Why do we need to add a second to 30 June and risk upsetting Internet?

There's quite a few things here that need comments.

First off, the "linking" you describe was actually the first definition of the second (day, hour, and all the way down to second). Remember, since the dawn of civilization we've used celestial bodies to keep time for us.

The phenomena of the Sun and the Moon had good enough precision for all practical purposes until very recently. It was then that the definition of the time unit was rebased to a different phenomenon. BTW, the second of time is not a physical constant. It is a unit of measure currently defined as a multiple of the frequency coming out of the transitions of a certain atom between two different states.

Of course, the motion of the Earth follows its own clock, different from atomic phenomena, so it ought to drift off after a while. Do we "need" to make adjustments then? That's debatable. We still prefer to keep the Earth's clock and our clock in sync, and therefore we make adjustments when necessary. But is it needed in an absolute sense? No, not really. It's just convenient.

As to the risk of "upsetting the Internet", this is an alarmist rumor the likes of the "Y2K bug". I actually happen to work in a related field - I've done infrastructure at a variety of tech companies in the Silicon Valley; the NTP infrastructure (Network Time Protocol - what keeps Internet services in time sync) is something I'm very familiar with. A jump of 1 second should be and will be tolerated by essentially everything that's out there now.

Will there be stragglers here and there that will hiccup when the time takes a tiny leap? Yes. And that's a bug, the engineers need to look at the software and fix it. Remember, a lot of existing services (like Google Search, or Instagram) have servers and instances right now that are off by more than 1 second (just the way there are cars coming out of the assembly line with missing bolts once in a while), and usually nothing bad happens as long as the drift is not too big); fix NTP, the time base jumps on that instance, and again nothing bad happens in the vast majority of cases.

Relax, nothing important will be "upset". In fact, I doubt that the vast majority of people will be affected in any perceivable way.


Understanding The Fluctuations In The CMB Maps

The anisotropies in the CMB are caused by four effects; three at the surface of last scattering (SoLS), and one after:

Temperature differences

Denser regions will be more compressed and thus hotter, on average. Hence, an overdensity will result in a hotter spot, with a fractional fluctuation $\Delta T/T_0$.

Gravitational redshift

Photons climbing up (or falling down) the gravitational potential $\phi$ of their SoLS will lose (or gain) energy. Hence, an overdensity will result in a colder spot, and an underdensity in a hotter spot. This effect is also called the non-integrated Sachs-Wolfe effect and changes the temperature by $\Delta T = \phi/3c^2$, where $c$ is the speed of light.

Doppler shift

The SoLS has a bulk motion $\mathbf{v}$. Generally, overdensities have gas infalling, and hence result in a colder spot. In the direction $\hat{\mathbf{n}}$, the temperature difference will be $\Delta T = T_0 \mathbf{v}/c$.

The integrated Sachs-Wolfe effect

If a potential well is getting shallower in time (which happens during dark-energy domination), photons receive a net blueshift in crossing the well and the CMB appears hotter. This gives a temperature difference of $\Delta T = 2\Delta\phi/c^2$, where $\Delta\phi$ is the change in the potential while a photon traversed the potential well.

The total temperature fluctuation is the sum of all these terms. In general, the gravitational redshift will dominate over the temperature differences, which in turn dominates over the Doppler shift.

A review of the CMB fluctuations is given by Challinor (2013).


Why can't we build a huge stationary optical telescope inside a depression similar to the FAST?

The surfaces of telescopes need to be configured to a fraction of a wavelength. If one is working in the FAST wavelength range of 10 cm to metres, then that is a relatively straightforward engineering problem. However, the difficulty and expense ramp up enormously when the working the wavelength is 500 nm. This would entail the construction of literally millions of precisely configured mirror segments to tile the inside of the structure and then some way of pulling them all into position and alignment (and keeping them that way) with a precision of 10 nm or so.
It helps that you aren't going to move the primary, and you can still track objects for 1-2 hours by having instrumentation at a movable focus. However, a second big problem is that the air above the mirror is turbulent and the size of the turbulent cells is much smaller than the size of the telescope. To take advantage of its large size, in terms of angular resolution, the individual parts of the telescope have to be individually controlled/deformed at a continuous rate of $\sim 100$ Hz. It is probable that to get the tenfold increase in resolution implied by the much larger diameter of this telescope would require higher frequency and more precise adjustment.
This sort of engineering is going on for effective mirror diameters as large as 39-m, but comes at a billion-Euro cost. Scaling up to something ten times bigger would almost certainly be a two order of magnitude increase in cost. So although there may be no showstopping physics, it would cost about as much as all other astronomy research combined.
Large interferometers, with sparse arrays of telescopes, are a much more likely proposition but come with their own very difficult problems.


Is there any planet/star bigger than VY Canis Majoris?

Part 1: Assuming "larger" means greater in diameter.

Stars
Estimates on the size of stars are just that, estimates, and estimates based on rather fuzzy observations. VY Canis Majoris has been bumped down to size. The current thinking is that there are seven known stars larger than VY Canis Majoris, the largest of which is UY Scuti.

Current models indicate that the first generation of stars were much, much larger than anything we see now. It will be quite some time before we can resolve a first generation star. So far, they are just theoretical objects.

Planets
Jupiter-mass planets are about as large as a planet can get. There are some exoplanets that are larger than Jupiter, but that's because they orbit much closer to their parent star than does Jupiter. This makes them puff up a bit. The reason Jupiter-mass planets are deemed to be the largest possible is that planets of this mass are presently assumed to have a core of degenerate hydrogen. A funny thing happens to degenerate masses when mass is added to them: They shrink in diameter. (The shrinkage becomes catastrophic as the mass approaches the Chandrasekhar limit.)

This means that assuming all other things are equal (temperature, composition), a planet more massive than Jupiter will be small in diameter than Jupiter is. Even if all other things aren't equal, a Jupiter-diameter planet is (give or take) about as large as planet can get.

Part 2: Assuming "larger" means greater in mass.

Stars
In terms of mass, VY Canis Majoris doesn't even make the top ten, not even close! The most massive known star is R136a1. Again, these are estimates, but mass is a bit easier to pin down than is radius (or diameter).

As is the case with physical extent, the first generation stars are presently modeled as being much, much more massive than anything we see now.

Planets
There's not much difference between the largest planet and the smallest brown dwarf. I would argue there's very little difference. It's a spectrum with no distinguishing characteristic that lets one say "this is a planet" and "that is a brown dwarf". Ignoring the distinction between super-Jupiters and brown dwarfs the largest is about 80 Jupiter masses. V1581 Cygni C is 79 Jupiter masses. (More massive than that and they start burning hydrogen, thus making them a small red dwarf.)

The current factor that is used to distinguish between brown dwarfs and super Jupiters is mass. Anything larger than 13 Jupiter masses is a brown dwarf, anything smaller, a planet. That boundary is very arbitrary.


Flatness of celestial systems like galaxies and Solar System

Flatness is caused by conservation of angular momentum. As solar system/galaxy collapses under its own gravity, its velocity increases to conserve the angular momentum. This high velocity leads to formation of disc shaped structure. The similar manner in which piece of dough becomes flat when it is rotated.


Can my Sky Watcher Infinity 76P "make the sky bright" & do I want it to?

To make the images in the time-lapse (and similar Milky Way photos), the photographer uses time exposures with wide angle lenses, may stitch together multiple images from longer lenses, and uses extensive post-processing to balance contrast and include foreground or bright objects in the scene.

The closest you can get to the same thing visually is to go someplace with very dark skies (middle of the ocean, or a wilderness with no nearby towns) and use the naked eye or wide-angle binoculars.  In dark skies, the Milky Way overhead is obvious and impressive, but it still won't match the color or vividness of the photos.

Your telescope isn't really useful for looking at the Milky Way as a whole, but it can probably give a satisfying view of the many star clusters that dot the Milky Way--the Pleiades and the Beehive are naked eye objects and so shouldn't be too hard to find with your telescope.  When these clusters are "resolved" into their individual stars, they are impacted less by the light-pollution you probably have in your backyard location.  I can't tell if the 76p eyepiece is one of the standard sizes that can be replaced (1.25", 0.965" OD).  If it is, an inexpensive eyepiece of about 25mm focal length might improve your views of astronomical objects.


What is direction of Sun's axis of rotation relative to the Earth's orbital plane around the Sun?

Well, Wikipedia says that it's 7.25°to the ecliptic and gives a reference.
Looking there clarifies the value in 3 dimensions:

North Pole of Rotation
Right Ascension: 286.13
Declination    :  63.87

Since those coordinates are relative to the Earth’s axis, being a practical measurement system for Earth ground observers to use in pointing their telescopes, you need to combine that with Earth's orbit.
As a useful visualization, mark that point on a model of the celesial sphere.  Compare that with the ecliptic pole: the difference between thise points is what you want to know.
I expect you’ll find they are separated by 7.25°, but by marking a model sphere you’ll understand the actual direction.


How does the Hubble telescope capture long-exposure photos?

The answer to this is that such images are not taken continuously. The HST did not stare at one part of the sky for 10 continuous days, but rather it stared at one part of the sky for short periods over a long time which, when added up, amounted to 10 total days of observations. To quote Wikipedia on the subject of how the Hubble Ultra Deep Field (HUDF) image was taken:

  The observations were done in two sessions, from September 23 to October 28, 2003, and December 4, 2003, to January 15, 2004. The total exposure time is just under 1 million seconds, from 400 orbits, with a typical exposure time of 1200 seconds. In total, 800 ACS exposures were taken over the course of 11.3 days, 2 every orbit, and NICMOS observed for 4.5 days. All the individual ACS exposures were processed and combined by Anton Koekemoer into a single set of scientifically useful images, each with a total exposure time ranging from 134,900 seconds to 347,100 seconds. To observe the whole sky to the same sensitivity, the HST would need to observe continuously for a million years.

In the above, ACS is the Advanced Camera for Surveys and NICMOS is the Near-Infrared Camera and Multi-Object Spectrometer.

