Should only spherical satellites be considered 'moons'? Since Titan and Ganymede fall into the same category as Deimos and Phobos or the components of planetary rings, the category of moons, it's like if all asteroids were considered planets. Tiny irregular satellites are called 'moonlets' but still fall into the category of moons.

Is there any distinction between the natural satellites and should there be one? From my point of view, only equilibrium-shaped spherical moons should be considered moons. While natural satellites are all satellites, the spherical ones should be moons. Otherwise it would be extremely weird that Deimos and Phobos fall into the same category as our Moon or the Galilean Moons. For example, Pluto should be considered having five natural satellites, and one of them (Charon) is a moon.

The need to distinguish between these potato-shaped and spherical moons hasn't arisen. As such, there aren't two different words to designate these two types of moons. And as long as the community of astronomers who study natural satellites doesn't feel the need for there to be an extra category, there likely won't be one created preemptively.

Perhaps, one day, we will discover that potato-shape and spherical satellites form in very different ways, of have very different properties (not just esthetic). Then the case might be made to sort them into two different categories.

On a side-note, defining arbitrary categories is never trivial. How spherical is spherical? The boundary has to be set somewhere, but there always will be edge cases where an object could reasonably fall in either category.


How can I interpolate 2 orbits My question is very similar to this one although the answer is not what I was looking for.
Basically I'm working in a space game (KSP) and I'm making it multiplayer. The clients send a packet with their orbital parameters at a interval (30 miliseconds by default) but between those packets I want to interpolate and draw an orbit between those 2 packets.

For normal vectors or speed I use linear interpolation:

private static float Lerp(float v0, float v1, float t)
{
    return (1 - t) * v0 + t * v1;
}

But It doesn't work well when I work with orbits

var inclination = Lerp(inclination, Target.inclination, lerpPercentage),
var eccentricity = Lerp(eccentricity, Target.eccentricity, lerpPercentage),
var semiMajorAxis = Lerp(semiMajorAxis, Target.semiMajorAxis, lerpPercentage),
var LAN = Lerp(LAN, Target.LAN, lerpPercentage),
var argumentOfPeriapsis = Lerp(argumentOfPeriapsis, Target.argumentOfPeriapsis, lerpPercentage),
var meanAnomalyAtEpoch = Lerp(meanAnomalyAtEpoch, Target.meanAnomalyAtEpoch, lerpPercentage),
var epoch = Lerp(epoch, Target.epoch, lerpPercentage),

What function can I use to interpolate them?

Several options are possible:

Don't interpolate at all. Just assume the planet stays fixed until the next update. The planet Earth moves $10^{-7}$ degrees in 30milliseconds. In the vastness of space that is a rounding error.

Given two objects, their position and velocity and acceleration, and the values of their Keplerian elements at times t0 and t1, you want to estimate their position, velocity and acceleration at any time in between.

You can use simple linear interpolation of their position. This has the advantage of being very quick, but it assumes a constant velocity. Over short time-spans this is a reasonable assumption. For most orbiting bodies this approximation should be accurate enough for differences in time of many minutes, as the changes in velocity are relatively small.

You can assume constant acceleration and interpolate using the SUVAT equation. This does very well if the objects don't move too far. For example it can model the flight of a ball in the Earth's gravitational field very well, giving a parabolic shape. This should give a fairly good approximation of the position over hours, or even days.

You can interpolate the orbital elements, linearly, and solve Kepler's equation to get the position of the body at any given time. For most bodies the Orbital elements change only slowly with time (except for the Anomaly) That is the shape of the orbit stays nearly the same, only the position on the orbit changes. Approximating the orbit of a planet by interpolation of its elements should give a good approximation for several hundred years of orbital time. However the process of solving the equation is slow.

Finally you can numerically integrate the n-body problem for the solar system. Depending on the time step you choose this can be accurate for an indefinite amount of time

So these can be summarised as:

Constant position
Constant velocity
Constant acceleration
Variable acceleration following an inverse square law.
Variable acceleration n-body graviational field.

For a game engine, speed is good. If the gaps in time are on the order of milliseconds (of real time) then simply not interpolating should be undetectable.


How much the moon movement can influence an equatorial mount setup? If I use an equatorial mount for my telescope to view the moon, is there a way to take in count the moon movement around the earth, or the movement is too small compared to earth rotation?

For most purposes, the difference between the Moon's motion across the sky and star's motion (the sidereal rate) is negligible. The Moon appears to move about 1/2 degree per hour relative to the stars, so while the stars travel 15 degrees per hour east to west, the Moon travels approximately 14.5 degrees per hour. Unless your mount is aligned accurately and has an accurate motor, you probably will not notice this difference.

Some telescope drives do have a "lunar rate" so that it will track the Moon more accurately. But these approximations do not take into account the fact that the Moon is also moving north or south in declination as it orbits the Earth; the lunar rate only compensates for its west to east motion. (Maybe a goto scope will track the Moon more accurately since it theoretically could be calculating the position of the Moon and pointing to a changing position.)


Sample bias' contribution to "Planet 9" hypothesis, why was this not thoroughly addressed before? Science Claim for giant 'Planet Nine' at Solar System's edge takes a hit
arXiv No Evidence for Orbital Clustering in the Extreme Trans-Neptunian Objects

The brightness of distant solar system objects varies as $1/r^4$ and so for elliptical orbits objects are more likely to be discovered near periapsis.
If discovery telescopes are pointed in a biased direction, resulting discoveries will show a bias in orbital elements due to this pointing bias.
That's a rough summary of the recent paper.
Question: Why was this sample bias not thoroughly addressed before? Was there perhaps insufficient statistics until now for such an analysis, or were there other complicating factors?

I think the underlying premise of the question -- e.g., "Why was this sample bias not thoroughly addressed before?" -- is somewhat incorrect. Previous papers, including papers by those making the "Planet 9" claim, have attempted to address sample biases; a secondary issue is that the new paper uses data unavailable to previous studies (and also ignores some of the previous data).
To begin with, the original paper by Batygin & Brown (2016) does include a brief discussion of the possible effects of selection biases (in Section 2). This was, admittedly, pretty cursory, but they have followed it up with more extensive analyses (Brown 2017; Brown & Batygin 2019). From the abstract of the latter paper: "To determine if observational bias can be the cause of these apparent clusterings, we develop a rigorous method of quantifying the observational biases in the observations of longitude of perihelion and orbital pole position. From this now more complete understanding of the biases, we calculate that the probability that these distant KBOs would be clustered as strongly as observed in both longitude of perihelion and in orbital pole position is only 0.2%."
So Napier et al. are not, as you suppose, the first to consider selection biases (as they indeed acknowledge). What, then, is the difference? Part of it is that Napier et al. argue for performing an analysis using detailed simulation taking full account of individual survey characteristics if they are known. In order to do this, they focus on objects found in three recent surveys. This means that they deliberately ignore the 6 objects used for the original claim of Batygin & Brown: "The six ETNOs considered in the Batygin & Brown (2016) (BB16) analysis were discovered in an assortment of surveys with unknown or unpublished selection functions, making it difficult to establish that the observed angular clustering was indeed of physical origin." Instead, they use 14 objects "detected by three independent surveys with characterized selection functions, all published since BB16."
(Note that the Batygin & Brown 2019 bias analysis also used 14 objects, but not the same 14 -- they included the original 6 objects that Napier et al. exclude, but did not have access to objects detected after their study, which Napier et al. do. We are in a situation where there are very few data points, and there is the potential for divergent results based on the small number statistics of different studies.)
Eric Jensen has already posted links to some Twitter discussions by Batygin on this topic; I can point you to a Twitter thread by the other original author (Mike Brown), as well as a recent blog post by him which tries to understand why they and Napier et al. get such different results.


What is the frequency distribution for luminosity classes in the Milky Way Galaxy? I'm working on a game concept that does some mild simulation of realistic stellar classes and luminosities. In particular, I'd like to roughly model the general frequencies of the classes and luminosities of the stars in the Milky Way.

Several sources, including Wikipedia's entry on stellar classifications, show a chart that includes the frequency distribution for spectral classification: the OBAFGKM categorization. So that's fine.

What I'm having trouble finding is any frequency distribution chart similar to that one but for the Yerkes luminosity categories: Ia+, Ia, Iab, Ib, II, III, IV, V, sub-dwarf and dwarf. I have a copy of the Hipparcos database, which contains a "Spectral Types" field, but it's highly incoherent text. Still, I could write some code to parse the values in that field to try to get a rough count of luminosity categories in those roughly 116,000 stars... but I'm a little perplexed that no such chart appears to exist already somewhere in Internetland. (Either that or my search-fu is weaker than usual.)

If anyone can point me to a chart of the frequency distribution for the luminosity categories noted above, or suggest a reasonably simple way for me to calculate those values myself, I'd appreciate it.

EDIT: Out of curiosity, I went ahead and did my own simple parsing of the spectrum fields from the Hipparcos dataset.

Out of 116472 rows, only 56284 (fewer than half) provided luminosity class data in the Spectrum field. Those 56284 rows broke down this way:

Ia0     16     0.03%
Ia     241     0.43%
Iab    191     0.34%
Ib     694     1.23%
I       17     0.03%
II    1627     2.89%
III  22026    39.13%
IV    6418    11.40%
V    24873    44.19%
VI      92     0.16%
VII     89     0.16%

Note: Around 1000+ rows gave an either/or value for luminosity class (e.g., "M1Ib/II"). In these cases, I counted only the first value provided. This probably skewed the results slightly compared to counting both luminosity classes.

I'm still very curious to know whether anyone else has produced or located a similar table of frequencies for the luminosity classes, if only to see how my very trivial analysis compares.

Here is how you do it "properly" for the Hipparcos data. As Warrick correctly points out, what you have done in your question is massively biased towards giant and supergiant stars, which actually form the very small minority of stars.

