How do you determine which star, between two, is more luminous? For example, if star 1 is 4x bigger than star 2 but both are the exact same temperature how do you determine which is more luminous?

One way to determine the luminosity of a star is to first determine its distance, d.  Then one can use the formula, Luminosity = Flux * 4 * Pi * d$^2$.  Flux is related to the apparent brightness that is measured with a telescope (and CCD or photometer).  mag = -2.5 * Log(Flux) + c.

If the star is relatively nearby, one can use parallax to get the distance.  If the star is in a binary system, one can solve for the distance from the period, orbital velocities and apparent size of the orbit.  If you know the velocities throughout one orbit, then one knows how far it has physically traveled.  Comparing this with the apparent size gives an approximate distance.  However, since one usually does not know the inclination, this causes uncertainty.  If it is an eclipsing binary, then you know that it is not very inclined and this works well.  On the other hand, if it is eclipsing, one can usually determine the radii of the binary pair and then get distances from Stefan Boltzmann Law.

Another method for estimating luminosity is to locate the star in the HR Diagram (Luminosity vs Color).   The color (difference between the magnitudes in two different filters) locates it horizontally, and stellar class locates it vertically in the diagram.  One needs a spectrum of the star to determine the class.

And we have one more trick up our sleeves.  If the star is a periodic variable such as a Cepheid or an RR Lyra, then there are relations between period of variability and luminosity.


Is spacetime hyperbolic? I have always seen answers to this point to measurements of how "flat" local geometry is, but my problem is, thats space not spacetime. I would say its safe to say that the geometry of space in our universe is indeed euclidean, but space isn't over time. The further I look, space should be less, the farther I go, space should be more. To me this seems as if spacetime is hyperbolic along the temporal axis, and that every "instance" cross-section is euclidean. So that ultimately is the question, is spacetime hyperbolic?

There are three kinds of different hyperbolic in this question and the previous answer by AtmosphericPrisonEscape: the behavior of Lorentz transformations, the spatial hypersurfaces in cosmology (which seem to be flat), and the hyperbolic appearance of spacetime when tracing world lines in a universe with accelerated expansion.

TL;DR: Spacetime is expanding exponentially and if you choose an unconventional hypersurface, that hypersurface is hyperbolic.

Lorentz transformations: Lorentz covariance is the mathematical expression of Einstein's principle of the constancy of the speed of light. Changing the frame of reference of an observer is done by a Lorentz transformation which preserves the light cone and is closely related to the concept of causality.  The answer to this question explains this for the Minkowski space, where a Lorentz transform is a hyperbolic rotation.

In this flat (Minkowskian) spacetime:

$$ ds^2 = c\, dt^2 - dx^2 $$

you can use a Lorentz transform to switch to different coordinates $dt$ and $dx$ while the line element $ds$ is invariant, analogous to the hyperbola

$$ 1 = a^2 - b^2. $$

This kind of hyperbolic is not the point of the question.

Cosmology: If we assume large scale homogeneity and isotropy (Copernican principle), we can choose isochronous coordinates.  There you have a synchronous time $t$ and spatial hypersurfaces for given $t$ ("time slices").  Freely falling observers are comoving, i.e. have fixed spatial coordinates $\mathbf x$.  Their world lines are perpendicular to the spatial hypersurfaces and their proper time matches the synchronous time $\tau = t$.  Galaxy clusters are such freely falling observers.

If you take these assumptions and allow time dependence of spatial coordinates on time, you end up with the FLRW metric:

$$ ds^2 = c\, dt^2 - a(t)^2 d\Sigma^2$$

where $d\Sigma^2$ describes the spatial hypersurface.  Now additionally to this being a hyperbolic equation, the hypersurfaces $d\Sigma^2$ can have curvature, especially constant positive (closed sphere), zero (Euclidean), or constant negative (open hyperboloid) curvature.

A flat universe is, for example, given by

$$ d\Sigma^2 = dr^2 + r^2 d\Omega,\\
   \text{where}\quad d\Omega = d\theta^2 + \sin(\theta)^2\, d\phi^2, $$

which is the metric of Euclidean 3-space in spherical coordinates.  Here the only relative acceleration between freely falling observers comes from the scale factor $a(t)$.

A hyperbolic spacetime could be

$$ d\Sigma^2 = dr^2 + \sinh(r)^2 d\Omega $$

which is a hyperbolic plane in spherical coordinates.  In such a spacetime, freely falling observers do not only measure a relative acceleration due to the scale factor, but an additional contribution due to the curvature of space.

Expansion: In the FLRW metric you see the scale factor $a(t)$.  The Friedmann equations describe the evolution of the scale factor.  It does not influence the qualitative shape of the spatial hypersurfaces, but in an expanding space their curvature becomes more and more flat (just like the surface of a sphere locally becomes more and more flat as the sphere expands).

A flat $d\Sigma$ and a constant $a$ gives you no acceleration between freely falling observers whatsoever.  An $a$ with constant change manifests as all observers flying apart, but no relative acceleration.  Finally, an $a$ with constant acceleration makes everything fly apart ever faster (currently observed).  The Friedmann equations tell us, that the behaviour of $a$ changed during the lifetime of the universe as different effects where dominating (radiation, matter, cosmological constant).

Currently we observe an accelerating universe, so $a$ is exponentially increasing (just at the right end of the illustration).  If you look at the world line of two Galaxies, their physical distance increases exponentially.  Therefore the circumference of a chunk of space increases exponentially, so the hypersurface spanned by a line of freely falling observers is actually hyperbolic (white grid in the illustration).  However, I struggle to put that into an equation that clearly shows the hyperbolicity, so don't trust that too much, yet.


Why aren't corrector plates aligned with the center of curvature in an SCT? According to wikipedia, Schmidt corrector plates must be aligned with the center of curvature to eliminate the spherical aberration. However, in compact SCT designs by Celestron and Meade, that is clearly not the case since the optical tube is usually shorter than the radius of curvature of the primary mirror. My question here is WHY?

In a schmidt camera, the corrector is placed at the centre of curvature, which is at a distance of twice the focal length from the primary mirror, and minimizes aberrations such as coma and astigmatism.  This makes for relatively long telescopes, particularly with large primary mirrors.

The sct design is similar, in that it uses a corrector plate, but this (along with the secondary mirror) is placed slightly closer to the primary mirror than its focal length.  This design allows much more compact telescopes, but at the expense of more aberrations.  As with most if not all optical designs, it is a compromise.


How is it possible that Saturn's gravitational acceleration felt by Mimas is stronger than Mimas' own surface gravity? The surface gravity on Mimas is $≈ 0.063\text{ m}/\text{s}^2$ and Saturn's gravitational acceleration at the distance of Mimas' orbit is:
$$\frac{{GM}}{{r}^2} = \frac{{6.674 \times 10^{-11} \times 568.34 \times 10^{24}}}{{(185.52 \times 10^{6})}^2} ≈  1.102 \text{ m}/\text{s}^2$$
How can this be? An object on Mimas' surface would be much more attracted to Saturn than it is to Mimas. Shouldn't Mimas itself get ripped apart or is my math wrong?

An object on Mimas' surface would be much more attracted to Saturn than it is to Mimas.

You are missing that Mimas as a whole accelerates gravitationally toward Saturn. What this means is that a point on the surface of the Mimas will feel the acceleration at that point toward Saturn minus the acceleration of Mimas as a whole toward Saturn. This is the tidal acceleration. It is equal to
$$a_\text{tidal} = \left|\frac{GM}{(R\pm r)^2}-\frac{GM}{r^2}\right| \approx 2 \frac{GMr}{R^3} = 2\frac{GM}{R^2}\frac{r}{R}$$
where $R$ is Mimas' semi major axis length and $r$ is Mimas' mean radius. The approximation assumes that $r\ll R$, which most certainly is the case given that Mimas's radius is about 1/1000 of the semi-major axis length of its orbit about Saturn. The result is rather small, about 0.002355 m/s2.


Trappist-1 planet albedo Is the albedo of any of the Trappist-1 planets know? Is it high enough so if you are on the surface of one of the planets, you can see the others?

I looked up the distances from the star and size of the planets and it seems a few of them will appear larger than the Moon in the night sky, unless the albedo is really low that is.

So far there have not been any published measurements of the albedos of the TRAPPIST-1 planets. While albedos have been measured for some exoplanets, the planets in question are irradiated far more severely than even the innermost TRAPPIST-1 planets.
Using the formulae in my answer to this question about computing planetary apparent magnitudes, we can make a back-of-the-envelope check. The definition of absolute magnitude becomes the apparent magnitude of a planet as the magnitude it would have at opposition if it were located 1 au from TRAPPIST-1, and the observer located 1 au from the planet.
Defining the absolute magnitude of the TRAPPIST-1 planets as the apparent magnitude if the planet were 1 AU from TRAPPIST-1, and the observer were 1 AU from the planet, the uniform disc model gives the following formula for the absolute magnitude $H$:
$$H = 5 \log_{10} \left( \frac{671\,000\ \rm km}{D_{\rm p} \sqrt{A_{\rm g}}} \right)$$
Where $D_{\rm p}$ is the diameter of the planet and $A_{\rm g}$ is the geometric albedo. The 834,000 km value is the equivalent of the 1,329 km value used for asteroid diameter estimation in our Solar System, adjusted for the fainter absolute magnitude of TRAPPIST-1 (using the V-magnitude 18.798 and parallax 80.4512 mas from SIMBAD, I calculate an absolute magnitude of 18.33).
The least favourable case for visibility of an outer planet from an inner one would be for planet h viewed from planet b: TRAPPIST-1h is both the smallest and outermost of the TRAPPIST-1 planets. To pick an unfavourable case for visibility, let's use a geometric albedo of 0.05 similar to the P-type asteroid 46 Hestia. Using a diameter of 0.752 times the Earth (NASA Exoplanet Archive), the absolute magnitude works out as roughly 12.5. The apparent magnitude is then given by
$$m_{\rm p} = H + 5 \log_{10} \left( \frac{d_{\rm p\ast} d_{\rm po}}{1 \mathrm{\ au}^2} \right) - 2.5 \log_{10} q(\alpha)$$
where $d_{\rm p\ast}$ is the star–planet distance, and $d_{\rm po}$ is the planet–observer distance, and $q(\alpha)$ is the phase integral at phase angle $\alpha$.
Again using the uniform disc model which gives $q(0) = 1$, the apparent magnitude of h from b, when b and h are located on the same side of TRAPPIST-1 is about -0.2 for the Hestia-like case, which is similar to Saturn viewed from Earth. On the other hand, the angular diameter in this configuration is about 3.7′, so unlike Saturn it would be visible as a disc rather than a point.


