Why do astronomers use supernova to measure distance in space?

Type Ia supernovae are not common; they are quite rare events, happening maybe once per 100-200 years in a big galaxy. Nevertheless they have two properties that make them fantastically useful for distance measurement.

They are (very close to) "standard candles". The physics of the supernova detonation, thought to be when a white dwarf accretes matter and exceeds the Chandrasekhar limit, is very "standardised". The bomb goes off in exactly the same way with the same amount of identical explosive. That means to a good approximation, measuring the apparent brightness of a type Ia supernova and comparing it with nearby examples means that the distances to these events can be accurately estimated.
They are really luminous and last 2-3 weeks. This means that they can be seen at enormous distances, they almost outshine the galaxies that they are in, and they last long enough for astronomers to discover them in automated surveys and still have time to follow them up and measure their light curves and estimate their peak brightness.

The limitations are that you can't choose which galaxies you measure the distance to. You have to wait for a supernova to go off and that might take 100-200 years or more for a particular galaxy. There are also continuing debates about just how standard a candle these objects are. It is possible that more distant galaxies that we see in the very early universe made stars with a different composition (far less elements heavier than helium) and that this alters things a bit. Other problems are associated with "de-reddening" the supernova light curves to account for the possibility of obscuring dust in the host galaxy. It is also possible that there is more than one way to produce a type Ia supernova (accretion of additional mass or merger of two white dwarfs).

I am puzzled by your question "Why can't the distances between our Sun and the celestial objects be measured directly?". Galaxies are much too far away to have a trigonometric parallax distance. Methods like using Cepheids or RR Lyrae variables don't work for very distant galaxies because we can't resolve the individual stars. You could estimate a distance using a measurement of redshift and a cosmological model for the expansion of the Universe, but really the point of using these supernovae was and is to test and improve the cosmological models.


Did we ever actually see the earth revolving around the sun? Is the geocentric model completely disproved?

What you're asking, basically, is whether there are any proofs for the heliocentric model of the Solar System.

A literal naked-eye observation of the Earth revolving around the Sun would be rather difficult, since human beings have never gone to another planet yet, and have only been to the Moon briefly, decades ago.

Here are several proofs; some of them are historically relevant also.

Kepler's laws of planetary motion

This became one of the earliest proofs, as soon as Newton figured out the law of universal gravitation, and the "fluxions" (what we would call today differential equations). When you assume a heliocentric model, and the inverse square law for gravity, then Kepler's laws in a heliocentric model come out of the equations naturally, as soon as you do the math.

This is like saying: "if it's heliocentric, and knowing that the law of gravity is correct, then Kepler's laws should be such-and-such". And then: "oh, but the theoretical calculations for Kepler's laws match the actual observations with great precision. Therefore, our hypothesis (heliocentric, inverse square law) must be correct."

It was the earliest strong indication that the heliocentric model is in natural accord with the basic laws of science, whereas the geocentric view was becoming more and more contrived as evidence accumulated.

Tycho Brahe in late 1500s provided the enormous mass of observations of planetary motion. Johannes Kepler in early 1600s used Brahe's observations to come up with his laws empyrically (and also arguing for the heliocentric model). Isaac Newton in late 1600s said "yes, Kepler is right, because of mathematics and the law of gravity, and here's the proof from calculus".

http://en.wikipedia.org/wiki/Kepler%27s_laws_of_planetary_motion

Stellar parallax

An early argument against heliocentrism was that, if the Earth was really revolving around the Sun, then very distant objects, such as the stars, would appear to be ever-so-slightly bobbing back and forth around their average positions. Since that's not the case, it was argued, therefore the Earth must be fixed.

You can see this argument in historical archives, proposed by theologians in the late 1600s, in favor of the geocentric model, and against the growing consensus then of "natural philosophers" (what we would call today scientists) that the heliocentric model was correct.

In reality, parallax does exist, it's just very small. It was measured experimentally in the 1800s, and was then quickly used to determine, for the first time, the distance to the nearest stars.

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

Aberration of starlight

The direction where we see a distant star also changes when Earth's speed vector changes during its revolution around the Sun. This is different from parallax; it's more akin to the way raindrops on the side windows of a car leave diagonal traces when the car starts moving (even though the raindrops fall vertically).

It's essentially a relativistic phenomenon (when applied to light), but it can be partially explained in a classic framework. It was actually observed before parallax in the late 1600s (the heyday of Newton), but it went unexplained until early 1700s.

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

Orbital mechanics of interplanetary probes

Landing a probe on Mars or Venus would simply not work if you assumed a geocentric model. A geocentric description of the Solar System might remain valid in a purely kinematic perspective (just the geometry of motion) as long as you remain on Earth. But the illusion breaks down quickly as soon as you start to consider dynamics (see Kepler's laws), and/or when you try to actually leave Earth (space probes).

Let me reinforce this point, since several answers and comments got it wrong: the geocentric and the heliocentric models are not completely interchangeable, or a matter of relativity. You could build an "explanatory" geocentric model, and it would be "correct", purely in a kinematic fashion (the geometry of motion), and only as seen from Earth. But the model breaks down as soon as you consider dynamics (forces and masses); it would also reveal itself as incorrect even from a kinematic perspective as soon as you leave Earth.

This is not just an artifice to simplify calculations. The dynamic calculations are wrong in a geocentric model.

In order to compute the very high precision trajectory of the space vehicle carrying the Curiosity rover and successfully place it on Mars, you must operate from a heliocentric perspective. The dynamics are all wrong otherwise. You would not miss the target only a little, in a geocentric approach, you would miss it by a lot - the vehicle would not even go in the general direction of Mars.

Miscellaneous

When observed in a telescope, Venus has phases like the Moon, and also grows and then shrinks in size, synchronized with its phases (it's large as a thin crescent, it's small when it's gibbous). In a geocentric model, the size changes could be explained by an elliptic orbit of Venus around Earth, but the phase changes synchronized with that are harder to explain. Both phenomena become trivial to explain in a heliocentric model.

It should be noted that this does not necessarily prove the heliocentric model, just the fact that Venus is orbiting the Sun, not the Earth. So it's an argument against pure (or strict) geocentrism.

Jupiter, when observed in a telescope, clearly has its own satellites. This was an early blow against a strict geocentric model, which assumed that everything must orbit the Earth. It opened the door to the idea that orbits could be centered on other celestial bodies too, and to the idea that things in orbit around larger objects could have their own smaller satellites (and therefore the Earth could orbit the Sun without losing the Moon).

The list could continue (and the full list is very long) but these arguments should suffice. You don't necessarily have to see something with your own eyes in order to know with certainty that it's there. In the case of Earth's revolution around the Sun, it was simply a matter of an overwhelming amount of evidence piling up in favor of it.

Geocentrism simply doesn't make any sense whatsoever in modern science and space exploration.


Is oxygen really the most abundant element on the surface of the Moon?

Yes, that's correct; it's also true for the Earth's crust. The reason is that "rocks" are typically made up of components containing combinations of silicon or one or more metals (e.g., magnesium, aluminum, iron) and oxygen, such as silica ($\mathrm{SiO}_{2}$); alumina ($\mathrm{Al}_{2}\mathrm{O}_{3}$); lime ($\mathrm{CaO}$); iron oxide ($\mathrm{FeO}$); and magnesium oxide ($\mathrm{MgO}$).
Examples of common lunar minerals formed from these components includes plagioclase feldspars (mixtures of NaAlSi$_{3}$O$_{8}$ and CaAl$_{2}$Si$_{2}$O$_{8}$), pyroxene (typically XYSi$_{2}$O$_{6}$, where X and Y are metals such as calcium, sodium, iron, magnesium, and aluminum), and olivine (made up of Mg$_{2}$SiO$_{4}$ and Fe$_{2}$SiO$_{4}$), along with oxide minerals like ilmenite (FeTiO$_{3}$). (Source)
Since in all these cases you have between one and two oxygen atoms for every non-oxygen atom, you end up with oxygen as the most abundant single element.


The image above (click for full size) shows the moon's path for much of July. The celestial equator is in purple and the ecliptic is in green. Note that:

On July 8th (meaning between July 8th at 0000 UTC and July 9th at 0000 UTC), the moon does indeed reach it's southernmost declination, but is still north of the ecliptic.
On July 12th, the moon, whose declination is increasing slower than that of the ecliptic, crosses the ecliptic from north to south, it's descending node.
On July 22nd, the moon is at its northernmost declination, but still south of the ecliptic.
On July 25th, the moon, moving south slower than the ecliptic, crosses the ecliptic from south to north, it's ascending node.

As the page you reference notes, the moon's northern and southernmost declination this month are +-19.4 degrees, about 4 degrees less than the tilt of the ecliptic. Thus, the moon changes declination slower than the ecliptic, at least for this month.


"Virtual Observatory" is a catchall phrase basically for a place any scientist can publish their data (ie. tables in their papers) or for hosting databases (such as the latest releases of major survey measurements) which comply with TAP protocols and so can be easily communicated with using a variety of astronomical data analysis programs.

There are many many virtual observatories, here, for example, is the German Virtual Observatory, which is run by the excellent people in Heidelberg: http://www.g-vo.org/

They have a link describing the data publishing process here: http://www.g-vo.org/pmwiki/About/PublishYourData

Usually what this entails is putting your data, such as tables from your paper, into a universal format, like VOTable. This type of table is readily accessed by astronomy data analysis tools such as TopCat (or Aladin, like you mentioned): http://www.star.bris.ac.uk/~mbt/topcat/

Programs such as TopCat can also pull data from these virtual observatories using the TAP protocol which is described a bit here in reference to the GAVO: http://dc.zah.uni-heidelberg.de/system/tap/run/info (note: this link is displayed incorrectly) and a bit here: http://www.star.bris.ac.uk/~mbt/topcat/sun253/TapTableLoadDialog.html

The main purpose of the VO type databases is to allow people to query them while referencing data they have on their computer by allowing SQL (ADQL) queries (crossmatching, selection cuts, etc.) The TAP protocol also allows programs such as Aladin and TopCat to communitcate, so you can do a crossmatch of your data in TopCat with a catalog on the VO and the data will be instantly displayed visually in Aladin as well. Vizier, as far as I know, is just a repository of databases and does not allow this type of querying and interaction.


