What would happen if we put a small part of White Dwarf or Neutron Star on Earth? I would know what would happen if we put a small part (let me assume 1 cm$^3$) of White Dwarf or Neutron Star on Earth. Will it explode or maybe it collapses to the Earth's core?

A piece of a white dwarf would explode, because atoms would expand to their normal size. This would hold for outer layers of a neutron star, too. For inner parts of the neutron star, that's not quite clear, but that kind of matter would probably decay rapidly, also resulting in an explosion.

If the material of a neutron star could be kept stable, it would fall down to Earth's core, up to the other side of the Earth, and oscillate a while that way, before being slowed down by friction, and stopped close to Earth's center. (1 cm³ neutron star matter has a mass of about 400 million tons.)

One cm³ white dwarf matter would have a mass of about 1 ton. Assumed it could be kept stable, solid rock could withstand this load.


Which star has the highest mass loss rate? We have known lists of stars with high luminosity, temperature, mass, size, and much more. However, I have not seen a list that compares each star's mass loss rates. So, which star (or stars) have the highest mass loss rates, and is it going to be more likely to be an LBV, RSG, or a WR star?

Some very hot and massive stars have strong stellar winds, known spectroscopically as "P Cygni profiles" after the type star. P Cygni 'erupted' in the 17th century to magnitude 3 and is now visible as a mag.5-6 star near the central star of the 'Northern Cross' that is Cygnus. An ordinary pair of binoculars will show it easily. So P Cygni stars have probably the highest mass-loss rate as I believe you are thinking of.


South of the tropics, where does the sun's noon shadow point? If I understand correctly, in countries to the north of the tropics the sun's noon shadow points north.

What happens south of the tropics? Is it pointing south?

Is it pointing south?

Yes, the shadow points south when you are south of the tropics (south of the Tropic of Capricorn).

In the tropics (between the Tropic of Cancer and the Tropic of Capricorn), the direction of the Sun's shadow depends on the location and the date.

As the angle between the sun and the earth's axis changes during the year, the apparent path of the sun in the sky changes.

Outside the tropics, this change is insufficient to ever bring the sun directly overhead.  Regardless of the date, the sun will be seen to be toward the equator at local noon.  (Same as saying the shadow points away from the equator).

Within the tropics, the changing path sometimes brings the sun directly overhead.  The position at local noon depends on the specific location and the date.

On the December solstice, the sun will be overhead for locations on the Tropic of Capricorn and will cross to the south at local noon for all locations in the tropics (and those further north).  During the June solstice, the situation is reversed for the tropics.  All locations south of the Tropic of Cancer would see the sun pass to the north.


How does a blackhole burp? Recently on the discovery channel I heard that a black hole burps matter out of it when it consumes more than it could process at a time. Given that even light cannot escape from a black hole, how does this "burp" matter escape?

The matter that the black hole is ingesting begins with some initial conditions outside of the black hole's event horizon, and due to numerous processes this surrounding matter loses energy and angular momentum and so falls into the black hole. The physics of this infall process can be quite complicated - especially if an accretion disk forms - but essentially, viscous friction between the particles that compose the infalling matter will cause it to lose momentum, among other possible processes.

There is a theoretical limit of how much matter a black hole (or any astrophysical object) may accrete: the Eddington luminosity limit. Basically, as the black hole swallows the matter it illuminates electromagnetic radiation due to the intense collisions the infalling matter is experiencing near the event horizon. Eddington found that once the luminosity of this emitted radiation reaches a certain limit, the accretor (in this case a black hole) will actually begin to expel the surrounding matter rather than continue swallowing it, due the fact that once the limit is reached the surrounding matter will be blown away by the radiation. See here and here for deeper explanations of this.

So the surrounding matter that is expelled (once the Eddington limit has been reached) did not ever reach the event horizon - it's expelled by radiation before it makes it that close to the black hole.

And here's a lovely image of a double burp! :)


Is the Moon not in hydrostatic equilibrium? Given the recently-announced observations from VLT/SPHERE that 10 Hygiea may be sufficiently round to qualify as the second main-belt dwarf planet, I found myself perusing Wikipedia's  Hydrostatic Equilibrium article, and came across the following passage, which currently has no linked references:

  The smallest body confirmed to be in hydrostatic equilibrium is the dwarf planet Ceres, which is icy, at 945 km, whereas the largest body known to not be in hydrostatic equilibrium is the Moon, which is rocky, at 3,474 km.

