What will future populations of stars be like? As stars slowly using up hydrogen and ejecting heavier elements into space, the future stars will form from nebulae that are more metal and helium rich.

The question is: how does the increased ratio of helium and metals affect star formation and stellar evolution?

Does more helium and metals in the core at the beginning means the star begins it's life in an advanced state so it dies faster?

(Adams & Laughlin 1997) discuss the effect of increasing metallicity in the future. A higher metallicity increases the stellar burn rate since the density increases but
the higher opacity reduces it a bit. The total effect is nonlinear; in their model the maximal lifespan happens for $Z\approx 0.04$ and beyond that it declines. Whether there are newer and better models of this I do not know.

They also point out that the maximum stable stellar mass decreases as metallicity increases, while the mass of the minimum mass mainsequence star declines, both as a result of opacity effects.

Overall, the metallicity growth is likely to continue across the stelliferous era. While extragalactic gas may fall in, it merely adds to the hydrogen pool and continued star formation. What could stop the metallicity growth is quenching due to a galactic merger or active galactic nucleus that blows away enough gas to stop further star formation.

There is no sign that  novae and supernovae are dumping enough heavy elements into the interstellar medium to affect formation of stars in the future by causing a shortage of hydrogen. It has recently been discovered that the Milky Way is at the centre of a vast cloud of hydrogen extending at least 300,000 light years in all directions, enveloping the Magellanic Clouds and containing enough mass to solve the dark matter problem without the aid of WIMPS. In addition, Australian astronomers have discovered much smaller but still massive clumps of hydrogen racing around the galaxy at speeds of up to 56 miles per sec, with no regard to the direction of rotation of the galaxy itself. What might happen 100 billion years down the road is too far ahead to speculate, but in the foreseeable future there is no fear of a shortage of hydrogen, so stars will continue to form in much the same way as they do now, and have closely similar lifetimes.


Correct relation between metallicity (z) and iron content ([Fe/H]) The Wikipedia entry on Metallicity states that:

$\log_{10}\left(\frac{Z/X}{Z_\mathrm{sun}/X_\mathrm{sun}}\right) = [\mathrm{M}/\mathrm{H}]$
where $[M/H]$ is the star's total metal abundance (i.e. all elements heavier than helium) defined as a more general expression than the one for [Fe/H]:
$[\mathrm{M}/\mathrm{H}] = \log_{10}{\left(\frac{N_{\mathrm{M}}}{N_{\mathrm{H}}}\right)_\mathrm{star}} - \log_{10}{\left(\frac{N_{\mathrm{M}}}{N_{\mathrm{H}}}\right)_\mathrm{sun}}$
The iron abundance and the total metal abundance are often assumed to be related through a constant A as [citation needed]:
$[\mathrm{M}/\mathrm{H}] = A\times[\mathrm{Fe}/\mathrm{H}]$
where $A$ assumes values between 0.9 and 1. Using the formulas presented above, the relation between $Z$ and [Fe/H] can finally be written as:
$\log_{10}\left(\frac{Z/X}{Z_\mathrm{sun}/X_\mathrm{sun}}\right) = A\times[\mathrm{Fe}/\mathrm{H}]$

This all sounds reasonable, but I have two questions:

Where does the $[\mathrm{M}/\mathrm{H}] = A\times[\mathrm{Fe}/\mathrm{H}]$ relation come from? I could not find a proper source.
This article actually says (see Eq 9) that the general relation is $\log_{10}\left(\frac{Z/X}{Z_\mathrm{sun}/X_\mathrm{sun}}\right) = [\mathrm{Fe}/\mathrm{H}]$ i.e., they seem to equate [M/H] with [Fe/H]. I assume that it is because the aforementioned A parameter is between 0.9 and 1, but again that leaves me with the need for a proper source to state that.

$$[\mathrm{Fe}/\mathrm{H}] = \log_{10} \left[\frac{f_\mathrm{Fe}(\alpha)}{f_\mathrm{Fe}(0)} \frac{m_z(0)}{m_z(\alpha)} \left(\frac{Z}{X}\right)_\ast \left(\frac{X}{Z}\right)_\odot \right]$$

You can use this relation to estimate [Fe/H] from [M/H] of a star. From [M/H] you can get the (Z/X) ratio.

$$\frac{f_{Fe}(\alpha)}{f_{Fe}(0)} \frac{m_z(0)}{m_z(\alpha)}$$

This is the Value of the constant that u get. The above values you can get from any standard paper. Check fergusion et al 2005


Could light be dark matter? Is it possible that light itself is dark matter?  I am speaking of photons (e.g. visible light, infrared, ultraviolet, etc...).  I realize light is understood to be massless, but it is obvious it at least contains energy because we can see with it (e.g. it energizes the cells in our retinas).  I wonder if light has a very tiny "net mass" (e.g. 0 mass * relativistically infinite speed).  I would think that light at least has a little mass, in proportion to its energy.  For example, take E=mc^2, then m = E/c^2 would describe how much mass it has.  If this is true, light should have a very little gravity too.  Although the effect would seem minimal, light is practically everywhere. Gravity from light would be more concentrated inside galaxies, and even more concentrated in the center of galaxies where there are many stars (like dark matter is).  It would be interesting to run the calculations, assuming light does have gravity, and see if this matches the gravitational observations of dark matter in the universe.  It would be funny and ironic if dark matter really is light.

Edit: Note that it appears light does have gravity as per the discussions here: How does light affect the universe?  If that much is true, I wonder if this is significant enough to account for dark matter?

Dark matter, is just a name for something we know nothing of. It was named to account for an extra gravity source for which there have been indirect observations, but yet we cannot explain.

The force of gravity exerted by light is negligibly small yet we have measured the gravitational pull of Dark Matter to be big enough to affect whole galaxies; it is what binds galaxies together.

Furthermore, we have included everything we can observe (all ordinary matter including photons) when we do the calculations for the amount of gravity there should be. So light is already there. 'Dark matter' is that extra gravity which we cannot account for.

Light has certainly a gravitational mass. Light is also dispersed into space with the inverse of the square of distance. That means that its weight is stronger all along the galaxies disk where the light sources are. Just like dark matter is supposed to be. Local stronger light in the galaxies is to be added to all the light from intergalactic sources and CMB, which is more or less the same across the universe, and should have an effect on galaxies too. The point is how much the light in the whole EM spectrum in the interstellar space weights. It's clearly negligeable into solar systems near stars, compared to regular mass, but maybe not if you add all the light in the huge space between stars. Plus add the gravity of light in "empty" space between galaxies (galaxies rotation should be also affected with the gravity of light & matter between galaxies).
I think astronomers must have taken this into account, but I can't find the calculated amount anywhere (I'm not an astronomer, maybe it's in many specialized astronomy books, I don't know). I suspect anyway that light weight is much, much less than required for being the dark matter, but I don't have the numbers. If anyone has them...


What is the density of stars in our region of the galaxy, up to say, 1000 light years distant? If we imagine a sphere of radius one light-year centered on Earth, it contains only one star - our Sun.

If we imagine a sphere of radius ten light-years centered on Earth, it contains about 10 stars, according to this wikipedia page.

If we imagine a sphere of radius one hundred light-years centered on Earth, about how many stars would it contain?

If we imagine a sphere of radius one thousand light-years centered on Earth, about how many stars would it contain?

Here's a crude way of estimating the answers:

The density of stars in the Milky Way disk decreases exponentially as you go out in radius and also exponentially as you go out in height above/below the disk midplane. The corresponding scale lengths are about 3 kpc (3 kiloparsecs = about 10,000 ly) radially and about 300 parsecs (about 1,000 ly) vertically ("scale height"); this means that if you go out in radius by about 10,000 ly, or up/down in height by about 1,000 ly, the density decreases by a factor of e (2.718...). Changing radius by about one tenth of the scale length or less, or changing height by about one tenth of the scale height or less, only changes the local density by about 10%, which isn't enough to worry about. That corresponds to about 300 parsecs (1000 ly) in radius, or about 100 ly in height.

So the for case of a 100-ly radius, you can assume the density everywhere is about the same as the local density suggested by the table you linked to. That density is 60 stars / ($(4/3) \, \pi \, 16.3^3$) = 0.0033 stars per cubic ly. For a sphere of radius 100 ly, you would then expect about 14,000 stars.

Naively, a sphere with a radius of 1000 ly would then have about 14 million stars. But a sphere that big extends above and below the disk midplane enough so that it samples about 1 scale height. So you can't just re-use the local density. Without getting into the proper approach of integrating the density functions in 3D, we can make a crude estimate as follows: If we used the density at one scale height (1/e times the local density), then the sphere would have only about 4.5 million stars. If we split the difference, that's about 10 million stars, which is probably not too terribly wrong. (I've been ignoring things like stars belonging to the "thick disk" and stars belonging to the halo, but they won't contribute a whole lot; also, the Sun isn't exactly at the disk midplane, but about 60 or 70 ly above it...)

Sol is about in the middle betwen the center and the classically defined edge. We have a density of about 1/4ly stars. It gets much denser within 5kly of the center of the galaxy, not sure how the density falls towards the edge. We also have the spiral arms which are density waves which are less dense outside of them...

Beyond this there isn't much more of an exact answer due to not be able to see beyond 100 or 1000pc of Sol.

As far as density beyond that, just look up the equation for volume of a cylinder, you can get that from the density in the wiki page if you find the volume of a sphere or cylinder (which is what you need beyond 1000ly radius) and then just divide.


Why does matter stay collapsed in the core, following a supernova explosion? Following a supernova explosion a star may turn into a white dwarf, neutron star, black hole, or just a stellar dust & gas leftover.

Excluding the latter case, why and how does the star's core matter stays collapsed, after such an event where matter is burst and scattered in space?

In order to "blow something up" you need to release more energy than its binding energy and have a way of trapping that energy so it can't escape in another way.

At the centre of a core collapse supernovae is a 10 km radius, $1.4 M_{\odot}$ ball of (almost) neutrons.
Its gravitational binding energy is $\sim GM^2/R = 5\times 10^{46}$ J.

This is almost exactly how much energy is released by the collapse of the core from a much larger size (i.e. the energy of the supernova is gravitational to begin with) and since some of that energy goes into dissociating iron nuclei and making neutrons (both endothermic processes) and most of the rest escapes in the form of neutrinos, then there can't be enough energy to unbind the core. Only a tiny fraction (1%) of this energy is transferred to the envelope of the original star, which since it has a much larger radius (by at least 5 orders of magnitude), is enough to overcome its gravitational binding energy and blast it into space.

