Nomenclature of Interstellar Objects and Stars I posted this question on the space exploration SE but people recommended that it would be more on-topic here. Could someone please help by giving some definite laws of nomenclature for celestial objects, like the provisional names given before a proper name is decided... I am copy-pasting the question here:
I tried searching the answer to this on Google for hours and visited the IAU website as well, but it didn't really clarify my doubt. I was reading about Black Widow Pulsar, which is an "eclipsing binary millisecond pulsar in the Milky Way" and after its common name, in brackets is given PSR B1957+20. How is this named? What nomenclature rules are used? Was this the provisional name given before it was named Black Widow Pulsar or is this the scientific name of the neutron star?
This is the case with most interstellar objects and stars. Like Ultima Thule has another name 486958 Arrokoth? Is this a scientific name? Please explain what nomenclature conventions are used to name the asteroids and stars.

Names are common enough for solar system bodies, but generally stars don't have names, they have identifiers from various catalogues.
In the case of Pulsars. they are named by their location in the sky.  Just like Earth has longitude and latitude, every point in the sky has a "right ascension" and "declination".  Due to the slow movement of the Earth's axis, to identify the location in the sky you also need to mention a date, though you can look up the positions of stars in 1950 (B1950) or 2000 (J2000). Also, for historic reasons, the angle of right ascension (or "longitude") is measured in hours and minutes (24hr = 360 degrees)
Now the Black Widow pulsar is located at 19hr 57 minutes of right ascension and +20 degrees of declination (in the B1950 coordinates).  So its designation is PSR B1957+20.


Direction of Earth’s poles relative to the Sun’s movement around the Galaxy The plane of the solar system is at 63 degrees to the plane of the galaxy, then either the North or South Pole of Earth would be on the ‘leading edge,’ i.e. pointing roughly in the direction of movement around the galaxy (+/-23.5 degrees depending on the direction of the tilt of the earth at the time). Would there be any effect on the atmosphere at the front of this direction of travel caused by additional collision / friction with space dust or solar wind? For example, if the South Pole was on the leading edge, could something like the ozone hole be exacerbated by interstellar particles, or could the auroras be more spectacular at the pole that points forwards? If so, is there evidence of seasonal variation in these effects as the tilt of the earth increases or decreases that pole’s angle towards the direction of the sun’s movement along the galactic plane?

Also, would the plane of the planets be shifted slightly to the rear of the midline of the sun because the sun is, in effect, dragging them behind it?

Finally, is there any evidence of increase or decrease in the above postulated effect depending on whether the solar system is above or below the galactic plane during its 30 million year oscillation? E.g. if there were more interstellar dust while passing through the mid-plane, would that leave a 30 million year signature in polar ice-core samples?

There is no difference in the number of cosmic particles.

You seem to imagine that there are static clouds of gas in the universe that the sun is moving through.  This is not the case. The gas in the galaxy is orbiting at roughly the same speed as the sun.

The sun does have some proper motion relative to the interstellar medium, but the interstellar medium doesn't penetrate the bubble formed by the solar wind.

There is no dragging. The gravity of the galaxy affects both sun and planets, we all move around the galaxy together. There is no shifting of the plane at all.

Interstellar particles are found in ice cores, a paper suggests that variation in the rate of cosmic radiation may cause hydromagnetic shocks that could transport metal rich dust into the inner solar system. No 30 million year periodicity is mentioned.


Why do M dwarfs emit more X-ray radiation than larger, more luminous stars? I have read that "The relative amount of harmful radiation (to life as we know it) that stars emit can be 80 to 500 times more intense for M dwarfs relative to our Sun". Source This seems counterintuitive to me since the HRD depicts M dwarfs to be less luminous. How are these types of stars able to produce such copious amounts of radiation? What are the physical mechanisms behind this? I cannot find an answer to this online or in papers. The information I find simply restates it but does not explain it. I could use some insight as this seems counterintuitive.

A large fraction of M-dwarfs are highly magnetically active because they have deep convection zones (or may be fully convective) and they have longer spindown timescales than higher mass stars, so remain at higher magnetic activity levels for longer.
The magnetic fields are generated by some sort of internal dynamo. The dynamo efficiency appears to increase with rapid rotation and deeper convection.
The magnetic activity manifests itself with a large surface coverage of equipartition magnetic fields (i.e. the magnetic pressure is similar to the gas pressure), with strengths of several kG. These fields protrude above the surface, and turbulence in the photosphere causes magnetic reconnection, acceleration of charged particles, and the evaporation of heated material into the corona, which emits the X-rays. Typical levels of X-ray emission for fast rotating M-dwarfs are $L_x/L_{\rm bol}\simeq 10^{-3}$, about 10,000 times greater than the Sun.
The large scale magnetic field is also responsible for slowing down the star's rotation, via the linkage of the ionised stellar wind to the field lines. At some distant point from the star, the wind decouples from the field, carrying away angular momentum. It is observed, through studies of clusters of stars with known age, that while sun-like stars have spindown timescales of 100 million years or less, lower mass M-dwarfs can remain rapidly rotating for a billion years or more.
Note that young ($\leq 100$ million years) sun-like stars also have $L_x/L_{\rm bol}\sim10^{-3}$; they just maintain these high activity levels for less time.


How official is TNO 486958, 2014 MU69's new name Arrokoth? The Trans-Neptunian object 486958 or 2014 MU69 has had several designations and nicknames. According to Phys.org's NASA renames faraway ice world 'Arrokoth' after backlash

  Ultima Thule, the farthest cosmic body ever visited by a spacecraft, has been officially renamed Arrokoth, or "sky" in the Native American Powhatan and Algonquian languages, following a significant backlash over the old name's Nazi connotations.

Wikipedia's 486958 Arrokoth says

  When Arrokoth was first observed, it was labelled 1110113Y, nicknamed "11" for short. Its existence as a potential target of the New Horizons probe was announced by NASA in October 2014 and it was unofficially designated as "Potential Target 1", or PT1. Its official designation, 2014 MU69 (a provisional designation indicating that it was the 1745th object assigned one for the second half of June 2014), was assigned by the Minor Planet Center (MPC) in March 2015, after sufficient orbital information was gathered. After further observations refining its orbit, it was given the permanent minor planet number 486958 on 12 March 2017.

Questions:

Exactly how official is the new name Arrokoth? Is there further approval necessary or are 486958 and 2014 MU69 and Arrokoth recognized equally as official names?
Are 486958 and 2014 MU69 and Arrokoth all likely to be used interchangeably in astronomical literature from now on, or will one designation emerge as the most commonly used name?

The International Astronomical Union describes this minor planet naming process:

Minor Planet Center assigns provisional designation
MPC assigns permanent designation
Discoverer suggests name
IAU working group reviews name
MPC publishes name

  Accepted names become official when they are published, along with their accompanying citations, in the Minor Planet Circulars, issued monthly by the Minor Planet Center.

The last page of the Minor Planet Circular published 2019-11-08 says:

  (486958) Arrokoth = 2014 MU69
      Discovered 2014 by M. W. Buie and the New Horizons Search Team.
      Arrokoth is the word for "sky" from the Powhatan language of native people from the Chesapeake Bay region.
  Institutions in this region played a prominent role in facilitating the discovery and exploration of this ancient and distant object.

so it's as official as it gets. As with these near-Earth asteroids:

(99942) Apophis = 2004 MN4
(101955) Bennu = 1999 RQ36
(162173) Ryugu = 1999 JU3

we can expect (486958) Arrokoth to displace 2014 MU69 as the prevailing identifier in new publications.


Do low-mass red giants last longer than high-mass red giants? It turns out that red giant stars are not defined by uniform mass.  Some can be only one-third as massive as our sun, whereas others can be eight times as massive.  So I'm calling the both sides "low-mass" and "high-mass", respectively.
I have been told that red giants last only as long as one billion years, but considering that lower masses equal longer lifespans, I have to ask--is it the high-mass red giants that have the one-G lifespan, or is it the low-mass?

Higher mass stars will have shorter lives.  Even though they have more fuel for nucleosynthesis, they burn this fuel much quicker than lower mass stars.
Generally, you should think of "red-giant" as an evolutionary phase and not a particular type of star.
Looking at the Scheller et al. (Aston. Astrophys. Suppl., 96, 269, 1992) data given in one of the comments, you want to compare the times denoted by points 5 and 6: ascent of the (first) red giant branch.  For a one solar mass star, the elapsed time ascending the giant branch is 634 million years.  By comparison, the time for the same phase for a five solar mass star is 0.289 million years.
When looking at the Hertzsprung-Russell chart (surface luminosity vs effective temperature), the RGB tracks at lower masses are all bunched up.  Therefore, it is very difficult to discern stellar mass (and age) from just luminosity and temperature.  By looking at chemical composition through spectroscopy, it is possible to infer mass by the different types of nucleosynthesis.


How exactly will DESI simultaneously capture individual spectra from 5,000 galaxies using optical fibers? The BBC News article Telescope tracks 35 million galaxies in Dark Energy hunt says:

The aim of the five-year programme is to shed light on Dark Energy - the mysterious force thought to drive an accelerated expansion of the Universe.
The instrument effectively contains 5,000 mini-telescopes. Each one can image a galaxy every 20 minutes.
In just one year scientists will have surveyed more galaxies than all the other telescopes in the world combined.

I wonder if instead that it would be better to call them 5,000 mini-spectrometers or 5,000 mini-slits, but I'm not yet sure how this whole thing works.

Inside DESI are 5,000 optical fibres, each acting as a mini-telescope. This enables the instrument to capture light from 5,000 different galaxies simultaneously, precisely to map their distance from Earth, and gauge how much the Universe expanded as this light travelled to Earth.

Question(s)

How exactly will DESI simultaneously capture individual spectra from 5,000 galaxies using optical fibers? Are the fibers used to route the light to a very, very long single slit which is carefully imaged on a 2D detector, or does it do something more complicated?
Considering that the exposure cadence is 20 minutes (per the article), how are all 5,000 fibers quickly repositioned between exposures? 20 minutes is only 1,200 seconds!

update: Wikipedia's Dark Energy Spectroscopic Instrument says:

The Dark Energy Spectroscopic Instrument (DESI) is a new instrument for conducting a spectrographic survey of distant galaxies. Its main components are a focal plane containing 5000 fiber-positioning robots, and a bank of spectrographs which are fed by the fibers.