You must form a volume limited sample. To do this, sort the stars by distance (1/parallax) and choose a cut off point. Your sample will always be incomplete, but the greater your distance cut off, the more incomplete it will be and it will become incomplete for more intrinsically luminous stars.

You have a philosophical problem to solve here in terms of what you are trying to achieve. The vast majority of stars in the Galaxy are faint M dwarfs with an absolute magnitude of $>10$. As Hipparcos is only complete to around magnitude 10-11, you will only get these M dwarfs in your subsample if you limit yourself to 10pc. But you will find that this sample contains no evolved stars (too rare) and no white dwarfs (too faint).

EDIT:
This has piqued my interest again, so I have a practical (approximate) solution, based on a two-part process. The first part involves a paper I wrote (actully an undergraduate experiment) based on the nearest 1000 stars to the Sun (from the Gliese & Jahreiss catalogue CNS3). This sample is roughly complete down to mid M-dwarfs, so everything else I say, and the results I give, only applies to a sample of stars more massive than that.

If you look at this volume-limited sample of 1000 nearby stars you can straight away say something about the relative numbers of different types of stars in the Galactic disk (saying something about stars anywhere else in the Galaxy is filled with much more uncertainty). A colour-magnitude diagram is shown below, and from this we see that:

The Sun is among the brightest of stars - brighter than 95% of other stars.

About 6% of the population are white dwarfs (though a few faint, old white dwarfs may still be missing from the sample). This makes sense. If you integrate a typical mass function assuming only stars more massive than about 1$M_{\odot}$ have even had time to become white dwarfs, then this is what you get.

Only 0.9% of the population are giants. The reason for this is that only a small proportion of stars are massive enough to have evolved into giants on the lifetime of the Galaxy. But once they are there, their lifetimes are short compared to the main sequence phase and most have become white dwarfs (see above).

There are a handful of objects, maybe 0.5%, that could be classified as subdwarfs, between the main sequence and the white dwarfs.

So, in broad terms: 92.5% of stars (above $\sim 0.2M_{\odot}$) are main sequence (class V), 6% are white dwarfs, 1% are giants (class III) and 0.5% are subdwarfs (class VI).

There are no very massive stars or supergiants in the immediate solar neighbourhood. That is because they are very rare. To get a better estimate we need to look at a larger volume limited sample. To do this I took all the stars (about 7000) from the Hipparcos catalogue that are closer than 50pc and assumed this was complete to just below a solar luminosity and assumed that those stars with an absolute magnitude brighter than the Sun (1949 stars with $M_V<4.5$ represent 5% of the total population in this volume. The absolute magnitude vs colour diagram for this sample is shown below.

Of these 1949 luminous stars, I find roughly 190 are giants - giving a giant fraction of $5\times 190/1940=0.5$%, in reasonable agreement with the nearby star sample based on smaller numbers. There are still NO supergiants, even in this larger sample. Thus supergiants have a frequency of $\leq 5 \times 1/1949 = 0.0025$%. i.e. less than 1 star per 40,000 is a supergiant.


Why is there a supermassive black hole at the center of every galaxy? According to https://en.wikipedia.org/wiki/Sphere_of_influence_(black_hole) the gravitational sphere of influence of a supermassive black hole is really limited compared to the size of its hosting galaxy, why then is there one at the center of every galaxy?

As some of the comments pointed out, there isn't a definite answer since this is still ongoing research. However, there are some things that might explain it:

Galaxy formation: Current theory says that dark matter is non-uniformly distributed in the early universe. Once radiation and baryonic matter decouple, matter flows towards the dark matter overdensities. We know that stars (Pop III) must form very early during this, because their light reionizes the universe. These early stars are thought to have a mass of up to 1000 solar masses. It's possible that some of the infalling gas collapses straight into a black hole.
Density profile: A galaxies' density rises sharply towards the centre. So, during the formation of the galaxy, the gas rushing inward might have formed a black hole without going through the process of star and supernova. Also, this means there's enough to feed the black hole once it's formed.
Dynamic friction: Imagine a self-gravitating system of heavy and light objects (like a star cluster or, indeed, a galaxy). A heavy object passing through an ensemble of lighter ones will lose some of it's kinetic energy to them, which means it will "sink" towards the centre of the gravitational well. This is well observed in globular clusters. Problem: The timescale for this to happen is far too long to matter for large galaxies.
There seems to be a correlation between the mass of the central black hole and the mass of the elliptical galaxy or the bulge of the spiral galaxy. So, some process makes sure that the central black hole gets to grow along with its galaxy.

Coming back to your question: There are two parts to it, one: why a black hole at all, two, why at the centre. The second part is simpler, it would get there eventually, and probably (let's say in a galaxy merger) quite quickly. For the first part, there is no definite answer but the points above are some indications as to why it's not too surprising that there should be one.

As a last point, black holes at the centres of other galaxies are not at all easy to find. Unless the black hole is active (quasar), there are at best very indirect detection methods (velocity dispersion in the central part etc.).


Black Holes, how do they form? Do black holes form by when the massive star dies and all the gravity falls in on itself, or something else. Please let me know, I'm really desperate!!

There are several major types of black hole, and each forms differently:

Quantum black holes - these aren't currently proven to exist, but many scientists think they are likely. These are microscopic black holes, which are believed to have formed in the Big Bang, though they can be created wherever there is enough energy.
Stellar-mass black holes. These are the remains of dead stars, and the main type of black hole people talk about. They form when very large stars collapse to the point where even light cannot escape at the end of their lives.
Supermassive black holes. These, as their name suggests, are extremely massive - with masses of millions to billions of times that of the Sun. Scientists aren't quite sure as to how these formed, but the main theory at the moment is that they started as several stellar-mass black holes, left behind by the first generation of stars, which fell into the centre of galaxies that were forming around them. As they did so, they came together, to form one large black hole. Another theory is that they started life as 'dark stars', a theoretical type of star held together by dark matter, which could be far larger than normal stars. These would have been extremely short-lived, and collapsed to create a supermassive black hole.


What are the periods of Saturn's rings? I don't mean necessarily a granular chart of all of the different rings at different distances, but is it days? Weeks?

Hours to days.

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

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


Local sidereal time interpretation I am new to this domain and trying to understanding the fundamentals of local sidereal time.
Through an online available calculator https://www.iiap.res.in/personnel/reks/software/javascript/calclst.php ,I am noting down the local sidereal time for my longitude of interest. I have repeated the exercise for same time for five consecutive days. What I observe is that the local sidereal time is advancing by 0.06/07 hours per day for the same time.Like 21.88 Hours for current date and time , 21.94 Hours for the same time next day ,22.01 Hours for next to next day and the pattern continues. Kindly let me know how this increment is happening?

If you extrapolate, you'll find that those ~4 minutes per day add up to 24 hours per year.
The difference is due to the Earth revolving around the Sun at a rate of almost 1 degree per day.
For every 365$\frac{1}{4}$ solar days, there are 366$\frac{1}{4}$ sidereal days.

Civil time is based on mean
solar time,
in which the Sun crosses the
celestial meridian
every 24h 0m on average.
In sidereal time,
24h 0m is the time between successive meridian transits by any given star, equivalent to 23h 56m 4s of solar time.

Sidereal time also corresponds to the
right ascension
of stars crossing the meridian at that time.
If a star in Orion has right ascension 5h 30m, then Orion is on the meridian at 05:30 sidereal time, which occurs 2 hours earlier in the solar day per month.


Do comets contain any significant amounts of iridium? I have Googled "iridium content of comets" and can't find anything that suggests comets contain iridium in any detectable quantities. Yet, whenever I read an article about the Chicxulub impactor, the article invariably says something to the effect that it may have been created by an asteroid or a comet. But Alvarez and company first offered the impact hypothesis for the K-T extinction by observing an iridium layer in global sedimentary deposits, which could only have come from an asteroid impact, since iridium is extremely rare in the Earth's crust. Later the Chicxulub structure was identified as the probable site of the asteroid impact. While a comet could have caused the Chicxulub crater, it could not have been responsible for the global iridium layer. So why do science writers offer the comet alternative? If Chicxulub was created by a comet, then another site must have been the source of the iridium layer.

I also googled "iridium content of comets", and the first result was https://news.dartmouth.edu/news/2013/04/dartmouth-researchers-say-comet-killed-dinosaurs

Now there is currently no consensus on the nature of the Chicxulub impactor. The observations of iridium and osmium suggest an asteroid. A minority opinion is that a comet may be responsible. The authors of the article in the linked news report suggest that measurements of iridium concentrations from ocean cores may be erroneously high, due to sorting in marine sediments. If these cores are excluded, the amount of iridium measured is lower, and consistent with some models of the composition of comets.

So, comets do contain some iridium. The amount of iridium is not certain. Some models of comet nuclei contain more dust and less ice, and so have more iridium. The exact amount of iridium in the KT boundary layer is also uncertain. The low end of estimates would be consistent with a cometary impact. The nature of the impactor is not yet settled science, but the balance of opinion currently favours a stony asteroid.


What is the cycle for when the maximum amount of the Moon's apparent area is sunlit? The Moon is defined as full when its ecliptic longitude is 180$^{\circ}$ distant from the Sun's.

But what is the cycle for when the maximum proportion of the moon's apparent area is sunlit, and for observers at what latitudes of the Earth's surface is this maximum realised?

Notes

1) After first guessing (on this expert Q&A site) that the answer is "one lunar month" and asking me whether there was "a reason" why the said cycle might be different from it, user UhOh has requested clarification of the term "moon's apparent area", pointing out that the moon has a rough surface. The difference of the apparent perimeter of the Moon's disc from a circle does not have a bearing on the answer to this question that is of anything like the magnitude of the contribution made by the 5$^{\circ}$ tilt of the plane of the Moon's orbit around the Earth relative to the plane of the Earth's orbit around the Sun. Please ignore mountains and craters. Please assume either that the Moon is a sphere, or if you wish then take it to be a spheroid with an eccentricity of about 1/900. The question is about the relative movements of the three celestial bodies concerned. (If further clarification is required of any other term used, please let me know.)