That, I think, is the basic answer to your question, but there's a few interesting and related points I'd like to point out. First, the HUDF image was actually comprised of many individual images rather than a single 10-day exposure image (as the quote indicates). This would necessarily have to be the case, even if Hubble had the ability to stare at the HUDF region of the sky for 10 days without interruption. Cameras work by taking in light. The Charge-Coupled Device (CCD) which actually collects the light (and converts it to an electric charge) can only hold so much charge before it becomes saturated. At some point, you have to "stop recording" and read out the CCD which clears the accumulated charge. That way, you can resume again with a fresh slate. A 10-day continuous exposure would have easily saturated Hubble's CCD. So even if Hubble did spend 10 continuous days observing this patch of the sky, it would have been broken into many, many smaller exposures. This means, even if the Moon did get in the way during part of such a 10-day exposure, they could have just not made any observations during that time.

The other related fact I wanted to point out is that Hubble has something known as the Continuous Viewing Zone (CVZ). This is a region of the sky which is never obscured by the Earth, Moon, Sun, etc. as Hubble orbits. As the name implies, it allows for continuous viewing, without interruption. The Hubble Deep Field (HDF) (the precursor to the HUDF) was observed in this region. However, there are reasons why you wouldn't want to observe in this region such as it being polluted by "Earthshine".


Low mass cut-off for the Dharma Planet Survey's detection of habitable planets around 40 Eridani A?

In general, finding the low mass cutoff of a particular RV survey is normally quite complicated as there are a lot of selection effects on many different variables (such as planet period, planet eccentricity, host star type, signal-to-noise, stellar activity level etc) that determine whether you can recover a planet signal. In the case of this paper, this would be particularly true as they are combining datasets from several instruments, each with different coverage, sensitivity and systematic noise properties. The way this is normally handled is through a "bootstrap" technique where you inject planet signals of known parameters and see what your detection code finds and study the recovery percentage as a function of the varied parameters.

Using the online Habitable Zone calculator for a Teff=5072 K, Stellar luminosity=0.47 * L_Sun (luminosity of the Sun) matching the host star, produces 0.678 AU as the inner edge of a conservative habitable zone. If we use this, the star's mass from the paper (0.78 M_Sun) and the 1 M_Earth mass you would like to find, then making use of the formulas given on the Exoplanet Archive website, gives us a period of 230.9 days. Plugging this into an equation for the radial velocity semi-amplitude (also on the Exoplanet Archive), produces an amplitude of about 12 cm/sec. This is assuming an edge-on orbit, so the sin i term from the inclination is 1; given the lack of detected transits in the paper, the inclination is likely lower than 90 degrees, which would lower the RV amplitude further.

This RV amplitude is far too small to be detected with current instruments. The best instrument we have right now is HARPS on the ESO 3.6m telescope, whose data is included in the paper. If I am understanding their modeling correctly, they get error bars of 181 cm/s on the HARPS data for this star in their analysis, 15x the expected signal. Also note that the calculated RV signal is somewhat best case (unless it is more massive); decreasing the inclination from edge-on (likely) and moving the planet further out in the habitable zone from the inner edge I calculated (also likely) will both decrease the RV amplitude further. Even for future instruments such as ESO's ESPRESSO which will go the 8.4m VLT and reach 10 cm/s precision, this would be a challenging measurement.


Converting equatorial coordinates to Cartesian coordinates for extragalactic distances

I find this discussion of cosmological distances by David Hogg to be very useful in answering questions like this.  In Section 4 he says:

The line-of-sight comoving distance between two nearby events (ie, close in redshift or distance) is the distance which we would measure locally between the events today if those two points were locked into the Hubble flow. It is the correct distance measure for measuring aspects of large-scale structure imprinted on the Hubble flow, eg, distances between “walls.”

That would suggest to me that it’s the distance measure you want for your problem, but I encourage you to read further in that article to be sure.


I contacted Dr. Danielle Adams, Deputy Director for Marketing and Communications at the Lowell Observatory. She was kind enough to reply, and generously provided the following (lightly edited for formatting):

  I spoke with one of our senior educators about the Clark knobs. The numbers below correspond to the numbers in the image posted in Why does this Lowell Observatory telescope have so many knobs? What do they all do?

  Slow motion control for adjusting Right Ascension (east-west movements) while the Right Ascension clutch (3) is locked down.
  Iris control that allows us to control the effective aperture of the telescope. There is an adjustable aperture that can be adjusted from 24" (wide open) to 6".
  Right Ascension clutch that when tightened prevents the telescope from being moved by pushing, allowing the telescope drive to operate.
  and 7. Slow motion controls for adjusting declination (north-south movements) while the Declination clutch (5) is locked down. The two controls are linked with bicycle chain so when one moves, so does the other. This allows operation of the control from either side of the viewing position.
  Declination clutch that when tightened prevents the telescope from being moved by pushing, allowing the telescope drive to operate.
  Sidereal clock that permits astronomers to point their telescopes to a given celestial object more easily.The sidereal day is about 4 minutes shorter than the 24 hour solar day. This is due to the fact that the Earth revolves around the Sun while it rotates about its axis.
  see (4)


Equirectangular
or equidistant cylindrical projection
is the simplest way to meet these requirements.
Plate Carrée is the case where distortion is minimal at the equator.
Dominic Ford's Map of the Constellations uses it,
as do the Deep Star Maps
by Ernie Wright of the NASA Scientific Visualization Studio.
In free software,
Stellarium calls it "Cylinder," and
Cartes du Ciel calls it "Cartesian."
If you plot your own, you could scale the horizontal dimension by
$\cos 30^\circ \approx 0.866$ to minimize distortion at
$\pm30^\circ$ declination.

Since you are not plotting the polar regions,
other cylindrical projections are possible.
The Mercator
projection is available in both Stellarium and Cartes du Ciel.
Stellarium also offers the
Miller cylindrical projection.


Is it true that open clusters have $E > 0$ and globular clusters have $E < 0$?

Stars in an open star cluster are - according to wikipedia - "loosly bound by mutual gravity". Whereas a globular cluster is more dense and will be "less loosly" bound - to use the same terminology.

Since the stars in the open star cluster still are gravitationally bound, the total energy $E$ must be less than zero. This criterium that $E<0$ is always correct for (gravitationally or otherwise) bound systems for exactly the reason you stated, the kinetic energy is 'less positive' than the potential is negative. Otherwise the particles (stars in this case) would break free from the gravitational interactions with the other particles - and that's the complete opposite of a bound system.

However, the fate of most open star clusters is that it will fall apart. This can be due to

mass loss of the cluster or,
internal processes like a slingshot due to close encounters between constituent stars.

From the get-go these open clusters are quite unstable, so $E$ will be close to zero, but still negative ofcourse because it's a bound system. But it is instable enough for the following to happen. Young, bright stars radiate a lot of energy, and this means that the radiation pressure on the surronding cloud from which these stars are born is high. The light from these young stars can blow large amounts of mass in the form of this gas and dust out of the system - which has an influence on the total energy of the system! Because the mass of the system drops, the total energy rises (read: becomes less negative) - wich makes it easier for the constituent stars to reach escape velocity. If you don't like thinking in terms of energy, the escape velocity is $v_e = \sqrt{2GM/r}$, so $v_e \propto M $. If $M$ drops, so does the escape velocity.

The second bullet speaks for itself, I think.

You can deduce that $E$ has to be negative from a conservation viewpoint. When the stars in the cluster are born from the same cloud, this cloud already is gravitationally bound, so $E<0$, and unless some energy is lost at some point, the cloud will stay bound. But like I said and as you said, most of these clusters don't stay bound for very long, and eventually will evaporate.


Is Venus in some way tidally locked to... Earth?

@BMF's comment links to (Gold & Soter 1969) Icarus 11, (3), November 1969, pp 356-366 Atmospheric tides and the resonant rotation of Venus. Since it is paywalled I'll add a short summary:

From the abstract:

  The observed spin-orbit resonance of Venus, whereby the same side of Venus faces the Earth at each inferior conjunction, cannot be explained adequately by gravitational interaction with the Earth alone. The expected solar tidal drag on the solid body of Venus would easily overwhelm the Earth's couple upon any reasonable permanent deformation of Venus. If there exists, however, a solar atmospheric tide, partly thermally induced and similar to that known on the Earth, its torque may counteract that due to the solar solid body tide at a particular rotation period. The small interaction with the Earth is then sufficient to lock the period to one of the resonances in the vicinity of that angular velocity.

The paper explains that if Venus' $J_{22}$ (transverse gravitational quadrupole moment) were about the same magnitude as Earth's it would be far too weak for this synchrony to be induced by gravitational interaction between these two planets because the tidal forces from the Sun would overwhelm the effect:

  The spin-orbit resonance between the two planets is undoubtedly due to the action of the Earth upon the net permanent de- formation (or transverse quadrupole moment) of Venus. But in order for this rather weak torque, which is significant only near inferior conjunction, to overcome the constant drag of solar tidal friction on the body of Venus, the permanent quadrupole moment required for Venus would have to be at least an order of magnitude larger than comparison with the Earth would suggest.

It then proposes that the tidal interaction with the Sun might be coincidentally cancelled by torque from Venus' atmospheric tidal effects. The paper discusses the history of understanding of transfer of angular momentum from a planets atmospheric tides to the planet itself:

  It was first shown by Lord Kelvin (Thompson, 1882) that the Earth's atmospheric tides transfer angular momentum and energy from the orbit to the rotation and tend to accelerate the latter. Holmberg (1952) suggested that the associated production of mechanical energy might be sufficient to balance the dissipation of oceanic and solid body tides, so that the Earth's present rotation rate would be one of stable equilibrium over geological time. Subsequent studies, however, appear to confirm the older view that the Earth is in fact slowing down; i.e., that the energy loss in conventional tides exceeds the input through atmospheric tides. Nevertheless, the notion that a planet's rotation rate might be stabilized in the balance between the opposing two kinds of tides is perhaps the key to the resonant rotation of Venus.

and concludes (in 1969) that this is probably what's happening. This is the extent of my answer, certainly in the fifty years since this paper it's been looked at in greater detail, but as far as I can tell there are no competing theories.