How gravity behaves for an Earth body and for a celestial body? Is there any difference in the gravity of the Earth for an Earth body or a celestial body? For example, will gravity be same for an Earth body, compared to the Moon?

Gravity is the same qualitatively.  It is the same type of thing: It pulls you down.  It can be described by Einstein's equations.  It can be very closely approximated my Newton's equations.  Force of gravity is proportional to mass and and inversely proportional to distance²

Newton proposed a law of Universal gravity.  The same law of gravity applies everywhere.  Although the details of what the law actually is have been tweaked (by Einstein) the principle of "the same law everywhere and at all times" seems to be true.

Gravity is different quantitatively. It is stronger in some places and weaker in others. The moon has a much smaller mass, and so your weight on the moon (ie the force of gravity on you) would be much less.


'Very High Noon' - Sun directly overhead Twice a year the sun is at a certain angle south (or north) of the equator between The Tropic of Cancer and The Tropic of Capricorn.
At that time the sun will be exactly overhead for people at the latitude (and longitude).
Is there a table or - better - a model that takes
Latitude (and Year) as input
and responds with
Date, Time, and Longitude as output.?
-The 'closest' 'midday' (solar time) (before or after) will be the closest people at that latitude - round the world - will come to experience the sun 'directly' overhead.
You could call it 'Very High Noon'.

vj

Let me give a slightly better answer. If you visit
http://ssd.jpl.nasa.gov/horizons.cgi with these parameters:

you'll see the Sun's astrometric right ascension and declination for a
48 hour period.

Let's look at one of the results and see what it tells us:

Date__(UT)__HR:MN     R.A._(ICRF/J2000.0)_DEC
**********************************************
$$SOE
2016-Nov-13 00:00     15 13 30.08 -17 57 18.0

This says the sun's declination is 17$^{\circ}$57'18.0" which means
the sun is (almost) directly overhead at that latitude.

The right ascension is 15h13m30.08s which means the sun is (almost)
directly overhead where the local sidereal time is this value.

To compute local sidereal time we first use
http://aa.usno.navy.mil/faq/docs/GAST.php to get GMST (I'm using the
approximation):

GMST = 18.697374558 + 24.06570982441908 D

To get D, the number of days since the year 2000, I used Unix's date
command as follows, and used other commands to finish the calculation:

: per http://aa.usno.navy.mil/faq/docs/GAST.php we compute the number of
: days since 2000 January 1, 12h UT (not 0h UT)
date -d '2000-01-01 12:00 UTC' +%s;: result is 946728000
date -d '2016-11-13 00:00 UTC' +%s;: result is 1478995200
: take the difference and convert result to days for formula
perl -le 'use POSIX; printf("%.10f\n", (1478995200-946728000)/86400)'
: result above is 6160.5 (trailing digits are 0)
: computing GMST and modding to get result between 0 and 24
perl -le 'use POSIX; printf("%.10f\n", fmod(18.697374558 + 24.06570982441908*6160.5,24))'
: result above is 3.5027478917
: convert solar RA to decimal
perl -le 'use POSIX; printf("%.10f\n", 15+13/60+30.08/3600)'
: result above is 15.2250222222
: how many hours east of GMST is the Sun (in degrees)? 1h = 15 degrees
perl -le 'use POSIX; printf("%.10f\n", (15.2250222222-3.5027478917)*15)'
: result above is 175.8341149575
: converting to minutes and seconds, we first find seconds from fractional part
perl -le 'use POSIX; printf("%.10f\n", 0.8341149575*3600)'
: result above is 3003 rounded to nearest second
: and now minutes/seconds from 3003 seconds total
perl -le 'use POSIX; printf("%.10f %.10f\n", floor(3003/60), 3003%60)'
: result above is 50 and 3

So you would think the Sun is overhead at 17$^{\circ}$57'18.0'' south
latitude (south because the declination is negative) and longitude
175$^{\circ}$50'03"E (east because longitude is positive) at 00:00 UTC
on 2016-Nov-13. Let's use HORIZONS to see if this is true (it turns
out it won't be, so we'll use a minute on either side of 2016-Nov-13
00:00 UTC to see what happens):

We can confirm the Sun's RA matches local sideral time within 1 second
of 00:00UTC from these lines in the output:

Date__(UT)__HR:MN:SC.fff   R.A.        DEC         Azi      Elev    Sid_Time
2016-Nov-13 00:00:00.000 * 15 13 30.08 -17 57 18.0 104.6639 89.7692 15 13 29.2615
2016-Nov-13 00:00:01.000 * 15 13 30.08 -17 57 18.0 104.9166 89.7730 15 13 30.2643

You'll notice the local sidereal time jumps from being 0.82 seconds
behind the sun's right ascension to 0.18 seconds ahead of the sun's
right ascension in the one second period (showing the two matched
sometime within that second, probably at 00:00:00.82 or so).

However, the Sun's highest position (noted by HORIZONS as 't' for
transit) occurs a little later:

Date__(UT)__HR:MN:SC.fff    R.A.        DEC         Azi      Elev    Sid_Time
2016-Nov-13 00:00:56.000 *  15 13 30.23 -17 57 18.6 178.5844 89.9415 15 14 25.4148
2016-Nov-13 00:00:57.000 *t 15 13 30.24 -17 57 18.6 182.4651 89.9415 15 14 26.4176
2016-Nov-13 00:00:58.000 *  15 13 30.24 -17 57 18.6 186.3230 89.9412 15 14 27.4203

when the local sideral time is almost a minute ahead of the sun's
right ascension. What's going on here?

Since the Earth is an ellipsoid, and not a sphere, it turns out that
"up", the direction perpendicular to a location on Earth, is NOT
directly away from the center of the Earth (it turns out gravity acts
in a direction that is neither towards the center of the Earth nor
perpendicular to the surface, but that's not relevant for this
question): see http://www.oc.nps.edu/oc2902w/c_mtutor/shape/shape4.htm
for details.

If you want really accurate calculations you'll have to use
http://naif.jpl.nasa.gov/naif/tutorials.html 's
http://naif.jpl.nasa.gov/pub/naif/toolkit_docs/C/cspice/surfnm_c.html
function.

Although this doesn't cause the discrepancy above, another
consideration is a location's elevation above the surface of the
Earth, which can also have slight effect (since "up" isn't always
towards the zenith). Happily, one thing you can ignore is refraction,
since refraction at the zenith is effctively zero.

My http://test.barrycarter.info/sunstuff.html is a very approximate
representation of where the Sun is overhead and is generated by
https://github.com/barrycarter/bcapps/blob/master/bc-sun-always-shines.pl


Astronomy as a profession for moderately good software engineer I'm a software engineer (a moderately good one). I like programming, mathematics and I love everything about astronomy.

What would be best way to go ahead?

A profession in astronomy can entail various things.  Unlike what other people are saying, you don't have to have a PhD in astronomy to do work in this field, to be an astronomer.  There are other options, but you will have to work hard.

Study as much physics and mathematics as possible.  Astronomy is essentially a branch of physics.  If you're still in school, then take classes in these areas.  If you're not in school anymore, then there are plenty of online resources to continue your education.  This will allow you to understand what people are talking about when they discuss technical things.
Decide what exactly you're interested in.  There are lots of different areas of study within astronomy.  Decide what appeals to you the most.  Do you like planets?  Galaxies?  Cosmology?  All of these different focuses require more specialized knowledge.
Your skills are already marketable.  Being an astronomer can mean lots of different things.  There are some astronomers who hardly ever look through telescopes, and spend most of their time reducing data.  A large part of astronomy is computer programming.  As a software engineer, you may already have the skills you need to work with astronomy research.
Keep an open mind.  There are lots of jobs in astronomy research where you might not be called an "astronomer".  For example, as a software engineer, you could find work with NASA supporting space research.  Your job title might not be "Astronomer", but you'll still be in the field.
Be prepared for hard work.  Being an astronomer can be difficult, not just because of the subject matter, but also because of the effort you have to put in to get a job.  It can be relatively difficult to find an employer looking for people with your particular skillset.  My recommendation would be to keep an eye on the NASA Jobs website to see when they're hiring.  You can also search for internships on NASA's OSSI site.
Try some amateur astronomy. There are lots of ways you can get involved in astronomy without having it as a job.  A quick Google search for "Amateur astronomy projects" yields many different results.
Join a local astronomy group. If you live in a reasonably sized city, there should be an astronomy group near you.  Find them and join them.
Don't give up! Having astronomy as a profession can be difficult, but it is also rewarding.  If you truly want to do this, you'll need to work at it.  Just don't give up!


Good source for the relationship between absolute magnitude, diameter, and albedo? In this answer I have rearranged an equation from somewhere, and now I can not relocate the source. My re-aranged form is:

$$M_{Abs} = 5 \left(\log_{10}(1329) -\frac{1}{2}\log_{10}(\text{albedo}) -\log_{10}(D_{km})\right)$$

This is the relationship between the absolute magnitude of an object and its diameter and albedo, assuming it is spherical.

Question: What is a good source for any form of this equation to which I can link in my answer?

The JPL CNEOS
asteroid size estimator
and various asteroid albedo papers cite
Harris and Harris 1997.
That paper is behind a paywall, but
Stuart and Binzel 2004
attribute this formula to it:
$$ H = C - 5 \log_{10} D - 2.5 \log_{10} p_V$$
where $H$ is absolute magnitude, $p_V$ is albedo, and $C$ = 15.618.


Is it possible to have a trinary star system with a planet that orbits their barycenter? I am trying to model a 3-star system with a planet. The three stars have the same mass. I have gotten the stars (Yellow, Pink, Blue) into a stable orbit. However, I am wondering if a "free-floating" planet can exist in this system. By "free-floating," I mean a planet that does not orbit any star; it orbits the barycenter.
Currently, I only know of metastable orbits that work for a while, but then the planet crashes into a star or gets ejected. Here are my setups:

Is there such a system that satisfies these conditions? If yes, is there a real-life example that astronomers know of?