2) Wouldn't it be great if there were a website where you could input your latitude and longitude and desired start date and end date and get a graph of the proportion of the moon's apparent disc that is sunlit from your location? But I don't think this exists (yet!)

I asked the same question on the Usenet group sci.astro and received the following helpful reply from Barry Schwarz that goes some way towards an answer, reposted here with permission. But the question remains hard. I have not seen this cycle addressed in any astronomical textbook, although it may be in one that I haven't looked at.

  Since the Sun's diameter exceeds the Moon's, the Sun always
  illuminates slightly more than 50% of the Moon's surface.  (To
  visualize, draw the external common tangents from a larger circle to a
  smaller one.)  Using 98M miles as the distance, 432K miles as the
  Sun's radius, and 1K miles for the Moon's, the Sun illuminates almost
  7 miles into the opposite hemisphere.

  Any observer a finite distance from the Moon can only ever see less
  than 50% of the Moon's surface. The further away the observer is, the
  closer the observable percentage is to 50.  (To visualize, draw the
  two tangents from an external point to a circle.)  However, when the
  Moon is in the plane of the ecliptic, an observer at the north pole
  can see approximately 12 miles over the Moon's north pole into the
  opposite hemisphere.  So he can see some 5 miles beyond the
  terminator.  Similar geometry causes his view of the Moon's south pole
  to fall short of the terminator.

  However, for a person standing in line with the ecliptic, his view in
  any direction falls 4 miles short of the terminator and that is the
  maximum you seek.

  When the Moon is within 4K miles of the ecliptic, an observer on Earth
  at the same offset from the ecliptic will have the view described
  above.

  When the Moon is more than 4K miles away from the ecliptic, all
  Earth-bound observers see some non-illuminated portion and therefore
  do not see the maximum you want.

  Not only is the Moon's orbit tilted with respect to the ecliptic, but
  the direction of the tilt wobbles like a child's top.  I have no idea
  how to factor that into an equation that would compute the  time
  between one visible maximum and another.


Primordial angular momentum? Angular momentum, and in particular the conservation of angular momentum, is of course very relevant in many processes in astrophysics, such as e.g. star- and planet formation, and disk formation in galaxies.

The concept of objects spinning up upon collapse is straightforward enough to understand, but this begs the question: where does the primordial angular momentum come from?

Tidal interactions between haloes, I suppose?

Your hypothesis is correct: it is the tidal interaction between neighbours which generates spin.

Think of your proto halo (galaxy, stellar disc etc) as an ellipsoid (set by it's inertial tensor). Should the matter around it not be spherically distributed, it will apply a torque onto that ellipsoid and make it spin. The assumption in cosmology is that even though the primordial angular momentum you mention may be null, tidal interaction between neighbouring proto structures is sufficient to allow then to exchange (hence acquire individually) angular momentum, while preserving a zero sum.

A Reference on this subject is given by this paper

This diagram may enlighten the issue: it represents the tidal torquing of the blue ellipsoid by the tides represented by the pink ellipsoid.

PS: the expression spin up is unfortunate it seems because it is in fact spin conservation at the later stage of collapse of the proto halo which makes it rotate faster while preserving it spin (= total internal angular momentum).


How do I get focussed images with my T-adaptor and 6mm eyepeice? I have a Celestron 5se + canon 650d. They are mated with the celestron t-ring and the 93625 t-adaptor.
I want to take magnified photos of planets with the camera. But if I add an eyepiece to the t-adaptor, everything gets blur.
Here are some scenarios that highlight the problem:

look at Jupiter with 25mm eyepiece in telescope: clear and focused image
look at Jupiter with 6mm eyepiece in telescope: clear and focused zoomed image
look at Jupiter with camera and no eyepiece inside the t-adaptor, in telescope: clear and focused image but small.
look at Jupiter with camera and 6mm eyepiece inside the t-adaptor, in telescope: large unfocused ball

I dont know how to solve this.
Clearly, the 93625 t-adaptor was made to accept an eyepiece and lock it with the thumbscrew, but why does it not result in focused images?

Using the eyepiece, and no camera lens, is a very different configuration to the camera+tube+scope (case 3).  The configuration you're after is "Eyepiece projection".

I don't know exactly what you have available, but I believe you will need to fit a T-threaded extension tube to the external thread on your T-adapter. Then fit the camera onto that.

Note - before you go out and buy a new extension you can probably do a hand-held test of this by the following:

Attach the adapter to the scope, with a low-power eyepiece in the adapter (low power makes the test easier to see.)
Point the scope at anything distant (you can do this in daylight, but NOT at the sun of course!)
Now hold the camera, by hand, a few inches behind the
eyepiece+adapter. Perhaps you can use a cardboard tube to keep the stray light out.
See if you can adjust the scope's focus, and the camera position, to get a sharp image.

If this worked, then you can get a T-threaded extension tube to go on the back of your current adapter, to hold the camera in position.

See for example this link:
http://www.astronomysource.com/2011/10/13/eyepiece-projection/


Which telescope would be better with a DSLR? I'm interested in astrophotography and since I haven't left my home in six weeks, my interest in photographing the skies has been growing. I'm on a tight budget and I've found these two options:

A Meade Infinity 70mm (f/10), a cheap refractor.
The Omegon Maksutov MightyMak 80mm, only for 20€ more.

The Meade has a T2 mount, the Omegon uses 1.25" but comes with a T2 adapter; meaning in both cases I'd need the T2 to Canon adapter.

I would also like to be able to use it for terrestrial photography once the lockdown is over, and that's a big factor in my case, even if the results are not excellent. I know that for astro the mount is extremely important, but I'm not thinking about deep sky, mostly planets and the Moon and, of course, terrestrial.

In most forums they always talk about expensive options, but these cheap ones are not often discussed for obvious reason.

Both of these look OK to me. The biggest plus in the Omegon looks like the extra 10mm. If you're on a budget, taking a cheaper refractor and later taking a decent budget tracker (e.g. SkyTracker Pro) is a good move. Gradually building up your set of gear is key. Good luck!

ps.
Don't think you're going to take "budget photos" because you're on budget gear. Good amateur astropics don't necessarily come from people with the best gear.


Accurate way to programmatically generate a main sequence star? I am working on a (personal) programming project, part of which involves generating random stars and their parameters.

In order to create a procedurally generated star, I want to randomly select as few parameter values as possible, and then extrapolate the rest using formulas.

For instance by interpreting the Stefan-Boltzmann law I can derive the star's luminosity based on the surface temperature and the star's radius:

$$
L = 4 \pi R^2 \sigma T_{e}^4
$$

There are however other parameters that I would like to generate, such as the mass of the star. What seems to be able to help me is the mass-luminosity relation:

$$
\frac{L}{L_{\odot}} = \left(\frac{M}{M_{\odot}}\right)^a
$$

By knowing $a$ (based on my randomly selected mass) it seems that I can derive the star's luminosity. By glancing at the Hertzsprung–Russell diagram (e.g. this one) I can see that I can perhaps somehow calculate the radius of the star, but I'm not entirely sure how.

There is an existing star "generator" which can be found here, however this one interpolates existing data using logarithms.

My full question would be, is there a way to "generate" a "regular", main sequence star, starting from one or two parameters and extrapolating the rest using formulas? Is there a sensible relation between the mass of the star and its radius that can be used to generate a "realistic" star? Or would most parameters need to be interpolated from existing data?

This is a tricky question, not immediate.

If we talk about Main Sequence stars, yes, you can constrain your fictional star by two parameters. The crucial parameters in the diagram, as you said, are the mass and the radius. This is why the HR diagram well represents the stellar population with two axis. Of course, for your scope, we need to do some assumptions. For example, the metallicity, that for your purposes can be fixed as the Solar one. Or, that the initial mass is already in a condition to contract.

You can see it this way: the luminosity is connected with the radius (because it depends on the irradiating surface), while the temperature is connected with the mass (the more is the mass, the more is the pressure at the center of the star).

If you have $L \sim RT$ and $L \sim M$, then you also have $RT \sim M$.
The degeneracy from one parameter can be overcame if you have your reference star (the Sun), which allows you to trace isoradii lines on the HR diagram. See here for example:

Or, that is equivalent, you can express your first equation (Stefan-Boltzmann), in Solar units as well (as shown here).

Try it out just by your self: choose a mass, or observationally speaking, choose a luminosity. Then trace your line on the HR diagram, and find your star. That star has a defined temperature (spectral class), Mass and Radius as well.

If you need the exact equations, feel free to ask please.

Nice aim, anyway. Have fun!


Can the supernova remnant SN 1572 be observed by amateur astronomers? The SN 1572 remnant, also called Tycho's supernova remnant, is beautiful in X-ray images. It seems to be rather dim in visible light. Are there any amateur photographs of this object? How long are the exposure times needed to get a nice picture?

As you say, SN 1572 is not very bright in the optical.  There are some Hα regions that have been observed with world-class optical telescopes, but they do not look like the X-ray and infrared images that you normally see.  In fact, images from the Palomar Optical Sky Survey 2 (with a limiting magnitude of ~22) do not reveal any nebular emission from this object at all:

Therefore, the optical emission must be fainter than 22nd magnitude.  I do not think there are amateur photographs of an object this faint and I believe a CCD on an amateur telescope would saturate from noise before revealing any of the nebular emission.


Biosignatures for anaerobic or other early microbiology? What are some possible atmospheric bio-signatures for a planet with anaerobic life? or even microorganisms before oxygen builds up in the atmosphere? Would there be a way of telling if methane, etc. is possibly not from abiotic causes?