The case of a type Ia supernova (an exploding white dwarf) is quite different. Here the energy source is not gravitational collapse, but from a thermonuclear detonation of all the carbon and oxygen that make up the white dwarf, to form iron peak elements. This exothermic process rapidly releases enough energy to unbind the original star (e.g. see here) and it is completely destroyed.

In a star, there are two opposing forces which usually balance each other Gravity is a force which induces collapse, while radiation pressure from the fusion reactions within resists the tendency to collapse. Small, sun-like stars, when they have used up most of their hydrogen fuel, will start "burning" helium and become red giants. When the helium runs out they will puff off their outer layers in a nova and collapse to form a white dwarf about the size of Earth. These white dwarfs are amazingly dense and heavy, because most of the mass of the original star has been compressed into a comparatively tiny volume. Further collapse is resisted by a force called electron degeneracy pressure.

Stars much larger than the sun will go on fusing elements beyond helium, building up layers of successively heavier elements until they reach iron. Fusion of elements beyond iron requires an input of energy rather than producing any, and the nuclear fires go out, so deprived of support from radiation pressure the outer layers of the star collapse, producing a supernova explosion. Electron degeneracy pressure is not enough to prevent a more drastic collapse than occurs with much smaller stars. According to the mass of the collapsing star, this will either result in the formation of a neutron star, which is like a gigantic atomic nucleus of incredible density about 6 miles across but containing a mass equivalent to several of our suns, or it will collapse further to form a black hole singularity in which matter enters a state not fully understood by science. Our sun, by the way, is 860,000 miles in diameter..


How variable does a star have to be, to be a variable star? Variable stars are stars whose apparent magnitude varies. But there are so many phenomena that can cause a star to be variable, that I would expect all stars to be variable. A rotating star has a starspot? It's variable! A planet is transiting? It's variable! A cloud of dust passes in front? It's variable!

This makes me wonder what exactly distinguishes a variable from a non-variable star? Must the variations in brightness be larger than a given magnitude? Must the variations in brightness be periodic? Do dips (or peaks) in brightness be observed more than once?

There is no lower limit, and as you say, all stars are somewhat variable.

However catalogues of variable stars exist, and they can record a wide range of levels of variability.  For example, the general catalogue of variable stars lists stars like Alpha Triangulum, with a variability of 0.01 magnitudes.

Ultimately a variable star is a star which has had its variability measured and studied and recorded in a catalogue.

Certainly sunspots (or "starspots") won't count, as they are statistically random and thus unpredictable, as well as causing a very small delta in total output.
edit:
I stand corrected.  There's a Wikipedia page (thanks, HDE) which describes several classes of aperiodic stars.  These stars do continue to undergo significant variation for at least many days, apparently.
The wiki page on general(periodic) variable stars identifies two  classes:

This variation may be caused by a change in emitted light or by
something partly blocking the light, so variable stars are classified
as either:
Intrinsic variables, whose luminosity actually changes; for example,
because the star periodically swells and shrinks.
Extrinsic variables,
whose apparent changes in brightness are due to changes in the amount
of their light that can reach Earth; for example, because the star has
an orbiting companion that sometimes eclipses it. Many, possibly most,
stars have at least some variation in luminosity: the energy output of
our Sun, for example, varies by about 0.1% over an 11-year solar
cycle.

After that, one descends into pulsars, binary stars, Cepheids,  and lots more.


Visibility of earth from moon during day-time of moon Inspired by this question. I am curious whether earth, besides being nearly fixed on one place on the moon's sky, is it visible during the day-time on moon too?

My understanding is that earth should be visible as moon has no atmosphere. Also, if the NASA didn't edit the following photograph, it suggests that the day-time sky of moon is all black and earth should be visible in it.

I think, Stellarium don't take into account the atmosphere once you are on other planet.

Yes, the Earth would be visible even during day on the Moon. Without a (significant) atmosphere diffusing light, the day sky would be much like the night sky on Earth. If you were to look straight up such that you couldn't see the ground, you may not even notice that it was day at all (save for the Sun shining down at you).  This could actually be rather hazardous as the Sun would be more intense, but your eyes may not be adjusted to high light like they are on bright Earth days.

Earth diameter = 6,371 km
Moon diameter = 1,737 km
ratio 3.6 : 1
Moon seen from Earth has size X.
so
Earth seen from Moon has size 3.6X x 3.6X making it look roughly 13 times bigger on the sky than Moon looks on Earths sky.
I wonder how come they forgot about this...


Does mean solar time and sidereal time sometimes indicate the same time? I'm trying to wrap my head around the different definitions of time. Since mean solar time depends on the Sun, and sidereal time depends on the stars, and since the position of the Sun relative to the stars changes over the course of the year, does this mean that the difference between the two times increases and decreases over the course of the year?

Once a year, mean solar time and sidereal time will be the same. Is that right?

The 3m56s mean difference between solar and sidereal days is due to the Earth's orbital motion around the Sun.
This adds up to 1 day per year; for every 365¼ solar days there are 366¼ sidereal days.

Sidereal time equals the right ascension of what's on the celestial meridian at that time.
The Sun is on the meridian at 12:00 apparent solar time.
These align at the September equinox, when the Sun is at RA 12h.
Mean solar time is ~7.5 minutes behind apparent solar time on that date and aligns with sidereal time ~1.9 days earlier.

the sidereal time is the time measured by the angle between the sight line of an observer and the direct ray of a fixed star to the observer, whereas the solar time is the time measured by looking at the position of the sun in the sky like you can say it's around noon when the sun is overhead... they  are just different versions of the same time... like metre and mile are of length 1 mile is not equal to 1 metre... similarly sidereal and solar time are different units of the same time... it's often said that the solar day is 4 mins less than a sidereal day


"Ideal" primary thickness in a Dobson I plan to buy a 12" or 14" dobson, the choice depending on my final budget and on the total weight of the instrument. I have several options for the primary mirror. I tried to find a formula that would give the acceptable range for mirror thickness depending on the diameter, maybe also on the material (BK7, pyrex, ....). Is there such a formula, or a back-of-the-enveloppe calculation, which could help me? If not, what do people go with usually (for these diameters, I mean)?

Thank you for your help.

If you're buying the instrument, as opposed to making it from scratch, my advice is to worry about the overall quality, as opposed to a single parameter such as mirror thickness. It boils down to whether you trust the vendor or not.

In any case, there is no simple formula. You could run the PLOP simulator and try a few different scenarios, and see what the overall RMS error is in each case. But the values will vary very significantly. The main factors are:

the diameter / thickness ratio
the number of support points in the mirror cell
the relative positions of the support points

All of the above assumes a traditional mirror with a flat back, and a classic mirror cell with multiple-of-three support points. But the analysis gets even more complex if it's a cell mirror (the back is not flat), or a honeycomb structure (the back is flat but there are empty spaces in the volume of the mirror). And then there are conical mirrors, etc.

It's difficult to second-guess the vendor, especially if you've never built an instrument. Just read the reviews, ask other owners, try and gauge the overall build quality.

Since you're just buiding a Dobsonian, you're not going for aspherics and in fact not going for particularly high-quality surface figure at all.  In particular, there's no need to worry about orientation-dependent sag (not that this is a significant factor for such a small diameter in the first place).
It's not clear whether you are buying or fabbing the primary.  If the latter, just choose a blank such that the center thickness, which is the thinnest spot on the mirror, is at least maybe 5 mm.  You also haven't specified the focal length of your primary, which certainly figures into the thickness and mass calculations.

If you just like having fun :-) ,  buy the blank as thick as you like and use an end mill to lightweight it by removing most of the material from behind, leaving a "web" to support the figured surface.


If nothing travels at the speed of light, except light, how can a black hole also pull light into itself? Is it because it has super massive gravity and energy? Or is it some sort of an anti-light component in it because it used to be a star?

Gravity is a force, and it need not have a "speed"1.

A gravitational body sets up a gravitational field around itself. Note that by the time the light wave approaches the body, this field has already been set up. The gravitational force does not need to "reach out" and "catch up to the light" — it's already there.

In fact, it's not even really a force (though we can treat it as one to an approximation). Gravity bends the fabric of spacetime around it, messing with the meaning of a "straight line". From different reference frames, different lines appear straight. To the light wave, the inspiraling path seems "straight" and thus it follows that path. To an external observer, the light wave is not going straight.

1. It does, but that's the speed at which changes to the field propagate, not the speed at which it "catches up" with other things. Changes in the gravitational field propagate at lightspeed.

Well, I have no astronomical background, so in layman's terms,

  If nothing travels at the speed of light, except light, how can a black hole also pull light into itself?

I believe OP is confused over how a gravity wave can capture light if both can travel at most at the speed C.

It does not work quite that way. Say you move a massive BH from position X to position Y and an observer is at 1 LightYear away from Y. Then, the observer will not feel the gravitation effect of BH until 1 year because that's how much it will take a gravity waves to reach to observer. That is, its changes in gravitational field "Delta" propagate at the speed of light.

As others have mentioned, gravity bends/shapes the spacetime fabric, so after one year, at observer, spacetime is bend according to the gravity of BH at Y. Thus, any light that passes near to the observer will bend as it follows the spacetime. Plus, the curvature of space BH create in their vicinity is so severe that space is wrapped completely around itself and hence light can not escape.

Refer to this for simple explanation.


What effect does Earth losing a part of its mass have, is there an impact on the Earth's orbit around the Sun? This is a random question that popped into my mind. I'm not a physicist or astronomist, however, I am genuinely curious. Also, this is not intended as a science fiction question, but an actual science question.

I had doubts about where this question should be posted - the Astronomy or the Physics SE page, in the end, I decided to post it here.

Sub-questions that I have, all being closely related to the one in the title:

Do the liftoffs of spaceships have any impact on Earth's orbit? As negligible as they might be?
Every bit of mass lost (sent into space) decreases the gravity of our planet, as tiny as the change might be. Is that correct?
Could the loss of Earth's mass over time (Hypothetically - centuries in the future, with tens of thousands of spaceships having left Earth.) change the orbit of Earth or produce any side-effects?

According to this answer to Is Earth getting heavier or lighter?:

tl;dr: The Earth receives 40,000 tons of dust from space every year, but looses 95,000 tons of Hydrogen and 1,600 tons of Helium every year as well. After all additional effects are balanced, the Earth looses about 50,000 tons a year.

According to this answer to Does launching a device into orbit change earth's orbit?