Objection on artificial moon for lighting

This has been called "fake news" However tides are caused by gravity, and the proposal doesn't seem to require very massive satellites, hence no tidal waves.


How was the Messenger spacecraft used to measure Mercury's orbital precession to such accuracy? Could this have been done using radar?

The measurements discussed weren't done by Messenger, they were done by round-trip delay-doppler radar ranging using the coherent transponder of Messenger's Radio_science_subsystem.

Park et al. (2017) discuss separating the effects of the solar quadrupole moment from the partially degenerate effects of the post-Newtonian $\beta$ and $\gamma$ parameters (where $\beta=\gamma=1$ in General Relativity). The "general" General Relativistic correction to the perhelion prececssion itself is given by:
$$\Delta \dot{\omega} \propto (2 - \beta + 2\gamma) $$
and the effect caused by the solar quadrupole moment is only 0.07% the size of this. Therefore just observing the precession does not yield an accurate value of $J_2$ without assuming the value of $\beta$ and $\gamma$, and even then would need precisons in position measurements of hundreds of metres over decades.

Park et al. instead show that separating $J_2$ from $\beta$ and $\gamma$
 requires a precise measurement of a periodic modulation (i.e. it isn't an accumulating error) in the argument of perhelion caused by the General Relativistic perturbations, which translates to modulations of order of 5-10 km in position. These also depend linearly on $\beta$ and $\gamma$ only, and if the amplitudes of these modulations can be measured to better than $\sim 0.07$% ($\simeq 3.5-7$ m) then $J_2$ could be determined independently of $\beta$ and $\gamma$.

Given that the position of the spacecraft can be tracked to a lttle better than 1 metre, this is possible. $J_2$ is indeed measured to about 4%, and half of this uncertainty is due to remaining degeneracy with the $\beta$ parameter and uncertainty in the adopted $\gamma$ parameter.

This suggests that precisions of the order of a few metres were required for this experiment and I don't believe this is possible by radar ranging to the planet itself.

Indeed on p.181 of the 1993 version of "Theory and Experiment in Gravitational Physics by Will, it is said that

  Unfortunately, measurements of the orbit of Mercury alone are incapable at present of separating the effects of relativistic gravity and of solar quadrupole moment in the determination of [the change in the perihelion precession rate]

and then on p.183

  The accuracy required for such measurements would necessitate tracking of a spacecraft in orbit around Mercury


How often does the supermassive black hole in the Milky Way become a quasar?

To answer your question, the black hole in the milky way was never a quasar, and as it currently stands will never become one.

Info about how earth and life on earth could be affected if this ever happened, can be found here.


Relation between mass of gas, mass of stars and dark matter halo mass

M(Galaxy) = Ms+Mh
_________ > M¹ :Mh¹ = M(Galaxy)
Mg+Ms
let say :
mass of star + mass of halo =assume as 1 terminology of process
mass of galaxy + mass of star = assume as 1 terminology of process to derive the mass of halo.
we are not given the amount,so this theories would be a posible ways to collect data for mass of halo.
create second terminology as M¹to support the terminology of Halo.
coz its happen in the galaxy itself.
Mass of Halo need to derive first in order to get posible
formula for range of star and mass of matter.
(Mh¹) = to M(Galaxy)


What's the highest frequency that's been imaged by a radio telescope?

Some extensive search e.g. with scholar.google.com led me to a manuscript from October 2020 entitled Design and Characterization of 275-500 GHz Corrugated Horns and Optics for a Wideband Radio Astronomy Receiver which already has all the information in the title: It looks like there is current development towards another 500 GHz receiver.
The question mentioned a 0.95 THz receiver which is even above that.
I also found an online course at nrao.edu which argues that 1 THz is the absolute upper boundary to far infrared in astronomy.


With the volume and mass of a planet concludes are possible of its material composition. But there is more, for example auroras been observed at Jupiter. This is only possible if Jupiter has a magnetic field just like our earth. A magnetic field is a strong evidence for an iron core.

The next point is, heavy materials accumulate in the center, so one could ask why there isn’t gold or even heavier stuff in the core of a planet. The most common material in the universe is hydrogen. Hydrogenis burned through atomic fusion into helium. All heavier materials are produced if a star dies in an explosion. So really heavy materials like gold are less likely and iron or nickel are more common so cores more likely made of these materials.


Why is it said that the far side of the moon's crust is thicker?

The near side and far side have very different topography. Compared to the near side, the far side is far more rugged and has far fewer maria. The near side in places has very thin crust (the maria).

Other than these surface features, the biggest indicator of a disparity in crustal thickness is the two kilometer offset between the Moon's center of figure and it's center of mass. This is typically attributed to a thicker crust on the far side of the Moon, consistent with the observations of the different surfaces of the near side and far side.

NASA discovered that the center of figure (geometrical center) and the center of mass are offset from one another in the late 1960s. The simple act of observing how a probe orbits an object gives clues regarding the interior of that object. Scientists know, for example, that Mars has a partially molten core from observing the orbits of the many probes that have been sent to Mars.

In the case of the Moon, NASA sent 16 unmanned probes to the Moon in the 1960s prior to the Apollo 8 mission. The first four of these, Ranger 6, 7, 8, and 9, were impactors. Their impacts the Moon indicated the Moon's center of mass was displaced from its center of figure by 2-3 kilometers. Later missions confirmed this, and also confirmed that the Moon's near side has five large mass concentrations ("mascons").  The Apollo 15, 16, and 17 missions carried laser altimeters that gave very precise measurements of the distance to the Moon's surface. These further improved the estimate of the Moon's center of figure / center of mass offset.

(Aside: Read Bizarre Lunar Orbits for a description of how those lunar mascons change how things orbit the Moon.)


I'm a bit unsure what you mean: Your spectrum doesn't include the red side of the quasar (seems weird), your Ly$\alpha$ line doesn't have a red side (even weirder), or you're only interested in the metal lines on the blue side?
Anyway, metals on the blue side are somewhat difficult: At high redshifts, they're blended with the $\mathrm{Ly}\alpha$ forest (or erased by the Gunn-Peterson trough), while at low redshifts they're observed in the UV which is absorbed in Earth's atmosphere.
So, you really need to know where to look. But if you know the position of the quasar's Ly$\alpha$ line, then you can look e.g. for

Fe II + Fe III at 1123 Å
O VI doublet at 1032/1037 Å (the strongest one, but unfortunately very close to Ly$\beta$
Semi-forbidden C III] at 988 Å (close to Ly$\gamma$)
O III at 831 Å
Ne VIII and O IV at 772 Å

where all wavelengths are given here in the rest frame.
Except for the O VI line, they're all rather weak. But if you have many spectra, you can try to stack them so that the noise and the Ly$\alpha$ forest lines cancel out, enhancing the metal lines. This has been done e.g. by Lusso et al. (2015).


Why can there be several seconds of lag between a binary neutron star merger and the emission of gamma rays from the same area?

The paper (section 5.1) discusses three possibilities in the context of a relativistic fireball model, where some of the kinetic energy in relativistic jets of material emerging from the explosion is converted into gamma rays. The gravitational wave emission is always "prompt" since any surrounding material is transparent to gravitational waves. In contrast, the gamma ray emission may be delayed because:

The gamma rays are produced by shocks. There is a delay in the jets reaching the radius where the shocks form, either generated internally in the fireball by components with different speeds or just through deceleration in the interstellar medium.

The fireball is initially opaque to gamma rays and it is not until it has expanded sufficiently that it becomes transparent, allowing the gamma rays to escape.

It may take some time for the jet to punch out of the dense inner regions of the explosion.


Why are all space observatories in Chile?

[disclaimer: I am also not a true observer, but I have visited observatories before, and I do amateur observations regularly]

Definitely not all observatories are in Chile. Chile, however (the Atacama desert in particular) is an excellent location for an astronomical observatory. The Atacama desert has one of the darkest night skies on Earth, which is excellent for astronomical observations.

As Warrick mentioned, the ideal observatory is located at the highest possible altitude, such that there is as little air as possible between the telescope and the stars. Something else to consider, though, is that if the telescope should be manned by humans permanently it cannot be located too high due both to the difficulty for humans to function at extremely high altitudes and the price tag of getting the telescope parts up there. The two observatories in the Atacame desert of Chile are both located at an altitude of 2.5 km, which is definitely not the highest point on earth (the Mauna Loa and Mauna Kea observatories mentioned by Warrick are both located at higher altitudes).

Another requirement is that the location should be as cold as possible. This is not to torture the poor observers operating the telescope, but because warm air tends to be more turbulent. More turbulence in the air means less precise observations. However, as temperature tends to decrease with increasing altitude, this is usually automatically satisfied. Having a stable temperature also helps. If the telescope heats up during the day, the temperature of the air inside will differ from that of the surrounding air at night. Since the refractive index of air depends (although very slightly) on temperature, this will cause further deviation of light rays from their ideal straight paths, leading to less accurate measurements. For this reason professional telescopes usually contain some kind of temperature regulation.

A combination of the above reasons is why observatories are in the locations they are. Theoretically, the best observatory location has been found to be Ridge A on Antarctica. The reason you always hear about Chile is not its location. There are other locations that are at least as well suited as the Atacama desert for astronomical observations.

The most important reason you always hear about Chile is probably that it is home to the VLT, or Very Large Telescope, which is the most productive ground-based telescope in the optical region (according to Wikipedia). It uses interferometry combining the observations from four individual telescopes. This makes the combined effective aperture of the VLT the biggest in the world. As both the most productive telescope and the telescope with the largest effective aperture it produces a lot of results, and those results are quite likely to be important enough to appear in popular science articles.

However Chile is definitely not the only possible location for observatories. Apart from Mauna Kea and Mauna Loa, the Observatory del Roque de los Muchachos on La Palma is the first example that pops up in my mind. It is home to the GTC, the worlds largest single-mirrored optical telescope.


What is germane doing in the atmosphere of Jupiter?