Wikipedia does not always serve as an authoritative and accurate source, especially for details. Five thousand robots sounds like the name of a Kraftwerk album, and "bank of spectrographs" is less than fully quantitative.

DESI will scan more galaxies in a single year than all the telescopes in the world combined, UCLA/ Lawrence Berkeley National Laboratory

This answer to What is the dimensions of the smallest object detectable by an optical fiber from a specific distance? shows several implementations of multiple optical fibers picking objects from a focal plane and bringing them to a spectrometer. However I don't think any of them come anywhere near to 5,000 separate fibers!

above: GIF from the video (with exciting music) A 2dF night at the Anglo-Australian Telescope

You can probably get most if not all of your questions answered by perusing the main DESI web site, which I encourage you to check out.

There is, for example, a nice video describing the assembly of the main focal plane elements (the fibers and the associated robot positioners) here.

But in simple terms: the circular focal plane is divided into ten wedges (or "petals"). Each wedge holds an array of 500 fibers and their associated robotic positioners. Each set of 500 fibers is collected at the back end and passed outside the telescope to a spectrograph, where the fibers are arrayed into a line to pass their light onto a slit. That is, there are ten separate spectrographs (that's your "bank of spectrographs"), each of which has a single slit with the light from 500 fibers passing into it. (Each spectrograph then has dichroics splitting the light from the slit into three different wavelength regimes and three different cameras, each with a different grating -- 360-555 nm ["blue"], 555-656 nm ["red"], and 656-908 nm ["infrared", though this is still handled by a CCD and it not really what modern astronomers usually call "infrared"]).

The fibers are pretty densely packed into the individual wedges, and so the robotic positioning is done by tweaking/wiggling each fiber a small amount around its default position within a circular hole (you can see glimpses of this in the video), rather than the wholesale movement of individual fibers across the whole focal plane that the 2dF instrument in your video link does.

One of the ten focal-plane wedges of DESI, with about 60 (of its eventual 500) fiber-plus-robot-positioner elements inserted.


How will microshutter arrays be used in the James Webb and future space telescopes? Question: How will microshutter arrays be used in the James Webb and future space telescopes? Are they acting as a sort of moving pinhole or slit, or is the pattern more complicated, like a coded aperture telescope? I can't imagine that would work though, as the application seems to include spectroscopy rather than just image reconstruction.

The NASA Goddard news item NASA to Demonstrate New Star-Watching Technology with Thousands of Tiny Shutters says:

The technology, called the Next-Generation Microshutter Array (NGMSA), will fly for the first time on the Far-ultraviolet Off Rowland-circle Telescope for Imaging and Spectroscopy, or FORTIS, mission on October 27. The array includes 8,125 tiny shutters, each about the width of a human hair, that open and close as needed to focus on specific celestial objects.

and

The microshutter technology gives scientists the ability to produce multiple spectra at once. This capability improves productivity on both sounding rocket missions, which offer only six minutes of observing time, or large space-based observatories, which can take up to a week to observe faint, far-away objects and gather enough light to obtain good spectra. With observing time at a premium, the ability to gather light from multiple objects at once is paramount.
Webb, scheduled to launch in 2021, will carry NASA's first-generation microshutter technology — four 365-by-172 microshutter arrays that together total 250,000 shutters. They will allow Webb to obtain spectra of hundreds of objects simultaneously.
What distinguishes the next-generation array on FORTIS from the one flying on Webb is how the shutters are opened and closed. Webb’s arrays employ a large magnet that sweeps over the shutters to activate them. However, as with all mechanical parts, the magnet takes up space and adds weight. Furthermore, magnetically activated arrays can’t be easily scaled up in size. As a result, this older technology is at a disadvantage for supporting future space telescopes larger than Webb.
Magnet Eliminated
To accommodate future missions, Goddard’s microshutter-development team eliminated the magnet. The shutters in the pilot 128-by-64 array that will fly on FORTIS open and close through electrostatic interactions. By applying an alternating-current voltage to electrodes placed on the frontside of the microshutters, the shutters swing open. To latch the desired shutters, a direct current voltage is applied to electrodes on the backside.
Without a magnet, the next-generation array can be dramatically scaled up in size — and that’s precisely what the team is attempting to accomplish. Particularly, Greenhouse and Li are using advanced manufacturing techniques to create a much larger, 840-by-420 array equipped with 352,800 microshutters, dramatically increasing an instrument’s field of view.
“The array that is flying on FORTIS is a technology development prototype for the big one,” Greenhouse said.

The role of the microshutters is NOT to act as pinholes or coded aperture and to produce a focused image, similar to a mirror or lens.
Its job is just to separate the the light from different sources. This is achieved simply by opening only the shutters corresponding to the sources of interest. See the following image:

This is important for spectroscopy as here the light is split in its wavelengths and dispersed in one direction by a grating or prism. That is why Webb's micro-shutter is only used for spectroscopy.  In order to not have a bunch of overlapping spectra you need some shutters to block everything you are not interested in.
The spread out image of multiple sources on the detector than looks like this:

In other words: As you can not detect a 3D image (2 spacial + 1 spectral) on a 2D detector, you have to make some sacrifice by blocking much of your field of view.

This sketch makes the process quite clear:

Further sources: Wikipedia Closed-source micro-shutter decription


How many solar system objects that can really qualify as a KBO are likely to enter Pluto's Hill sphere each day? A comment below Which celestial body is able to come closest to Pluto? says that

Technically, many, many small Kuiper belt objects enter Pluto's Hill Sphere (or sphere of influence SOI) every day.

Pluto is small, but it is very far from the Sun, so maybe its Hill Sphere could be surprisingly large, and if we called each particle of dust out there a "small Kuiper belt object" then this can be true.
Question: But realistically, how many solar system objects that can really qualify as a KBO are likely to enter Pluto's Hill sphere each day?
For the purposes of this question let's see if there is some way to ascertain what a KBO might be, and not focus on the helpful comment too much.

The Hill sphere radius of Pluto is about $r$ = 6 million km.  Most of the Kuiper belt is in prograde motion around the Sun (like Pluto).  Pluto's average speed is under a lazy 5 km/s for an orbital period of about 248 years.  If the difference in orbital speed between Pluto and an average KBO is just 1 km/s, then Pluto will "sweep out" $\pi r^2 *86,400s *1km/s = 9.8 \times 10^{18} km^3= 2.9 \times 10^{-6} AU^3$ of the Kuiper belt per day.
The Kuiper belt is described as a donut extending from 30 to 55 AU.  The volume $V$ of a torus is $V=2\pi^2Rr^2$, where $R=42.5$ and $r=25$ are the major and minor radii, so the approximate volume of the Kuiper belt is $5.24 \times 10^{5} AU^3$.  Of course, KBO's are distributed more densely near the center of the torus, and exist increasingly rarely outside the torus.
So, Pluto might sweep out about $V_1 = 2.9 \times 10^{-6} AU^3$ of the $V_2=5.24 \times 10^{5} AU^3$ volume Kuiper belt per day.   That means there would have to be approximately $V_2/V_1 \approx 1.8 \times10^{11}$ (or 180 billion) total objects in the Kuiper belt for one to enter into Pluto's Hill sphere each day, on average.  It might even need to be much higher since Pluto's orbit is more inclined then the distribution of the Kuiper Belt, as shown here:

We only know of about 2000 KBOs according to this NASA site.  Pitjeva and Pitjev postulate only millions of objects in the Kuiper belt, so it may be unlikely that many KBOs enter Pluto's Hill Sphere radius each day.


Is the composition of stars in future made of more and more heavy elements? In the beginning stars only consisted of the hydrogen element and due to nuclear fusion of those elements in stars and supernova's more heavy elements were created. Because of that, like our Sun, the stars were made not only of hydrogen but also oxygen and carbon etc.

Now is it expected that in future more stars will be made of more heavy elements or are there causes/laws which prohibit stars forming of e.g. stars made of elements without hydrogen.

The initial stars were made of hydrogen and helium.

These enriched the interstellar medium (ISM) with some chemical elements right across the periodic table, when massive primordial stars ended their lives as supernovae.

Subsequent generations of stars continue to enrich the ISM, if their lives are short enough.

So the general gist of what you suggest is true, but the rate of enrichment is very slow. The Sun was born 4.5 billion years ago, but sun-like stars born within the last few million years also have a metallicity consistent with the Sun.

Why is this? Well, star formation is quite inefficient; only about 10 percent of a collapsing gas cloud ends up in stars. Then a star only processes about 10-20 percent of its mass as nuclear fuel; and then only a fraction of this is returned to the ISM in the form of winds and supernova ejecta.

To get a more enriched ISM requires bursts of massive star formation in dense environments. Such conditions occurred early in the life of the Milky Way and perhaps will again when we collide with the gas-rich Andromeda galaxy.

In conclusion, star formation will continue into the future; the stars will very slowly get more metal-rich; there is nothing to prevent more metal-rich stars forming; perhaps the main consequence would be a small reduction in the maximum possible stellar mass (due to the increased gas opacity and consequent radiation pressure).


When will the next series of mutual eclipses of Jupiter's moons begin? When I was young (I won't tell you when) I saw one of the four Galilean moons of Jupiter suddenly disappear while watching through a small refractor. It wasn't a coincidence, I'd seen the prediction in some announcements in Sky and Telescope and it happened like clockwork. The orbital plane of the Galilean moons had rotated so that it included the Sun, and the moons were Eclipsing each other, or so I remember.

Will this happen again? If so, when will they start, and where can I find predictions of individual eclipses?

Mutual eclipses and occulations of Jupiter's moons occur during Jovian Equinox, twice in each orbit of Jupiter, and so roughly every 6 years.

The last series of events was in 2015, so the next will start in 2021. There will be a series of events between January and August.

The site http://lnfm1.sai.msu.ru/neb/nss/nsszph517he.htm can generate ephemerides for these events.


Sun, Moon: Relative Distance from Crescent? This is more a question about the history of astronomy.