I'd long assumed that the Moon was in hydrostatic equilibrium, given its spherical-to-the eye shape and ranking among the 20 largest solar system objects. If the quoted statement is true, why isn't it in hydrostatic equilibrium?

The paragraph notes previously that Mimas is not in hydrostatic equilibrium for its current rotation. A quick search for the moon and hydrostatic equilibrium turned up M. Burša (1984) "Secular Love Numbers and Hydrostatic Equilibrium of Planets". According to this paper the Moon, Mercury and Venus are all far from hydrostatic equilibrium. The discrepancy is much smaller for the Earth. The paper goes on to note that the rotation periods required for the flattenings of these objects to be explained as hydrostatic are 3.7 days, 4.7 days and 17 days respectively, all substantially faster than the current rotation periods.

According to the abstract of C. Qin's presentation "Formation of the lunar fossil bulge and its implication for the dynamics of the early Earth and Moon" the usual hypothesis in the case of the Moon is that the shape is a "fossil bulge", a relic from when the Moon was spinning faster early in its history. Possibly this explanation may also apply to Mercury and Venus, which also have been spun down by tidal forces.


Is the moon's orbit circularizing? Why does tidal heating circularize orbits? The moon has an elliptical orbit (of course), but as it inches away from the Earth each year, does its orbit become more circular and how come? Io's orbit is becoming more circular due to tidal heating effects, but why?

Short answer: YES. I will assume you mean the eccentricity in the orbit of the Moon around the Earth.

In general, tidal forces in binary systems (like the Earth-Moon system, or a binary star, etc...) effect the binary in three main ways: in order of longest timescale to shortest timescale

1) circularization of the orbit (eccentricity goes to zero, binary separation goes to minimum).

2) alignment of the binary components' spin angular momenta with the orbital angular momentum (the directions of $S$ and $L$ are the same).

3) synchronization of the binary components' rotational frequency with the orbital frequency.

  but why?

There's different ways of answering "why," and here is a great conceptual answer given by the God Father of astrophysical tides himself, Z. P. Zahn:

  A fundamental property of closed mechanical systems is that they conserve their
  total momentum. This is true in particular for binary stars, star-planet(s) systems,
  whether they possess or not a circumstellar disc, if one can ignore the angular
  momentum that is carried away by winds and by gravitational waves. Through
  tidal interaction, kinetic energy and angular momentum are exchanged between
  the rotation of the components their orbital motion and the disc. In the absence
  of such a disc, which is the case that we shall consider here, they evolve due
  to viscous and radiative dissipation to the state of minimum kinetic energy, in
  which the orbit is circular, the rotation of both stars is synchronized with the
  orbital motion, and their spin axis are perpendicular to the orbital plane. How
  rapidly the system tends to that state is determined by the strength of the tidal
  interaction, and thus by the separation of the two components...

So, basically, the tidal torque drives dissipation, and this dissipation brings the binary to a minimum kinetic energy state, i.e. circular orbit, synchronized spins with the orbit, aligned spins with the orbit.

Here's a contour plot of the timescales provided by equations #$9 - 13$ from Hut's 1981 seminal paper, assuming the separation does not change much relative to the other quantities, for a binary composed of a black hole and a Wolf-Rayet star, which is a similar system to a planet-satellite system, where the timescales are parameterized in terms of the mass of the WR star and the separation of the binary:

The dotted black line is the merger timescale for the binary due to gravitational waves, meaning below that line you are binary that mergers within the lifetime of the universe. The synchronization timescale is independent of the initial spin of the WR star, which is why there is only one synchronization line in the plot, but the alignemnt timescale does depend on the initial spin of the component feeling the tides. Points below the contours achieve that process (below the red dashed line are synchronized). The quantity $f_{B}$ is the break-up fraction parameter, is between $0$ and $1$, and the choice of which determines the initial WR star spin as a fraction of its break-up spin. (the tidal timescale contours were made by equating the WR lifetime with the tidal timescale). I did not include the circularization timescale because it is essentially always less than the sync. timescale (seen simply from the fact that generally, $S << L$).

Lastly, it must be said that a WR star is quite different from a moon/satellite, however the effects of tides on each is similar enough to make a conceptual comparison (since a WR star is much less massive than the BH, as is the moon than the Earth, the hierarchy of tidal timescales is the same, but you must keep in mind that the WR's apsidal response to the tidal gradient is very different than a rocky moon's).