How much specific elements sink inside a planetary caldron with variations of heat and thermals and chemistry, depends not just on their density but also the element's chemistry.   This question discusses Uranium in Earth's core and the top answer suggests (I believe correctly) there's essentially no Uranium in Earth's core, it's largely in the crust because Uranium readily oxidizes, so it forms lighter elements that don't sink but rise with the silicates and basalt, which is (roughly) similar density.   Uranium floats in the Earth's mantle because it binds with Oxygen.

Free Oxygen (O2) should destroy any Germane, but there's likely so little free Oxygen in Jupiter's atmosphere that the Germane is able to be present (in a fraction of parts per billion - it's very rare in Jupiter, but it's detectable).

Also, given Jupiter's internal heat, thermodynamically unfavorable doesn't matter much.   There's enough heat that chemistry tends to operate more in an equilibrium balance, and not always in the exothermic direction, like we usually see in most natural reactions at temperatures we see on Earth.

Finally, and at risk of stating the obvious, Germane is one of those "weird" heavy gases, due to the atomic electrical symmetry, and because it's a gas, not a solid and given Jupiter's significant thermal currents, it mixes in Jupiter's atmosphere.   It likely freezes out of Jupiter's upper atmosphere when it gets too cold, but it's present in the lower atmosphere, present enough to be detected anyway.  (I think it's less than 1 ppb).    - That's a brief answer anyway.   I've not read any of the recent articles on this.  They're all pay articles.    I invite anyone who has read them to give a more detailed answer.

Edit

  So what the heck is it doing there, considering free germanium would
  sink and free hydrogen would float?

I want to add that gases behave differently than solids or liquids.   In chemistry, sometimes things mix and form solutions (like water and alcohol), sometimes they don't and for layers (like oil and water).

Gases are particularly good at staying mixed and not layering, especially where wind and updrafts are present.

  why aren't scads of other simple components mentioned in discussions
  of Jupiter's atmosphere, like methylamine

There is Methylamine in Jupiter's atmosphere.   Saturn's too.   This research study mentions "bands" of it.    This one also discusses it.

  urea, and oxamic acid, to name just a few?

Urea is a solid at room temperature.   It decomposes when melted so it's not stable as a gas.   It melts at 133-135 C.   It might never be stable on Jupiter as a gas since it's not even stable as a liquid above 135 C.

Oxamic Acid, isn't even stable by itself, it's water soluble.    It's also Oxygen abundant, and free Oxygen is rare on Jupiter.

So those last 2 aren't good examples, neither is a stable as a gas at all, Germane is stable in Jupiter's atmosphere, though it's present in very small quantities.

An important point to consider with trace gases in planetary atmospheres is formation and half life.  Methane, for example isn't stable in Earth's atmosphere.   It reacts with Oxygen, though at a few hundred parts per billion, it's rare enough that it doesn't react a lot.   Methane is also constantly being produced in the stomachs of mammals (gas) but it also gets destroyed, reacting with oxygen whenever it gets near a flame or lightning, combining with Oxygen to form CO2 and Water.   It has a half life of about 10-15 years, but because it's constantly being created by incomplete digestion and decomposition, it's always present in our atmosphere - in about 1 part per million.

CO2 has a longer half life in our atmosphere, mostly it isn't destroyed, it needs to be rained out.   It's also constantly being resperated, destroyed in photosynthesis and re-created by respiration.

Point of all this is that any trace element in a planet's atmosphere, must have a source of creation and a half life of, destruction if you will.

A curious exception on Earth is Argon.   Argon is a byproduct of radioactive decay and as a noble gas, it doesn't react with anything, so it gets created and it just stays in our atmosphere.   It gets created slowly but because it's never destroyed, it's currently about 1% of our atmosphere.

GeH4 follows the same basic guideline.  It gets created inside Jupiter and it's stable enough in Jupiter's atmosphere that it's equilibrium between creation and destruction leads to about 1 part per billion.    I don't know nearly enough to say what it's atmospheric half life is.

As you said, there are "scads" of elements in Jupiter's atmosphere.   There re countless more that are NOT in it's atmosphere.   For an element to be present in an atmosphere it needs some degree of stability and some form of creation.

Jupiter is nearly 90% hydrogen.   It's not practical for it's hydrogen to float around the heavier elements.   Think of it as a sea of hydrogen, with other stuff in the hydrogen, some mixes and floats, some sinks, some chemically reacts, but the hydrogen is always present.

Anyway, I only meant to touch on some of the basics and not give so long an answer.

  I mean literally, how did you figure it out? What did you google? I
  ask because I'm dealing with hundreds of compounds and I'm hoping you
  found a nice centralized location for information like that.

We're getting a little outside the guidelines here, but briefly, I looked up the molecular structure of each of the compounds you suggested and they didn't look like gases to me.  They looked too complex and crooked shaped, so I was suspicious from the get go.

But the answer to your question is simply that I checked Wikipedia, which can be long (Urea) or very short (Oxalic acid), and not always 100% reliable.   Once there I did a word search for "boiling point" and "Melting point" when boiling point didn't show up.   For any molecule to be a gas it needs a boiling point, so that's the first quick and easy thing to search.

You can also look for phase diagrams, which are great, but rare for anything but the most common molecules.

For Oxalic acid, I was wrong, it is a solid.   Wikipedia only mentioned it in an aqueous solution.   This site gives it a melting point but no boiling point, which suggests to me, it doesn't make it as a gas, or, at least, not easily.    Maybe under very specialized circumstances, perhaps in a pressure tank filled with noble gases, but I'm just speculating.

Googling "Melting point" "boiling point", next to the name of the element isn't briliant, but it can provide a quick answer.

Longer addition:

Gases tend to be smaller molecules, noble gases or when more massive, they need to be symmetric.  A chemist probably wouldn't put it that way, but that's the gist of it.

Take line 2 of the periodic table, Lithium (number 3) to Florine (number 9) - ignoring noble gases for now. These atoms have 4 electron pair orbitals in their outer shell which want to form a tetrahedron.   CO2 has 2 double bonds - making it a straight line, or non-polar is the chemistry term.   I used the word symmetrical in my initial answer.  Non-polar is, I think, more correct.

That non-polarity or symmetry is why CO2 doesn't bond easily with other things, because it's a straight line, with equal charges on each end.   because it doesn't bond easily with other molecules, it's a gas at relatively low temperature.    CH4, while a tetrahedral, not a line, is also non-polar and has an equal charge on all sides, so it also remains a gas at quite cold temperatures.

Water (H20) is different.   Oxygen shares 2 single bonds or pairs of electrons with Hydrogen, it fills it's other 2 outer electron pairs itself, so it's bent shape and it's polar.   This polarity gives H20 a side with a positive charge and a side with a negative charge.   That polarity helps it bond with itself.  That's why water stays wet or stays ice in much warmer temperature than similar mass molecules like CH4 and heavier molecules like CO2.

For a gas, non-polarity helps.  A straight or tetrahedral, or flat triangle or six sided polyhedral (kind of a cube shape) and a few carbon chains are, as far as I know, the only options for non-polarity.    Very light, polar molecules can be gas molecules too.  All this varies somewhat with temperature and pressure, so there's no exact answers.

Common gases at room temperature. - note, this is not a complete list.

I'll resort them and add a few.

Elemental gases

H2 (hydrogen)       2.02
N2 (nitrogen)       28.01
O2 (oxygen)         32.00
F2 (fluorine)       38.00
Cl2 (chlorine)      70.91
BR2 (bromine) - at about 60 degrees C.
I2 (Iodine) - at about 180 degrees C.
AT2 (Astatine) - at about 337 degrees C.

It's worth noting that all of these, with the exception of Hydrogen and Nitrogen, bond more readily with other elements and are unlikely to be present in a planetary atmosphere.   (Photosynthesis creating O2 being an exception to this rule).

Noble gases

He (helium)     4.00
Ne (neon)       20.18
Ar (argon)      39.95
Kr (krypton)        83.80
Xe (xenon)      131.30
Radon (radioactive, short half life, but it is a gas)

These can be present in a planet's atmosphere as they are mostly chemically inactive.  (Xenon is slightly reactive).

Gas molecules (non-polar)

CH4 (methane)       16.04
NH3 (ammonia)       17.03
C2H6 (ethane)       30.07
PH3 (phosphine)         34.00
CO2 (carbon dioxide)        44.01
C3H8 (propane)      44.10
C4H10 (butane)      58.12
BF3 (boron trifluoride)         67.80
SF6 (sulfur hexafluoride)       146.05

(I'll say more about these shortly).

Gas molecules, polar

HCN (hydrogen cyanide)      27.03
CO (carbon monoxide)        28.01
NO (nitrogen oxide)         30.01
H2S (hydrogen sulfide)      34.08
HCl (hydrogen chloride)         36.46
N2O (dinitrogen oxide)      44.01
NO2 (nitrogen dioxide)      46.01
O3 (ozone)      48.00
SO2 (sulfur dioxide)        64.06
CF2Cl2 (dichlorodifluoromethane)        120.91   (only a little polar)
H20 (a gas under right temperature and pressure)

There are more polar molecules than non-polar, but outside of the somewhat odd CF2CL2, all polar gases are relatively light, SO2 being the most massive, molecular weight of 64.

You mentioned Methylamine which is basically ammonia (NH3) where one of the hydrogens is replaced by a methyl (CH3) group.   NH2CH3.

DiMethylamine (CH3)2NH is also a gas at about 7 degrees C and up (boiling point).

Playing around with temperature and variations on the gas molecules (replacing H with CH3, replacing H with NH2, replacing H with OH, but remember, Oxygen tends to be spoken for, like a perfect 10 at a dance, so that's not a good one, unless there's life and a source of oxygen (photosynthesis).

Similarly the "Ane" series, more accurately called the Group 14 hydrides.   Group 14:   carbon, silicon, germanium, tin, and lead, and the hydrides, Methane, Silane, Germane, Stanane, Plumbane.   All of these are polar and all are gas molecules.   Most are very reactive with Oxygen.  Methane requires a flame, but the other 4 react with Oxygen quite easily.

And as temperature goes up, you add new gases, but heat tends to destroy complex chemistry, so there's a bit of a trade-off.    There's no easy answer as to what could be a gas and what couldn't, but starting with the building blocks and swaping might be a place to start.   That doesn't always work though.     CO2 is non polar and a gas.   SO2, even though Silicon is in the carbon group, is polar and bent.  It's not a gas  (it's closer to sand) with a very high melting point.