Anaerobic life has much less impact on the atmosphere than aerobic life. Anaerobic life could exist in the deep sea and have no gas exchange with the atmosphere at all.

Methane would get everyone excited. Biotic methane might be expected to have a different isotope mix compared with abiotic methane. But it is hard enough to distinguish isotope mixes when you have a rover on a planet.

More complex compounds would be very exciting, but a very sensitive detector would be needed to find the very low concentrations required.

It may be that the marker of abiotic life is the detection of anomalously low CO2 levels. Most abiotic atmospheres have very high levels of Carbon dioxide, but life will progressively reduce this level. A low level of carbon dioxide, in an atmosphere containing water, especially if coupled with methane would look bery interesting.


Has any of the orbiters or rover on Mars captured image of comet NEOWISE? We are seeing stunning pictures of comet NEOWISE from earth (and ISS too), are there any pictures captured from Mars by an orbiter or rover, or any other spacecraft?

Not yet, as far as I can tell.  The comet looks very different from Mars and while I haven't done the math, I don't think it has an apparent size from the rovers' perspectives to make it look like anything more than a dot.  Since Neowise is moving away from the Sun, the length and brightness of its tail is diminishing rapidly.


How good is the HST compared to ground telescopes? The Hubble Space Telescope (HST) is a remarkable device. It produced (produces) images whose resolution and clarity were a step change ahead of anything that could be captured from the ground. Is this still the case, or have adaptive optics and ever larger mirrors caught up with the image quality the HST can achieve? Is it still true that the HST's images are better than anything produced on Earth?

Adaptive optics techniques have only been successfully applied in the far red and near infrared. HST produces images with spatial resolution equivalent to ground based adaptive optics images, but it works at near infrared wavelengths and visible wavelengths.

The spatial resolution of HST at visible wavelengths has not really been matched with ground based telescopes. "Lucky imaging" has shown some promising results, but even here, this is usually applied at far red wavelengths.

Except now see this How did VLT's adaptive optics obtain this resolution for Neptune? Is it really working in visible wavelengths? Progress occurs! (Though it now seems my statement above should only be slightly modified, since the picture contains no information shortward of 550nm and is largely dominated by light at 600-920 nm).


Is the Object in this Photo Mars? A little bit of context:
I dabble in amateur astronomy, however only very lightly. I originally wanted to make some observations of Mars because I know it is supposed to be very close these days but unfortunately in the Netherlands where I live we are having very bad weather. However, I think I spotted it today in the evening coming home (sadly I did not managed to get home quickly enough for a telescope/binoculars) and it was quickly gone behind clouds. However, I am not sure if it actually was Mars. It seemed too bright. My place is close to very active airport (Schiphol) and sometimes with just naked eye planes can look like celestial bodies (when they move away from me it looks like they don't move at all and I only had chance to look at the object for 2-3 min).
Information about Location & Object:
I took photo of the object with my phone (see below).
It was very bright - I recently (during summer) did some observations of Venus & Jupiter and this object outshined them both (although they were not visible today due to clouds so I am comparing it to my imperfect memory).
I immediately looked up my location on my phone: 52.294 N 4.879 E elevation 0 meters and the local time was 21:42 GMT+2.
The object was in the direction 110° East.
Now, I already looked it up and Mars should be in that direction, but I worry it might just have been a plane.
Furthermore, it looks like it is little bit too high. When I did observations of other planets in the past they were much lower in the sky. I know that the position changes due to Earth's rotation and orbit around the Sun, but I can't remember seeing them so high up and as far as I understand planets are all roughly around the ecliptic (although this was not place where I usually go stargazing so I might just be getting confused). Hence I would like to know, is the object in the picture Mars or we can't tell or it is something else?

Is the Object in this Photo Mars?

Almost certainly yes if it didn't move like a plane or hover and make noise like a quadcopter.
in-the-sky.org' skymap shows Mars at 117 degrees East and 25 degrees altitude at 21:43 local time in Woerden in central Netherlands yesterday. Clicking on Mars yields more information including a charts showing that it's close to opposition.
I took a photo from about 25 degrees north almost the same day I think :-) It was higher in the sky for me but just as bright and similarly cloudy. The big photo is intermediate focus. The two below it are best and bad focus. Even when defocused to 200 pixels in area it show brightly, when in focus it is saturated and so just looks like a few white pixels.


During an eclipse, how big is the shadow of the moon on the earth? This picture was taken from the ISS during a solar eclipse.

You can see the shadow of the Moon on the surface of the Earth. But how big is this shadow? How many kilometers is its diameter?

But how big is this shadow? How many kilometers is its diameter?

That's a photo of the umbra and penumbra on the surface of the Earth taken from Space. It's a little distorted because it's not directly under the ISS but far off near the terminator.

It's hard to pin down the size of the penumbra because it's fuzzy and fades near the edges, but if you could see the very edges then it would be twice† the diameter of the moon or very roughly 6900 kilometers total.

†Actually this is only true by coincidence because the angular diameter of the Sun happens to be the same as that of the Moon. @Flaffo's answer does a good job of explaining how to calculate the diameter of the umbra, and that math could probably be extended to calculate the diameter of the penumbra as well.

The umbra has a well defined diameter but the size varies a lot due to variation in the distance from the Moon to the Earth since it's orbit is not circular. Sometimes the Moon is so far away that it can't fill the Sun and there is no umbra at all, that's called an annular eclipse.

Wikipedia says:

  Typically, the umbra is 100–160 km wide, while the penumbral diameter is in excess of 6400 km.

Source: Geometry of a Total Solar Eclipse

You can see an example of a very detailed simulation of just the umbra moving across the Earth's surface in the NASA Goddard video Tracing the 2017 Solar Eclipse The precise 3D shape of the Moon generates the shadow and it then moves over the contour of the Earth's topography. If you think the shape is weird I agree! See answers to The Moon's shadow could not possibly look like this — could it? (also see What are the “Moon L, B, C” angles shown in this solar eclipse simulation?)

Here's a screenshot:


Is there any way to distinguish the doppler redshift from a distant galaxy from the probably much greater cosmological redshift? Is there any way to distinguish the doppler redshift from a distant galaxy, caused by its initial recessional velocity relative to this galaxy, from the probably much greater, cosmological redshift, caused by the expansion of space?

The two effects are observationally indistinguishable for an individual galaxy. The redshift associated with universal expansion is often interpreted as a recession velocity and there is nothing in a spectrum that tells you what fraction of the redshift is contributed by which effect.
If a distant galaxy is associated with a cluster of galaxies, then you can make an attempt by assuming that the average recession velocity of the cluster represents the cosmological redshift and that the perturbations from that for individual galaxies are their peculiar velocities.
Nevertheless, the cluster will have it's own peculiar velocity as an additive term to all these individual peculiar velocities. Fortunately, that doesn't really matter when it comes to studying cluster dynamics.
Since the cosmological redshift increases with distance (Hubble's law), whereas peculiar velocities do not, then at large distances cosmological redshift can be assumed to dominate. See What does "located in the Hubble flow" mean?


How did matter first come into existence? How did matter come to be?
Why is there matter if matter and antimatter cancel each other?

Actually, I also believe that before anything should have come into existence nothing should have been in existence. So everything should have got created out of nothing as we get +ve numbers and -ve numbers from zero as follows.
0 => + 4 - 4 or 0 => +5 - 3 - 2
Wherein, the +ve numbers represent the matter and the -ve numbers represent antimatter. Of course though splitting of zero can be explained but how we explain it in physics is an unresolved mystery.


How are rogue planets discovered? Planets are usually found by observing a star and waiting for the light level to drop when a planet passes in front of it, but what about rogue planets that don't have host stars?

The most successful methods are by gravitational microlensing and direct observation in the infrared or far infrared.

The transit method doesn't work well for rogue planets, because usually at least three transits are needed to confirm a planet.


What would happen if space-time began to contract? Assume that the expansion of the universe stops and it starts to contract again. What would happen?

Would only distances between objects get smaller, or are there other effects we can expect?

Given that the expansion of the universe shows us that the distance between galaxies (or rather moreso between galaxy clusters) is increasing, without any change in the distances within or between atoms, or between stars within galaxies (apart from normal stellar motion), and that galaxies themselves are also not expanding, we could assume that a contraction would manifest as a decrease in the distances between galaxies/galaxy clusters.


Does Light Year is their past? I just wanted to know if what we can only see out-side of our planet is just the past of other objects?

If 1 Light Year is equals to 365.25 Days, then I saw an object outside our planet that is 1 Light Year away from us, does it means I just saw its past 365.25 days ago?

Light year is a measure of distance. Light in vacuum travels at a (large) finite value, $3 \cdot 10^5$ kilometers per second to be almost exact. We can exploit this speed as a function of time. If an object is 5 light years away, it means light must travel at that large speed for 5 years to reach Earth from where it left.

The Sun is about 8 light minutes away. It takes light 8 minutes to reach Earth from the Sun. The sunlight you see is from 8 minutes ago, as it was when it left its source — the Sun.

If there were a telescope powerful enough to clearly see objects hundreds of millions of light years away, and if some hypothetical advanced alien race used such a telescope right now from hundreds of millions of light years away, they would see dinosaurs on Earth since that light had enough time to reach its destination. We aren't visible to them yet because our light is still traveling on its way to destination.


Air temperature range recommendation for observations in a summer night with an entry level telescope? Recent nights where I try to make my first observations with the entry levell telescope Seben with 114 mm mirror it seems that for lenses less than 6 mm details go lost due to warm air refraction (as observed with the Moon contour).

Everywhere they say, it shouldn't be too warm: is it possible to recommend a rough temperature range where you could go for say 4 mm and still see details?

(Side note: the overall visibility index was about 77%).

The air is only part of your problem here.  There are several factors involved.

it is absolutely true that the air can cause a loss of detail in the image you see.  But even if there's no atmosphere between you and the moon or anything else you're looking at, there is only so much detail you can see.  Technically speaking, you can magnify as much as you want ad infinitim.  But there is only so much USABLE magnification  you can get out of a telescope.  This is due to the diffraction of light.