As you pointed out, the Moon's orbit is slightly eccentric, but the moon is also tidally synchronized with the Earth's rotation (which are both sync'ed to the orbital rotation). This should make sense since it takes longer for tides to circularize than to synchronize.

I'm sure somebody can post an example of using tides in an actual planet-moon system, rather than my black hole-star system. :)


Perceiving movement of stars Are there any stars visible to the naked eye whose position, over a normal human lifetime, can be seen to have changed (relative to other stars in the vicinity)?

To answer this, I interrogated the revised Hipparcos catalogue produced by van Leeuwen (2007), accessed at the CDS, Strasbourg.

There are 4559 stars in the catalogue with a Hipparcos magnitude $<6$ and which you might class as "naked eye" objects, though you'd need good eyes, a good site and plenty of patience to be dealing with stars fainter than about 5th magnitude.

Sorting these objects by total proper motion, we find there are three objects with proper motions between  4.0 and 5.3 arcseconds per year:

Star           Mag       Total PM (arcsecs/yr)

HIP19849       4.56       4.1

HIP108870      4.83       4.7

HIP104214      5.37       5.3

Let's say a human lifetime is 80 years. This means that these stars move an angular distance of between 328 and 424 arcseconds on the sky with respect to the average position of all the other stars.

The angular resolution of the human eye is about 30-60 arcsecs (for really good vision and a completely dark adapted eye). So the separation between two stars could be estimated to some fraction of this.

I think this would be very difficult though unless these stars were very close to another naked eye star, so that you could see the separation changing with time.

So, I did another search. I looked for naked eye stars that were within 10 arcminutes of another naked-eye star and which had a high relative proper motion that would make them move significantly with respect to each other.

The best candidate pair is HIP71500 (which is basically stationary) and HIP71502 (which is a high proper motion star), magnitude 5.58 and 5.85, which are separated currently by about 15 arcseconds, but after 80 years, this will have changed by 52 arcseconds. What this means is that for someone with acute naked-eye vision (is it possible to have that over your lifetime), a star that was apparently single to the naked eye will gradually become a (just) resolved visual binary.

Nice bit of trivia!


When will a day on Earth & Mars be the same length? I read on Wikipedia that:

  As on Earth, the period of rotation of Mars (the length of its day) is slowing down. However, this effect is three orders of magnitude smaller than on Earth because the gravitational effect of Phobos is negligible and the effect is mainly due to the Sun.[19] On Earth, the gravitational influence of the Moon has a much greater effect. Eventually, in the far future, the length of a day on Earth will equal and then exceed the length of a day on Mars.

So, how far into the future exactly will it be when the length of a day on Earth and Mars will be equal? Also, for how long will their day lengths be within say, one second of each other?

By "day", I mean the Mean Solar Day.

Since the rate of lengthening of the day on Mars is three orders of magnitude less than that of the Earth for a first go at calculating how long it will be before the day lengths are equal we can ignore the change in day length on Mars. The difference in day lengths is $\approx 40$ minutes. The Earths day lengthens by $\approx 1.7$ ms/100 years. Therefore the day lengths will be equal in: $\frac{40\times 60}{1.7\times 10^{-3}}\times 100 \approx 141 \times 10^6$ years


Could Tabby's star be beamed by a neutron star's jet causing irregular star spots? There have been some other questions here about what might cause the light curve dips of star KIC 8462852 as observed by Kepler Space Telescope. Here's one I haven't heard before:

Could it be that a neutron star, from Earth yet unobserved, or another supernova remnant in the relative vicinity of Tabby's star, hits it with its polar jet stream? Being in the focus of an irregular jet of charged particles would disturb the star's magnetic field, somehow causing star spot patterns, and upsetting its photosphere perhaps otherwise too.

Can this be excluded?

I think this scenario is unlikely.  There is no evidence Tabby's star is near a neutron star.  Astronomers are monitoring Tabby's star very closely and there has not been any detection of the radio emission, gamma rays, or x-rays you would expect to see from a neutron star.  Also something as heavy as a nearby neutron star would have a lot of gravity probably show up in radial velocity measurements of the star.  Globs of dust, broken up comets, or more mundane stellar variability are a far more likely explanations for the weird dips seen in the star's light-curve.


Are there any ways in which the light time difference between the center and the edge of the solar disk is noticeable? In the ABC News video New space discovery has astronomers buzzing at about 02:30 Dr. Hakeem M. Oluseyi explains that the shortness of FRBs (before dispersion) suggests a compact source size and illustrates this by using the Sun as an example.