It recently occurred to me that you can estimate the ratio of earth-moon and earth-sun distance in the following way: We can look at the triangle observer-moon-sun. Call the angle at the observer $\alpha$, the angle at the moon $\beta$.

At the crescent to half moon stage, sun and moon are both visible, so one can determine the angle $\alpha$ (at the observer).
From the illuminated fraction of the moon, one can estimate $\beta$. (Precision is presumably lower here.)
Thus, the third angle is fixed, and the shape of the triangle is known (but not the overall scale).

So it should be possible to derive that the sun is much further away than the moon. If we could get an independent determination of, say,  the earth-moon distance, the distance to the sun would be known.

This seems rather straightforward. Has this been used historically? What accuracy was (could have been) achieved? For example, I seem to recall that in the "celestial spheres" approach, people have at least ordered the heavenly bodies according to distance, but has this method been taken into account?

This is exactly how the ancient Greeks attempted to estimate the relative distances to the sun and moon.

The difficulty is the angles involved are rather small

Aristarchus measured the Moon-Earth-Sun angle at half moon to be 87 degrees. (or 3 degrees less than a right angle) Which implies the Sun is about 19 times further than the moon. And from this (and an estimate of the sizes of the Earth and moon made at a lunar eclipe) he estimated the distance of the sun.

The actual angle is more like 89.85 degrees (which is very hard to distinguish from 90 degrees) (0.15 degrees less than a right angle).

So this method can be done, but it requires very exact measurement to succeed.


What is the force of a CME on objects in space? I've been reading a little bit about coronal mass ejections lately and they seem really interesting, and they cause a lot of amazing things to happen. However, one thing I haven't seen is the physical force they can have on objects. CMEs seem to have a lot of mass and they're moving at a high rate of speed so I imagine they must have some effect on objects in space, right? What kind of effect can they have on smaller objects like meteoroids, comets, or satellites (as far as orbit, rotation, anything else)?

This isn't a complete answer, but it's a partial one:   http://pwg.gsfc.nasa.gov/istp/outreach/sunearthmiscons.html

  Despite the electromagnetic havoc they play with satellite electronics
  and with Earth's magnetic field, the solar wind and CMEs barely exert
  a pressure or force that you could measure. In fact, they couldn't
  ruffle the hair on your head. The solar wind has fewer particles per
  cubic centimeter than the best vacuums scientists have ever created on
  Earth. Our own air is billions of times denser than the solar wind,
  such that a cubic centimeter of air has as many particles as a cube of
  solar wind measuring 10 kilometers on each side.

So, for your question on changing orbits - not very much.   Granted everything moves a little under any pressure.

Now, the lighter the object is to it's surface area, the more a CME can move it, that's why a solar sail, very light weight and large area, might work.    CMEs can also move gases, like the tail of a comet can be pushed visibly away from the sun (I read that, but misplaced the article), or lighter, high atmospheric gas can be blown away from smaller planets.

Also, no 2 CMEs are exactly alike.   A smaller and/or a younger star will emit much stronger CMEs, so around a red dwarf for example, the effect might be more noticeable for tiny asteroids and the like.


Has the Earth's wobble around the Earth-Moon barycenter ever been observed by a spacecraft? Pluto's motion around the Pluto-Charon barycenter has been imaged by the New Horizons spacecraft:

Has anything like this been imaged of Earth? Yes, the barycenter lies inside of Earth, but it is 3/4th of the distance from the center — more than enough for a noticeable wobble over the course of one orbit.

Has the Earth's wobble around the Earth-Moon barycenter ever been observed by a spacecraft?

Absolutely yes if you can count the wobbling motion of the spacecraft that tracks Earth's wobble about the Earth-Moon barycenter.
DSCOVR sits in a heliocentric orbit which is in a 1:1 resonance with Earth, otherwise known as a "Lissajous orbit associated with Sun-Earth L1".
Its EPIC camera is pointed carefully at Earth all the time in order to keep a 0.45° to 0.53° Earth nicely framed in its only slightly larger 0.61° field of view.
If the Earth-Moon average distance is 385,000 km and they have a mass ratio of 81:1, Earth's wobble should be roughly 4,800 km or about 38% of its diameter. If DSCOVR didn't track this wobble the Earth would appear to move +/- 0.18° in EPIC's field of view such that parts of Earth would disappear outside the frame around the times of first and third quarter of the Moon as seen from Earth.
One can download roughly daily data dumps from EPIC which contain data fields for the pointing vectors of the camera. One can subtract a smoothly varying rotation of $2 \pi$ per year and see both Earth's eccentricity and a monthly wobble due to center-of-mass motion of the Earth. I've looked at EPIC metadata in the past, and while I haven't done this specific analysis this can't not be true!

Why is DSCOVR in a Lissajous orbit? Wouldn't a halo orbit completely avoid the Sun exclusion zone?
Where are post-2017 EPIC images of Earth and the DSCOVR spacecraft coordinates available for download?

From this answer:

above: Image from NASA/NOAA from Bad Astronomy


Do astronomers have any idea what percentage of our galaxy’s stars move in retrograde orbits? I recently noticed that none of the planets in our solar system orbit the Sun in a retrograde direction. This is likely because they formed moving in the same direction as the planetary disk did in respect to the Sun.

Are there stars that move retrograde to the Milky Way galaxy's rotation? If so, where are they concentrated, and why?

About two-thirds of the Milky Way’s estimated 100 billion stars reside in the disk, and these all move in direct orbits. Most of the remaining third are in the football-shaped central bulge, and while their orbits are more varied, on average they too are direct. A few percent of our galaxy’s stars occupy a vast, spherical halo. These travel on randomly distributed orbits — accordingly, half are direct and half retrograde.

Source: Do astronomers have any idea what percentage of our galaxy’s stars move in retrograde orbits?

And while formation of disk galaxies is a complex process, which is not yet fully understood, the reason that almost all stars move in the same (direct) orbits is the conservation of angular momentum when gravitational force pulls gas clouds to come together. It is the same mechanism that is responsible for the planets to orbit the star in the same direction.


Affect of leap seconds on the prediction of an eclipse There seem to be irregularities in the motion of the earth, so we add "leap seconds" to adjust our time accordingly. Since 1972 there have been 26 leap seconds I think. If we extrapolate that backwards 0.5 leap seconds per year, it would be 1 leap minute in 120 years and 5 leap minutes in 600 years.

Would this be enough to throw off a prediction of an eclipse? In other words, could someone in 1416 predict an eclipse right now given the unpredictable errors caused by leap seconds?

It has been known that changes in earth rotation affect the location of eclipse paths for over a century.  The most active research in this has been Morrison and Stephenson.  See, for example http://adsabs.harvard.edu/abs/2004JHA....35..327M


Would DCBH form supermassive stars first before BH formation In reading some literature it seems ambiguous (intentionally?) as to the path that Direct Collapse Black Holes (DCBH) take. Some papers do talk about pristine Halos collapsing to form supermassive stars (SMS) ($10^{3}$) that then undergo direct collapse. However most that I've seen do not.
So my question is, do all DCBH models assume a short lived SMS phase or can there be a runaway collapse/accretion in the core of a H cloud that collapses into a intermediate mass black hole or larger? Are there multiple routes for DCBH? This is all assuming $z>15$ and $\Lambda$CDM.

This article summarises the research:

How does a cloud of gas become a black hole? According to a 2017 study, such a transformation requires two galaxies with very different personalities: one of them a cosmic overachiever that's forming lots of baby stars and the other a low-key heap of starless gas.

As new stars form in the busy galaxy, they blast out a constant stream of hot radiation that washes over the neighboring galaxy, preventing the gas there from coalescing into stars of its own. Within a few hundred million years, that starless gas cloud could accrete so much matter that it simply collapses under its own weight, forming a black hole without ever producing a star, Basu found.

I assume that some fusion does happen during the collapse, but presumably it never catches up with the increasingly intense gravitational pull of the densening gas cloud, so never supports the gas against gravity


Scientific possibility of stationary moons for a writer of fiction I tried to do some research to find an answer for this question before posting this but didn't find anything that I could sink my teeth into.

I was wondering:

if it would be possible for a planet to have a stationary orbit Moon, so only one side of the planet would see the moon ever
what that would do to the oceans and tides of the continent that is on the opposite side of the planet.

Conversely how would oceans and tides be different without the moon?

For my question here I'd like to ask about only the planetary science aspects. For the speculative and hypothetical aspects I can ask a separate question in Worldbuilding SE.

The possibility you mention is feasible if the moon is in the equivalent of a geostationary orbit, where the orbital period of the moon is exactly the rotation period of the planet and it orbits in the equatorial plane of the planet.

Such a moon would always keep the same face to the planet if it were close enough to the planet (which in turn depends on the spin period of the planet) such that its spin is tidally locked to the same orbital period.

The presence of significant tides may be required to keep this configuration stable, without a tidal elongation of both objects it is likely that it would drift away from synchronous rotation. This in turn probably needs the moon to have a significant size/mass compared with the planet.

The tides on such a planet would essentially be "fixed". i.e. the tidal forces would be constant (with the exception of the possibly quite significant tidal forces due to the star it was orbiting). That would nean no water sloshing around due to tides. The "sea level" at point nearest to, and opposite, the moon would be higher than if there were no moon at all.

Something like this has been asked in another Astronomy SE question (Can a natural satellite exist in a geostationary orbit?) and, as pointed out there, Pluto and Charon are an example of the system you are looking for.


What is a "semester series" for eclipse prediction? - Solar saros series 147 and 152 Yet again, I just stumbled across another baffling eclipse question (actually, two now). I was interested in two eclipses (Saros series 147 and 152) that will both be occurring in 2021. Each of these eclipses will occur relatively near the opposite poles of the earth. The Wikipedia link for saros 147 provided the following additional information:

  "Solar eclipses of 1964-1967[edit]

  This eclipse is a member of a semester series. An eclipse in a
  semester series of solar eclipses repeats approximately every 177 days
  and 4 hours (a semester) at alternating nodes of the Moon's orbit.[1]
  "

These are both fairly young saros series. Saros 152 is comprised of 70 eclipses, while saros 147 contains a total of 80 eclipses. Observations: They are 10 eclipses difference in length - They each began 180 years apart. So, following on from that, I checked out their ending dates and found, much to my amazement, that both series end during the exact same year - exactly 177 day apart - in the year 3049 !!!