So, the disappointing answer is, sometimes you can go down the column in the periodic table and find another and another gas simply by replacement and sometimes you can't.  In the case of SIO2, the bonds are very different than CO2, and the melting point is over 3,000 degrees.   Explained in detail in this question here.

Some of the Hexaflouride series is interesting.  4 of them are stable gases at Earth's temperature and generally non reactive enough to breath and sound like James Earl Jones when you talk, but they're not likely to be found in a planet's atmosphere cause they're vulnerable to photodisintigration and not likely to be reformed in significant numbers   Other molecues are more likely to form.

And, ofcourse You could have a planet with a 3,000 degree surface temperature and all sorts of elemental gases, see cool periodic table with temperature slide) and at 6,000 degrees, all the elements are basically gases, but temperature that high, destroys any complex chemistry so you don't get complex molecules.  Also, at temperatures that high, the exo/endo thermic direction no longer applicable.  Molecules tend to form back and forth in an equilibrium ratio.  That bit I remember from School.

On the Germane in Jupiter question, it does raise some other questions, what about the other Group 14 hydrides?   What about Saturn?  It's possible that Saturn has too little metallic hydrogen to create much Germane.

The trick in general is both formation and stability.   If a gas has too short a half life (like if it's vulnerable to photodisintigration like a lot of carbon chains are), then it's not likely to last.

Jupiter also has storms and wind powerful enough to bring up some elements that aren't necessarily gas - like dust in a sense, up high into it's atmosphere, some carbon chains, sulfur and phosphorus and ammonium hydro-sulfide, (which, despite having a boiling point, it's more accurate to say it separates into Ammonia and Hydrogen Sulfide at 56.6 degrees C, in the cold upper clouds of Jupiter the two elements could combine, into something like a dust.   Technically it's a salt.   Source.

Apologies if I got carried away.   I love thinking about planetary atmospheres.   Can't wait till the J.W.S.T. gets a picture of some in other solar systems.

My kitten is trying to delete what I wrote, so I'll post it now - will tidy up later.


Basically because Jupiter is a gas giant and thousands of times more massive than the Earth. Everything on Jupiter is on a gigantic scale. The Great Red Spot for instance could accommodate several Earths, so that explains why you couldn't have a storm on Earth which came anywhere near that size. As well as heat from the sun, which is what drives the weather on Earth, Jupiter has a massive internal source of heat.


Could the Sun have liquid iron in orbit?

As this source stats, galvanized building iron has an albedo of 0.351.

In general, we can assume, that a half of its total surface is shined by the Sun, with an average 45% angle. That means that it gets the radiation of the 6000 K Sun on $\frac{1}{2} \cdot \frac{1}{\sqrt{2}} \approx 35\%$, as if it would be shined directly on its whole surface. It radiates back 35% of it, the remaining 65% heats it. That means, in its equilibrial temperature, the iron meteor radiates $\approx 23\%$ in all direction due to its temperature, than it gets from the Sun.

Furthermore, being $n$ solar radii away from the Sun, it also gets the per-surface thermal energy of $\frac{1}{n^2}$. Here I committed yet another "approximation", because the Sun is not a pointlike radiator if $n$ is small, and the final result will be probably small, but it won't be too big.

The thermal radiation is proportional roughly with the 4th power of the temperature.2 Thus, $n$ solar radii away, it gets $\frac{1}{n^2}$ per-surface solar heating, it radiates away $0.23\frac{1}{n^2}$, what makes its equilibrial temperature $\sqrt[4]{0.23 \cdot \frac{1}{n^2}} = 0.69\frac{1}{\sqrt n}$ of the Sun's.

The average surface temperature of the Sun is 5778 K, the meltig point of iron is 1881K. Thus,

$$5778\ \mathrm K \cdot 0.69\frac{1}{\sqrt n} = 1881\ \mathrm K$$

$$\frac{1}{\sqrt n}=\frac{1881\ \mathrm K}{5778\ \mathrm K\cdot 0.69}=0.47$$

$$n=\frac{1}{0.47^2} \approx \underline{\underline {4.5}}$$

The answer is: 4.5 solar radius from the solar center, which is $\approx$ 1.8 solar diameter from the surface. It is far deeper than the orbit of the Mercury. As a comparison, the Parker Solar Probe will near the Sun 8 solar radii away. A 2000 K maximal temperature is predicted on its heat shield (note, that probe won't have an equilibrial temperature).

Another remark: liquid things in vacuum evaporate. The evaporation for molten metals near their melting point is slow, but it exists. Such an object, if it does not have something to keep the iron vapor around it (f.e., gravitation), will slowly disappear.

1Surprisingly, I did not found better data with google.

2Also this is not exact, because it is true only for absolute black body radiators, but the spectrum of most ordinary materials does not cause a huge error in the calculation.


At the time when the universe was ~380,000 years old it was completely filled with a plasma and light. Light could only travel very short distances (I forget the specifics, but think a couple meters at the most) before being absorbed. So the entire universe was a fog, if you were there and there was an object 10 meters in front of you, you couldn't see it because all the light it emitted or reflected would be reabsorbed before it reaches your eyes. There are two things to keep in mind: 1) This fog disappeared across the entire universe simultaneously (plus or minus some time, but effectively simultaneously); at the same point in time after the Big Bang, 380,000 years after the Big Bang, the fog dissipated due to recombination. And 2) Light takes time to travel.

So let's say you are sitting in the fog-filled universe a few seconds before recombination happens. Light travels about 1 foot per nanosecond. So if the fog disappeared and there was a street lamp 10 feet in front of you, you wouldn't receive the first light from that lamp for 10 nano seconds. All of the light the lamp was emitting before recombination was just reabsorbed, but after recombination the first photons to leave the lamp no longer have anything to run into, so they continue on until hitting your eyes. A street lamp 20 feet away wouldn't be visible for 20 nanoseconds after recombination. The light of the plasma itself is effectively a street lamp at every location in space, so due to the 2 facts above, you would see successively farther street lamps in a shell centered on you.

After recombination happened and the entire universe became transparent. The fog is gone, but the farther you look the farther back in time you look, so you can only see back as far as when the fog existed, and the farthest you can see is the fog itself. After 13.8 billion years you can see all the way out to a shell of radius 13.8 billion light years around you (ignore expansion of the universe for now), and what you see is the light of the plasma (the fog) that was filling the universe 13.8 billion years ago, the CMB.

This scenario is the same if the universe is infinite in extent. We are not the center of the universe, we are simply the center of the observable universe, as is anyone else. Aliens looking towards Earth but are at the current location of edge of the CMB in some direction will not see Earth, they will see the last light of the CMB itself at our location when it was emitted 13.8 billion years ago; we are the edge of their observable universe.


Exact ecliptic position of Earth and Moon from Mars as photographed by Curiosity Rover?

NASA Photojournal item PIA17936 says:

Researchers used the left eye camera of Curiosity's Mast Camera (Mastcam) to capture this scene about 80 minutes after sunset on the 529th Martian day, or sol, of the rover's work on Mars (Jan. 31, 2014).

Between sols 528 and 532, Where is Curiosity? shows it in a place called Dingo Gap at Martian longitude 137.41°, latitude -4.63°.
At that location on Mars in Stellarium, sunset on 2014-01-31 occurs around 21:00 UT, so the picture was taken around 22:20 UT.
JPL HORIZONS gives these J2000 heliocentric positions at that time:

lon. (l)
lat. (b)
dist. (r)

Mars
168.784°
1.613°
1.662 au

Earth
131.767°
-0.001°
0.985 au

Moon
131.732°
0.010°
0.983 au

and these Mars-centric positions:

lon. (λ)
lat. (β)
dist. (Δ)

Earth
22.935°
-2.535°
1.058 au

Moon
22.826°
-2.522°
1.059 au

In other words, the Moon should appear about 0.11° west of Earth in the Martian sky, or to the right if Mars's north celestial pole is up:

But looking toward the western horizon (zenith up) from a tropical latitude, celestial north is to the right, so the Moon appears below Earth in the photo.


Why is the Earth's magnetic field stronger than its neighbors?

I've learned that the Earth's core is hot due to decay of radioactive elements.

This is unproven, non-standard geophysics. There are several arguments against this. One is that all of the long-lived radioactive isotopes are isotopes of uranium (two isotopes, 235U and 238U), thorium (232Th), and potassium (40K). The problem: Uranium, thorium, and potassium are strongly lithophilic ("rock-loving") elements. These elements dissolve very nicely in molten rock, but not so much in molten metal. The presence of long-lived radioactive isotopes is enhanced in the Earth's crust, slightly depleted in the Earth's mantle, and by all rights should be strongly depleted in the Earth's core.

Another problem is that any significant amounts of uranium and thorium in the Earth's core have been ruled out due to neutrino detectors. Potassium-40 has not been ruled out because the neutrinos from 40K decay are too low in energy to detect, but that brings us back to problem #1.

The only hope for this conjecture is that potassium somehow becomes siderophilic at high pressure. There are some experimental results, most of which are highly controversial, that this might be the case.

Yet another problem is the conjecture of high radioactivity in the Earth's core was motivated by explaining the Earth's magnetic field. A number of recent papers say that there is zero reason for this motivation. The Earth's magnetic field is fully explainable without resorting to the chemically unsupported hypothesis that 40K somehow becomes a siderophile at high pressure.


If a comet strikes a glacier, does it make a print on the land below?

While it's a clever suggestion, the answer is almost certainly that they are not related because the Nastapoka Arc "crater" is too big.   It's 450 km in diameter.  By comparison, the Chicxulub crater is 180 km.   Even if an impact through mile of ice could, somehow, spread the impact out wider, the proposed Younger Dryas Impact object was almost certainly too small to leave a 340 km crater.

There's also the problem that glacial ice carves the landscape and a neat circular crater is unlikely, even if the size had been a closer fit, an impact crater that was run over by a glacier should look disturbed by the glacier, not a neat circle visible on the surface.

What geological evidence might remain for a large impact on a glacier is an interesting question given that the ice would significantly buffer any impact and continue to reshape the landscape below it as the glacier moves and/or melts and possibly floods the region.