I'd never though about this until just now, but the radius of the Sun is indeed about 2.3 light-seconds. It takes light from the edge of the Sun's disk over 2 seconds longer to reach us than light from the central area.

Are there any ways in which this light time difference is noticeable, or needs to be taken into account?

@Astrosnapper's answer to Helioseismology of the Sun, what is actually measured? begins with:

  The two main methods to detect the solar p-mode oscillations, which have a period of about 5 minutes (so frequency of ~3 milliHertz) are line of sight Doppler shift and change in irradiance.

3 mHz is the peak of the power spectrum, but a 2D histogram of shows significant power at frequencies beyond 8 mHz, corresponding to a period of only 128 seconds.

What does this mean? A dopplergram is collected over an extended period of time, maps of line-of-sight velocity determined via doppler shift of a given spectral line, and fit to a model based on spherical harmonics. Below is an example of a dopplergram and a histogram of power as a function of frequency and spherical harmonic. The shading left-to-right is the doppler shift associated with the rotation of the Sun, I'm not sure if the orbital motion of the Earth and other kinematic effects have been subtracted.

I went with 8 mHz and calculated the time and phase delay due to the different distances of parts of the 695,700 km radius Sun at a given instant of time. (script):

I found some older data and a nice description Measurements of Frequencies of Solar Oscillations from the MDI medium-l Program by E.J. Rhodes, Jr., A.G. Kosovichev, P.H. Scherrer, J. Schou & J. Reiter which comes from The Michelson Doppler Imager aboard SOHO.

above: Fig. 1c: Full disk Dopplergram. (full-scale version). Full-disk Dopplergram obtained with MDI on 9 July 1996. The large left-right variation is the signal from solar rotation. The "texture" is a combination of the motion from the sound waves that permeate the Sun and the large convective cells called "super-granulation". The bright feature in the lower right quadrant is an active region. Source

above: Fig. 1b: Power spectrum obtained from 144 days the MDI Medium-l data for the modes averaged over the azimuthal order m. The power concentrates in ridges corresponding to solar acoustic (p) modes. The lowest weak ridge corresponds to the fundamental (f) mode. Source


How does ALMA produce stable, mutually coherent ~THz local oscillators for all of their dishes? The ALMA (Atacama Large Millimetre/submillimetre Array) radio telescope's band-10 capabaility is now operational, per this answer. That's confirmed by NRAO's First Science with ALMA’s Highest-Frequency Capabilities.

According to this site ALMA's bands and frequencies are as follows:

band  wavelength   noise   frequency
        (mm)        (K)      (GHz)
 1    6.0 - 8.5     26     35 -  50
 2    3.3 - 4.5     47     65 -  90
 3    2.6 - 3.6     60     84 - 116
 4    1.8 - 2.4     82    125 - 163
 5    1.4 - 1.8    105    163 - 211
 6    1.1 - 1.4    136    211 - 275
 7    0.8 - 1.1    219    275 - 373
 8    0.6 - 0.8    292    385 - 500
 9    0.4 - 0.5    261    602 - 720
10    0.3 - 0.4    344    787 - 950

900 GHz (0.9 THz) is quite a high frequency for a radio receiver! Each ALMA dish down-converts received frequency to a baseband of a few GHz before they get digitized and sent to the correlator for digital interferometry, but you still need an ultra-stable local oscillator (LO) for downconversion, and all of the LO's of all of the dishes need to be mutually coherent. That's quite a feat considering they can be tens of kilometers apart!

Question: How does ALMA produce stable, mutually coherent ~THz local oscillators for all of their dishes?

As suggested in this partial answer to the question How does the Event Horizon Telescope implement the interferometry? hard drives collect digital data down-converted by local atomic clocks, probably using GPS as a secondary reference, then brought to a central location for post analysis.

I am guessing that they spend a significant amount of time trying to reconstruct coherence at the millimetre (picosecond) level, but that option is not available for ALMA as data is continuously recorded at some large fraction of 24/7, and the huge volume of data at a location where magnetic hard drives don't work is overwhelming. So they need to get it right the first time.