So, since this additional information, my new questions concerning these eclipses have become:

  1) What exactly is a "semester series" for eclipses (any detail)? The day spacing definition is obvious but doesn't explain much else.

  2) Does this mean that some or all eclipses in series 147 and 152 are in
  the same semester series?

Any eclipse is a coincidence of a new or full moon (repeating every 29.53 day lunation or synodic month)
and a lunar node (moon crossing the ecliptic, repeating every 27.21 day draconic month).
A semester
is 6 synodic months, close enough to 6.5 draconic months that eclipses can occur at that interval in a series 3 or 4 years long.

The interval between any two solar or lunar eclipses can be expressed as $aI + bS$,
where $S$ is the saros interval of 223 lunations,
$I$ is the inex interval of 358 lunations between adjacent saros series,
and $a$ and $b$ are integers.
The semester interval of 6 lunations is $5I - 8S$; for each eclipse in saros $n$, there may be an eclipse in saros $n-5$ one semester earlier, an eclipse in saros $n+5$ one semester later, or both.
As the eclipses in a given saros series are about 18 years apart, each is in a different semester series.

For example, one semester series runs from 2015 to 2018 and includes one solar eclipse in each of saros series 120, 125, 130, 135, 140, 145, 150, and 155.
Another runs from 2018 to 2021 and includes one solar eclipse in each of saros series 117, 122, 127, 132, 137, 142, 147, and 152.


Is the flat earth cosmology scientifically justifiable today? With binoculars or telescopes you'd be able to see to other countries if the earth were flat. And looking out at the ocean we can see where the earth curves and things like the top of ships, not the whole ship. Unless those who think the earth is flat believe they are looking at the end of the world when looking out at the ocean? How do they arrive at this position?

I realize this queston might be closed, and perhaps this should've been a comment, but it got too long:
I was in a radio show last spring with a flat-earther. I really tried to understand him, rather than just ridicule him. We went to the beach with his 87x magnifying camera, to watch the coastline 19 km away. If my model of the geometry of Earth were right, we would be able to see the buildings on the other side of the belt, but not the shore. If his model were correct, I thought we would be able to see the shore, but it turns out they have an ad hoc explanation that has to do with the vanishing point on a flat Earth lying below the ground, thus making sufficiently distant objects disappear.
So, apparently this is their explanation.
When after five minutes I still didn't understand this, he told me it was no wonder that I didn't understand; it took him two  years to understand.
Well, big surprise, we couldn't see the shore. We could in fact see a bit more of the buildings than we should be able to if there were no atmosphere. I assigned this to refraction (which can actually move the horizon several tens of kilometers away, or even ~100 km over water in the springtime when the air is getting warm but the water is still cold), but since he didn't believe in refraction, that didn't convince him.
We then talked a bit about what mechanism might produce a flat Earth. The conversation sort of ended when I realized he didn't believe in gravity either…
Is a flat Earth scientifically justifiable today?
The question title has now been edited to a more easily answerable one. The answer is — unsurprisingly to most people on this site, I suppose — a big, resounding "NO"! The flat Earth theory is not a theory, in the scientific sense of the word. It's not even a hypothesis. You could call it a model, and in many cases, it's a fine model, namely on scales much smaller than the horizon distance. You may assume a flat Earth if you want to calculate the area of the parking lot. This is analogous to Newtonian dynamics being a fine model for small velocities and gravitational fields. It works fine if you want to calculate the force of an apple falling on Newton's head. But everybody knows it become less exact, the larger and faster the apple.
I suspect that the use of the word "theory" in everyday language influences how people think about science. "It's just a theory", they say, without realizing the enormous amount of work, experiments, verifications, and falsifications that go into building a scientific theory.
Flat-Earthers in general don't realize, I think, that in science you can't just assume that something works in another way, without this assumption having severe consequences for many other theories.
Why not?
The list of reasons that a flat Earth is incompatible with a range of well-established theories is long. Here's a subset:

Gravity

Gravity explains why a large mass becomes round. FE'ers have no explanation of why it could be flat.

Electromagnetism

The Sun and Moon are circulating 5000 km over Earth, held up by electromagnetic forces. This is completely impossible within the frames of the accepted theory of electromagnetism, but they don't propose a new theory of EM.

Astronomical distances

Standard trigonometry reveals that in their model, the Moon would look different from different places of Earth. It does, but much, much less than if it were 5000 km away (because it is in fact 400,000 km away)

Meteorology

The guy I talked to had no explanation for why cyclones rotate clockwise and counter-clockwise in the Southern and Northen hemispheres, respectively. The Flat Earth Society explains this by "The Wind Currents are put into gradual motion by the attraction of the Northern and Southern Celestial Systems, which are grinding against each other as gears at the equator line". I… I don't know what to say about this…

Constellations

The Southern Cross can be seen from Australia and South America. I'm in Oslo at the moment, which according to their world map is right in the middle between Australia and South America, yet I can't see the Southern Cross.

Satellites

FE'ers generally don't believe in Moon landings and space stations, but many satellites can be seen even with a relatively small telescope.

The disappearance of ships and other objects below the horizon.

Simple trigonometry shows that on a flat Earth, you should be able to see as far as the airmass allows you. On a round Earth, this is complicated by refraction, enabling us to see farther than if there were no atmosphere. But still, you can easily see how the Sun or Moon sets or rises partly below the horizon.

Lots of other inconsistencies

Stellar parallaxes.
The different weight of the same object at equator and the poles.
Eclipses (in particular the times at which they occur).
The movement of Jupiter's moons (which can be seen with a regular binocular).
Tides.

The problem is that there is no way to discuss these things with FE'ers. If you give an argument that they can't disprove, they will either claim that you're just wrong, or that "you think so because that's how you were schooled", or that you're a part of the conspiracy.
In particular the last point is important. Since the FE model, were it right, would imply that most of all physics would be wrong, that means that either

all physicists in the world are stupid, which makes it pretty amazing that we've made planes fly, computers work, bridges hold up traffic, compasses show us the way, as well as predicted stuff like Venus transits and the weather tomorrow. Or

all physicists are a part of a conspiracy, which would be equally amazing how none of the millions of physicists through history has confessed. Where's my bribe?!

Why am I not just ignoring them?
You may say that by writing this, I am just feeding the trolls, and giving people the impression that flat-Earthers are a problem for science. But in my opinion, the attitude that they hold toward science is much larger than just a few crackpots claiming what is obviously false. It's a symptom of something sad in our society, namely that "science is just one opinion, and my opinion is just a valid".
I acknowledge that scientists can tend to be cocky and should try to be more open to understanding why people don't believe us. But what makes a few people believe in a flat Earth is the same thing that
makes people deny that human activities affect the climate,
or makes them believe that vaccines are evil,
and makes some politicians get away with using a term like "alternative facts".


Accretion disks - why are they disk-shaped, rather than spherical? So, the theory (briefly) about accretion disks is that stuff falling into the black hole will most likely not fall straight into it but will miss it and go into orbit, perhaps for a long time, perhaps for a short time, before finally falling in.  That makes sense.

But, why disks?  Wouldn't the matter falling into the BH be coming from all directions, and thus fall into orbits in all planes around the BH, not just one?  Or, is it that once there's a hint of "plane preference" in the debris, this will tend to be self-reinforcing, and anything subsequently falling into orbit around the BH will hit something in the disk, and so end up (perhaps after many collisions) being swept around in the disk as well?

There's two different effects here, and they're both related to viscuous forces in the accreting matter.

First, if the infalling matter has some nonzero angular momentum vector $\mathbf{L}$, then consider a plane perpendicular to this vector. Due to its conservation, we can't get rid of rotation along this plane, but the component of angular momentum orthogonal to the plane is zero. As the accreting matter heats up and radiates away energy, that energy is taken from the gravitational potential, which is lessened by bringing the matter particles closer together. Thus, such interactions tend to flatten the matter into the plane orthogonal to its angular momentum.

But there's something more that happens for rotating black holes, as a combination of its gravitomagnetic field and viscous forces in the matter, that drives the matter to the plane of the black hole's rotation, regardless of its initial angular momentum. This is called the Bardeen–Petterson effect [1]:

  In an accretion disk, viscosity causes each ring of gas to spiral inward toward the hole with a radial speed $|u^r|\ll\text{(orbital velocity)} = (M/r)^{1/2}$. If this were the only effect of viscosity, the warping of the disk would consist of a differential precession. Successive rings, each of smaller radius than the preceding, would have interacting longer with the [gravitomagnetic] field and thus would have precessed through successively larger angles... . As a result the disk would be radially twisted, but it would not be driven into the equatorial plane. However, the precession changes the disk's shear $\sigma_{jk}$ and thus changes the viscous forces; and these altered forces drive the disk toward the equatorial plane of the black hole.

This is only significant closer to the black hole, at $r\lesssim 100M$. At larger radii, the disk shape is due to the direction of the angular momentum of the matter instead.

[1] Black Holes: The Membrane Paradigm, ed. by Kip S. Thorne, Richard H. Price, and Douglas A. Macdonald.


Why can't supermassive black holes merge? (or can they?) The CNet article Astronomers discover two supermassive black holes in a death spiral links to Discovery of a Close-separation Binary Quasar at the Heart of a z ~ 0.2 Merging Galaxy and Its Implications for Low-frequency Gravitational Waves (available in ArXiv) and says:

Supermassive black holes are usually found at the center of galaxies, including our own, and during a galaxy merger they end up beginning a dance of death, spinning around each other in a near-endless waltz, until finally merging. However, researchers are currently unclear as to the time it takes for black holes to merge -- or indeed, if they merge at all.
"It's a major embarrassment for astronomy that we don't know if supermassive black holes merge," said Jenny Greene, a professor of astrophysical sciences at Princeton and co-author of the study. "For everyone in black hole physics, observationally this is a long-standing puzzle that we need to solve."
This puzzle is dubbed the "final-parsec problem." Some astronomers believes that once two supermassive black holes get close enough together, reducing their distance to 1 parsec (3.2 light years), they may dance for an eternity.