Mile high ice glaciers are relatively rare events, both localized and periodic, and large meteor impacts are also rare events.   A combination of the two, large enough to leave clear evidence through the ice, would at best, be a hugely rare event and there's a good possibility that the right combination of impact and ice hasn't happened at all over the last 2.5 million years (northern Hemisphere) or on Antarctica (3% of the surface area of the earth) - last 20-30 million years.

I think the safer bet is that there's no impact crater from the proposed Younger Dryas impact specifically because of the thickness of the glacier.   It remains a debated hypothesis, so there's some uncertainty, but Ice should significanly buffer all but the largest impacts.

For Younger Dryas, there should be other physical evidence such as debris from the impacting meteor or comet, but debris cast over a glacier that first re-freezes over and later melts can be spread very far and be very hard to identify.   Much of it could have washed into the ocean.   A somewhat related example, it took 105 years before debris from the Tunguska event was identified.

(and this question should probably be moved to Earth Science).


The best way to find a particular star/planet/etc. is almost certainly to use a star map of some sort.

For example, doing a Google Image search for "star map sirius" would get you quite a few pictures such as the one in this article about Sirius that give a better picture of how Orion's belt "points" to it.

Alternately, there are a number of computer programs that offer a real time views/maps of the night sky.

Also check out Google Sky.


What does gravity-wave noise resemble locally?

For orbiting objects, gravitational wave intensity depends on the second time derivative of the quadrupole moment of an isolated system's stress–energy tensor (which for the wimpy objects in our vicinity is proportional to the mass times the line of sight acceleration) and drop off with the distance. So you can easily estimate the GW intensity of any object.  Hint: It's small for nearby objects.

As requested, I have tried to estimate the GW amplitude and I found the numbers somewhat surprising.  I looked at the Moon, Jupiter, Io, the LIGO neutron star and a pair of supermassive BHs if they were at the Galactic center.

I got the mass of each object in Suns, the orbital radius in miles, the orbital period in years and the distance in AU.  The angular acceleration is computed as r * omega2. Multiply by mass to get GW radiation at source. Divide by distance to get GW amplitude observed at Earth. (All units relative and arbitrary.)

Object             Acceleration   Mass*Acceleration  Est. amplitude
Moon               3 x 10**7      1                  4 x 10**2
Jupiter            3 x 10**6      3 x 10**3          7 x 10**2
Io                 1 x 10**10     5 x 10**2          1 x 10**2
LIGO Neutron star  1 x 10**19     3 x 10**16         4 x 10**6
Supermassive BHs   1 s 10**9      2 x 10**15         3 x 10**6

(The supermassive BH numbers are really quite arbitrary, since you can assume any separation you like and pretty much any mass.  I picked 1 AU and 1,000,000 suns mass.  You can easily get much larger numbers, but the frequency will typically be too low to be seen by LIGO.)

The other really important thing to remember is that in spite of Io being about as bright as the sources LIGO has observed, the frequency of Io's radiation is about 10-5 Hz. LIGO is only really sensitive in the 10-1000Hz range and has no chance at all of ever seeing IO or Jupiter or the Moon or even a pair of supermassive BHs that are not in the last milliseconds of merger.  If eLISA gets built, it can probably see at least some supermassivs BH pairs.


What is a quaternary mirror and why does the E-ELT need one?

You may know that a standard Newtonian telescope has two mirrors, they are called the primary and secondary mirror.
The E-ELT has five mirrors: The quaternary mirror is simply "mirror number four", counting in the direction the light enters the scope.
It's complex because that's where the adaptive optics sits:

The quaternary mirror has an approximate diameter of 2.4-m (2380x2340mm). It is a flat adaptive mirror, with up to 8000 actuators, thereby allowing the surface to be readjusted at very high time frequencies. [...]
This mirror will correct in real time for high order wavefront errors (e.g. atmosphere, wind shake, low spatial frequency telescope errors) and small amplitude residual tip-tilt corrections.

Source: The European Southern Observatory's page on the E-ELT optics


Why is the time between equinoxes different?

This is because the summer and winter solstices (approx. June 21st and December 21st) do not correspond to the aphelion and perihelion (approx. July 5th and January 4th). Therefore, the average distance from the Sun is longer in the period from the Summer Solstice to Winter Solstice than vice versa, so the Earth is moving slower (on average) and it takes longer.

I think the values you mentioned are incorrect (switched around); 181.0 is a more likely value for Winter to Summer (February is a few days shorter than August).


What is the orbital velocity of Pluto around the Pluto-Charon barycenter?

From the NASA Pluto Fact Sheet:

Mass of Pluto $m_1 = 1.303\times 10^{22}\ \mathrm{kg}$
Mass of Charon $m_2 = 1.586\times 10^{21}\ \mathrm{kg}$
Mean distance from Pluto $r = 19596\ \mathrm{km}$
Orbital period $P = 6.3872\ \mathrm{d}$

The orbit is nearly circular (the fact sheet gives an eccentricity of 0.0, other sources give very small values). So treat as circular motion.
The radius of the orbit of Pluto around the barycentre is given by:
$$r_1 = r \left(\frac{m_2}{m_1+m_2}\right) = 2126\ \mathrm{km}$$
Then use the usual formula for circular motion:
$$v = \omega r$$
where $\omega = \frac{2\pi}{P}$, giving a value of around $24.2\ \mathrm{m\,s}^{-1}$.


Can the diminishing energy of the CMBR be the source of dark energy?

No - the decreasing energy in the CMB is already well modeled in the Friedmann equations. The term in the density parameter that is proportional to $a^{-4}$ is the contribution of radiation energy density to the evolution of the universe, the term proportional to $a^{-3}$ is matter density (mostly dark, but includes ordinary matter), $a^{-2}$ is the contribution of the curvature of space-time itself, and the term without any factors of $a$ is the contribution of dark energy. The size of the radiation density, today, is already a small fraction of the matter density (about 0.03% of the matter density, 0.01% of the density of the universe overall - ordinary matter is about 5% overall).

The last time the energy density in the radiation fields was the same size as what's in the matter fields was around $z=3,300$.

I also disagree with @J.Chomel's answer - the energy stored in the radiation field is decreasing. Then energy density in the radiation field scales like $a^{-4}$, and the volume scales like $a^3$. Since the total energy is the energy density times the volume, the total energy scales like $a^{-1}$, just as you would expect with the number of photons being fixed, but the energy in each photon scaling as $a^{-1}$ as the wavelength increases.


But as far as I know entropy is the amount of disorder.

Entropy is a measure of the number of possible microscopic states consistent with an observed macroscopic state1, $S = k_\text{B}\ln N$. Fundamentally it has nothing to do with disorder, although as an analogy it sometimes works. For example, in simple situations like an $n$ point-particle gas in a box: there many more ways to put point-particles in a box in disorderly manner than an orderly one. However, the exact the opposite may be true if they have a positive size and the box is crowded enough. Overall, disorder is just a bad analogy.

1 Even that's not quite true, but it's better than disorder. Specifically, it's a simplification under the assumption that all microstates are equally likely.

  A black hole is denser than a star. For a density that high, I assume a certain amount of order (inverse entropy?) is required.

If an object is crushed inside ideal box that isolates it and prevents any leaks to the outside, the crushed object still has information about what it was before. And an event horizon is about as an ideal box as there can be.

Classically, black holes have no hair, meaning that the spacetime of an isolated black hole is characterized by mass, angular momentum, and electric charge. So there are two possible responses to this: either the black hole really has no structure other than those few parameters, in which case the information is destroyed, or it does have structure that's just not externally observable classically.

Thus, if information is not destroyed, we should expect the number of microstates of a black hole to be huge simply because there's a huge number of ways to produce a black hole. Roughly, at least the number of microstates of possible collapsing star remnants of the same mass, angular momentum, and charge (though this is idealized because a realistic collapsing process sheds a lot).

  For a density that high, I assume a certain amount of order (inverse entropy?) is required.

Quite the opposite; black hole are the most entropic objects for their size.

In the early 1970s, physicists have noticed an interesting analogies between how black holes behave and the laws of thermodynamics. Most relevantly here, the surface gravity $\kappa$ of a black hole is constant (paralleling zeroth law of thermodynamics) and the area $A$ of a black hole is classically nondecreasing (paralleling second law). This is extended further with analogies of the first and third laws of thermodynamics with $\kappa$ acting like temperature and $A$ as the entropy.

The problem is that for this to be more than an analogy, black holes should radiate with temperature that's (some multiple of) their surface gravity. But they do; this is called Hawking radiation. So the area can shrink as long as there is a compensating entropy emitted to the outside:
$$\delta\left(S_\text{outside} + A\frac{k_\text{B}c^3}{4\hbar G}\right)\geq 0\text{.}$$
Thus, semi-classically, the entropy of a black hole is proportional to its surface area. In natural units, it is simply $S_\text{BH} = A/4$, which is huge because Planck areas are very small.

Thus, we know that in a semi-classical approximation, a black hole must radiate with temperature proportional to its surface gravity and entropy proportional to its area. It's natural to wonder the next step: if a black hole has all this entropy, where is the structure? How can it have so many possible microstates if it's classically just a vacuum? But going there takes us to the land of quantum gravity, which is not yet firmly established, and is outside the scope for astronomy.


Does the ecliptic precess? Does the Sun's rotation precess?

As you mentioned, the Earth's rotational axis undergoes a slow precession. This results from tidal torques caused largely by the Sun and the Moon. There would be no precession if the axial tilt with respect to the orbital angular momentum vectors was zero or if the Earth was spherical.

The Sun is almost spherical it's oblate to about $10^{-6}$, and is almost perfectly aligned with the ecliptic, it has a relatively slow rate of rotation, and all of the bodies that might cause a precession are far from the Sun, and because the gravitational gradient torque is an inverse cube relationship. All of these factors combined means the precession rate of the Sun is ridiculously small. So the short answer to your question is:

Yes, the Sun precesses.

Planetary orbits are a little different, I'm not exactly sure how to state the process for their precession without waving my hands about the room, so I won't attempt it. However, I will say that, as all things in the universe, I am sure the ecliptic is a dynamic system unto itself.

