and latitude are heliocentric,
but the origin of their rectangular coordinates is near the center of our galaxy.

The catalogue documentation,
section 6. Distances, says:

  The Galactocentric coordinate system (XX, YY, ZZ) is right-handed
  with the Sun at (0.0, 8.5 kpc, 0.0)
  and the ZZ axis directed toward the north Galactic pole.

From a heliocentric point of view
(U, V, W)
= (8.5 - YY, XX, ZZ).
Then $\tan{l} = V / U$ and $\tan{b} = W / \sqrt{U^2 + V^2}$
as you'd expect.


If the Universe is infinite, would heat death be impossible?

Heat death of the Universe doesn't mean that there is no energy in it, it means it has reached thermodynamic equilibrium and hence no useful energy in it. This could in theory occur in an infinite Universe.

Of course thermodynamics is statistical in nature, so an argument could be made that some extremely unlikely event in a region of the Universe could destroy thermal equilibrium in that region and that the larger the Universe the more likely that some region like that exists.


Why nobody cares to give the nearest stars better names?

There are a thousand or so stars within 50 light years, most are very dim red dwarfs, and most are unexceptional. If we invent "names" then we would have to learn the names of all the stars.

If we want to name all the stars that can be seen with a telescope, we would have to come up with a billion or so names.  The idea of naming stars rapidly becomes impractical - nobody is ever going to learn thousands of names which were chosen on whim.

So what could be a "better name" than just making up a name based on one's heroes, friends, gods, or jokes?

Instead we can make a list of stars, and name them systematically. We could use the part of the sky that the star appears in, and some letter or  number. Or we could use the coordinates that the star appears at to label it. Or we could just give each star a number. This way we don't need to invent names, we just use our system. When we do this we get "names" like "omicron ceti" (using the letter "omicron" and the part of the sky "ceti")
or "HD 128621" (star number 128621 in the list prepared by Henry Draper). or WISE 0410+1502 (a brown dwarf discovered by the Wide-field Infrared Survey Explorer, and located at 04hr10min, 15deg02min in the sky)

Cataloguing is simply a better way of naming stars as there are so many of them.

When a star is exceptional in some way, then there may be a reason to give it a name - Barnards Star (moves very fast), Sirius (very bright). But most stars are unexceptional, and there are so many of them.


Multi-messenger astronomy: what is the potential of simultaneous detection of gravitational waves and neutrinos from a supernova?

This article basically seems to answer the question. They quote from an earlier study:

  "Although no CCSNe have currently been detected by gravitational-wave detectors, previous studies indicate that an advanced detector network may be sensitive to these sources out to the Large Magellanic Cloud (LMC). A CCSN would be an ideal multi-messenger source for aLIGO and AdV, as neutrino and electromagnetic counterparts to the signal would be expected. The gravitational waves are emitted from deep inside the core of CCSNe, which may allow astrophysical parameters, such as the equation of state (EOS), to be measured from the reconstruction of the gravitational-wave signal."

Since we know from SN1987A that neutrinos from a supernova can be detected at that range, that seems to be a "yes". The biggest uncertainty seems to be how much gravitational wave energy would be emitted by the supernova, and at what frequencies, which depends on a relatively detailed understanding of exactly how the matter moves around in the explosion, one simulation of which is illustrated in the (rather awesome) video in the article.


What factors influence the Widmanstätten pattern in iron meteorites?

It’s been a while since I studied this, and meteorites are very much not my field, so I’ll share my current understanding but would welcome corrections from those more informed :-)

These meteorites are mostly iron with a small amount of nickel. For a given composition of iron and nickel, there are different crystal structures that can be formed. Of relevance here are two crystal structures called “taenite” and “kamacite” (in the iron-carbon system where very similar patterns can be formed, these would be called “austenite” and “ferrite”, but they refer to the same crystal structures).

Whether taenite or kamacite is more stable depends both on the temperature and the amount of nickel present. At high temperatures, when the meteorite is young, it will be all taenite. However, as it cools, we eventually reach a temperature where, energetically speaking, the universe would like a mix of both taenite and kamacite.

However, just because it’s now energetically favorable for kamacite to form, there still needs to be a mechanism by which some of the taenite crystal can transform into kamacite.

The Widmanstätten patterns only form at very low undercoolings, in other words when the temperature is only just in this range where kamacite is starting to become stable.

What you’re actually seeing in this pattern are the plates of kamacite that have formed. The exact pattern you see will depend which slice of the meteorite you decide to polish, and is related to the underlying symmetry of the crystal structure.

The reason you don’t see these patterns be so large on earth is for two reasons: one is simply that the grain sizes themselves are much smaller, and so the size of the microstructure is limited by that. The other reason is that the cooling rate is much faster, so the patterns can’t grow as large.

When these plates nucleate, the extent (length/width) of the Widmanstätten plates can grow rapidly, but they will only get thicker at a rate that's limited by the diffusion of nickel atoms between the iron atoms. This is because the sides of the plates have a coherent interface with the parent taenite phase and diffusion is slower, whereas the tips/edges of the plate have an incoherent interface where diffusion is much more rapid.

I believe the rate of the plate growing longer is related to the degree of undercooling, so at very slow cooling rates like those in space, the plates grow longer much more slowly. This then gives time for atoms diffuse on the plate sides for the plate to grow thicker. And therefore, eventually, we have this beautiful microstructure large enough to see with the human eye.

That’s my understanding, however I think there isn’t yet consensus on exactly how Widmanstätten patterns form, the role that diffusion plays, and also the role of other additions in mediating this (e.g. the presence of phosphorous in the case of iron-nickel meteorites).

I hope this helps!


What would the CMB look like somewhere else in the universe?

