2 Earth-like planets in co-orbital configuration I am trying to set up 2 Planets in a co-orbital configuration around a Jupiter size planet similarly to Saturn's moons Janus and Epimetheus. I'm trying to do this in Universe Sandbox 2 but when I place these planets they mess with each others orbits too much. Is it possible to have such a system and how would it be done?

Diagram showing what Epimetheus and Janus do.

Because this has gone unanswered, and using this very nice answer as a reference, I can answer this part way.  Source article here.

  it was found that if the mass of second body becomes larger than
  1/1200 of the first body, horseshoe orbits become unstable.

So Earth mass orbiting Jupiter (1/318) won't work according to the research above.  You could try Mars mass objects orbiting Jupiter (1/2970) and you might get that to work for a fairly extended number of orbits.   Horseshoe orbits are never stable over the long term the way trojan orbits can be, but they can be stable over numerous orbits if they are properly set up.

I tried to work out the specifics for higher mass horseshoe objects, but my attempts at calculations have just sent me in circles so far.   If I make any progress I'll post an update.  You'd have to move the Mars-mass moons quite far from the planet and you'd need to keep the variation in their semi-major axes quite small.   Janus and Epimethius have only 50 km variation in their orbits, which is 1 part in 3000 of their orbital radius.

I'd explore variations in that range, and I'd start at the outer edge of stable region of Jupiter's hill sphere, and if you can't get it to work, decrease the mass some more, maybe lunar mass.


Can life on earth be supported by only a Red Dwarf star or a White Dwarf(state still hot not cooled enough yet)?(Provided it is near to the star) Can a red dwarf star or a white dwarf have enough energy to support life on earth? How it will change the climate and the seasons of earth if it is our only star? (energy source)(Expecting it to change since earths orbital mechanics will also change)

Certainly a red-dwarf star can have enough energy for a planet around it to be in the goldilocks zone.   There are some difficulties with red-dwarf stars and Earth like planets.   The planet would need to be very close to the star and as a result, tidally locked.   The orbital period would be quite short, so there would be no seasons and one side of the planet would be in permanent sun the other side in permanent shade.

Red-Dwarf stars, unlike our sun, grow colder with time and they can be quite magnetically active, sometimes called flare stars, though not all red-dwarf stars are flare stars, many start out that way.   The proximity to that much solar-ejection could strip planets of their atmosphere early in the planet's life.

None of that makes it impossible.  It's certainly possible for a Red-dwarf star to host an Earth-like planet, but there's some difficulty.  Orange or Yellow stars are, on average, considered better candidates.

A white dwarf, however, likely not possible.   A white dwarf follows the red-giant stage of a star.  It would be too small and give off too little energy and any planets that survived the red-giant phase would likely be very distant.   Life and/or an Earth-like planet around a white dwarf is virtually impossible.


Infrared Region Telescopes Why do we need to build telescopes that can view in the infrared? What exactly do we need to see in the infrared region? I know this seems like a really broad question but I'm just curious about it.

Three main reasons.

Lots of things in the universe are cool, much cooler than the Sun. Therefore they emit most of their radiation in the infrared spectrum. Examples might include brown dwarfs, giant exoplanets, circumstellar discs.
Lots of things in the universe are distant and highly redshifted. Thus even if they emit their radiation at visible wavelengths, we receive it at infrared wavelengths.
Lots of things in the universe are hidden by dust, which effectively absorbs visible light. Dust is much more transparent to infrared radiation, so we can see beyond it. An example would be the centre of our Galaxy.


At what speed does something have to travel away from us for it to red shift enough that it becomes invisible to the human eye? Are there stars, galaxies etc that we cannot see because they are traveling too fast and their spectrum is shifted below our visible range? From what I understand, red shift is caused by stars and such moving away from the viewer. At what speed does a star etc have to travel to be invisible to us?

What if you are in one galaxy on one side of the universe that is expanding looking at a galaxy on the other expanding edge side. They would be traveling faster than the speed of light away from each other. Would they be invisible to each other?

The following deals only with redshift caused by motion (Doppler effect).

Wavelength (wl) shift for an object moving away from a stationary observer is calculated by following formula: shift = wl x V/C
with V speed of the moving object and C speed of light
(source: https://en.wikipedia.org/wiki/Doppler_effect)

Let's say we want a Sun-like star to become invisible to the human eye. Our Sun's emission starts around 250 nm and human vision ends around 700 nm (source: https://en.wikipedia.org/wiki/Sunlight#/media/File:Solar_spectrum_en.svg).

So we want a minimum shift of 700 - 250 = 450 nm for wl 250 nm.
Formula yields V/C = 450 / 250 = 1,8
which
(1) is impossible because nothing moves faster than light
(2) makes the classical formula irrelevant

The relativistic formula (for objects moving at speeds > C/10) is:
shift / wl = SQRT( (1 + V/C) / (1 - V/C) ) -1
with SQRT the square root
(source: https://en.wikipedia.org/wiki/Relativistic_Doppler_effect)

Rounding (shift / wl) up to 2, the relativistic formula yields V/C = 8/10

The speed V would have to be even higher for a hotter (white to bluish) star with UV emission starting below 250 nm.

So my answer to the question is:
very close to the speed of light (80% in above calculation)
and therefore not a real-life scenario (see below).

"Among the nearby stars, the largest radial velocities with respect to the Sun are +308 km/s" (source: https://en.wikipedia.org/wiki/Doppler_effect), that is 1000 times smaller than C = 3e5 km/s. Rounding down to 300 km/s, it would produce a maximum shift of 700 x 300 / 3e5 = 0,7 nm, several hundred times smaller than required for invisibility.

Besides, the star would still appear on an infrared image.


Saturn observation : how to calculate requirements for Parallax method In September, Saturn is 9.9au away from Earth. If we use the parallax method, what should be the base line between the two people?

So I know that $D = B / \theta$, but how do I find out theta? I'm pretty sure the answer is pretty obvious, but my mind really is not working right now, so any help is appreciated

Thank you in advance.

Are you treating $\Theta$ as the only unknown, i.e.  you believe you know the baseline distance exactly?    If so, then  you compare your angular sighting of a distant star vs. Saturn with the same for the other observer.    The devil is in the error analysis.  Your two angles will be absurdly close to each other in magnitude, which means a very small error/uncertainty in value leads to a large error in the calculated distance $D$ to Saturn.

You should find some first-year-level text on measurement & statistical error analysis to get a better feel for the difficulty here.


What is the current accepted theory as to why Venus has a slow retrograde rotation? According to this NASA overview, the planet Venus is unique (amongst the major planets), Venus has a slow retrograde axial rotation, taking 243 Earth days to make one rotation (which is longer than its orbital revolution).

What is the current accepted theory as to why (and how) Venus developed this anomalous slow retrograde axial rotation?

There seem to be a few, and none are accepted by the whole scientific community. The main ones:

Venus was struck by a large body during its early formation
The spin axis flipped, as can happen with a gyroscope
The spin slowed to a standstill and then reversed, caused by the sun's gravity, the dense atmosphere and friction between core and mantle

That final one seems to be the most recent, being proposed by Alexandre Correira and Jacques Laskar in 2001. Their research seems to imply that the conditions on Venus and its distance for the sun make a retrograde spin slightly more likely than a forward one.


If an event occurs at a point in space time what will an observer experience first, it's gravity or the light from it? Let's assume a person is close enough to an object in space time to be able to observe it's gravitational effects and the light coming from it.
If that object suddenly pops out of existence (hypothetical situation), what will the person observe first,

The gravitational field's absence
The object disappearing (light stops coming)

P.S.- pardon me for the noobness of this question, physics isn't one of my strongest subjects.

it's an interesting question. It could ask about any violent transient effect that produces light and gravitational waves, though "popping out of existence" is the ultimate step function.

electromagnetic (EM) radiation is slowed very slightly by the presence of interstellar medium (especially by free electrons) and it's wavelength dependent. The image below from this answer shows the lower frequency radio waves arriving later than the higher frequency ones due to such an effect.

The farther away an object is, the larger the time difference, so this technique can be used to determine a likely distance for a given object.

So if you relax the term "light" to include a wide range of wavelengths which many astronomers now do, you can see a quite a big delay.

Answers to How fast can and do gravitational waves travel? say that gravitational waves generally travel at the speed of light, and answers to Have more recent LIGO/VIRGO gravitational wave measurements narrowed down the speed of gravity further? addresses experimental confirmation. I've never heard of an "index of refraction" for gravity that deviates from unity and could slow it with respect to the speed that an EM wave would travel between the same two points, so I think it's probably safe to assume that it will always come before the EM waves.


Does Jupiter rotate at the same speed at every depth? Jupiter rotates in less than 10 hours. Its rotation is differential, with the upper layers of the atmosphere spinning a bit faster at the equator than at higher latitudes. Do the inner layers of Jupiter rotate at the same (angular) speed as the outmost layers of the atmosphere?

No. The deep interior of the planet rotates as (nearly) a rigid body (Guillot et al. 2018), while the outer part of the planet undergoes differential rotation. Therefore the interior and exterior rotation cannot match at all latitudes simultaneously.


Evolution of the Hubble parameter In the lambda-CDM model describing an accelerating Universe, the Hubble parameter is currently decreasing with time. Will it continue to decrease forever?

The solution to the Friedmann equation in a flat universe is
$$H^2 = \frac{8\pi G}{3}\rho + \frac{\Lambda}{3},$$
where $\rho$ is the matter density (including dark matter) and $\Lambda$ is the cosmological constant.

As the universe expands, $\rho$ of course decreases, but $\Lambda$ remains constant.