About the ecliptic (dashed line) and the galactic plane (spotted strip) see the attached image.

If the ecliptic and the galactic plane were aligned, the two would line up on top of one another, as the Sun's and other planet's paths would be roughly the same path.

There is still residual angular momentum within the solar system, which is how we arrived at our current orientation amongst the stars.


When did scientists discover that the Sun has a life cycle and that it is going to die?

Although various astronomers have speculated that the Sun was a star (some were imprisoned or even burnt alive for such heresy), this was not known definitively until 1838 when Friedrich Bessel used parallax to calculated the distance to 61 Cygni.
In the late 19th century, Lord Kelvin provided rather small (less than 100 million years) for the ages of the Sun and of the Earth. He did not know about fusion. The nature of what powers the Sun was not known until the 1920s based on observations of the mass of a proton versus that of a helium-4 nucleus. The exact mechanism wasn't known until 1938 when Hans Bethe developed models by which hydrogen fuses to form helium.
This model provided the basis for knowing that the Sun will eventually die as it will eventually run out of fuel. Working from this, a number of scientists developed models of what happens inside stars of various masses based on the growing knowledge of stellar nucleosynthesis.


Exactly what in Saturn has been located to within 4 km precision?

I suppose it is the Saturn system's barycenter which has been located gravitationally. At this level of precision I suppose that the moons', especially Titan's tidal effects must be considered. ..., so I wonder if these too are being considered at the 4 km precision level, and overall exactly how the point located is defined?

Short answer: No.
As pointed out in the comments by @AtmosphericPrisonEscape (Oct 5 '16 at 14:42) and @userLTK (Oct 6 '16 at 9:25) - it is a barycentric measurement.
Other references:
Solid-body tides on the Moon
Is Io's orbit or rotation affected by its volcanism?
The Barycenter coordinates are an average, a mean, small perturbations are smoothed in the calculations by various factors such as: measurement accuracy, use of the International Celestial Reference Frame (ICRF), even the Cholesky whitening (Source: "Measuring the mass of solar system planets using pulsar timing" (21 Aug 2010)) used in some of the calculations.

"In astronomy, barycentric coordinates are non-rotating coordinates with the origin at the center of mass of two or more bodies. The International Celestial Reference System is a barycentric one, based on the barycenter of the Solar System.
...
Relativistic corrections
In classical mechanics, this definition simplifies calculations and introduces no known problems. In general relativity, problems arise because, while it is possible, within reasonable approximations, to define the barycenter, the associated coordinate system does not fully reflect the inequality of clock rates at different locations. Brumberg explains how to set up barycentric coordinates in general relativity in his book "Essential Relativistic Celestial Mechanics".
The coordinate systems involve a world-time, i.e. a global time coordinate that could be set up by telemetry. Individual clocks of similar construction will not agree with this standard, because they are subject to differing gravitational potentials or move at various velocities, so the world-time must be slaved to some ideal clock that is assumed to be very far from the whole self-gravitating system. This time standard is called Barycentric Coordinate Time, or TCB.

Example: Motion of the barycenter of the Solar System relative to the Sun, 1945–1995.
In the paper "Uncertainties in the JPL Planetary Ephemeris", by Folkner, on page 47, is shown this chart of the uncertainty in right ascension, declination, and distance of the barycenters of the Earth and Saturn systems calculated for the years 1950 to 2050:

Celestial mechanics and the n-body problem take a long time to solve.
The paper associated with these measurements of Saturn is titled: "VLBA Astrometric Observations of the Cassini Spacecraft at Saturn" (1 Dec 2010), on page 2 it says:

"These observations provide positions for the center of mass of Saturn in the International Celestial Reference Frame (ICRF) with accuracies ∼0.3 milli-arcsecond (1.5 nrad), or about 2 km at the average distance of Saturn.
...
The DE 422 post-fit residuals for Saturn with respect to the VLBA data are generally 0.2 mas, but additional observations are needed to improve the positions of all of our phase reference sources to this level. Over time we expect to be able to improve the accuracy of all three coordinates in the Saturn ephemeris (latitude, longitude, and range) by a factor of at least three. This will represent a significant improvement not just in the Saturn ephemeris but also in the link between the inner and outer solar system ephemeredes and in the link to the inertial ICRF.".

I might return for an edit if there is interest in this Q&A.


Is it plausible to use other stars for the proposed FOCAL mission instead of the Sun?

Gravitational lensing works from anywhere beyond the focus, so in that sense, we could use any star as a gravitational lens. The problem is that the field of view is tiny. We only get any useful information from alpha centauri as a gravitational lens if the target object is almost exactly behind alpha centauri from our point of view. To look in a slightly different direction, we need to move a huge distance sideways.

In fact we do use gravitational lensing from other bodies for astronomy. We use very large clusters of galaxies to amplify the images of other galaxies which lie behind them, and we watch for brief flashes as planets and stars line up for a moment as a way to detect planets (called "microlensing").


Explain to a 4-year old why people below the equator are not tilted or suspended upside down

Explain that "down" is more or less toward the center of the Earth, and that direction changes with location. (For a four year old, skip the "more or less" part. It's just toward the center of the Earth, period.) The direction of "down" doesn't change much on trips across town, but it does change with larger trips.

If you have taken your son on a long trip (just a few hundred miles or kilometers will do), show your son those two locations on the globe. The direction of "down" at that remote location and the direction where you live point in slightly different directions for a trip of a few hundred miles, vastly different directions for a longer trip.  Ask him if he felt tilted sideways at that remote location.


Isn't the date of the Big Bang a bit bogus?

I'd like to correct a couple misonceptions that might be confusing you.

  If I understand correctly, the date of the Big Bang is an extrapolation of acceleration of the universe's expansion through time based on the erroneous assumption that the universe is approximately 13 billion years old.

This is a false assumption. The observable universe is much, much larger than a sphere with a radius of 13 billion light-years. The radius is actually ~46 billion light-years (see, for example, Gott et al. (2005)).

  The age of the universe, as far as I can tell is based on the speed of light and the fact that the furthest we could possibly see into the depths of the universe is approximately 13 billion light-years.

Look at pela's answer here, if you don't want to go off Stack Exchange (or see Wikipedia for another starting point). See how the age of the universe is calculated. As an approximation,$$t_\mathrm{Uni} \sim \frac{1}{H_0} = 14\,\mathrm{billion\,years}.$$This only relies upon observations of the movements of stars.

  That the expansion rate of the universe is equal to the speed of light.

The universe is not expanding at the same rate at all distances from us. Objects further from us recede at greater speeds.


absolute speed of bodies as observed by a distant observer

In general relativity, then notion of space and time is local. As underlined in the OP, it means that at a radius $r$ of a star of mass $M$, time will run at $$\sqrt{1-\frac{2GM}{rc^2}}$$ the rate at an infinite distance of the object. For example imagine that $1-2GM/rc^2 = 1/4$, so that at a distance $r$, time runs at half the rate compare to an observer at infinity. If you are initially with an observer at infinity, you travel to $r$ (we neglect the time distortion during the travel), stay here an hour and then go back. When you'll reach the observer once again, your local clock will be one hour late, because 1 hour at $r$ will have passed in two hours at $r=\infty$.

So this distortion of space and time is only local. A massive object, like a star is only impacting in its neighborhood space and time. A distant observer is not impacted by the distortion, so that time and distances are unaffected.

To answer your question, an observer far away (for example on Earth) measuring by some meaning the speed of the two stars will actually see $c/2$ for both of them.


Do black holes produce thermal radiation?

Hawking radiation is the black body radiation that black holes emit. It is a well established theoretical result but it has never been observed.  For all currently known astrophysical black holes it will be totally insignificant and will never be observed.  The reason for this is that the equivalent temperature of the black body is inversely proportional to the mass of e black hole.  For all currently known black holes the temperature would be a tiny fraction of a degree above absolute zero.

Quantum mechanics is a probabilistic approach to physics, and deals with all kinds of probabilities and uncertainties. There is one such uncertainty of time and energy, given by (Δt) (ΔE) ≥ ℏ/2

One of its predictions, which has been experimentally confirmed, is that a particle may take an energy E from “nothing” and yet not violate the Law of Conservation of mass-energy if it “returns” the energy within a time given by (Δt)≥ ℏ/(2E).

One of its many implications is that random vacuum fluctuations constantly take place everywhere, i.e. even if a certain region of space has nothing or is purely vacuum, it can still constantly generate and annihilate particles due to the E-t uncertainty.

fast forwarding

When such a fluctuation causes a particle and its anti-particle to be generated on the edge of the event horizon of a black hole, one of the two enters the black hole and the other escapes. Now for returning the energy to the system, the particle relies on crashing into its anti-particle causing an “annihilation” and conversion of mass to energy. In the scenario I described, this can never take place since the pair of particles have been separated effectively forever. The particle that enters the event horizon has been trapped in an infinite potential gravitational well, so can be thought to have negative energy. So anti-intuitively (wait, what part of all this is even intuitive!), the particle left lurking outside the absorbs some energy from the black hole, (thus conserving mass-energy) gets converted into a “classical” photon and escapes the vicinity of the black hole.

Thus little by little, such events cause a black hole to effectively “evaporate” and diminish.

The apparant paradox about the fact that what happens to the information that goes into a black hole is a matter of debate. No one knows exactly how this Black Hole Information Paradox can be solved, however certain speculations are provided by Kip Thorne and Stephen Hawking, which include Firewall theory and Ads/CFT.


There are many cosmological distance measures and what you need depends on what you want to do. If you want to position the object in 3D space where it is today, you need its proper distance. If you want to e.g. display the object with a certain angular size on the sky, and you want to figure that size out from how far away the object is, you need the angular diameter distance.

To calculate any of the measures, you need the object's redshift $z$. You will probably need to store the comoving distance (= proper distance today) and redshift so that you can convert to other distance measures quickly without redoing the integral.


I can't give a good answer but I did find the documentary and out of curiosity, watched some of it, skipped forward and found the hologram section you're talking about, it's at the end about 1:18:40 in.  (I'm not going to post a link to the video cause I'm not sure it's correctly copyrighted by those who are showing it)

The video itself isn't bad, but it's basically a science channel level program, which is written to appeal to viewers and explanations are often over-simplified.   This video in particular seemed to try to cover so much ground that it explains very little.   But on the positive side, they have Brian Greene and Leonard Susskind discussing the hologram idea and those guys are both real physicists who've talked about this before.  I don't think it's a main stream idea, just an idea that a few people like.

