Gravitational Waves and the Big bang The Ligo website says "Detecting the relic gravitational waves from the Big Bang will allow us to see farther back into the history of the Universe than ever before."

I find this puzzling. How can a Gravitational Wave originating at about the same time and place as the particles that make up their detector (ie in the Big bang) be detected?
To use a simple analogy - an explosion in an electronics warehouse causes a flash of light and pieces of electronics gear to be scattered. These pieces are then assembled to make a camera which is able to photograph the explosion. I am confused!

Answers suitable for a high school physics class greatly appreciated.

Firstly, Big Bang didn't happen at a point in space, away from which we are traveling. Big Bang was the creation of space. This space has been expanding ever since, so that the distances between everything increases, but Big Bang happened right where you are, where the Andromeda galaxy is, where GN-z11 is, and so on.

The Universe has evolved ever since. If we want to know how galaxies look 13.8 billion years (Gyr) after Big Bang, we can just look around in our neighborhood. If we look too far away, we don't see 13.8 Gyr old galaxies, because the light has taken some time to reach us. Thus, if we want to see galaxies that are 12.8 Gyr old, we simply look 1 billion lightyears away; if we want to se 10 Gyr old galaxies, we look (roughly) 3.8 billion lightyears away, and so on.$^\dagger$ In this way, we look back in time.

If we look away far enough, we would in principle be able to look 13.8 Gyr back in time. However, we face a problem in that until the Universe was 380,000 years old, it was opaque to light (why this is so is another story). It wasn't opaque to gravitational waves, however. And since lots of GWs are thought to have originated during the epoch called inflation, which were responsible for the expansion of the Universe, and which took place a fraction of a second after the creation of space, we say that GWs offer the possibility of looking all the way back to Big Bang.

$^\dagger$This is somewhat imprecise, since galaxies weren't created instantly after Big Bang, and since all galaxies weren't created at the same time. But for the sake of the arument, let's pretend they were.


Brown bands in Andromeda galaxy? Many images of the Andromeda galaxy depict a yellow-red-ish bright center and then those brown "dust layers" around the center. As far as I know the bright light comes from stars, but what is this "brown dust"?

It is the remains of dead stars.

Large stars are able to fuse light elements (like helium) to make heavier ones (like Carbon, Oxygen and Silicon, but also all the elements found on Earth). When these stars die, some of these heavier elements are ejected into space, where they condense to form dust particles. The dust particles are small, typically a few micrometres long (similar in size to a bacteria). The dust particles are mixed with gas and ices and form nebulae. If the gas and dust can become compressed enough, then clumps of dust and gas can collapse to form new stars, and planets. Every carbon atom in your body started out in interstellar dust.

The dust is therefore composed of normal matter, and in fairly familiar forms, such as "silicon oxide" (which is what much of the rock on Earth is made of) There are also carbon grains, and metals: aluminium, magnesium and iron (usually as oxides).


Why does east and west appear to be mirrored in the celestial sphere? I'm wondering why East is left and West on the right of the celestial sphere if you look at images of it?
Example below:

(source)

You're viewing the celestial sphere from the inside looking outwards, as opposed to maps of the Earth's surface which are viewed from the outside looking inwards.

Imagine drawing a map on a sheet of glass. When you view the map from the other side, it will appear mirrored.


The size of Copernicus’ and Ptolemy’s corrections I understand that both Copernicus and Ptolemy needed epicycles, but one might assume that a heliocentric system would need much smaller corrections to account for the non-circular motion of the planets.  This is however not mentioned as an important reason to favor Copernicus’ system. Is this not the case?

I gather the worse problem for Copernicus was Mercury.  Not only is Mercury's orbit quite elliptical, it has relatively high precession due to the gravitational influence of other planets.

I found a quote attributes to Copernicus in this document (sorry, only direct link available) where he says :

  It is interesting to note that Copernicus had difficulty explaining Mercury’s motion and once commented that ”this planet has .  ..  influenced many perplexities and labours on us in our investigation of its wanderings".

This is apparently originally from this source :

Baum, Richard and Sheehan, William.In Search of Planet Vulcan:  The Ghost in Newton’s Clockwork Universe. Plenum Trade, New York. 1997.

The gravitational perturbation of the planets, particularly by Jupiter, and the concept of barycenter for even ideal elliptical orbits required the work of Newton (following Kepler's concept of elliptical orbits) to explain.  Up until Newton there was no solid theory for the motion at all and people were "just" (poor choice of words) trying to find the simplest model.  Copernicus provided a simple model, but also a less accurate one.


At what cosmological redshift $z$, does the recession speed equal the speed of light? How is it calculated? At what cosmological redshift $z$, does the recession speed equal the speed of light?

What equations are used to calculate this number (since at large redshifts, $z=v/c$ won't apply)?

[The interested reader may refer to answers on Physics SE.]

From Friedmann Equation, distance as a function of redshift is:

$$d(z)=\frac{c}{H_0}\int_0^z \frac{dx}{\sqrt{\Omega_{R_0}(1+x)^4+\Omega_{M_0}(1+x)^3+\Omega_{K_0}(1+x)^2+\Omega_{\Lambda_0}}}$$

The Hubble-Lemaître Law:

$$v=H_0 \cdot d$$

We want $\boxed{v=c}$ The distance that fulfils this condition is known as Hubble Distance, (or Hubble Radius, or Hubble Length):

$$d_H=\frac{c}{H_0}$$

Combining both, we obtain the condition:

$$\int_0^z \frac{dx}{\sqrt{\Omega_{R_0}(1+x)^4+\Omega_{M_0}(1+x)^3+\Omega_{K_0}(1+x)^2+\Omega_{\Lambda_0}}}=1$$

For $\Omega_{R_0}\approx 0 \quad \Omega_{K_0}\approx 0 \quad \Omega_{M_0}\approx 0.31 \quad \Omega_{\Lambda_0}\approx 0.69$

The condition is:

$$\int_0^z \frac{dx}{\sqrt{0.31(1+x)^3+0.69}}=1$$

Searching by trial and error, we find that the value of redshift that fulfils the condition is: $$z=1.474 \approx 1.5$$

I hope I am not breaking any rules by repeating here the solution I wrote on Physics StackExchange
I did it because the creator of the thread asked me to do it, there.

Best regards.


What would a moon with a six-month orbital period look like from the earth? This is a kind of follow-up question to "What orbital period would produce one New Moon (and one Full Moon) each year?"
Given the six-month orbital period that is needed to produce one New Moon and one Full Moon each year, what would such a satellite look like as it made its progress through the sky (assuming we are in a system that is as close to our current situation as possible)?
In other words, I am wondering things like:

How long would it take the moon to cross the sky?
What kind of moonrise and moonset times would we expect, and how would they vary with the seasons (if at all)?
What would be the effect on tides?
What other observable effects (observable by mere mortals going about their business, that is, rather than astronomers or astrophysicists) would result from a moon with a six-month orbital period?

Grateful for any insights here.

A moon with a sixth month period would be about 1.3 million km from Earth. That puts it close to the edge of the Earth's Hill sphere, and probably isn't stable in the long term.  So the biggest thing you might notice is "goodbye moon" (and having the moon in an Earth crossing orbit wouldn't be pleasant in the longer term, a collision would melt the crust)
Your questions:

One day.  The moon's motion is mostly due to the rotation of the Earth.
The moon would be very distant and small but moonrise would happen just as it does now.
Tides would be much reduced.
The cycle from new moon to full moon would take much longer. Nights at full moon would be much darker than they are now. The moon is less likely to be tidally locked, so might show different faces at different times.


How is the total mass of all the Solar System's small asteroids and meteoroids estimated? This diagram from wikipedia (shown below) graphs asteroid Diameter against Number of asteroids, for the Solar System.  (Graphic is courtesy of the Marco Colombo, DensityDesign Research Lab).

Presumably the Diameter data for the larger asteroids is strongly supported by observations.  But for progressively smaller bodies direct observational data is presumably less available.

The pattern shows a strong, but not perfect, linear inverse correlation between log(Diameter ) and log(Number).  This seems quite reasonable for large bodies. It also sounds plausible that such a pattern continues into the realm of small asteroids and meteoroids.

But what is the physical justification for extending the log:log pattern into the realm of small bodies which are too small to see from Earth?  Presumably the current estimate of the total mass of the Astroid Belt (4% of the mass of the Moon) is consistent with this distribution model.  But can we confidently rule out the possibility that the linear log:log pattern breaks down for small objects allowing, for example, a much larger number (and mass) of small bodies to be present in the Asteroid Belt, sufficient say for the total combined mass to equal 100% of our Moon's mass?

Your question body is different from your question title and it seems you really want to ask what you did in the question body so I'll address that.

Short Answer: The simple power law which applies for larger asteroids and comets actually doesn't extend that well to smaller bodies and shouldn't be trusted too much in that range.

Long Answer:

You're right to be suspicious of using a simple power law to relate asteroid size to quantity at small sizes. Especially because there are known deviations from that power law at certain radii. As wikipedia notes:

  The number of asteroids decreases markedly with size. Although this generally follows a power law, there are 'bumps' at 5 km and 100 km, where more asteroids than expected from a logarithmic distribution are found.

Who's to say there aren't other bumps or deviations at radii that we can't observationally confirm. Our confidence in any power law's applicability for small asteroids and comets basically comes from modelling.

This power law you're asking about is essentially something referred to as an "Initial Mass Function" (IMF). A lot of work has been done to define an IMF for stars, relating the mass of stars to the number of stars at that mass. This stellar IMF follows a power law more or less, but strongly deviates for very low masses. At the surface, it would be reasonable to assume an IMF for asteroids and comets likewise deviates.

The way you answer this, is through models. See for example Cuzzi, Hogan, and Bottke (2010) (that link is to the formal journal paper, but for some reason some plots are obscured so you can also look at their "unofficial" arxiv version). In this paper, the authors attempt to construct a representative asteroid/comet population by modeling their formation via accretion of protoplanetary dust grains into asteroids, comets, KBO, and planetesimals. They make certain assumptions about the initial gas disk around the Sun and bake in physics of how these dust grains formed into larger conglomerates. Their end goal is to produce an IMF that identifies how many asteroids/comets exist at a given size. They back up a lot of their modelling with data from observations so they stick to reality as much as possible.