Thus the Hubble "constant" actually decreases from its current value $H_0$ and asymptotically tends towards $ H = \sqrt{\Lambda/3}$ as time tends towards infinity.

As $\Lambda = 3H_0^{2} \Omega_\Lambda$, and measurements suggest that $\Omega_{\Lambda} \simeq 2/3$, then $\Lambda \simeq 2H_0^2$, and the Hubble parameter will therefore decrease to approximately $\sqrt{2/3}$ of its present value if the cosmological constant stays constant.

Of course if $\Lambda = \Lambda(t)$, (ie not the basic $\Lambda$-CDM model) then the behaviour will be different.

EDIT: Another useful form of the solution (for the case of a constant vacuum energy density) is

$$H^2 = H_0^2 \left(  \frac{\Omega_r}{a^4} + \frac{\Omega_M}{a^3} + \frac{\Omega_k}{a^2} + \Omega_{\Lambda}\right),$$
where $H_0$ is the Hubble parameter now, $a(t)$ is the scale factor of the universe, $\Omega_r$ is the current (i.e. $a=1$) ratio of the radiation density to the critical density and $\Omega_M$, $\Omega_k$ and $\Omega_{\Lambda}$ are the equivalent densities for the matter (baryonic and dark), curvature and (constant) vacuum energy densities.

As $a$ increases you can see that all three of the leading terms get smaller and the Hubble parameter decreases at all times. When $a$ is very large, $H$ approaches $\sqrt{\Omega_{\Lambda}} H_0$ as before.


Why isn't the dark energy getting decreased? AS the articles on the web suggest, Dark energy is the reason behind the expansion of universe. If so, why isn't it getting used up in doing so. And consequently, the rate of expansion should reduce!

AS the articles on the web suggest, Dark energy is the reason behind the expansion of universe.

If some articles on the web suggest that, they are mistaken. Dark energy affects the acceleration of the cosmic expansion, but it is not necessary for the universe to be expanding, so it cannot be "the reason behind the expansion".

How the dark energy density changes as the universe expands depends on its current energy density and its equation of state--the relationship between its energy density and pressure. For cosmological constant provides the simplest dark energy model in cosmological models--it's the $\Lambda$ in the ΛDCM model--for which the pressure is exactly the negative of its energy density: $p = -\rho$. In general relativity, both energy density and pressure gravitate, and in three spatial dimensions expansion the acceleration is proportional to $\rho + 3p$ (cf. also Friedmann equations). As a result, a cosmological constant with positive energy density $\rho>0$ leads to accelerated expansion, because the contribution of its negative pressure means the overall effect is repulsive.

  If so, why isn't it getting used up in doing so.

Somewhat ironically the answer to your question for cosmological constant can be interpreted like so: it doesn't "get used up" because of a kind of energy conservation.

As a given small volume expands, it keeps a constant energy density $\rho$ and a constant pressure $p$. Is this situation sensible? Suppose its volume increased by $\Delta V$, so its energy content increased by $\rho\Delta V$. If the pressure is constant, this is some kind of isobaric process, so the work done is $W = p\Delta V$. But if $p = -\rho$, then everything is consistent because the dark energy does negative work as it expands to exactly balance the increase in its energy.

In reality, you shouldn't take the above reasoning too literally. There are many issues in interpreting anything like "how much energy" there is in general relativity and whether or not it's conserved, some of which I've covered in this answer, though some of which can be alleviated by the stressing of the "small volume" qualifier. Still, "it doesn't get used up because the equations of the theory say it shouldn't be getting used up", while perfectly true, would not be a very interesting answer.

Additionally, it should be clear from the above that the cosmological constant case of $p = -\rho$ is very itself special. The reason for its simplicity is that it's completely Lorentz-invariant: in different local inertial frames, the energy density and pressure of a perfect fluid get intermixed and transformed into different values... except in the case where the pressure is exactly the negative of energy density. Every observer would agree about the cosmological constant, so one can interpret it as an intrinsic energy density of the vacuum. In fact that's how it was originally introduced; other forms of dark energy are a generalization of $\Lambda$.

  And consequently, the rate of expansion should reduce!

With a slightly more general equation of state $p = w\rho$, the cosmological constant corresponds to $w = -1$. Other cases give a perfect fluid, which might indeed "get used up" in the sense of getting more dilute, and for positive energy density, $w<-1/3$ is necessary to outwardly accelerate cosmic expansion. But something curious happens if $w<-1$: we should expect from the work done argument for the energy density to accumulate instead of remaining constant. This kind of "phantom energy" would, if positive, produce a Big Rip scenario, in which the cosmic expansion accelerates to an infinite value in finite cosmological time, ending the universe by tearing everything apart.


How did Kepler come to the potencies in his Third Law? How did Kepler come to the conclusion,  that exactly the square of the period and the third potency of the great semi-axis of the ellipse is proportional?
Why is only square divided by cubic = constant?  Why do other dimensions not work? Kepler certainly did not simply try all potencies.
Isn't this a similar problem to Fermat's last theorem?

If you plot the log of the period against the log of the semi-major axis then it is obvious that $P^2 \propto a^3$. Any other power law relationship simply wouldn't fit.

The following passage (from https://www.mathpages.com/rr/s8-01/8-01.htm ) seems relevant:

Is it just coincidental that John Napier's "Mirifici Logarithmorum Canonis Descripto" (published in 1614) was first seen by Kepler towards the end of the year 1616? We know that Kepler was immediately enthusiastic about logarithms, which is not surprising, considering the masses of computation involved in preparing the Rudolphine Tables. Indeed, he even wrote a book of his own on the subject in 1621. It's also interesting that Kepler initially described his "Third Law" in terms of a 1.5 ratio of proportions, exactly as it would appear in a log-log plot, rather than in the more familiar terms of squared periods and cubed distances. It seems as if a purely mathematical invention, namely logarithms, whose intent was simply to ease the burden of manual arithmetical computations, may have led directly to the discovery/formulation of an important physical law, i.e., Kepler's third law of planetary motion. (Ironically, Kepler's academic mentor, Michael Maestlin, chided him − perhaps in jest? − for even taking an interest in logarithms, remarking that "it is not seemly for a professor of mathematics to be childishly pleased about any shortening of the calculations".) By the 18th of May, 1618, Kepler had fully grasped the logarithmic pattern in the planetary orbits:


What is it like at the edge of the universe? If the universe is expanding, then if you were to be close to the edge of that expansion, what would it look like? What would be happening? Would it be chaotic? Would objects be appearing or disappearing? I don't even know if it's possible to have any knowledge of this, but I'd like to know if there are any theories.

The edge of the Universe in space-time can only be the moment of its creation (or end if the Universe is closed). What it would be like at that time depends on what you would consider close to the edge. Your question is about the edge of the expansion so I will presume that you mean during the Inflationary era (until about $10^{-32}\textrm{s}$ after the creation of the Universe).

There would be no objects as matter as we know it today did not exist (according to the big bang model). The major interacting forces (gravity, strong and weak nuclear force, and the electromagnetic force) were unified at the time of the Inflationary era. There would be no protons or neutrons as the strong interaction has not decoupled yet from the other forces. The temperature would be extremely high $>10^{27}\textrm{K}$.


What is minimal aperture/magnification to recognize phases of Galilean moons? Given you take a (non-amateur) refractor telescope, what are aperture and magnification required to recognize phases of the four Galilean moons?

What did Galileo see:
http://www.astro.umontreal.ca/~paulchar/grps/site/images/galileo.4.html

From Earth the resolution of is limited by the atmosphere.  Turbulence in the atmosphere make it impossible to resolve the moons of Jupiter to more than a few pixels wide.

Amateur refractor telescopes tend to be small and less powerful than reflector telescopes. I'm not aware of any amateur refactor telescope that could resolve a moon of Jupiter to more than a point of light. The eye is unable to resolve an object that is less than about 60 arcseconds and it would need to be closer to 1000 arcseconds before many surface details became visible. The Moon, for comparison, is 1800 arcseconds.

Professional equipment can partially overcome the atmosphere by using adaptive optics, but this is not a feature that is generally available to amateurs.

Jupiter and its moons orbit outside the orbit of Earth, so from our point of view Jupiter is never partially lit less than 99%. The moons are sometimes eclipsed by Jupiter, but they too are never partially lit less than 99%.  With amateur equipment, Jupiter and its moons are never anything other than "full".

Even with the Hubble telescope, the resolution is not sufficient to identify phases of the moons.

What Galileo saw was Jupiter, resolved as a disc and 4 moons orbiting it which appeared to him as "stars" (he called them the Medicean stars). The four moons were not always visible, sometimes a moon was hidden in the shadow of Jupiter, or behind Jupiter from our point of view. At other times a moon would be too close to Jupiter to be visible, or outside the field of view of his telescope.

What he realised was that these were bodies in orbit around Jupiter, and that they obeyed the same rule for orbits that Kepler had proposed for the planets.


What's so interesting about Coma-type galaxy clusters? When the dynamical mass of a proto-cluster of galaxies is ~(1-2)x1014 Msolar, the proto-cluster is expected to evolve into a coma-type cluster.

In almost every paper I read about the confirmation of proto-clusters, the authors of the paper are (almost) always interested in proto-clusters (of galaxies) that will eventually evolve into a Coma-type cluster.

Why are astronomers and astrophysicists so interested in studying those type of structures (Coma-type galaxy clusters)? I mean is it interesting that a proto-cluster will eventually evolve into a Coma-type galaxy cluster? Are these structures rare in the universe?

I think the answer to this isn't that Coma-type galaxy clusters are interesting, it's that galaxy clusters are interesting. Chiang et al. (2014) talks about the importance of studying protoclusters as a means of understanding how galaxies evolve over time.