Here's a magazine article on the subject and an excerpt from Brian Greene's book.

and Here's Leonard Susskind giving a talk on the subject at Stanford, not at a class, more of a lecture cause he mentions that not everyone in attendance might be a physicist, but I've since watched his lecture and he makes a few of these points more clear, at least to me.

And Here's a stack exchange question and answer on a similar subject.    There's also no shortage of articles that can be found on the Holographic principal or the "universe is a hologram", though I think the first is a better set of words to search and the 2nd might lead you to some sub-par articles.

Philip Gibbs - inactive's answer (also in the link above) on what the Holographic Principal implies is better than anything I could say about it, so I'd start there.

As a rule, it's a good idea to post links and if the video is 2 hours long, the specific time as well.   YouTube links can also be removed so they're not ideal.

Edit to answer questions.

Had to think about this one a bit, and sometimes trying to answer "why" is opening a can of worms, but . . . here I go.

  Why do they think that it is our Universe that is the Holograph?

I'm not a fan of the theory myself and I'd guess that not all physicists agree with it either.   One big problem with the theory is that it can't be proven, which really makes it more of a guess based off a mathematical model and not a good theory.  One definition of a good theory is one that can be tested.   If it can't be tested true or false, it's really more of an idea.

As for why Brian Greene likes the theory, you might need to ask him or search his interviews, but maybe he likes it so can sell books - I don't know.  Maybe he likes the idea of the Universe being a Hologram.   He's obviously a pretty smart guy.  You don't get a PHD in physics and get recognition and invited to ted talks and all that by being dumb, but smart guys are wrong all the time when it comes to new theories.  There's no evidence that this theory is true.

Now, I should point out that the Holographic principal as it applies to black holes and the information paradox, all that stuff is good and valid theoretical physics rooted in ideas we understand such as entropy and time dilation and red-shift.   The Holographic principal seems a very reasonably deduced feature of black holes that also, cleverly answers the information paradox, but whether it can be extrapolated to apply to the entire universe is a whole different ball game and to me, it's basically a guess.  Not much more.

  Why couldn't our Universe be reality and a Holograph exists of it
  somewhere else?

Well, I could say "you're right", but the truth is, you're just flipping around the question.  The more relevant questions are whether the theory is true at all, and as I said above (and Susskind said in his video), whether it can be tested.   Susskind's video is really quite good though he starts slowly, but if you're really interested in this it's worth watching.

  What makes them think that this is the Holograph instead of the
  reality?

One thing about Physics to keep in mind.  It's often surprised us.  Physics has given us answers that physicists didn't expect many many times, perhaps more than any other field of study.  Physics is probably more likely to yank the rug of reality out from under us than any other field and with that in mind, it's not a subject to study if you want to feel secure in what you believe and know.    In the days after Newton, Physicists were very confident in what they knew and felt that nothing very new was out there, though the constant speed of light was a puzzle they'd not expected.

Then Einstein came along and pretty much threw everybody's understanding out the window when he started to figure things out and not long after that, the quantum physicists came up with ideas that made Einstein's head spin and to which he directly disagreed.   After that the big bang was unexpected, background radiation was unexpected.  Dark matter was unexpected (except for maybe Fritz Zwicky) and dark energy was quite the surprise when it was discovered.

Being regularly surprised, of-course, doesn't make the hologram theory any more true or false, cause nobody knows, but good physicists these days try to keep an open mind to mathematically valid new ideas.

my 2 cents as a hobbyist.  Hope that wasn't too long and off the point, and please, take what I say with a grain of salt.   I'm not an expert.


How can I clean my lenses without causing scratching or damages to the lens surface?

Clean your eyepieces in the same way you would clean a camera lens. An air blower, soft (i.e. camel hair) brush, lens tissue or microfibre cloth, and a solvent or alcohol will do the trick.

When I was an undergraduate I did some work with lasers and very expensive dielectric coated mirrors. The techniques here will show you how to carefully clean optics. Never completely wet your lens tissue or cloth. One drop on a lens tissue is usually enough, or a couple of drops on a microfibre cloth.

Essentially, you need to:

Use a blower to blow off as much dust as you can.
Then use a soft brush.
Dampen a small square of lens tissue with pure alcohol or acetone (see warning below). Gently draw or wipe the tissue across the surface once, and then discard. Repeat if necessary.
You may need to fog the lens with your breath to help remove water-soluble gunk.

Be insanely careful if you are using a solvent such as acetone. Although a good for cleaning optics, it's not very nice to paint and plastic so it should be a last resort for very stubborn marks well away from the edges of the lens (so you'll never touch the paint with it). Methyl or isopropyl alcohol are safer but less effective, or a generic lens cleaner from a camera store.


Stellar encounters are very rare outside dense stellar environments. Such dense stellar environments are only found in a star's birth cluster, where all the stars are coeval to a close approximation.

Thus the switching and capture you refer to is probably common in the first few million years of a star's life, but would occur with stars of a similar age. Thereafter, once a stellar system is "in the field", encounters with other stars hardly occur at all, but are not impossible.

I am not aware of any binary where the components are thought to have a very different age.


minimum size for a black hole to capture a photon

There isn't really a minimum size. All black holes have an event horizon from which nothing, not even light can escape. The (Schwarzschild) radius of this event horizon is 3km for a black hole of a solar mass and scales linearly with mass. There is a region just outside the event horizon called the photon sphere, within which light can briefly orbit the black hole, but not in a stable fashion. It will either fly off in some other direction or spiral into the black hole.

You can check out this Physics SE question to see a derivation of the impact parameter (the effective radius at which light will be absorbed) and you will see that it is a small numerical factor ($\sqrt{27/4}$ for a non-rotating black hole) times the Schwarzschild radius (the event horizon of a non-rotating black hole), which is indeed proportional to the black hole mass.


What happens if two or three stars collide with each other?

The collision of another star with our sun would be a very effective way of ending all life on Earth. Fortunately stars are so far apart that the the chance of it happening in the life of the sun is very close to zero.

However in the densely packed cores of globular clusters the rate of collision is much higher, and there may be a collision somewhere in the milkyway once every 10000 years or so. Three way collisions don't happen (or are exceptionally unlikely) but would be similar to two two-star collisions.

The result varies depending on the size of the stars, their density, the relative velocity, and whether a collision is head on or not. Generally a denser star is likely to come out of a collision in better shape. So a red giant, could have most of its very rarified outer layers pulled off and the core escaping.

On the other hand, if two main sequence stars, like our sun, collide, they first compress each other. Some of the stellar envelope escapes perpendicular to the collision (think of jam being squeezed out of a sandwich). But within a few hours the two stars have mixed and merged, and   a new, larger star formed.

The mixing adds lots more hydrogen to the core of the star, which causes it to rejuventate. The first evidence of stellar collisions is in globular clusters, in which there are some large blue stars that seem younger than all the other stars in the cluster (Blue stragglers)

A white dwarf and main sequence star collision is bad news for the star. The collision causes the core of the star to heat up and a massive release of thermonuclear energy. Plenty more than is needed to completely disrupt the star. The star is completely destroyed in the process. However the white dwarf is relatively unscathed.

White dwarf-white dwarf collisions can, depending on the size result in gravitational collapse to a neutron star.

The paper Stellar collisions, mergers and their consequences is a very readable summary.


Build a high quality telescope from scratch

Amateur telescope making (ATM) is a popular past-time for astronomers.

You can grind your own mirror, or you can buy ready-made mirrors.

You can coat your mirrors with silver using chemical solutions yourself, or have a coating of aluminium applied using a vacuum chamber by an outfit offering such a service.

The main materials are a thick round slab of glass, usually Pyrex, and a smaller elliptical flat mirror. Using this you can make a Newtonian reflector. The support tube or structure for the mirrors can be one of a huge variety of designs, or you can be inventive yourself.

There are websites to guide you in this, such as Mel Bartel's, Stellafane, and AmateurTelescopeMaker.com.


Answer posted under the wire during closure. Others may be able to write an additional and/or different answer as soon as two more reopen votes are cast.

Would it have been possible to send a radio signal towards ʻOumuamua?

Yes. It is certainly true that at any time, a large radio telescope or even a small transmitter can always sent a signal toward something. New Horizons can receive our signals now even though it's in the Kuiper belt because it knows when and where to listen and at what frequency.

As we could not get any radio signals from asteroid ʻOumuamua, couldn't we have sent a powerful radio signal to it and then check if we can get any radio signal in response?

The asteroid was quite close to us in the 2nd half of 2017, but now it's much farther than New Horizons. However if it had advanced technology it would know we were very active in the electromagnetic radio spectrum and it might be listening. If so, we certainly could have sent it a signal that it could receive.
Checking for a response is also hard because we'd need to know what frequency and when. We'd have to allocate resources, etc.
For the frequency, see also

https://scifi.stackexchange.com/q/197905/51174
https://scifi.stackexchange.com/q/16348/51174

Below: Plots of the distance of ʻOumuamua (bumpy, blue) and the New Horizons spacecraft (straight, green) from Earth as a function of years since 2017-01-01.


What is the cause of the variation from high and low mean obliquity periods of Mars?

This figure comes from numerical simulation by Laskar et al. (2002), see Fig. 2. An extended figure from a longer simulation can be found in Laskar et al. (2004), Fig 10.

As you can see more clearly from the extended figure, the simulation doesn't particularly suggest that this is periodic behavior. The question is, then, what happened ~5 million years ago that changed up Mars' obliquity like that?

Laskar mentions the event in the 2002 paper, but doesn't bother to offer any particular physical cause. These are numerical orbital simulations after all, not recorded observations. As unsatisfying as it might be, I think your answer is that evidently the planets aligned, so to speak (albeit probably not in a literal line).

Fig. 8 in the 2004 paper gives a tiny bit of context in terms of uncertainty, it shows that the simulation is only precise for the last 10 million years, and then goes chaotic.