The CMB is nearly featureless, but the operative word is nearly. The tiny temperature fluctuations that we've measured should look about the same anywhere within tens to hundreds of millions of light years of Earth or so. This would allow you to get an absolute orientation relative to the CMB and thus relative to Earth.

It's fairly easy to measure velocity relative to the CMB, so if you measure your velocity relative to the CMB somewhere else (within that 10-100 million lightyears of Earth, anyway) and if you know your orientation, you can compute your velocity relative to Earth.

But I don't see any way to easily measure the distance from Earth except very approximately. (The CMB that you see from any point is different because it comes from a different sphere. If you move far enough from Earth, the temperature fluctuations you see are no longer the ones we see from Earth because you're sampling a different part of the 300-year-old universe.)

Later:
I've been thinking about the position problem and there's a solution, but it's complicated.  The CMB samples the universe of 13.7 billion years ago and from any point in it see the distribution of matter in a shell that old. If we move to another point, in effect, the shell shifts also and samples different parts of the gas. The key point is that the faint ripples in the gas's temperature that we see as mottling in the CMB are three-dimensional bubbles.

So if you shift a short distance in space, the CMB will shift the most in the direction of motion (both in front and behind) and the least at right angles. Furthermore, the amount of shift will depend on the angular size of the fluctuation you're looking at. Low-angular-frequency fluctuations come from physically large ripples int he gas and you need to move a long distance to move the shell you are sampling out of the bubble. High-angular-frequencies ripples come from smaller bubbles and shorter shifts in position will take the shell you observe out of one bubble and into another.

Waving my hands just a tiny bit, it should be possible to use statistical methods to estimate the amount and direction of movement by comparing two detailed CMB maps. A major limitation of the method is that the finer the angular resolution, the sooner you lose all correlation between the two views fo the CMB, but the coarser the angular resolution, the coarser your measurement will be.

Numbers? As long as we're not talking about jaunteing many billions of light years, we can ignore general relativistic issues. A fluctuation with an angular size of 1 degree corresponds to a CMB fluctuation which appears to us right now to be around 250 million light years across and consequently ought have a good chance to be visible for moves within roughly that distance. This suggests that the CMB method ought to be able to give you a rough position up to maybe a half-billion or even a billion light years, but will drop off rapidly in accuracy after that.

Note that for shorter hops, it ought to be possible to actually recognize galaxy super-clusters and get a position that way.


Where can I find a set of data of the initial conditions of our solar system?

Horizons Ephemeris generator can give you a planet's position and velocity vectors at a specified time.

This is one set of possible options:

Clicking Generate Ephemeris on that page will give you position and velocity vectors:

Above the position and velocity vectors are the Julian date as well as the more conventional date.

Wikipedia can give the masses of the sun and planets.


Calculating object velocity at perihelion

The vis-viva equation is commonly written like this:

$$v^2 = GM\left(\frac{2}{r} - \frac{1}{a}\right)$$.

For $r=a(1-e)$:

$$v = \sqrt{GM\left(\frac{2}{a(1-e)} - \frac{1}{a}\right)} = \sqrt{GM\frac{1}{a}\left(\frac{2}{1-e}-1\right)} = \sqrt{GM\frac{1}{a}\left(\frac{1+e}{1-e}\right)}$$.

The derivation of the vis-viva equation is not at all trivial and can be found here.

The product $GM$ is also called the standard gravitational parameter and for solar system bodies is often known more accurately than $G$ and $M$ separately. For the Sun $GM_☉$ is about 1.327E+20 m³ s⁻² which in different units is 1.327E+11 km³ s⁻² or about 1.0 AU³ year⁻².


Why don't small stars end up as a black holes?

The actual answer has nothing at all to do with the temperature.

Even low-mass stars would form black holes if they ran out of nuclear fuel to burn, and simply cooled whilst being supported by "standard" gas pressure in their centres.

That is because that gas pressure would be proportional to the temperature, but the star is able to cool so it would need to shrink and use gravitational energy to stay hot, but of course eventually it would disappear inside its event horizon and become a black hole.

The real reason this does not occur is electron degeneracy pressure. This is a quantum mechanical effect, related to the Pauli exclusion principle, that does not allow two electrons to occupy the same quantum state. As the gas gets squashed in the core, the electrons are forced to fill higher and higher momentum states in order to "avoid" each other. Because these electrons have large momentum they also exert a large pressure - degeneracy pressure. This degeneracy pressure does not depend on temperature and so the star (it is a white dwarf when a low-mass star arrives at this stage) can cool and remain in hydrostatic equilibrium without shrinking any further. Thus it avoids becoming a black hole (or even a neutron star) and simply cools at nearly constant radius.


tl;dr: This is the famous "Blue Marble" image taken by one of the Apollo 17 astronauts. This particular rendering is labeled in Wikipedia as

Image redone with colors

From Wikipedia's The Blue Marble:

The Blue Marble is an image of Earth taken on December 7, 1972, from a distance of about 29,000 kilometers (18,000 miles) from the planet's surface. It was taken by the crew of the Apollo 17 spacecraft on its way to the Moon, and is one of the most reproduced images in history

The question asks for:

Source of red in earth's photographs
What is the red in this picture of earth?

This is not asking what land mass appears red, it asks for the source of the red color.
That's a good question and I think it has a disappointing answer in (at least) two parts:

This is an old image and has been reproduced many times. Below are another versions of it. As it continues to be reproduced and "color corrected" each time the original colors may be lost in reproductions.
If the image was from a weather or other Earth-observation satellite, they would be false colors generated for coding the real wavelengths that the satellite imaged into human interpretable colors, since rarely if ever is a space-based Earth-imaging camera fitted with regular RGB color filters that match Human's sensitivities. Instead, filters are optimized to tell the scientists that will be reviewing each satellites data best whatever it is that those particular scientists are looking for. Very often "red" in images is used to code infrared wavelengths, but since some satellites have several or even a dozen or more wavelength channels, anything can happen.