Question: If it turns out that supermassive black holes can't merge, or have a tough time doing so, what might the reasons be?

The main problem is angular momentum. In order for two gravitationally bound objects to merge (whether black holes, supermassive black holes, planets, stars, etc.), they must shed enough angular momentum for their orbital separation to become small enough.  Average orbital separation (semi-major axis) is determined entirely from the angular momentum of the orbit (at least in classical mechanics; I don't know if that is true for relativistic situations such as merging black holes when they get close together).  Removal of angular momentum requires interactions with other objects.

When two galaxies merge, their supermassive black holes both have angular momentum.  Through a phenomenon known as "dynamical friction," gravitational interactions with other stars sap the black holes of much of their angular momentum, until they are brought within a few parsecs or so of each other.  At this point, the black holes have flung out all of the stars that were in the region and there is (presumably) nothing left for dynamical friction to sap their angular momentum.  Once the black holes are close enough (I don't know off the top of my head how close), emission of gravitational waves will sap the orbiting pair of their remaining angular momentum and a merger becomes inevitable.

So to answer your question, the reason for supermassive black holes to be unable to merge is that they are too close together for there to be any remaining material (stars, gas, etc.) in the center of the galaxy to remove angular momentum from the orbiting pair, having already removed the material themselves, but are not close enough for emission of gravitational waves to remove angular momentum fast enough for their merger to occur any time soon (in an astronomical sense).


Appropriate battery for overnight viewing to drive telescope and GOTO? I have a Plettstone 18" Dobsonian telescope with GOTO and clock drive that runs off a 12 volt cigarette lighter adapter.  What is an appropriate battery to use to run this overnight while viewing all night long and then charge during the day for the following night's use?

In the past I've used deep cycle lead acid batteries, but these are heavy and can fail if not carefully maintained with a float charge.  Also, I don't need cranking power of a car, just sustained energy to drive the scope.  A typical Optima YellowTop deep cycle is rated 75 ampHours.  I don't know what the draw in Amps of my telescope is, but imagine it is less than an amp.

I've also seen these Battery Tender Lithium Iron Phosphate types but wonder if their 10-12 ampHour rating is enough and whether they are meant to be drained over a whole night?  Some say if you get below 8 volts the battery becomes broken, but I doubt the telescope would still run at such a low voltage.

What would a reasonable ampHour rating be?  What kind of battery is a good fit for astronomy equipment overnight?

Before buying anything, get out your Amp-meter and measure what the motors pull in clock drive and GOTO mode, starting from a variety of scope positions.
An 18 incher will have a lot of mass, and Dobsonians are not known for their excellent bearings, so the amount of juice you need for a night of observing could be considerable.
Do you track most of the time, or GOTO...GOTO...GOTO?
How many hours is your typical observing session.
Multiply your average tracking amperage by the approximate number of hours you spend tracking, then do the same for your GOTO work.
Add the two numbers (amp hours) together, and double the result.
That should tell you about how big a battery you need.

The doubling is done to provide leeway, and because taking a standard lead acid below 50% is a sure way to kill it after only a few recharges. If you go Li, you can probably cut the multiplier.


Nebula and galaxies using 70mm scope Telescope: Orion 09843 SpaceProbe 3,3 inch aperture(76mm), 700mm focal length, eyepieces - 25mm and 10mm focal lengths. Please tell me whether it is possible  to see nebula and  galaxies.

You can see some of them with a naked eye. Many of the Messier objects, including galaxies and nebulae are observable even without any telescopes. As for others, if Charles Messier could catalog them in 1771 with a 4 inch refractor, you should see most using your 3.3 inch reflector too given good observation conditions, knowing where to point and using a properly collimated and otherwise maintained telescope.

Messier was from the Northern hemisphere and you seem to be closer to the equator judging by information you provide in your profile, so you shouldn't have problems seeing these objects either, and there are equally interesting objects also in the Southern skies that you could observe. Some of the easiest recognizable on the Northern skies are the Andromeda galaxy, or Messier 31, and the Orion Nebula, or Messier 42. You should also be able to discern quite a lot of detail of other astronomical bodies, including e.g. Pleiades, or Messier 45 which is also visible to the naked eye and many astronomers use to determine observational conditions and their visual acuity by.


The recent results on Hubble constant measurements In recent news there is this announcement
In the introduction they say:

Distance measurement discrepancy: a $4.4σ$ tension on the value of $H_0$

and I understand the discrepancy is with the value derived using the cosmic microwave radiation (CMB). There will be more publications to follow.
These measurements are local, i.e. of the universe as it is now.
By coincidence I was looking up baryogenesis and came upon the Sakharov condition  for generating the asymmetry between baryons and antibaryons which must have happened by the time the CMB map solidified.

Any of the three necessary conditions that a baryon-generating interaction must satisfy to produce matter and antimatter at different rates: (i) baryon number violation; (ii) C-symmetry and CP-symmetry violation; (iii) interactions out of thermal equilibrium.

iii) seems to me to be relevant to a measurement of the Hubble constant using the CMB data because it would affect what is presumed in the standard cosmology :

The CMB is also used to determine the Hubble constant, where the temperature is analyzed as a function of frequency – a power spectrum – and a best fit analysis is made to constrain the Hubble constant.

If interactions out of thermal equilibrium are responsible for baryogenesis, maybe the assumptions entering the fit should change, and maybe the discrepancy between the two methods tells something about baryogenesis.
As  a particle physicist I am not up to astrophysical calculations. Is anybody familiar with this topic who could say  whether this is a possible suggestion?
Edit:added
This link says that the Sakharov conditions are necessary, and the standard model obeys all three:

During a first order electroweak phase transition bubbles of the broken vacuum form in an unbroken phase. The expansion of these bubble walls is shown to lead to considerable departure from thermal equilibrium.
Since Sakharov’s conditions are satisfied, the baryon asymmetry of the Universe could be created during the electroweak phase transition. Unfortunately, it has been shown that the amount of CP violation coupled with the strength of a first order phase transition is not sufficient to create enough baryon asymmetry in the SM

Surely the standard model must be  used in estimating by fits to the CMB the Hubble constant, but not the extra baryogenesis mechanism necessary for the observed asymmetry. These could affect the thermal equilibrium estimates.

The Carnegie-Chicago Hubble Program. VIII. An Independent Determination of the

Hubble Constant Based on the Tip of the Red Giant Branch. I hope this helps.
https://arxiv.org/pdf/1907.05922.pdf


Is it possible to see an image with radio telescope of something not so far away in detail? Can an image of a person using a radio telescope see a person from far away? Could a radio telescope be made to see someone walking on Mars?

https://space.stackexchange.com/questions/30477/could-we-see-someone-walking-on-mars-from-earth

In brief: you cannot resolve details of an object on a scale less than roughly a wavelength or two.  The distance to the object only marginally affects this problem.

So take a look at the wavelengths your radiotelescope is capable of receiving and compare with dimensions of interesting "features" on a human body.

Further, as the comments pointed out, unless the human is either emitting radio-wavelength energy in significant amounts (unlikely) or is being "painted" by a large emitting source (as active radar systems do), you won't get any signal in the first place.


How should I interpret this result obtained using lomb scargle? I used lomb scargle for periodic analysis.
The time interval used is constant at 500 seconds.

#My data (x=time, y=intensity values) is :
x= (
00:28,
00:37,
00:45,
00:53,
01:02,
01:10,
01:19,
01:27,
01:36,
01:45,
01:53,
02:01,
02:11,
02:19  )
y=(18264.682317  ,18720.708081  ,18476.409461  ,19592.046320
,19166.021184 , 18984.852405 , 20971.112057  ,20946.024653
21590.194169 , 21771.175308 , 21474.776018 , 20156.969939
21688.756080 , 22570.152358  )
--> The time interval of x data is equal to 500 seconds.
and lomb scargle code is:
frequency,power = LombScargle(x,y).autopower(nyquist_factor=1)
ax.plot(frequency,power,'-')
        #ax4.axvline(x=best_frequency)
ax.set_ylabel('(4-2) band R₁(2.5) Intensity - Lomb Scargle Power')

How should I interpret the lomb scargle results shown in the picture(below picture)?

Why does the maximum appear near zero(below picture)?

Not being familiar with Lomb-Scargle (never having used it myself) the first thing I would say is that your graph is not really suitable for this kind of analysis.  If I graph it in my spreadsheet I get what looks more like a "bumpy line" not a periodic pattern :

The purpose of this algorithm is to take data that has a periodic pattern and find try and determine the frequency spectrum of that periodic data.  If you use data that is not periodic or does not properly exhibit a range of data values that allow discovery of the period then you unsurprisingly do not get sensible results.  So what I'm saying is that I suspect your data is not going to give you useful results with this algorithm.  I think you need more data points over a longer timescale.
Note my limited understanding is based on how similar algorithms are used for analyzing light-curve data.  This is well explained in some detail in this paper on Arxiv.  THe first section has no mathematics at all and illustrates a typical use case of the algorithm.  Make particular note that you need data covering more than than the period of the light curve to get a sensible result.  Note also the way the first "maximum peak" in the power spectrum is the period.  Figure (1) is that paper shows the raw data (a jumbled mess covering multiple periods) and figure (2) shows the way the period is identified and how the raw data can be graphed using this discovered period to properly align out of phase values.  I am dubious that your data is suitable, possibly because the underlying data is not periodic or is only a partial window of a period.
Section 8 of that paper has several pieces of advice about effectively using this algorithm, so I would recommend you read that even if you have to skip the mathematics in the middle of the paper.


How ready are we to deal with a asteroid/comet/giant-meteor on a collision course with Earth? I keep hearing astronomers saying there are observatories scanning the skies all day and night looking for near-Earth objects, extrapolating their trajectories.

My questions are

If we were to get the news of an on-coming object, how early we will find out or how much time we will have?

Currently, people don't even think about it; Governments are cutting spending in their space exploration programs, let alone deflecting an impact-er.

Under certain circumstances, would the notice from the watchdogs be enough for us to build something that would prevent the loss?