I'll let you peruse their results since they actually produce many different IMFs using a variety of different assumptions and starting conditions. However, I think one can summarize their results in a way which addresses your main question. As they state it:

  Asteroids were born big

What they mean is that asteroids (and comets), rather than forming by successive growth via accretion of many small particles, seem to just clump together into larger bodies, completely bypassing the formation of many smaller meter-sized asteroids and comets (that's not to say there aren't meter-sized objects). You can actually see this in their various plots (fig. 4 for example). Much like the stellar IMF, the power law drastically changes once you get to small sizes and the power law which applied for larger sizes is no longer applicable. At which size the general power law no longer becomes applicable depends upon certain tunable parameters. Cuzzi et al. shows many different results with their tunable parameters set to many different possible values.


What are these dark spots on Mars' north pole? I was exploring Mars in Google maps and I found these dark spots, like tree shadows on the north pole. I don't have any idea what they might be.

View this in Google maps.

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


How can we tell the age of a rogue planet? Is it possible to find out how old is a planet, especially a rogue planet? I know that people measured the radioactive decays to determine Earth's age with some amazing accuracy, but what about interstellar planets be, they gas or terrestrial?

At the moment there is basically only one way. That is to associate the planetary-sized object with a cluster of stars or moving group of stars of known age.

That's basically it. If the planetary-sized object really can't be associated with another object, then only limits can be placed on its age by comparing it's luminosity to theoretical planet cooling models. But to use these models you need to know the mass!

So at the moment, the only "free-floating" planetary mass objects known (they could all be low-mass brown dwarfs actually) are those for which an age is estimated by association and therefore a (planetary) mass has been deduced from cooling models.


Why hasn't the CMB already receded away faster than the speed of light? If there are supposedly galaxies, or whole other parts of the universe, that we can't ever see due to them receding away from us faster than the speed of light, then why isn't or hasn't this occurred with the cosmic microwave background?

Let me try this from a building blocks point of view.

Imagine the Universe is basically uniform, that is, it's pretty much the same everywhere at any given time, or, at any specific age.  We think this is essentially true.

Next, imagine that the Universe is ridiculously, impossibly large, much larger than we can see, because when we look at things very far away, we also look back in time and we can only see as far as the universe is old.   We can't see most of the Universe because most of it is much too far away.   Given the observed expansion, we'll never see most of the universe.  The light from objects too distant will never reach us.   We assume there are countless galaxies out there, similar to the galaxies we can see, but we'll never see them.

Lets skip the very early parts when the observable universe was the size of a baseball, or the size of the solar system, because that happened long before the cosmic background radiation and it's not relevant to your question, and we can begin in the, lets say, around 370,000 years of age, when the Universe was much less spread out than it is today.

At that age, about 370,000 years after the big bang, the Universe was still hot everywhere, kind of like fire thinly spread out everywhere.   There were no galaxies or stars as they hadn't coalesced yet.  Just spread out clouds of fiery gas still spread out everywhere, but beginning to clump together under gravity.  Basically just hydrogen and helium nuclei and free electrons, hot enough to be in a plasma state, not too different than flame, and flame is opaque.   It's not fully opaque like a solid object, but flame diffuses light, see more details here.

so, 370,000 years of age, the entire universe was kind of a glowing fog.  You couldn't see very far in any direction.  You could maybe see an AU or a few hundred AU before the opaque plasma blocked your view if you happened to be around at that time, but basically everywhere you looked, you'd just see the heat-glow of the hot universe.

This all changed around the time when the universe was 379,000 years old, during a time called recombination.   The universe expanded enough and became cold enough that most of the free electrons rejoined with the atomic nuclei and the plasma became atoms.   As a byproduct of this, the universe went from a glowing fog to transparent for the first time.   This didn't happen all at once, it was more of a transition into transparency, but for the purposes of this question we can think of this as happening at 379,000 years.

But, because when we look in any direction, the farther away things are the older they are, so at the edge of this transparency, there's always going to be the cosmic background radiation, that was true shortly after the 379,000 year recombination and it's been true ever since.   What has changed, is the expansion of the universe keeps stretching the wavelength of the CMB and it keeps moving further away as we move forward in time.

What a person would actually see, when the Universe was 380,000 years old, is the cosmic background radiation in a sphere around them, about 1000 light years distant and in visible light, because the expansion wouldn't have affected it much over just 1000 years.   The CMB is kind of like a sphere of time in addition to a sphere of background radiation.   It's the furthest thing we can ever see in the EM spectrum, about 379,000 years younger than the age of the universe.

The thing to keep in mind about this sphere is the universe is much bigger than we can see.   The glow was everywhere, but the glow we see at any given time is the age of the universe minus 379,000 years old, it's a kind of observable universe bubble, past which we can't see.

The first stars didn't show up until about 180 million years later and the first galaxies, another 20 million years after that, so it was a long time between recombination and the first stars.  If a person was around 180 million years after the big bang, they could observe the first stars forming, but only stars that were close to them.  Because looking further out looks back in time, so the first stars anyone could observe would be in their neighborhood, but over time, we saw stars form further and further away, but never as far as the CMB.

But what we can see doesn't change what the largely uniform universe did.  Stars and Galaxies began to form everywhere, both inside and outside our visible edge of the CMB.   Every star and every galaxy is younger than the CMB, but we can infer, even though we'll never see them, that a great many stars and galaxies formed outside or our visible edge of the universe because the universe is much bigger than we can see.

That's kind of a step by step overview, perhaps more wordy than it needs to be.   Let me know if that helps.


Is the material that could have formed a "Super-earth" divided among the four terrestrial planets I know "super-earths" are quite common in other planetary systems, but there are none in the Solar System. Is it due to the material that could have created a single large super-earth being used to create four smaller terrestrial planets in the Solar System?

I think the answer to your question lies with Jupiter, be it directly or indirectly, the gas giant is now believed to have had a large influence on the way the inner planets formed.

Many planetary scientists believe that the reason we only have 4 small inner planets today and not any "super-Earth" sized ones is because of Jupiter, and how it interacted with the circumstellar disk and protoplanets at the time of the solar system formation. There may well have been one or two super-Earths that formed in the early solar system, but models suggest that they would have been sent crashing into the Sun by the inward advances of Jupiter.

I found this blog article which explains it in more detail. And there is also another MNRAS article that goes into more detail about the physics and orbital mechanics behind what might have happened in the early solar system.


What implications do younger Earth and Moon have on Late Heavy Bombardment genesis hypotheses? The Late Heavy Bombardment (LHB) or the Lunar Cataclysm is when the inner solar system, including the Earth and the Moon, underwent multiple and sustained heavy impacts early in their history (around 3.7-4 billion years ago). This event formed many of the major impact scars we see on the Moon, (presumably as well as much of the impacts observed on Mars and Mercury). Two main theories are based on perturbations in either the Kuiper Belt, or in a primordial larger asteroid belt.

What implications would a younger Earth and Moon, discussed in "Moon and Earth may be younger than originally thought" (Lars Borg et al., 2011) have on hypotheses on what caused the LHB?

From the article: "In fact, its age may be 4.36 billion years old." This is still older than the estimated age of the Late Heavy Bombardment, so it wouldn't affect that hypothesis much. In fact, it may actually help the hypothesis, as it means the Moon would have more of its interior molten (though probably not by much). It's suggested that the magma from underneath the crust filled the lunar mare after the giant impacts through fissures in the crust. (Wikipedia)


Can a binary star optically "orbit" a planet? I was watching a science fiction anime from the mid 70's. In it, there was a star orbiting a planet. Since this seemed impossible, and I had never heard of it, I looked on the internet to see if this could be possible in a very unlikely situation or in theory.

On a similar question on Quora, after many no, no, no, no answers from astronomers/astrophysicists, some of them with very detailed models and theories with different scenarios and equations, one answer read,

  Absolutely.

  If there were two stars of identical mass orbiting each other in equilibrium, then you could throw a planet of any size at the barycenter of the two stars. Then optically you'd have not just one, but two stars orbiting a planet. The planet would be equidistant from the two stars at all moments in time, and would experience identical gravitational pull in two diametrically opposite directions and hence never deviate from the barycenter. This system would not be in a stable equilibrium though, and a passing celestial object of any mass could disrupt this system. Hence, this is probably something that is unlikely to be observed.

Is this possible?

Planets don't orbit stars. Stars don't orbit planets.

Whenever there are two bodies bound by gravity, they are both orbiting their common center of mass. For example, both the Earth and the Moon orbit their common center of mass - but that's pretty close to the center of the Earth actually, so it seems like the Moon orbits the Earth.

For a star to seem to orbit a planet, that would mean the planet is much heavier than the star. As far as we know, that's impossible. Their common center of mass would be much closer to the star, so it would seem like the planet is orbiting the star, as usual.

If you have two stars orbiting each other (actually, orbiting their common center of mass) very closely, then you could have a planet circling around both. If the planet is circling around too closely, its orbit would not be stable.

But if the planet's orbit is far enough from the pair of stars, the orbit could be stable a very long time. The minimum distance it at least 2x ... 4x the distance between stars, ideally much bigger. We have discovered such planets. Kepler-47c is a gas giant in the circumbinary habitable zone of the Kepler-47 system.

This wiki page has more details:

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

EDIT: If you have a closely-bound binary star, with a planet right in between them, in the barycenter of the two stars, that's not a stable system. Any slight perturbation would pull the planet out of there, and then the net force would be pulling it further out. It would eventually fall into one of the stars. It's not self-stabilizing.

I can't even imagine any conceivable mechanism that would put the planet in there to begin with.


How to do logarithmic wavelength binning (in Python) in order to find redshift? In Tonry and Davis (1979), p.1513, they formulated a cross-correlation method for extracting velocity redshifts:

Theory of Correlation Analysis
a) Introduction
Let $g(n)$ be the spectrum of a galaxy whose redshift and velocity dispersion are to be found and let $t(n)$ be a template spectrum of zero redshift and instrumentally broadened stellar-line profiles. These spectra are discretely sampled into $N$ bins, labeeled by bin number $n$; the relationship between wavelength and bin number is
$$n = A \ln \lambda + B$$
Because the spectra are binned linearly with $\ln \lambda$ a velocity redshift is a uniform linear shift. The spectra are assumed periodic with period $N$ [...]