The link from the article is: https://l.hdnux.com/350x235p/photos.demandstudios.com/getty/article/41/249/87467861_XS.jpg
Here's a bigger one: https://images.unsplash.com/photo-1564053489984-317bbd824340?ixlib=rb-1.2.1&ixid=eyJhcHBfaWQiOjEyMDd9&auto=format&fit=crop&w=2014&q=80
Here it is again at https://www.innovatorsmag.com/enter-earth-school/ and the credit says "Photo by The New York Public Library on Unsplash" and while they are amazing I didn't know that The NYPL has a branch in Space!
and here it is cropped below, and then analyzed. It looks like the center of the image is roughly 29 degrees South latitude, and 37 degrees East longitude.
It's not from a geosynchronous weather satellite, it's from something that's far from Earth but in an unusual orbit.
Earth's axis is tilted 23 degrees to the ecliptic, and for example DSCOVR's halo orbit around Sun-Earth L1 is big and Wikipedia tells us that it

orbits the L1 point in a six-month period, with a spacecraft–Earth–Sun angle varying from 4 to 15°

so that means that at about 1.5 million km from Earth it could have easily taken this image.

Click for larger sizes. left: credit: The New York Public Library and CC BY right: approximate subsurface point of spacecraft is at -27 degrees latitude, so not from a geosynchronous spacecraft.
But it wasn't taken by DSCOVR!
Going to NPYL at unsplash:  https://unsplash.com/@nypl then searching for "earth from dscovr" I found https://unsplash.com/photos/yEauzeZU6xo and the caption reads:

Full Disk Earth, Apollo 17, 1972

So after all this hullabaloo about narrow band filters from satellites this turns out to be a hand-held photograph from a large format 70-millimeter Hasselblad camera by an Apollo 17 astronaut using Kodachrome or whatever!
So the image is nearly 50 years old and has been reproduced a zillion times in a zillion ways, and the exact colors seen reproduced today on some random website are meaningless!
From https://commons.wikimedia.org/wiki/File:The_Earth_seen_from_Apollo_17.jpg#/media/File:The_Earth_seen_from_Apollo_17_(AS17-148-22727).jpg

Image redone with colors 2,400x2,400 (3.33 MB)
"The Blue Marble" is a famous photograph of the Earth taken on December 7, 1972 by the crew of the Apollo 17 spacecraft en route to the Moon at a distance of about 29,000 kilometers (18,000 statute miles). It shows Africa, Antarctica, and the Arabian Peninsula.

From NASA Johnson Space Center's https://eol.jsc.nasa.gov/SearchPhotos/photo.pl?mission=AS17&roll=148&frame=22727

Image Caption: This classic photograph of the Earth was taken on December 7, 1972. The original caption is reprinted below:
View of the Earth as seen by the Apollo 17 crew traveling toward the moon. This translunar coast photograph extends from the Mediterranean Sea area to the Antarctica south polar ice cap. This is the first time the Apollo trajectory made it possible to photograph the south polar ice cap. Note the heavy cloud cover in the Southern Hemisphere. Almost the entire coastline of Africa is clearly visible. The Arabian Peninsula can be seen at the northeastern edge of Africa. The large island off the east coast of Africa is the Republic of Madagascar. The Asian mainland is on the horizon toward the northeast.


Electrons and protons do not annihilate each other, since protons have a much greater mass than electrons. Only matter and antimatter does, since a particle and its antimatter counterpart have the exact same mass.

However, under the right conditions, an electron and proton can combine. This forms a neutron, since the charges will cancel each other out. This process is called electron capture, and it's the reason why neutron stars form.

I'm not an expert on black holes (I haven't studied GR), but I know that a lot of black holes form from neutron star collapses.


Delta Velorum - non-observed variable star

Yes it varies in the visible spectrum.
The paper  The nearby eclipsing stellar system δ Velorum describes why this is difficult target.  Surprisingly it is because it is so bright.
The absolute brightness of a star will vary significantly due to absorbtion in the atmosphere.  If the sky is slightly hazy, then the stars will be dimmer.  To measure the magnitude of a star, then, you don't directly measure how much light is coming from it, but compare it to nearby stars.  But δ Velorum is one of the brighter stars in the sky, so there are fewer nearby comparison stars.  This, combined with the long period of the orbit makes observations harder. This probably explains the shortage of observations.


Can we violate law of physics in any other planet?

If there is a way to violate a law of nature, it will become a law of nature as soon as it is discovered, studied and formalized into a scientific theory. Therefore, many new discoveries in science do violate (then current) laws of nature, but will not do so for very long. Often there are all sorts of medals and prizes involved too.


Confusing satellite-like object seen in night sky

At first blush, this sounds like it was a large, low-Earth orbiting satellite, like the ISS for example. Depending upon the angle above the horizon the transit time could be very low s you described. Your position would determine whether the satellite was illuminated by the Sun or not at that local time.


How to tell which of the ecliptic cross horizon angles is to the east?

These steps can be used to find which of the two horizons is to the east:

The local sidereal time equals the right ascension ($\alpha$) of the meridian. From equation (13.3) that transforms the ecliptical into equatorial coordinates, the ecliptic longitude of the meridian ($\lambda_m$) can be calculated. Since the ecliptic latitude is zero ($\beta=0$) on the ecliptic,

$$\frac{\tan(\alpha)}{\cos(\epsilon)} = \tan(\lambda_m)$$

Note that $\lambda_m$ must be in the same quadrant as $\alpha$. If this is not the case, add 180 degrees. Of the two ecliptic longitudes calculated to intersect the horizon, the one that is approximately 90 degrees greater than the adjusted $\lambda_m$ is the one that is in the east.