  Galaxy clusters are extreme products of structure formation. They are ideal laboratories to study galaxy assembly, quenching, and sub/super-halo galaxy environments. It has become clear that to obtain a complete picture of cluster formation, we also need to find and study their progenitors at high redshifts that  were  still forming.

In their paper, they discuss the discovery of a large number of new protoclusters. They bin their protoclusters by mass and designate them as "Fornax-type" ($M_{z=0} = 1-3\times 10^{14}\ M_{\odot}$), "Virgo-type" ($M_{z=0} = 3-10\times 10^{14}\ M_{\odot}$), or "Coma-type" ($M_{z=0} > 10^{15}\ M_{\odot}$). Clearly, they're using present-day clusters as analogs to describe what these protoclusters may turn into as they continue to evolve. The fact that a protocluster is considered a Coma-type cluster, is only based on the mass. Astronomer's aren't so much interested in studying Coma-type protoclusters specifically, they're just studying protoclusters and the high mass protoclusters are considered Coma-type protoclusters.

It's very possible that you were preferentially seeing references to Coma-type clusters because Virgo- or Fornax-type clusters are harder to find (being of smaller mass) and so there is an observational bias that preferentially makes Coma-type protoclusters easier to find and subsequently publish about.


Complex life in binary black hole - Sun(s) system I was wondering if there could theoretically exist systems in which an Earth-like planet was in the habitable zone of one (or two) Sun-like star(s), while actually not orbiting around any star.
To be more precise, I started asking myself whether a planet like earth could be placed in the inner orbit of a black hole, while one or a pair of Suns would be revolving around the black hole too, on an outer orbit.

My question is not whether such a system could actually exist somewhere, but only if it could theoretically exist, even if it is probabilistically not likely to exist.

So I found this sets of issues:

Planet must be far enough from the black hole ( > Roche Limit )
Harmful radiations from black hole's accretion disk
Planet-Star relative distance must not vary too much...which I guess in turn means that the planet-blackhole distance must be small compared to the average blackhole-star distance

Given these limitations, how would you choose the stars and blackhole masses and their relative distances to the planet, in order to have it suitable for complex, perhaps intelligent forms of life? Do you see other issues to this purpose?

Below I add what I've been able to collect so far (more like a collection of unsolved questions..)

So, about the issue #1, I thought about a rather lightweight blackhole, with M < 10 Solar masses. I once read in a thread that when considering a central body = 10 solar masses, and a satellite with a density of 5000 kg/m^3 (Earth's average density), then the Roche Limit would be as little as 0.8% of an AU. That would be conveniently small, though I guessed that using Earth's average density may not guarantee that the whole planet is kept together (it may lose its atmosphere...). So, just to be sure, would you rather increase this distance to 1/20 of an AU, perhaps less, perhaps more? I'm asking here because I'm not able to conduct these calculation myself.

Then again, I'm not an expert and I don't know if staying out of the Roche Limit is a sufficient condition for orbiting around a blackhole of this small size without things becoming nasty. I know of issue #2 though,and about that I read someone suggesting that If the planet were subjected to tidal lock, then at least the face looking away from the blackhole would be spared from the harming X rays. But how close a planet should get to a blackhole in order to experience a tidal lock within a reasonable amount of time? Would it likely be tidally locked if it stayed at 0.05 AU away from a 10 solar masses black hole? I guess the answer to be yes, but I wanted some validations from the experts here.

Then about issue #3 again, I thought that If there were two Suns (making another binary system of their own), then there would be also a larger insulating flux, perhaps allowing us to keep the Suns more far away (by a quick calculation I think I found that two Suns at 1.4 AU would give the same flux as one Sun at 1 AU). Another possibility is to have a brighter Star class, at expense of its life-time. So for simplicity let's say we can go up to 2 AU (maybe even more) without the Planet getting too cold, or the Stars too short-lived.

Summarizing, I now have a 10 Solar Masses black hole and a (likely) tidally locked earth-like planet orbiting at 0.05 AU from it. Tidal forces should still be small enough not to suck the planet's atmosphere. Perhaps having a planet slightly bigger than the Earth would also help in this account. Then we have a couple of Stars orbiting together around the black hole, at a distance of 2 AU. Relative distance from the twin Stars to the Planet is 2 +/- 0.05 AU. I guess that would be within the limits of the habitable zone, right?

One last curiosity: assuming the black hole has an "average" spin, would it be realistic if the two Stars had a tilted orbit with respect to the black hole spinning axis?

I'm sorry for the many questions, by any means please don't assume I'm expecting a single person to answer all of them.

The Roche limit for a solar-mass (or 10-solar-mass) black hole is much larger than the Schwarzschild radius, or even the last stable orbit (which is a few times the Schwarzschild radius). So the Roche radius is a good limit (and, yes, you're right about the limit for an Earth-like planet and a 10-solar-mass BH.)

You needn't assume the BH has an accretion disk; I suspect most of the ones wandering around in our galaxy don't at the moment, just because accretion disks accrete (go into the black hole) over time, and so you only see them when there's some relatively recent source of gas to create the disk. (Most of the Milky Way black holes we know about are in close binaries, where the companion star is close to the Roche limit and is a regular source of gas.) Now, in principle the stellar wind(s) from the star(s) in your system could contribute to a very weak, tenuous accretion disk, but in practice it's unlikely to produce enough radiation to be of any concern, especially as you're assuming a planet with an atmosphere.

So there's no need to worry about whether or not the planet is tidally locked to the black hole, unless you want to specify the length of the day-night period. Of course, if the planet is only a few tenths of an AU away from the BH, it probably will be tidally locked before very long.

The variation in distance from the star(s) you suggest should be fine in "habitable zone" terms. In fact, given that the planet orbits the BH quite rapidly (about 1.3 days for a 0.05 AU orbit around a 10 solar-mass BH), it won't spend much time at either extreme, so you could probably have it go inside and outside of the formal habitable zone and still be habitable, as long as the average insolation is in the habitable range.  This would create some interesting day/night/seasonal effects, particularly in the tidally locked case: e.g., at a certain time of "year", the "eastern" hemisphere would experience noon when the stars were only 1.95 AU away, while for the "western" hemisphere noon would happen when the stars were 2.05 AU away. Half a year later, the situation would be reversed.

One last curiosity: assuming the black hole has an "average" spin, would it be realistic if the two Stars had a tilted orbit with respect to the black hole spinning axis?

That really depends on how the system formed in the first place. If the initial pre-BH star and the other star(s) formed from the same collapsing cloud, they would probably have similar spin and orbital angular momenta; if the spin of the BH is entirely inherited from its progenitor (which goes supernova to form the BH), then the orbital plane might be similar to the BH's spin. But if the system got put together later (maybe an encounter between a binary consisting of the BH and another star and the current binary pair, in which the BH's original companion star gets ejected), then the two stars could have a tilted orbit.


Why can space telescopes see through a planetary nebula? I recently read the book "An Introduction to Planetary Nebulae" by Jason J. Nishiyama. Although I'm not an astronomy student, I could at least understand the written texts and less the physical formulas. So, I got a nice basis knowledge now -- or that's what I like to think.
Now, I browsed through some pictures of planetary nebulae. Mostly from Hubble and such. I know that they synthesize their images by combining infrared + visible light + uv-light, but I still can't comprehend why these telescopes are able to see INSIDE a nebulae when the shape of mentioned nebulae are spherical? For example, the planetary nebula: NGC 3242 (Ghost of Jupiter); NGC 7293 (Helix Nebula); NGC 6543 (Cat's Eye).

You can see the exposed white dwarf. And it looks like a thin shell envelopes that star, while the space between the star and the shell is scarce, like see-through. But wouldn't that image imply that the planetary nebula is disk-shaped? If the nebulae is spherical, then shouldn't there be "more nebulae" between the star and the shell? Or shouldn't the inner space be tinted in the color of the nebulae, because we look at a white dwarf that's behind a shell? I've drawn my problem with Paint too, sry for my artistic skills: Pattern A is what I thought it should be; Pattern B is what I think it looks like.

Let's say we have a spherical shell of some material emitting light, much bigger than the star it surrounds. If we look right down the center of the shell, our line of sight takes us through only a small amount of gas - not enough for there to be significant emission. On the other hand, if we look at the nebula near the outer edge, our line of sight takes us through substantially more material:

Inspired in part by an image in this piece by Phil Plait.
If you look through more light-emitting material, you'll see stronger emission than if you look through less material, so in this toy model, we would perceive the edges to have strong emission. So a spherical nebula wouldn't lead to uniform emission throughout our cross-sectional view of it.
This simple spherical model holds for some cases, but certainly not all, particularly because many nebulae aren't spherical (Wikipedia makes the claim that only ~20% are truly spherical, which I haven't been able to verify.) For example, we know that the Ring Nebula isn't actually a sphere because the central surface brightness is much lower than the projection effect I described above would predict. Something else clearly must be going on. In this case, the discrepancy is solved by the fact that the Ring Nebula is a bipolar nebula, with its axis along our line of sight, with a dense ring of gas around its equator. Therefore, there's more material near its edges than its center from our perspective, and so we see more emission.


Can radio telescopes such as arecibo image the subsurface of asteroids or planets? This was done with the moon: Radio Dishes Peer Beneath Moon's Surface

It is important to understand how this process works. The method described in your article is known as Bistatic Radar. In effect, a transmitter sends out a signal (generally a radio telescope in the microwave region) which hits the surface of some body and bounces off to be received back on Earth by a second, separate radio telescope. Now, because a microwave's wavelength is so long ($\sim 0.1-100\:cm$) the microwaves don't bounce off from the exact surface, but instead are able to penetrate slightly into the subsurface before being reflected. This means the receiver gets a reflection of the subsurface of the body.