As long as "what happens in GEO stays in GEO" and there's no mass lost to deep space, the Shuttle and the Earth will mostly orbit around their common center of mass.
When the Shuttle is moving (relative to Earth) in the prograde direction (Earth's motion around the Sun), the Earth will be moving slightly slower than it normally would, but a half-orbit later when the Shuttle is moving retrograde, the Earth will be moving slightly faster than it normally would.
When the Shuttle returns, to 1st order (and maybe 2nd order as well) the Earth will be in the same place it would be if the Shuttle had never taken off.
It's quite a bit like the Earth and Moon orbiting around their common barycenter, and it being pretty much the EM barycenter that moves in a nice elliptical orbit around the Sun, except that as @RussellBorogove points out very effectively, it's a quite a bit smaller effect than that.

So as long as the launched mass stays in orbit around the Earth, the Earth's orbit is essentially unaffected. If the spacecraft goes off into deep space, then there will be an extremely tiny change in the Earth's velocity (rather than due to its mass change) by conservation of momentum.
If the Earth loses mass by natural process isotropically, the orbit will be essentially unaffected. Of course the Earth and Sun orbit around their common center of mass, so if the Earth gets lighter it will have an extremely tiny second-order effect on the orbit, but that's nothing compared to the effect that the outer planets have on the Sun.

Yes, the blast off of a rocket from Earth obeys Newton's 3rd Law: for every action there is an equal but opposite reaction. Think of it in terms of a man firing a gun. The bullet goes one way, the man's shoulder, absorbing the recoil, goes the other way, though only a very short distance because the man is heavier than the bullet. The same principle applies to the rocket fired into space: the rocket goes one way, the Earth goes the other. That is the theory, but because the mass of the Earth is trillions of times greater than that of the rocket, the movement of the Earth in the opposite direction to the rocket is so tiny as to be undetectable and unmeasurable.

Now for the decrease in Earth's mass. Again you must bear in mind that the Earth has trillions of times more mass than a large rocket, so not only is the mass loss trivial, but most of the objects sent into space eventually come back down to Earth. Also. there is a constant rain of space debris into Earth's atmosphere: space dust, meteors, meteorites and occasionally asteroids, so the Earth gains more mass than it loses.

Could loss or gain of mass eventually change the Earth's orbit? No, not to any measurable extent. Just as the mass of the Earth far exceeds any mass that leaves us, it also far exceeds the amount which arrives, so it doesn't have any detectable effect on our orbit.


(How) Can one determine if an exoplanet is synchronously tidally locked or not? Mercury is asynchronously tidally locked so that all of its surface regularly sees the Sun. But an exoplanet which is synchronously locked to its star, could we tell that it is? It would have a hot side and a cold side. This should affect transit and occultation on some wavelengths, but does the symmetry of the movement degenerate any possibility to disentangle tidal locking from that?

Mercury seems to cool down during night time, and gas giants maybe distribute their temperature atmospherically. So their tidal locking might be hard to discern. But there are terrestrial exoplanets with much shorter orbital period which should have a cool side and a hot side permanently. That's my speculation. My basic curiosity is:

Is Mercury a freak or a standard in terms of tidally locked planets being asynchronously so? And by observational support, not just models.

No. Not with the limitations of our current technology. 99% of anything you've heard describing the state, makeup, nature of or appearance, of any exoplanet, is wishful thinking and conjecture. Our current technology is still rather limited. Today, we are able to come up with reasonable guesstimates of the sizes of exoplanets and their distances from their host stars, all with large degrees of uncertainty. That's it - size and distance - nothing else - at least for today (2017).

When we look at a star and see that its light has regular periodic dips, we measure the length of, the strength of, and the period of those dips (that is all we can see). Then we do a lot of math and guess at the size (mass) of the exoplanet and its distance from its host star. Even the size (both mass and diameter) of these planet-hosting stars are mere guesses. Only 16 of all the visible stars have ever been directly imaged (resolved to a disc).

However, people do love to dream. Some scientists LOVE to embellish. Some scientists like to walk around like peacocks with their plumage spread in display, hoping folks won't see their behind behind their tail feathers.
(I got to tell you - I really enjoyed thinking up that last sentence.)


Why do we continue to find a discrepancy in the Hubble Constant? After taking a look at some recent papers (examples below) that attempt to derive the value of the Hubble Constant, they all conclude with dramatically different values (speaking in context). Why is this? Each of the techniques seems sufficiently rigorous and equally well rooted in understood physics. Is this simply some form of error or is it due to some yet-to-be-discovered physics?

Papers:

A gravitational-wave standard siren measurement of the Hubble constant, H0 = 70
Planck 2015 Results, H0 = 67.8
New Parallaxes of Galactic Cepheids from Spatially Scanning theHubbleSpaceTelescope: Implications for the Hubble Constant, H0 = 73.48

I do understand that this is a hot topic right now in cosmology, however I just don't yet understand why we're having trouble with getting our measured values of the Hubble Constant to converge.

To make a long story short, the measurements from Planck and the Hubble Space Telescope disagree, and the reason behind this isn't known.

First, let's look at the values with the uncertainties. We then have three different results that are, perhaps, not as inconsistent as they originally seemed:

$70.0^{+12.0}_{-8.0}\text{ km s}^{-1}\text{ Mpc}^{-1}$ from LIGO.
$67.8\pm0.9\text{ km s}^{-1}\text{ Mpc}^{-1}$ from Planck.
$73.48\pm1.66\text{ km s}^{-1}\text{ Mpc}^{-1}$ from Hubble Space Telescope (HST) observations.

The LIGO measurement is consistent with the others, as the authors happily note in their conclusions. However, there is a discrepancy between the results from Planck and the results from the HST. This isn't a new problem; the Hubble group previously determined a value for the Hubble constant of $73.24\pm1.24\text{ km s}^{-1}\text{ Mpc}^{-1}$ in 2016. The uncertainties in the measurements describe one standard deviation; it doesn't mean that all values beyond the interval of the measurement plus or minus one standard deviation are impossible.

Now, the difference between Planck results and the 2016 HST data is $3.4\sigma$, which is a lot. The group came up with several explanations for the discrepancy:

Systematic errors in the data collection, including in the use of the Cepheid variables used for the measurements.
Small-scale variations in the Hubble constant, which appear unlikely to have propagated through the data.
Systematic errors in the Planck results.
Corrections needed to the $\Lambda$CDM model of cosmology, used by Planck.

In the 2018 paper, the authors note that the conflict with the Planck data is even stronger - $3.7\sigma$ (and we're now getting results that say it could be $4.4\sigma$; I should check those out). They haven't been able to point a figure at any single problem that could be behind the discrepancy; they do optimistically say that it could be the sign of new physics.

The Hubble constant discrepancy with Planck lambda CDM results has been reconciled by the Quantum Mass-Energy theory (QME)= 67.76 (km/s)/Mpc ... See research paper QME Theory Universal Hubble Constant Ho = 67.76 (km/s)/Mpc Allows Full Reconciliation of Space Telescopes with Satellites Results.

Abstract:

  By applying the principles of Quantum Mass-Energy (QME) theory, the vastly different Hubble constant (Ho) values from various sources such as satellite measurements, space telescope observations, QME theory, LIGO, BOA, WMAP, and sky surveys are scientifically reconciled into a single true universal value of Ho(Universal)=67.76 (km/s)/Mpc.


What is the current accepted theory as to why Titan has retained its atmosphere? Titan (moon of Saturn) is unique in that it possesses a very thick atmosphere.  However, Titan is certainly is not the largest of the moons - Ganymede being larger.

What is the current accepted theory as to why Titan has retained its thick atmosphere?

It's not really known why only Titan has a thick atmosphere while moons like Ganymede don't.

One thought is that temperatures may have been too high (well above ~40K) in the Jovian subnebula of our solar system due to the greater gravitational potential energy release, mass, and proximity to the Sun, greatly reducing the $NH_{3}$-hydrate inventory accreted by both Callisto and Ganymede. The resulting $N_{2}$ atmospheres might have been too thin to survive the atmospheric erosion effects that Titan has withstood.

EDIT: some links:

http://perso.utinam.cnrs.fr/~mousis/papier3.pdf (though it's from 2002).

http://adsabs.harvard.edu/abs/2012ttt..work...25N (this paper critiques the previous idea of a $CH_{4}$ supported atmosphere)

I would also think the NASA/JPL Cassini pages would have any up-to-date info in regards to Titan.

It has retained it's atmosphere because there is enough water (or ice) to keep it still. That's also how Earth can retain it's atmosphere. If it is a planet or moon like Venus, then the clouds and the greenhouse effect would block the atmosphere, and therefore retain the atmosphere.


What does it mean 'black holes is created when center of a very massive star collapses in upon itself'? I read many times that black holes is created when center of a very massive star collapses in upon itself but what does it mean ? Does weight of center of massive star increase so much that itself can't bear weight or I can say space itself cant bear weight of such center of massive star that it collapse. I tried to read answer on google but want to know in layman language as I am not astro physics researcher or student. Thanks

This Wikipedia page on Stellar Evolution is a good starting point with lots of links to more info.

A star is a huge sphere of mostly hydrogen.

Like any large mass it has a gravitational field and that field acts on star itself.

So it would always want to collapse to as dense a sphere as possible under it's own gravity.

Now as the density increases, particularly at the center, the pressure also increases and so does the temperature.  When these are high enough the core of the star will be able to fuse hydrogen (and later other materials) into heavier elements.  Doing so releases huge amounts of energy and this is what creates all that light we see.

But the energy released also prevents the star from collapsing more.

And it can stay that way (more or less) for billions of years.  It is this energy produced by fusion at the core that balances the gravitational force (weight) of the rest of the star trying to crush the core.  So fusion that generates more energy than it requires to happen is essential to keep a star "alive".

After a long time, the hydrogen at the core runs out and the star starts to burn heavier elements at the core (helium, carbon, etc. ).  It does that until those run out and (roughly) we reach elements like Iron remaining.  At that point fusing those elements no longer generates enough energy to stop the star collapsing and it starts to collapse again, but very, very rapidly.

Exactly what happens next depends mostly on the mass of the star.  It can end up as a white dwarf, neutron star or black hole.

But the final collapse is a very rapid process and quite violent.

Once the useful fuel at the core runs out, the stars dies a spectacular death one way or another.

What is a black hole (in simple terms) ?

Gravity is a result of the way mass (or even energy) can distort space.

Beyond a certain point this distortion becomes so great that everything on one side becomes trapped inside a "surface" called the event horizon.  An event horizon is what a black hole actually is - no event horizon, no black hole.

Within the black hole, space is distorted in a way that's basically impossible to describe without some mathematics.  The effect is that there is no direction you can move in and no speed you can travel at which will not lead you to the center of the black hole - the singularity.  Don't even try and visualize it, it just isn't a common sense idea.