They bin the linear spectra into logarithmic wavelength scale. Similar work was done by Baldry et. al (2014) and I have inserted relevant excerpt from TD79. I understand that log wavelength binning is a pre-requisite of cross-correlation procedure.
I have been trying to follow this approach and trying to rebin the linear spectra (3000 to 9000 Angstroms) in equal intervals of log wavelength. Could someone please give me a guidance about how to do this logarithmic re-binning  in Python?
Here the text as picture:

I think you could just take the natural log of your wavelength scale, then re-bin this onto a new log of wavelength scale generate using np.linspace().
Your new scale would be generated using the maximum and minimum values of your raw log of wavelength scale then it is up to you to define the amount of bin in the new homogenous log scale.
To re-bin the raw log of wavelength scale onto the homogenous log of wavelength scale you would assign fractions of the old flux values into the new bins where the fractions are determined by the fraction of the old bin (that the flux value was in) that overlaps the new bin.
In the end you will need this to be on kilometres per second or a unit of velocity. I expect you would simply multiply the scale by the speed of light although I have experienced slight loss of precision when I do this.


Globular Cluster diameter I am doing a homework on globular clusters and could not find trustworthy data on the average diameter of a globular cluster. Most clusters I included had a diameter of 120-150 ly.
But is that the average diameter of such globular clusters?

Globular clusters vary in size quite a bit. They contain from a few hundred thousand to over a million stars. Probably best way to determine size is to measure its extent across the sky by telescope and then calculate that distance in km. One could also model the size by assuming the number of stars, the average size of the stars, and the spacing based on this size.  One could build a globular cluster of 300,000 stars or so and determine the approximate size.

You ask if 120-150 ly is the average diameter of such globular clusters. I'm confused by exactly what you mean but you have sampled a small range of sizes considering globular clusters can be well over 300 ly in diameter.

From Wikipedia, and based on comment from @AtmosphericPrisonEscape, you have "Almost all globular clusters have a half-light radius of less than 10 pc, although there are well-established globular clusters with very large radii (i.e. NGC 2419 (Rh = 18 pc) and Palomar 14 (Rh = 25 pc)).[64]"


What's the theoretical maximum density of a galaxy? In watching this Hubble video on Omega Centari it got me thinking about how dense a galaxy can get. Do we know how dense a galaxy can get?

Are there any good resources on reading more about very dense galaxies?

As a general rule, galaxies aren't very dense at all, though precise numbers are impossible, rough estimates can be made.

Mass of the Milky-way:  with it's dark matter halo, about 6 x 10^42 kg

Volume of the Milky-way (per Wikipedia) about 3.3 × 10^61 cubic meters, or, 3.3 x 10^52 cubic km.

That gives us a rough density of the milky way of about 1 kg for every 5 billion cubic km.   That's obviously not very dense.   There are probably denser galaxies out there and larger galaxies, but the overall density is going to be very low, even for the densest galaxies.   Space is, generally speaking, mostly empty.

Even in regions that are so called "densely packed with stars", if you measure the density in the traditional way, mass to volume, you end up with very low density, 1 kg per thousands if not millions of cubic km.   On earth, that density of air would essentially be a vacuum.


How much iron does a star fuse before going supernova? I understand stars fuse atoms into heavier and heavier elements. When the star starts to fuse iron, the amount of energy released is less than the amount of energy it to takes to fuse the atoms. This creates an energy deficiency and the core is unable to maintain its outward pressure against gravity's inward collapse. The star collapses, rebounds against electron degeneracy pressure (I believe?) and goes BOOM. But how much iron is actually fused? And, for how long is the star fusing iron before the collapse?

Also, as a little bonus, I read somewhere recently that the presence of nickel is possible prior to going supernova. I can't find where though. Was that person wrong?

Wikipedia says the iron-nickel core is inert, and doesn't mention any endothermic iron fusion reactions, just the core collapse, which converts the core to neutrons (and a not insignificant proportion of protons & electrons), and releases a lot of neutrinos. The nickel is radioactive, decaying to cobalt & iron.

The silicon burning phase leading to nickel only lasts about 5 days. The inert core collapses when it exceeds the Chandrasekhar limit of about 1.2 M$\odot$.


How was the Sun formed? Refer to the below minutephysics video, in which they say that in order to reach the Sun from earth, we must first stop revolving around the Sun ie make our tangential velocity zero, and then head straight for the sun, otherwise it is nearly impossible to reach the sun because even a very huge push towards the sun will either send us in an elliptical or hyperbolic orbit:-

https://www.youtube.com/watch?v=LHvR1fRTW8g

If even a little bit of orbital speed tangentially, prevents something from falling into the sun, then I wonder how does the sun formed itself at the beginning? Because, most collapsing material probably would be having at least some tangential velocity with respect to the center.

Where you're wrong:

  even little bit of orbital speed tangentially prevents something from falling into the sun

You need to have huge tangential velocity to escape the sun attraction! The closer you are, the higher the velocity must be for celestial body to avoid falling on the sun.

  While Venus is orbiting the Sun at 35.02 Km/s, the Earth is also orbiting the Sun in the same direction at 29.78 Km/s. [Reference]

It's not easy to understand, because there is no clear analogy of how stars form at our level on earth.

Interstellar gas clouds eventually collapse to form stars are very large, very massive, very cold, and very tenuous. The star is at the beginning a little portion of the cloud that is more dense than the rest - it results of local and large scale instabilities (some call it an element of chaos).

Little by little, the density at the center of the cloud increase - at the same time, it has to overcome opposite forces arising from more probable friction. If the gas is cool enough compared to the density, at some point you start having more and more matter, which results in more and more gravitational attraction for surrounding gas.


Where is the Moon for the last five or six days? The date today is March 18th 2021 Thursday it's 6:00 p.m. I'm in the area of Sarasota-Bradenton Florida. I haven't seen the moon and it's been clear skies for roughly the last five or six nights... I see all the consolation of the Stars where they're supposed to be but I cannot find the Moon. It makes me nervous and I'd like to know why?

Two things: You either have a new moon or waxing crescent moon that is difficult to see or even invisible to the eye. Or (this is more likely), you viewed the sky before moonrise.
Try viewing the night sky in a few days (it will be clearly visible), or view the sky later in the night.


Would humans hear gravity waves from a binary BH fusion nearby? The frequency of the gravity waves detected by aLIGO recently were in the audible sound spectra. Actually spanning over half of the keys of a piano, I've been told, during 2/10 or so seconds.

Would a human ear sense a gravity wave if powerful enough? The waves going through empty space from the event horizons, through the walls of a spacecraft and through the ears of astronauts. Would one actually hear a "zip" sound because space makes waves inside our ears?

As your ear is extremely small compared to the wave length of the gravity waves, remember they travel at the speed of light not the speed of sound. The structures in the ear would not respond, also the ear would change in size (by a  ridiculously small amount) itself, thus nullifying the effect anyway. LIGO works by comparing the light travel time of two laser beams at precise right angles, your ear is not directional like that.


Trouble with plotting a graph for the Planck function Here is my problem statement:

For a temperature of $7.3 \times 10^5 \ \text{K}$, make a graph of the Plank function (Eq. 3.24), plotting $\log_{10}{\nu} B_\nu(T)$ vs $\log_{10} \nu$ for $\log_{10} \nu$ between 15.5 and 17.5. How does the behaviour of your graph of a blackbody compare with that of Fig. 28.14 for the continuous spectrum of the quasar 3C 273?

Here is the equation 3.24

\begin{equation}
    B_\nu (T) = \frac{2h \nu^3/c^2}{e^{h\nu/kT}-1}.
\end{equation}

I have put together an excel spreadsheet and created a graph of $\log_{10}({\nu} B_\nu(T))$ vs $\log_{10} (\nu)$, however, it is just a straight line (which is what I would have expected although the graph in the book is significantly different). I am wondering if I am misunderstanding the problem statement. What I did to get my data points for $\log_{10}{\nu} B_\nu(T)$ is I first computed $\nu$ values from the given range of $\log_{10} \nu$ values, then I computed B(T) values with the above-given equation. Then I substituted the product into logarithm : $\log_{10}({\nu} B_\nu(T))$, thus obtaining my $y = f(x)$ values. and then I plotted it against  $x = \log_{10} \nu$. Am I doing things wrong or is this fine?

Here's what I got for $\nu B_{\nu}(T)$ using Python, which is easy enough to read that maybe you can check against your Excel calculation. I also plotted just $B_{\nu}(T)$ for comparison.

Neither looks like a straight line and they shouldn't. If you print out $\log_{10}(k_B T / c)$ you get about 16.2 which is right in the middle of your range. They've specified a range of frequencies centered roughly on the peak of the Planck distribution.

def Bnu(nu, T):
    top = 2 * h * nu**3 / c**2
    bot = np.exp(h * nu / (kB * T)) - 1.
    return top/bot

import numpy as np
import matplotlib.pyplot as plt

h  = 6.6261E-34 # m^2 kg / s
c  = 2.9979E+08 # m / s
kB = 1.3806E-23 # m^2 kg / s^2 K

T   = 7.3E+05  # K

print "log10(kB T / h): ", np.log10(kB * T / h)

nu  = np.logspace(15.5, 17.5, 201) # Hz

lam = c / nu

print nu.min(), nu.max(), 'Hz'
print 1E+09*lam.min(), 1E+09*lam.max(), 'nm'

B = Bnu(nu, T)

plt.figure()

plt.subplot(1, 2, 1)
plt.title('B(nu, T)', fontsize=16)
plt.plot(nu, B)
plt.yscale('log')
plt.xscale('log')

plt.subplot(1, 2, 2)
plt.title('nu x B(nu, T)', fontsize=16)
plt.plot(nu, nu * B)
plt.yscale('log')
plt.xscale('log')

plt.show()


Where can I watch the announcement from LIGO about gravitational waves? Today the LIGO consortium are having a press conference to update us on their search for gravitational waves. How should I watch the announcement?