There are 3 stars of the same mass.
There is a planet that orbits the barycenter of these three masses.
The planet is orbiting in a "Hot Jupiter" setup (the planet orbits close to the barycenter)

This configuration should be possible if the planet is located sufficiently far from the barycentre (i.e. sufficiently far outside the outermost orbit of the triple): close-in orbits will end up being destabilised.
So far there are no confirmed exoplanets orbiting the common barycentre of three (or more) stars: while planets are known in triple star systems, they are usually found orbiting just one of the stars. There might be a triple star containing a circumbinary planet but can't find it right now (there's definitely a quadruple with a circumbinary planet though).
If you're willing to consider candidate planets, and to relax your requirement that the stars have the same masses and that the planet is as close to the barycentre as possible, Phuong et al. (2020) suggest the presence of a protoplanet located outside the dust ring of the system GG Tauri A, at a separation of ~290 au. GG Tauri A is a T Tauri triple system, comprising the primary GG Tau Aa (~0.6 solar masses) separated at ~35 au from a close (~4.5 au) binary GG Tau Ab1/Ab2 (~0.38 and ~0.3 solar masses).
The configuration is as follows (apologies to users of assistive technology):
            ┌ GG Tau Aa (~0.6 Msun)
            │
      ┌35 au┤
      │     │      ┌ GG Tau Ab1 (~0.38 Msun)
      │     └4.5 au┤
      │            └ GG Tau Ab2 (~0.3 Msun)
290 au┤
      │
      │
      │
      └ GG Tau Ac (candidate protoplanet)

Masses for Aa, Ab1 and Ab2 from Di Folco et al. (2014)
The protoplanet is inferred from spiral structure observed in the disc. The paper suggests that other spiral structures may be induced by additional protoplanets (GG Tau Ad and Ae) located even further out. As mentioned before, these protoplanets are not currently considered confirmed.
GG Tauri A is itself a member of a quintuple system, the binary GG Tauri B is located at a projected separation of ~1500 au from GG Tau A. GG Tau Ba is a ~0.12 solar mass star, while GG Tau Bb is a ~46 Jupiter mass brown dwarf (White et al. 1999).


Milky Way Galaxy from Earth I've seen a lot of beautiful pictures of the Milky Way from Earth such as this:

...but I can't understand what the cloudy ribbon at the bottom the horizon is. Is it a super-large nebula? Or is it due to the Gegenschein? What happens in the ribbon?

The Milky Way Galaxy is a large spiral galaxy with some characteristic features worth mentioning:

1) The bulge - This refers to the collection of tightly packed stars located in the central region of the galaxy.

2) The spiral arms or the disk - This region extends from the inner region of the galaxy (where it meets the bulge) to outskirts of the galaxy, and contains stars, dust, and lots of gas. It's also very thin. The reason why a disk forms is largely thought to be due to slight asymmetries in the accretion of material early on in the galaxy's history. Over time, and due to conservation of angular momentum, the material collapses down into a disk. Incidentally, this is where the sun lives.

That ribbon is the disk of the Milky Way. It looks cloudy because of the dust and gas which scatters light from the rest of the galaxy. The reason why gas accumulates here is because as things begin to collapse into a disk, the gas collides with itself, and sort of 'sticks' together. In other words, where a bunch of stars may very happily pass through a bunch of other stars without many of them colliding or being disturbed, gas has a much harder time doing this. If you want to learn more about what happens when galaxies and clusters collide, take a look at the Bullet Cluster.

The majority of the other stars you you see which are not part of that structure are much more local. The reason why the rest of the sky looks rather transparent in comparison is because you are looking out of the plane of the galaxy. There is simply much less stuff (namely gas and dust which obscures your sight in optical wavelengths) that you are looking through. Because of this, most optical telescopes look in these directions when studying the rest of the universe.

For more about the classification and structures of galaxies, see the Hubble sequence.


Calculating Postition of ISS / Hubble in space How do I calculate the positions of above satellite in space ?

The orbit of the ISS does vary. Drag from the very thin atmosphere slowly causes the orbit to lower, and the speed to increase. So the ISS needs to be boosted about a couple of times a month.

Its position can be predicted, at least in the short term, and websites such as Heavens Above publish current position, predictions, and orbital information.


How much is usually deposited in a Mars regional dust storm? Is there any indication of how much dust is deposited, especially in areas that are not open where wind might produce a cleaning event?

Philip R. Christensen wrote an article back in 1986 called Regional dust deposits on Mars: Physical properties, age, and history. The abstract says

Dust deposited during global storms is subsequently removed only from dark regions, resulting in a net accumulation in the low‐inertia, bright regions. The thickness of these current dust deposits is 0.1–2 m.

This rather rough approximate for global storms seems to be a upper boundary, because e.g. Bruce Cantor states in the abstract of the article MOC observations of the 2001 Mars planet-encircling dust storm, the dust storms observed by the two rovers do deposit significantly less material (but can still cover solar cells):

The storm caused substantial regional albedo changes (darkening and brightening) as a result of the redistribution (removal and deposition) of a thin veneer of surface dust at least 0.1–11.1 μm thick.

However, the abstract also says that local storm events might significantly redistribute material:

The redistribution of dust by large annual regional storms might help explain the long period (∼30 years) between the largest planet-encircling dust storms events.

Both articles are paywalled, so I could not dig deeper here, so let us explore a bit a very local phenomenon, namely the dust devil. Dust devils have been frequently observed on Mars.
S.M. Metzger, Matt R. Balme et al. give an estimate of mass transport per dust devil in their article In-Situ Measurements of Particle Load and Transport in Dust Devils:

We find that dust devils [...]  can  loft  about  1  kg  of  PM10  material  to  heights  of  several  hundred  meters over their lifetime. Furthermore, applying the 5-15% TSP/PM10 ratio to the TSP data  from  the  five  large  dust  devils  sampled  in  1996  suggests  that  larger  examples  can  deliver  up  to  20  kg  of  fine  particulates  to  great  height.

References

Hall and Lemmon: Mass eject from Martian dust devils as functionof height and devil morphology


how long a light year on earth will seem First of all, sorry if this is not the right StackExchange to ask this question, but i don't know a better place (open to suggestion in case).
So, the question:
Suppose that for some reason, we can travel at the speed of light, and lets suppose that there is no acceleration involved, so it's something like uniform rectilinear motion.
Let's say that today someone leave with this spaceship at the speed of light, the after one year (for him), he will have travelled 9460730777119.56 km, but how much will be spent on the Earth since he has leaved?
So, in other words, I read about a special planet 31 light years from Earth, and I was wondering, if we can travel at the speed of light and someone travels for 31 years to that location, how much older would I be from when he has leaved the Earth?

You can never reach the speed of light but can get very close to it. Thus, when you travel with, say, 99.99% speed of light it will take you a little more than 31 years to reach the star 31 light years away. This is, however, the time it takes seen from earth. In the spaceship, the time taken will be dilated by
$$\gamma ={1\over \sqrt {1-{v^2\over c^2}}}$$
Which is roughly a factor of 70.7. This means in the space ship it is only 0.44 years or 160 days.


Realistic space environments You've seen it in all kinds of science fiction, be it books, games, movies, and such. Space is depicted all sorts of ways, with asteroids far too close together, nebula thick enough to hide in and much more. Yet for a writer that want's to describe space accurately, this is a problem.

So I need stellar features and reasons why they would form/exist

Here's a rough list of what I'm talking about:

Thick asteroid style fields, averaging around 100 meters apart
Nebula regions that are actually thick enough to obscure light, like clouds
Dust cloud regions thick enough to obscure light
Any other kind of cloud of material that's similar to the two above
Naturally occurring fields of explosive gas or a physical material that does the same
Radiation levels high enough to eat through ship plating
Naturally occurring anomalies of any sort, such as black holes, rogue planets and stars

An example would be a protoplanetary disc. Could this allow for material in close proximity so you have rocks in orbit that are close together? What about clouds of material? Would a forming star would have a thick cloud of material around it?

Anything naturally occurring or even unnaturally occurring is fine, so long as space magic or whatever isn't involved.

If I missed anything you can think of, please elaborate. This topic has been difficult for me to find information on.

Extra question: Could a cloud or clouds of opaque gas exist, even for a short time in space? With or without other material present works.

Thick asteroid style fields, averaging around 100 meters apart

That's a very high density. It's not any "asteroid" field that would occur naturally and would be stable for a long time. I can only think of a few environments that would be close to that:

the ring around a big planet, like Saturn
the debris field shortly after a major planetary collision, like the one that created the Moon. It would be a very dangerous place to be in, because not all fragments would move at the same speed; there would be quite an ongoing amount of collisions.
maybe, and this is a very cautious maybe, the pile of material coalescing into a protoplanet

All of the above would be short-lived, on a cosmic scale. It would be a coincidence that you're there to see it, and you'd probably have to explain the coincidence somehow.

"Asteroid fields", the way most people imagine them nowadays (very high density), are largely fiction, originating in the iconic space chase sequences in the Star Wars movies, and the like.

  Nebula regions that are actually thick enough to obscure light, like
  clouds

On what scale? Quite a few nebulae obscure light, but we're talking on the scale of distance between two neighboring stars. There are some nebulae that glow from within but we don't quite see the star - that's an example.

If you're talking about blocking light on the scale of mere kilometers, that's not possible "in outer space". Not for nebulae, not for dust clouds or anything else. That would be a density approaching the atmosphere of a planet. Consider the Earth - you can see just fine kilometers away through its atmosphere, unless fog gets in the way.

  Naturally occurring fields of explosive gas or a physical material
  that does the same

Even if you could find something like that, which is pretty much impossible, it would be highly unstable, and would eventually decay into the lower energy state - i.e. it would tend to explode. Such things might exist for the duration of a cosmic blink of an eye. It would be quite a coincidence that you could find something like that out there. A coincidence that would require some explanation, and very likely would not be a "mere coincidence". I think that's called a plot point. ;)

Such things just don't occur naturally.

  Radiation levels high enough to eat through ship plating

It would melt off, rather than being eaten through. Radiation tends to be absorbed; it doesn't tend to chip material away. And when it's absorbed, it yields its energy as heat.

Also, there is no material that could shield nuclear radiation so well that it would melt off on one side, while blocking it to non-lethal levels on the other side. At least not within any reasonable thickness.

Anyway, such conditions might be found near extreme places such as the polar jets of massive black holes - but only while the accretion disk is very active.