This even works with light.

Now an event horizon is not a surface you can feel.  You can actually travel across an event horizon (but only inward, not outward !).  So this change in the way space-time is distorted is not something that is sudden or obvious to the person doing the falling.

Back to our star collapsing.

As it collapses (without energy from fusion to stop that), it becomes denser and denser.

Now if it is large enough it will collapse to a black hole, because the core will become so dense from the collapse that it will be enough to form an event horizon.

Our Sun is not large enough to do that.

A star is always a balance of gravity trying to collapse it inward and nuclear fusion trying to blow it up. Usually, the forces are in balance, as the gravity causes more nuclear fusion from hydrogen all the way to iron. However, iron takes an exponentially higher amount of energy to fuse, so gravity wins and crunches the star inward. This weight forms the black hole and also generates a final burst of nuclear fusion, causing the burst of light that people see, and also forming the elements beyond iron. Hope that this answer helps!


Compton scattering of high energy radiation I was reading up on compton radiation, and a question came up which I could not quite answer.

Compton scattering is known to be less efficient at high energies, due to the fact that the Klein-Nishina cross-section falls off at the high energy end. This means that the Universe is more transparent to Compton radiation at high energies.

What other processes could stop high energy photons from propogating through the Universe (i.e., a process that preferentially stops photons at the high-energy end)?

The answer is pair production. Once photon energies exceed 1.02 MeV it is possible to spontaneously create an electron-positron pair in the presence of an atomic nucleus (to conserve momentum). In general for high energy photon interactions with matter you need to consider the photoelectric effect, Compton scattering and pair production. The former is more important at lower energies, the latter at higher energies.

The picture below shows the mass attenuation coefficients for Aluminium and Iron as a function of energy, with the contributions of these three processed identified.

The Compton cross section starts to get reduced significantly only when the photon energy is significant compared to the electron rest energy.  So at that energy scale, for a more isotropic radiation field, you would have to start worrying about some of the photons having enough energy to combine with another photon and create an electron-positron pair.


Astromaster Celestron 130 EQ I brought a new Astromaster 130 EQ and collimated it using a laser. However, whenever I try to observe moon I get a perfect picture but with other planets I see only a bright glowing ball which increases and decreases in size as I adjust the focusing knob. Also, I see secondary mirror shadow in between that bright glowing ball. Kindly help me where I have gone wrong. I am using all kinds of eyepieces from 20mm to 06mm. I have a Barlow lens 2x also.

Subject Size
In angular dimensions, planets are tiny compared to the Moon.  The Moon is roughly 30 arc-minutes (about 1/2° ... varying slightly from apogee to perigee).  In real life they are, of course, MUCH larger than the Moon... but they are also MUCH farther away so they appear much smaller.
In comparison to the Moon's approximately 30 arc-minutes size, Mars is nearing it's closest approach (making it appear larger than usual) but is a still a mere 23 arc seconds.  Saturn is 17 arc-seconds across.
This means you are trying to resolve fine detail on an extremely tiny area and this is pushing the limits of what your telescope handle.
Collimation
Other factors such as optical collimation (alignment of the optical axis of the primary and secondary mirrors).  Portable Newtonian reflectors tend to need somewhat frequent collimation.  Collimation is easy to perform and will improve optical quality.  Also, if the temperature of the telescope mirror has not had time to acclimate to the ambient temperature, the mirror itself will reduce the quality of what you can see (e.g. if a telescope were stored in a warm house but taken outside on a cool evening).  Smaller telescopes will acclimate faster because the mirror has less mass.
Seeing Conditions
Seeing conditions will also have a large factor.  Weather fronts such as cold fronts, warm fronts, or the presence of the jet stream will impact seeing quality.  There is little you can do about this.  (Large telescopes can use 'Adaptive Optics' but this isn't a common thing in amateur astronomy.)
The best analogy I've come across to describe seeing conditions is to imagine placing a coin in the bottom of a pool or fountain.  If the surface of the water is perfectly flat -- no waves -- then you can easily see the coin.  With magnification you can see details on the coin.  But if a second person starts tapping the water surface to create waves... you can see the coin ... but not details.  The coin is optically distorted by the waves.  The atmosphere does the same thing when we use telescopes.  It need not be physical "waves" but can be things such as mixing warm & cool air in the upper atmosphere.
Infinity Focus
Everything in space is effectively focused to "infinity" for your telescope.  If anything is focused, then everything is focused.  This means if it makes things easier for you, focus your instrument on the moon and achieve the sharpest focus possible.  Then re-point the telescope to a planet and should not need to re-focus the telescope.  You will likely want to do this using your 6mm eyepiece (because changing eyepieces usually requires re-focusing the instrument.  There is such a thing as "parfocal" eyepieces -- which all share the same focus and you can swap eyepieces without refocusing.  Most eyepieces are not parfocal with other eyepieces.
Maximum useful magnification
There are some formulas for working out the maximum useful magnification of an instrument.
One such (rather simplistic) approximation is that the maximum magnification is 2x the aperture when measured in millimeters.  As you have a 130mm aperture, this would imply that 260x is the maximum magnification.  HOWEVER (huge caveats) this assumes flawless optics in flawless seeing conditions -- which almost never happens.  A more practical limit is roughly 1x of your aperture.  This becomes subjective as to when you decide that the magnification is making things worse instead of better.
It turns out mathematically... that the focal length of an eyepiece needed to provide magnification that matches 1x of your telescope's focal length in millimeters... will always be the same as the focal ratio of the telescope.
Since you have an f/5 telescope, a 5mm eyepiece will provide 130x magnification.
Both your 20mm and 6mm are fine and should not over-magnify the image ... you can use a 2x barlow with the 20mm, but you will not likely enjoy the views of the 6mm eyepiece combined with the 2x barlow.
There are more technical ways to discuss maximum focus such as the very simple Dawes' Limit or using Rayleigh criterion.  These methods discuss maximum resolving power of an instrument based on aperture and wavelength of the light.  But even these methods have a low-bar.  That is to say... their standard is whether two points of light with some amount of angular separation could be resolved as two points of light and not just one merged point of light.  They wont look sharp ... they'll just be blurry points that have noticeable separation.
One conclusion that can easily be drawn by looking at the formulae for maximum resolving power of an instrument... is that the bigger the aperture of the instrument, the greater the detail it will be able to resolve (which is why professional astronomers want to strive for larger and larger telescopes).
On a personal note, my first telescope had a 3.5" aperture.  I could see that Saturn had "rings" and I could see that Jupiter had "bands".  Then I got a 5" telescope... and on nights with better seeing, I could just barely make out that the ring system on Saturn had a dark band in it (the Cassini division) and that Jupiters belts sometimes had texture.  Over the years I would eventually get larger telescopes and, unsurprisingly, larger aperture telescopes allowed for viewing greater amounts of detail.  Have a reasonable expectation as to what you should be able to resolve with a 130mm (roughly 5") instrument.
Atmospheric Dispersion
When looking at anything low in the sky you may notice some "color fringing".  If a planet has a slightly blue edge on one side and a slight red edge on the opposite site, then this is typically a result of the atmosphere acting like a prism ... it's separation the light spectrum because the object is very low in the sky.  I live at fairly northern latitudes and this is true of observing outer planets during the summer months (when Earth's northern pole is tilted away from the outer solar system and toward the Sun ... from my latitude.  But southern hemisphere observer's are tilted toward the outer solar system, so they see these same planets very high in the sky where atmospheric dispersion doesn't impact them as much.  Basically outer planets appear high in the sky when observed during your hemisphere's "winter" season and low in the sky when observed during your hemisphere's "summer" season.
This dispersion will tend to blur the surface details of any planet.  It isn't really the fault of your instrument (there is a device called an Atmospheric Dispersion Corrector which is basically an adjustable prism and it is used to reverse the dispersion created by the atmosphere).

Actually you only think the Moon appears to be perfect in your eyepiece. Take a good look at Lunar images made from above the atmosphere.
Campbell gave you an very good explanation for what amounts to eyepiece viewing with any telescope is disappointing. This will of course get all the traditionalists up in arms but you have the ability to make up your own mind.
You have done nothing wrong, you have only discovered a seldom discussed truth about amateur astronomy. To me the view is all that matters. I have dumped all my traditional eyepieces into a junk box and exclusively view with electronic eyepieces.
No doubt someone will cry foul and say this answer has nothing to do with the question but your question indicates you want to see better with your telescope and that is what I am responding to.
By the way I started viewing with an eyepiece over sixty years ago and acquired all the skills many claim make them the best observers. I just never let the experianse get to my head. I used some very good equipment and but did not like the view because I always knew some sort of technology would come along and give me the view I wanted, and it has.
Now watch the fur fly. If you find any humor in the responses to my post, do a Google on Vogon :)


Is there a star over my head? Say I'm standing up straight, and I draw a straight line from my core through the top of my head (perpendicular to the ground). What is the probability that that line intersects with a star?

EDIT: I'm not trying to exclude any stars. This should include stars which we've observed and stars which we haven't yet observed but can predict due to other things we've determined (like the overall star density of the universe). Also it should include all stars regardless of naked eye magnitude limit.

Summary

There's a 1 in 500 billion chance you're standing under a star outside the Milky Way, a 1 in 3.3 billion chance you're standing under a Milky Way star, and a 1 in 184 thousand chance you're standing under the Sun right now.

Big, fat, stinking, Warning! I did my best to keep my math straight, but this is all stuff I just came up with. I make no guarantees it's completely accurate, but the numbers seem to pass the sanity check so I think we're good.

Caveat the First: The numbers for stars other than the Sun are based on data with a great deal of uncertainty, such as the number of stars in the universe and the average size of a star. The numbers above could easily be off by a factor of 10 in either direction, and are merely intended to give a rough idea of how empty space is.

Caveat the Second: The numbers for the Sun and the Milky Way are based on the assumption that you are standing (or floating) at a random point on Earth. Anyone outside the tropics will never have the Sun over their head. People in the northern hemisphere are more likely to have Milky Way stars over their head, with the best odds being people near 36.8° N, because at that latitude straight up passes through the galactic center once a day.26

Note: You can mostly ignore everything in this answer and just look up the solid angle of the Sun to get the same result. All the other stars are really far away and very spread out. The difference in solid angle subtended is five thousandths of a percent more when we add the rest of the universe to the Sun.

Background

Let's try to get a somewhat realistic, hard number. To do that, we'll need some assumptions.

As pointed out in Michael Walsby's answer1, if the universe is infinite (and homogeneous2), there is only an infinitesimal chance of there not being a star overhead, which normal math treats as exactly zero chance. So let's presume the universe is finite.