The event will be live streamed on youtube here: https://www.youtube.com/user/VideosatNSF/live

The will also be tweeting: https://twitter.com/ligo

A video of the conference will also be available afterwards if you miss it. It starts at 10:30 EST, which is 3:30 GMT.


Must time pass more slowly, relative to our inertial reference frame, inside galaxies that are currently located at half way to the Hubble Horizon? Trying to understand simultaneity of events and time-dilation on a Universe scale, I would like to know must time pass more slowly, right now, relative to our current inertial reference frame, inside galaxies that are currently located at say "half way" to the Hubble Horizon because such galaxies are moving at around .5c? In this case no acceleration was experienced to generate the speed difference, just space expansion.
I can't see how it could not be true, (I think the Lorentz transformation still hold) but I think if it is true it leads to problems.

The point of the unification of space and time in relativity is that there's no sense in asking what's happening "right now" at a different position. It makes as much sense as asking what's happening "right $y$" at a different $x$ in Euclidean geometry. If you fix a Cartesian coordinate system then "right $y$" is mathematically well defined, but being mathematically well defined doesn't make it any more sensible to talk about. The laws of nature don't have any notion of distant simultaneity and don't care about your coordinates.
There are situations in which there's a coordinate-independent, geometric difference between two curves (worldlines) in spacetime that justifies saying that one is shorter (less elapsed proper time) than the other in some absolute sense. One such case is when the curves meet at two spacetime points and you're only interested in the length between those points (as in the twin paradox/effect). Another case is gravitational time dilation, where the curves are analogous to circles of constant latitude on the earth.
Worldlines moving with the Hubble flow are more like lines of constant longitude on the earth. The distance between them varies (as a function of latitude), and as a result if you draw rhumb lines from one of them to another (analogous to lightlike worldlines), they'll arrive at a different separation than they departed at (analogous to red/blueshift). But the situation is symmetric, and it makes no sense to say that one longitude line is longer than another.


Using Tully Fisher to measure Distance Problem Can anyone explain by looking at the solution in the pic, how did the cos (i) came about?
I guess they are getting this from the major-minor axis info in the question, but I am not sure about the derivation.

Also v=300/sin(i), what is this formula?

Many thanks :)

With the Tully Fisher relation, the inclination angle is determined from the major-minor axis ratio where $sin(i) = \sqrt{1-q^2} $ and q is the inverse of your ratio.

You need to apply the sine of the angle to the rotational velocity because the measure is based on our line of sight observation which must be converted to the true 3D velocity.


Travelling, flashing Star? I have recently gotten into observing the night skies and learning about the stars/ star systems that are visible to me using an app on my phone. For the last month or so Venus and Mars appear in the early evening followed by some of the regular constellations (Orion, the two Canis'....) and later in the night skies Jupiter followed closely by the moon with Saturn peering out before daybreak. Yesterday night at around 20:15hrs I was looking up at the clear skies as I happened to be outside. I took a peek at the bright stars Sirius, Canopus and Rigel before concentrating on the Pleiades. As I was looking at that, one of the stars seemed to disappear in front my eyes. I thought it an illusion and carried on looking and about 6-7secs later a star appeared to the right of where I thought I saw one disappear initially. The light stayed for about 2secs and off it went again. I felt like someone was flashing a torch or even space station personnel having a call and response (baffled). One thing I was sure of was that it wasn't a plane as I am an Aeronautics man by profession and am familiar with all aircraft strobes. Another 6-7secs later it reappeared even further to the right than the second time round. For the next 11mins or so it came on and off at those regular intervals and yet still traveling to the right until it was out of my sight. tried to record a video of it with my phone but it wouldn't pick it up nor could it the other brighter stars. I was almost convinced by the end that it was some celestial matter but for the flashing light at exact intervals. Is it one of satellites in orbit? If so, is it normal for it to be equipped with flashing light/ strobe?

It is highly likely that what you have seen is an Iridium communications satellite. Because of the material covered, it can reflect sun light and appear very bright and sometimes flashing.


Where can I find thousands of actual images of stars in the galaxy? I'm trying to write a program to differentiate between images of stars and nebulae. However, I'm not able to find a large enough dataset of actual images of stars that I can use to train my deep learning algorithm. Is there any database that has thousands of actual images of stars that I can use to train my deep learning algorithm? Thanks.
EDIT: I need individual images of stars so that my program can learn the identifying features of a star and be able to differentiate a star from another celestial object.

The online version of the digitised sky survey will give you as many images of the sky as you want. Be aware that these are based on photographic plates. Also be aware that every imaging system has its own peculiarities in terms of the shape of the point spread function (i.e. the shape of a star in the image) and features such as diffraction spikes.
https://archive.eso.org/dss/dss


Would two Earth-like planets on opposite sides of a Sol-like star at about 1 AU be a stable configuration? Would two Earth-like planets on opposite sides of a Sol-like star at about 1 AU be a relatively stable configuration? I understand that there are tadpole-like orbits and the two planets may not stay exactly opposite each other, but oscillate around those position.

But given perturbations from other planets, say a Jupiter-like one at say 5 AU, might they eventually crash into each other, or would one end up moving to a higher orbit and the other to a lower orbit?

It would not be a stable configuration, but it wouldn't be too far from the stability.

The are 5 points where a third body could have a semi-stable orbit in a 2-body system:

The Earth is the blue point. The "anti-Earth" on the other side would be in the third Lagrange point ($\rm{L}_3$).

However, this is not a stable orbit. Any small deviation, including the effect of possible fourth bodies, would change the system so, that it results yet more deviation. Only $\rm{L}_4$ and $\rm{L}_5$ are stable, and even this has some requirements (the third body can be at most $\approx$ 3.5% of the second).

Corresponding this, the space on the other side of our Sun was examined by some space probes already, and there is nothing.


Why do we only see the clouds directly infront of the moon glow from moon light? Surly the moon is not projecting a direct beam of light which only lights up the small area of clouds around the moon. What phenomenon is happening that only allows us to see that center focus glow and not all the clouds glowing?

This is an experiment that you can do at home:

Put a paper fixed in front of a light bulb
Move your head around and see how the bright spot on the paper accompany your moves.

But, why is that? How is it that not all the paper is lighted the same?

When light travels through a translucent material some of it gets reflected back, some bounces inside of it and some goes through. As always the shortest path is the easiest, so the light coming straight "bulb-paper-eye" gets through easily (brightest spot) and as you go further from the center you are not seeing direct light anymore, but diffused light, so it gets dimmer and dimmer because to get to your eye it has to bounce in the right direction. That doesn't imply that the rest of the paper is not getting light, it is just not bouncing in your direction and the same happens with the clouds.


What color is a red dwarf star? I read recently that a red dwarf star might actually look white.   I realize there's a wide range in red dwarf surface temperature, 2,500 to 3,500 degrees K, compared to 5,778 for our Sun.   Metal at 2,500 degrees K is nearly white hot with just traces of yellow.   Is that true for 93% hydrogen 7% helium as well?   Would the color of red dwarf stars be closer to yellow/white rather than red based on our vision, if we were to be close enough to get a good look?

The spectrum is of course heavily weighted towards red wavelengths, but your answer is mainly to do with the physiology of the eye, not astrophysics.
This site has folded stellar spectra through the appropriate response and arrived at "orange-ish" for M-dwarfs.
A "white-hot" piece of metal is not really white. Sunlight appears white - a blackbody spectrum at around 5800 K. The composition doesn't play a great role, though the spectrum of heated pure metal is probably closer to a blackbody than is an M-dwarf. I suspect that when people talk about "white hot metal", what is going on is the eye being overwhelmed by a large flux of photons. The intrinsic spectrum is clearly not that of a "white" object (such as sunlight). However, it could be that the emissivity is a bit lower in the red part of the visible spectrum so that when at least some green/blue light is produced (recall that the Wien tail is an exponential fall on the short wavelength side) that the eye's color sensory cells may be triggered sufficiently to fool it that the light is white. Either way that's physiology, not astrophysics.
Postscript: Most people can make out the orange-ish colour of M-type stars like Betelgeuse and Antares. These are (super)giants, but their spectra are not so different from M-dwarfs.


Sky photography through compositing images in software? I've read about and seen some rather sophisticated rigs for night sky photography that allow for ultra-long-exposure photos by counteracting the rotation of the sky by rotating the camera with it. The problem is mechanics for such rigs is rather complex, expensive and not entirely within reach of an amateur.

I was considering a different approach:

Set the camera on a firm tripod or whatever, leaving it motionlessly.
Use intervalometer in the camera firmware to take multiple photos of moderately long exposure, short enough that the stars don't blur into lines.
Save each as RAW so that the trace amounts of light captured in each of them don't get lost.
Either using more bright stars as markers, or even calculating the misalignment of the photos basing on movement of the sky and lens geometry composite the images in software, to obtain good images of the night sky.

Obviously I wouldn't get very far with high zoom images, as the field of vision would shift away long before I'd get enough light in for a good photo, but obtaining a photo of a fair section of the sky from a relatively wide angle lens should be doable.

Is such approach used or even viable? Is there such software "out there" already or would I need to write my own from scratch?

What you are talking about is making a 'mosaic'. This technique is commonly used in astronomy to stitch together many images that are off-set from each other of some (apparently) large astronomical object (e.g., a star cluster or a nearby galaxy). In addition to covering a large field of view, making a mosaic generally also improves signal to noise where the images overlap.

One way to realign the images is to uses cross-correlation to find the optimal shift for consecutive images in order to align them. This can be done in common data analysis languages such as python (free), matlab or IDL (not free).

I have not used black-box software to do this, but here is a list of programs that might help: http://astronomyonline.org/AstronomySoftware.asp#CCD_Control_and_Imaging_Software:_

Hopefully some astrophotographers here can recommend some software.


Dealing with damp and dew whilst stargazing at night or in the early morning Sometimes when using my telescope late at night or early in the morning, especially in the more humid of seasons, I notice i sometimes get dew buildup on the lenses.