The short answer is "not very, but we're getting better".

In the case of particularly large Earth orbit intersecting bodies, kilometers across and bigger, they're fairly well known and tracked. We'd probably get months to years of warning. Technologically, we've probably got the capability of diverting such an object if we have enough warning, but it'd likely be the most complex human endeavor ever. Blowing it up is unlikely to help, we'd need a nice long shove from as far away as possible to divert it.

The bigger problem is smaller bodies in the tens to hundreds of metres. Some of these are known about and tracked, but plenty more aren't. The Chelyabinsk meteor was about 20 metres across and, in terms of size/density/speed/impact angle wasn't that far off being a city killer. We never saw it coming. (Ironically, a very similar but unrelated 30m object was being tracked through a near-miss that same day.)

I'd suggest that a city-killer impact with little or no warning is still quite possible. Something capable of taking out, say, a small country we'd probably have some warning of, but probably not enough. A mass-extinction event we'd probably have lots of warning for, but doing anything about it would be a very big challenge.


Why does the cloud of Orion Nebula appear more distinct when I am not directly looking at it? I have a 6cm aperture newtonian refractor telescope through which I saw the Orion Nebula yesterday night. As all you can hope with a telescope of such a small aperture, I saw some faint cloudy structure, but when I looked away from the cloud through the telescope it appeared to be more distinct and dense. Why is it so?

This is called the Averted vision. It involves not looking directly at the object, but looking a little off to the side.

How does it works?

Because the retina is less sensitive at the center, and this is very remarkable on the very faint objects (like nebulas or galaxies)

This is a common technique used by astronomers.

More info here

How to make the most of Averted vision?

To maximise the effects of your Averted Vision, you first need to know which eye is your dominant eye. To do so:

Point with your finger a distant object
Then blink an eye, then the other eye.
The eye in which your finger is perfectly aligned with the object is your dominant eye.

Then, for the best Averted vision:

Look at the top-right of the object if your dominant eye is the right one
Look at the top-left of the object if your dominant eye is the left one


Were magnetic monopoles created before, during or after inflation? In the case that magnetic monopoles exist, when were they created? I thought that all particles were created after inflation, but I've read some comments recently that hint at a pre-inflationary creation of magnetic monopoles.

Before. The point is that there were too many predicted, and no concept of inflation yet. Here is a reference that doesn't just gloss over it.

Here is the original paper by Guth, and physics.SE has a nice drawing of the same timeline shown on that page. (Original here)

You ought to try that physics.SE thread and continue there for more details.


What would happen to the Galilean moons and Titan if Jupiter and Saturn disappeared? The Moon reportedly doesn't "need" the Earth to revolve around the Sun. If the Earth wasn't there, the Moon would continue its current path from a heliocentric reference frame around the Sun. My question is whether that would happen to some of the moons of Jupiter and Saturn too. If Jupiter or Saturn weren't there, would Ganymede or Titan continue to freely revolve around the Sun, or what would happen to their orbits?

Thank you.

They might escape from the solar system, if the angles are right. If not, they'll probably wind up in elliptical orbits around the Sun. We'll use a simplifying assumption that the orbits are circular to make the calculations easier; All objects mentioned have orbital eccentricity of less than 0.05.

The mean orbital velocity of Jupiter is 13.1 km/s, which makes Solar escape velocity at Jupiter's mean distance from the sun 18.5 km/s, a difference of 5.4 km/s.

The mean orbital velocities of the Galilean Moons of Jupiter are:

Io: 17.3 km/s
Europa: 14.3 km/s
Ganymede: 10.9km/s
Callisto: 8.2 km/s.

Which means, if the geometry is right, with the direction of travel of the moon in the same direction of the direction of travel of Jupiter at the time of the disappearance, the resulting vectors could add up to any of the Galilean moons exceeding solar escape velocity if Jupiter suddenly vanishes, and exiting the solar system.

Doing some trigonometry to find the limiting angle:

Io: Reaches Solar escape velocity  if the moon was traveling within 106° of Jupiter's direction.
Europa: Escapes if traveling within 95°
Ganymede: Escapes if direction within 79°
Callisto: Escapes if direction within 62°

There's also a tiny, tiny chance of Europa or Io being in the right positions and having the right direction to hit the Sun if their directions at the time of disappearance were exactly right, but the chance of that is extremely unlikely.

For comparison, Saturn has a mean velocity of 9.5 km/s, (Solar Escape Velocity at that distance is about 13.4 km/s) and Titan has a mean velocity relative to Saturn of about 5.5 km/s.

Titan could escape under the right conditions (Within 56° of Saturn's direction of travel), but doesn't have enough Saturn-relative velocity to have a chance of crashing into the Sun without help.


The deduction of the Hill Sphere formula This question may be a little lazy, but can anybody give me a proof of the Hill sphere formula? Acording to wikipedia, the formula for the radius, $r$, is

$$r\approx a(1-e)\left(\frac{m}{3M}\right)^{1/3}$$

where a body of mass $m$ is orbiting a much more massive body of mass $M$ with a semi-major axis $a$ and eccentricty $e$.

Hill sphere is named after John William Hill (1812–1879) and its simple logic follows from the presence of three bodies (let's assume Sun is the largest mass with Earth as the secondary mass and a satellite of negligible mass orbiting the Earth as the third mass), where the radius of the Hill sphere will be the largest radius at which a satellite could orbit the secondary mass (Earth in this case). If its orbit exceeds the Hills radius, then it will fall to the gravitational influence of the first body (sun) and hence will no longer be a satellite of the secondary body.

One could write Newton's equations using the idea that the satellite has the same angular velocity as the secondary object. This is that, the angular velocity of the Earth around the sun equals to the angular velocity of the satellite around the sun. A demonstration about the derivation is given in the following link as well as that of the Roche limit:

http://www.jgiesen.de/astro/stars/roche.htm


How many black holes exist? How many black holes exist and what is their distrubtion over mass?
There is a list of black holes, but I'm interested in an estimate of all black holes in the visible universe.

There exists probably about one supermassive black hole per galaxy, hence about 100 billion ($10^{11}$) in the observable universe.

If we take 1% of the stars ending as a black hole as a very rough order of magnitude estimate, we'll get about 1 billion stellar black holes per galaxy, hence about $10^{20}$ stellar black holes in the observable universe, give or take one order of magnitude.

Stars of more than about 15 stellar masses (O class stars) will probably end up as black holes. About 0.00003% of the stars are O-class. The lifespan of a star with 15 solar masses is about 11.5 million years. So, we need to multiply the 0.00003% at least by a factor 1000, since the age of the universe is more than 1000-times the life of an O-class star, resulting in 0.03% of all stars over time should have been O-class stars.
(Most stars are red dwarfs, and have a longer life-span than the current age of the universe.) Star formation, especially of heavy stars, in the early universe has probably been higher than today.
Therefore I'm estimating the number of stellar black holes as about 1% the number of all stars.

Intermediate-mass black holes as presumed centers of globular clusters are harder to estimate, since there isn't much observational evidence. If the hypotheses turns out to be correct, and we take 1000 globular clusters per galaxy, we get about $10^{14}$ IMBHs.

There is no observational evidence for the existence of small black holes (smaller than a few solar masses).


Shouldn't tidal locking be impossible for a satellite that has a considerably eccentric orbit? This popped up into my mind just now. The Moon is tidally locked to Earth, and also has a significantly eccentric orbit. This means that its orbital velocity near periapsis is considerably faster than its orbital velocity at apoapsis.
Thus, for the Moon to always show one side to Earth, its rotation around its own axis has to slow down and speed up depending on where it is in its orbit, but doesn't that violate the law of conservation of angular momentum? And if its rotation around its axis was constant then it would get out of sync and not be tidally locked, no?

You are right, that would be weird if the Moon speeds up and slows down this way to always show the exact same side to the Earth. That's why it doesn't.
At some point in the orbit the Moon's rotation (its phase) lags behind and at some point it is too fast. It is (pretty much) constant and does not adjust to the varying orbital velocity. That's (one reason) why it wobbles. And this wobble is what's called libration.
But the Moon is still tidally locked as its rotation period is still equal to its orbital period.


How do we know how the Sun (or any star, for that matter) will evolve? The evolutionary path of the Sun has been described in some detail, and, subtle differences aside, it's been described as such for decades - the Earth-engulfing red giant stage, the helium flash, etc.. It would seem that the knowledge necessary to make such a prediction has been around since the 60s, if not earlier.

How do we know it will happen, and on the timescales described? If we make some measurements/estimations of the mass and composition of the Sun, and use our understanding of nuclear physics - energy of fusion and so forth, and an understanding of Newtonian physics (or do we have to invoke Einstein?), do the predictions just inevitably fall out?

Could the predictions be wrong? Is there any significant measure of uncertainty, perhaps due to sensitivity of certain calculations to initial conditions, etc?

Is the red giant future of our Sun a firm conclusion, or just a likely possibility among a range of others and commonly talked about merely because it seems the most likely one?

The thing you are not taking into account is that our predictions for the future evolution of the Sun are not based on our understanding and observations of the Sun alone. They're based on our understanding and observations of all the stars we've ever looked at. The thing is, stars of the same mass and composition must necessarily evolve in the same way. So we can look out at other stars and find some which are more or less the same as our own, but might be farther along the evolutionary track. We can study these stars and see what is happening to them. By conglomerating a lot of data about many stars at many different points in their evolutionary life cycle, we can put a coherent picture together for our own Sun.

Scientists have spent a lot of effort to create stellar evolutionary models. To do this, they put together the multitudinous information gleaned from the tens of millions of stars we've observed, add in a lot of math and physics representing the physical processes occurring in these stars and what you find is that the "predictions just inevitably fall out". There are many stellar evolutionary code bases out there you can find. For example, Pols et al. 1998 applied stellar evolutionary code to clusters of stars. A figure pulled from their paper is shown below. It shows a Color-Magnitude diagram of the M67 open cluster of stars, along with predictions of the evolutionary track of those stars. There's of course some variance as every star is not exactly the same, but overall, you can see that the stars follow the predicted evolutionary track pretty well.