In the case you linked, they used the Arecibo Telescope to provide the transmitted signal to the Moon and the Green Bank Telescope to receive the signal. A similar process could be done to other bodies besides our Moon. However, you run into the problem that the farther away an object is, the harder it will be to detect the returning signal.

I believe that most objects within our solar system (and certainly all objects outside it) are outside the distance where this method will work. I was able to find an instance where someone used this method for an asteroid that passed by the Earth at a distance 11 times that of the Moon. I'm not sure what the limit is to how far away something must be before this method no longer works, but I imagine it wouldn't work even for Mars unless we seriously upgrade our technology and power.

Of course your question seems to presuppose the transmitter and receiver are both on Earth. If your transmitter is much closer, say a satellite orbiting that planet, then it is certainly feasible. Here is a source which discusses the concept of using Bistatic Radar probing via a satellite, which also goes into the history a bit on what has been done before.


Type II supernovae as a distance indicator Could someone explain in brief exactly how type II supernovae could be used as distance indicators? as in simply the formulae used and how they are employed? I know there are certain methods based on expansion velocity etc but the open supernova catalog ( https://sne.space/ ) gives the absolute and apparent mag for type II supernovae too, why would we then need to use methods dependent upon expansion velocity and other complications when we can simply calculate distance using $10^{m-M + 5}$?

The open supernova catalogue calculates absolute magnitudes from the observed apparent magnitude, luminosity distance and redshift decrements.  They note that they don't account for spectral energy distribution shape, so absolute magnitudes may be approximate if there is significant absorption by dust.
So the catalogue isn't recording a measured or modelled absolute magnitude derived independently of the distance, (which is required if the SN are used as standard candles), but a calculated absolute magnitude derived from the apparent magnitude and the redshift.


Size of Saturn's ring material How big are the chunks of rock ice that make up Saturn's rings? Are there many objects larger than pebble size?

The vast majority of the particles in Saturn's rings are small, on the order of $\sim10^{-1}$ m or lower. The columnar number density, according to data from Voyager 1 and Earth-based observations, can be approximated as a function of particle radius by a power law for all particle radii $a$ in meters such that $0<a<1$, as can be seen on this log-log plot (Fig. 15.5, Cuzzi et al. (2009); taken from Fig. 8 of Marouf et al. (1982)):

Though it is only acknowledged in the original paper, the vertical axes for the three different ring regions have been shifted upward different amounts to fit all three on the same graph.

After $a=1$, there's a deviation from the law, and then a steep dropoff at about $a=3$. Obviously, particles larger than this exist, and they certainly play an important role in ring structure, but they're relatively rare.

Obviously, the trends show that smaller particles are much more common, and thus while there are indeed particles larger than pebbles - some as big as boulders, perhaps, or bigger - they are certainly few in number. Most particles are extremely small, smaller than pebbles.

This data covers observations from ring semi-major axes of $\sim$ 75,000 km to $\sim$ 135,000 km - a fairly big spread, covering most of the rings and ending near the Roche Division. The paper doesn't have one single graph of particle number density of a given size at a given semi-major axis, but it does have several subdivided plots (Fig. 15.1 and 15.2) of optical depth as a function of distance from the center of Saturn, which should give you some helpful data on total number density, if you want to make some basic assumptions about mean particle radius. This data is a bit newer, from Cassini, but the Voyager 1 data is just as helpful.


Observatories with public access in Europe This is a follow-up to Observatories that allow public access with a smaller (geographical) scope.
I am interested in getting to know observatories in central Europe (e.g. France, Germany, Poland, Switzerland, Austria) where you could could go with a class of school children or university students.
Maybe you could add more entries to my list, please?

The Effelsberg radio telescope has a visitor programme.
The Heidelberg observatory has some outreach programme, but is close to a city with its lights.
Similar problems with light pollution has the Paris Observatory.

I tried to find some European website with such a list, but I was not successful yet.

There are plenty of possibilities - and often, even when not publicly advertized, many observatories will be very open to find an arrangement when approached, especially by a group of interested audience. So this list IMHO would be the complete list of possible observatories found in central Europe.
The wiki page on astronomical outreach resources in Europe has such list of observatories (well, for the whole world, but filter it) - mostly the professional ones; of course it's not complete, e.g. the Thuringian observatory at Tautenburg is missing (who also offer regularily public tours) as does the GEO600, the gravitational wave observatory near Hannover.
There are various lists for amateur associations, many have their own observatory, e.g. this one lists many in the DACH countries sorted by post code.


What detail can Hubble see on Mars? I'm researching a scene for a sci-fi novel in which the near-future protagonists observe earth through a station-mounted telescope in Mars orbit. My goal is to understand how much detail they reasonably could discern.

The Hubble telescope is probably a reasonable comparison for my purposes. I found images Hubble took of Mars during a close approach, but I don't know if those represent the best resolution possible or simply the resolution that was selected or available at the time, or indeed, if the whole planet was imaged at a higher resolution than the photo published in popular media outlets.

Can a Hubble-like telescope observe significantly greater detail than displayed in the article below? If so, what might reasonably be resolved? Large cities? Individual buildings?

From Vox.com's Hubble can see galaxies impossibly far away. Here’s what happens when it looks at Mars and Saturn.

above: Cropped from Source NASA, ESA, and STScI

above: Cropped from Source NASA/Hubble

Forget about magnification. People who know telescopes don't think in terms of magnification. What matters is the angular resolution, or the resolving power: the angular size of the smallest details that you could see in an instrument.

Rule of thumb: the resolving power of a telescope with a diameter of 10 cm is 1 arcsecond when using visible light. The numbers are inversely proportional. A 20 cm telescope resolves details 0.5 arcsec in size. A 1 meter telescope resolves 0.1 arcsec.

Hubble has an aperture (diameter) of 2.4 m, so its resolving power is 0.04 arcsec.

The minimum distance between Earth and Mars is about 55 million km and it only happens very rarely. The maximum distance is 400 mil km. The "average" distance is 225 mil km (but actual distance varies all the time).

Let's apply the tangent of 0.04 arcsec at 55 mil km:

https://www.wolframalpha.com/input/?i=tan(0.04+arcseconds)+*+55000000

It's 10 km. It would only be able to see the major geographic features.

To see buildings (down to the scale of 10 m), it would need a 1000x increase in resolution. That means an aperture of 2.4 km. None of the classic telescope designs can provide that. It would have to be some kind of interferometric design - a large, flat field where several mirrors are placed several km apart and are coupled optically to function as a single huge mirror (well, sort of - this is more of an intuitive explanation).

It would be similar to the Navy Precision Optical Interferometer near Flagstaff, Arizona.

Some of the wide, flat parts of Valles Marineris might provide a good location for the interferometer. Acidalia Planitia would provide even more space for building huge interferometers, and should be a good place to build structures in general - flat to beyond horizon; it's the place where much of the book/movie The Martian set their story. But any big, reasonably flat field would work.

All of the above assumes the distance of closest approach between Earth and Mars. In practice, the distance is greater than that, so aperture must increase. You're contemplating an interferometer with a base of dozens of km if you want to distinguish structures such as buildings.

Conceivably, the interferometer could be built in orbit, but you must ensure that the distance between mirrors is maintained with extraordinary precision. On the planetary surface, the ground provides the required rigidity. In space you'd have to... I dunno, use space magic.


Can there be planets, stars and galaxies made of dark matter or antimatter? We know that the universe has more dark and anti matter as compared to normal matter. Can there be dark matter galaxies or antimatter galaxies?

Dark matter galaxies are possible but very speculative. On a theoretical level, they are hard to form because dark matter interacts only gravitationally (see Anders Sandberg's answer), which makes it hard to lose energy and become bound structures. On an observational level, they would be hard to detect. Gravitational lensing can do something, but since one cannot actually see the galaxy, it's also hard to say where the dark galaxy is -- if there is one at all.
Still, people have studied the idea, so it's not impossible.
Antimatter galaxies: At some level the idea that there are antimatter galaxies out here is appealing. First it can solve the baryon asymmetry problem at a stroke. It's also the case that an antimatter star would shine. From long distance, it would also be practically indistinguishable from a "normal" star.
However, there are strong reasons to believe that there are no antimatter galaxies. That's because antimatter annihilates with normal matter, which leaves experimental signatures. If any part of the Earth were made of antimatter, it would immediately vanish in a flash, so we can be sure that the Earth is mostly matter. Similarly, if the Sun were made of antimatter, we would be quickly annihilated (thanks to the antimatter solar wind radiating from the anti-Sun), so we can be sure the Sun is also mostly matter. Similar arguments allow us to conclude that the Milky Way is almost entirely matter, the Local Group is almost entirely matter, etc, all the way up to the largest structures in the sky.
If antimatter galaxies exist, they are probably outside our observable universe, at which point some will argue it's no longer science.


Mechanics of Supernovae Where does the energy from a supernova come from?

My understanding is that the iron core collapses into a large ball of neutrons - is that effectively a star-sized nuclear explosion? What is the reaction product called?

Why does the effect go supercritical, instead of finding some equilibrium rate of burn? I've read discussion that the reaction energy does not appear to be enough to explain the energy seen in a supernova? How big is the gap?

The energy derives from gravitational potential energy. The core of a bit more than a solar mass collapses from the size of the Earth to a 10km radius. Some of the gravitational energy (a tiny percentage) released is transferred to the overlying envelope and blasts it into space. Further energisation takes place due to radioactive decay.

A ball of neutrons is not a star-sized nuclear explosion   ?

I don't understand what you mean by reaction product? The proto-neutron star consists mainly of neutrons(!) and neutron-rich nuclei.