Could the unseen companion of HR 6819 be a close binary system?

The absolute minimum mass of the unseen companion is $4.2 M_{\odot}$ (and my reading is that this really is an unlikely minimum, requiring an abnormally faint primary for its spectral type and an orbital inclination of 90 degrees, but let's go with it) and the argument is that a main sequence star of this mass would make its presence known in the spectrum. But let's suppose that instead of a $4.2M_{\odot}$ main sequence star, we had  a pair of A stars with $2.1M_{\odot}$ each, in a tight binary. The luminosity of ths combined binary would be less than 25% that of a $4.2M_{\odot}$ single star, so might have gone undetected.

The orbital period of the B3 giant primary with the putative black hole is 40 days. Assuming the minimum mass of the system as a whole is $9.2M_{\odot}$ (for an inclination of 90 degrees), then Kepler's third law yields a binary separation of 0.48 AU.

The radius of a $2.1M_{\odot}$ A star (like Sirius A) is about $1.7R_{\odot}$, so the minimum separation of a pair of such objects in a binary system is about $3.4R_{\odot}$, or 0.016 AU.

As a rule of thumb, as long as the separation of the closely orbiting stars is less than about ten times that of the wider binary then that ought to be stable to tidal disruption (though it also depends a bit on the masses too).

There are examples of systems where there is a wide binary, where one of the components is itself a triple featuring another close binary, in the catalog of binary stars by Tokovinin (1997). These would be multiples with a level 112 in his notation and there are a dozen examples in this catalogue. Perhaps the most famous is Tau CMa, which consists of an O supergiant in a wide, just resolved, 250 year orbit with another bright B-star. However, the O-type star is in a 154 d binary system with a pair of eclipsing B-stars that orbit each other every 1.28 d.

So I think you have a legitimate case that this may not be what the authors claim.

Edit:
I was interested enough by this to email Rivinius and put this scenario to him. The reply raised one or two interesting objections.

Firstly, he suggests that the reduction in luminosity of the secondary (compared with a main sequence single star of double the mass) might not render it invisible in the spectrum. The reason is that the A-type stars have narrower (metallic) lines that could be observed than the B-type giant. He cites the star 66 Oph as an example where this has been done. I am not totally convinced by this, or at least not convinced that it has been demonstrated.

The second point is better. If the inclination is anywhere near edge-on (as required for the unseen companion to be at its minimum mass), then in a close binary you would expect eclipses or at least significant ellipsoidal modulation due to distortion from spherical symmetry. Using Spica as an example (a 4d binary) Rivinius argues that even with dilution that this modulation would have been picked up in the SMEI and TESS light curves discussed in the paper.

If the binary were more face-on to reduce this ellipsoidal modulation, then of course the companion mass would increase dramatically from the 4.2 solar mass minimum, making even a binary companion impossible to hide.

So my conclusion changes. I'm reasonably convinced by the second argument - that a close binary companion would be revealed through ellipsoidal modulations.


Where to get Polaris Right ascension value from

There are several published star catalogs that list right ascensions for stars, along with a lot of other almanac-type data.  Wikipedia's a good resource for common stars (e.g. Polaris).

Right ascensions change over time, as stars move through space.  Star catalogs will have a year associated with them, which is the year that the measurement was taken.


Are there any planets or objects orbiting in Goldilocks zone?

From the Wikipedia article Circumstellar habitable zone, which is just another name for the Goldilocks zone:

  Estimates for the habitable zone within the Solar System range from 0.38 to 10.0 astronomical units, though arriving at these estimates has been challenging for a variety of reasons. Numerous planetary mass objects orbit within, or close to, this range and as such receive sufficient sunlight to raise temperatures above the freezing point of water. However their atmospheric conditions vary substantially. The aphelion of Venus, for example, touches the inner edge of the zone and while atmospheric pressure at the surface is sufficient for liquid water, a strong greenhouse effect raises surface temperatures to 462 °C (864 °F) at which water can only exist as vapour. The entire orbits of the Moon, Mars, and numerous asteroids also lie within various estimates of the habitable zone.

See also these questions:
 - What is the Habitable zone all about?
 - Why the Circumstellar Habitable Zone is defined as it is, if life could be possible outside of it?


Will biological process be disrupted under strong gravitational time dilation?

Remember this is a theory of "relativity". Now, time dilation due to gravitation effects is rather outside our normal experience. But there are relativistic effects that that you experience all the time.

I'm sitting on a train. Relative to my computer I'm not moving. Relative to the track I'm travelling at 100km/h. Relative to the sun I'm moving about 30 km/s, as the Earth orbits. Relative to a distant galaxy I'm moving close to the speed of light.

But in my frame of reference, I'm not moving at all. So, even though I'm moving at nearly the speed of light, relative to some distant galaxy, I don't notice any slowing down of my watch. My watch is moving with me, in my frame of reference.

It would be the same if you were experiencing intense gravitational time dilation. For example, if you were falling around a black hole. Your watch would be in the same frame of reference as you, so there would not be any observable difference, locally.

If you tried to stop yourself falling, the force required would crush you. Small black holes have such extreme variation in gravity that your head and feet would be pulled apart by the tidal difference, and if you passed the event horizon, you can never come back. So you would be dead if you tried the experiment in practice. However "gravity" itself can't hurt you, only variation in gravity or your own attempts to act against gravity.