Update: The problem was also addressed by (Ingersoll and Dobrovolskis  1978) in Venus' rotation and atmospheric tides and by (Bills 2005) Variations in the rotation rate of Venus due to orbital eccentricity modulation of solar tidal torques J. Geophys. Res. 110, E11007

More recently Auclair-Desrotour, Laskar & Mathis (2017) Atmospheric tides in Earth-like planets A&A 603, A107 have addressed the general topic in great detail and Venus is examined in great detail, but I can't tell if they address this subject directly or not.

The paper also describes the rapidly evolving studies of Venus by Earth-based radar observation in the 1960's such that by 1969 Venus' apparent synchronicity became obvious at about the same time the Moon landings began. Radar studies of Venus started early, as discussed in

When did planetary scientists realize Venus' surface pressure was almost 100x that on Earth? How did they find out?
How was the radius of Venus measured so accurately (± 3 km) via radar in the mid 1960's, before Venera 4 and Mariner 5?

  FIG. 1. Radar determination of the retrograde sidereal period of Venus. With increasing accur- acy, the values have converged toward the reson- ant period of --243.16 days. The next nearest resonant periods are at -201 and -308 days. Figure adapted from Eshleman (1967); see also Shapiro (1967).


What is the difference between solar wind and solar radiation?

Okay we can state it as follow:

Solar Radiation:
It is the radiation from the sun which includes all wavelengths of electromagnetic radiation coming from the sun.
Solar Wind

  The solar wind is a stream of charged particles (a plasma) released from the upper atmosphere of the Sun. It mostly consists of electrons and protons.

And I can simplify it as like it is something like when you stand in front of an operating fan and this fan is rotating to spread the air all over the place, so this will make an air wind around it, this is the wind you can sense. Back to sun, Solar wind is just like that it is the sun rotating particles produced out of the sun energy around the sun shaped by the sun's rotating magnetic field.
Solar Flare

  A solar flare is a sudden brightening observed over the Sun's surface or the solar limb, which is interpreted as a large energy release of up to 6 × 1025 joules of energy.

And for more clarification, it is an event happen from the sun due too a high charged accelerated particles, this event could have some damages on the earth like the big flare happens every 11 years.


Period coincidence of Planet Nine with furthest Kuiper belt objects

I did a little digging and I found this article.

It's worth noting that the initial publication wasn't about resonances but the fact that all 6 objects came from the same side of the sky, from somewhat similar directions, which, given that they were the 6 most distant at the time, is statistically unusual and it was both similar direction and similar inclination to the plane, so the statistical improbability was high enough to investigate.  Orbital resonance wasn't the primary argument, it was tacked on later as a secondary.

From the article:

  Their analysis also offered suggestions as to what kind of resonance
  the planet has with the KBOs in question. Whereas Sedna’s orbital
  period would have a 3:2 resonance with the planet, 2010 GB174
  would be in a 5:2 resonance, 2994 VN112 in a 3:1, 2004 VP113
  in 4:1, and 2013 GP136 in 9:1. These sort of resonances are
  simply not likely without the presence of a larger planet.

  “For a resonance to be dynamically meaningful in the outer Solar
  System, you need one of the objects to have enough mass to have a
  reasonably strong gravitational effect on the other,” said the
  research team. “The extreme Kuiper belt objects aren’t really massive
  enough to be in resonances with each other, but the fact that their
  orbital periods fall along simple ratios might mean that they each are
  in resonance with a massive, unseen object.”

  But what is perhaps most exciting is that their findings could help to
  narrow the range of Planet 9’s possible location. Since each orbital
  resonance provides a geometric relationship between the bodies
  involved, the resonant configurations of these KBOs can help point
  astronomers to the right spot in our Solar System to find it.

now, I want to point out that it's them, not me who did the typos on 2 of the objects, I'm pretty sure (based on Wikipedia and the chart below) that it should be 2004 VN112 (not 2994 VN112) that has the 3:1 and 2012 VP113 (not 2004 VP113) that has the 4:1.

This chart below also has a typo, it should be 2010 GB174 not 2012 GB174.  I guess nobody proofreads letter/number assignments.   Correct names will help if anyone wants to run the numbers see how see how close the resonances come to each other.

As to the degree of statistical certainty?   If you're a good enough statistician, you can run the numbers.   3/2, it seems to me, would be the most common resonance based on Jupiter's abundant Hildas and the Plutinos and only 1 of the 5 potentially resonant objects has a 3/2, but 3/2 is also very far from the sun and hardest to detect, so that's not a deal breaker.

I'd guess that the degree of certainty from these orbital resonances is OK but not great - but that's entirely a guess.   I also don't know enough about astronomy to know how close numbers have to be to provide some evidence of orbital resonance.  Jupiter's resonant moons, for example, if you check their orbits, are off by 1 or 2 percentage points from each other.


What instruments are used to measure the distance to the Sun?

They look like radio telescopes, because that is what they are.
Some radio telescopes have been outfitted to transmit rather than receive, requiring using a different radio telescope to receive the reflected signal. Other radio telescopes have been outfitted to transmit as well as receive in a pulsed manner. These can be used for radar astronomy without the help of a separate transmitter.


Can looking at the sun through solar viewing eclipse shades everyday harm your eyes?

It depends on the quality of glasses and intensity, nature and degree of harmfulness of sunrays which you are watching to. Every eclipse has different intensity so as it can harm you.

When there was a solar eclipse around 3-4 years ago, it was a full solar eclipse. That time I watched it from my school with total naked eyes. My eyes were not harmed from that, but my school principle did.

But I still recommend you to not to watch it directly, use projectors which would allow you to get the view without directly watching the sun. Only use original products which are being certified.


Is it possible that we see our solar system as a star?

I think the original answer has confused you somewhat - it is spacetime that is expanding faster than light - when viewed from sufficiently far away - we certainly are not moving away from the Sun at anything even remotely close to the speed of light (in fact the expansion of spacetime is not a motion at all in the true sense, but a change in the underlying metric).


A commenter points out that accelerations in the galactic gravitational potential are extremely small:

  The Sun's galactic orbital speed is roughly 200 km/s and its distance from the galactic centre is around 8 kpc. Using $a=v^2/r$ that gives an acceleration of roughly $1.6×10^{−10}\rm\,m/s^2$.

A century is only $\pi×10^9$ seconds, so in a hundred years the $\Delta\vec v$ for the Sun and its neighbors is only half a meter per second --- and that's mostly a difference of direction, not magnitude.  We can't measure three-dimensional stellar motions to a precision of a half-meter per second, and the career of a professional astronomer is usually shorter than a century.


Interstellar movie: What is the "portal" to the other galaxy?

Yes, it is a wormhole indeed. This has been indicated quite clearly in the movie as well, when Dr. Romily explains with a pen and paper to Dr. Cooper. The explanation goes like this :

Imagine a sheet of paper to be 2-D space, then a line joining two points on the sheet of paper is the shortest distance possible to reach that point, but if due to some disturbance, the space is bent ( achieved by folding the sheet of paper), you can pierce a hole in the sheet after aligning the two points together. That is a 2-D wormhole (which is a circle indeed). So what happens when you consider a 3-D wormhole? It becomes a sphere.

Now, you can see the other end of a 2-D wormhole (which is effectively the other hole in the sheet visible from the first hole, and one can see beyond the hole in the other direction as well). Same happens with the 3-D spherical wormhole where you can see to the other side of the wormhole as well.

Now, the means to bend spacetime : Space time can be bent by having a very big mass placed inside the space time (read Einstien's General Relativity ). So, it is safe to assume that the wormhole is a space time disturbance created due to a massive object, but it is not a blackhole.


So it looks something like this? https://cloud.planetmaker.de/index.php/s/RkmkBNfb8YyDHSE

That sounds like the circum-horizontal ring (yes a name as simple as that). It (probably) is caused by hexagonal ice crystals floating horizontally. Where it touches the 22° halo you might see sun dogs which can be slightly colourful brighter patches.

See also https://www.meteoros.de/themen/halos/haloarten/ (German) and http://www.atoptics.co.uk/halosim.htm


Does the sun have cycles causing temperature changes on Earth?

The Sun's magnetic field is way too weak to have any measurable effect on the Earth's orbit. The sunspot cycle does produce a small but detectable signature in the global average temperature (about $0.2\sideset{^{\circ}}{}{\mathrm{C}}$). Over a billion-year timescale the Sun is getting hotter as helium builds up in its core.


Is there a publicly available Gliese Star List

The Gliese catalogue can be found at CDS here.

Given an RA and Dec then you can use the fortran program found here to get the constellation.


Eventual outcome of tidal acceleration and deceleration

So that's question 1: Is it really possible for tidal acceleration to
  throw a satellite into space before the planet is tidally locked, and
  if so, does either objects' size impact it, or just fluidity and
  relative periods?

During the locking process, the angular momentum of one is fed into the other until the periods match.  The final destination depends on the initial momentum of both objects and the moment of inertia.  As an example, if you make the moon less massive, or the earth more massive or with a faster spin, then the final configuration of the system would have a larger orbit.

You could certainly create a system where the final theoretical configuration has a distance to the satellite that is beyond the primary's Hill sphere.  Such a system would be expected to eventually separate.

Fluidity might change the rate at which the final destination is reached, but not its configuration.

  Will all non-retrogade tidally decelerating satellites eventually
  crash into their planet, or is it possible for the planet to catch up
  given the right starting conditions?

Each object is transferring the same amount of angular momentum.  But the response of each object depends (at least initially) on the relative magnitudes of their moments of inertia.  For a system where the satellite is (relatively) small, or where the difference in rotation rates is large, impact is inevitable.  The primary cannot speed up fast enough to keep up with the increasing speed of the descending partner.

But when the difference in periods is small, and the primary is small enough/far enough, then the objects would lock before impact.