How can I minimize damp buildup whilst using my telescope?

Simply dabbing away the liquid with a dry cloth doesn't seem to do the trick as I am still left with some water tracks and residue that I just can't soak up, this then means that all of the light is refracted providing a very fuzzy and blurry image.

What preventative measures can i take to stop damp build up in the first place?

Condensation will build up because the temperature of your optics has fallen below the dew point.

You simply have to keep the temperature your telescope above the dew point, which can be quite difficult on humid nights. Dew shields can help, as can dew heaters (which literally just gently heat your telescope and/or eyepiece).

  Hypothetical: "I'm looking at the graph of dew point vs air temperature on the link above. I don't understand how my optics can get colder than the air temperature!

At night, our source of heat (the Sun) has disappeared. The unlit side of the Earth (and everything on it) begins radiating its latent heat away into space (which has a temperature of ~3 Kelvin). Your telescope is trying to get into thermal equilibrium with outer space.


how to find the approximate [stellar radius]? (distance from center to most distant star) of a circular flat disk galaxy based on its stellar mass? I'm doing a high school programming project on drake's equation (simple galactic model) and wondered if there was a (simplified) way to find the radius (measured by the distance between the center and furthest star) of a galaxy based on its stellar mass. I found a few websites but I don't understand many of the terms and symbols in the equations shown. would love a simplified equation if possible.
In the project, I am modeling a 2d circular disk galaxy with no rotation.
I have a Y10 level understanding of maths.

What you are looking for is called the "galactic size-mass relation" for galaxies dominated by disks.
Theres an interesting research paper by Rebeca Lange and others where this relation (equation 3 in the paper) has the form
$R = \gamma M^{\alpha}(1+M/M_0)^{\beta-\alpha}$
Where $R$ is the radius of the galaxy in kiloparsecs ($1$ kpc $= 3260$ light year), $M$ is the mass of the galaxy measured in solar masses ($1$ solar mass $= 1.988\cdot 10^{30}\; kg$), and $\gamma$, $\alpha$, $\beta$ and $M_0$ are parameters that have been tweaked to adjust the relation to the real data.
In their paper they propose values for $\gamma$, $\alpha$, $\beta$ and $M_0$ for different kinds of measurements and galaxies. For the late-type galaxies (which are the ones that you want to study), these parameters are shown in table 2 with values that have a certain range.

$\alpha$ between $0.13$ and $0.17$
$\beta$ between $0.91$ and $1.00$
$\gamma$ between $0.08$ and $0.18$
$M_0$ between $19\cdot 10^{10}$ and $48\cdot 10^{10}$ solar masses

To make an approximate guess let's take the average values they present for each parameter. Then you will have the relation:
$R = 0.12\cdot M^{0.15}\cdot(1+3.8\cdot 10^{-12}M)^{0.81}$
Again with $R$ in $kpc$ and $M$ in solar masses.
With everything into account we are ready to plug-in some numbers. For example, the Milky Way galaxy is known to have a mass of $M = 10^{12}$ solar masses. This means that according to our model the Milky Way disk has a radius of $R = 26.97 \; kpc$. Well it turns out that we think the disk to extend between $170,000$ and $200,000$ light years in diameter, which translates to $26 \; kpc$ to $31\; kpc$, so we proved that the model works at least for our galaxy, even if our prediction is in the lower bound of the range.

Now, the devil is in the details; $R$ is not actually the radius of the disk as you might think. The radius is an ambiguous term here since disk galaxies don't have a sharp edge where stars stop to exist. The $R$ in the formula is actually the so-called "effective radius", wich is in fact just the radius which contains half of the light emmited from the galaxy (this is an unambiguous way to measure the size of the galaxy). So, when talking about the radius $R$ I would take this into consideration, and use maybe a $1.2R$ estimate (20% larger) as my "true" radius for the galaxy.
Another important thing to consider is that the mass of the milky way has been measured including the mass of the dark matter and the mass of the gass. If you only want stellar mass, then the calculation will yield a galaxy that might be smaller.
These two are unimportant points if you are making just a highschool project and they don'tchange drastically your models either way, but I have to warn you about a consideration here. A galaxy that has a mass of $x$ solar masses is not a galaxy that has $x$ stars (planetary systems). That would certainly happen if all the stars where like the Sun, but as it turns out the vast majority of stars are less massive than the Sun, thus an $x$ solar mass galaxy will have much more than $x$ stars. If you are going to play with the Drake equiation that might be important. Finally if you search for the Milky Way data you will see that we think it has a mass of $M = 10^{12}$ solar masses, as I said, and we think it has around $2\cdot 10^{11}$ stars, so what's happening here? Shouldn't a galaxy with mass $x$ solar masses have more than $x$ stars? The problem here is, as I said previously, that we are considering the mass of the entire galaxy including gas, dust and dark matter, but we are only concerned about stellar mass.

I'm very interested in your project. Would you mind to tell me more about it? I'm just curious.


Why is this shooting star wavy or squiggly toward the end of it's visible streak I was taking photos of some stars when I noticed this shooting star was wavy toward the end (start?) of it's trail. My cousin suggesting that it could be the CCD in my camera or as it spiralling as it enters the atmosphere. This is a heavily cropped image

It's shutter vibration. It begins at one end of the trail (which I'm guessing is the start of the exposure) and quickly damps out.


How far away would a supernucleus comet be visible? If an instrument like LSST quickly spots something as soon as it reaches magnitude 24½, how far away would a 60km first-time visitor from the Oort cloud be when it is spotted? I'm thinking that a huge iceball will make a visible coma sooner, but have no idea what actual numbers would be.

And, given that distance and a parabolic orbit, how much time would that give before it crosses Earth's orbit?

Even 100 meter NEOs could cause significant damage, and there are far
  more of them. LSST will be able to detect objects as faint as 24.5 in
  magnitude in a 30s visit, enabling it to detect 140m NEOs as far away
  as the Main Belt asteroids.
  [...] During its survey of the sky, LSST can find 90% of the PHAs over
  140 meters in diameter.

Earth is 1 AU from the Sun, so that would be 1.2 to 2.2 AU away from us:

  The Asteroid Belt is located in an area of space between the orbits of
  Mars and Jupiter. That places it between 2.2 and 3.2 astronomical
  units (AU) from the Sun. The belt is about 1 AU thick. The average
  distance between objects in the Asteroid Belt is quite large. If you
  could stand on an asteroid and look around, the next one would be too
  far away to see very well.

Timewise, it would be about 8 years away:

  The bottom right panel shows an example of a very different hazardous object - the 3 km large comet C/1996 B2 Hyakutake, which passed within 0.10 AU from Earth
  in 1996. The apparent magnitude vs. distance diagram indicates that the current
  generation of wide-angle imaging surveys, such as SDSS with V < 22,
  could have provided an early warning about 3 years before the closest
  approach (if they had sufficient cadence and sky coverage). With LSST,
  the warning time would be ∼8 years, with over 500 observations over
  that period.

Fortunately, your comet is 60 km, making the comet tail easy to spot:

  While the solid nucleus of comets is generally less than 50 km across,
  the coma may be larger than the Sun, and ion tails have been observed
  to extend 3.8 astronomical units (570 Gm; 350×106 mi).

This table of impact risks shows that the most likely (5.8e-02) impact is between 2095 and 2113.


How do we get radio signals of the big bang? From TV shows I have known that we still get some radio signals of the big bang and they are every where. My question is is the universe expanding so fast that even electromagnetic wave has failed to cross it after big bang?

Or is it possible that there is another dimension that we don't know of? Like the earth is nearly flat and in small portion it's surface can be called 2d. But it possible to go "straight" and come back at the same spot as it is in "3d space". So is it possible that the space is 4d and somehow light has crossed the cosmos several times?

The best estimates at the moment are that the Universe is "flat". By that we mean that a pair of parallel laser beams will travel as far as you like and they will always remain the same distance apart and parallel.

Ironically, the best evidence for this comes from detailed measurements of the cosmic microwave background by the WMAP and now Planck satellites.

The problem you are having is you (I think) are imagining that the big bang happened at a point in space 13.7 billion years ago, and that radiation has been travelling outwards from that point. This is incorrect. Every point in space that we see now was, about 13.7 billion years ago, part of the big bang. As the space in the Universe expanded, its contents cooled and due to the trapping of free electrons by positively charged protons to form hydrogen atoms, the light that was emitted by the hot gas everywhere in the Universe at a time about 400,000 years after the big-bang was able to travel unabsorbed and unscattered. This radiation has travelled (at the speed of light) in all directions.

Now, when we observed the microwave background, we are seeing light that has been travelling in a straight line for just short of 13.7 billion years. That light was in the near infrared and visible part of the spectrum when it was emitted, but because of the expansion of space, the wavelength has also been stretched by a factor of 1100 to the microwave/short-wave radio part of the spectrum.


What mechanism causes oscillations of the solar system's orbit about the galactic plane? In a recent paper (news release here) Lisa Randall and Matthew Reece propose that a dark matter disk coinciding with the galactic plane together with the solar system's oscillations through the galactic plane could explain the 35 million year periodicity in mass extinctions.  They propose that the solar system passes through this dark matter disk in the galactic plane every 35 million years, disrupting bodies in the Oort cloud and causing some of them to collide with Earth.  An image of the proposed cycles from the news release is below.

My question is: What mechanism is responsible for this 35 million year oscillation about the galactic plane?  Is there a companion body to the sun? Are we orbiting around one of the arms of our galaxy?  Is this a well known phenomenon or are they proposing the 35 million year oscillation as well as the dark matter disk?

The cause for the oscillations perpendicular to the galactic plane is the gravity of the non-spherical mass distribution (needed for a plane Kepler ellipse) in the Milky Way. Simplified, there is a dense galactic plane. The density is not exactly known; therefore there is some uncertainty (a few million years) about the precise oscillation period. Details see this article, subsection 3.3.

The idea of a correlation of mass extinctions with this oscillation isn't new, it originates back probably to about 1970, or earlier.

"The outer solar system probably does not contain a large gas giant planet, or a small, companion star", see this press release.