Black hole in the centre of our galaxy is not active (which means it is not feeding much matter) as compared to black holes in the centre of other galaxies. So it does not emit that much gamma rays as other supermassive black holes does but still emits some amount of gamma rays. Along with gamma rays cosmic rays are also emitted which are ward off by radiation pressure of our sun's heliosphere.


Starting a fire in a cold planet that was full of flammable gas

The outer parts of Neptune are mostly hydrogen and helium. There are small amounts of other gases such as methane, ammonia and water vapour. However, there is no oxygen at all.

If you took some of Neptune's outer layer back to earth and mixed it with our air, it could burn. Even very cold hydrogen can burn (it soon heats up!)  This couldn't happen on Neptune, because a fire needs both fuel and oxygen to burn.

It is very unlikely that any planet would have large amounts of both fuel and oxygen in its atmosphere. Oxygen is very reactive and will react with any flammable gases to produce (mostly) water and carbon dioxide. Oxygen is not stable in a hydrogen atmosphere over a period of millions of years.  If we find oxygen in a planet's atmosphere, we can be fairly sure that something on the planet is making it.


Is there a natural process by which hydrogen is generated from heavier elements in the cosmos?

There are a couple of relevant questions one would want to ask:

1) Do protons decay, and if so, what do they decay into? The answer appears to be no, or at least the theoretical lifetime of the proton must increase as a results of these experiments. If they do, eventually the universe could end up in a state of radiation (and dark energy, and dark matter, unless they also decay).

2) Is Hydrogen a bi-product of any natural decay process? Below is a table of all known nuclides.

As you can see, the majority of the elements (not necessarily by number or mass in the universe) do decay through some type of process. There exists a 'stable' ridge (called the island of stability, surrounded by the sea of instability) of elements which will happily exist forever.

The question is, which modes of decay produce protons (Hydrogen nuclei)? Well, there is proton decay (not the proton itself decaying), which is colored in red, though I have to admit that I don't know exactly what this refers to. Fission bi-products are gamma rays (high energy photons), neutrons, and daughter nuclei (see Decay chain). Though, I should mention that free neutrons produced from this type of radioactive decay are not long lived, decaying into a proton and an electron (this process takes on average approximately 11 minutes). By this logic, also isotopes which decay by emitting neutrons, colored purple, would also eventually produce protons. $\beta^{-}$ and $\beta^{+}$ refer to the beta decay process, where the minus sign refers to the emission of an electron and the plus sign refers to the emission of a positron (the anti-particle of the electron). $\alpha$ decay is the emission of a Helium nucleus, which is stable.

Now, given that there are ways for heavy elements to naturally produce protons, the question I would ask is what is the rate of these processes in the universe compared to fusion processes occurring at the centers of stars. I'm not sure that I could give you an answer to this question (or even point you to the appropriate material), but in principle these rates are known. I'd imagine that it'd be quite a lot of bookkeeping to get it correct.


Why are some universal entities round and others are flat?

Preliminary note: there is a continuity of more or less flattened shapes. No macroscopic objects are really spherical, or really flat. Take Jupiter: it has a rotation period around its axis of 10 hours, and you can clearly see that it is an ellipsoid. It's the centrifugal force, combined with the gravitational attraction, that shapes these objects.

Now to your question: Planets and stars are dense bodies, and rotate as single objects. On the other hand, the Oort cloud, Saturns rings, stellar clusters and galaxies are made of independent particles, free-floating in lots of empty space. Depending on the formation history and gravitational environment (e.g., Saturn and its moons) of these entities, the orbits of these individual particles can be more or less co-aligned, and form a more or less spherical clouds.


Are there stars that don't emit visible light?

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

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

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


Think about it as if you were baking bread (or cake, whatever you prefer). When you bake a bread with raisins, it rises in all directions (due to the yeast). Every raisin in a rising loaf of raisin bread will see every other raisin expanding away from it. Now suppose the raisins as galaxies and the yeast as the Hubble constant.
An illustration from Hyperphysics (Georgia State University) to illustrate this further:

When you now suppose you are the raisin at the edge, from your perspective, you are standing still and the other raisins are receding away. It is all about the perspective, the body which you take as a reference (like with relativity).


Why isn't there a large gap in the number of captured moons Jupiter and Saturn have?

Gravitational mass is not the be-all and end-all. Two other metrics are the Hill sphere and gravitational sphere of influence. These are widely used metrics that describe the volume over which a planetary body dominates gravitationally over than of the Sun. Saturn has a larger Hill sphere and a larger gravitational sphere of influence than does Jupiter.

This suggests that Saturn might have many more moons than does Jupiter. Saturn's much greater distance from the Sun compared to Jupiter's suggests that discovering a new Saturnian moon is over three times harder than is discovering a new comparably sized Jovian moon. This might be why the ratio is as close as it is.


the signal period of known pulsars is 1 ms-15 s (0.07-1,000 Hz since F=1/T).

That's the frequency of how often the signals occur, not the frequency of the waves they emit. Just compare it to a lighthouse, which has a signal period of a few seconds but a wave frequency of $10^{15}$ Hz.

The first known pulsars, for example "Little Green Man" or PSR B1919+21, were radio pulsars; your graph shows those waves are detectable from Earth's surface. I stumbled upon this article which lists the frequency of some of the pulsars:

  After the discovery of pulsars at 81.5 MHz (Hewish et al. 1968), the first pulsar detected at 40 MHz was PSR B1133+16 (Arecibo Observatory, 1968, unpublished). In 1969–1970 the pulsars PSR B1133+16 and B0809+74 were detected at 25 MHz (Bruck 1970).