Presumptions

Specifically, let's presume the universe only consists of the observable universe. (Look up the expansion of the universe3 for further information.)
Further, let's presume the contents of the observable universe are measured at their current (presumed) positions, not the position they appear to be. (If we see light from a star from 400 million years after the universe began, we would measure it as being about 13.5 billion light years away, but we calculate that it's likely closer to 45 billion light years away due to expansion.)
We'll take the number of stars in the observable universe to be $10^{24}$. A 2013 estimate4 was $10^{21}$, a 2014 estimate5 was $10^{23}$, and a 2017 estimate6 was $10^{24}$, with each article expecting the estimate to increase as we get better telescopes over time. So we'll take the highest value and use it.
We'll take the size of the observable universe7 to be $8.8\cdot10^{26} \text{m (diameter)}$, giving a surface area8 of $2.433\cdot10^{54} \text{m}^2$ 9, and a volume10 of $3.568\cdot10^{80} \text{m}^3$ 11.
We'll take the average size of a star to be the size of the Sun, $1.4\cdot10^{9}\text{m (diameter)}$ 12. (I can't find any sources for average star size, just that the Sun is an average star.)

Model

From here, we're going to cheat a bit. Realistically, we should model each galaxy separately. But we're just going to pretend the entire universe is perfectly uniform (this is true enough as we get farther away from Earth in the grand scheme of the cosmos). Further, we're going to start counting far enough out to ignore the Milky Way and Sun entirely, then add them back in later with different calculations.

Given the above presumptions, we can easily calculate the stellar density of the observable universe to be $\delta = \frac{10^{24}\text{stars}}{3.568\cdot10^{80} m^3} = 2.803\cdot10^{-57} \frac{\text{stars}}{\text{m}^3}$ 13.

Next, we need to calculate the solid angle14 subtended by a star. The solid angle of a sphere is given by $\Omega=2\pi\left(1-\frac{\sqrt{d^2-r^2}}{d}\right)\text{ sr}$ 15, where $\Omega$ is the solid angle in steradians16 (sr), $d$ is the distance to the sphere and $r$ is the radius of the sphere. Using $D$ as the diameter, that converts to $\Omega=2\pi\left(1-\frac{\sqrt{d^2-\left(\frac{D}{2}\right)^2}}{d}\right)\text{ sr}$. Given the average diameter presumed above ($1.4\cdot10^{9}\text{m}$), this gives an average solid angle of $\Omega=2\pi\left(1-\frac{\sqrt{d^2-4.9\cdot10^{17}\text{m}^2}}{d}\right)\text{ sr}$ 17.

At this point, we could set up a proper integral, but my calculus is rather rusty, and not very sharp to begin with. So I'm going to approximate the answer using a series of concentric shells, each having a thickness of $10^{22}\text{m}$ (about a million light years). We'll put our first shell $10^{22}\text{m}$ away, then work our way out from there.

We'll calculate the total solid angle of each shell, then add all the shells together to get the solid angle subtended by the entire observable universe.

The last problem to fix here is that of overlap. Some stars in the farther shells will overlap stars in the nearby shells, causing us to overestimate the total coverage. So we'll calculate the probability of any given star overlapping and modify the result from there.

We'll ignore any overlap within a given shell, modeling as if every star in a shell is at a fixed distance, evenly distributed throughout the shell.

Probability of Overlap

For a given star to overlap closer stars, it needs to be at a position already covered by the closer stars. For our purposes, we'll treat overlaps as binary: either the star is totally overlapped, or not overlapped at all.

The probability will be given by the amount of solid angle already subtended by previous shells divided by the total solid angle in the sky ($4\pi\text{ sr}$).

Let's call the probability a given star, $i$, is overlapped $P_i$, the solid angle subtended by that star $\Omega_i$, and the number of stars $n$. The amount of non-overlapped solid angle subtended by a given shell, $k$, is then $\Omega_{kT}=(1-P_1)\Omega_1+(1-P_2)\Omega_2+\ldots+(1-P_n)\Omega_n\frac{\text{ sr}}{star}$. Since we've said stars in a shell don't overlap each other, $P_i$ is the same for all $i$ in a given shell, allowing us to simplify the above equation to $\Omega_{kT}=(1-P_k)(\Omega_1+\Omega_2+\ldots+\Omega_n)\frac{\text{ sr}}{star}$, where $P_k$ is the probability of overlap for shell $k$. Since we're treating all the stars as having the same, average size, this simplifies even further to $\Omega_{kT}=(1-P_k)\Omega_k n\frac{\text{ sr}}{star}$, where $\Omega_k$ is the solid angle of a star in shell $k$.

Calculating Solid Angle

The number of stars in a shell is given by the volume of the shell times the stellar density of said shell. For far away shells, we can treat the volume of the shell as being its surface area times its thickness. $V_\text{shell}=4\pi d^2 t$, where $d$ is the distance to the shell and $t$ is its thickness. Using $\delta$ as the stellar density, the number of stars is simply $n=\delta V_\text{shell}=\delta 4\pi d^2 t$.

From here, we can use the calculation for solid angle of a shell (from Probability of Overlap, above) to get $\Omega_{kT}=(1-P_k)\Omega_k \delta 4\pi d^2 t\frac{\text{ sr}}{star}$.

Note that $P_k$ is given by the partial sum of solid angle for all previous shells divided by total solid angle. And $\Omega_k$ is given by $\Omega_k=2\pi\left(1-\frac{\sqrt{d_k^2-4.9\cdot10^{17}\text{m}^2}}{d_k}\right)\frac{\text{ sr}}{star}$ (from Model, above).

This gives us $\Omega_{kT}=\left(1-\frac{\Omega_{(k-1)T}}{4\pi}\right)2\pi\left(1-\frac{\sqrt{d_k^2-4.9\cdot10^{17}\text{m}^2}}{d_k}\right) \delta 4\pi d^2 t\text{ sr}$. Given that each shell is $10^{22}\text{m}$ away, we can substitute $d_k$ with $k 10^{22}\text{m}$. Likewise, $t$ can be substituted with $10^{22}\text{m}$. And we already calculated $\delta=2.803\cdot10^{-57} \frac{\text{stars}}{\text{m}^3}$ (from Model, above).

This gives us
$\Omega_{kT}=\left(1-\frac{\Omega_{(k-1)T}}{4\pi}\right)2\pi\left(1-\frac{\sqrt{(k 10^{22}\text{m})^2-4.9\cdot10^{17}\text{m}^2}}{k 10^{22}\text{m}}\right) 2.803\cdot10^{-57}\frac{\text{stars}}{\text{m}^3} 4\pi (k 10^{22}\text{m})^2 10^{22}\text{m}\frac{\text{ sr}}{star}$

$=\left(1-\frac{\Omega_{(k-1)T}}{4\pi}\right)\left(1-\frac{\sqrt{k^2 10^{44}-4.9\cdot10^{17}}}{k 10^{22}}\right) 2.803\cdot10^{-57} 8\pi^2 k^2 10^{66}\text{ sr}$

$=\left(1-\frac{\Omega_{(k-1)T}}{4\pi}\right)\,2.213\cdot 10^{11}\,k^2\left(1-\frac{\sqrt{k^2 10^{44}-4.9\cdot10^{17}}}{k 10^{22}}\right)\text{ sr}$

From here, we can just plug the numbers into a calculation program.

$\Omega_{T}=\sum_{k=1}^{k_\text{max}} \Omega_{kT}$

Where $k_\text{max}$ is just the radius of the observable universe divided by the thickness of a given shell. Thus $k_\text{max}$$=\frac{4.4\cdot 10^{26} \text{m}}{10^{22} \text{m}}$$=4.4\cdot 10^4$$=44000$

$\Omega_{T}=\sum_{k=1}^{44000} \Omega_{kT}$

Results

Because of the large numbers involved, it's difficult to just run this in a program. I resorted to writing a custom C++ program using the ttmath library18 for large numbers. The result was $2.386\cdot 10^{-11}\text{ sr}$, or $1.898\cdot 10^{-12}$ of the entire sky. Conversely, there's about a 1 in 500 billion chance you're standing under a star right now.

Note that we ignored the Milky Way and the Sun for this.

The C++ program can be found at PasteBin25. You'll have to get ttmath working properly. I added some instructions to the top of the C++ code to get you started if you care to make it work. It's not elegant or anything, just enough to function.

The Sun

WolframAlpha helpfully informed me the Sun has a solid angle of about $6.8\cdot 10^{-5}\text{ sr}$, or about 2.8 million times more than all the stars in the universe combined. The solid angle formula above gives the same answer18 if we provide the Sun's 150 gigameter distance and 0.7 gigameter radius.

The Milky Way

We could get an approximation for the Milky Way by taking its size and density and doing the same calculations as above, except on a smaller scale. However, the galaxy is very flat, so the odds greatly depend on whether you happen to be standing in the galactic plane or not. Also, we're off to one side, so there are far more stars towards the galactic center than away.

If we approximate the galaxy as a cylinder with a radius of $5\cdot 10^{20}\text{ m}$ (about 52000 light years) and a height of $2\cdot 10^{16}\text{ m}$ (about 2 light years), we get a volume of $1.571\cdot 10^{58}\text{ m}^3$ 20.

Current estimates of the galaxy's radius are closer to 100000 light years21 22, but I'm presuming the vast majority of stars are a lot closer than that.

There are estimated to be 100 to 400 billion stars in the Milky Way21. Let's pick 200 billion for our purposes. This puts the density of the Milky Way at $\delta = \frac{200\cdot10^{9}\text{stars}}{1.571\cdot 10^{58}\text{ m}^3} = 1.273\cdot10^{-47} \frac{\text{stars}}{\text{m}^3}$ 22, or about 4.5 billion times denser than the universe at large.

This time, we'll take shells $10^{17}\text{ m}$ thick (about 10 light years) and go out from there. But we need to re-organize the math into a spherical form, so we'll presume the galaxy has the same volume, but is a sphere. This gives it a radius of $1.554\cdot 10^{19}\text{ m}$ 24, or 155.4 shells. We'll round to 155 shells.

$\Omega_{T}=\sum_{k=1}^{155} \Omega_{kT}$

Using our formula from above (Calculating Solid Angle), we can start substituting numbers.