We are probably not orbiting around a spiral arm.

The dark matter disk is a hypotheses, an idea to investigate. Usually only a small fraction of this kind of hypotheses can be confirmed definitively later, most of them can be ruled out after some time, some remain unsolved, some can be refined to match observations.


Why is Enceladus the only geologically active moon among its neighbours? My understanding of why Enceladus is geologically active is that tidal forces from Saturn and - to a lesser extent - from the nearby larger moon Dione provide heat to the moon's interior, just like Jupiter does for Io.

Should the other mid-to-large-sized moons Mimas, Tethys, Dione and Rhea not also exhibit similar activity from the same forces? Granted, Dione and Rhea are further from Saturn and thus less susceptible to tidal stresses, but what about Tethys and Mimas? What special attributes does Enceladus possess that has allowed it to be geologically active in modern times? Is it purely down to the fact that the interior of Enceladus is more ice than rock compared to the other moons, or are there more factors to consider?

This puzzle is something called the Mimas paradox. The two bodies are nearly twins, but Mimas is substantially less active than Enceladus. There are currently several proposed solutions:

Mimas cooled quickly, before it entered its resonance with Tethys (which contributes to tidal heating, just as Dione does for Enceladus). A "hot start" from rapid radioactive decay (of aluminum-26 and iron-60) shortly after Enceladus' formation could have allowed its subsurface ocean to stay liquid.
Enceladus is rockier and therefore has a higher density. This would have led to quicker cooling, and would also mean that tidal dissipation could be higher for Enceladus than for Mimas. It does seem a bit paradoxical - one would imagine that more rock would mean less water ice and therefore a smaller subsurface ocean - but this isn't a problem.

It's worth noting that Tethys is thought to be made almost entirely of ice. If the rocky-Enceladus hypothesis is true, then that could be a major factor in the lack of geological activity on Tethys. However, Tethys likely once had a subsurface ocean. Dione currently does, and experiences tidal heating; however, its semi-major axis is large enough that the effects are not as strong as on Enceladus.


Heavy star and redshift A heavy star should look redshifted, due to gravitational time dilation. How is that incorporated into the calculations of distances to the stars, or is it negligible?

How about an entire region of space that is denser or more energetic than our region? Would that not appear further away, than another region that is less dense than our region?

If we were in a region of low density and energy, and the density of space increases as we look further away, is that something that could theoretically explain Einsteins "blunder"? Can one be sure that this is not the case?

If the universe were collapsing, and gravity does indeed propagate. Wouldn't all distant regions seem to be affected more by gravity than the region of the observer, and thus ever more redshifted? (this due to less gravity in "front" of an approaching region of space than "behind" it, relative to us)

I'm curious of how these things are found to be important or unimportant on a cosmic scale.

I will make a small calculation here, but please proceed to the results if you may like to.

Calculation

Stars are spherical and static, so metric near their surface (photosphere) and outside on is Schwarzschild. Hence time-time metric component on the surface is:

$$g_{44}=1-\dfrac{R_{grav,*}}{R_*}$$,

where $R_*$ is the radius of the star and $R_{grav,*}$ is its gravitational radius.

Then, if the velocity of the star is much smaller than the speed of light, gravitational redshift in the lowest order does not depend on this velocity. Therefore, the emitting star can be assumed at rest.

The light from the star propagates along isotropic geodesic in Schwarzschild metric. The geodesic is described by lagrangian:
$$ \mathcal{L} = \dfrac{1}{2}g_{\mu\nu}k^{\mu}k^{\nu}$$,
where $k^\mu=(\vec{k},\omega/c)$ is the 4-vector of the light wave and $\omega$ is the frequency of light. Since the metric is static $\dfrac{d\mathcal{L}}{dk^4}=g_{\mu 4} k^4=g_{44}k^4=\textrm{const}$. Therefore:

$$(1-\dfrac{R_{grav,*}}{R_*})\omega = \textrm{const}$$

for the light as it travels towards us. So:

$$\omega_{obs}=\omega_{emitted}(1-\dfrac{R_{grav,*}}{R_*})\Longleftrightarrow \lambda_{obs}=\dfrac{\lambda_{emitted}}{(1-\dfrac{R_{grav,*}}{R_*})}$$,
where $\lambda$ is the wavelength.

Redshift is simply $z=\dfrac{\lambda_{obs}-\lambda_{emitted}}{\lambda_{emitted}}$. Assuming $z\ll 1$ one has a simple formula:
$$
z_0=\dfrac{R_{grav,*}}{R_*}
$$

If $z_0$ turns out to be comparable to unity, one should compute
$$
z=\dfrac{1}{1-z_0}-1,
$$
which then gives the correct value of redshift. Note that redshift does not depend on $\lambda$.

Nice numerical forms for this would come from $R_{grav,*}=2.95\textrm{km}\dfrac{M_*}{M_\odot}$:

$$z_0=0.295\dfrac{10\textrm{km}}{R_*}\dfrac{M_*}{M_\odot}\Longleftrightarrow z_0=4.24\cdot 10^{-6}\dfrac{R_\odot}{R_*}\dfrac{M_*}{M_\odot}$$

It is also nice to express the redshift in $\textrm{km}/\textrm{s}$:

$$
z_0=8.84\cdot 10^4 \dfrac{10\textrm{km}}{R_*}\dfrac{M_*}{M_\odot} \textrm{km/s}\Longleftrightarrow z_0=1.27 \dfrac{R_\odot}{R_*}\dfrac{M_*}{M_\odot}\textrm{km/s}
$$

Summary and discussion

In summary, when redshift $z$ is small, it is approximated by $z_0$, the numerical expressions for which are given just above. If $z_0$ turns out to be not small, one can calculate $z=\dfrac{1}{1-z_0}-1$, which then gives correct redshift.

Stars

One can see that:

For normal stars like the Sun ($R_*\sim R_\odot, M_*\sim M_\odot$) the redshift is of order $1\textrm{km/s}$. It is almost important since stars in solar neighbourhood move normaly at the velocity of a few tens of $\textrm{km/s}$
For white dwarfs ($R_*\sim 10^4 \textrm{km}, M_*\sim M_\odot$) redshift is a few times $100 \textrm{km/s}$ and gets very important when doing proper spectroscopy. Therefore one typically accounts for it.
For neutron stars ($R_*\sim 10 \textrm{km}, M_*\sim M_\odot$) redshift is very important $z\sim 0.4$, but neutron stars are general relativistic objects anyway, so it would be expected beforehand.

So, in summary, when measuring light from individual stars one does have to account for gravitational redshifts to get accurate results, and particularly so when studying white dwarfs.

Groups of objects

Now, the same formulae are correct to an order of magnitude when applied for larger volumes of space, with $R_*$ and $M_*$ now meaning the size of the volume and the mass inside it. However, as typical interstellar distances are of order of parsec and $\textrm{pc}=3\cdot10^{13}\textrm{km}$, resulting $z$ will be very small even for such dense groups as globular clusters ($z$ is of order $10^{-8}$ in this case). So, groups of objects do not affect the redshift.

Cosmological overdensities

Nevertheless, cosmological scale underdensities of order few $100\textrm{Mpc}$ in size may affect the apparent redshift of distant objects, as we would be inside the underdensity. However, such an underdensity would have to be significantly symmetric around us in order to explain the lack of corresponding anisotropy in cosmic microwave background. Therefore, it is considered unlikely.


Multi-messenger astronomy: what is the potential of simultaneous detection of gravitational waves and neutrinos from a supernova? Thanks to the efforts of the aLIGO team, gravitational wave astronomy is a reality. At the same time, neutrino detectors like Hyperkamiokande are becoming much more sensitive.

My question is: what are the prospects for the pseudo-simultaneous detection of gravitational waves and neutrinos from the same supernovae? What sort of stuff could we learn from such an event, both about supernovae and neutrinos? In particular, what are the prospects for estimating the neutrino mass?

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

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

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


Why are the natural satellites (moons) of all planets solid? Why are the natural satelites (moons) of all planets - including the moons of the gas giants - solid or rocky, and not gaseous?

To answer this, we have to consider the definition of an atmosphere. A popular way of looking at it is to think of an atmosphere as a layer of gases surrounding a body. by that definition, we can say that gas giants are really just planets that are massive enough to accrete substantially large atmospheres (because deep down, they have a rocky core - although not a core similar to the "rocky" inner planets in our solar system). So now we can reduce the question to "Why do most natural satellites not have extremely large atmospheres?"

This comes down to mass. The more massive an object is, the more gravitational pull it has on objects around it. This means that a very massive object traveling through a region of gas would attract the gas more than a less massive object would. In the early Solar System, the more massive bodies far enough out in the protoplanetary disk managed to gain a substantial amount of matter via core accretion, as gas and dust were collected by a rocky core. Moons are unable to do this because they are very low-mass; therefore, they cannot become gas giants (in fact, many moons around giant planets may be the remains of the material the planets never accreted).

Gas dwarfs, low-mass gas planets with small envelopes, are certainly possible. However, they would have to have radii at least 1.5-2 times the radius of the Earth, and masses scaled appropriately. Moons are nowhere near that size.


What gases are needed to make a star? What gases do you need to start the creation of a star, and why do you need these gases? What are their functions?

It depends on the kind of star you want to make.

Basically, what you really need is hydrodgen; as soon as enough hydrodgen is gathered somewhere and collapse, you end up with a star.

But, if you just have hydrodgen, you will get a very massive star, and with a history different than the stars observed in the local Universe. Why so?

hydrodgen is a poor coolant, therefore the Jeans' mass (the minimum mass for a density structure to gravitationally collapse), which strongly depends on temperature, will be much higher if the temperature is higher. In practice, it means that the formed stars will be much more massive.
you need at least a small fraction of carbon to launch the CNO cycle in stellar cores (CNO cycle is one of the two ways to burn hydrodgen and to turn it into helium); stellar cores are then hotter and denser, and stars become hotter and more luminous.

So if you want to make a star as we typically observe in our galaxy, you will need also some molecules (hopefully there are plenty of them) and also some carbon. Else, you'll get big, massive, hot and luminous stars. Like the first generation of stars (called Population III stars, you can see here if you want more dirty details) in the early Universe.