When the light from the moon or any other object enters your telescope tube, it begins diffraction.  If you want to better understand the physics behind it, there's a fairly easy-to-undersatand Khan Academy video about it.   Essentially, what is happening is interference patterns form among the light waves and the result is a blurring of detail.  This blurring is less noticable at lower magnifications, and gets worse the more you magnify.  While the specifics are a little different from the result of blowing up a picture, the effect is similar.  I put together this example of the effect.  In my example, I started with a picture I took of the moon in May or June and each image in the series is doubled in size then cropped.  I go from a full image of a waxing gibbous moon to zoom in on the crater Herschel.  As the magnification increases, you start to see the image getting fuzzier and fuzzier.  In the case of a picture, this is because there is only so much data to work with, depending on the geometry of the image sensor in the camera. Technically, it is also the case with the light that there is a fininte amount of information/data to work with, but the mechanics are a bit different.  In the end, the result is the image blurs past a certain point and you can't get a clearer image out of it.

(Technically speaking, using speckle interferometry, a camera CAN increase detail when multiple images are captured and analyzed together, but there's still a finite limit, and this doesn't apply at all to visual observing.)

Your telescope has an aperture of 114mm, as you indicated.  I did a quick search for Seben 114mm telescopes and the only ones I found showed a focal length of 1,000mm.  As it turns out, it appears this is not a true Newtonian telescope, but a Bird-Jones Newtonian (or Jones-Bird, depending on who you ask), which is actually a type of catadioptric design and has some inherent issues, which I'll circle back around to.

They eyepieces I found listed that come with this telescope include an SR4mm and a H6mm. The SR4mm is a symmetrical Ramsden design eyepiece with a 4 mm focal length.  The H6mm is a Huygens design with a  6mm focal length.  The Huygens eyepiece design was invented in about 1662 by Christian Huygens, and was an improvement upon the original basic single-lens eyepiece design of the first Lippershey and Galilean telescopes, and Kepler's minor improvements in 1611.  Huygens added a second element and produced an eyepiece that provides better eye relief (the distance from the eyepiece to your eye, and less chromatic aberration (the separation of colors of light caused by refraction.  These eyepieces, however, have fairly narrow apparent fields of view and perform poorly in telescopes with focal ratios under f/10.  The next major upgrade in eyepiece design was Jesse Ramsden's invention in 1782.  Ramsden's eyepiece provided better apparent fields of view, but decreased eye relief and increased chromatic aberration.  When used in longer focal lengths, this is less of a problem,b ut still a problem.  The Ramsden eyepiece is also susceptible to "ghosting" which is caused by reflections within the eyepiece.  These eyepieces aer fairly inexpensive to manufacture, however, which makes them common in lower-cost telescopes such as yours.

laying aside, for the moment, the inherent issues with these eyepieces, the sizes you specified are 4 mm and 6 mm.  Based on the design types, I am going to assume an apparent field of view (AFOV) of about 40° for both.  WhenI do the calculations, I find that the 6 mm gives you a magnification of about 167X with a TFOV (true field of view) of about 14 arcminutes. The 4 mm gives you 250X magnification with a TFOV around 10 arcmin.

As I mentioned above, however, aperture limits usable magnification.  In particular, it limits the angular size of details you can resolve.  A 114 mm aperture can provide details no larger than 0.93 arcseconds in size.  At the distance to the moon (about 384,400 km) this means the smallest details you can see are about 1.73 km in size.  But this is the theoretical limitation.

In the mid 1800's, William R. Dawes produced a method of calculating how far apart two stars must be to be able to determine that they are, in fact, two separate stars. His formula, which provides a measure in arcseconds we call the Dawes' Limit, is one of the most commonly referenced ways to describe how small a detail you can see with a given telescope aperture.  For your telescope, that would be about 1.02 arcseconds.  By this measure, you could not see details any smaller than about 1.9 km in size on the moon.  A third measure, known as the Raleigh Criterion, which is also measured in arcseconds, suggests the smallest details visible in a 114 mm telescope would be about 1.21 arcseconds in size or 2.25 km at the distance of the moon.

Based on these formulae, then, if there was a particular crater on the moon that was about 2km in size, you could probably just make it out.  You could magnify this as much as you like, but you won't see more detail, just a blurrier and blurrier hint of a crater.  Eyepiece size/design won't change that, nor will your focus ability.

Depending on who you ask, for a given inch of aperture, you can only get about 25 to 60x of usable magnification - meaning that anything over that level is going to be so blurry as to not be worth viewing.  On a typical night, that would be toward the lower side, and on a night with very good air and sky conditions, toward the higher end.  Based on this, I'd say on a typical night you shouldn't expect more than about 135X magnification, and on a really nice night, maybe 225X.  Technically, the 4mm gets you much higher than that, and with the Barlow lens, double that (to around 500X) but there's absoloutely nothing you can do to improve the sharpness of the image - the limits of physics dictate that the telescope simply cannot do it.  Anything over 135X on an average night is likely to be fuzzy and get fuzzier still as you try to increase magnification.

But wait, there's more!

As I mentioned, your telescope isn't a true Newtonian.  A true Newtonian uses a mirror with a curvature defined by the formula of a parabola.  These kinds of mirrors are a bit harder to manufacture, so many manufacturers of lower-priced telescopes cheat a little and use a mirror with a spherical curvature.  The spherical curvature is a constant-radius curve, and this makes it MUCH easier to manufacture.  However, it introduces a problem known as spherical aberration  Luckilly, spherical aberration and most other aberrations are decreased by increasing focal length.  The Bird-Jones design uses a shorter-focal length spherical mirror and adds a correcting lens (almost identical to a Barlow lens) in the path of light to correct the errors. If you look into your focuser without an eyepiece, you'll probably notice a lens between your eye and the secondary mirror - this is the Bird-Jones corrector.  It adds more glass to the optical path. The more glass, the more chance for distortion and further aberration.  Additionally, this is not likely to be a very high quality optical component, making it more likely to produce a poorer image.  But worse, it makes the process of collimation very difficult indeed.

I'm guessing, though I may be wrong, you have little or no understanding of collimation.  Simply put, collimation is the proper alignment of the optical components in the telescope.  If the mirrors and lenses are not properly aligned, it introduces further errors in the path of light, which manifest themselves in image aberration and distortion.  When collimation is off, the image quality degrades and sharpness is lost.  Unfortunately, this built-in lens makes collimation more difficult, which means it's likely that your collimation is off somewhat, which will FURTHER degrade your image quality at higher magnifications.

So, what can you do?

Honeslty, not much.  While warmer air will cause issues inside the tube of the telescope, this can be reduced through acclimatization - letting the telescope come to ambient temperature.  If you can collimate your telescpe accurately, that will improve your image.  And replacing your eyepieces with better designs (even lower-price Plossl eyepieces are significantly better) will help. Even then, you're not going to get sharp detail at higher magnification.  It is simply not possible. To get sharper detail, you need a larger aperture and better air conditions.

My personal recommendation for a beginner telescope is an 8" Dobsonian. Though somewhat more expensive, the 8" aperture provides a lot more detail.  And a Dobsonian with a true Newtonian OTA is far easier to collimate and get a clearer image out of.

For now, keep using your 114, but accept its limitations.  And start looking to your next scope.

I would also recommend finding a local astronomy club you can join and learn more about how scopes work, how to maintain them, and ways to better enjoy your observing sessions.

Good luck and clear skies!


Can dark matter be found in the shape of planets, galaxies etc.? If dark matter has gravity just like normal matter, does that mean it can also form planets, solar systems and so on?
 Any answer will be appreciated.

Planets and stars, no. Globular clusters and galaxies, yes.
Small scales
To condense into such relatively compact objects as planets, stars, and even the more diffuse star-forming clouds, particles need to be able to dissipate their energy. If they don't do this, their velocities prohibit them from forming anything.
"Normal" particles, i.e. atoms, do this by colliding. When atoms collide, they're excited, and when they de-excite, they emit radiation which leaves the system, carrying away energy. In this way, an ensemble of particles can relax into a less energetic system, eventually condensing into e.g. a star. Additionally, the collisions cause more energetic particles to donate energy to the less energetic ones, making the ensemble reach thermodynamic equilibrium, i.e. all particles have the same energy on average.
Dark matter is, by definition, unable to collide and radiate, and hence, on such small scales as stars and planets, particles that enters a potential well with a given energy will maintain that energy. They will thus accelerate toward the center, then decelerate after its closest approach to the center, and finally leave the system with the same energy as before (if it was unbound to begin with). This makes it impossible for collisionless matter to form such small objects.
Large scales
On the scale of galaxies, however, various relaxation mechanisms allows dark matter to form structure. This is the reason that in pure N-body simulations of the Universe, such as the Millennium Simulation, you will see galaxies. The sizes of these structures depend on the resolution, but are measured in millions of Solar masses.
The relaxation mechanisms include:
Phase mixing
This is sort of like galaxy arms winding up, but in phase space rather than real space.
Chaotic mixing
This happens when particles come so close that their trajectories diverge exponentially.
Violent relaxation
The two mechanisms listed above assume a constant gravitational potential $\Phi$, but as the systems relaxes, $\Phi$ changes, giving rise to an additional relaxation process. For instance, more massive particles tend to transfer more energy to their lighter neighbors and so become more tightly bound, sinking towards the center of the gravitational potential. This effect is known as mass segregation and is particularly important in the evolution of globular star clusters.
Landau damping
For a perturbation/wave with velocity $v_p$, if a particle comes with $v\gg v_p$, it will overtake the wave, first gaining energy as it falls into the potential, but later losing the same amount of energy as it climb up again. The same holds for particles with $v\ll v_p$ which are overtaken by the wave.
However, particles with $v\sim v_p$ (i.e. that are near resonance with the wave) may experience a net gain or loss in energy.
Consider a particle with $v$ slightly larger than $v_p$. Depending on its phase when interacting with the wave, it will be either accelerated and move away from resonance, or decelerated and move closer to resonance. The latter interact more effectively with the wave (i.e. be decelerated for a longer time), and on average there will thus be a net transfer of energy from particles with $v \gtrsim v_p$ to the wave. The opposite is true for particles with $v$ slightly smaller than $v_p$
You can read more about these mechanisms in Mo, Bosch, & White's Galaxy Formation and Evolution.