The collapse occurs because electron degeneracy pressure is circumvented by neutronisation (inverse beta decay), which removes electrons from the gas. At that point, the collapse occurs on a freefall time of $<1$ second. There is no "burn-rate" as such; just a collapse and then a "core bounce" as the equation of state hardens because of neutron degeneracy pressure and the strong nuclear force between closely packed neutrons.

The "observed" energy (i.e. the electromagnetic output and ejecta kinetic energy)  is only of order 1% of the energy released from the gravitational collapse. Most of it emerges in the form of unseen neutrinos. So there is no energy gap and plenty of energy to power what is observed.


Accurate catalogue of the Stars according to Hemisphere I am in the need for an accurate catalogue of the stars in the Hemisphere I am currently in (or the one that I chose) and their positions at any time of the year.
The more accurate the catalogue is, the better. I don't want any live-time-location app, but a simple place whence I can refer where a star or any other celestial object is.
What do you mean by a 'live-time-location app'?
To answer this question asked by @DrChuck in the comments below, I will take the example of an Android app called 'Skymap'. This app takes your live location and uses the phone's internal compass to determine what direction the phone is pointed in, and shows the celestial objects (visible/non-visible) in the direction according to the time-settings of your phone.
What data are you looking for?
Another question asked by @DrChuck in the comments below. I am looking for the following info :-

Catalogue of all stars visible from all over the Hemisphere.
Their respective positions.
Catalogue of other celestial objects like Galaxies, etc.
Special events like Solar Eclipses visible from the Hemisphere.*
Position of any celestial body during any time in the future or past.*
(*  - Not necessary)

Um, can we make things, well, simpler?
Some things listed below are really not necessary :-

It does not matter if it does not list the objects that cannot be seen through the naked eye.
It does not have to be really complicated. Anything a normal person can use is fine.

P.S. - Any help with any appropriate tags for this question will also be appreciated.

Note that astronomers generally are not concerned with hemispheres, so it is unlikely that you will find a catalog that is only for a specific hemisphere. Catalog generally cover the entire sky. Also, most catalogs are for either stars or deep sky objects (galaxies, nebulae, and so on). I have not seen a catalog that includes both stellar and non-stellar.
For stars, I suggest the Yale Bright Star Catalog. It includes around 9000 stars, some of which are slightly too faint to see with the naked eye.
For nonstellar objects, I suggest the NGC Catalog.


How can a red giant grow so big? Supposedly when the sun becomes a red giant, it will grow big enough to swallow Earth.
However, this requires the radius of the sun to expand by a factor of approximately 215×, meaning its volume would have to expand 10,000,000×.
Maybe it's just me, but something about this doesn't feel intuitive :-) especially given that the sun isn't gaining much mass. Exactly what about fusing elements heavier than hydrogen implies that the reactants and/or products take up 10,000,000× as much volume? Nobody ever explains this when explaining the stellar lifetime, and I don't understand why it should be the case. (In fact, I would have expected fusion to result in a decrease in volume, given that nuclei are combining...)
Edit:
It seems there are 2 types of red giants, some that occur during hydrogen fusion, some helium.
If the answer is different for these two types, I'd at least like to know the answer for the helium type (though of course I'd appreciate one that goes further and addresses both).

In my mind, none of these explanations really cover the actual reason that red giants expand.  Indeed, this subject seems like an area where people just make up anything that sounds plausible, but it's often quite wrong (Fraser Cain mentions both light pressure and a higher volume in the fusing shell, but light pressure plays no role at all, and the volume of the shell isn't much different than the core, which is way smaller than the Sun's core). So let's get the story straight.

Many of the descriptions do include some of the key elements, including the fact that you have a shell of hydrogen fusion going on atop an inert degenerate helium core.  But the key reason for the expansion is that the way this shell self-regulates its fusion rate is quite a bit different from the way the Sun's core is self-regulating its fusion now.

Now, the Sun's core self-regulates its fusion rate to match the rate that energy (in the form of light) is diffusing out through the mass of the Sun.  The way it does this is essentially the same in all main-sequence stars: they regulate their core temperature, which is why the core temperature of more luminous main-sequence stars is a little higher.  But this is not at all the way the shell fusing in a red giant self-regulates its fusion rate-- it cannot regulate its temperature, because the temperature is handed to it by the gravity of the degenerate core its sits upon.  (This sets the temperature via the virial theorem, that is the key way the degenerate core affects the shell-- it sets its temperature.)  Since the shell doesn't regulate its own temperature, the temperature tends to be quite high, especially as the core gains mass (this is why the luminosity rises with time).  Fusion is very sensitive to temperature, so being stuck with a very high temperature makes the fusion rate go berserk.  The rest of the star cannot support this spectacular fusion rate, so something else happens.

The star puffs out, and in so doing, we find the way the shell regulates its own fusion rate: it lifts weight off the shell.  That reduces the pressure in the shell, which compensates for the high temperature and brings the fusion rate down to what the rest of the star can manage (which is set by the rate light can diffuse through the shell).  So there's the real reason-- the star must find a way to lift weight off its crazy-high-temperature shell to keep the fusion rate from going nuts, but the upshot is that the fusion rate is still pretty high, and gets higher as the mass of the core rises, forcing the shell temperature up ever more and forcing the star to puff out even more.


How do they know the TNT equivalent of the meteor explosion over Vermont? I've seen several reports of this meteor explosion, below are two.
Was there infrasound recordings of the pressure wave, or were weak, local seismic signals recorded, or it this just a ballpark estimate based on circumstantial evidence?

Bangor Daily News: Meteor exploded over Vermont with force of 440 pounds of TNT
Sputnik News: Yield of Fireball Explosion Above Vermont Rated at 200 Kilo(grams) of TNT - Video

According to NASA Meteor Watch Facebook page, they:

obtained infrasound measurements from 3 nearby stations - the
amplitudes and durations of the signals put the energy of the fireball
fragmentation at 440 pounds (200 kilograms) of TNT.

Here is an image of the infrasound:

From this news story,

A man in Canton, Maine said there was a delay of several minutes from
the time he saw the fireball until heard a loud boom. He said it shook
his home unlike anything he experienced before.

It all makes me wonder How loud was the asteroid that killed the dinosaurs?
Further reading on infrasound detectors monitoring space:

What is used to monitor objects burning up in the atmosphere on descent?
this answer to What local pressure changes would a starship launch on Mars cause?

and others:

What efforts have been made to separate the microseismic events related to wind-related ocean waves from the components of volcanic origin?
Project Mogul: Does upper atmosphere indeed have a sound waveguide channel?


Whereabouts of the Pleiades I assume they should be in the Milky Way galaxy. I have read about it a little but did not find information on where they are.

Which galaxy do the Pleiades belong to?

With the exception of the Andromeda galaxy and the Magellanic clouds1, every star, star cluster and nebula that is visible to the naked eye is part of the Milky Way. The Pleiades is a star cluster in the Milky Way.

Objects in other galaxies are too far and too dim to be visible with the naked eye.

1The Triangulum galaxy (M33) may also be visible to the naked eye from a dark site to trained observers with excellent vision


It is possible to combine a reflector and a refractor telescope together? I know it sounds crazy, but is it possible to
connect a reflector telescope and a refractor telescopes (like monocular for bird watching) together to build a better telescope?

...is it possible to connect a reflector telescope and a refractor telescopes (like monocular for bird watching) together...

Yes, a properly focused telescope (for someone who's not near-sighted) will have infinity conjugate focal points; parallel rays in make parallel rays out.
If the telescope is set up for 50x magnification (f=600 mm objective, f=12 mm eyepiece) then a 1 degree field of view (FOV) in the sky will be a 50 degree apparent FOV at the output of the first eyepiece.
Alas, a second 8 x bird-watching monocular may only have a 7 degree input FOV so we will indeed get 50 x 8 = 400x magnification, but of a 0.14 degree field.
That's just what you would get looking through a single telescope set up for 400x (say f=1200 mm objective, 2x Barlow and f=6 mm eyepiece).
In other words, you could do this, but...

...to build a better telescope?

I'm not sure it would be better in any possible way if it's for looking through.
However, if you have an old CMOS imager webcam from the 1990's with giant pixels and you want to blow up the primary focal plane of the first telescope to get some useful resolution, this might be a suboptimal solution. You could remove the eyepiece of the second bird-watching monocular and put the image sensor with too-large pixel size there and get better resolution than if you'd put it at the focal plane of the first telescope.
A better way to solve this unlikely problem might be to simply use positive or eyepiece projection with the first telescope's eyepiece moved outwards so that it's final focal point moves in from infinity to say roughly 8 times its focal length.
That would have the same effect but without adding the monocular.


Is diffuse energy distribution considered in the galaxy rotation curve explanations? Problem number 23 in chapter 1 of Purcell's "Electricity and Magnetism" Second edition page 37 shows that 90% of the energy stored in the electrostatic field of a spherical charge distribution of radius R is contained in a spherical region of 10R.

The energy by this description is diffuse and only partially resides within what we would call the physical bounds of the particle. Gravity has the same form as the electromagnetic force and I would expect would have this same character of distribution of energy (i.e. 90% of the energy of the gravitational mass contained within a radius of 1R is stored within a sphere of 10R).

Is this true of gravitational energy and has this little detail been considered in the galaxy rotation curve explanations? I would think that something this elementary would be well accounted for in galaxy rotation descriptions. Much of the electromagnetic and gravitational energy seems diffused into the area surrounding mass and I think that galaxy rotation curves require more diffuse gravitational energy distribution don't they?