$\Omega_{kT}=\left(1-\frac{\Omega_{(k-1)T}}{4\pi}\right)2\pi\left(1-\frac{\sqrt{d_k^2-4.9\cdot10^{17}\text{m}^2}}{d_k}\right) \delta 4\pi d^2 t\frac{\text{sr}}{\text{star}}$

$=\left(1-\frac{\Omega_{(k-1)T}}{4\pi}\right)2\pi\left(1-\frac{\sqrt{(k\cdot 10^{17}\text{ m})^2-4.9\cdot10^{17}\text{ m}^2}}{k\cdot 10^{17}\text{ m}}\right) 1.273\cdot10^{-47} \frac{\text{stars}}{\text{m}^3} 4\pi (k\cdot 10^{17}\text{ m})^2 10^{17}\text{ m}\frac{\text{sr}}{\text{star}}$

$=\left(1-\frac{\Omega_{(k-1)T}}{4\pi}\right)\left(1-\frac{\sqrt{k^2\cdot 10^{34}\text{ m}^2-4.9\cdot10^{17}\text{ m}^2}}{k 10^{17}\text{ m}}\right) 1.273\cdot 10^{-47} \frac{\text{stars}}{\text{m}^3} 8\pi^2 k^2 10^{51}\text{ m}^3\frac{\text{sr}}{\text{star}}$

$=\left(1-\frac{\Omega_{(k-1)T}}{4\pi}\right)\cdot\,1.005\cdot 10^6\,k^2\,\left(1-\frac{\sqrt{k^2\cdot 10^{34}-4.9\cdot10^{17}}}{k 10^{17}}\right)\text{ sr}$

Plugging this into the program gives $3.816\cdot 10^{-9}\text{ sr}$, which is $3.037\cdot 10^{-10}$ of the total sky. The odds you're standing under a star in the Milky Way are about 1 in 3.3 billion.

Solid Angle Totals

Solid angle is:

Sun, $6.8\cdot 10^{-5}\text{ sr}$
Milky Way, $3.816\cdot 10^{-9}\text{ sr}$
Universe, $2.386\cdot 10^{-11}\text{ sr}$
Total, $6.800384\cdot 10^{-5}\text{ sr}$ (the extra digits are basically meaningless, adding about five thousandths of a percent to the Sun's solid angle)
Milky Way plus Universe, $3.840\cdot 10^{-9}\text{ sr}$ (about 0.6% more than just the Milky Way)

References

1 Michael Walsby's answer to this question, is there a star over my head?. https://astronomy.stackexchange.com/a/33294/10678
2 A Wikipedia article, Cosmological principle. https://en.wikipedia.org/wiki/Cosmological_principle
3 A Wikipedia article, Expansion of the universe. https://en.wikipedia.org/wiki/Expansion_of_the_universe
4 A UCSB ScienceLine quest, About how many stars are in space?, from 2013. https://scienceline.ucsb.edu/getkey.php?key=3775
5 A Sky and Telescope article, How Many Stars are There in the Universe?, from 2014. https://www.skyandtelescope.com/astronomy-resources/how-many-stars-are-there/
6 A Space.com article, How Many Stars Are In The Universe?, from 2017. https://www.space.com/26078-how-many-stars-are-there.html
7 A Wikipedia article, Observable universe. https://en.wikipedia.org/wiki/Observable_universe
8 A Wikipedia article, Sphere, section Enclosed volume. https://en.wikipedia.org/wiki/Sphere#Enclosed_volume
9 A WolframAlpha calculation, surface area of a sphere, diameter 8.8*10^26 m. https://www.wolframalpha.com/input/?i=surface+area+of+a+sphere%2C+diameter+8.8*10%5E26+m
10 A Wikipedia article, Sphere, section Surface area. https://en.wikipedia.org/wiki/Sphere#Surface_area
11 A WolframAlpha calculation, volume of a sphere, diameter 8.8*10^26 m. https://www.wolframalpha.com/input/?i=volume+of+a+sphere%2C+diameter+8.8*10%5E26+m
12 A nineplanets.org article, The Sun. https://nineplanets.org/sol.html
13 A WolframAlpha calculation, (10^24 stars) / (3.568⋅10^80 m^3). https://www.wolframalpha.com/input/?i=%2810%5E24+stars%29+%2F+%283.568%E2%8B%8510%5E80+m%5E3%29
14 A Wikipedia article, Solid angle. https://en.wikipedia.org/wiki/Solid_angle
15 Harish Chandra Rajpoot's answer to a geometry.se question, Calculating Solid angle for a sphere in space. https://math.stackexchange.com/a/1264753/265963
16 A Wikipedia article, Steradian. https://en.wikipedia.org/wiki/Steradian
17  A WolframAlpha calculation, 2*pi*(1-sqrt(d^2-(1.4*10^9 m/2)^2)/d). https://www.wolframalpha.com/input/?i=2*pi*%281-sqrt%28d%5E2-%281.4*10%5E9+m%2F2%29%5E2%29%2Fd%29
18 Website for ttmath. https://www.ttmath.org/
19 A WolframAlpha calculation, 2*pi*(1 - sqrt(d^2 - r^2)/d), where d = 150 billion, r=0.7 billion. https://www.wolframalpha.com/input/?i=2*pi*%281+-+sqrt%28d%5E2+-+r%5E2%29%2Fd%29%2C+where+d+%3D+150+billion%2C+r%3D0.7+billion
20 A WolframAlpha calculation, pi * (5*10^20 m)^2 * (2*10^16 m). https://www.wolframalpha.com/input/?i=pi+*+%285*10%5E20+m%29%5E2+*+%282*10%5E16+m%29
21 A Wikipedia article, Milky Way. https://en.wikipedia.org/wiki/Milky_Way
22 A Space.com article from 2018, It Would Take 200,000 Years at Light Speed to Cross the Milky Way. https://www.space.com/41047-milky-way-galaxy-size-bigger-than-thought.html
23 A WolframAlpha calculation, (200*10^9 stars) / (1.571*10^58 m^3). https://www.wolframalpha.com/input/?i=(200*10^9+stars)+%2F+(1.571*10^58+m^3)
24 A WolframAlpha calculation, solve for r:  (4/3)*pi*r^3 = 1.571*10^58 m^3. https://www.wolframalpha.com/input/?i=solve+for+r%3A++%284%2F3%29*pi*r%5E3+%3D+1.571*10%5E58+m%5E3
25 My C++ program code on PasteBin. https://pastebin.com/XZTzeRpG
26 A Physics Forums post, Orientation of the Earth, Sun and Solar System in the Milky Way. Specifically, Figure 1, showing angles of 60.2° for the Sun, and 23.4° less than that for Earth. https://www.physicsforums.com/threads/orientation-of-the-earth-sun-and-solar-system-in-the-milky-way.888643/

According to Olbers, of paradox fame, if the universe is infinite a line of sight in any direction should eventually reach a star. Why then was the night sky so dark when in theory it should be bright as day? Leaving that particular question aside, we have no proof that the universe is infinite, but it is sufficiently large that a line in any direction should sooner or later reach the surface of a star. Whether the line in question would have to travel only tens of light years to reach the star or many billions depends on where you are standing and at what particular moment you choose to draw the line. If you happened to be on the equator at the right time of year and the right time of day, the line might only have to travel a little over eight light minutes to reach a star. In the universe, as opposed to on paper, you cannot have really straight lines, only an approximation, because the universe is curved and any line defining device, a laser beam perhaps, would be acted on by a force and bent.


Why is the ring of light around the M87 black hole bigger than the photon sphere? Today we know that a black hole represents a region of space where the gravitational forces are very intense, this consequence of a concentration of matter in a very small region of space. On April 10 of this year, scientists from the EHT project (event Horizont Telescope) presented to the world the image that I believe everyone has seen. The first image of a black hole in the galaxy M87. We know that what we are really seeing is the "shadow" of the black hole.

But the "photon sphere", where light can orbit a black hole, is actually at $1.5R_s$, (Schwarzschild radius), whereas the observed ring is a factor of 2 wider than this. Why we not see a ring of light at the radius of the photon sphere?

The apparent radius of something residing in a Schwarzschild metric, when viewed from infinity is given by
$$ R_{\rm obs} = R \left(1 - \frac{R_s}{R}\right)^{-1/2}\ ,$$
where $R_s$ is the Schwarzschild radius $2GM/c^2$.

This enlargement is due to gravitational lensing and the formula is correct down to the "photon sphere" at $R =1.5 R_s$.

Most of the light in the EHT image comes from the photon sphere. It is therefore observed to come from a radius
$$ R_{\rm obs} =\frac{3R_s}{2}\left(1 - \frac{2}{3}\right)^{-1/2} = \frac{\sqrt{27}}{2}R_s\ .$$

There are small (<10%) corrections to this for a spinning black hole governed by the Kerr metric.

A sketch which illustrates why gravitational lensing leads to an enlarged image can be found in Pela's answer to a related question.

If the light is orbiting the black hole, then it isn't escaping to our telescopes!

What matters is how close to the black hole the plasma of the accretion disk goes, since that's what produces the light. There are lots of papers on this, and the numbers are usually around 5 Schwarzchild radii, together with effects like gravitational redshift relativistic beaming and so on, and the geometry of the paths followed by light, this leads to the image we see.


What natural mechanisms could lead to the unlikely case of the same rogue asteroid or planet passing through our solar system twice? I find he following answer to the question What's the soonest Oumuamua could return? unsatisfying and oversimplifying and the last five words "Therefore it will never return." unsupported at a minimum.

  Oumuamua as an object is remarkable, because it has a positive net energy, which means it is not bound to the gravitational well of our sun. Therefore it will never return.

So I'd like to ask: *What natural mechanisms could lead to the unlikely case of a rogue asteroid or planet passing through our solar system twice?"

Stick to accepted scientific facts and theories
Let's give an ample time limit of 10 billion years between visits.
For the purposes of this question, the solar system includes the Kuiper belt (50AU) but not the Oort cloud
Don't include intervention by artificial means or intervention (divine or otherwise)

Space is big. Really big. You just won't believe how vastly, hugely, mind-bogglingly big it is (Douglas Adams)

The wikipedia page notes that 'Oumuamua "has circulated the Milky Way several times" so it was unlikely to have formed around any nearby star. While it would take 'Oumuamua  600,000 years to travel here from a local star, the time before it can be expected to have another stellar encounter would be much greater, hundreds of millions of years. And the next star that it meets is unlikely to be a currently local star, it could be almost any star in the galaxy.

Once an interstellar object has left the solar system it will travel in a roughly straight line, (slowly curving with the general gravity of the galaxy). For it to return to the Earth it would have to have a gravitational encounter with another star (or more likely a series of encounters) that send it back towards the solar system. Even given a vast mind-boggling long period of time like 10 billion years the chance of this happening is minute.