Are stars expected to become dimmer before a supernova? With the recent news about the "fainting" of Betelgeuse and the speculation that this might be a precursor to a supernova, I'm wondering if there is any theoretical/observational basis for this interpretation, or whether this is a case of reading too much into a greater-than-average change in an intrinsically variable star. Why would a star become dimmer in the run-up to a supernova?

The connection between the dimming and a putative supernova relies on the interpretation that the decrease in luminosity may be due to circumstellar material, ejected in the years/decades/centuries immediately preceding a supernova. There are several mechanisms that could lead to this sort of mass loss (see slides 24-25), including

gravity-wave driven envelope loss (in red supergiants, during neon/oxygen core burning)
pulsations due to pair-instability in the days to decades prior to the supernova
turbulent eruptions à la luminous blue variables

These then lead to Type IIn supernovae, with narrow lines arising from interactions with the previously ejected circumstellar material. If the dimming is due to extinction by circumstellar dust from eruptions due to any of these mechanisms, it could mean that a supernova is imminent on a timescale of days to years.


Virial ratio behaviour I think I didn't understand something about virial theorem for an $N$-body system, for instance the behaviour of virial ratio $T/\Omega$, with $T$ kinetic energy and $\Omega$ gravitational potential energy. As far as I understood, if the system is stationary, than virial ratio is $0.5$. This means on the other hand that the system is not expanding nor collapsing. But the behaviour of virial ratio is something like the plot in this question. Here the system starts from $T/\Omega < 0.5$, so it starts  collapsing and than reach a maximum. I would expect this maximum to be at $T/\Omega \sim 0.5$, since once the collapse stops the system is not accelerating nor decelerating, but it's about $0.75$. If the system is stationary at $0.5$ how can that maximum (and the next minimum too) be explained? I don't understand how virial ratio has a single value since it actually oscillates until converging to a specific value.

The ratio $T/\Omega$ tells you about the acceleration of the system - or more specifically, the second derivative of its moment of inertia - it does not tell you about the velocity.
If the system collapses because it has $T/\Omega<0.5$, then when it reaches $T/\Omega=0.5$ it stops accelerating. That doesn't mean it stops collapsing. It overshoots in the same way that if you compress a spring it goes beyond the equilibrium position before bouncing back once the deceleration becomes large.
NB: I answered the question (of which your question is a subset) that you linked to in the same way.


Notable differences if all stars would disappear except the Sun I am not an astronomy expert, but as of today, all other stars than the Sun have little to no effect on life on Earth. So I was thinking, would there be any differences for the Earth at all if all stars would disappear except the Sun?

There are two main things we should look at here, because there are two main ways these distant stars can affect us: light and gravity.

Light

If all the stars in the universe disappeared (besides the Sun), the most of the luminous objects in the universe would simply vanish. There would still be objects emitting radiation - stellar remnants, accretion disks, etc. - but the notable sources of light would be gone.

Would this be an issue? I can't give you any numbers, but I would assume that there wouldn't be much of a difference. Night/day cycles would continue as normal, because the rotation of the Earth also would not be affected. The behavior of stoma in plants should remain the same, as should all the other natural rituals of life.

Gravity

This is where I'm slightly more concerned. Stars have mass, and so they influence other objects with gravity. There are varying estimates of how much of the Milky Way is made up of stars. For example, McMillan (2011) estimated the stellar mass of the Milky Way to be about 6.43$\times$1010 solar masses, while the virial (total) mass was estimated to be about 1.26$\times$1012 solar masses.

The extra matter consists of gas, dust, planets, stellar remnants, and, of course, dark matter (a lot of it). The mass of the Milky Way would be reduced by about 5% - not a terrible amount, but certainly non-negligible. I don't know how this would impact the motion of the Solar System; detailed simulations based on density models would be needed.

However, any effects from stars wouldn't reach us any time soon. Light and gravity both travel at the speed of light, so we'd have at least four years before we started seeing the effects (4 light-years is the distance to the nearest star, Proxima Centauri). We wouldn't feel the effects from the rest of the galaxy for many more years.


If a black hole has the same mass as me, then why dont i suck everything at my center of mass? So, a black hole is very dense that you can get really close to its center of mass that it has a strong pull, well, if i have the mass of a black hole that has a really small event horizon, why doesnt my center of mass create an explosion from hawking radiation, or suck myself

Because you are somewhat larger than your Schwarzchild radius.
In order to turn into a black hole and start experiencing exciting things like Hawking radiation, you'd need to be compressed down into a ball about $10^{-25}$ meters in diameter, about one ten-billionth the size of a single proton.  At that size, you'd have a hard time sucking anything in: your gravity falls off quickly with distance, and even most subatomic particles will just bend their paths slightly as they pass you by.


Accuracy of Laplace Method for determining orbital elements Recently I had to implement Laplace method and apply it to 3 observations of Mars (10 days in between two observations). The results were pretty good, with discrepancies with real data well below 5% in most of the orbital elements.

I supposed that Laplace method was very accurate, so I tried to apply it to an asteroid (CASLEO, 5387). The 3 observations were also 10-days spaced but now the results widely diverge from the real orbital elements.

I expected it to be less accurate than in the Mars case since the observations covered less fraction of the orbit, but some of the elements were quite far from expected.

So my question is: What is the accuracy of the Laplace Method? Is it really that sensitive to the distance?

In principal, Laplace's method (and Gauss's method) should be very accurate for determining orbits.

Of course "garbage in garbage out" applies, and if the observations were inaccurate the results may also be inaccurate. However, you should not get great discrepancies. I would consider careful checking of your calculation, and perhaps comparing your results with those of an independent orbital determiner, to check for a mistake in the maths.


Is the universe expanding faster than speed of light? The celestial bodies like stars and galaxies are moving away from each other. How fast are they moving apart? Is that speed more than the speed of light?

According to Hubble's law

  For [proper] distances $D$ larger than the radius of the Hubble sphere $r_{\mbox{HS}}$ , objects recede at a rate faster than the speed of light ... $$r_{\mbox{HS}}=\frac{c}{H_0}$$

With $H_0=67.8\mbox{ }(\mbox{km}/{\mbox{s}})/\mbox{Mpc}$ the Hubble constant according to the Planck mission, $c=299,792,458\mbox{ m}/\mbox{s}$ the speed of light, and one Megaparsec $1\mbox{ Mpc}=3.0857\cdot 10^{22}\mbox{ m}$, we get
$$r_{\mbox{HS}}=\frac{c}{H_0}=\frac{299,792,458\mbox{ m}/\mbox{s}}{67.8\cdot 10^3\mbox{ }(\mbox{m}/{\mbox{s}})/3.0857\cdot 10^{22}\mbox{ m}}=1.3644\cdot 10^{26}\mbox{ m}.$$
That's $$1.3644\cdot 10^{26}\mbox{ m}/1.3644\cdot 10^{26}\mbox{ m}= 4,421 \mbox{ Mpc}$$
or $r_{\mbox{HS}}=14.422$ billion lightyears (1 pc = 3.26156 ly).

Hence objects further away than this proper distance (what you would measure with a chain of rulers) recede faster than the speed of light. The distance is also called Hubble length.

You may find 13.8 billion lightyears elsewhere. That's calculated with $H_0=70.4\mbox{ }(\mbox{km}/{\mbox{s}})/\mbox{Mpc},$ according to the estimates of 2010, based on WMAP data, or with even older data. The exact value is not known.

Since the Hubble constant isn't really constant over time, the universe is thought to be a little younger (13.8 billion years) than the 14.422 billion years you get by just dividing the Hubble distance by the speed of light.


How Does a Refractor Telescope Work? As I understand it, the difference between a reflector and a refractor is that a reflector uses mirrors on the back to reflect the light to the secondary mirror then the secondary mirror reflects the light to your eye focusing the image with each reflection. How then does a refractor use a lens to sharpen the image? Does it just sharpen the image and shoot it to the prism at the back, and into your eye?

All telescopes have in common that they gather and focus light from far away objects. They use a primary opical element, such as a concave mirror or a (planar- or bi-)convex lense (or lense system), and they use an eyepiece with another lense system (for viewing) or a camera in their primary focus.

A refractor telescope does not sharpen the image per se. The convex lense concentrates the light rays, not unlike a magnifying glass. To actually focus the enlarged image on your retina, you need an eyepiece, which is another bi-convex lense (in its most simple form). This will re-align the light rays after they passed through the primary focus. See this image for a visual explanation:

Source: Wikipedia

The above image also explains why the image of a refracting telescope appears upside down. You don't need (or want!) any prisms in this kind of setup.

On the other hand, a reflector telescope uses a concave mirror plus an eyepiece. There are different configurations, but one of the most simple and most common ones is the Newtonian telescope:

Source: Wikipedia

So instead of refraction of light by a lense, we use the reflection of light on a mirror, to enlarge the image. Focusing on the retina is again done by an eyepiece in the same way as with the refracting telescope.

The advantage of refracting telescopes is that there is no obstruction in the optical path inside the telescope. This is not the case with reflector telescopes. They usually have a secondary mirror in the middle of the optical paths, hence reducing light gathering performance.

On the other hand, reflector telescopes are often much lighter, and cheaper to assemble. Also, very compact models of reflector telescopes can be built.

Also, simple refractor telescopes will produce colourful fringes on object edges, called chromatic abberation, which is due to the glass used in the lenses. This can be compensated by multiple lenses, but this will make the refractor even heavier and more expensive.


How do scientists determine the age of stars? I was Googling about the oldest galaxies in the Universe. Everywhere is written that their age is known by the light. Referring to this line "Since light travels at a set speed, if you look at a star ten light years away, you're really seeing what it looked like ten years ago" on this site.

So my question is, as we know light is continuous, how do scientists know that a particular ray has started its journey some 1000 light years back? How is the age of stars determined?

Measuring the ages of stars is an incredibly difficult process. It is not something that can be summarised adequately in an SE answer. Please take a look at these references.

Soderblom et al. (2013); Jeffries (2014).