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

  Naturally occurring anomalies of any sort, such as black holes, rogue
  planets and stars

Black holes are pretty ordinary, not anomalous in any way. There are huge black holes in the center of galaxies, and regular-size blackholes sprinkled throughout.

Not sure what you mean by "rogue planets and stars". Planets not belonging to any star system are actually thought to be pretty common. Stars not belonging to any galaxy are also not extraordinary, but of course the distances between them would be gigantic, since any such star, by definition, would be alone in intergalactic space.

  An example would be a protoplanetary disc. Could this allow for
  material in close proximity so you have rocks in orbit that are close
  together? What about clouds of material?

Yes and yes. But at that point it would evolve rapidly, coalescing into a planet. Not rapidly on the human scale, but rapidly on cosmic scale. To you, as a human being, it would seem quite stable.

  Would a forming star would have a thick cloud of material around it?

The transition between a disperse, rarefied cloud of gas, and an actively burning star with a clearly defined "surface" is very rapid. However, for a very brief time (maybe thousands of years? not sure), you might find something with a dense center, surrounded by a thin cloud; but even then, density near the central area would increase rapidly. Perhaps you need to ask an actual expert in stellar evolution; there are models that show exactly how each phase looks like and how long it would last.

If you're looking for large expanses of gas or dust at densities comparable to Earth atmosphere, such a thing is simply not stable on a cosmic scale. It either dissipates if there's not enough gravity to hold it together, or it's part of an active process of collapse as a protostar or protoplanet.


Why is the boundary of friends-of-friends (FOF) halo corresponding to iso-density contour? The friends-of-friends algorithm (hereafter FOF) is commonly used to find halos in cosmological simulations. (For more information, please refer to here and here)
I found that some literature argues that the boundary of the FOF halo is corresponding to an iso-density surface. For example, Springel et al 2001 states

It
places any two particles with a separation less than some linking
length $b$ into the same group. In this way, particle groups are
formed that correspond to regions approximately enclosed by
isodensity contours with threshold value $\rho\propto1/b^3$.

More et al 2011 says

One could expect that for a given value of $b$, the FOF algorithm defines the boundary of a halo as corresponding to a certain isodensity surface, at least in the limit of large number of particles.

However, I can't understand the connection between the boundary of the FOF group of particles and iso-density contour. How to derive this connection in formula form? Does this connection need other hypotheses?

If you have enough particles to sample your field, then a given (local) density corresponds to the same mean distance between your particles. The FOF algorithm keeps linking particles to a halo until a particle is farther away than the linking length. That means that the halo boundary will be ~exactly on the border between regions of interparticle distances smaller and larger than the linking length, corresponding to a boundary between larger and smaller densities, respectively. That is, an iso-density surface.
In contrast, a halo finder using the spherical overdensity (SO) algorithm will always result in a spherical boundary, which will only be an iso-density surface in the case of a completely symmetrical halo.
This is the reason that FOF halo finders tend to include more filaments outside the region identified by SO halo finder. This may or may not be desired.


Why do the Pleiades look clearer when viewed indirectly? I've noticed that the Pleiades look bright in the sky, but when I look directly at the constellation, it suddenly goes dim and it's more difficult to make out the individual stars.

Why does this happen, and is there a way to "beat the system" by both being able to look directly at a star and see it clearly.

What you describe is a technique called averted vision and takes advantage of the arrangement of cones and rods, two types of light sensitive elements within the eye:

  The retina contains two types of photoreceptors, rods and cones. The
  rods are more numerous, some 120 million, and are more sensitive than
  the cones. However, they are not sensitive to color. The 6 to 7
  million cones provide the eye's color sensitivity and they are much
  more concentrated in the central yellow spot known as the macula. In
  the center of that region is the " fovea centralis ", a 0.3 mm
  diameter rod-free area with very thin, densely packed cones.

Shorter, the cones are sensitive to color and work best in bright light, while the rods are most effective in low light conditions. And since the rods are concentrated off center of retina, our night vision is best off center. This averted vision technique can be used with a naked eye, as well as with any optical equipment like binoculars or telescopes, although it can be a bit of a bother with eyeglasses and contact lenses. Especially with the latter, using this advert vision technique can cause additional irritation of the cornea.

As per "beating the system" part, with a bit of practice, you could train yourself to view dim objects better with the corner of your eyes at will, and not merely by chance or for only short periods of time. It might seem hard with first few tries, but it really isn't and you soon get used to not looking directly towards the observed object, countering what we tend to do naturally, direct our eyes towards the object of our observations. Practice makes perfect, but there is no long term and unwanted side effects to doing this, and you'll still be able to direct your eyes towards whatever you'll want to see clearer, just as you did before.

There is one other thing to add to techniques to observing the night skies. Namely, our eyes need some time to adapt to darkness, and with our more frequent staring in computer monitors, television and other illuminated sources, our pupils need even longer to dilate and let the maximum of light to fall on the retina. So for best results, rest your eyes and do not expose them to anything except red light for 30 minutes prior to observing. This allows your pupils to expand to their maximum diameter and build up the levels of optical pigments, which are rapidly lost if exposed to bright light.

It is also important to observe with both of your eyes open to avoid fatigue. If you find this difficult, you may try covering the other eye with your hand or an eye patch. It should also be mentioned that both smoking and drinking cause deterioration in your sight. Your blood-sugar levels are also important and a low level of blood sugar causes lowering of sensitivity of eyes.

Lastly, your eyes will tire using this technique, just as they would by focusing for longer periods of times as you normally would. Look away every now and then, preferably not towards too strong sources of light, but still at least slightly brighter than those you previously observed. Also stretch your legs and blink fast a few times. Stretch your back too, and relax your bottom. You might find it funny, but this technique of relaxing and contracting muscles of that part of our physique has been used by trained spies to cheat polygraph tests that also measure pupil dilation. How's that for "beating the system", huh? It is by far the fastest way of relaxing your eyes, at least that I know of.


How cold is interstellar space? The vastness of space brings me a sense of chilliness even though I have never experienced it, although I wish to. Just how cold is interstellar space (on average)? How is this even measured? I mean you can't just stick a thermometer in space, right?

You can stick a thermometer in space, and if it is a super-high-tech one, it might show you the temperature of the gas. But since the interstellar medium (ISM) is so dilute, a normal thermometer will radiate energy away faster than it can absorb it, and thus it won't reach thermal equilibrium with the gas.
It won't cool all the way to 0 K, though, since the cosmic microwave background radiation won't allow it to cool further than 2.7 K, as described by David Hammen.
The term "temperature" is a measure of the average energy of the particles of a gas (other definitions exist e.g. for a radiation field). If the gas is very thin, but particles move at the same average speed as, say, at the surface of Earth, the gas is still said to have a temperature of, say, 27º C, or $ 300\,\mathrm{K}$.
The ISM consists of several different phases, each with their own physical characteristics and origins. Arguably, the three most important phases are (see e.g. Ferrière 2001):
Molecular clouds
Stars are born in dense molecular clouds with temperatures of just 10-20 K. In order for a star to form, the gas must be able to collapse gravitationally, which is impossible if the atoms move too fast.
The warm neutral medium
The molecular clouds themselves form from gas that is neutral, i.e. not ionized. Since most of the gas is hydrogen, this means that it has a temperature of roughly $10^4\,\mathrm{K}$, above which hydrogen tends to get ionized.
The hot ionized medium
Gas that accretes onto the galaxy in its early phases tend to have much larger temperature, of roughly $10^6\,\mathrm{K}$. Additionally, the radiative feedback from the hot stars (O and B), and the kinetic and radiative energy injected by supernova explosions ionize and heat gas bubbles that expand. This gas comprises the hot ionized medium.
Cooling
The reason that the ISM is so sharply divided into phases, as opposed to just being a smooth mixture of particles of all sorts of energies, is that gas cools by various physical processes that have a rather temperature-specific efficiency.
"Cooling" means converting the kinetic energy of particles into radiation that is able to leave the system.
Hot gas
Very hot gas is fully collisionally ionized and thus cools mainly through free electron emitting Bremsstrahlung. This mechanism becomes inefficent below $\sim10^6\,\mathrm{K}$.
Warm gas
Between $10^4\,\mathrm{K}$ and $10^6\,\mathrm{K}$, recombinations (i.e. electrons being caught by ions) and collisonal excitation and subsequent de-excitation lead to emission, removing energy from the system.
Here the metallicity$^\dagger$ of the gas is important, since various elements have different energy levels.
Cool gas
At lower temperatures, the gas is almost fully neutral, so recombinations cease to have any influence. Collisions between hydrogen atom become too weak to excite the atoms, but if molecules or metals are present, it is possible through fine/hyperfine lines, and rotational/vibrational lines, respectively.
The total cooling is the sum of all these processes, but will be dominated by one or a few processes at a given temperature. The figures below from Sutherland & Dopita (1993) shows the main cooling processes (left) and the main cooling elements (right), as a function of temperature:

The thick line show the total cooling rate. The figure below, from the same paper, shows the total cooling rate for different metallicities. The metallicity is a logarithmic scale, so [Fe/H] = 0 means Solar metallicity, and [Fe/H] = –1 means 0.1 times Solar metallicity, while "nil" is zero metallicity.

Since these processes don't cover equally the full temperature range, the gas will tend to reach certain "plateaus" in temperatures, i.e. it will tend to occupy certain specific temperatures. When gas cools, it contracts. From the ideal gas law, we know that the pressure $P$ is proportional to the product of the density $n$ and the temperature $T$. If there's pressure equilibrium in the ISM (which there isn't always, but in many cases is a good assumption), then $nT$ is constant, and thus if a parcel of hot ionized gas cools from $10^7\,\mathrm{K}$ to $10^4\,\mathrm{K}$, it must contract to increases its density by a factor $10^3$. Thus, cooler clouds are smaller and denser, and in this way the ISM is divided up in its various phases.
So, to conclude, interstellar space is not as cold as you may think. However, being extremely dilute, it is difficult to transfer heat, so if you leave your spaceship, you will radiate away energy faster than you can absorb it from the gas.

$^\dagger$In astronomy, the term "metal", refers to all elements that are not hydrogen or helium, and "metallicity" is the fraction of gas that consists of metals.


Question about redshift and effective gravity at the rotating neutron star surface For a rapidly rotating neutron star, if consider the star as a sphere, redshift at the equator surface will be larger than at the pole. But if consider the star as an obsolate sphereiod (the ellipsoid varies according to the rotation velocity), then how about the redshift at equator compare to the pole surface? Whether the obsolate decrease the redshift at equator?
Similar is the effective gravity, if consider the obsolate effect, how about the gravity difference between the equator and the pole compare to that a sphere?