What is length of the universe on lights perspective According to Einsteins theory of special relativity, when something is travelling close to light speed, travel path is contracted. When the traveller is light itself, does it see the travel path contracted to zero? In another word, does light see length of the whole universe is zero across its travel path? How?

Sort of.  The Lorentz factor is

$$
\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}
$$

whereupon a stationary object in the stationary reference frame of length $L$ has a length of $L' = \frac{L}{\gamma} = L\sqrt{1-\frac{v^2}{c^2}}$ in the moving reference frame.  As the velocity $v$ increases toward $c$, we get

$$
\lim_{v \to c} L' = L \lim_{v \to c} \sqrt{1-\frac{v^2}{c^2}} = 0
$$

That is to say, if a spaceship were to move very rapidly from here to the center of the galaxy and back again, a clock aboard that spaceship would advance much less than an initially synchronized clock here at home.  From our point of view, the reason is that the clock on board the spaceship slowed down (time dilation).  From the point of view of someone on the spaceship, the reason is that the distance to the center of the galaxy was shortened (length contraction).  And in the limit (which cannot actually be reached by anything with mass), the clock on the spaceship becomes arbitrarily slow from our point of view, while the distance to the center of the galaxy becomes arbitrarily short from their point of view.  Of course, we can't actually ask the photon how it experiences the trip (even in principle), since the trip is instantaneous for it.  So the way to express it is to say what the limiting behavior is, as $v$ approaches $c$.

Wikipedia has a useful article on special relativity.


How do I calculate the positions of objects in orbit? In a model of the solar system, given the Sun is at the point $(0, 0, 0)$, an given the six orbital elements for each object in orbit ($a$, $\epsilon$, $i$, $\Omega$, $\omega$, $M_0$), how can I calculate the relative position to the sun at a given time $t$?

You must solve the Kepler's equation:

$M = E - e \sin E$

(where $M$ is the mean anomaly, $e$ the eccentricity of the orbit, $E$ the eccentric anomaly).
As you can see, this is a transcendental function, so you will probably need to iterate through a calculator, in order to solve it.
Many of useful formulas to understand the parameters dependence are reported on the Wikipedia page as well.

Here is another page with some examples. Just use the formulas and try to play with numbers, at some point you will realize at least orders of magnitude of the binary system elements.


What is Propeller Effect exactly? I find the original reference seems to be here. I still do not understand the mechanism. It is necessarily occur in neutron binaries? Anyone can introduce it in detail? This paper is old.

The magnetic field of the rapidly rotating neutron star interacts with the material coming from the other star in the binary. This results in a transfer of angular momentum, spinning the neutron star down but accelerating the material out of the system. The effect is somewhat like a garden sprinkler seen from above, with material flung out ("propelled") of the system.

I think there needs to be a quite delicate balance between the binary period (or separation) in the system and the strength of the magnetic field and rotation rate of the neutron star to form the propeller effect. It seems to happen most in the early stages of the X-ray binary when the neutron star is rotating rapidly. After the neutron star spins down by the torques exerted on it, the material from the secondary star in the binary forms either an accretion disk or is directly accreted onto the neutron star down the magnetic  field lines.

The effect has also been seen, possibly only the once, in Cataclysmic Variable  (CV) systems, which consist of a low-mass star (typically a K or M dwarf) and a magnetized white dwarf. The "poster child" for this type of propeller system in CVs is AE Aquarii A magnetic propeller in the cataclysmic variable AE Aquarii; Wynn et al 1997

The propeller theory seems to be alive and well as there over 800 citations to the original 1975 paper. A recent highly cited review discussing propellers and accreting neutron stars seems to be Alipar 2001


How far apart are stars in Terzan 5 cluster? I've just learned about the Terzan 5 globular cluster, which according to Wikipedia has one of the highest star densities in our Galaxy - $10^6\ M_\odot/\mathrm{pc}^3$.

Problem is, I've no idea how to interpret this number.

So, the question is, what are the distances between stars in that cluster on average?

You have to assume a characteristic mass for the stars in the cluster. Usually, to get an approximate I go with stars of mass $1 M_\odot$.

Then, given the density of $10^6\ M_\odot.\mathrm{pc}^{-3}$ you have a density of stars of $\rho_\star = 10^6\ \mathrm{pc}^{-3}$. In a box of $1\ \mathrm{pc}^{3}$, you have $10^6$ stars, so the characteristic distance between stars is simply given by the density of stars to the minus one third:
$$ d_\mathrm{carac} = \rho_\star^{-1/3} \approx 10^{-2}\ \mathrm{pc} \approx 2000\ \mathrm{AU}.$$

Addendum: if you want to do a more correct calculation, you should take into account the different masses of stars to estimate the correct number of stars, using for example a Chabrier initial mass function and try to get the mass function at the age of the cluster. Because of the fact that stars of about $\geq 1 M_\odot$ reach their red giant branch and lose much of their material in less than the age of the cluster, it will decrease the mean star mass of the cluster to something between $0.1$ and $0.8 M_\odot$ ($0.3 M_\odot$ according to Rob Jeffries). However it won't change the order of magnitude of the mean distance between stars.


Does time slow down because the universe is expanding at an accelerating rate? If the universe is expanding at an accelerating rate, such that the galaxies' moving away from each other is accelerated, then time should also slow down.
And when universe will accelerate to the speed of light, then time should stop.

Does time slow down because universe is expanding at an accelerating rate?
Can any physicist explains if this is true or not?

Yes, time does run slower for far-away objects, as observed from our point of view; this is a prediction of general relativity. And yes, because expansion accelerates, this time dilation slowly, very slowly, becomes more pronounced (this would happen even if the expansion didn't accelerate, but just continued at the same rate).

This time dilation  is a well-known effect, and is always taken into account when doing observations. For instance, when observing distant supernovae, one is often interested in how their luminosities decrease as a function of time. This is called their lightcurve. In order to compare lightcurves at different redshifts, they are usually converted to their restframe, i.e. how they would look if you were "standing next to the supernova" (e.g. Goldhaber et al 2001).

However, time dilation does not work exactly as you seem to think. At galaxy at a redshift of $z$ has its time dilated by a factor of $1+z$, so time runs twice as slow for a galaxy at, say, $z=3$ than for a galaxy at $z=1$. Galaxies with redshifts larger than $z\sim1.5$ recede faster than the speed of light, and time does not at all stop here. Only for $z\rightarrow\infty$, i.e. at the beginning of time at Big Bang, does the time dilation approach infinity.


Does the Abell 1689 supercluster have angular momentum? The Abell 1689 is a large group of galaxies, containing a lot of elliptical galaxies :

https://en.wikipedia.org/wiki/Abell_1689

Usually, an elliptical galaxy don't have large angular momentum around its center of mass (spin angular momentum, if you prefer).

My question is about the galactic group, globally as a whole.  Does it have angular momentum around its center of mass, or should it be negligible ?

I have not studied this galaxy cluster specifically. However, I can speak about systems like galaxy clusters that have a large number of bodies that are in close enough proximity to have a non-negligible gravitational effect on each other. For any such n-body system, where n is large (a solar system, a galaxy, a galaxy cluster,...), it can be assumed that, 1. there is a net angular momentum of the system, and 2. it is reasonable to assume that this angular momentum is non-negligible. I state 1 and 2 based on the following (very general model of solar system and galaxy cluster formation).

Such n-body systems form by the collapse of a very large amount of gas, dust, and other minerals. When fully extended, these clouds are cold, the the velocities of any given molecule or atom in them is comparatively slow. However, it is very improbable that the total angular momentum of the cloud of gas is zero, and more importantly, that it is close to zero. From this point, it is easy to reason that, so long as no angular impulse is applied to the system and no mass is added or removed from the system, the gas cloud can collapse or disperse in whatever way is consistent with the conservation of angular momentum, and the angular momentum of the system will not change. Thus, a galaxy cluster like Abel 1689 should have a non-negligible angular momentum.

But I have skipped over the most important logical step in the above argument. I simply assumed again that the angular momentum of the gas cloud would not be even close to zero, effectively shifting the level of the explanation up one level. Unfortunately, the best answer to "why do the gas clouds have non-negligible angular momentum?" requires explaining where the gas clouds came from. The extent of my understanding here is that the gas clouds that formed galaxy clusters only existed during the very early universe. Once they collapsed into galaxies, there is no process that could have caused a galaxy cluster (or even a galaxy) to disperse so completely that it forms a new very low density gas cloud that can re-collapse. Thus, the gas that made the galaxy cluster must have had an angular momentum if the galaxy cluster system is to have angular momentum.

I do not know if the sum total of the angular momentum in the universe is said to be zero or some finite value. However, I do know that even if it is zero, localized regions of matter in a system that has zero angular momentum do not also need to have zero angular momentum. Thus, the Abell 1689 system could be one such local region with non-zero angular momentum.

Continuation: According to The Cosmic Perspective: Seventh Edition, when considering galaxies that have not yet collided with another galaxy, the initial angular momentum of the protogalactic cloud is a large contributor in determining if a galaxy will be spiral or elliptical galaxies.