Such an object will probably orbit the galaxy several times between stellar encounters, perhaps it has an encounter every 250 million years. And it is equally likely to encounter any of the disc stars (conservatively 50 billion stars). This means that the mean time between encounters of the sun is on the order of ten quintillion years.

It isn't impossible.  It is quite easy to design a free-return flight to the moon.   There is no reason why a free-return flight to another star can't be done. However, space is big "therefore ʻOumuamua will never return".

tl;dr: It's a hard problem. While this answer says

  However, space is big "therefore ʻOumuamua will never return".

I say never say never!

  What natural mechanisms could lead to the unlikely case of the same rogue asteroid or planet passing through our solar system twice?

I can think of three mechanisms.

1. Stellar and solar-system pinball

Under normal situations an asteroid or planet could not swing around a massive object, make a U-turn, and come straight back nearly along it's same path after a 180° deflection. However deflections by two or more large bodies can do it. Those two bodies could be

Two unrelated stars: This would make a triangular path; Sun, star-1, star-2, Sun again.
A double star: Three-body unbound orbits are crazy and chaotic and just about anything can happen including a 180°U-turn. This is my favorite since it only requires one encounter and star systems of two or higher are pretty common.
A star and its planet: Stars with Jupiter-size (and larger) planets are pretty common, and a hyperbolic orbit around a star doing much/most of the bend with the planet doing the rest of the 180°U-turn is possible.
other variations on this theme.

2. General Relativity (or Neutron Stars)

A black hole: As explained in answer(s) to Could a trajectory around a large mass ever deflect by more than 180 degrees due to general relativistic effects? "...there is no limit on how many "laps" a hyperbolic orbit can make before it returns to infinity." That's in the limiting case of a zero-mass object. For large mass ratios such as an asteroid or planet versus a black hole, the maxiumum number of turns an unbound orbit can make is about 0.41 times the mass ratio, which would be greater than a million for an Earth-sized planet and a vanilla stellar-collapse black hole.
A neutron star: I am not certain yet if the GR effects due to a neutron star are strong enough to produce a 180°U-turn without a destructively close approach; I'm currently working on this...

3. Galactic roller-coaster

The solar system orbits the galactic center at about 220 km/s with a period of about 250 million years. That speed is an order of magnitude or possibly two faster than the orbital speeds of asteroids and planets within solar systems. Jupiter orbits at 13 km/s and (486958) 2014 MU69 (Ultima Thule) moves at 4.6 km/s. So no mater which direction our rogue friend leaves the solar system, it's going to be moving around the galactic center with a nearly parallel orbit to that of the solar system, and it's going to be doing a complicated dance above and below the galactic plane as our solar system is doing, due to the restoring force of the galactic disks's gravity. This can bring us and our rogue friend back together in a way that would at first surprise us both until we remember how stars move in the disk.

Overall probability?

For all of these the probability is small, but the single collision options (double stars, solar systems, black holes, neutron stars) may make the probability better than the two unrelated star sequential scattering scenarios, and the galactic plane oscillations don't require a near approach to any other body at all.

So I think the statement:

  ...Therefore it will never return.

could be considered cavalier and possibly scientifically unsupported.

It's a hard problem, so never say never!


Will the Heat Death scenario ever occur or just the Big Freeze? According to the Wikipedia page regarding the ultimate fate of the universe, heat death is a different scenario than the big freeze.

It states that in the big freeze, temperature will asymptotically approach absolute zero. However, the following section about heat death states that heat death may only occur if temperature reaches an eventual temperature minimum.

If temperature never truly reaches absolute zero (the temperature minimum), can heat death even occur in the first place?

The Wikipedia article is somewhat confused on this point (at least one of the references does not support the sentence it is supposed to support).

The classic concept of the heat death of the universe was that eventually it will run out of free energy and there will hence be no possibility of any work. A more modern view of heat death is an approach to maximum entropy where the system will be in equilibrium. This does not rule out random fluctuations bringing it away from equilibrium, but they are so rare it does not matter.

Note that it is entirely possible to have a heat death at any temperature.

The big freeze would happen if expansion of the universe led to temperatures decreasing towards absolute zero. However, if there is accelerating expansion this generates a constant finite temperature horizon radiation, about $10^{-29}$ K. This means that the big freeze will never happen.

Had there not been horizon radiation it would in principle have been possible to avoid heat death by storing some background heat in a perfectly reflective box, waiting until the universe become colder, and use it to perform work. This can be repeated indefinitely. (Not really: the timescales quickly become long enough that random fluctuations destroy your device, but that it a separate issue).

Due to the universe not only expanding but gaining speed, it's pretty well accepted (for now at least) that galaxies will be sitting alone, or maybe with other gravitationally bound neighbor galaxies, as the rest of the universe goes dark  all around.  Eventually our home galaxy would burn out it's last star. Thats the big freeze,  and not happening for a while... The end stages at least. Its seems like heat death is a variation to the big freeze, except expansion somehow stops for heat death? and im not sure what the universal temperature is thought to plato at, but after all thermodynamic process' end, the universe' energy will be lost to the void and entropy will reach its maximum as universal equilibrium is reached (uniform temperature and lack of activity) but tthe time it would take for the last black hole to evaporate is so long it's hard to comprehend, 10^100 or a googol years, if im not mistaken.
   heat death and the big freeze seem like different ways of explaining the same event, the former being on a longer time scale, but both seeming headed for the same fate. and wiki and google agree most of the time, but a few articles off a google search said the opposite. I think maybe you read some info from contradicting sources is all, if there is a difference in these two, its that expansion stops for heat death, allowing a marginally warmer universe than one with infinite expansion. there is no reason to think thats going to happen with our current understanding of dark energy.


Does the Sun impose back it's tidal forces onto the Earth (such as the Earth's to the Moon)? It is known that each year the Moon is moving away from the Earth due to tidal forces.

What about the Earth relative to the Sun?

Yes - the earth and sun do have tidal forces like the moon and earth. There are two main reasons this is happening.

The sun is always losing mass due to nuclear reactions, the sun is always converting a tiny amount of its mass into energy. This means the pull on the earth is weakened.
The tidal forces that happen between the earth and moon also happen between the sun and earth, except on a much smaller scale.

The sun is also losing its angular momentum at a rate of 0.00003 seconds per year

The theorized reason I saw on either the NASA or Discovery Channel for the Moon's orbit spiraling away from earth is that it's formation resulted from a collision between the Earth and an asteroid, and the resulting momentum carries it away.


Does the gravity of the planets affect the orbit of other planets in our solar system? When one planet passes near another during its trip around the sun, does their gravitational pull is strong enough to disrupt noticeably each other's orbit ?

It does - although the term 'disrupt' may be a bit too strong to describe the effect; personally, I think 'influence' would fit better.

An interesting consequence of such iterations is something called orbital resonance; after long periods of time - and remember that the current estimate for our planet's existence is 4.54 billion years - the ebb and flow of tiny gravitational pulls cause nearby celestial bodies to develop an interlocked behavior. It's a double-edged sword, though; it may de-estabilize a system, or lock it into stability.

Quoting the Wikipedia entry,

  Orbital resonances greatly enhance the mutual gravitational influence
  of the bodies, i.e., their ability to alter or constrain each other's
  orbits.

Another gravity-related effect (although, as pointed out by Dieudonné, present only on our solar system between bodies that have very close orbits like the Earth-Moon and Sun-Mercury systems) is known as Tidal locking, or captured rotation.

More about orbital resonance on this ASP Conference Series paper: Renu Malhotra, Orbital Resonances and Chaos in the Solar System.

Yes it does, especially with the gas giants.

Jupiter's magnetic field is so strong it yanks on almost everything in the Solar System. Life on Earth also relies on the Jovian magnetic field, because if Jupiter's magnetic field was not there, the gravity would feel much weaker.

After the discovery of Uranus, astronomers predicted another gas giant beyond Uranus due to Uranus being yanked on by something large, which led to the discovery of Neptune by Urbain Le Verrier and John Galle in September 1846.

The Saturnian magnetic field primarily affects Uranus and Neptune.


Difference in energy released in stellar mass black hole merger and supermassive black hole merger Super massive black holes are usually millions of times more massive than stellar mass black holes. I have read on SE and over internet that 3M☉ worth of energy is released in black hole mergers (not sure which type of merger). I have tow related questions:

What is the difference in energy released from a stellar mass black hole merger vs. a supermassive black hole merger?
How much energy will be released in the super massive black hole merger when Milky way and Andromeda will merge in few billion years?

Too long for a comment, but it's an incomplete answer.

The gravity waves detected were from a stellar mass binary black hole merger, sometimes abbreviated bbh for binary black hole.

The two black holes are thought to be about 36 and 29 solar masses with a final combined mass of about 62.  So, roughly 5% of the black hole mass turned into into gravitational wave energy.   Source.

and from Wikipedia article (which can be taken with a grain of salt)

  As the orbiting black holes give off these waves, the orbit decays,
  and the orbital period decreases. This stage is called binary black
  hole inspiral. The black holes will merge once they are close enough.
  Once merged, the single hole settles down to a stable form, via a
  stage called ringdown, where any distortion in the shape is dissipated
  as more gravitational waves.3 In the final fraction of a second the
  black holes can reach extremely high velocity, and the gravitational
  wave amplitude reaches its peak.

  The existence of stellar-mass binary black holes (and gravitational
  waves themselves) were finally confirmed when LIGO detected GW150914
  (detected September 2015, announced February 2016), a distinctive
  gravitational wave signature of two merging stellar-mass black holes
  of around 30 solar masses each, occurring about 1.3 billion light
  years away. In its final moments of spiraling inward and merging,
  GW150914 released around 3 solar masses as gravitational energy,
  peaking at a rate of 3.6×1049 watts — more than the combined power of
  all light radiated by all the stars in the observable universe put
  together.3[4][5] Supermassive binary black hole candidates have been
  found but as yet, not categorically proven

Supermassive bbh are thought to have been observed, see here and we may see one in our lifetime, not sure how often they happen.  But beyond that, I have no way of knowing how much more energetic a supermassive BBH merger would be other than to say, I would think, a whole lot, as it would have to scale upwards by some formula.

With the case of Andromeda and the Milky Way, we're talking Andromeda's 100 million solar mass black hole and the Milky way's 4-6 million solar masses.  Maybe someone here can estimate, but if the energy output is even 5%-10% of the smaller of the two, that's still a few hundred thousand solar masses of energy, an absolutely crazy energy output.  Even if it's far smaller than a few hundred thousand solar masses, it's still enormous energy.