The redshift factor is given in (Asaoka & Hoshi 1987) and depends on latitude and radial coordinates. The equilibrium surface may however be an ellipsoidal surface of constant redshift and frame dragging (Marsh 2014).


Contradiction between Theory of Relativity and Newton's Law of Universal Gravitation? Newton's Law of Universal Gravity states that an object will accelerate constantly as it falls. Let's assume an object falling into a black hole. According to Newton's Law, when it reaches the event horizon it must be traveling at c, because c is the escape velocity at the event horizon. Therefore, after it passes the event horizon, it must continue to accelerate past c, yet theory of relativity states that it's never gonna happen. So, in this condition, what will the object do?

A) Don't take Newton's law too seriously in the relativistic regime.

  According to Newton's Law, when it reaches the event horizon it must be traveling at $c$, because $c$ is the escape velocity at the event horizon.

This is true, assuming the particle freefalls from rest at infinity, but the details are quite different from Newtonian theory. For orbits in Schwarzschild spacetime, the effective potential is
$$V = -\frac{GM}{r} + \frac{l^2}{2r^2} - \frac{GMl^2}{c^2r^3}\text{,}$$
where $l\equiv L/m$ is the specific angular momentum of the orbit. The last term offers relativistic corrections that have no analogue in Newtonian gravity, so in general Newton's law cannot be counted on when near the horizon of the black hole.

But for radial freefall, $l = 0$, and we a total orbital specific energy:
$$\mathcal{E} = \frac{1}{2}\left(\frac{\mathrm{d}r}{\mathrm{d}\tau}\right)^2 - \frac{GM}{r}\text{,}$$
which has exactly Newtonian form. In particular, a particle freefalling from rest at infinity will have
$$\left|\frac{\mathrm{d}r}{\mathrm{d}\tau}\right| = \sqrt{\frac{2GM}{r}}\text{,}$$
so at as it nears the horizon ($r\to 2GM/c^2$), we have $|\mathrm{d}r/\mathrm{d}\tau|\to c$, just as Newtonian theory would predict.

However, the motion is only roughly analogous to what's prescribed by Newton's law, because there are very important differences in the interpretation of those quantities. Specifically, (1) the $\tau$ refers to the proper time of the orbiting particle, not coordinate time $t$, and certainly not any absolute time as Newtonian theory proper, and (2) the $r$ refers to the Schwarzschild radial coordinate, which is defined such that a sphere at this coordinate have surface area $4\pi r^2$, and this is not the radial distance to the center.

B) Coordinates have no intrinsic meaning.

  Therefore, after it passes the event horizon, it must continue to accelerate past $c$, yet theory of relativity states that it's never gonna happen.

Suppose I fling a particle along a straight ruler, and that ruler is marked in feet. So if, for example, my particle goes five marked units along the ruler in a nanosecond, that's clearly a problem, because the particle must be superluminal ($c\approx 1\,\mathrm{ft}/\mathrm{ns}$).

Now suppose the ruler is not marked in feet, and I don't tell you how it's marked. Is it a problem that the particle goes five units in a nanosecond? Without knowing what the marks are, the statement of this sort of "coordinate speed" is not even meaningful. Finally, suppose I tell you how the ruler is marked, but its marks do not actually measure the distance along the ruler. Then is it a problem that it goes through five of those marks in a nanosecond?

The moral is simply this: by themselves, coordinates are simply labels to identify events. One can pick coordinates such that the coordinate speed is arbitrarily large or small, and it doesn't matter, because coordinates are not physical things. The universe does not come with its own coordinates; they're just labels we put on things.

However, what can give coordinates meaning is the metric tensor, which relates coordinates to actual lengths or durations, so that, for example, the Schwarzschild radial coordinate has a set meaning in terms of surface areas. Finally, the Schwarzschild coordinates are pathological near the horizon. They are not defined across the horizon (technically, the Schwarzschild coordinates are actually two entirely different coordinate charts disconnected by the horizon). You can see this in the Schwarzschild metric (units of $G = c = 1$):
$$\mathrm{d}s^2 = -\left(1-\frac{2M}{r}\right)\mathrm{d}t^2 + \left(1-\frac{2M}{r}\right)^{-1}\,\mathrm{d}r^2 + r^2(\mathrm{d}\theta^2+\sin^2\theta\,\mathrm{d}\phi^2)\text{,}$$
where the metric coefficients in the Schwarzschild chart are undefined at the horizon $r = 2M$.

C) "So, in this condition, what will the object do?"

It will cross the event horizon and meet its end at the singularity.

Here's the same Schwarzschild geometry in Gullstrand-Painlevé coordinates:
$$\mathrm{d}s^2 = -\mathrm{d}t^2 + \left(\mathrm{d}r + \sqrt{\frac{2M}{r}}\,\mathrm{d}t\right)^2 + r^2(\mathrm{d}\theta^2+\sin^2\theta\,\mathrm{d}\phi^2)\text{.}$$
In this coordinate chart, space is moving towards the singularity at the estate velocity $\sqrt{2GM/r}$, and the particles are carried along with it. In a related coordinate chart, Lemaître coordinates, particles freefalling from rest at infinity have zero coordinate speed, with $\mathrm{d}r/\mathrm{d}t = 0$, where here $r$ is the Lemaître radial coordinate. (Edit: corrected example mistakenly attributing this property to the GP chart.)

You can pick any coordinate chart that's well-behaved at the horizon, and any such chart will show it moving across the horizon with no trouble. But the Schwarzschild coordinates do not give such a chart, because they're undefined at the horizon.


Why is liquid water considered a requirement for life? I`ve heard about different extremofiles living on earth. Is water a requirement for life because we have not seen life survive on earth without it, is it a chemical  reason which makes life without water seem impossible? Could life exist with water, but in gas form? What makes the liquid part so important? Can life be based on other  liquids on potential exoplanets?

The reason for which water is needed for Life as we know it (and we have not been able to successfuly imagine a different model from that of DNA that has ability to self-replicate, react to the ambient, and evolve), is that water is the "universal solvent".

That means that most chemical reactions needed for life happen on a liquid environment, more specifically a water-based environment for most of them.


Why do I see rainbows around the moon? Regularly, I see rainbows around the moon. Here is approximately what I see:

Right at the edge of the moon, I see a bright blue ring of light. Further out, I see an inner rainbow. This is not as bright as the ring, but is still clearly visible. Even further out, I see an outer rainbow that is very faint.

I see the blue ring during every full moon. I see the rainbows more frequently in late fall and winter than in spring or summer. But every year, I see these moon rainbows during a full moon.

Since I see these moon rainbows more frequently in late fall and in winter, are these moon rainbows caused by moonlight reflecting off of ice crystals in cirrus clouds high up enough that I can't see the clouds? Also, why do I see a bright blue ring right at the edge of the moon during every full moon?

There are a couple of atmospheric phenomena that can create rings around the moon.

Corona are caused by water droplets diffracting the moon's light, they are fairly small, and close to the moon. They are coloured but not as brightly as a rainbow. Our eyes are also not so good at seeing colours in dim light.

The 22-degree halo is a larger ring, as big as two hands stretched out at arms length. Solar halos are clearly coloured, lunar halos are dimmer and I haven't seen lunar halos that are bright enough to appear coloured to me.

An eye condition, astagmatism, can also create rings, such as the "thin blue ring" (though I am not a doctor) You may be seeing a combination of all three at various times.


What is the mean separation between a star and its closest neighbor in the Milky Way halo? My question is inspired by this previous question, Are there any stars that orbit perpendicular to the Milky Way's galactic plane?. In the great answer posted by @ConnorGarcia, one commenter said he would be really jealous of anybody that could see the view of the Milky Way from close to the galactic poles, since a star stream passes close to them. I think, on the flip side, they would envy us due to their near complete isolation. Space travel outside their home system would be much harder, for example. Here on the galactic plane, near the Sun, I guess the average distance to the closest nighboring star is about ~10 Ly or less. The Alpha Centauri system is about 4 Ly away, for example. Regarding the galactic halo, I guess if we picked a random star there, its closest neighbor would be probably located > 100 Ly away. Is this estimate reasonable? Are there reliable figures on that?

I did some rough calculations, and 100 light-years doesn't seem to be a bad guess. If we assume that the average mass of a halo star is $\sim0.3M_{\odot}$, as would be expected for a typical IMF, and that the total stellar halo mass is $\sim10^9M_{\odot}$ (Deason et al. 2019), then we should expect there to be $\sim3.3\times10^9$ halo stars.$^{\dagger}$ The halo extends to somewhere near 100 kpc, so if we take it to be roughly spherica-ishl, this leads to an average number density of
$$n\approx\frac{3.3\times10^9\;\text{stars}}{\frac{4\pi}{3}(100\;\text{kpc})^3}\approx7.9\times10^{-7}\;\text{pc}^{-3}$$
and a mean inter-star distance of $l\approx n^{-1/3}\approx108\;\text{pc}$. So my back-of-the-envelope result differs from yours by only a factor of 3.
Now, the halo has a distinctly non-uniform density. It follows a power law of $r^{-n}$, with an index somewhere near $n\approx3$. Different surveys have yielded values ranging from $2.5\lesssim n\lesssim 3.5$ (ish), depending on the methodology and groups studied. The upshot is that in the outer halo, you'll certainly see much larger separations, whereas closer to the Galactic plane, you could see densities perhaps within a couple orders of magnitude of disk populations.

$^{\dagger}$Two things: First, the halo population's mass function should be different than a typical IMF, because it's composed largely of aging Population II stars - but I think $0.3M_{\odot}$ is still reasonably close. Second, I also checked the number by actually integrating via a Kroupa IMF (normalizing by total halo mass), (and again, maybe not a totally accurate distribution) and found a value within a factor of 2 of 3.3 billion.


How can it be known that Venus does not have plate tectonics? This answer provides some insight into Venus' surface geology:

  Water may be necessary as a lubricant for plate tectonics. Whether or not this is the case, Venus does not have plate tectonics. It instead has a stagnant lid geology, punctured occasionally by extreme vulcanism (Siberian traps level vulcanism, and then some).