Why do the rings around the Gas Giants get their 'ring' shape?

There are different theories as to how planetary rings form, and so there are different answers to your question.

One theory of planetary ring formation is that a small (or large, depending on the ring size) moon wandered into the gas giant's Roche limit (the radius from the celestial body inside which no second celestial body can stay intact due to tidal disruptions by the larger body). The moon would have been orbiting the planet, and would thus have been moving in a flat . The resulting debris continued to move in a plane; if it had been angled, collisions would have flattened it.

Another theory is that the debris is left over from the early Solar System. A protoplanetary disk formed around the Sun; it contained all the material that makes up the planets. Over time (and through a lot of collisions!), bits of the material slammed together and formed planets. The trouble is, not all of that material became planets. Some became asteroids, some became comets, and some became moons (although the theory of the formation of the Earth's moon is a bit different). Some of this material could also have coalesced around planets.

Now, the protoplanetary disk would already have been in a disk shape. Why? Think of a rotating object - like a planet. You might have read that the Earth is not a sphere. Aside from the obvious deviations of mountains and valleys, some of the deformation comes from the Earth's rotation. Any rotating body will bulge a little around its equator; a spherical body will become an oblate spheroid. The same thing happened for the protoplanetary disk - it became a disk because it was rotating! Any material that coalesced around planets would already be in a plane - that's why the planets' orbits aren't at odd angles to each other. So this material would be in a flat shape if it was captured by the planets, and would become a circular ring because of rotation. If, at that point, there was material that wasn't in a flat plane, it would become flat by the same process that made the protoplanetary disk flat.

There are stellar systems where the protoplanetary disk isn't in the same plane as the star's rotation. This could affect planetary orbits; however, such systems have not been thoroughly studied because of their rarity.


Hours to days.

The orbital period is proportional to the 3/2 power of the orbital radius, and the orbital period of the moon Methone (and thus of the particles in its associated ring arc) is very close to 24 hours. Working from the orbital radii given in this table, I get:

D ring: ~5 hours
C ring: ~6-7 hours
B ring: ~8-10 hours
A ring: ~12-14 hours
F ring: ~14.5 hours
Janus/Epimetheus ring: ~16-17 hours
G ring: ~19-20 hours
Methone & Anthe arcs: ~24 hours
Pallene ring: ~27 hours
E ring: ~21 hours to ~90 hours
Phoebe ring: 90-500 days


What does the word "shock cooling" mean in supernova physics?

Shock cooling in a SN is the scenario where kinetic energy carried by the shock transforms into other energy forms: adiabatic expansion, radiation, acceleration. A shock is actually a term meaning discontinuity. For example, a density profile of steady-wind medium is $r^{-2}$ while that of external part of ejecta is $r^{-n}$ where $7 \le n \le 12$ as typical values in core-collapse supernovae, but at the interface between the medium and the external ejecta, where they interacts, its density profile spikes before asymptotically decaying to the characteristic profile of each region. The interface is the shock.

The mechanism of shock cooling in SN involves processes which are not completely understood. What we know is that the shock is formed as a double-shock structure: forward and reverse. The forward shock propagates outward, while the reverse one goes inward in mass coordinate. (Note that in spatial coordinate both shocks propagate outward). Kinetic energy in each shock is transformed by interactions. At a forward shock, typically the energy is transformed into acceleration of relativistic particles, which in turn radiate via synchrotron or inverse Compton processes, and collisionally interacts with the medium. At a reverse shock, the shock energy thermalizes ejecta, which in turn thermally radiates.


For long exposure pictures you need to have a motorized mount for your camera. The earth's rotation will lead to streaks otherwise. An affordable way to do this is to use a standard tripod with a star tracker on top. There is a variety of products like:

iOptron Sky Tracker
Vixen Polarie

They cost around 400-500 USD and are fairly small. You need to align them to the sky's north pole (using the star Polaris in Ursa Minor), and then they will rotate your camera as the earth rotates during the night.

Also, you may want to start with your 50mm lens first. The greater the focal length, the higher the magnification, and also the more visible the tracking errors and misalignments of the sky tracker and camera system will be. Be prepared to spend quite some time trying to get good results!


When did people first measure that the Earth was closest to the Sun during January?

Hipparchus, not Kepler

Kepler got the conic sections right, and Newton gave us the mechanics.  But the question is about when people knew that the Earth was closer to Sol in one part of the year than others, and Hipparchus knew that, even though he wasn't too hot on the values of the orbital radii.  Hipparchus' version of the eccentric model had Sol's (purported) circular orbit around the Earth not centred upon the Earth, but 1/24th of an AU away.  Therefore Sol (purportedly) orbited at varying distances from Earth.  This was, after all, the whole point of the eccentric model, to explain non-uniform apparent motion through variation in distance.

Perigee and apogee were known in the times of Hipparchus and Ptolemy.  Hipparchus even worked out when the furthest point (apogee) was.  Ptolemy furthermore made an error based upon knowing that his placement of the apogee in Gemini was the same as that of Hipparchus 280 years beforehand, declaring that perigee and apogee were fixed.

They of course were not.  Hipparchus placed apogee at 5.30° Gemini.  Astronomers in the 9th century in Baghdad applied the same calculations to their measurements and placed it at 20.45° Gemini.

As for how this was observed, it was not done by measuring the Sun's appearance at all (although Hipparchus did do that).  Ptolemy and Hipparchus had a geometric model of a true geocentric circular orbit versus the (purported)  eccentric circular orbit of Sol.  It incorporated the equinoxes and the solstices.  By observing the times of the equinoxes and solstices, the lengths of the periods between them, they were able to determine trigonometrically all of the other orbital parameters, which included placement of perigee and apogee.

That points of closest and furthest approach existed was known in the 2nd century BC, as was their angular locations relative to the solstices; they've been in the models from then onwards.  That they moved around took about 11 centuries after that to discover.  The correct conic sections and the idea of both bodies orbiting around a barycentre came somewhat later, but that wasn't the question.

Further reading

Hugh Thurston. (1994) "Appendix 2: Calculation of the Eccentric Quotient for the Sun, and the Longitude of its Apogee". Early Astronomy. New York NY: Springer.  DOI 10.1007/978-1-4612-4322-9_1
James Evans (1998). The History and Practice of Ancient Astronomy.  Oxford University Press. ISBN 978019509539.
Viggo M. Petersen and Olaf Schmidt (June 1968). "The Determination of the Longitude of the Apogee of the Orbit of the Sun according to Hipparchus and Ptolemy".  Centaurus.  Volume 12.  Issue 2.  Pages 73–96.  DOI 10.1111/j.1600-0498.1968.tb00080.x


What planets and moons have we seeded with life (if any)?

This would seem to be constrained to planets and moons earth has landed on or impacted at all, as we have no way of telling after the fact whether a lander was perfectly sterilized. That list is as follows:

Planets:

Mercury: USA's MESSENGER in 2015 de-orbited and crashed upon completion of mission. Contamination is unlikely because of high impact and temperatures.
Venus: 13 missions by the USSR and 1 by the USA (1966-1985). Contamination unlikely due to harsh environment.
Mars: 12 missions, multi-national origins. Contamination plausible.
Jupiter: The Galileo missions (2), by the USA Contamination unlikely, because of pressure.

Moons:

Earth's moon: Far too many to count, myriad landings. Contamination plausible.
Saturn's moon Titan: Huygens probe, landed by the ESA (2005) Contamination Plausible.

Asteroids/Comets:

Eros: 2001 landing by USA. Contamination possible.
Itokawa: 2005 landing by Japan. Contamination possible.
Comet 9P/Tempel 1: 2005 landing by USA. Contamination is unlikely because of high impact speeds.
Comet 67P/Churyumov–Gerasimenko: 2014 landing by ESA. Contamination possible.


Will a planet's core always be very hot?

We would expect all large bodies, such as planets, to have a hot interior.

The interior of a planet is heated through two main mechanisms: Heat released from gravitational energy as the planet is forms, and heat released from the decay of radioactive elements, particularly uranium.

The outer layers of the planet are a very good insulator, so the any heat is trapped in the interior for a long time.

In gas giants, gravitational energy is the main source of heat, Saturn has an internal temperature of about 21000°C, Jupiter is higher, 35000°C in some models. Terrestrial planet's cores are heated by radioactive decay. Earth has a core temperature of about 6000°C. Smaller bodies have cooler temperatures. Dwarf planet Pluto may have a core temperature less than 1000°C, and asteroids would have reached thermal equilibrium, and have roughly the same temperature throughout.

No planet in the universe can be much more than about 10 billion years old, and this is not old enough for a planet like the Earth to have formed a completely cold core.


Is there a preference for orbital planes of planetary systems to be aligned with the galactic disk?

Unless I'm missing something we know very little about the orbital planes of individual systems.

Most exoplanetary systems (all of the transiting ones, whereas it is just a strong bias for the doppler-discovered planets) are discovered because they have high orbital inclination. This means that the planet's orbit takes it across the stellar disc as we view it. However, the entire arrangement can be rotated around the line of sight by any angle and this would give exactly the same observational signature.

That being the case, the only test I can immediately suggest is that if the planes of planetary systems were preferentially aligned with the galactic plane (ours isn't), then it would result in a lower planetary detection rate for stars in directions perpendicular to the Galactic plane in comparison with stars lying in the Galactic plane.

There may be some mileage in pursuing this, though I don't know of any results. For example the main Kepler field was directed just out of the Galactic plane, but the K2 fields are at a variety of Galactic latitudes along the ecliptic plane. If one could normalise somehow for the types of stars observed in each field and the differing data qualities then it might be possible to look at the "planetary occurrence rate" as a function of Galactic latitude.

The "all-sky" ground-based surveys are less suitable for this because they tend to avoid the Galactic plane because it is too crowded with targets.