I'm also not sure if it's a dangerous form of energy output as it basically squashes things back and forth maybe a couple times before returning them to normal.  I have no idea what the safe distance is from that kind of merger and gravity wave creation.    Interesting question though.  Fun to think about.

Super massive and stellar mass black holes are different in mass and size.. Since we cannot describe nor we can be certain in either of their mass or size .. It's completely impossible (for the time being) to answer whether or not how much difference there will be regarding their energy release and the amount it releases..Yes,the energy is high  3 solar mass .. but the energy cannot be determined regarding there are various factors which leads to the determination of Mass and energy.


Expansion of the universe speeding up I know that the expansion of the universe is speeding up, but is there an equation to measure this? Also, what is the cause of this?

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

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

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

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


Lunar phases if Earth was tidally locked to the Moon I know that the Earth will never become tidally locked to the Moon, so this is purely hypothetical.

If I understand correctly, in this hypothetical scenario an Earth day is now a lot longer at around 29 current days, and you can only ever see the Moon from half of the Earth.

So standing on the "lucky" side of the Earth you should be able to see all the lunar phases during a single night. The Moon become gradually more illuminated, peak as a full Moon and then darken as the night ends.

Is this correct?

tl;dr: An observer on the "Moon side" would see only half the phases during the fortnight-long night: from waxing half-moon to waning half-moon. They would also see the Moon during the day (early morning, late afternoon), as we currently can, but they would never see the New Moon or crescent moons at night since these would be only visible during the day.

Background
While the Earth being tidally locked to the Moon would indeed mean that the side of the Earth facing the Moon always sees the Moon, the phases of the Moon result from the angle between the Earth, the Moon and the Sun, and probably wouldn’t change significantly.

The Moon orbits around the Earth-Moon barycentre with an orbital period of 27.3 days and a synodic period (i.e. from Full Moon to Full Moon) of 29.5 days. An observer on the Moon-facing side of the Earth would certainly see the Moon go through its different phases, but it would still take 29.5 days. [For the purposes of this question I'm ignoring the increasing lunar period due to the Moon receding from the Earth.]

As the question notes, one obvious outcome of Earth becoming tidally locked to the Moon is that Earth’s “day” becomes the same as the Moon’s synodic period: 29.5 days. Daylight will last for about 15 days, baking everything other than at the poles. The 360-hour night will be freezing, even in the tropics. Local weather will be utterly different.

Another notable outcome is there will only be really tiny “solar” tides spaced about 15 days apart (due to the Earth’s slow rotation in relation to the Sun). The loss of significant tidal flows would probably have a dramatic negative effect on marine life.

The Earth tidally locked to the Moon is a nightmare scenario!

The core question

  So standing on the "lucky" side of the Earth you should be able to see
  all the lunar phases during a single night. The Moon become gradually
  more illuminated, peak as a full Moon and then darken as the night
  ends.

No: you would only see only half the phases during the fortnight-long night: from waxing half-moon to waning half-moon. You would also see the Moon during the day (early morning, late afternoon), as we currently can, but you would never see the New Moon or crescent moons at night since these would be only visible during the day.

We currently see the New Moon because when it's roughly overhead for some at midday, it's not far above the horizon just before sunrise or just after sunset for others. Since the New Moon phase currently lasts about 3-5 days (depending on how you define "new"), but the Earth rotates in just one 24-hour day, most habitable places on Earth get to see this phase every lunar month. However, in the tidally-locked-Earth scenario, the phase still lasts 3-5 days but the Earth rotates once every 29.5 days. Those viewing the Moon just after sunset will only ever see the Moon in that position above the horizon – it more or less "sits" there, apart from libration and nodal period effects – and they will miss some of the other phases because the Sun will be too bright at, say, the waning half-moon.

EDIT - Two additional things to note:

(1) Earth's phases
While the Earth becoming tidally locked to the Moon is hypothetical, the Moon is already tidally locked to the Earth, which means the above descriptions can be applied in reverse for a real observer on the Moon:

a lunar "day" (sunrise to sunrise) is 29.5 Earth days;
the Earth's position in the lunar sky is relatively constant over the lunar day, within about 15º (due to a slow but noticeable libration);
the Moon's orbit is elliptical, so over much longer periods the Earth would alternately recede and grow larger (a "Super Earth");
unlike the Earth-locked scenario, all of the Earth's phases – from Full Earth at lunar midnight to New Earth at lunar noon - would be visible, because there's no atmosphere and hence no "bright daytime sky" to block out the Earthlight.

Note also that the famous Earthrise experience of the Apollo 8 astronauts would not be available to an observer on the Moon's surface. The Earth "rose" above the limb of the Moon because the astronauts were in orbit. Nonetheless, someone located on the border of the "dark side of the Moon" would see the Earth on the lunar horizon, and libration could result in the Earth alternately setting below the horizon and rising again over a lunar day. For that observer, however, earthrise and earthset occur at roughly the same point on the horizon!

(2) Will the Earth really end up tidally locked to the Moon?
Tidal braking is slowing the Earth's rotation by 23 milliseconds (and increasing the Moon's distance from us by 38 metres) per 1,000 years. At that rate, the Andromeda Galaxy will have collided with the Milky Way long before the Earth's rotation has slowed down enough to approach tidal locking.

In fact, this AstronomySE answer suggests that as the Earth's oceans "are responsible for almost all of the tidal deceleration" (David Hammen's comment), the evaporation of the oceans in a billion years will eliminate the main drag on the Earth's spin and tidal locking may therefore never occur.

The moon will look as it normally does with regards to phases, but will be in a fixed position in the sky.

Practically speaking, that's all that will change. (At least in regards to the moon's appearance that is, having a 29 day-long day would wreak havoc on the earth's geology.)

The moon will only be visible from one half of earth's hemisphere, as you speculated.

It's also worth noting, in response to Chappo's answer, that not being able to see a new moon at night is something that happens currently, not just in this scenario. You can't see it at night because it's in the same rough direction as the sun. Similarly, you can see a crescent moon at night, but only if you're near the edge of the moonward hemisphere and the earth is blocking out the sun.


Did the scientists really observed gold and platinum being created from kilonova? I read that after LIGO and Virgo interferometers detected gravitational waves signal from a kilonova, many ground based telescopes quickly turns their mirrors towards the spot and "sees" gold and platinum ,etc other heavy elements over the next few days. Is it really possible to directly observe heavy elements being formed consider that the event occurs at a distance of around 130 millions light year away?

No they didn't. The deduction that (very) heavy elements are being produced is somewhat indirect and involves creating a theoretical model of the expanding "fireball" and comparing that model against measurements of brightness vs time at a range of wavelengths and against spectra (mostly in the infrared) as the afterglow faded.

Normally, when we view a stellar spectrum, one can see individual absorption lines that correspond to particular, identifiable elements. In the afterglow of this explosion that is not possible. It is moving so fast that the absorption features that were seen are smeared out by the Doppler effect and the atomic physics required to predict the strength of absorption is too uncertain to make accurate identifications of particular elements.

What is likely though is that the heavy elements, known as lanthanides are providing the large opacity in the "photosphere" of the fireball and are responsible for the rapid fading of the visible signal, the rather more lengthy decline in the infrared and the appearance of extremely broad absorption features that develop in the infrared. The heating of the ejecta by the radioactive decay of very neutron-rich nuclei is also a vital component of explaining the development of the afterglow.

See for example Pian et al. (2017) and Chornock et al. (2017).

Yes and no, as the previous answers suggest. It is confirmed that matels heavier than Fe are created in merging neutron stars. However, better models are needed to infer the creation of specific metals from spectra.


Can radio-based telescopes produce visible light images? It's a fairly simple question (at least in my layman's mind). Can a radio telescope that has no mirrors or lenses to capture visible light (to human eyes) still produce imagery like an optical telescope such as the Hubble? I'm talking just ROYGBIV light (no infrared or UV).

Can a radio telescope that has no mirrors or lenses to capture visible light (to human eyes) still produce imagery like an optical telescope such as the Hubble?

Of course, either via false color or pseudocolor. A false color radio image maps various frequencies (typically three) in a multispectral radio image to visible colors. A pseudocolor radio image maps various intensities in a single spectrum gray scale radio image to visible colors. The same techniques are also used for gamma, X-ray, UV, IR, and microwave imagery. We can't see in those frequencies, so something needs to be done to the captured imagery to make them visible to us.

If the question is asking whether a radio telescope can be used to capture visible light, the answer is no. I'll start with the receiver. It's designed to capture long wavelength radiation. Photons in the visible spectrum will have minimal, if any, impact on a radio telescope receiver.

An even bigger problem with some radio telescopes is depicted below.

Source: https://en.wikipedia.org/wiki/Haystack_Observatory

The above is the Haystack Radio Telescope at the Haystack Observatory in Westford, Massachusetts. The radome protects the telescope against weather, but also hides the antenna from optical light. A couple of other radio telescopes at the Haystack Observatory show yet another problem:

Source: https://en.wikipedia.org/wiki/Haystack_Observatory

The fixed zenith telescope antenna in the foreground and the fully movable telescope antenna in the background are made of a wire mesh, making them see-thru in the visible portion of the spectrum. This wire mesh construction is quite common for long wavelength radio antennae because it allows the wind to blow through it. Shorter wavelength radio antennae are more likely to be solid, but those are oftentimes painted.

A wire mesh with wires separated by centimeters is very smooth to a meter long radio wave, as solid surface smooth at the millimeter level to a centimeter long radio wave. A surface with millimeter-scale roughness would make for an extremely lousy optical mirror. There is no reason to make a radio telescope antenna smooth to the visible portion of the spectrum, and there are many millions of reasons (i.e., many millions of dollars) not to do so.

The answer is either "no, of course not,"  or "yes, in a pathological sense."  A radio "telescope" captures photons which are capable of forcing electromagnetic responses in metal -- which is to say, the induced voltage caused by an E-M field. The statistical odds of a visible-range photon causing an induced signal of detectable level in a metal antenna are ridiculuously small.


Why did the Event Horizon Telescope take so long to take a photo of a black hole? The Event Horizon Telescope is in the news now because they are about to release the first ever picture of a black hole. They have been working on this since 2006. Why did it take 13 years to produce the first picture? What was the critical mass that was recently hit that made the picture now complete?

A single telescope does not have the right angle size to picture the black hole that far away. The scientists had to use 8 telescopes to take a picture of the same point in the sky and merge the data from all sites using super computer. The data retrieval must point to the same object at the same time. Given the diverse condition of the 8 locations, there are only 5 days in a year when all of them have clear enough skies to point to the same location in a year.

Event Horizon Telescope