Regarding elliptical galaxy clusters, TCP has this to say: "Observations of galaxies in clusters support the idea that at least some elliptical galaxies result from collisions and subsequent mergers. Elliptical galaxies dominate the galaxy populations at the cores of dense clusters of galaxies, where collisions should be most frequent. This fact may mean that any spirals once present became ellipticals through collisions."

Thus, TCP suggests in this paragraph that at least certain clusters contain dense cores of elliptical galaxies that resulted from collisions and mergers of galaxies that had markedly different angular momenta, resulting in a new galaxy that has much less angular momentum.

But this doesn't really answer your question either. I have looked also for scholarly articles to find some consensus on what the typical angular momentum characteristics are, and found very little. Most discussions of galaxy cluster formation discuss filament formation and collapse. The one article of note I found was from Tom Crawford at the University Of Chicago:
askanastronomer.org

"The most obvious and uncontroversial class of structures beyond the galaxy is the galaxy group or cluster. These structures consist of tens to thousands of individual galaxies that are bound by gravity to one other and orbit a common center. That center is not defined by some huge central object, as the Sun defines the center of our solar system, but instead by the center of all the mass in the cluster, including dark matter. In fact, the dark matter dominates the mass budget of galaxy clusters, outweighing the normal matter by about a five-to-one ratio."

This description suggests more than just "random" motions of gravitationally bounded objects (galaxies), but does not definitively say anything about the net angular momenta of these systems.


Why is Chandrasekar Limit expressed relative to Sun? In 1931, Chandrasekhar was able to show that there is a certain critical mass (Chandrasekhar's limit) beyond which a white dwarf cannot exist, since the electronic fluid at that point cannot support its weight no matter how compressed it is.

The core of such a star will simply collapse inward.

The critical mass, Chandrasekhar showed, is 1.4 times that of the Sun.

When I first heard this, I was totally confused. Sun is a star too. The Sun will eventually enter the Red Giant Phase, and then it will eventually become a white-dwarf.

How and why did Chandrasekar express his limit relative to Sun?

The mass of the sun is just a unit of convenience in a astronomy. The sun's mass and luminosity, in particular, are relatively easy for us to measure precisely, and when we're talking about stars, provide a convenient scale where the numbers won't be too "astronomical" (be too high a power of $10$ to picture easily). You could derive the Chandrasekhar limit in grams, kilograms, or slugs from the relevant physics, if you wanted. The relevant equation (from the Wikipedia article) is:
$$M_{\mathrm{limit}} = \frac{\omega_3^0 \sqrt{3\pi}}{2} \left(\frac{\hbar c}{G}\right)^{3/2} \frac{1}{(\mu_e m_\mathrm{H})^2},$$
where $\mu_e$ is the average molecular weight per electron (stellar composition dependent), $m_\mathrm{H}$ is the mass of hydrogen, and $\omega_3^0$ is a numerical constant that is approximately $2.018236\ldots$.

It should be noted that the Chandrasekhar limit is a limit on the mass of the final white dwarf, not of the object that will produce the white dwarf.


Relationship between temperature of nebula and size of star I was wondering, in the process of star formation, does the temperature of the nebula that produces a star play a role in the size of that star? I mean, it's only logical that the size would depend on the amount of matter around.But then, if the temperature is lower, that means that the net density is lower, so the star would 'absorb' less matter. Or is that not a factor, and the growing star would gobble up anything on its vicinity? I've encountered some contradicting sources and I'm trying to double check.

This is such a complex issue that I'm not going to try to be comprehensive. One major thing you are missing is that stars tend not to form in isolation, especially massive stars, but in groups and clusters.

The basic unit in star formation is the Jeans mass.
$$ M_J \propto T^{3/2} \rho^{-1/2}$$
where $T$ is the temperature, $\rho$ is the density and $M_J$ is the minimum mass for a cloud to gravitationally collapse.

So from this point of view, a higher temperature cloud must be more massive in order to collapse in the first place.

But as the cloud collapses, the density goes up and the Jeans mass decreases - and the cloud fragments into smaller sub-units and ultimately into a cluster of stars.

Exactly how this pans out in practice is basically the whole research field of star formation. It depends not only on relatively simple things like the temperature of the gas, its density, the equation of state; but also on much more complicated ideas like magnetic field support, competitive accretion between stars, feedback from newly-formed stars and protostellar outflows and the rate at which turbulence is generated and then dissipated.

There almost certainly is no simple relationship as you propose. In fact the outcome of star formation seems to produce a mass function (number of star per unit mass) that is only weakly dependent on the initial conditions, but may differ a bit in low density and high density environments.


How to calculate the point on the globe with maximum magnitude with given Besselian elements and given time for a partial solar eclipse? I am currently working on improvement of an Android app that calculates solar eclipses. For many days now I have not been able to solve a mathematical problem concerning solar eclipses, so I decided to try to ask for help.
I am working with the book "Elements of solar eclipses 1951-2200" by Jean Meeus. Calculating the central line of an eclipse from given Besselian elements already works good, so I am able to calculate the path of the central line of total and annular eclipses and the curves of equal magnitude.
My problem now is that Meeus only describes the calculation of the central line, so for a given time, that means for given Besselian elements I can calculate the corresponding point of the central line on the globe. Using these algorithms I can not calculate the line of maximum magnitude for partial phases of total and annular eclipses or for partial eclipses. So my question is:
How can I calculate latitude and longitude of the point on the globe with maximum magnitude with given Besselian elements for a time when eclipse is partial, that means, when no central line exists. As far as I understand the problem, this should be the point on the globe that is nearest to the shadow axis, which in the case of a partial eclipse or partial phase of an eclipse does not intersect the globe.
I would very much appreciate any help.

Jean Meeus, Morsels 2, chapter 69 talks about the Nelder-Mead simplex algorithm for minimising a function of two variables - in this instance magnitude as a function of latitude and longitude. Maximising the magnitude equates to minimising the negative of the magnitude. Apparently the method first appeared in magazine Byte in the 1960s, there are various references on-line.
Maximum magnitude always occurs near the polar circle, so try starting with three points equally spaced round the 60th parallel. They would have to be converted to Cartesian X Y with centre at the pole (giving an equilateral triangle), and the result converted back to latitude/longitude at the end. The sign of the element Y will make it obvious which polar circle to use.
Because the shadow is conic, not cylindrical, the resulting point will not (in general) be exactly on the horizon.


Creating a 2D matrix from 2 FITS images of unequal dimensions for Radio/FIR correlation I have two images of NGC 6946:

item       description        size (pixels)    scale (arcseconds/pixel)
 1.     70 micron MIPS FITS     241 x 241           4.5
 2.     20 cm VLA FITS          512 x 512           2.0

I need to create a matrix with 2 columns (70 micron vs 20 cm), where each column will store the values at each pixel in such a way that the central pixel coordinate of each image coincides, so that I can plot for Radio/FIR correlation.

How should I perform such task in python?

If you have the WCS for both images, you can use the astropy reproject module. Once you have projected image 1 onto image 2, the (x,y)th pixel from the output will correspond to the (x,y)th pixel from image 2.

Then you can flatten both images with numpy, and assemble them into a 2-column array.


Using parallax to find distance to the moon I want to use the parallax method (from scratch) to find the distance between the earth and the moon. To get a significant parallax, I need to choose a large baseline. I live in India and my friend in the US and so the baseline will be about $13,000$ Km. That's quite a large so I suppose I would get a significant parallax.
What I don't think of, is taking of angle measurement?

How can I measure the angle $\alpha$?

I am sure that measuring the angle compared with distant stars is the best approach, but you can in fact work out the angle even without looking at distant stars, if you assume the Earth is spherical and you know its radius, and you know both the length and the direction of the great circle between you & your friend.
Here's a terrible drawing:

So here, the radius of the Earth is $R$, and the great-circle distance between you and your friend is $l$.  So the angle between you and your friend is $\theta = l/R$ (radians).
Now, you need to agree both to measure the Moon's angle above the horizon at a particularly good moment: that moment is when it crosses the plane defined by you, your friend, and the centre of the Earth.  In other words, you need to measure its angle above the horizon as it crosses the great circle between you.  This is annoying, because you have to know the direction of the great circle, but it's also nice because you don't need a clock.
So, when you measure its angle above the horizon you get two angles, $\psi$ and $\phi$.  And now you have the quadrilateral I drew below the main diagram, and you know three of its angles and two of its sides.  The total angle in a quadrilateral is $2\pi$, so this tells you the fourth angle, which is $\alpha = \pi - (\theta + \psi + \phi)$.  And you can then go on to solve the quadrilateral for the distance to the Moon, $d$ (I am leaving this out because I can't remember how to do it!)
Note that this will only work if the Moon is above the horizon for both of you when it crosses the great circle: it may not be I think.


In my mid-forties, I believe I have seen Pluto for the first time with the naked eye Early this morning going out on the balcony, I looked up on a star chart app to verify it was Jupiter I was seeing.  Then I noticed the alignment of Mars, Saturn, and Pluto on the app.

Never being able to identify it before, I stared at where Pluto should be and I'm pretty sure I saw it.

My only question is - since it's said that planets shine and stars twinkle, it did seem that Pluto was flicking a bit.  Is this normal?  Something to do with the relatively low luminosity and greater length of space?

Pluto is something like magnitude 14. The limit for the human vision is somewhere between magnitude 6 (widely accepted) and 8-ish (highly trained observers with perfect vision in ideal conditions using special techniques - and it's a bit controversial anyway).

There's zero chance that was Pluto. It was definitely a fixed star.


Direction of Meteors Do meteors always streak in the same direction? While I know it's not likely to be exactly the same direction, but generally speaking. Would one say meteors always travel, for example, east to west?

Not in general, but sometimes there is a dominant direction.
When the Earth passes through a stream of debris from a comet or asteroid,
a
meteor shower
occurs, and those meteors appear to come from a point in the sky called the radiant.
Since the parent bodies have different orbits, different shower radiants appear in different parts of the sky.
For example, the Geminid radiant culminates earlier in the night than most, and the Perseid radiant is well north of the Ecliptic.
As the Earth can encounter meteors from any direction at any time, there are usually a few sporadic meteors unrelated to any active radiant.
Meteor observers have identified some
not quite random
directions for these.