The question can be more broadly addressed by looking at visual binary systems - i.e. those where you can resolve the components and study the orbit. There, it is possible to say something about the orbital planes of individual systems. A paper by Agati et al. (2015) has looked at this. They collected 51 systems with the requisite information, that are close to the Sun. They found only weak evidence for any anisotropy of the orbital poles and this weak alignment was closer to the ecliptic pole than the Galactic pole.


What supernova has created the iron currently found in Earth core?

Iron is mainly made or is the product of decays from nuclear processed material inside supernovae.

As the Earth (and solar system) is around 4.5 billion years old, then the stars that manufactured the iron that is currently in the Earth's core died more than 4.5 billion years ago. Note that the solar system formed out of the gases that had been enriched by the supernova explosions of hundreds of millions of stars all mixed up together.

There are basically two categories of star that can have exploded as supernovae and disseminated this iron into the interstellar medium and from which the solar system could then have formed. The first is massive stars ($>8-10M_{\odot}$). These can produce iron and nickel in their cores as a result of silicon burning in the final stages of their evolution. There is then a brief core-collapse followed by an explosion and the resulting supernova can scatter some of this processed, iron-rich material into space.
The remnants of these ancient supernova explosions could be a neutron stars or a black holes. These are almost untraceable/unobservable now, but there should be around a billion of them in our Galaxy.

The second category is the progenitors of what are called type Ia supernovae. These are thought to arise from the thermonuclear explosion of a white dwarf star. White dwarfs are the end point of the evolution of less massive stars. Iron-producing type Ia supernovae $>4.5$ billion years ago, would have begun as stars with masses between about 1.5 and 8$M_{\odot}$. These would have burned hydrogen, then helium to produce a degenerate carbon and oxygen core. This core then simply cools and fades away as a white dwarf star in most cases. In type Ia supernovae, some event later in their lives, either mass transfer from a companion, or merger with a companion, caused the white dwarf to exceed its Chandrasekhar mass and triggers an instability that leads to the rapid total consumption of the star in a thermonuclear explosion. The products of this explosion include a large amount of Nickel, that then radioactively decays into Iron. Nothing is left of the white dwarf.

EDIT: Having established this we can begin to look at your edited question. Firstly, the gas and stars in the Galaxy are basically orbiting the Galactic centre. The orbital period at the Sun's radius is about 230 million years, so it has completed many Galactic orbits. Not only that, but it could have migrated in orbital radius as well. There are claims and counter claims in the literature and the issue is not settled. The Sun could have moved in or out by a significant fraction of its current Galactic orbital radius.

The high mass progenitors of core collapse supernovae will have been born (and died) very close to the Galactic plane. The same is not so true of type Ia supernovae, which had longer lived progenitors that could have moved significantly from the Galactic plane before exploding, and indeed would themselves have orbited the Galaxy many times. The gas expelled in a supernova explosion spreads out (over thousands of years) over tens of light years and becomes mixed into the interstellar medium. The interstellar medium is itself stirred and mixed by the energy input from these supernovae, but also due to the heating and winds of other stars, the tides of the galaxy and spiral arms. The interstellar medium appears to be quite homogeneous in terms of chemical composition, though radial gradients exist with scale lengths of order ten thousand light years.

In conclusion, what you ask is almost impossible to answer. The solar system iron almost certainly came from countless supernovae with a variety of progenitors, that would have exploded any time between almost the birth of the Galaxy 11-12 billion years ago (in fact the supernova rate was likely higher then) up until the Sun's birth. The biggest contributors would come from those stars inhabiting an annulus of many thousands of light years across, centered roughly on where the Sun was born, which is itself uncertain.


What would I see with a telescope of 70mm aperture , 300mm focal length and maximum magnification of 150x

It's 70mm aperture, so the maximum useful magnification will be 165x.
If you have a 4mm eyepiece, you will have 75x magnification, and at that, you can expect to see rings of Saturn, moons of Jupiter in good conditions. Venus will show phases. I can't say much about Mercury.
You can see many surface features of the Moon. The view will be particularly good at 150x. You will have to use a 2x Barlow for that. Remember, the view will be good in good "seeing".
If you are in dark sky away from light pollution, you can expect to see many globular clusters and nebulas particularly, the Orion Nebula.


Which eyepieces I can use it for best viewing experience using my existing telescope?

Every scope has a minimum and maximum useful magnification (minimum on your scope is 18X and maximum is 152X - from specs on Telescope.com). Anything outside of those numbers will compromise your ability to see well through the scope. To figure out how best to make use of your scope within the restrictions of useful magnification, you need to start with how to calculate magnification from focal length.

The magnification is equal to the focal length of the telescope divided by the focal length of the eyepiece. For example, your 25mm eyepiece would allow you to get 28X magnification (700mm / 25mm = 28X). Using the same formula, your 10mm eyepiece would allow you to get 70X magnification. For an easy way to get close to 152X, you could purchase a simple 2X Barlow. (700mm / 10mm) * 2X = 140X. For now, start with your 25mm eyepiece and get used to the night sky through it. You can always step it up a bit with other eyepieces later.

So you want to see Jupiter? I think you should just barely be able to make out the dark bands in the atmosphere around 100X (so maybe a 6mm or 7mm eyepiece for your scope). You will definitely be able to see the Galilean Moons though. And speaking of moons, you should get a moon filter to cut down on the brightness when looking at our Moon. I've found it's best to observe the edge of the Moon where light turns to dark. The shadows created along the craters and mountains are magnificent to see. Enjoy!


Are there other proofs of the expanding universe apart from the redshift?

Yes, there is direct, non-red-shift evidence of expansion.

The past temperature of the Cosmic Microwave Background Radiation (CMBR) has been directly measured and found to be substantially higher than it is today.  Its reduction in temperature over time is direct evidence of expansion.  Here are the details:

According to this paper, the CMBR was measurably hotter in the past (less-technical synopsis here).  The researchers observed absorption lines in a gas cloud located in a distant galaxy and found that the pattern of lines seen could only be explained if the CMBR temperature at time of absorption was between 6 K  and 14 K (it's 3 K now).  This temperature is consistent with the expected temperature for that galaxy's redshift (9 K).  Note that the temperature was measured from the specific pattern of lines seen and not from how much the lines had been red-shifted; this measurement would yield the same temperature even if there were no red-shifting.  Since a hotter temperature implies a higher density, this cooling of the CMBR over time is direct evidence for expansion of the universe.

Additional comments

What is the relationship between red-shift and absorption lines?

Inspired by a conversation with uhoh in the comments:

In my answer I refer to a "pattern" of "absorption lines".  For those not versed in the topic allow me to explain.

When a light shines through a cloud of gas specific frequencies of light get absorbed.  When this light is then shone through a prism the blocked frequencies will appear as black lines in the spectrum (see illustration below).  The exact lines that appear and their positions in the spectrum (the "pattern" of "absorption lines") depends on the elements present in the gas and the gas's environment.  The effect is most clearly seen with a light that emits photons at all frequencies; this kind of light is known as black-body radiation.  Although emitting light at all frequencies, a black-body radiator will emit the most light at a particular wavelength; the location of this peak is referred to as the black-body's temperature.  The CMBR discussed in the question is an example of black-body radiation.

Source: Doppler Shift, Edward L. Wright
(Excellent site BTW, the FAQ is worth a look for more info on red shifts and cosmology in general)

As light is travels through (expanding) space it's wavelength and the wavelengths of the absorption lines stretches at a fixed rate for all frequencies.  Let's say that at time of emission/absorption a spectrum shows lines at wavelengths of 1, 3, and 5 nm1.  After the photons have traveled for a certain amount of time, all of the spectrum's wavelengths will appear to have doubled2.  The line formerly at 1 nm is now seen at 2 nm, the one formerly at 3 nm is now seen at 6 nm, and the one originally at 5 nm is now seen at 10 nm.  Though their absolute frequencies change over time, the ratio of the lines' wavelengths (and frequencies) relative to one-another remains constant.

The precise amount that a given object's spectrum is shifted directly correlates with its distance.  As seen in the diagram above, close objects (like the Sun) show no red-shifting.  As one looks at objects further and further away one sees increasing amounts of red shift3.

In the discussion in the answer above, it's this pattern of relative positions in the lines that is affected by the CMBR temperature at time of absorption and not the degree to which the lines have been shifted.

1 To put it technically, this point is at $z=0$ where $z$ indicates the magnitude of the shift, positive for red shifts (moving away) and negative for blue shifts (approaching).  A more in-depth discussion of this topic (including the precise definition of $z$) can be found here.
2 The wavelength-doubling (frequency-halving) point is at $z=1$
3 It should be noted that since there is some uncertainty in the rate that the Universe is expanding at, red-shifts do not refer to precisely-known distances.  Thus astronomers and cosmologists rarely refer to the distances to distant objects in absolute terms of, say, light-years or parsecs, preferring, rather, to use the amount of red-shift observed (the $z$ mentioned above).

The mechanism behind red shifting isn't that the photons themselves are changing rather it is that the very space that the electromagnetic waves are moving through is expanding.  (Photons are both particles and waves; no, it's not exactly intuitive.)  This constant stretching of space stretches the light's wavelength giving rise both to the effect of red-shifting and the increase in a given photon's red shift over time.

Douglas Hofstadter, CC A-SA 3.0

How does red-shift relate to the CMBR?

In the comments Alchimista asked "Isn't CMBR actually the quintessence of red-shift anyway?"
(I'm assuming you're using the common, and not cosmological, meaning of "quintessence")

Yes, the current CMBR temperature (3 K) is generally agreed to be the result of relatively high-energy photons (3000 K) emitted about 380,000 years after the Big Bang that have had their wavelengths stretched over time by the expansion of the Universe towards the red (i.e. cooler or lower-energy) end of the spectrum.  This expansion was inferred by Hubble et al. from the observation that smaller and dimmer galaxies (as seen from Earth) have a greater shift in their spectra.  The farther the apparent distance, the greater the observed shift.  Using this apparent distance-correlated red shift we can infer that the Universe was smaller in the past and thus denser with a higher temperature for the CMBR.  Based on observed red shifts of distant galaxies, we can then deduce, but not directly measure, what the CMBR temperature was at each distance.