What are the observations that lead to this conclusion? The existence of plate tectonics on Earth was first determined by putting a lot of pieces of the first-hand observational puzzle together. There's much less data available from Venus.

There's much less data available from Venus.

Some data exists. As mentioned in HDE 226868's answer, maps of Venus's surface exist. Like Earth's atmosphere, Venus's atmosphere is transparent to some low frequency electromagnetic radiation such as those used by radar. These observations are consistent with a planet that has stagnant lid tectonics and inconsistent with a planet that has active plate tectonics; more on this below.

In addition to these remote observations, the Soviet Union successfully sent several spacecraft into Venus's atmosphere, some of which landed and briefly operated on the surface of Venus. The initial attempts failed because nobody thought the surface conditions on Venus would be as brutal as they are. To make landing successful, the Soviet Union had to significantly downsize the parachutes and they had to use materials and avionics that could withstand very high temperatures.

Every piece of evidence gathered to date regarding Venus is inconsistent with a planet with active plate tectonics:

Venus surface temperature and pressure are well above water's critical point. Venus cannot have any liquid water on its surface. Water is widely (but not universally) thought to be critical as a lubricant that enables plate tectonics to occur.
The radar observations show a planet with a nearly universal surface age, about half a billion years old. This is inconsistent with a planet with active plate tectonics but consistent with a planet with stagnant lid tectonics.
Venus's atmosphere is very thick, much thicker than the Earth's atmosphere, and is dominated by carbon dioxide. Plate tectonics recycles carbon dioxide at subduction zones. Stagnant lid tectonics does not. A planet with plate tectonics will see a gradual reduction in the amount of carbon dioxide in its atmosphere over geologically long periods of time. A planet with intermittently active stagnant lid tectonics will instead see a gradual increase in the amount of carbon dioxide in its atmosphere over geologically long periods of time.
Multiple physics-based models of a planet with very high surface temperatures suggest that such planets will have rather thin and rather ductile crusts that can readily repair themselves against damage caused by subsurface tensions.

Note very well: The stagnant lid tectonics of hot terrestrial planets such as Venus  and possibly Titan (Titan is "hot" because its geology is ice-based rather than rock-based) is rather different from the stagnant lid tectonics of cold terrestrial planets such as the Moon and Mars. The surfaces of the Moon and Mars are very old. The surface of Venus is much younger in comparison. Venus has undergone at least one somewhat recent nearly global resurfacing event. This has not happened on the Moon or Mars. Plate tectonics appears to require a planet whose surface is neither too cold nor too hot, and that has a good amount of liquid water on the surface.

Some references:

David Bercovici and Yanick Ricard, "Plate tectonics, damage and inheritance," Nature 508.7497 (2014): 513.
DOI: 10.1038/nature13072.
A. Davaille, S. E. Smrekar, and S. Tomlinson, "Experimental and observational evidence for plume-induced subduction on Venus," Nature Geoscience 10.5 (2017): 349.
DOI: 10.1038/NGEO2928.
James F. Kasting and David Catling, "Evolution of a habitable planet," Annual Review of Astronomy and Astrophysics 41.1 (2003): 429-463.
DOI: 10.1146/annurev.astro.41.071601.170049.
Mikhail A. Kreslavsky, Mikhail A. Ivanov, and James W. Head, "The resurfacing history of Venus: Constraints from buffered crater densities," Icarus 250 (2015): 438-450.
DOI: 10.1016/j.icarus.2014.12.024
Ignasi Ribas et al., "Evolution of the solar activity over time and effects on planetary atmospheres. I. High-energy irradiances (1-1700 Å)," The Astrophysical Journal 622.1 (2005): 680.
DOI: 10.1017/S0074180900182427.
Accessible pdf: https://iopscience.iop.org/article/10.1086/427977/pdf.


Would all neutron stars inevitably collapse into black holes due to quantum tunneling? The materials inside a neutron star are densely packed (extremely close to each other) and as time goes by say about an eon later, most of the neutrons would eventually tunnel into each other and voila! the neutron star becomes black hole? sound suspicious to me but can it happen?

The materials inside a neutron star are densely packed (extremely close to each other) and as time goes by say about an eon later, most of the neutrons would eventually tunnel into each other

Based on what ?

The material's collapse was halted from an astonishingly high speed collapse by the neutron degeneracy pressure that exists precisely because the universe doesn't want to let two neutrons fuse.  The enormous forces involved in the collapse and the enormous forces keeping the mass compact are not high enough to overcome this.  That's why it exists in the first place.

Over time "tunnelling" works in many, many ways. If two neutrons were to fuse in this manner the list of possible by products is enormous, and a black hole is most unlikely, IMO.  We've a lot of experience of whacking particles into other particles in extreme conditions ( think CERN ) and we haven't seen any sign of a black hole yet ( and believe me, they're looking for them ).  So I think the odds of a fuse into a black hole are extremely low.  Over time it's more likely the neutron star would "evaporate" into other particles.  But over those time scales you're looking at the cold death of the universe and protons decaying and so on.

And if, for argument sake, a mini black hole formed from two neutron's "fusing" it's life expectancy would be minuscule as tunnelling effects ( predicted by Hawking ) suggest it would "evaporate" in the blink of an eye.  So isolated fusing events, even f they happened, would be unlikely to produce a black hole that grew, but instead one that just blew up as fast as it was created.

  and voila! the neutron star becomes black hole? sound suspicious to me but can it happen?

No evidence I am aware of suggests neutrons will fuse in the manner you're describing.

"Neutron Star" is probably something of a misnomer.  Apart from neutrons, current models ( "educated guesses" ) suggest that the core of such a body might be a lump of something called quark-gluon plasma, which is essentially mush made from neutrons.  This is what you get when you fuse neutrons together -  a form of mush made up of the stuff that neutrons and protons are made from - quarks.

So theory and evidence suggests that you don't get a black hole from just waiting for a neutron star to change into one.


What (actually) is Jupiter doing to this year's Perseids meteor shower? After reading the Astronomy Magazine article Perseid meteor shower set for its best show in nearly 20 years and NASA's Look Up! Perseid Meteor Shower Peaks Aug. 11-12, I understand that Jupiter's gravity is affecting this year's possible peak rate of Perseid's meteors.

I think it is - in simple terms - deflecting some of them out of their general orbit, a bit more towards earth, so that we intercept more than usual. But I am having difficulty visualizing this. Is this deflection in the plane of the ecliptic, or out of plane? At our closest approach, how far away are we from the orbit of this dust?

A good diagram would really help!  I found this at space.com credited to Jeremie Vaubaillon, but it's not helping me with the 'big picture'.

Also found this here and credited to NASA/JPL:

To put it in simple terms, a comet's debris trail or dust trail is like a ring of debris that generally follows the entire orbit of the comet, more concentrated close to the comet, more spread out further away, but it can be thought of, for simplicity, as an entire ring of debris, kind of, a miniature asteroid belt with a fixed elliptical path.   The picture below isn't perfect, cause it gives the impression that orbits are all the same when there's some spreading out, but it's the gist of it.  (Note, the debris field and comet in the picture below isn't Swift-Tuttle/Perseid)

The reason for this is that any debris that's knocked off the comet at a low velocity has pretty much the same orbit as the original comet, and over hundreds or thousands of orbits, it tends to fill up the entire comet's orbit with mostly tiny particles, some larger ones.  The debris field or dust trail from the comet generally follows in the comets orbit.

This picture is a bit more accurate as to what the Earth is passing through.

https://lintvwpri.files.wordpress.com/2016/08/perseid_base4.png?w=650&h=366

The more compact the dust trail or debris field, the more meteors, but the smaller the target the Earth has to hit, so odds are greater that the Earth will miss the field entirely.  The more spread out, the greater chance the Earth hits the debris field, but more spread out means we get fewer meteors.

Earth's orbit crosses the Perseid (eliptical ring of debris) or Swift-Tuttle's debris field about the same time every year because Earth's and Swift-Tuttle's orbits nearly intersect.

This year, because of the placement of Jupiter, the debris field (which we can kind of think of as a ring), got bent by Jupiter, and Earth is hitting closer to the center of the debris field which means, more meteors.   I can't say on which plane the field got bent, cause it depends on it's orbital relation to Jupiter and I couldn't find that, but it was the ring of debris that got bent, so Earth will hit it more directly this year than usual.

Because the ring of debris is pretty much permanent, Jupiter bends the ring every 11 years as it passes closest to it, and occasionally that bend works in earths' favor and gives us more meteors, so its a rare event that might not happen again for 11 years, it might be over 100 years before Jupiter does this for us again.

See article.

  Typically Earth just grazes by Swift-Tuttle’s debris field, but our
  planet will be even closer to the particle stream this year thanks to
  some help from our neighbor Jupiter. The gas giant occasionally gets
  close to the comet stream, and its immense gravity pushes the debris
  nearer to Earth.


What colours can rocky planets (or moons) have I am currently working on a simple program to randomly generate and display rocky exoplanets (for a space based strategy game), but I am having some problems figuring out what colour the rocks of rocky planets may or may not have.
Specifically, I am thinking of what natural colours may be so dominantly found on the surface of a planet or moon, that the whole planet or moon, seen from space would seem to be (completely or partially) that colour, which is not made out of ice, and which neither has oceans nor life.
As far as I can tell from looking at pictures of the not lifebearing rocky planets, dwarf planets and moons in our solar system, it seems that red and yellow colors are dominant on larger bodies such as Mars, Venus or Io (maybe because their atmosphere cause oxidation of metals on the surface, or in Io's case volcanic activities as pointed out by Phiteros):

All while smaller bodies such as the Moon or Ceres are completely grey:

Therefore my current approach is to make sure the red component of the RGB colour of the planets is the biggest, the green the second biggest and the blue the smallest.
However just the fact that there aren't any rocky planets or moons in our solar system made of blue and green rock, doesn't mean that they can't possibly exist.
Therefore my question is: is my assumption that rocky planets tend to be mostly red or yellow correct, or are there any good logical arguments for that predominantly blue, green or purple rocky planets could exist.
Edit
Maybe i was a little to vargue on defining colour; i was thinking about the »diffuse colour« (sometimes called »albedo colour«) input, used in the openGL shader that i am using to display the planets; this colour is however the same as something not very reflective (like planets) has when under white light.

I'll begin by saying that for exoplanets, the answer to your question is an active area of research, and currently there are no observational constraints on what rocky exoplanets would look like to the human eye.

Also, this may be a lot more complicated an answer than what you want for a computer game.