Why can I see the whole moon during various non-full-moon phases? Here's an image that describes the phenomenon I'm asking about.  The very thin sliver to the left is of course the surface of the moon as lighted by the sun.  But the rest of the moon is also faintly lighted -- by what?

I'm guessing that some amount of "earthshine" -- sunlight reflected from the earth -- is the culprit here, but I've found no authoritative confirmation online. It would be great to have a link to a source that explains what's really going on.

The dark parts of the Moon are partially illuminated by "Earthshine". That is sunlight which is reflected from Earth to the Moon. Just like the ground on Earth is lit up a bit by the Moon at night, so is the ground on the Moon lit up a bit by the Earth in the lunar night. And even more so because Earth is much larger in the lunar sky, than is the Moon in our sky. And the surface of Earth is actually more reflective per area unit than is the surface of the Moon.


What are hot and cold stellar systems? My idea of the matter is: hot stellar systems are which form of objects which have a velocity standard deviation comparable to their mean velocity. Is this correct? If it is, is there a $\frac{\sigma_v}{<v>}$ boundary, which draws a line between hot and cold stellar systems?

I was wrong, hot systems are where members have velocities which have magnitudes in all directions in the same order (causing them to have roundish shape), while cold dynamical systems are where random motions do not have the same average magnitude in all directions (ie disc galaxies).


Are brown dwarf masses notoriously difficult to measure? The Next Big Future article Rogue Exoplanet 12.7 times bigger than Jupiter is 20 light years away

  SIMP J01365663+0933473 has a magnetic field more than 200 times stronger than Jupiter’s. The object was originally detected in 2016 as one of five brown dwarfs the scientists studied with the VLA to gain new knowledge about magnetic fields and the mechanisms by which some of the coolest such objects can produce strong radio emission. Brown dwarf masses are notoriously difficult to measure, and at the time, the object was thought to be an old and much more massive brown dwarf.

Is this true? Are brown dwarf masses notoriously difficult to measure? Naively, it seems to me if I had brightness at a few wavelengths I could deduce a temperature, and if I had a distance, at least for a blackbody I'd know the area as well.

Are one or both of these in fact difficult to get, or is there something else I'm missing?

Yes, they are difficult to measure. The only way to measure a brown dwarf mass is if it is in a binary system. The binary frequency of brown dwarfs is low - perhaps 10-15%.

Having found your binary system, then it has to be wide enough to resolve with your telescope, but not so wide that Kepler's third law means it takes centuries to do an orbit. The low mass of brown dwarfs does not help here. To have an orbit of say a decade or less means they have to orbit close to the other object. This in turn means adaptive optics are required.

Having said all that, the masses of quite a few brown dwarfs have been measured, so it is not extraordinarily difficult.

The object talked about in the article is not in a close binary, so it's mass cannot be measured, it can only be estimated indirectly. But here come the difficulties. To estimate the mass requires accurate calculations of the evolutionary tracks of brown dwarfs. The measurements of brown dwarf masses in binaries indicates that we do not have these. Secondly, to use the evolutionary tracks requires a luminosity (and therefore a distance), but because brown dwarfs are cooling objects, you also need an age. In general, unless the brown dwarf can be associated with a group of other stars with known age then this isn't possible.  Measuring the temperature of a brown dwarf isn't helpful because the cooling tracks are very close together in the Hertzsprung-Russell diagram, and in any case, it is very difficult because brown dwarfs are nowhere near blackbodies and their atmospheres are still poorly understood.

In this case, it seems that the brown dwarf has been newly associated with an (assumed to be) coeval group of stars at 200 Myr. Thus its luminosity can give a (highly) model-dependent mass. I cannot examine the details as the claimed published research cannot be found!


Is "Zero Shadow Day" necessary to perform Eratosthenes' experiment? Eratosthenes experiment to find the circumference of the Earth is usually said to be performed on the Zero Shadow Day at Syene. Is it necessary? Can this experiment be performed on any day for two locations lying on almost same longitude?

It's not necessary that there be zero shadow.  The altitude $\alpha$ of an object when it crosses the local meridian on the celestial sphere is $\alpha = \delta + 90^\circ - \lambda$, where $\delta$ is the object's declination and $\lambda$ is the latitude of the observation point.  This means that if you carefully measure the altitude of the same object at meridian crossing from two separate locations, and find altitudes $\alpha_1$ and $\alpha_2$, then the difference between these altitudes is
$$
\alpha_2 - \alpha_1 = \delta + 90^\circ - \lambda_2 - (\delta + 90^\circ - \lambda_1) = \lambda_1 - \lambda_2.
$$
Thus, the difference in the measured altitudes is equal to the difference in latitude between the two locations.  By measuring the distance between these locations on the Earth's surface, one can then infer the circumference of the Earth.
The condition that there be "zero shadow" simply means that one of $\alpha_1$ or $\alpha_2$ is equal to 90° (i.e, the sun is at the zenith.)  Effectively, it allows you to know what one of the angles is without physically measuring the height of your gnomon and the length of its shadow.  But so long as you're willing to measure the solar altitude in two different locations, it is not necessary that the sun be at the zenith.
Note that this method can be used to find the difference in latitude between any two points on the Earth, so long as the object's declination doesn't shift significantly between the meridian crossings at the two points.  The reason it's helpful to have the observations at two points on the same longitude line is that it leads to a simple relation between the great-circle distance between the points, the difference in latitude, and the radius of the Earth.  If the observation points aren't on the same line of longitude, then you need to know the difference in longitude as well to perform this experiment;  and Eratosthenes didn't have Harrison's chronometer at hand.


How many judges sat on Galileo's trial over heliocentrism and who were they exactly? I saw several chronologies of the trial online but most of them only mention the name of Urban 8 as involved in the actual court proceedings. How many judges were involved and who were they exactly?

Note 1. Reading the small print at wiki I found the names of the three of the ten cardinals who did not vote to convict: Francesco Barberini, Laudivio Zacchia and Gaspare Borgia.  I still don't know the names of any cardinals that voted to convict.

Note 2. Material related to the trial can be found in "Hidden agenda of the Galileo trial?" on the History of Science and Mathematics Stack Exchange site.

The seven who convicted were
Felice Centini, Desiderio Scaglia, Antonio Barberini, Berlinghiero Gessi, Fabrizio Verospi, Guido Bentivoglio and Marzio Ginetti. And as you note Francesco Barberini, Laudivio Zacchia and Gaspare Borgia did not. These three didn't vote to acquit, they were just absent from the signing of the condemnation - we don't know why

All information from The Galileo Case: Trial, Science, Truth page 142
By Mario D'Addio. Gracewing Publishing, 2004 ISBN 9780852446652


Were all first generation stars of same mass and size? If the density of the matter from which the first generation stars were made was homogeneous, so during the star formation nuclear fusion would have started around when they all reached to same critical mass and core temperature, once fusion started the outward pressure would have counteracted on inflow of matter so all stars should have been of similar size

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

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

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


Hulse-Taylor binary pulsar - what is the rate of mass/energy loss from the source? Following on from an earlier question about the very interesting Hulse-Taylor binary pulsar.  The high-frequency (radio) beam from the spinning pulsar sweeps across Earth about 17 times per second.  The total power of the gravitational radiation (waves) emitted by the binary system is calculated from GTR to be $7.35 × 10^{24}$ watts at present (declining as the orbital period and radius diminish).

Weisberg & Taylor, 2004 report that the beam has "a flux density of about 1 mJy at 1400 MHz."  (mJy = milliJansky, thanks Stan Liou). Assuming that this flux is uniform across a conical beam with cross-sectional radius 5 arc degrees  and assuming that the source is 21,000 light years from Earth and has mass = 1.44 Solar Masses:-  How much energy is emitted (per second) from the source in the beam? Also (if it is possible to estimate a reasonable range of values) what might be the rate of steady mass loss from such a source?

For reference, the relevant bit from the paper is:

  The observable pulsar is a weak radio source with a flux density of about $1\,\mathrm{mJy}$ at $1400\,\mathrm{MHz}$. ... Our most recent data have been gathered with the Wideband Arecibo Pulsar Processors (“WAPPs”), which for
  PSR B1913+16 achieve $13\,\mathrm{\mu s}$ time-of-arrival measurements in each of four $100\,\mathrm{MHz}$ bands, using $5$-minute integrations.

Those are millijanskys, aka milli flux units, so that the flux density is about
$$1\,\mathrm{mJy} = 10^{-29}\,\frac{\mathrm{W}}{\mathrm{m}^2\cdot\mathrm{Hz}}\text{,}$$
and hence the detected irradiance is on the order of $10^{-27}\,\mathrm{W}/\mathrm{m}^2$. Since we're about to make rather uncertain assumptions anyway, I won't bother worrying about doing more than an order-of-magnitude calculation.

  Assuming that this flux is uniform across a conical beam with cross-sectional radius 5 arc degrees and assuming that the source is 21,000 light years from Earth ... - How much energy is emitted (per second) from the source in the beam?

A spherical cap has surface area of $A = 2\pi Rh$, and here $R = 21\,\mathrm{kly}$. Now, I'm unclear what cross-sectional radius means if measured as an angle, but I take it to mean that the opening half-angle of the cone is $\frac{\vartheta}{2} = 5^\circ$, in which case
$$A = 2\pi R^2\left(1-\cos\frac{\vartheta}{2}\right) \sim 10^{39}\,\mathrm{m}^2\text{.}$$
Thus, the power would be $P\sim 10^{12}\,\mathrm{W}$, but note that in addition to the assumptions you've just listed, we're only talking about a particular radio band.