Methods split into -

Fundamental - e.g. radioisotope dating - only possible for the Sun

Semi-fundamental - kinematic traceback of stars to their birthplace; observing the depletion of lithium in a star.

Model-dependent - comparing the luminosity and temperature (and composition) of a star with that predicted by stellar evolutionary models. Or equivalently, its surface gravity and temperature.

Empirical - looking at how things like rotation and magnetic activity decline with age and using this to estimate the age of a star.

Statistical - using the metallicity or galactic space motions of a star to provide a (crude) estimate of its age.

There is also a lot more detail given in this SE answer.
Without using absolute magnitudes or isochrones, how might we tell a star's age and evolutionary status?


Can we see the color of nebulae? Someone asked me this question:

  To see the color of the Great Nebula in Orion, how large the aperture of the telescope should be.

I don’t even know whether this is possible. I understand that the colorful patterns of the nebulae are from the cameras. Besides, my own experience shows that when I target the telescope at them, I see nothing but some white blur.

So is this possible when we watch the deep sky objects with a really huge telescope, under the finest conditions?

Very large (for amateur) scopes in very dark places, with healthy eyes on a good night have credibly reported seeing color in nebula. The Orion is about as good as it gets,
lots of photons, my eyes are not particularly good, but have been out with others who claim noticing color in the Trapezium in a thirty-ish inch reflector


Ensemble average VS spatial average for a cosmological density field I'm studying statistical properties of a cosmological density field and I have some doubts; in the following quote and pictures you can see what my problem is:

  A fundamental limitation arises in cosmology – because there is only one universe to observe, i.e. there is only one realization of the stochastic process that generates the fluctuations whose consequences we observe. Therefore we cannot measure ensemble averages
  or expectation values, as we would in a repeatable laboratory experiment. What we can do when observing a fluctuation on a given scale $\lambda$ is to average over many distinct regions of size $\sim\lambda$. An ergodic-type hypothesis allows us to replace the ensemble average by a spatial average over these regions. (see pag 307 from Relativistic cosmology-George F.R. Ellis, R. Maartens, M.A.H MacCallum; Cambridge University Press (2012))

So here we talk about an average over many distinct regions of the Universe (someone call them "fair sample" of the Universe if I understood correctly) and an ensemble average is replaced by a spatial average over these regionS. Thus I should average the field over region 1, region 2 and so on and I would get many mean values of the field.

But in this presentation things seem different:

Here instead we talk about an average over an ensemble replaced by a spatial average over one realization of the random field (one of the possible  universes I imagine). In this case I have only one realization: our Universe, so I should average my field spatially over the entire Universe. And what about the regions mentioned above?

I'm confused and I would appreciate an explanation.

Thanks.

Addendum

So theoretically if i wanted to get the average of a stochastic field or of a function, we say $\delta(\vec x)$ or $f[\delta(\vec x)]$, in a point $\vec x_1$ I should do the average over the ensemble of all the universes using (see slide): $\int{\delta}\cal P(\delta) d\delta$ (it is obviously impossible because we have only one Universe). I have imagined this thing thinking about a stack of planes representing the different universes of the ensemble and drawing the point $\vec x_1$ in each plane. Now i can take the value of $\delta(\vec x_1)$ in the first universe, in the second, and so on and finally make the average of these values; if I repeat this procedure for all the points I'll get the average of $\delta (\vec x)$, i.e. $\langle\delta\rangle$ over the ensemble of the universes. Ok it's fine...but it is not possible because I have only one Universe, so I have to appeal to Ergodic Hypothesis (really a spatial version of the Ergodic Hypothesis).

  An ergodic-type hypothesis allows us to replace the ensemble average by a spatial average over these regions

It seems that I have to divide the Universe in regions (we can imagine squares for simplicity in two dimensions, I thought the Universe like a chessboard) and I can replace the ensemble averege with a spatial average over any single region. Any square of the chessboard is a fair sample of the Universe, so I have many squares that can "reproduce" the Universe (I divided my sky in many regions and I can consider these squares like the realisations of the "Universe experiment", the outcomes of the experiment). Thus I can do:

$$\langle\delta\rangle=\frac{1}{\sum_i V_i}\sum_i \langle\delta\rangle_{V_i}$$

where $\langle\delta\rangle_{V_i}=\int_{V_i}\delta(\vec x) d\vec x^3$.

I can't see the contradiction. One version says average over the regions, the other says integrate over one realisation (containing many regions). I suppose you could argue that one is performing a summation,whilst the other is an integration, but if the regions are small compared with the size of your observable universe then they amount to the same thing.
$$\frac{1}{\sum_i V_i} \sum_i \lambda_i  V_i \simeq \frac{1}{V}\int \lambda(r)\ dV$$


Lunar terminator illusion Sometimes, you can see the Moon high in the night sky, which is illuminated from the side where Sun goes down but slightly from above. It is paradoxical when Sun hangs at the lower level above the horizon than the Moon. Sun can even go down behind the horizon, yet, the Moon is illuminated slightly from above.

The paradox is defined in Wikipedia and captured in this video

There is a video from "Vsauce"

http://youtu.be/Y2gTSjoEExc?t=280

which pretends to explain the phenomenon but fails, IMO. From his explanation I only get that light beam should deviate from the direct route to move along the orbital around the Earth, like if it is the Sun or Moon orbit seen from the Earth. But why? The nebular objects orbit because they have mass and gravitationally bound. Why should light beam travelling from Sun to the Moon bend the same way as if it orbits around the Earth? I do not get the explanations.

The popular explanation of the illusion does not hold the water

http://youtube/watch?v=Rsz2QNPprB8&lc=z134gjggimumexwgr04cdxbrnlattlwp5n40k

The light ray does not bend (in fact it does a little bit, but that's small and doesn't matter now). However, what you see is not a path of a ray of light, but a projection of this path on the celestial sphere around you. This projection is an arc of a great circle with you in the centre.


Locate planets and stars for beginners I am a beginner in astronomy. In fact, I just bought my first refractor telescope with 60 mm aperture and 600 focal lengths with 4mm, 12.5mm and 20 mm eyepiece and 3x barlow.

Could I get info on android app or book where I can see real-time position of planets and other amazing celestial bodies, I am in the northern hemisphere.

There are lots of apps you can download on your laptop / smartphone :

Stellarium : the one i use on my computer. Useful for beginners. You will see planets, some of the best known object, ...
Skymap for android : used it before, quite nice.
SkySafari : don't use it but some people told me it was great.

All of them are free. However, you can just search on google, there are plenty of online solutions.


Which unit is most suitable in comparing needed light intensity for night reading on the Moon, vs on Earth? Which unit is best to use, and why, when determining how many times easier it is to read a newspaper (or any other printed text) on the moons surface at night under a full earth, compared with reading it outdoors on  earth during a cloudless full moon night: Albedo, Bond Albedo, Apparent Brightness, Apparent Magnitude, Geometric Albedo, Lux, or something else?

Readability requires taking into account the sensitivity of the human eye to various wavelength. This is called Illuminance.
Physics knows Lumen (flux weighed by human perceptibility) and Lux (flux per unit area weighed by human perceptibility).
Quoting that, different light sources have roughly these values:
Source             Lux
Full daylight   10,000
Overcast day     1,000
Very dark day      100
Twilight            10
Deep twilight        1
Full moon            0.1
Quarter moon         0.01
Starlight            0.001


Why are Uranus' satellites named after characters in Shakespeare? Oberon, Titania, Ophelia, Puck, Miranda... Uranus' satellites are named for characters in the works of Shakespeare and Pope. Yet the names of Jupiter and Saturn's moon come from the Roman and Greek mythology.
Who decided that the moons of Uranus would be named after English poets? What process led this to come to be?

From W. Lassell (1852) in Astronomische Nachrichten vol. 34: "Beobachtungen der Uranus-Satelliten" (Observations of the satellites of Uranus):

In these tables I have for facility of reference employed a nomenclature of the satellites of Uranus proposed at my request by Sir John Herschel and selected by him from fairy mythology. The most distant of the two bright satellites discovered in 1787 by Sir W. Herschel is denominated Oberon, the other Titania, and pursuing still the order of distance, the interior ones now discovered are named Umbriel and Ariel.

This shows that it was John Herschel (son of William Herschel, the discoverer of Uranus, Titania and Oberon) who originally decided the nomenclature scheme at the request of William Lassell, the discoverer of Ariel and Umbriel. Interestingly it appears that the scheme was originally intended to be fairy mythology, which is not how the names are currently assigned. Nevertheless there is some debate around who exactly was responsible because the actual correspondence has been lost.
This pattern was broken by Gerard Kuiper, the discoverer of Miranda, see Kuiper (1949) in Publications of the Astronomical Society of the Pacific, vol 61, "The Fifth Satellite of Uranus". Unlike the previous satellites, Miranda is named after a human character (the daughter of Prospero) in Shakespeare's The Tempest:

Miranda was chosen as the name for the fifth satellite.
Uranus' own children, the Titans, are not suitable for mythological reasons; they have been assigned to the son of Uranus,
Saturn (Kronos), who gained supreme power after wounding
his father. Sir John Herschel named the four bright satellites
Ariel, Umbriel, Titania, and Oberon.
Oberon and Titania are
the king and queen of the fairies in Shakespeare's Midsummer
Night's Dream; Ariel and Umbriel occur in Pope's Rape of the
Lock, while Ariel is also found in Shakespeare's Tempest. In
the Tempest Ariel is "an airy, tricksy spirit, changing shape at
will to serve Prospero, his master," while Miranda is "a little
cherub that did preserve me" (Prospero).

The convention subsequently seems to have been formalised with the discovery of the inner satellites by the scientists working on Voyager 2. The IAU set up a nomenclature committee in the wake of the naming of the Mare Moscoviense (which had the somewhat dubious justification of declaring that Moscow is a state of mind). The IAU nomenclature committee decided that the common factor behind the existing nomenclature was the source works (Shakespeare's plays and Alexander Pope's The Rape of the Lock) and named the satellites according to this pattern. Some details of this can be found in this podcast from the Folger Shakespeare Library that includes Tobias Owen, who was involved in the naming process.