If you want to hunt around for this in the astronomical or planetary science literature, what you're trying to do is determine the reflectance, or albedo spectrum, of the planet. That's what tells you how much of the incoming light is bounced back off the surface for you to see as function of wavelength. Note that this assumes the planet is not hot enough to thermally emit in the optical, which will only start to matter if it's >3000K or so. This albedo spectrum is determined by the surface properties of the planet: what it's made out of, how shiny it is, the scattering properties, etc.

Once you have an albedo spectrum, you then can multiply it be the incoming radiation from whatever primary star you've chosen. That's what will make a planet seem slightly more orange around an M-dwarf, and more "normal" around a Sun-like star.

To get to how a person would see it, you'd then need to then multiply that resulting reflected-light spectrum by the spectral sensitivity of the human eye, and then convert that final observed spectrum into a color.

For the planetary albedo spectrum, there aren't a lot of sources in the astronomical literature. The best I can find is this paper. You'll probably have more luck looking through Solar System planetary stuff.

For the stellar spectra, you can try here, but that presumes you're comfortable with astronomical FITS files. For stars hotter than >4500K (the Sun is 5800K), you can probably just assume that the incoming radiation into the planet's surface is blackbody of whatever temperature you want.

Unfortunately, I am not familiar with the spectral sensitivity of the human eye, or how to go from an input observed spectrum to a single color value.


Planetary reference systems and time I am researching into how coordinate systems of solar systems objects are created by reading some of the reports written by the Working Group on Cartographic Coordinates and Rotational Elements (e.g. 2009). However, I am finding it difficult to completley understand the role of time in defining reference systems.

When observing at a planet from Earth, for example Jupiter, there are a variety of factors that make it difficult to construct a reference system (including no solid surface and planetary precession), so we use geometry to define a reference system. However, because our perspective is dynamic, meaning Jupiter's surface changes and the planet's move, we say that at time J2000 we know the precise orientation and position of Earth and therefore can say from position defined at J2000 this is the coordinate reference system for Jupiter.

So, does incorporating time (e.g. J2000) mean we can say a coordinate reference system is based on the situation of an object, Jupiter in this example, at a given moment?

Look at J2000 Solar System Barycentric (spatial reference) and Barycentric Dynamical Time (TDB).  Together they are good space and time coordinate system.  NASA/JPL has some good info and data on these.

In terms of time, TDB is rescaled so that from the Earth, it appears to be close to the same as TT (roughly ~UTC). The rescaling is due to the fact we are in a gravitational well, as well as moving at 30 km/s with respect to the solar system barycentre, due to relativity. The original unscaled is called Barycentric Coordinate Time (TCB) and differs by ~0.5seconds / year.


Is this image captured by the Hubble Telescope an original image? Is this image an original picture which is not photoshoped? I see a huge number of Hubble Telescope images on Google and I'm curious whether they are real images or not. If not, where I can find the real images captured by the Telescope which are not photoshopped?

All images provided from large telescopes are a rendering of the data for viewing pleasure. Usually sensors are sensitive in a wide range of wave length, even outside what humans can perceive and color is added by using filters. Hubble uses infrared filters and the so-called Hubble palette (false color).

So, if you mean, do these pictures show what a human observer will see? No, they don't. They are a visualisation of the data collected.

All things observed at low light conditions are black and white, as humans lose color vision at low light.


Attraction to barycenter If an object was a distance from a planet away that equaled the distance from the object to a moon of the said planet, would it be drawn into the direction of the barycenter as it is the common center of mass?

The Newtonian gravitational acceleration of a test object (a small object with negligible mass) toward a point mass is given by

$$\vec g =\frac {GM}{||\vec r||^3} \vec r \tag{1}$$

where $\vec g$ is the acceleration vector, $G$ is the Newtonian gravitational constant, $M$ is the mass of the point mass, and $\vec r$ is the displacement vector directed from the test object toward the point mass.

Equation (1) cannot be used to compute the net gravitation acceleration of a test object toward a pair of point masses by pretending that the pair of point masses is equivalent to a single point mass located at the barycenter of the two point masses. Instead, what needs to be done is to apply equation (1) to the point masses individually and then form the vector sum of these individual accelerations. This in general will not point toward the barycenter.

There are two classes of points where the net gravitational acceleration vector to a pair of point masses is directed toward the barycenter of the point masses:

Points along the line connecting the point masses, less for two regions where the net gravitational acceleration vector points away from the point mass, and
Points away from this line and whose projection onto this line is exactly midway between the point masses.

For a test object located anywhere else, the net gravitational acceleration vector does not point toward the barycenter.


What happened before, Big Bang or inflation? I always thought that the cosmic inflation happened after the big bang (10-36 sec) and lasted for a very tiny fraction of time).
Now, I recently came across a couple of articles (links below) claiming that Big Bang happened after the inflation stopped and when the reheating phase started.
I'm a bit confused as I always imagined the Big Bang as "when everything started", so how can something be happened before that?

Ethan Siegel (2019) "What Came First: Inflation Or The Big Bang"
Gabriele Veneziano (2017) "A Quantum Universe Before the Big Bang(s)?"

The big bang (a singularity 13.8 billion years ago that was the beginning of time) didn't happen at all. Inflationary cosmology consists of:

ΛCDM (aka "big bang") cosmology starting at a cosmological time of some small fraction of a second, preceded by

an inflationary epoch lasting for an unknown amount of time (but at least 50-60 e-folds), preceded by

there may be no way to know, because inflation erases any hints of what came before.

Times quoted as "after the big bang" are ΛCDM model times and don't count from the true beginning of time. There's a common misconception that if inflation ended $10^{-32}$ seconds "after the big bang", that it must have lasted for no longer than $10^{-32}$ seconds. In reality, it has to last longer than that in order to accomplish its goal of producing smooth starting conditions for the ΛCDM epoch, and in most (all?) specific models it lasts for far longer than the minimum time. There are eternal inflation models in which it lasts forever.
We don't know how old the universe is. We only know that the current era of (non-inflationary) expansion started about 13.8 billion years ago.


Is our sun in a star cluster? Sorry for the absolute begginer question here, but is our sun a part of some globular cluster?
It is something related to Virgo supercluster?

No the sun is not part of a cluster.

There are several types of clusters that we see in the sky. The most familiar is the "open cluster", like the Pleiades. These are a group of stars that formed together and have remained close.

As the stars drift apart they can become part of a "moving group", a collection of stars that don't appear to be a cluster, but since they share the same age and direction of motion we can tell they used to be a cluster.  Many nearby stars are part of the Ursa Major moving group, but the sun is not one. It just happens to be in the same part of the Milky way. The sun was probably part of a cluster shortly after it formed (4.6 billion years ago) but that cluster has long ago broken up. We don't (yet) know of any other stars that seem to have come from the same cluster.

Globular clusters, like M13 and Omega Centauri, are larger and have many more stars tightly packed together. They are all rather distant, and the brightest look like slightly fuzzy stars (in fact Omega Centauri was originally thought to be a star).

Of course a galaxy is a group of 100 billion stars. We don't normally think of galaxies as a star cluster, because they are so much bigger and the stars in them don't form at the same time.

Then there are clusters of galaxies, the Virgo cluster is a cluster of galaxies, and the local group of galaxies is on the edge of this cluster. But the Virgo cluster is not a star cluster.


Why doesn't Earth's leading tidal bulge cause the Moon to start spinning in the opposite direction than its original spin? Please refer to the image below:

My question is, why doesn't Earth's leading tidal bulge (encircled in the green circle 1) pull on the moon's tidal bulge (encircled in green circle 2), leading to a force $F_{1,2}$, and consequently a torque on the Moon's tidal bulge, leading it to start spinning as shown in the image. This spinning direction would be opposite of the original direction of the Moon's spin, which got retarded and eventually stopped due to tidal interactions over billions of years.
Please note: I know that Earth's tidal bulge causes the Moon as a whole to increase its velocity which is causing the Moon to be drifted away from the Earth, but that is not what my question is.

why doesn't Earth's leading tidal bulge (encircled in the green circle 1) pull on the moon's tidal bulge

It does....

leading to a force, and consequently a torque

There's nothing here that will give an ongoing torque.
Because of the Earth's high rotational speed and lag in response to forces, the earth's bulge stays displaced from the direction of net gravitational pull.
You have drawn the moon's bulge directly in line with the Earth's center.  But in fact it would be pointed at the net gravitational pull direction, which includes the earth's bulge.  Since the moon's rotation does not carry its distortion away from that line, there is no net torque in either direction.  (I'm ignoring libration and long-term changes in rotation due to orbit raising).
So, the Earth's bulge is continuously displaced due to rotational mismatch, while the moon's bulge is not.  If the moon's bulge were directed at earth's center, then there would be a temporary torque and the moon's orientation (in this non-inertial frame) would appear to pendulum back and forth, not a consistent torque that would get it spinning.  (Again, I've completely ignored libration effects and assumed a circular orbit for simplifications).


Relation between mass of gas, mass of stars and dark matter halo mass As we all know that Mass of the galaxy can be written as the sum of the mass of gas(Mg), the mass of stars (Ms), and mass of dark matter halo(Mh), as
M(galaxy)=Mg+Mh+Ms,
But is there any formula which relates Mg and Mh and, Mh and Ms??
Does anybody know?

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


where is the boundary between dark night and grey night Is there a clear cut between them?

How long and how bright will the moon be for a dark night?

I googled but could not find a definition for them. People just talk about them empirically?

What a nightmare - there is no accepted definition.

A common method is to use the fraction of the night (between the times of astronomical twilight) that the moon spends above the horizon. The problem is that this does not correlate perfectly with fractional lunar illumination or sky brightness. It does provide a good method of deciding when a night is "dark", but you get a  bit of a mixed bag of "grey" nights.

It is found that the sky brightness is pretty much determined by fractional lunar illumination (FLI - the fraction of the moon's visible hemisphere illuminated by the Sun) once it exceeds about 0.15; so this is the definition that is currently used at the Isaac Newton Group of Telescopes.

Dark $0\leq$ FLI $<0.25$; Grey $0.25\leq$ FLI $<0.65$; Bright $0.65 \leq$ FLI $\leq 1$, where the FLI is calculated at 0h UT (close to midnight at the ING).

See  http://www.ing.iac.es/PR/newsletter/news6/tel1.html for further discussion (in which I only just noticed I got a mention in the acknowledgements ;) ).