Laskar mentions that his data are free to anyone who requests it, although I can't say how accessible it would be. In theory, someone could go in there and look for correlations (e.g. maybe there was some kind of strong resonance with the Earth and/or Jupiter around that time). I looked through ADS a bit, and found a newer paper by Edvardsson and Karlsson (2008) that pretty much supports Laskar's results in over the past 8 million years, but my brief search didn't turn up anything looking into what exactly caused that anomaly.

Finally, from context it appears that Touma and Wisdom (1993) were the ones to first discuss that obliquity drop, although their paper unfortunately appears to be hiding behind a Science paywall.


Why wouldn't massive torrents freeze on Mars?

... so why is this the case?

It may in fact NOT be the case!
Interestingly Phys.org's new item Early Mars was covered in ice sheets, not flowing rivers: study just showed up. It begins:

A large number of the valley networks scarring Mars's surface were carved by water melting beneath glacial ice, not by free-flowing rivers as previously thought, according to new UBC research published today in Nature Geoscience. The findings effectively throw cold water on the dominant "warm and wet ancient Mars" hypothesis, which postulates that rivers, rainfall and oceans once existed on the red planet.

and says later:

Grau Galofre's theory also helps explain how the valleys would have formed 3.8 billion years ago on a planet that is further away from the sun than Earth, during a time when the sun was less intense. "Climate modelling predicts that Mars' ancient climate was much cooler during the time of valley network formation," says Grau Galofre, currently a SESE Exploration Post-doctoral Fellow at Arizona State University. "We tried to put everything together and bring up a hypothesis that hadn't really been considered: that channels and valleys networks can form under ice sheets, as part of the drainage system that forms naturally under an ice sheet when there's water accumulated at the base."

and links to the new paper in Nature Geoscience Valley formation on early Mars by subglacial and fluvial erosion


How does a cross dispersed spectrum look like in a reality (from echelle spectrographs)?

You are doing it incorrectly if you are trying to cross-disperse your Fig.2. You should be cross-dispersing your Fig.1.
A "cross-dispersed" spectrum requires requires the dispersive elements to be at right angles.
Your Fig.1 shows the dispersed spectrum from the echelle with overlapping orders.
To separate the orders you view Fig.1 with the cross-dispersing element, but with the cross-disperser dispersion direction placed along the y-axis of your pictures. A schematic would be something like this, where the grating rulings are indicated with lines. (I believe this picture is originally from Optical Astronomical Spectroscopy, by C. R. Kitchin).

The slit orientation would be parallel to the rulings on the echelle, but you need to make the slit short enough that the orders can be separated by the cross-disperser.
That should produce something like that shown below, if you illuminate it with a continuum source. Exactly what you see will depend on the dispersive power of the cross-disperser (should be low and often, prisms are used), how large your CCD is and how long your slit is.


Why do dark objects look white from a distance? (Moon, Ceres, but not Earth!)

Here's my take on it. I'm not sure what the mystery is - there appear to be two contributing factors.

Both the Moon and the Earth simply reflect/partially absorb the light that is incident upon them from the Sun. The overall of albedo is of little consequence, since the reflection of any light will result in an object that appears bright against the night sky.  The moon reflects quite uniformly across the whole of the visible range - it actually reflects red a bit better than blue. On the other hand the Earth's albedo is strongly wavelength dependent depending on what is doing the reflecting on the visible hemisphere. See the image below from this site. This injects some perceived colour into reflected earthlight.

The second point is the way the eye works at low illumination levels. Although you might think so on a dark night, the moon is not very bright. It is faint enough that the eye starts to switch over to mesopic or even scotopic vision where the light levels are too low to stimulate the colour-sensitive cone cells and just stimulate the colour-insensitive rod cells.

In the case of the moon, this has the effect of removing the red tinge it would have otherwise and giving the moon a "silvery"/white appearance. This is known as the Purkinje shift.

How the Earth appears to the human eye from the moon I could not say. It is possible that it is bright enough to stimulate the cone cells and thus appear as coloured as the photographs suggest. Either way, you can't tell anything from a photograph other than that the light from the Earth definitely is coloured, whether we perceive it to be or not. Photographs of the moon however do not show any strong colouration. In terms of brightness you also can't tell anything from a photograph showing just the Earth. The Earth is much brighter than the moon in terms of reflected light per unit area and integrated over its disc as seen from the moon.


Amateur observing targets for binary star systems?

Okay! Here are some of the popular targets, search for the season of visibility yourself please.

The above list describes the angular seperation between the stars and also the magnification required to resolve it, both explicitly. So, this list is not only for amateurs, but there are some intermediate ones which you can try if you have a good enough telescope.


Were all first generation stars of same mass and size?

If the universe had been entirely homogeneous then no stars would have formed. Star formation requires regions of over-density to collapse.

There are plenty of ideas of what determines the spectrum of masses that stars are born with (known as the initial mass function) - mainly concerning turbulence, fragmentation and accretion. The commencement of nuclear fusion is not a factor, since nothing really changes at the surface of a star when this happens (the luminosity is the same).

The reasons for the observed initial mass function in the present-day universe (which is remarkably similar from place to place) are not fully understood, so unsurprisingly, neither is it a settled matter in the early universe where it cannot even be observed. However, it seems likely on theoretical grounds that stars formed from pristine, primordial gas could be much more massive and less susceptible to fragmentation into smaller objects. Even if this is the case, simulations still suggest that stars would be born with masses that vary by orders of magnitude (e.g. Stacy (2016) and recent references therein).


How much mass per year may the black hole at the center of the Milky Way gain by capturing neutrinos?

Not enough to worry about.
Most neutrinos we detect on Earth come from the sun.  The black hole at the centre of the galaxy is not so close to a star as the Earth is so you would expect the neutrino flux to be lower than on Earth,  let's assume it is similar: about $10^{11}$ neutrinos per $\text{cm}^2$ per second. But each neutrino is light, even including its kinetic energy it has only a few hundred KeV (again based off solar neutrinos for the sake of a rough estimate). Converting that to mass ($m= E/c^2$) gives the mass of neutrinos per cm² per second.  It's about $5×10^{-20}$ kg/cm²/s
The black hole is relatively small in diameter, even if you account for the fact that the effecive radius is larger than the actual event horizon radius, since things will tend to be bent towards the black hole, the cross sectional area is about 10²⁵ cm², so roughly 500000 kg of neutrinos might fall into the black hole every second.  (Although this is very much in the spirit of a Fermi estimate, this is probably an overestimate)
But the mass of the black hole is about 8,000,000,000,000,000,000,000,000,000,000,000,000kg
So 500,000kg is negligible.


How far can we detect lightning in radioastronomy?

Frequencies of terrestrial whistlers are 1 kHz to 30 kHz, while radio telescopes work at 30 megahertz to 300 gigahertz.
Radio telescopes would need to be 1000x larger in order to resolve the direction of extraterrestrial whistlers.


Be careful when saying "nearby stars". The scales involved in "nearby" are beyond what most people imagine. If our Sun was the size of a soccer ball located in California, the nearest star would be another soccer ball located in Greenland.

Let's say the average distance between "nearby" stars is a few light years. Let's say an average supernova is like the recent one in the M82 galaxy. That supernova, at, say, 10 light years, would be less bright and less hot than the Sun as seen from Earth. To make that supernova appear as bright and as hot as the mid-day Sun, you'd have to be at 1/3 light years from it (the supernova would have to be in the middle of the Oort cloud).

Interstellar distances are stupendous. Even a supernova cannot do much to compensate for distance.

But what if the stars are very close to each other? They would pretty much have to orbit each other, forming a double star. Moreover, both would have to be on the verge of going supernova. That's a pretty unlikely situation.

Even then, there are many kinds of supernova. There's type Ia, where a dwarf orbiting a red giant leeches too much material from the giant and eventually explodes:

Then there's the core collapse supernova, where a star is simply too bloated and too old to sustain enough pressure in the core; the core collapses into itself due to gravity and re-ignites fusion, effectively exploding like a bomb and ejecting the outer layers:

Is a nearby stellar explosion going to increase instability in its about-to-blow-up companion, or is it going to reduce its chances of exploding? The answer is not trivial.

Bottom line: I wouldn't say it's mathematically impossible, but it's a combination of quite unlikely scenarios. In a "normal" galactic neighborhood like ours, it's never going to happen.


Why does it make sense to say the universe has no centre?

The answer to this question lies in cosmology.

It's because of relativity in positions.  Every galaxy appears to be moving away from us and the universe we see is only the observable universe.  Further points are more in the distant past, when you observe the deep field you observe the past.  This may seem irrelevant to your question, but the idea of a "center" to a universe where any position in the universe, be it from our galaxy or from a galaxy in the Sloan Great Wall is the true center is not consistent with observation.  That's because galaxies/clusters of galaxies are moving away from each other, relative to each other.

Every galaxy is slowly being pulled apart from other galaxies, save for those in gravitationally bounded clusters.  (The whole architecture of the universe, with massive voids of absolute nothing and bunched up clusters of galaxies reflects this quite a bit)  If the universe had a central point, this would be easily observed because there would be variation in the color shift of galaxies that could be analyzed to deduce a central point where the galaxies are moving away from.

That is not what is observed.  Instead, pretty much most galaxies (save for exceptions like Andromeda or the Triangulum, who prove the rule since those galaxies are gravitationally connected to our galaxy via the local group) are moving away from us.  The Great attractor being not a path to the center of the universe, but the center of the recently discovered Laniakea super cluster.  So if there's attraction between galaxies, it's because of gravitational interactions.  The universe is also without any real edge unless you count the edge of the observable universe.  The observable universe being more a historical record than what the contemporary universe around us looks like and is a byproduct of light speed being so limited.  The speed of light in a vacuum is also the speed of information.  The whole universe is far more massive by comparison.

So to speak about a center of the universe would require the following:

A region of space that galaxy is moving away from or towards. (Which does not exist, instead relative trajectories and gravitational clustering is all regional at most.)
Edges that are defined enough to pinpoint a center (If such edges exist, they are impossible to observe due to the limited speed information travels in the universe)
Absolute positioning to pinpoint where this center would be. (All positions are relative, the galaxy, the sun and our planet all are moving at extremely fast velocities we don't notice thanks to relativity.)