Once you have a well calibrated stellar evolutionary program, it's not too difficult to apply that program to our Sun. There will be some unknowns and some variance, e.g., we might not know the physical extent of the red giant phases' expansion, but we know it will happen.


Would dark energy save the earth for a while as the sun heats up? Would the expansion of space (e.g. dark energy) move the earth further away from the sun over time, possibly saving us for perhaps a few billion years as the sun heats up?

Scientists now tend to think that the sun is heating up very gradually (e.g. over billions of years):
http://tvblogs.nationalgeographic.com/2010/07/21/hotter-and-hotter/

Dark energy is the unexplained accelerating expansion of the universe, as described here:
http://en.wikipedia.org/wiki/Dark_energy

Put another way, my question is, is dark energy working at such a rate that it would move earth (and other planets) further from the sun and prolong life on earth as the sun heats up?

I don't think this is a bad question, and I think Pela is essentially correct, the effect is very small.

Dark energy doesn't expand objects, for example, cause those are bound together, but dark energy should expand space between the sun and the earth though the effect is tiny.

If we take this number, 68 KM per second per 1 million light years (link below), that's 245,000 KM per hour per million light years or about 1 KM every 4 hours per lightyear.   With 63,000 Astronomical units per light year, between the sun and the earth, we can expect the earth (assuming it's in a perfectly stable orbit around the sun - which it's not, but lets say it is), a velocity of 1KM/30 years.   And that's over the age of the universe.   The earth/sun is only 1/3rd the age of the universe, so dark energy has accelerated the earth maybe 1 KM/90 years velocity since the formation of the solar system.   That's a pretty small push.  A velocity change of a bit a meter per year over the age of the solar-system.   The effect of the solar wind and the Jupiter and Venus orbital tugging are probably greater than that.

Source:  http://www.space.com/26279-universe-expansion-measurement-quasars-boss.html

Now, over a few billion years, a small velocity change of 1 meter per year would add up - so I think dark energy is part of the solar system's long term orbital calculation.  I might be wrong on that, but I think it has a measurable effect over billions of years even in distances as small as planetary orbits.

But would it save the earth - I doubt it.  The sun's heating up accelerates over time.   It will at some point in the future, move too fast and get too hot for the earth, so, we'll have to save ourselves, either by moving the earth or terraforming mars, or building space colonies or some combination of all 3.   The good news, we have tens of millions, maybe hundreds of millions of years to figure that stuff out.

Dark energy has been useful though.   It's probably kept us from crashing into Andromeda and/or the great attractor - however far away that is.   Certainly delayed those collisions (We're never going to crash into the great attractor, I gather, thanks to dark energy), and crashing into Andromeda, when it happens, might harmless anyway, but I'm just kind of thinking out-loud.

I do think it's a good question as there probably is some small expansion due to dark energy between the earth and the sun.   If my numbers are wrong, I welcome correction.


Would an object placed at L1 be seen in front of the Sun from Earth? Does anything placed at the Lagrangian point L1 between the Earth and Sun obscure the Sun like a transit of the innermore planets? The James Webb telescope is to be placed at L2, but if it was placed at L1, would it be in front of the Sun when looked upon from Earth? When you looked from the equator into the Sun, the telescope would be in the very center of the solar disc, right?

Objects are not place at the Sun-Earth L1 or L2 Lagrange points. They are instead placed in pseudo orbits about these points. These pseudo orbits intentionally avoid being directly in line with the Earth and the Sun for two key reasons. One reason is that these points are directly in the line between the Sun and the Earth. An object at the Sun-Earth L1 would be hard to see, while an object at the Sun-Earth L2 point would be in the Earth's umbral shadow.
A much more important reason is the incredible amount of propellant that would be needed to maintain their position at those unstable Lagrange points. The pseudo orbits, while also unstable, are much less unstable and require considerably less propellant for stationkeeping.


How to differentiate elements that have same spectral lines in a star? Suppose that iron and calcium have a spectral line with the same wavelength. How would you determine which element is present in the atmosphere of a star if you found this line in its spectrum?

If you only had that single line, you would be unable to differentiate them.  This is especially true of redshifted objects where we couldn't tell the actual wavelength of an isolated peak.  But a good spectrograph will give you information over a range of wavelengths.

Each element has multiple lines.  It is the overall pattern of relative strengths that is definitive.

As an example, the stronger lines for a few elements are shown below:

(From https://cnx.org/contents/LnN76Opl@10.1:H5KhIDcK@3/Spectroscopy-in-Astronomy)

You can see there's a cyan line that is similar in both hydrogen and calcium.  But the other lines are so different, you could tell if one or both were present, even if they were overlaid on a single chart.


Do we see stars from other galaxies (not Milky way) in the night sky with the naked eye? I saw reports that only Andromeda Galaxy and Big/Small Magellanic Cloud galaxy stars could be visible with the naked eye (how many stars would be from them out of a few thousand visible?)

This question appeared because there are some very "sparse" galaxies (very few stars in them) and scientists say night sky in them is practically all black (starless)...

So is it true that 99,9% visible stars in the night sky (naked eye) are from the Milky Way only?

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

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

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

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


Size of the Universe before the Big Bang I can't understand how all the matter contained in the Universe, was contained in an area the size of a grapefruit before the Big Bang ? How could so much matter have been compressed in such a tiny space ?

Nobody knows what the big bang was, but using the laws of physics as we understand them, and using what has been observed studying quantum mechanics and very high energy collisions, a lot of what's defined as moments after the big bang are basically working backwards and deduction.  Quantum physics models explain very accurately why we have the hydrogen/helium ratio that we see and why the cosmic background radiation formed.

The simple answer to how it was possible to get all that stuff into such a small space is two fold.   One, when it's that dense, it's no longer matter, but energy, and two, it was that small for an incredibly short period of time - too short for gravity to win, which, when you have that much mass in that small a space, you'd think gravity would win, but the primordial grapefruit expanded faster than the speed of light and gravity is limited by the speed of light, and perhaps limited in other ways by quantum gravity, but we don't know how to study that yet.

So you can have that a universe amount of mass in a grapefruit or watermelon size if it's that small for a very very short time, close to impossibly hot and faster than light expanding.

Somewhat different explanations could be that the Universe is infinite, or curved in on itself, in which case the gravity, if the young universe is uniform, balances out to zero everywhere until the uniformity breaks.   This is similar to the gravity at the center of the Earth.  With equal mass in every direction, gravity becomes zero.

The cosmic background radiation suggests that the young universe was very close to uniform and that, plus rapid expansion circumvents the "why didn't it become a black hole" problem.   Various shapes of the universe don't change the rapid expansion model, which is the most popular model today.

There are also models that involve additional dimensions which explain the big bang, but I think those have fallen out of favor, perhaps because extra dimensions have never been observed at CERN and because I've not read about that model recently, though if memory serves, Stephen Hawking put the 5 dimensional big bang idea in one of his books that he marketed to the general public.    Extra dimensions make all kinds of things possible, like putting all that mass in a small space.


What else could the Event Horizon Telescope Observe? The Event Horizon Telescope was made possible in order to observe the details of supermassive black holes. This took a huge amount of work installing extra telescopes and developing the hardware and software needed to do VLBI at such short wavelengths. They have achieved spatial resolution around 25 micro arc seconds, comparable to the astrometric data from the Gaia space probe, but optimised for very different purposes.

So my question is what else could they usefully observe? What scientific targets have interesting detail at that angular scale and emit enough radiation at that wavelength to be observable?

How about Betelgeuse?

Betelgeuse is about 640 light years away compared to 54 million light years for M87.  It has an angular diameter of 0.042 to 0.056 arc seconds while the quoted resolution of the EHT was 0.000025 arc seconds so you'd expect some detail on its surface.

Betelgeuse seems to be undergoing a number of rapid changes right now.  It is a young but very massive star and is racing through its evolution.  It is only ten million years old but is expected to explode as a type II supernova within the next million years.

The best image of Betelgeuse is this from ALMA

I guess the problem with this question is that there are many, many researchers and teams who would like EHT time to investigate their stuff, but those individual telescopes are already committed to other projects.  The EHT people managed to pull this off because of the sensational nature of the target, a black hole!  No other subject in astronomy would get so much attention, so many institutes working together, and bring in the funds and time


How many observations does it take to determine the orbit of a TNO Several observations of a distant solar object must be taken before determining its orbit, but in the case of objects discovered beyond Neptune's orbit, just how many are required over what period of time (or what portion of arc) before a reasonably accurate orbit can be determined, sufficient, say, to be added to the Minor Planet Database or a JPL ephemeris?

I'd be interested to know to what degree of certainty orbital elements for objects beyond 30 AU can be determined when we have only observed them for (so far) short portions of their orbits. What about 60 AU, or 90 AU?

As an example, the Wiki for VNH004, a TNO that is a photo target for the New Horizons probe, states:

  The asteroid only was observed 12 times by the Mauna Kea and Las Campanas Observatory over a period of about 33.8 days between May 29th and July 2nd, 2011, so its current orbit is extremely uncertain.

How uncertain? Here are orbital elements along with +/- values:

  Heliocentric elements and errors
  Epoch . . . . . .  2455710.5000  =  2011/05/29
  Mean Anomaly: . . .       309.75557 +/-    31.234
  Argument of Peri: .     112.06593 +/-    54.010
  Long of Asc Node:  .    247.74234 +/-     3.336
  Inclination: .  . . . . . . . .   3.81106 +/-     0.396
  Eccentricity: . . . . .     0.29886369 +/-    0.8050
  Semi-Major Axis: .    37.47217640 +/-   14.2156
  Time of Perihelion: 2467404.0720 +/-    2926.4
  Perihelion: . . . . . . .   26.27310354 +/-   31.7695
  Aphelion: . . . . . . . .    48.67124926 +/-   35.3678
  Period (y) . . . . . . . . . . 229.3885 +/-    130.53

An orbital period of over 229 years, plus or minus over 130 years. How many more observations would be required to half the margin of error?

The main problem with determining an object's orbit is we only know the position with certainty in two dimensions.  The distance to the object is largely unknown.  This accounts for the large uncertainty in the period of newly discovered TNOs.  Many possible orbits could fit the early observations, and therefore the uncertainty is large.