Below is the full detection range for the telescopes:

  Therefore joint observations (simultaneous or not) will be highly beneficial, providing a continuous frequency coverage from 8–240 MHz.

which is roughly equivalent to wavelengths of 40m to 1m.

X-ray pulsars like Centaurus X-3 were discovered later, and not by telescopes on the surface:

  Centaurus X-3 was first observed during experiments of cosmic X-ray sources made on May 18, 1967. These initial X-ray spectrum and location measurements were performed using a sounding rocket.


You are at the centre of the Big Bang. It happened here and here and here. In fact everywhere in the universe was the centre of the Big Bang.

Think of it this way - if there was somewhere that exists now that wasn't at the centre of the Big Bang then that point existed outside space (a nonsense) or the Big Bang was not the beginning of space (and so not the Big Bang).

Perhaps the strongest evidence in support of the theory is the cosmic microwave background which comes from all directions precisely because the Big Bang took place everywhere in the Universe.


Verifying refractor lens about scratches

Price vs. quality.   You're unlikely to see any significant degradation in the final image if those are scratches.  However,  that is not how you should make your purchase decision.  You want to inspect the entire body for any evidence of dents, bends, etc. which indicate rough usage, not to mention misalignment of the optical elements.

If you can't try out the 'scope in person, make sure there's a 100% refund guarantee before having it shipped to you.


I realise that the following is not a great answer to the question, but was a bit too much to fit into a comment.

The official word on exoplanet naming is

http://www.iau.org/public/themes/naming_exoplanets/

The rules for the scientific designation of an exoplanet are clear and I think will remain so in the published literature for the forseeable future.

However, here is an earnest contribution to the literature which sees the need to change things, especially to cope with planets in multiple star systems.

http://adsabs.harvard.edu/abs/2010arXiv1012.0707H

But perhaps the big news is that the IAU is setting up to allow "popular" names to be given to exoplanets and are running a competition to name a batch. These names will not replace the scientific designations, but will be recognised by the IAU.

http://www.iau.org/news/pressreleases/detail/iau1404/


Does the sun's rising/setting angle change every few months?

The Sun rises and sets at a different point on the horizon every day. The change is small, so without careful observations, it may take several days or weeks to be fully aware of the change. Mathematically, the position of rising/setting can be found from the following formula:
$$\cos(\theta) = -\frac{\sin(declination)}{\cos(latitude)}$$
where

$\theta$ is the angle measured from due south to the point on the horizon where the object rises. (The angle is the same for rising and setting, if you ignore the change in the Sun's declination during the day.)
declination ranges between approximately +23.5 and -23.5 degrees for the Sun during 6 months (southern hemisphere winter to summer) and then from -23.5 to +23.5 during the next 6 months.
latitude is the observer's latitude.

From 35° S latitude, the winter Sun would rise at $\theta=119$ from due south (or 61° from due north, approximately ENE). The summer Sun would rise at $\theta=61$ from due south (or 119° from due north, approximately ESE). The range is 119-61=58° along the horizon.
At latitudes closer to the equator (0° latitude), the difference is smaller. The range is approximately 47° along the horizon when at the equator. At latitudes closer to the poles, the range becomes larger. At the Arctic and Antarctic Circles, the range is the largest that it can be: 180° (at least mathematically). At these high latitudes, the diameter of the Sun and refraction become important to calculate the precise location of rising.


How is the height of the Great Red Spot measured?

The Great red spot looks oval from above but is actually saucer shaped as the storm curves towards the center and the deepest point lies at the center. It is 16350 km wide, so the height is negligible compared to the width. The storm has its roots beneath the surface of the atmosphere. NASA’s Juno spacecraft calculated that the storm penetrate about 300 kilometers into the planet’s atmosphere.
Scientist also calculated the "thickness" of the storm which I believe the portion of storm that is above the atmosphere. The calculated thickness value agreed well with measurements at the cloud level since the Voyager mission in 1979 and was same during the Juno space mission indicating the thickness is more or less constant and is not affected from the apparent shrinking of the spot. The scientists calculated that the Great Red Spot is about 105 miles (170 kilometers) thick.
The cloud cover around the center of the spot vary by about 30 km and at the extreme end, the clouds hovers at a higher position which is about 8 km above the surrounding clouds. The difference is apparent in the false color Galileo image:

The deepest clouds are dark blue, higher clouds are light blue, high thin hazes are pink and high thick clouds are white.
Scientists were able to calculate the dimensions and dynamics of the Jupiter spot by performing a simple experiment. They investigated the dynamics of large Jupiter vortices by performing numerical simulations from a simple laboratory experiment involving a salt-water-filled Plexiglas tank that measured 20 inches by 20 inches by 28 inches (50 by 50 by 70 centimeters). You can find more information in ref. 4.
References:

https://earthsky.org/space/juno-probes-depths-of-jupiters-great-red-spot
https://www.space.com/jupiter-great-red-spot-shrinking-thickness-steady.html
https://www.sciencedaily.com/terms/great_red_spot.htm
Lemasquerier, D., Facchini, G., Favier, B. et al. Remote determination of the shape of Jupiter’s vortices from laboratory experiments. Nat. Phys. 16, 695–700 (2020). https://doi.org/10.1038/s41567-020-0833-9
Space: From Earth to the Edge of the Universe, Carole Stott, Robert Dinwiddie, Giles Sparrow, Dorling Kindersley Ltd, 2010


What is the probability that we are the first ever intelligent life in the whole universe?

We know that intelligent life exists on one planet. We do not fully understand the processes by which life (a) starts and (b) becomes intelligent.

Therefore, the probability of these things happening elsewhere ranges from 1 to as close-to-zero as needs to be to ensure we are the first and perhaps only intelligent life in the universe.