You appear to be asking whether gravitational potential energy acts as a significant contributor to the total mass-energy of a galaxy - a question that is only relevant within the context of General Relativity, where indeed other forms of energy besides the rest-mass energy contribute to the gravitational effects.

It turns out that GR is not required to interpret galaxy rotation curves because the influence of the other forms of mass-energy (or more formally, other contributors to the stress-energy tensor) are very small.

The contribution of gravitational potential energy can be assessed using the ratio $GM/rc^2$, which is approximately the ratio of gravitational potential energy to rest mass energy. For example, if $10^{12}$ solar masses exists within a 100 kpc radius (reasonable numbers for the Milky Way); then the ratio is $5\times 10^{-7}$. Of course, if there is no dark matter then this would be even smaller.

I should also add that this tiny correction, like the gravitational potential energy, is negative. That is, it reduces the gravitational influence of the mass/energy in a galaxy. The (positive) total kinetic energy of the Galactic components (also tiny) would also need to be added, but since bound systems have a total energy (kinetic plus gravitational potential) $<0$, there is still a net reduction in the gravitational mass.


Position of the Sun in the constellation Pisces around the year 1500 I am reading a book on ancient astronomy and it claims that around the year 1500 the position of the sun, at the time of the spring equinox, was on the tail end of the second fish in the constellation of Pisces. Now I understand this might be more guess work than anything else but I am wondering if this claim could be even remotely true. I don't even know where to begin in calculating this. Any help is appreciated.

The position of the sun, relative to the stars, at the time of the equinox changes slowly and predictably.  This means that we know to a high degree of precision exactly where the sun was at the time of the equinox in 1500.

It is pretty easy to calculate  The position of the sun at the time of the equinox changes by about 1 degree every 71.3 years. The sun is currently in pisces at the time of the Equinox.  520 years ago it would have been 7.3 degrees closer to Aries.

Dbachmann / CC BY-SA (http://creativecommons.org/licenses/by-sa/3.0/)


Absolute magnitudes of stars Where to find comprehensive list of absolute magnitudes for stars of different spectral types?  I need published paper or web page with data and reference to the source.

VizieR is an online source for all sorts of astronomical data published in scientific papers. As you mentioned, The HIPPARCOS catalogue contains visual magnitude data.

Open the query page for the main HIPPARCOS catalogue
Select the fields you want (defaults are ok for you)
Hit submit to see the results
You can limit the number of results and format under Preferences on the left

This table gives you the measured visual magnitude, i.e. the Apparent Magnitude ($m_V$, V column). To convert that into Absolute Magnitude ($M_V$) you need to know the distance to the star. This can be calculated using the Parallax field (Plx column).

Here's the formula for you:

$$M_V = m_V + 5 * log_{10}( Plx / 100 )$$

You can easily dump the data into Excel or something, put a formula into an extra column and calculate the Absolute Magnitude.


How do we know that the Universe is still expanding now? Yes, I know that most galaxies have a red shift and that means they are moving away from us.

The problem is that the farthest galaxies are 13.8 billion light years away. That means that the info is 13.8 billion years old.

What has been puzzling me for years is that the red shift observation shows that the galaxies were receding 13.8 billion years ago. We don't have direct data on what they are doing now. It is natural that the universe was expanding with a tremendous speed at that time because it was near to the big bang, but how do we know that it still is?

If the furthest galaxies are moving with higher speed than the nearer, that means to me that their speed is reducing with the time.

You're right that in principle we cannot know if the distant galaxies are still receding from us. When, however, we think that they indeed are, it hinges on the cosmological principle, viz. that the Universe is homogeneous and isotropic, i.e. "looks the same everywhere and in all directions". This is a philosophically motivated statement, which is continuously tested observationally. So far, there has been no significant experimental evidence that this is not correct.

When we observe the velocities of galaxies throughout the Universe, we see a smooth velocity as a function of time, or distance, with some intrinsic scatter which is due to the gaalxies' peculiar motion, i.e. their velocities through space, which is usually of the order of 100–1000 km/s. That is, distant galaxies move away from us fast, and nearby galaxies move away from us slowly. As you say, the expansion rate decreased with time, but only until a certain point, after which it started increasing again (due to dark energy).

The fact that we see that nearby galaxies are still receding, together with the cosmological principle, implies that the distant galaxies should also still be receding. We know of no reason that they shouldn't. Of course this is no proof, but in physics, there are no proofs, only verifications and falsifications. To postulate that the distant galaxies no longer recede, and to be taken seriously, you would need to come up with an experiment that could verify this hypothesis, preferably along with a mechanism that would be able to cause this phenomenon (this looks a bit condescending when I type it; this is not my intention).

The expansion rate is currently — and hence locally — around $70\,\mathrm{km}\,\mathrm{s}^{-1}\,\mathrm{Mpc}^{-1}$. Due to the aforementioned velocity scatter, we cannot probe expansion below ~1 Mpc, but at this scale, space doesn't really expand anyway due to the mutual gravitational attraction of the Local Group. This means that in principle, the expansion of the Universe could have come to a halt within the last few million years, and still be in accord with the cosmological priciple. However, this would mean that the evolution of expansion velocity as a function of time would have a sharp kink. Again, this is not impossible, but we know of no physical mechanism which would be able to cause this.


Stars at near break-up rotation rates Accretion discs are ubiquitous in astrophisics. As a direct corollary, they are important for the following question.

Consider the following model, representing one of the most simple models for accretion discs. A central object is a star (pre-MS, WD or NS, but not a BH) of mass $M$, surrounded by a thin flat disc of material, which continuosly feeds the star at a rate $\dot{M}$, such that $M/\dot{M}$ is much larger than thermal and dynamical timescale of the star (i.e. accretion rate is slow).

Everywhere in the accretion disc its local motion is nearly circular and nearly Keplerian. Therefore, at the interface of the star and the disc the disc will always tend to make the star rotate at nearly-Keplerian velocities. From the other hand, if the stellar outer parts were to rotate at nearly-Keplerian velocities, these parts would become gravitationaly detached from the star, which would have significant consequences for the stellar shape and structure. Surely, though, the process is going to be slow and the acquired angular momentum will be redistributed within the star.

Now the question: What will be happening to the star if it approaches nearly break-up velocities due to such a spin-up? This involves a few subquestions: How close the rotation rate can actually get to the critical one? If it can get close enough, how would the whole process look like? That is, what would happen in the short term to the star when the effects of rotation will start to affect its structure? What would happen to the star in the long term?

I would like to keep this problem as a purely hydrodynamical one. That is, assume, that the only laws involved are hydrodynamical and gravitational ones, with some constant accretion rate supported. In reality magnetic fields would also play an important role for some stars, and stellar winds could also possibly be important.

Examples of the decribed systems are numerous. It might concern cataclysmic variables, millisecond pulsars, pre-main sequence star in a protoplanetary disc, and many more.

I have not the qualification to answer the question in its whole but the question is interesting (I worked on Be Stars which are episodically surrounded by an decretion disk and which rotates at nearly critical velocities. The phenomenon in Be stars is different from accreting stars. The only consequences of subcritical velocity is a flattened envelope and modification of its inner structure and of the oscillations modes found in these stars
(if you have time and curiosity, a good example of flattened star with Keplerian rotating decretion disk is Achernar, a Be star observed using interferometry --> Have a look at Meilland et al. 2007: www.aanda.org/articles/aa/pdf/2007/10/aa4848-06.pdf)

Anyway...

I found this paper about critically-rotating accretors. May be you will find answers to your questions here or in its references (use nasa ads site for your query: http://adsabs.harvard.edu/).
http: //arxiv.org/pdf/1306.1348v2.pdf
It seems that in the introduction, there is some answers to your questions about reaching the critical velocity.

The accreted mass can increase the rotation rate until the star reaches the critical velocity.

It is said : "For a typical 6 + 3.6 M⊙ system, with initial period
Pinit = 2.5 days, in the absence of spin-down mechanisms, only
3 per cent (0.12 M⊙) of the total amount of matter transferred
by RLOF (more than 5 M⊙) is enough to spin the gainer up
to the critical rotation."

But we still don't know if the gainer can really reach the critical velocity. Some papers are dealing with break-up mechanims which does not allow the gainer to reach critical velocity: spin down by tides, magnetic breaking, limitation of the accretion angular momentum through interaction with the accretion disk, stopping of the accretion mechanism...

I am sure that you will find many papers on nasa ads that will give you answers to your questions.


Is the dot on the photo - Mars? The below photos were taken yesterday (August 4, 2018) at about 10 p.m. from the plane over the Black Sea.

We were seeing the light dot on the left for at least 10-20 minutes. It looked like a bright star, but there were no other stars.

How likely that it was Mars?

This was not Mars.  Mars is close to opposition now, so it almost opposite the sun, and visible as an obviously red "star" in the Southern sky at midnight.

Your photo shows that the sun has just set, and you note that other stars were not visible.  This is the planet Venus. Currently shining at magnitude -5 it is 100 times brighter than the most bright stars. It is common, shortly after sunset for it to be the only visible "star", and as it orbits closer to the sun than the Earth, it is always seen in the sky close to the sun, as your photo shows. The when the Planet Venus is an "evening star" it is seen in the west. This is consistent with your plane travelling in a roughly northerly direction.

It is hard to be sure what time zone, you are in, or exactly where you are, but Venus will be visible shortly after sunset in most parts of the world.


How does gradual crossing over of the Roche limit transform a planet or moon? Some cartoon simulations (an example) of an object (assume here a large moon with Earth-like density) which crosses the Roche limit of a much more massive giant planet, illustrate a circle suddenly disintegrating and forming a ring. But such a process occurs gradually over geological time scales. I doubt that there would be a single bad day when the whole moon disintegrates (as in -Oh, there Mt. Everest broke off and flew away!)