Reasons for orbital inclinations I see that there was already about Pluto's orbital inclination, but I would like to ask a more general question.

Why do planets have orbital inclinations?  And is there any particular reason, other than chance, for Mercury's inclination being so small?

Thank you!

The planets formed from a disc of dust and gas. Since the orbiting the proto-sun formed into a disc the planets, which formed from this matter are in a disc too, but not a perfect disc. There was some scattering of the matter in the primordial disc, and so the planets are also slightly scattered. Interactions between the planets have also scattered the planets out from the disc too. The formation of the planets caused them to have small, but not exactly zero inclination

The Earth has the greatest inclination (except pluto) however, by convention, we measure inclination relative to the Earth, all the planets have a positive inclination.

So there is no special reason that the planets have slight inclinations. It is just what would be expected due to random effects during their formation.


On a lunar map, why is it labelled East on Earth, West on Moon? source
Why is east and west reversed for the moon? Why does north and south remain the same?
Please explain like I'm five.

Imagine you're lying down outside, looking up at the sky, with your body aligned so your head is pointing North, and your Feet are pointing South. If you look to your Left, you'll be looking East, and see the East part of the ground and sky.  If you look to your right, you'll be looking West, and see the West part of the ground and sky.
Now imagine a friend  holds a globe above your head, centered on your home city, with the north pole pointing North, and the South pole pointing South.
When you look at your country on that globe, the parts East of your city will be on the right side of the globe, and the points west of it will be on the left side of the globe.
The same thing happens with the Moon. When you're standing on the Earth, looking up at the Moon near the meridian,  from the Northern Hemisphere (so it appears generally south of you), The parts of the Moon that are Lunar-Surface-East of the Moon's center will appear on the right side, and the parts of the Moon that are Lunar-Surface-West will appear on the Left.


Fate of lone spiral galaxies? I know Hubble's galaxy classification scheme and the bifurcation from elliptical galaxies into two types of spiral ones. I am also aware of different theories on how spiral galaxies form e.g. through collision with another one. I was always wondering in how much any kind of galaxy morphological classification actually relates with the dynamics of a galaxy. The following paragraph of the Wikipedia article seem to support my uncomfort:

To this day, the Hubble sequence is the most commonly used system for classifying galaxies, both in professional astronomical research and in amateur astronomy. Nonetheless, in June 2019, citizen scientists through Galaxy Zoo reported that the usual Hubble classification, particularly concerning spiral galaxies, may not be supported, and may need updating.

Differently put: Would a lone elliptical galaxy eventually become a spiral one? And if so, how to define the transition point? For me, it does not appear obvious how to set a threshold after which density fluctuations within an elliptical galaxy deserve to be called spiral arms. More technical: Which confidence band around the average radial density distribution of a given galaxy has to be breached in order for the galaxy to be called spiral?
Related

Evolution of galaxies with time
How small can a spiral galaxy be?
How do disk-shaped galaxies then form at the sites of spherical galactic halos? Do the halo stars start orbiting or just sit in place?

To start, if I'm reading your question correctly you've got the general galaxy evolution model backwards, at least as far as morphology goes. At a very high level the picture goes like this: most if not all (large) galaxies form as spirals, then at varying points in their lives they merge with other large galaxies (either ellipticals or other spirals) and the product becomes an elliptical galaxy. This happens faster and more often in dense environments like galaxy clusters, and takes longer to occur for more isolated galaxies like the Milky Way (which will merge with Andromeda in a few billion years).
Question on isolated spirals
If what you were asking about is isolated spirals, well no I don't think an isolated spiral would ever turn into an elliptical galaxy by itself (at least not on life-age-of-the-Universe timescales). It may eventually start to resemble an elliptical galaxy (in that its stellar halo may grow) depending on how much it gets harassed by other small dwarf galaxies. No galaxy is truly isolated, there is always a hierarchy of smaller galaxies nearby and they gravitationally interact with the host galaxy. For example the Milky Way has the Sagittarius dwarf galaxy, and the Large and Small Magellanic clouds which will merge with us soon (it's basically happening now for Sagittarius, and the Magellanic clouds will be a few billion years). None of these events will be impactful enough to turn the Milky Way into a true elliptical, that will only occur with it merges with Andromeda.
Question on morphology versus internal dynamics
To specifically answer your other question on how morphology relates to the internal dynamics of a spiral galaxy: it most definitely does, but the relationship is complicated. When you see a spiral galaxy with a central bar, or spiral arms, these features are being generated by dynamical instabilities within the galaxy. This could happen thanks to a number of causes, which may operate from within the galaxy itself, or from outside the galaxy. One of the most common means of exciting a bar or spiral arms in a disk galaxy is by interactions with smaller dwarf satellite galaxies that orbit around the larger spiral galaxy and gravitationally perturb it. If this part of the answer sounds like a cop out, well it sort of is. Exactly how disk perturbations like bars and spiral arms are started, how they work, what the heck they do to the galaxy, these are all truly bleeding edge research topics that astronomers are working on right now.
Hope that helps!


The multiverse spacetime paradox? If there are indeed multiple universes, where do they exist? We "know" that outside our universe, consisted of spacetime, there is nothing - i.e. 0 dimensions, no space and no time. So "where" are other universes, how would the distance between our space and other be measured inside nothing?

Read Edwin Abbot's book Flatland.

In this story we visit a vision of a two dimensional "Universe" embedded in a higher dimensional space. Now by analogy we may suppose that our Universe is embedded in a higher dimension space with plenty of room for other (3+1)-Universes.

I do not claim this is a viable model of "reality" just a thought experiment to show "where" other universes might be if there are indeed multiple universes. Also Flatland is a fun read.


Apparent size of the super blue blood moon Is it normal that super blue blood moon is much smaller than before the eclipse? If so, why? I'm quite positive it's not cloudy here. On my TV, they broadcast a live super blue blood moon from the 4th floor, and it looks big and red. You think they use a telescope?

Yes, they probably use a telescope if it's much bigger.

The 'super' refers to the fact that the moon is closer to the earth than normal during a full moon (it's near to the perigee in its orbit), but the human eye cannot see the difference between the moon size an hour before and an hour after the perigee.

Also, there's a phenomenon called moon illusion which makes the moon appear larger near the horizon, which might or might not be the case in your situation.

Incidentally, the fact that it's a blue moon has nothing to with it; that's just the moment it takes place on the calendar.


How exactly 'polar aligning' a Telescope with equatorial mount helps? What is true purpose of polar aligning? Last night, I looked at Mars. It was moving fast.  I still need to adjust RA axis to track fast-moving Mars. Why the telescope didn't  track it by itself?

The Earth's rotation makes the stars and planets appear to move westward across the eyepiece field of a stationary telescope at about 900 arcseconds per minute.
An equatorial mount can counter this by moving only on its polar axis at a constant rate.
An altitude-azimuth mount must move on both axes at variable rates in order to track.

In early October 2018, Mars appeared to move about 1.1 arcseconds per minute relative to the stars.
In an equatorially mounted telescope, properly aligned and tracking at the sidereal rate, Mars would take about 13 minutes to move through its own apparent diameter of 15 arcseconds.
Without a tracking motor, you'd have to move the RA axis by hand, but you could leave the declination axis alone for a while.

If the mount has a tracking motor designed to be usable in either hemisphere, it may have a switch to reverse direction.
If that switch is in the wrong position, stars appear to move westward twice as fast as if the motor were switched off.

If an equatorial mount is misaligned, stars appear to drift in declination, depending on their position in the sky, up to 15 arcseconds per minute per degree of polar alignment error.
For example, if you're in the northern hemisphere and the mount's polar axis is 5 degrees too far to the east, objects near the meridian appear to drift southward at 75 arcseconds per minute, whereas objects 4 hours off the meridian drift only half that fast.
If the polar axis is more than 30 degrees off the celestial pole, it might as well be an alt-az mount.


very bright object (visible even after sun risen) Brevard Co. FL. At first I thought it was a comet it appeared to have a tail. Starts out low in the east and gains elevation as the sun rises. As the sun is just above the horizon it is still visible at about an 80 degree elevation. What am I seeing?

This is the planet Venus.  It is always bright (when visible) but exceptionally so at the moment.  Venus shows phases (similar to the moon) and is currently nearly half illuminated.  That, combined with its relative closeness (about 75million km) is why it is particularly bright at the moment.

It is often visible in broad daylight, but can be difficult to spot because of the bright sky background.


What is the 'scale factor' equation for a dark-matter dominated universe? The Friedmann equations can be solved exactly in presence of a perfect fluid with equation of state
$${\displaystyle p=w\rho c^{2}} \qquad p=w\rho c^2$$
where ${\displaystyle p}$ is the pressure, ${\displaystyle \rho }$  is the mass density of the fluid in the co-moving frame and $w$ is some constant.
In spatially flat case ($k = 0$), the solution for the scale factor is
$${\displaystyle a(t)=a_{0}\,t^{\frac {2}{3(w+1)}}}$$
where ${\displaystyle a_{0}}$ is some integration constant to be fixed by the choice of initial conditions. This family of solutions labelled by ${\displaystyle w}$ is extremely important for cosmology. E.g. ${\displaystyle w=0}$ describes a matter-dominated universe, where the pressure is negligible with respect to the mass density. From the generic solution one easily sees that in a matter-dominated universe the scale factor goes as
$${\displaystyle a(t)\propto t^{2/3}}$$ matter-dominated
Another important example is the case of a radiation-dominated universe, i.e., when ${\displaystyle w=1/3}$. This leads to
$${\displaystyle a(t)\propto t^{1/2}}$$ radiation dominated
Note that this solution is not valid for domination of the cosmological constant, which corresponds to an ${\displaystyle w=-1}$. In this case the energy density is constant and the scale factor grows exponentially.
So, '$a$' is proportional to $t^{2/3}$ or $t^{1/2}$ for matter- or radiation-dominated universes, respectively... But if '$w$' is negative-one then '$a$' is proportional to $t^t$?
I mean, what is the exponent in this 'exponential growth' phase where the '$w$' 'constant' is $-1$?

You answered your question actually. You have said

Note that this solution is not valid for domination of the cosmological constant, which corresponds to an w=−1

then you are saying

But if 'w' is negative-one then 'a' is proportional to tt

The $w=-1$ does not apply to $a(t) \propto t^{2/3(1+w)}$