Therefore the answer to your question ranges anywhere from zero to one.

A more sensible, informed calculation could arise if we ever create life in the lab or if life with an independent genesis is found elsewhere (in the solar system or beyond).


Why does the 70-solar-mass black hole in the LB-1 system challenge current astrophysics?

There're several pieces of information one needs to understand this.

Although stars more massive than 70 solar masses exist, when they become black holes, they usually lose mass in the process. The exact amount of mass lost depends on the metallicity (which is a technical term that describes how much "metals" - the astronomer's definition of metals is anything heavier than hydrogen & helium - is in the star). The companion star that we see today is at solar metallicity, so it's probable that the original star (the one that became the black hole) was also at solar metallicity. Unfortunately, that means it shouldn't leave a 70-mass black hole remnant. From the paper:

  This [70 solar mass black hole] would strongly challenge current stellar evolution models, which only allow for the formation of black holes up to $25 M_{sun}$ at solar metallicity.

Where, then, did this 70-solar mass black hole come from? The paper discusses a few alternatives. The obvious one is that two smaller black holes simply merged to form this one. Problem with that is, you still need two 35-solar mass black holes, and 35 is clearly > 25. (In principle you could also have a 25-solar mass black hole merge with a 45-solar mass black hole, but that still leaves the question of where the 45-solar mass black hole came from in the first place.) Note also that this black hole probably didn't arise from neutron stars merging, since neutron stars have a mass limit of about 2 solar masses. Finally there is three black holes merging into one, but this is unlikely: mergers are already rare events, and having two mergers must be even rarer.

Here're a few more unlikely explanations I can think of:

Perhaps it's a primordial black hole. Problem with this is that primordial black holes are something of a unicorn - a last resort explanation - because they work whenever you need a missing mass. If you can explain an observation without invoking primordial black holes, that's much preferable. See also the next bullet point.
Perhaps the black hole formed elsewhere, in an environment that isn't at solar metallicity, then travelled to this place and captured its companion star. Problem with this is that stellar capture is not a likely process. In the same way we can say that the Earth probably formed around the Sun; it didn't form elsewhere and was captured by the Sun later. There's another problem, which is that the star is observed to have an eccentricity of almost zero (this means its orbit is roughly circular). Newton's laws predict an elliptical orbit. There are processes that will drive the orbits towards circular, but the time taken is long. A (rare) stellar capture event that leads to almost exactly a circular orbit is even more unlikely.

The paper discusses a couple of more credible alternatives:

Perhaps this was a triple system where one star became a black hole, and then this black hole "fell into" one of the other stars and ate it from the inside.
Perhaps there was a "fallback supernova". This is when a star goes supernova, but the ejected material somehow falls back onto the stellar remnant. This has never been directly observed, and this might be the first.
Perhaps something is wrong with the measurement, e.g. an unaccounted for systematic effect. This is the most mundane explanation.

In any case, the system is now an attractive target for telescopes.

Update: there are now several articles claiming an error in the analysis, and that there is no 70-solar mass black hole in this system.


What measure should I use to help optimise the design of a telescope in a cubesat?

I make telescopes and telescope mirrors. You're asking a hard question, since it depends on so many things. But it's an interesting project, so I'll try to help as much as I can.

The hard limit for performance is aperture, or diameter of the primary optics. Bigger aperture = higher hard limit for performance. On a nanosat you only have room for so much aperture, so performance will necessarily be limited.

Secondly, you'd need some kind of compact design. A reflector for sure, but not newtonian. Some kind of Cassegrain variant, you can make those pretty short and stubby. Unless you make it so that it "unfolds" in space, but that's going to be tricky.

Also, the whole optical stack has to be very rigid and must hold its shape through launch and while in space. The optical system must remain collimated at all times.

Finally, you probably have some weight limits. That will make it harder to create a rigid system.

One measure of performance is the resolving power, which indicates the angular size of the smallest details resolved by the telescope. The relationship is linear. An aperture of 100 mm provides a resolving power of 1 arcsec. An aperture of 200 mm provides a resolving power of 0.5 arcsec. And so on.

Telescopes operating at ground level are limited by atmosphere in several ways.

One is the so-called "seeing", or air turbulence. This blurs the image and reduces the effective resolving power. Since small telescopes have less resolving power to begin with, they are less affected. A 100 mm aperture is unaffected most of the time. A 200 mm aperture is affected most of the time.

Your telescope will be above the atmosphere, so the effects of seeing will be zero. It will operate at full resolving power all the time.

Another way the air influences telescope is via light pollution and other forms of luminescence and reflected light from the air. This reduces contrast and "erases" the faintest objects from the image; bright objects such as planets are not affected at all.

Your telescope will not be affected by light pollution, so its performance on faint objects (such as galaxies or nebulae) will always be 100% of the maximum for that aperture.

Bottom line: you are going to have a pretty small telescope, but operating completely free of the effects of the atmosphere.

Since it's operating in microgravity, the optics will not deform under their own weight, so you can make them quite thin and lightweight.

The instrument will operate under extremes of temperature. You will need some kind of dynamic focusing to compensate for expansion / contraction. Be careful, with some catoptric designs such as Cassegrain you also need to pay attention to the distance from primary mirror to secondary elements, etc. Using carbon fiber for the skeleton would reduce expansion / contraction a great deal, and would also provide excellent stiffness per unit of weight.

Quartz mirrors could deal very well with huge temperature variations, but quartz is expensive to process to optical precision. Borosilicate glass (the material used for most amateur telescopes) should also work well; Supremax 33 for example is cheap and fairly good. It depends on the optician who will make the mirrors. If you find someone who could give a discount on quartz, go for it; otherwise use borosilicate and don't worry about it.