Wouldn't the tidal tugging cause volcanism and gradually melt the moon as its (somewhat eccentric) orbit spirals inwards across the limit during millions of years?

How would the melting moon deform? Would it really assume an oval shape, elongated towards the primary planet? Since the near side wants to orbit faster than the far side, wouldn't that cause the moon to spin at an accelerated rate, even if it initially was tidally locked? Would the melting, deforming and spinning save it from disintegrating further to within the Roche limit?

Some examples of Roche limit crossings I know of:

Comet Shoemaker-Levy 9 may have broken up suddenly because it had
such a high velocity relative to Jupiter.
Phobos will cross its Roche limit to Mars in ~50 million years. It
will hardly be a sudden process. I suppose its very low density and
mass won't allow any volcanism and melting, though.
KOI1843.03, an exoplanet candidate with a density of ~7 g/cm³,
is already inside the Roche limits for its less dense compounds which
it has shed (as I interpret it).

A large moon (the size of a planet) would be modeled by the fluid model, meaning friction and tensile strength are too weak to modify the shape of the moon significantly, shape is determined by rotation, vortices, self-gravitation, and tidal forces.
Smaller moons, like Phobos would likely follow a rubble pile model.
On circular orbits tidal heating doesn't play a major role.

The fluid model has been studied extensively. The occuring shapes are known as equilibrium figures.
For a full understanding, the parent body should be considered together with the moons, as one system. (Fluid) moons getting too close to the Roche limit gradually deform (tidal heating depends on inner friction), and start to shed mass at some point. There exist solutions with vortices, and vortex-free solutions, depending on the properties of the fluid. The lost mass may either form an n-body system outside the Roche limit, or form a ring system, mostly inside the Roche limit.
Closer to the planet, it may either impact, and fuse with the planet to a the overall spheroid of the planet, or - if the planet is already rotating fast, and the mass of the moon was sufficiently high - increase the rotation of the planet, such that it may undergo e.g. the Maclaurin - Jacobi series, up to the point, that it cannot accrete more mass. There are several more possibilities in detail.

Rubble piles with tensile strength are more likely to undergo sudden shape changes or disruptions than rubble piles without tensile strength. Behaviour of rubble piles without tensile strength depends on friction. The less friction, the closer to the fluid model.
The more tensile strength of the rubble pile, the more it behaves like the ridgid model. Ridgid objects may survive well below the Roche limit for fluid objects, but disrupt suddenly when tidal forces get too strong.


How fast are we moving relative to the CMB? The cosmic microwave background radiation should provide kind of a global reference frame, because you can determine your speed relative to it using the redshift.

Is it known how fast we are moving in relation to the CMB? If you subtract the various orbital motions (Earth around the Sun, Sun around the Galaxy), are we standing still in the expanding universe, or traveling in a certain direction?

Yes, our (i.e. the Sun's) motion in the "global", or comoving, reference frame can be measured accurately from the dipole of the cosmic microwave background. The latest results from the Planck Collaboration et al. (2018) yielded a velocity of
$$369.82\pm0.11\,\mathrm{km}\,\mathrm{s}^{-1}
$$
in the direction
$$
\begin{array}{rcl}
\ell & = & 264.021º\pm0.011º\\
b    & = & 48.253º\pm0.005º
\end{array}
$$
(in Galactic coordinates).

Since Earth orbits the Sun with some $30\,\mathrm{km}\,\mathrm{s}^{-1}$, there's a small, biannual correction to this result. On much larger timescales ($\sim100\,\mathrm{Myr}$) our motion round the Milky Way alters our comoving velocity with the order of $\sim100\,\mathrm{km}\,\mathrm{s}^{-1}$.


Orbits in a binary star system I know of three sets of stable orbits in a binary star system: orbiting closely around star A, orbiting closely around star B, or orbiting distantly around both stars (and their mutual center of gravity) at once.

Is there a fourth set of stable orbits, around the mutual center of gravity, but inside both stars' orbits?

The point you appear to refer to, is called the Lagrangian point $L_1$.
This point is a saddle in the field of gravity, hence not to be considered to be stable in the strict sense.
Two other Lagrangian points, called $L_4$ and $L_5$, can be stable, provided the considered orbiting objects are of small mass in comparison to the two main bodies of the system, and if the masses of the binary components are sufficiently different.

According to theorem 4.1 of this paper, $L_4$ and $L_5$ are stable in all directions, if and only if the mass ratio of the two main binary components $\frac{m_1}{m_2}\geq\frac{25+3\sqrt{69}}{2}\approx 24.9599$.
According to theorem 3.1 of the same paper all Lagrangian points are stable in z-direction, which is the direction perpendicular to the orbital plane of the binary system.
(Credits for this corrected version go to user DylanSp.)


Source of red in earth's photographs What is the red in this picture of earth?  At first we thought it might be pollution, but Jupiter also has red.
Picture source:  https://education.seattlepi.com/primary-movements-motions-earth-4701.html

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.


Light beam (1 photon) in the limited universe? If our universe is not infinite, what happens with the light beam (or photon) when it will travel through the whole universe?
For example, observable universe according to wikipedia has diameter 93 billion light-years. What will happen with photon after 93 billion years if I generate it now? Will it return back to me? Or it will never reach the "edge" of the universe because the universe is expanding faster?

UPDATE 1

Of course, I assume that photon will no be absorbed :) because it is just a theoretical thinking. And return back I mean not exactly to the same point, but approximately within some range (let say 5% of the universe size).

Do I correctly understand that there are a few options for the case when the universe is finite (means "closed geometry"):

Photon will never return back because the universe will collapse before it (in case if there are no dark energy).
Photon will never return back because universe expanding faster than light speed.
Photon will come back because universe expanding slower than light. And what
of these 3 options is most probable from a current scientists point of view?

A finite universe is said to have a "closed geometry", or to be "positively curved", meaning that, in principle, you may travel in a straight line and eventually return back to your starting point. In the 2D analogy, the surface of Earth is positively curved, and if you travel 40,000 km straight, you're back where you started.

A finite universe that does not contain dark energy of some sort, however, is bound to collapse, and you can show mathematically that even a photon will not make it back to its starting point before the universe has collapsed. Only if you consider a photon emitted exactly when the universe is created (Big Bang), it will make it back exactly when the universe ends (Big Crunch).

If the universe contains a sufficient amount of dark energy, this may prevent collapse. This may cause it to expand too fast for the photon to "get back", but you could fine-tune its components such that the expansion/collapse is slow enough for the photon to get back.

One such model is "the loitering universe", which gained some interest in the 1960's, as an explanation for an excess of quasars around redshift $z\simeq2$ (e.g. Petrosian et al. 1967). If the deceleration of matter is almost exactly balanced by the acceleration of a cosmological constant, you can have an (arbitrarily long) period of time where the universe is static.

However, observations indicate that our universe is not such a universe (Planck Collaboration et al. 2018).

Note also that since the universe is not empty, a photon may be absorbed on its way, not returning even if the expansion allows it. However, in our universe, photons that were emitted shortly after Big Bang (the so-called CMB photons) have traveled mostly freely through space ever since, with only 5% interacting with matter along its way.


How do I calculate the inclination of an object with an amateur telescope? Suppose I would like to calculate the inclination of a satellite from the ecliptic. Would it be possible to do this with an amateur telescope? How would I go about doing so?

Note: A good answer should tell what kind of telescope an amateur would need, what measurements they would need to make, then what calculations they would have to perform to get the inclination (or the "instantaneous angular measurement from the ecliptic at the time of measurement").

Any telescope can be made to give you the information that you are looking for.  The first thing that you will need to know is the location of the ecliptic which varies throughout the year.  Or are you looking to find in relation to the the celestial equator?

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

Either way, you would start of the same by find the Declination (D) and Right Ascension (R.A) of the object in question.  You do not need a telescope to find this, unless you are unable to see the object with the naked eye.  If the telescope has a polar mount and is properly set up.  You can read the R.A. and D off of the mount.

Otherwise it will require some trigonometry and knowing where you are (at least your latitude)

You will need to determine the Altitude and Azimuth of the object.  Which is really just the direction from North and the angle from horizontal that your telescope is pointing.  You could do this with a compass and protractor or even an astrolabe.  Knowing this you can then convert to R.A and D with the following formulas:

$$RA = \arctan(\frac{- sin(Az) \times \cos(Alt)}{\cos(Lat) \times \sin(Alt) - \sin(Lat) \times \cos(Az) \times \cos(Alt)})$$

$$Dec = \arcsin(\sin(Lat) \times \sin(Alt) + \cos(Lat) \times \cos(Alt) \times \cos(Az))$$

You are looking for the inclination to the ecliptic, so you are mostly concerned with the declination.  The ecliptic changes in declination throughout the year from 0 at the equinox to +/- 23.5 at the solstice.  So your inclination from the ecliptic would be the Declination of your object +/- the Declination of the ecliptic.


Slow moving star like objects in the night sky So yesterday at about 2am I was watching the sky when I saw a small star like object moving very slowly across the sky. My initial thought was that it was a satellite due to the fact it was moving so slowly and wasn’t flashing. However  then I saw another one moving in the same direction about 10 minutes later. I had to keep letting my eyes adjust to the darkness to see it as it was so small but in total, I saw 4 of them moving in very similar directions, almost like they were following a path. There was about a 5-10min gap between each one and I could see it the entire time until it dipped over the tree line. I live in a very rural area so there was no light pollution or anything to interfere with the view. Now I know that around about now we are able to observe the Perseid meteor shower from the comet Swift-Tuttle but these object were very slow moving and I don’t really know how slow a meteor can move, but isn’t it strange that all of them were moving in the same direction? I also did see some normal meteors but as you’d expect, they only lasted for a second or under. Anyone have a clue what’s going on? Thanks.