What the authors of the above paper did was make a direct measurement of the CMBR's temperature at a specific time in the past.  The measured temperature is higher than it is today which implies a denser and thus smaller Universe.  The researchers further found that the directly-measured temperature fit neatly with that inferred from the observed red shift of the galaxy being studied.

In a nutshell, the chain of inference is swapped:

For reasoning based on red-shifting:
Increasing red shifts with apparent distance (directly measured) ⇒ Expansion ⇒ Denser Universe in the past ⇒ Higher CMBR temperature in the past.
For a direct measurement of past temperature (as with this paper):
Higher CMBR temperature in the past (directly measured) ⇒ Denser Universe in the past ⇒ Expansion ⇒ Observed red shift.

These two inference chains based on different sets of evidence neatly complement and support one-another.

One thing to note is that the CMBR wasn't created by expansion (at least not directly) rather it's expansion that explains its current temperature and uniformity.  Per the Big Bang theory, the early universe was very dense; so dense and hot that all matter was a plasma of subatomic particles, opaque to photons. At about 380,000 years after the Big Bang the Universe had cooled (through expansion) enough that protons and electrons could combine to form neutral Hydrogen gas (which is transparent). The CMBR is the light that was set free at this time and has been cooling ever since.


Why is it okay to watch a sunset but not an eclipse?

During a sunset, the Sun is lower in the sky than during most of the day - much lower. Therefore, light from the Sun travels through about 120 miles of dense atmosphere, compared to the roughly 2 miles it travels through from straight up. Here's a rough sketch (not to scale) to demonstrate this. It is clear that $B>A$:

Light scatters in the atmosphere; in fact, one type of scattering is why the sky is blue. The longer travel distance means that there is much more scattering of ultraviolent light, which in turn means that the light you see is less intense.

Okay, you say. But doesn't the eclipse still block a lot of light? Well, unless there's 100% coverage - totality - there's still plenty of light coming from the uncovered part of the Sun, and that matters. The uncovered part is as bright as it normally is, and looking at that part is just as dangerous with or without the eclipse.

There's one more thing to consider, which is that people watching a sunset don't look at the Sun; they look at the clouds and sky around the Sun. If you look directly at the Sun, your eyes will be damaged, no matter what's happening, an eclipse or a sunset.


Why does the sun have a convection layer and a radiation layer?

The onset of convection is controlled by temperature gradients (see the famous experiment from Rayleigh & Bénard. When those become too strong, convection sets in.

How strong a temperature gradient gets, depends on the local efficiency of energy transport inside the star. But for the sun (and other lower mass stars), the surface radiates energy away efficiently, dropping the temperature there compared to the rest of the star, making the outer layers convective. Lower temperatures (not gradients) compared to high-mass stars also play a role in controlling the energy transport for lower mass stars. Cool temperatures will allow the formation of $H^{-}$ (negatively charged hydrogen ions) that give high opacities and thus prohibit energy transport.

For high-mass stars it is the enormously fast fusion processes in the core that 'overheat' this region, again letting convection take over.


What is the ultimate fate of a cluster of galaxies?

As you point out, in an accelerating Universe, large scale structures will become more and more isolated. So at a certain point you will have gravitationally bound superclusters separated by very large voids and less and less filamentary structures.

Once isolated, we can then study the dynamics of these independent superclusters. On very large time scales, galaxies will collide and merge. After collisions you will tend to form elliptical galaxies. So I think that you will end up with a big single elliptical galaxy.

Then we can be interested in the future of the stars of these galaxies. First we currently see that the star formation rate has already peaked some billion years ago. So as the number of galaxies collision that usually targets star formation, the star formation rate will slowly continue to decrease. Moreover, as heavy elements (all elements apart from hydrogen and helium) are formed in stars, the future generation of stars will have more and more heavy elements. From the nuclear point of view, the most stable element is iron, so on very very large time scales light elements will be converted into iron, whereas heavy elements will decay into iron.

This is a little speculative, but I think that on large time scales, more and more interstellar gas and stars will fall at the center of the gravitational potential of super elliptical galaxy. So as the density will increase at the center, you will mechanically form a heavier and heavier supermassive black hole. Another interesting point is that we do not currently know whether proton are stable or not. So on time scales larger than $10^{30}$ years (see more details here) protons may naturally decay into lighter subatomic particles.

So maybe as $t\rightarrow\infty$ we will end up with supermassive black holes and light particles. But as you mentioned, black holes will themselves slowly lose mass by Hawking radiation. In the same time, the expansion rate may have increased significantly leading ultimately to isolated particles in an expanding Universe.

Note: this is an hypothetical scenario and there are a lot of unknowns


How many of the whole zodiac constellations can be seen in one single night?

Running a simulation in Stellarium, on March 21st, you can just about see Pisces at about 7pm, but it is setting, and at 5am Aquarius is just rising, before being hidden by the sun. The sun is "in Pisces" on the 21st (the astrological houses have not been updated) but it is a pretty big constellation!

This means that you can see, or partly see all the zodiac constellations.  Although Pisces and Aquarius are not fully visible, but you can see the other 10 at some point during the night.


Will just a glimpse (less than a second) of sun during partial solar eclipse damage eyes?

Glancing at a partial solar eclipse is about as dangerous as glancing at the Sun on any other day. If you look at the Sun a few minutes after sunrise or a few minutes before sunset, when the Sun's altitude is low, the light is filtered through a lot of air, and most of the ultraviolet is scattered, so it's a lot less dangerous than looking at the Sun in the middle of the day. However, during an eclipse, people tend to look at the Sun for more than just a few minutes, and that's not healthy for the eyes.
The big danger of looking at a solar eclipse without proper protection occurs when it's a total (or nearly total) eclipse. When the eclipse is near totality, the sky is dark, so your pupils dilate, letting in more light. But the light rays coming from that small sliver of Sun are just as bright as normal, and have the same proportion of UV (ultraviolet) as full sunlight. Those UV rays can cause a lot of damage to the retina very quickly, and that damage can be permanent.


What is the boundary in the atmosphere seen in this video?

Yes this is neither aurora, nor reflection of any light.
This is airglow, that comes from upper layers of atmosphere.
The brightest emission is green 558nm light from excited atomic oxygen in a layer 90-100 km high. At that height most of the oxygen is in atomic state, not molecular, and is constantly being exposed to solar UV radiation.

Here you can find more details.
 airglow formation


Yes, it would be possible. There are two roads here:

Visible light
In case of detecting light in the range of visible wavelength, perhaps you would consider that rare occasion when there is a solar eclipse.
And it may also be possible at times of 'early' dawn and 'late' dusk.

Invisible light (outside visible range, beyond the IR and the UV)
Now, other than that, we have telescopes for radio, infra-red, UV rays and X-rays which wouldn't require us to look into the 'eyepiece'. Astronomers have detectors for that and light received is stored in 'pixels'.

You might also want to look at a similar question in Physics-SE here.
As pointed out by Jeremy in a comment below and also pointed out here, it is actually a far fetched idea. You need to be really far away from the sun to use it as a G-lens. The telescope has to be around 50 billion miles away and to cover that distance within one generation, probably we haven't achieved that technologically advanced level yet.


How to disentangle a very distant star's relative velocity vs. redshift distance

Conrad is almost right. It is true generally that if a Galaxy is close enough to take spectra of individual stars (e.g. luminous supergiants) then it is not far enough away to be regarded as part of the "Hubble flow" and so applying Hubble's law to this star, or its host galaxy, would not yield a reliable distance in any case, but would reflect the "peculiar motion" of that galaxy.

To put some numbers on this. Galaxy peculiar motions tend to be a few 100 km/s, as do the individual velocities of stars with respect to their galaxies. Taking a Hubble constant of 70 km/s per Mpc, we see that we need to be at distances of 15 Mpc before Hubble recession velocities ($v = H_0 d$) become large compared with peculiar motions. At these distances we cannot observe individual stars - they are too faint and unresolved from the bulk of the Galactic light.

The exceptions are supernovae. The redshifts of individual supernovae, that briefly outshine their galaxies, can be measured right across the universe. Here you are correct that the measured redshift is a combination of cosmological redshift due to the expansion of the universe and a velocity of the star relative to the Hubble flow at that distance. There is no way to distinguish between these two unless velocity measurements could be obtained for other objects in the same galaxy. Given the rarity of supernovae, we might wait a long time for this.

But does it matter? Even if we look at a "low redshift" supernova at $z=0.1$, its Hubble recession velocity is 30,000 km/s and far in excess of any peculiar velocity contribution at the level of $\sim 1$%.


Can you give some details about this process, is the orbit changing
  shape, eccentricity?

This isn't a simple answer.   Also, do you want to know this for a simple 2 body system or the more complex, essentially 3 body system of Sun-earth-Moon.

If the Earth and Moon were alone in the cosmos, far away from everything else, then over time, the tidal forces would circularize the Moon's orbit as it pushes the Moon away.  The reason for this is that the same force that slowly pushes the Moon away from the Earth pushes hardest when the Moon is closest, and least hard when the Moon is furthest away, so over a long period of time that circularizes it's orbit.  All 2 body systems where the moon is being pushed away tends to circularize.  This can be observed in our own solar-system.   Two body systems that are very far away from orbital disturbances, like Pluto-Charon or Neptune-Triton or even, Sun-Neptune, are highly circular orbits.

But when you bring the Sun into the picture, everything changes.  The Moon orbits the Earth and the Earth Moon system orbits the sun, and in a very real sense, the Moon also orbits the sun, but it's within the Earth's stable region of it's hill sphere, so the Moon's orbit around the Earth will remain stable for billions of years.   But while it's stable, it's measurably inconstant.  The moon's orbit wobbles a great deal, being stretched and contracted based on it's position in relation to the Earth and Sun.  3 body mechanics is quite complex, even with stable orbits.  Isaac Newton was aware of this but he wasn't able to work it out.  Mathematicians who came later were able to figure out orbital interactions with more than 2 bodies, called perturbation theory.