You will also need some kind of inertial elements to change the orientation of the telescope in space. That's not a trivial task.

Bottom line: you begin with a performance budget limited by aperture. That's the ideal performance. Then real world issues will push real performance below that level: collimation errors, temperature shifts, optics quality, focusing errors, etc.

Performance of a real instrument is affected by everything.


Is there matter forming in the inflationary space?

A note: don't talk about inflation for the current expansion. Inflation is the brief period immediately after the BB when the universe expanded at an outlandish rate.

Matter was formed when the universe was so compact that its temperature was billions and even trillions of Kelvin hot. The energy of the vacuum in any given volume was so high that matter-antimatter pairs were created, which usually quickly annihilated. Sometimes, however those short-lived particles collided, releasing large amounts of energy. For instance, to create an electron you need an energy of 0.5 MeV. The graph below shows that at a temperature of $10^{16}$ K collision energy could be as high as 1 TeV.

Because of its expansion the universe has cooled so much that its temperature is less than 3 K, which isn't nearly enough to create any new particles.

The graph shows how temperature and energy drop as the universe expands in the first seconds after the BB. Note that at 1000 seconds (about 20 minutes) the energy is too low to create electrons (0.5 MeV), let alone heavier elements.

What is created in the expansion, though, is dark energy. Its density is constant, so when the universe expands to double its size the amount of dark energy will also be doubled. (That's the reason why the expansion goes ever faster and faster, with nothing to stop it.)


Opening VLA Export Format data on Windows

The VLA data downloaded from the archive is raw data. It contains the output of the correlator in its own format, data on weather, water vapor in atmosphere, turnings of the antennas, wellbeing of each antenna, all calibrators, information which part of the data might not be healthy and why, comments by staff, if any, who examined the data, etc.
CASA allows the user to run the data through the data reduction pipeline, which is designed to process these outputs and deliver “the product”, i.e. calibrated data.
The product is a large file containing the UV data (Fourier transform data) on calibrators and targets. To get images this data has to be imaged and cleaned. Imaging implies doing the inverse Fourier transform. Cleaning implies application of CLEAN algorithm which takes care of the fact that the telescope beam has non-trivial shape.
Unless one knows everything about the telescope hardware and software, it is impossible to make any use of the raw data from the archive. The purpose of CASA is to eliminate necessity of the telescope’s inner working knowledge as well as necessity to write ones own imaging code.
CASA is essential tool. Try NRAO’s Helpdesk. They might know a workaround the Windows issue.
If you are determined to play with data yourself, ask Helpdesk to point you to the documentation about VLA’s correlator and its output format.


Where did the grand design spiral galaxy designation come from?

The term can be traced back to the theoretical work of C.C.Lin in explaining the spirals as due to density waves in galactic discs.

A paper from 1970 discusses the "Existence of 'Grand Design'". Note the use of "scare quotes", suggesting that the term was new and would not be widely recognised by his readers.

Chia-Chiao Lin was an important Chinese/American Mathematician. It is reasonable to believe that the term was coined by him or one of the mathematicians that he worked with at MIT. And was in use since about 1964, when the first of a series of papers "On the structure of Spiral Galaxies" was published.


Mechanisms of binary/multiple star formation

I don't have enough reputation to comment...

I think this might help you understand the formation or binary and more stars systems. This of course is not the only possible method but it might explain the systems with big mass differences.

As the initial rotation speed increases (marked in the videos as beta) you will see how the protoplanetary disk breaks up and form more than one protostar.

http://www.astro.ex.ac.uk/people/mbate/Animations/prestellar_discs.html


Question regarding the path of sun crossing zenith before the summer solstice

Question 2: Doesn't the Sun take the longest arc across the sky on the day it passes through zenith?

Answer: There are two effects. (In the remaining description, I will be referring to an observer in the northern latitude.) As the Sun moves northward from 0 degrees declination, the radius of the circle that it travels is smaller. In other words, the total length of the 20 degree declination circle is smaller than the 10 degree declination circle which is smaller than the 0 degree declination circle (the celestial equator). But at the same time, the fraction of the declination circle that is visible above the horizon is greater. At some combination of declination, these two effects result in the maximum length of the declination circle above the horizon. If the declination is north or south of that optimized solution, the length of the arc is smaller.

Skipping the calculations for now and going right to the solution, here is a graph showing the length of the arc for various latitudes and declinations. This graph plots the results relative to the length of the arc along the celestial equator which is 180 degrees regardless of your latitude (ignoring refraction and the Sun's diameter).

If your latitude is between 0 and 23.4 degrees, the longest arc is before the Sun reaches the zenith. For example, an observer at 20 degrees latitude witnesses the longest arc when the Sun is at about 13 degrees declination which is 7 degrees before the Sun reaches the zenith. If you live north of 23.4 degrees, the Sun is never overhead. North of 30 degrees, the longest arc is when the Sun is highest in the sky.

I can go into more details and the math, but I do not know if I should do it here or maybe in a new question.


How is the difference of bolometric magnitudes not dependent on the stars' radii?

The R in that equation is the distance from the star to observer, not the star radius. The light emitted from the star is distributed uniformly on a sphere of radius R, and when the light arrives to the Earth, that sphere will have a radius equal to the distance Earth-star.
Therefore, the second relation for the two fluxes is about the apparent magnitudes (which describe the brightness of an astronomical object observed from Earth),
$$m-m_\odot = -2.5 \log F/F_\odot$$
The first relation is alright. The absolute magnitudes are related with the luminosity of the star (the overall energy flux emitted by the star) and they are not dependent on the distance to the observer.