Unlike meteors, which burn out in a second or two, satellites can be visible for a few minutes at a time.
In twilight they may be visible all the way across the sky; at darker times of night the Earth's shadow cuts their visibility short unless they're in high orbits.

One good resource for such things is Heavens Above.
From Hereford, UK, their Daily predictions for brighter satellites list a few matching your description on the morning of 2019-08-07.
For example, the polar-orbiting Russian intelligence satellite Cosmos 1975 passed from WSW to N between 02:00 and 02:04.


How can I estimate the M/L ratio from a fitted SED on a galaxy? This answer (here in the forum) provides background to my question: https://astronomy.stackexchange.com/a/5953

I am trying to understand the process in which a stellar population synthesis (SPS) is created, in order to calculate the stellar mass of a galaxy. (This is my main goal: understand how the stellar mass is calculated via SPS and broadband photometry)(Section 2.5:  https://ned.ipac.caltech.edu/level5/Sept14/Conroy/Conroy2.html)

Ok.
I have already understood the process to obtain a stellar population library (SPL). Such SPL will be fitted to observational data in order to create a SED (Spectral Energy Distribution)

Ok. Now I have the fitted SED on a galaxy, here are my questions:

How can I estimate the M/L ratio of that galaxy?
How can I estimate the L?

I feel it is an easy and direct answer, but I don't get it.

The luminosity L isn't estimated, it's observed. You make models of stellar populations with many different parameters and those give you a set of SEDs. For each set of parameters, you will have an SED covering a range of wavelengths (a continuous line).

From your observations, you will have an SED of a few points depending on how many wavelengths you observed. You use $\chi^{2}$ fitting or some other technique to compare all your model lines with those few points and see which fits best. This gives you the best fit value for the parameters in your model, one of which is M/L.

You have the actual luminosity of the galaxy from the observations (at some wavelength) so you multiply the estimated M/L by the measured L to get an estimate of M.


Why is the Earthian atmosphere so thin? Venus is somewhat lighter then Earth, yet has a much thicker atmosphere. One would imagine that the following should be true:

During the formation phase, all inner planets had captured as much gas as they could hold per gravitational/thermodynamic equilibrium. After all, even puny Mars managed to capture a sizable atmosphere.
The rate of atmospheric escape should be much higher for Venus:

Venus receives more heat from the sun, thus higher Jeans escape rate
Venus has negligible magnetic field, so some of its atmosphere must be lost to direct "blow-off" by solar wind

Yet, it is Earth that seemingly misses a great deal of atmospheric volume. So the question is: what are the current theories regarding the "thinning" of the Earthian atmosphere? When and why the atmospheric gases had left the planet?

The short answer: Atmospheric gases never left Earth, they're in it!

The long answer to this question isn't just about the planets' current states, but rather the processes that led them there. Let's start at the very beginning (a very good place to start).

The very early years

When our solar system began forming 4.6 billion years ago, most of the mass from the collapsed portion of a molecular cloud (see Nebular hypothesis) collected in the center to form the Sun. The mass that didn't collapse into the Sun left a protoplanetary disk—a cloud of dust and gas—surrounding the new star. Gradually, dust particles began to come together through accretion, attracting more and more particles into fledgling planets.

Close to the Sun, where Venus and Earth both reside, it was too hot for many particles to condense, so the planets in this region formed with metals and silicates, which have high melting points. This is why the four planets in the inner Solar System are called the "rocky" or "terrestrial" planets. The earliest atmospheres on these planets began forming with the gradual collection of gases from the solar nebula, mostly hydrogen.

Goldilocks and the two planets

At this point in the evolution of the two planets, they looked pretty similar, but there is one major difference: the distance to the Sun. Earth, it seems, got lucky to be in the "Goldilocks zone", where the temperature is just right to support life. Being in this zone has two major implications: liquid water and, as a consequence, active plate tectonics. (See this article for an in-depth view into why the two are related.)

Carbon sinks

On Earth, there is considerable water contained in liquid oceans. On Venus, this is not the case. It is simply too hot that close to the Sun, so all the water evaporated into the atmosphere. (Venus likely contained liquid water in its early stages, but it all evaporated after a billion years or so.) It is also likely that the young Earth once had a dense, punishing atmosphere like Venus's today. However, surface oceans and plate tectonics both provided ample pathways for gases to be absorbed into the surface of the Earth. Oceans and plate tectonics offer huge amounts of carbonate storage, allowing the transfer and balance of carbon compounds to and from the atmosphere.

So now we have two things reinforcing the differences between Earth's and Venus's atmospheres:

Evaporation of liquid water: It is too hot on Venus for liquid water to exist. All water has evaporated, resulting in a denser atmosphere. On Earth, water can reside on the surface, lessening the amount in the atmosphere.
Carbon sinks: Liquid water and plate tectonics allow Earth to absorb considerable amounts of gas, allowing the atmosphere to be thinned of certain compounds like carbon dioxide. There is no such pathway on Venus, forcing all gas to remain in the atmosphere.

With no major mechanisms for gases to be absorbed by the planet, Venus is experiencing a runaway greenhouse effect.

Atmospheric escape

You mention Jeans escape. It is true that this effect is greater at higher temperatures; however, it is much easier for smaller molecules to escape than for larger ones. Hydrogen and helium, being the two smallest elements, are the most greatly effected by this phenomenon. In comparison, carbon dioxide, which makes up the majority of Venus's atmosphere, is not much affected by Jeans escape.

You also mention solar winds. While these play an effect, especially on planets without a magnetic field, this phenomenon is not as powerful as you might think. Ultraviolet light (i.e. photoionizing radiation) causes ionization in the uppermost region of the atmosphere. These charged particles now form a shell (called the ionosphere) that deflect solar winds, much like a magnetic field would. On Venus, the thick atmosphere provides more particles for ionization, resulting in a more powerful deflection. (Compare this to Mars, where solar wind is the primary non-thermal escape mechanism due to the thin atmosphere with few ionized particles.)

The primary atmospheric escape mechanism for Venus is actually a bit more complicated. In the absence of a magnetic field, it is easier for charged particles to escape. In particular, electrons are most susceptible due to their small mass. As electrons escape, the net charge of the ionosphere leans positive, causing the ejection of positive ions, mostly H+.

Conclusion

While Earth and Venus formed similarly, Earth got lucky. It has pathways to remove gases from the atmosphere, while Venus does not. Additionally, the two planets do not experience significantly different rates of atmospheric escape. This results in the atmospheric densities where know today: 66 kg/m3 for Venus and just 1.2 kg/m3 for Earth.


Thinking about the Twin Paradox, what would a Twin Traveling at .8c, calculate as the amount his EBT aged at the mid-point in a 4 lightyear trip? Referring to the twin paradox, where we have an Earth Bound Twin “EBT”  and a Traveling Twin “TT” who travels away from Earth at velocity= .8c, lets say to Alpha Centauri, when the TT is half way there, what would the TT calculate as how much his EBT’s aged at that point?
To simplify, round the distance to Alphas Centauri to say 4 lightyears.

I am thinking .8c yields a lambda of .6 and so the TT’s  4 lightyear trip, reduced to 2.4 lightyears due to space contraction. Therefore the midway point is only 1.2 light years. Traveling at .8c, means the TT must himself be 1.2/.8…. 1.5 years older at the midway point.
Since his EBT, relatively, is “traveling” at v=-.8c during this time, his EBT must appear to age relatively slower compared to him.  Using the lambda of .6, the TT would calculate as the amount aged by his EBT to be 1.5*.6=.9 years.
Is this correct?

I find it easier to just write down the coordinates of the events of interest.
Given coordinates $(t, x)$ in the frame of the Earth and choosing units such that $c=1$, the Lorentz transform to the frame of the traveller is $(t', x')$, where
$$\begin{align}
t' &= \gamma(t - \beta x) \\
x' &= \gamma(x - \beta t)
\end{align}$$
In your scenario, $\beta = \frac{v}{c} = \frac{4}{5}$, giving $\gamma = (1 - \beta^2)^{-1/2} = \frac{5}{3}$. The relevant events, in the reference frame of the Earth, are then:

Traveller leaves Earth at $(0, 0)$.
Traveller reaches halfway point at $(\frac{5}{2}, 2)$.
Earth is at the origin for all $t$, giving coordinates $(t, 0)$.

Transforming these into the traveller's frame, these become:

Traveller leaves Earth at $(0, 0)$.
Traveller reaches halfway point at $(\frac{3}{2}, 0)$ — as expected, the space coordinate is zero because the frame is moving with the traveller.
Earth is located at $(\frac{5}{3}t, -\frac{4}{3}t)$ — note this is expressed in terms of $t$, the time coordinate in the reference frame of the Earth. As expected, the Earth falls behind the traveller as time advances.

Equating the time coordinates in the traveller's frame gives $\frac{5}{3}t = \frac{3}{2}$, which gives $t = \frac{9}{10}$ years for the time that has passed on Earth, as viewed in the traveller's frame (who would have experienced a duration of $\frac{3}{2}$ years up to this point).


Why do pictures of the Milky Way look like a spiral? When I see pictures of the Milky Way they show a spiral shape. Why is this?

The objects in a galaxy orbit it's center as a planet does with it's star, they keep the same distance. They look like a spiral for something that is called density waves that make some parts of the galaxy (the arms) more dense.

You have to imagine a star system like ours orbiting the galaxy in your picture following those radial lines, so sometime the system will be inside of and arm and sometimes outside of it.

Thanks to this extra density in the arms is where most of the star formation takes place.