So, your question depends on whether you mean a 2 body system or a 3 body system.

The Earth's eccentricity, for example, changes on roughly a 100,000 year cycle, mostly due to Jupiter, but that effect is pretty slow, on a roughly 100,000 year cycle.

The Moon's apogee and perigee happen every 27.3 days (Sidereal orbit) and it's eccentricity changes measurably every orbit.  In a sense, it's not very accurate to say the Moon even orbits around the Earth in an ellipse, it's too squashed and stretched for the elliptical model to be accurate.

While the Earth takes about 100,000 orbits to go through one elliptical period, the Moon needs only 6 or 7 orbits to do that.   Note how measurably it's perigee and apogee change every orbit.   See table.

There's a fair bit of information available on the Moon's irregular orbit, but I suspect that has almost nothing to do with your primary question.   In general, in the 3 body system, as the Moon moves away from the Earth it's orbit grows more erratic, so in a sense, because of the sun, the Moon's orbit grows more eccentric, over time, where, if the sun wasn't there, it would grow more circular.

  is the average orbital speed decreasing, by what value? is the spin
  angular velocity increasing, by what value?

On average, as the Moon moves away it's orbital speed decreases.  More distant orbits have less kinetic energy, but again, with a 3 body system it's more complex.

OK, I think I see what you're asking on Spin.

  from Wiki "angular momentum is slowly being transferred from Earth's
  rotation to the Moon's orbit."

and

  There can be no transfer, since both earth and moon are slowing down

You should look at the Earth-Moon as a closed system.  For a closed system, angular momentum remains constant, but angular momentum in this sense, is around the barycenter.   The Moon's angular momentum, in the sense that Wikipedia means isn't the Moon's spin but the Moon's orbit.  As the Moon moves away from the Earth, it's velocity decreases but it's angular momentum around the barycenter increases because angular momentum is a product of both velocity and distance from the center.

Jupiter, for example, has about 71% of the total angular momentum of the solar system.   See article, which I think is related to your question.

Finally,

  wiki says that average distance from earth increases 3.8 cm per year,
  can you tell how much of that is due to the expansion of the universe?

Expansion is best understood if objects aren't orbiting around each other.  If the Moon was stationary and the Earth was stationary, and the only way to do that in our thought experiment is to give one of them a mass of zero so there's no orbit and no gravitational attraction.   We're breaking almost every rule in the book with this thought experiment, but bear with me.

If we take the measured expansion of space to be 74.3 km/s over 3 million light years.   Source.

Average lunar distance 384,400 km.  And, 9.461 trillion km in 1 light year, so about 2.8 x 10^19 km in 3 million lightyears.  Roughly 74 trillion lunar distances. which works out kind of neatly, 74.3 km/s divided by 74 trillion gives us one trillionth of a KM/s, 31.5 million seconds in a year gives us about .0000315 km/year expansion between a stationary, no gravitational attraction moon and Earth, which equals about 3.15 cm per year of expansion of space over 1 lunar distance, which is curiously not far off from 3.8 cm per year increase in distance due to tidal forces, but the similarity is purely coincidental as well as irrelevant.

(too long?)


How old is the oldest light visible from Earth?

The oldest light in the universe is the cosmic microwave background. Roughly 380,000 years after the Big Bang, protons and electrons "recombined"1 into hydrogen atoms. Before this, any photons scattered off the free electrons in the plasma filling space, and the universe was essentially opaque to light. Once recombination occurred, however, photons were able to "decouple" from the electrons and move through space unimpeded. This relic radiation is still observable today; it has been redshifted and cooled.

We can detect light from very distant objects, and we have. It makes more sense to talk about distance in terms of redshift; the larger the redshift, the farther away an object is. There are a number of extremely high-redshift objects, some of which have had their measurements confirmed, and others of which have not. Candidates include

MACS0647-JD (redshift $z=10.7^{+0.6}_{-0.4}$), a very distant galaxy.
UDFj-39546284 (redshift from $z=10$ to $z=12$), another galaxy or protogalaxy (although its distance is uncertain, and its exact nature is disputed).
GN-z11 (redshift $z=11.09^{+0.08}_{-0.12}$), a very luminous distant galaxy.

All of these objects would have formed some hundreds of millions of years after the Big Bang, however, so the light we see from them is much "younger" than that of the cosmic microwave background.

1 I've never liked the usage in this context, as this was the first time they combined; the "re" is kind of misleading.


number density of Dark Matter Halos Versus number density of Galaxies

No. You need to specify structure formation and galaxy evolution for details, but in general, you will expect many galaxies in a dark matter halo. If you need a relation, the halo occupation distribution might be a start. The Wikipedia page references three good texts on the matter.


"Perfectly" is a funny word.

Perfect circles are a mathematical abstraction. Real objects are not "perfect".  So supposing a "perfectly spherical planet" is to suppose something that does not and could not exist. All real planets are made of atoms and anything made of little clumps of matter cannot be perfectly spherical.  Even if you built a planet that was as spherical as possible, it would be distorted by its rotation and the tides. So there are no perfectly spherical planets.

Now you say "put in a perfectly circular orbit". This is like drawing a line that is exactly $\pi$cm long. Again you are supposing something that does not, and could not exist.

What we can do is consider a mathematical model of gravity. If you model the sun and the planet as "particles" (ie point masses) and you model gravity with Newton's law of universal gravity, and if you give the model the system with the exact amount of energy to give a perfect circle, then the system will remain in a perfect circle, it will never become elliptical.

If you use general relativity to model gravity, then the release of gravitational radiation will mean that no circular orbits are possible, all orbits will spiral inwards, however it would not become elliptical. Something similar will happen with quantum models of gravity.

So your question can only be answered in the context of a mathematical model of gravity.


Why do pair production and photodisintegration have such different effects in collapsing supermassive stars?

A bit of wild speculation that I'll need to check.

Photodisintegration produces simpler particles. These simpler particles are able to "neutronise" (i.e. undergo inverse beta decay/electron-capture) with electrons, because the energy threshold for neutronisation is much lower for free protons and alpha particles than it is for say oxygen or iron nuclei. This process removes free electrons from a core that is supported by electron degeneracy pressure and it collapses.

Pair production converts photon energy into electrons and positrons. The net effect will be to take kinetic energy out of the gas and turn it into rest mass. This will lower the pressure, but an increase in density/temperature can restore it.

I think the process of neutronisation/electron-capture is the key difference. The threshold electron energy for neutronising a free proton is 1.3 MeV, but is about 11 MeV for a proton inside an oxygen nucleus, so photodisintegration is required to trigger collapse (or the Fermi energy of the electrons reaching the neutronisation threshold).


The mass of Vesta was determined by observing its effect on 197 Arete. The model used is that Vesta will cause the eccentricity and semi-major axis to change. By observing the asteroid before and after their encounter, and determining the asteroid's orbital elements before and after, one can find the angle of deflection. Model the motion as a hyperbolic encounter and one can find the combined mass. Since 197 Arete is much smaller than Vesta, this gives a good estimate of the mass of Vesta.

The mass of asteroids can also be found by their perturbations of Mars, providing there are sufficiently accurate observations. Viking radar data (with an accuracy of 7m) allowed for good estimates of the mass of the major asteroids. The Dawn mission to Vesta and Ceres made very high accuracy measurements of the gravity field around them and so their mass.

The math involved is essentially just Newtonian mechanics. You know the positions and velocities before the encounter. There is an energy conserving encounter with a force of $$F = \frac{m_1 m_2 G}{r^2}$$ Finding the angle of perturbation allows you to find the reduced mass $$\frac{m_1 m_2}{m_1+m_2}$$


Why does the Earth have a z-component at the start of the J2000 epoch?

You made the same fundamental mistake that Anton Gromov made in his question on the sister Space Exploration StackExchange network site: You used the solar system barycenter rather than the Sun as the frame origin. Had you used the Sun, the apparent discrepancy would have dropped by almost two orders of magnitude.

A much lesser flaw is that you apparently used midnight rather than noon. The J2000.0 epoch is 12 noon Terrestrial Time on 1 January 2000. This doesn't change things by much, but it is nonetheless important. That half day offset has bitten me more than once.

The z coordinate of the Earth's position would not have dropped to zero had you made the above corrections (Sun-centered rather than solar system barycenter, and 12 noon rather than midnight). It would not have dropped to zero even if you had used the Earth-Moon barycenter rather than the center of the Earth.

One reason it would not have dropped to zero is that the Earth's orbit about the Sun is not an ellipse. The Earth's orbit doesn't even lie on a plane! The key cause of this non-planar motion is that the Moon's orbit about the Earth is inclined by about 5.15° with respect to the ecliptic, making the Earth bob up and down with respect to the ecliptic1. Because the Earth's orbit is not truly elliptical or even planar, the ecliptic plane at some epoch is a time-averaged plane centered about the epoch time that makes the Earth's position with respect to the Sun (or the Sun's position with respect to the Earth) average to zero.

A lesser reason is that the J2000.0 mean ecliptic and mean equinox frame was defined over 36 years ago. HORIZONS uses DE431, which was released 6 years ago. JPL's ability to model the solar system improved significantly in the intervening three decades.

1 Strictly speaking, it makes the Sun appear to bob up and down with respect to the ecliptic. The concept of the ecliptic plane dates back to the ancient Greeks, and it retains a vestige of this definition to this day.


How much did we know about asteroids at the beginning of the 20th century?

Well as late as 1942, with a reference to book The Little Prince, by Antoine de Saint-Exupéry, we didn't even know the correct shape of asteroids, believing them to be round like planets. It was in 1971 that we got a look at the moons of Mars that we got a look at what an asteroid would really look like, although we are now thinking Phobos and Deimos were formed with an impacted. The best information for what your asking can come from XKCD's What If

  Before the 1970s, it was common for science fiction to assume small
  asteroids would be round, like planets.[4] The Little Prince took this
  a step further, imagining an asteroid as a tiny planet with gravity,
  air, and a rose. There's no point in trying to critique the science
  here, because (1) it's not a story about asteroids, and (2) it opens
  with a parable about how foolish adults are for looking at everything
  too literally.

http://what-if.xkcd.com/68/

So we had we had fifteen asteroids discovered by 1851, and until 1891, all asteroids had been discovered by telescope and eye alone, at which point Max Wolf pioneered the use of astrophotography to detect asteroids. Max Wolf himself ended up credited with the discovery of 248 asteroids.