As time goes on, and more observations are made, more and more of these possible orbits are excluded and the orbit fit becomes more exact.

So to answer your question, it really depends on how exact you want the answer, and how good your observations are.  The more observations you take, and the better they are, the closer your orbital fit becomes.  And more importantly, the bigger the arc of the orbit your observations cover, the more accurate the orbit becomes. It is impossible to give a specific number.

To give a practical example, Pluto takes 248 years to orbit the Sun.  It was discovered in 1930, so we've only observed it for 85 years, or about 1/3rd of one orbit.  One of the problems the New Horizons probe sent to Pluto had to overcome was that we didn't know, at launch, exactly how far away Pluto was, so therefore we didn't know exactly when New Horizons would pass the dwarf planet. That made it difficult to tell the probe where to point to take pictures at closest approach.


Are planets moving away from the sun? I saw on the tv show Through the Wormhole (hosted by Morgan Freeman) that the planets in our solar system have been continuously moving away from the sun for millions of years. however, when I try to Google about it, there isn't much written about it.

Is it a proven fact that the planets are moving away from the sun?

I asked a somewhat similar question but just about the Earth.  Here and in my question there's some links that you might be interested in.   Jupiter for example is thought to have moved closer to the sun during the late heavy bombardment, then back outwards.   So I suspect that through the wormhole didn't get that point quite right.   Observations by the Kepler Spacecraft in the hunt for exoplanets suggests that a number of planets in other solar systems have moved closer to their suns over time.

In general, there's a few parts to this question.

On the sun losing mass:

Yes, it is, but only a little bit.  See Here.  Their math isn't complete as they don't factor in coronal mass ejections or take into account the sun's expected increase in luminosity, but the gist of their point remains true.   The sun is only losing a tiny percentage of it's mass.  The effect on planetary orbits is pretty small.

Sunlight also creates a pressure against the planets, but this effect, isn't very much due to the comparatively enormous mass of the planets compared to the light and coronal mass ejection material that hits the planet.

Tides

Tidal effects can work either way depending on the rate of rotation compared to the planet's orbital rotation.  If the star turns ahead of the planet, the tidal force should push the planet slowly outwards, like the Earth is currently doing to the Moon.   If the planet moves around the star faster than the star turns, the tidal effect is reversed and it slowly pulls the planet in.  At our Sun's current rate of rotation it's moving ahead of all the planets, so there should be a small tidal force that pushes the planets outwards, but very very slowly.

Orbital debris and effective loss of angular momentum during planet formation.

If the orbit is littered with dust and comets, asteroids and such, the sun can gain mass faster than it loses it and the planets can effectively slow down as they move through the dust or orbital debris with more oblique orbits.   This may be the explanation for hot Jupiters and so many observed planets very close to their suns in other solar systems.

Planets can affect each other.

See here and a curious possibility here.   Planet to planet effects are tiny unless they are in resonance where the effect can be amplified and become significant over time.

Hope that's clear enough.  I can tidy up and provide additional links for reference if needed.


What safety concerns should an amateur astronomer take? I'm an amateur trying to take a shot at recreational stargazing. I've read about how looking a some eclipses without the appropriate instruments (i.e, with naked eyes) is bad for the eyes. To what extent is this true? Is staring at the phenomena bad or is even a quick glance dangerous?
Are there any more phenomena like eclipses I have to watch out for to protect myself?

The only thing you need to protect your eyes from is the Sun. A lunar eclipse, no matter what phase it is in, is not dangerous to look at. A partial or an annular solar eclipse are dangerous because the Sun is still visible. A total solar eclipse is perfectly safe to look at—however, as the phenomenon starts and ends with a partial eclipse (as the Moon slowly blocks the view, then slowly moves away), those parts of the (otherwise) total solar eclipse are dangerous to look at directly.
DO NOT LOOK DIRECTLY at the Sun with the naked eye, even though some people have done it for VERY SHORT times (seconds, at most) without permanent damage. Again, DON’T DO THIS! The problem is the eye doesn’t have pain receptors, so you wouldn’t know when you had been looking at the Sun for long enough to be hurting your eyes. This can range from a small “blind spot” in your field of view, that goes away after a few days or weeks, to complete, permanent blindness.
There are safe ways to view the Sun through an optical instrument (which, because it collects more light than the eye, would be even MORE dangerous to look through without protection!), and some of them are really affordable financially. The best, however, is a solar filter that goes in front of the objective, NOT at the eyepiece (as the temperature there is so high, it could melt and make you permanently blind INSTANTLY)! Such filters come in many different qualities and many different prices, but are generally rather affordable as well. They are made by applying a special metallic coating to an optical-grade piece of glass or even plastic (Mylar).
DO NOT LOOK AT THE SUN THROUGH AN EXPOSED PHOTOGRAPHIC EMULSION (yes, those still exist!) and DO NOT LOOK AT THE SUN THROUGH A POLARIZED FILTER (unless you’re also using a safe filter as described above).


Does all of time exist and if so where is it? I've recently learned about the theory of relativity and time.  I'm trying to understand it more.

My (basic) understanding is that there is no distinction between past present and future and that all of time already exists.

If this is true, what prevents us from seeing the past and the future? And/Or what causes us to only be able to see the "present"?

Your question is difficult to answer exactly, because it touches the question of consciousness, which is problematic in general. But the answer is basically this:

Theory of relativity describes the space-time as four-dimensional space that exists as a whole. You are described by a world-line, which contains all states you go through from birth to death. There is many slices, which describe "you" in different times. What you can see is given by what information reached your sensory inputs (your senses) and how they are processed by your brain. The brain creates a model of reality based on your sensory inputs, which is stored in space-like slice of your world-line in the space-time, which we call present. This is what "you" can see. So the reason basically is in how the brain processes the data it is getting.

Along you world-line, there is many versions of you at different ages and each version feels its own present and the versions more in the future also have memories of the past (because the brain stores the past sensory inputs). There is no single "present time" in the theory of relativity, only the space-time as a whole.

But you should't forget that the theory of relativity is still only a very well working model. Even within the theory of relativity, the question whether space-time exists as a whole, or whether there is "present state", which evolves in time, is rather philosophical than scientific, because there is no observable difference between the two.


Why is the letter *z* customarily used to denote redshift? What events led to the nearly universal acceptance of the letter "z" as the denotation of redshift? What did the letter originally stand for?

It may be that someone will come along with a better answer, but the first example I can find for the use of "z" explicitly for redshift is is William de Sitter's 1934 paper.

I'm tempted to think this is one of the earliest uses if not the first use, certainly looking at de Sitter's 1933 book where he touches on this briefly, he does not use the letter "z". However he did use "z" in a 1930 paper to represent the ratio between the radius of the Universe and initial radius, which in this context is actually the same quantity as 1 + z and so why he would've used "z" in his 1934 paper.

If I had to hazard a guess, there is no particular reason for the choice of "z" for redshift, it was just a convenient letter to use for the variable.


How Does Black Hole Evaporation Look From the Inside? Imagine you're inside a black hole event horizon. The curvature of spacetime is such that the direction towards the singularity becomes timelike, as reverse movement becomes literally impossible.

How would Hawking radiative evaporation look like from inside the black hole? From the outside perspective, we know that black holes form (from our time dimension perspective) and then evaporate at a rate that is an inverse function of their mass.

Hawking radiation is created when the vacuum state at the horizon looks as if it is transformed into a non-vacuum state for a distant observer, resulting in a negative energy 'input' to the black hole. If the energy loss this way is not compensated by inputs from outside, the black hole event horizon should gradually shrink, resulting in eventual evaporation. How if at all would that be perceptible by an observer inside the black hole horizon?

Similar question, but no accepted answer.

An infalling observer cannot remain stationary inside the event horizon of a black hole.

According to them, there is a short period of proper time before they meet whatever lies at the centre of the black hole (a singularity in GR).

If the observer looks back, they do not see an infinite amount of time pass by in the outside universe. In fact for the type of black hole where they might (a) survive and (b) have time to make any observations (i.e. a supermassive black hole) then they would see the outside universe only for a relatively short time longer than their experienced proper time and certainly for nowhere near as long as the Hawking radiation evaporation timescale. So the possibility that the black hole is (slowly) evaporating is of no consequence to their final moments.

The necessary material and diagrams to understand the previous paragraph have been covered before on both Physics and Astronomy SE.

https://physics.stackexchange.com/questions/82678/does-someone-falling-into-a-black-hole-see-the-end-of-the-universe

Would time go by infinitely fast when crossing the event horizon of a black hole?


How small can a spiral galaxy be? The smallest observed spiral galaxies I can find are NGC 2976 and NGC 4605, both with a diameter of 20 kly, but I don't have anything like an exhaustive source to search. I've also found a mention that the Small Magellanic Cloud (7 kly diameter) might have had more spiral structure before it was distorted by the Milky Way.
From looking at these galaxies, spiral structure seems to be less clear the smaller the galaxy is, which makes me think there might not be a clear lower limit. Still, I'm curious how small a galaxy with recognizable spiral structure could be, either based on observations or theory.

The first Google-hits suggest that Segue 2 is the smallest spiral galaxy, but that is actually incorrect since it is the smallest spherical galaxy yet discovered. It has a radius of roughly 34 parsec or 110 light years. The original publication on that topic dates back to 2013, but it was again in the news in April 2020.
Then, there are lecture notes on a course Astronomy 626 Galaxies which shows the following diagram:

Also, the lecture notes claim:

Typical values for the scale length [of spiral glaxies] are: $1 {\rm kpc} < h_R < 10 {\rm kpc}$

I agree that this does not answer your question fully, but I got the impression that the lower bound should be around $1 {\rm kpc}$.
References

https://www.nasa.gov/press/2014/september/hubble-helps-find-smallest-known-galaxy-containing-a-supermassive-black-hole
https://www.treehugger.com/just-how-big-is-the-smallest-galaxy-in-the-universe-4862842
http://www.sci-news.com/astronomy/article01147-segue-2-dwarf-galaxy.html
Evan N. Kirby et. al: Segue 2: The least massive galaxy
https://physics.stackexchange.com/questions/148418/