So while I don't know when Mark Twain made that quote but as far as refering to asteroid, I'll quote wikipedia for the term 'asteroid'

  Traditionally, small bodies orbiting the Sun were classified as
  asteroids, comets or meteoroids, with anything smaller than ten metres
  across being called a meteoroid.[21][22] The term "asteroid" is
  ill-defined. It never had a formal definition, with the broader term
  minor planet being preferred by the International Astronomical Union
  from 1853 on. In 2006, the term "small Solar System body" was
  introduced to cover both most minor planets and comets.[23]

and

  The minor planets beyond Jupiter's orbit are sometimes also called
  "asteroids", especially in popular presentations.[30] However, it is
  becoming increasingly common for the term "asteroid" to be restricted
  to minor planets of the inner Solar System.[26]

http://en.wikipedia.org/wiki/Asteroid

Mark Twain could have been referring to quite a few objects with the unclear definition of asteroids.


Do the actual false colours in the M87 black hole picture convey information?

See this article recently published by the EHT collaboration describing how they generated the image. Specifically, chapter 5 of that article describes the origin of the image. To quote the article (specifically the caption of figure 3):

  The image is shown in units of brightness temperature, ${T}_{{\rm{b}}}=S{\lambda }^{2}/2{k}_{{\rm{B}}}{\rm{\Omega }}$, where S is the flux density, λ is the observing wavelength, $k_{B}$ is the Boltzmann constant, and Ω is the solid angle of the resolution element.

The false colors in the image convey the surface T brightness (a sort of measurement of the intensity or flux from that area) of the material surrounding the black hole's event horizon. Brighter color means higher brightness temperature. There is not information about the physical T of the in falling materials as the collected radiation is of synchrotron type and not that of a black body. As you say, the image could have been shown in grayscale or any other color scale. The fact that it is shown in an orange-ish color is merely the convention chosen by the scientists who generated the image.

BUT there's a lot more information in the image than just how bright the material is. For example, the fact that the material is brighter on one side than on the other can tell us about the geometry of the material surrounding the black hole, or about how that material is rotating around it. If you have the time, check out this article and read up on the actual journal articles published by the Event Horizon Telescope collaboration linked at the bottom.


Why doesn't star size distribution conform to space rock size distribution?

You are comparing distributions in a way that they are not easily comparable and the eye is misled: watch the scaling of your axes of the plots you compare! In order to compare, you want to make sure that you use similar, either log-log for both graphs or linear-linear or something else - but identical in both graphs.
Mind also that the size distribution of the small solar system objects is NOT an exponential, but a power law distribution. A power law distribution is a sloped straight line in a log-log plot and the exponent of the power law is found in the slope in such plot.
The stellar size distribution usually is found under the keyword "initial mass function". And when you look at that (see the plot therein), you will find, that the size distribution of stars also follows a similar power law as does the size distribution of smaller bodies as found in our solar system. This ignores the evolution of stars - but that is what you want to do in order to compare the sizes of objects as they are created.
The slopes of the size distributions course differ as different processes play a role in formation of stars and the objects forming around a star-being-born.


The galaxy is kept together by the combined mass of the matter in the galaxy, of which the supermassive black hole is a negligible part. There are galaxies that don't have a central black hole (such as the Triangulum galaxy), and they are also held together by their combined mass.
In particular, the dark matter of the galaxy is what provides most of the mass that holds the visible matter together. Galaxies usually have far more dark matter than visible matter. This mass is distributed in a roughly spherical shape, with the greatest density towards the centre of the galaxy, and dropping off further from the centre. This "dark matter halo" is substantially larger than the visible disc.


Could martian meteorites come from Olympus Mons?

Since this has gone unanswered, I'm going to give it a shot.

Probably the answer is no.   We know that on some moons, volcanic eruptions do send material into space.

Enceladus for example, feeds Saturn's E-ring.   It regularly shoots about 200 KG per second into space.

The other highly volcanic moon in our solar system is Io which shoots material high into it's orbit, and it has some of the largest if not the largest observed volcanic eruptions.   Source:   http://www.space.com/26732-jupiter-moon-io-volcano-eruption-photos.html

But I couldn't find any references that Io's volcanoes shot material into orbit.   short article on Jupiter's rings

The difference between Enceladus and Io is size.   Enceladus's escape velocity is 0.239 km/s (860.4 km/h) Source in link above, and it's gravitation is just over 1% of Earth's.    Io's escape velocity is 2.558 km/s and it's gravity is similar to the Moon's, about 18% of earth's.  If Io has a hard time sending volcanic material into orbit, on Mars, it would be nearly impossible.

Io can shoot material up to 300 miles into the air, Source:  https://en.wikipedia.org/wiki/Volcanology_of_Io  but that's still only about 1/7th to 1/8th of it's radius, probably well within it's Sphere of Influence.  On Mars, it's Sphere of influence is significantly larger and it's gravity twice as much.   If a super large Volcano shot material 300 miles into the air above Mars, that's still well below escape velocity.  (I could run the Maths if you like, but in a nutshell, escape velocity changes with the square of the distance, so, ignoring air resistance, Io's enormous eruption reaches 14% of it's diameter or 28% of the radius is about 40% of escape velocity - assuming I got my math right.   On the Mars, an equivalent enormous eruption would be what, 20% of necessary escape velocity?  Less?

Also, while Olympus Mons is enormous, it didn't all erupt at once.   Source:  http://www.space.com/20133-olympus-mons-giant-mountain-of-mars.html

Olympus Mons is over a hot spot on Mars, subject to regular volcanic eruptions (regular meaning, perhaps every few million years), but because Mars doesn't have plate movement, the eruptions happen in the roughly the same location and that, in combination with virtually no erosion and low gravity, is why Mons is as large as it is.   It probably does erupt impressively when it goes, but I'd be pretty skeptical that it erupts violently enough to send magma into space.

Also, most of the Martian meteors on Earth are from a recent impact, about 5 million years ago:    http://www.space.com/24962-mars-meteorites-single-impact.html   Olympus Mons may have erupted about 5 million years ago, so the timing works, but I still think it's unlikely.

(I could add some links on how far Earth volcanoes can send magma/pumice if interested, I looked that up as well).

To get that kind of velocity, you need significant amounts of concentrated energy.   A gun works by creating a controlled explosion with only one place for the energy to go.   Rockets work by sustained acceleration over time.  While it seems quite different, a meteor strike is similar in some ways to an enormous gun.   When a meteor strikes the surface of a planet, it hits with so much force in a localized area that the ground get's crushed and then rebounds with enormous force in a localized area.   A volcano simply can't (I think) generate that kind of concentrated energy unless it's very low gravity like Enceladus, 1.1% of the gravity of Earth.

That probably needs tidying up a little.   I'll try to clean it up later.

A somewhat related conversations on this subject if interested:   http://www.reddit.com/r/askscience/comments/1gkiz3/could_a_volcano_eruption_theoretically_be/


It is the remains of dead stars.

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

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


The problem may be focus and or a faint image in the viewfinder, try looking at a distant object in daytime.

Of course unless you are using a dslr type camera the viewfinder will not be looking through the main optics anyway so wont show anything.


Is it possible that quasars are not powered by black holes but some other kind of compact, massive object?

The solution to your question is surprisingly simple, I think:

A quasar that puts out energy around Eddington luminosity or higher, must accrete at a certain rate, corresponding to the energy output.

When consulting a lecture and a random paper from the archive on this topic, it is evident that those accretion rates correspond to 1-10 solar masses per year or higher. A neutron star accreting 1-10 solar masses in one year will become a very massive black hole quickly enough.


How often do they move most/all of ALMA's dishes?

The list of array configurations is given in the Proposers Guide and ten configurations have been defined for the 43 12-meter antennas. They have the form C43-x where x goes from 1 (most compact; 15-161 meter separation) to 10 (least compact; 244-16200 meter separation). The changes between configurations are shown in the Configuration Schedule; over the ~12 months of Cycle 6, it looks like they change configuration 14 times. There is a month long "Antennae Relocation Shutdown" in May 2019 as they transition from fairly compact to least compact - presumably because it takes a longer time to move the 43 dishes out multiple kilometers

As an example, here is a snapshot of the planned schedule for a year starting 01-Oct-2018 as it stands now:


Why are stars still distant from one another?

The initial star formation regions were regions that have a high enough mass density to form a star. The density of the early universe was not constant at different locations. Some regions had high enough density to form a star, and some didn't.
When a star forms it draws in matter from a large distance away. This forms an accretion disk and leaves a temporary relatively empty space around the star at a large distance away. There is no way a star can form in this low mass region. Once you get further away from this region the matter density may return to a level where a star can form. But this region is far from a star.
Yes, gravity from a particular star is pulling at other stars but there are many stars pulling on the particular star of interest. So the net force on the star may be very small.


Do we see stars from other galaxies (not Milky way) in the night sky with the naked eye?

Stars in other galaxies are too dim for our eyes to see them as individual points of light. The brightest stars in the Magellanic clouds are magnitude 10, and we can only see magnitude 6 stars (bigger magnitude means dimmer)

So instead we see the Magellanic Clouds and the Andromeda galaxy as patches of light. The galaxy is visible, but its individual stars are not. The only time a "star" was visible in the Large Magellanic cloud was a supernova in 1987.

In a very sparse galaxy the night sky would be very dark, with few visible stars. Other galaxies might be visible, but they would be dim patches of light.

100% of the naked-eye stars you see are in the Milky Way. You can see Andromeda, but you can't see any individual stars in it.