ALMA produces stable, mutually coherent ~THz LO (Local Oscillators) for all the antennas by...
Using a single central LO and piping it to every antenna via fiberoptic cable!  The fiber expands and contracts due to temperature fluctuations so a laser system is used for calibration among the antenna.  If you can believe it, they manually lengthen or shorten the fiber to each antenna to adjust for the contraction/expansion. The LO for the high frequency observations for ALMA requires femtosecond accuracy.
Check out this youtube for a more thorough explanation than I could provide:

Notes:

I think this is an excellent question, since solving the LO synchronization problem for high frequency observations was one of the most difficult technical and engineering problems associated with bringing the full capabilities of ALMA online.

Building and designing interferometers decades ago, we always remembered that light travels about a foot per nanosecond.  To get femtosecond accuracy for synchronization, they have to adjust fiber lengths by the micrometer.  Picoseconds?  Ain't nobody got time for that!

The video shows and explains details of how the LO signal included in the modulated laser is split and then distributed by optical fiber to each element in the array. From the explanation on the video's page:

Complex electronics accurately stitch ALMA’s individual wave detections together into one dataset. The first step in this process is to have exact measurements of where and when the antenna picked up its waves. On each antenna is a clock that timestamps the data using a kind of atomic metronome, or rhythm-keeping device, kept near the supercomputer. The timekeeping waves from this central oscillator beam out to each of ALMA’s antennas. Onboard the antennas, a local oscillator injects this timekeeping beat into a microscopic mixer with the waves coming through the receiver, and a mixed-down signal is digitized and sent back along the fiber into the supercomputer.
The timekeeping signals generated by the central local oscillator are sent through optical fiber, which presents an additional challenge. The length of the optical fiber can vary with temperature, but in order to achieve the incredible level of accuracy that ALMA requires, the fiber length must not change by more than one millionth of a millimeter over the entire 15-kilometer-distance to the farthest antenna. Stabilizing the length of the optical fiber is another job of the central local oscillator.


What's the significance of finding neon on lunar atmospehere? new explanation of moon light? Now that NASA's LADEE spacecraft has confirmed the presence of Neon in the moon's atmosphere could the official explanation of the moon's light end up being change from reflecting the sun's light to some sort of electromagnetic interaction with the neon caused by solar winds, or something like that?

The moon's "atmosphere" isn't much of an atmosphere, and is sometimes called an exosphere.  The particle density in the moon's atmosphere is low enough to be considered a very good vacuum for Earth experiments.
The discovery is mentioned in this article on phys.org, which specifically says

There's not enough neon to make the moon visibly glow because the moon's atmosphere is extremely tenuous, about 100 trillion times less dense than Earth's atmosphere at sea level.
...
The behavior of a dense atmosphere is driven by collisions between its atoms and molecules. However, the moon's atmosphere is technically referred to as an exosphere because it's so thin, its atoms rarely collide.

As far as the question "what's the importance" in your title, it seems mainly to be a confirmation of something that has been predicted for a while, as well as providing more data concerning exospheres: the most common type of "atmosphere" in the solar system.


Black hole at Galactic Center of the Milky way and Bar of stars? The Milky Way is a barred spiral galaxy, with a supermassive black hole at its center. Here

As it is a barred spiral galaxy, it has a central bar-shaped structure composed of stars.Here

So my questions are -:

Do the bar and supermassive black hole overlap or cross each other?
Do the stars from the bar get sucked inside the supermassive black hole?

Lets get a sense of scale. The bar is about 30000 light years long, and contains the central bulge of the galaxy, a roughly spherical mass of stars with a diameter of 10000 ly.  It's big.

The supermassive black hole is pretty big for a black hole, but only about 80 light seconds across. Compared to the size of the bar, it is tiny.

The supermassive black hole is found at the centre of the bulge, at the centre of the bar, and entirely within it.

Sometimes a star does fall into the black hole. The black hole is currently inactive, that means that there is not much matter falling in every year. Each year, roughly 1/10000 of the mass of the sun falls into the black hole. This is mostly matter from the solar winds of nearby stars.


Suggestions for a beginner-friendly telescope I'm hoping someone can point me in the right direction for a beginner-friendly telescope for my family.

My husband has mentioned wanting a telescope off and on for a few years, but he's never delved into the idea enough to give specifics. We have an autistic 7-year-old who has developed a Special Interest in astronomy, and since it's our practice to delve into his SI's with him, it seems like a good time to take the plunge.

Pertinent information...

We live in a major metropolitan area -- in the suburbs about 30-45 minutes from downtown Chicago. We will be able to drive to a less light-polluted area from time to time, but realistically, we'll be seeing what we can see from the yard or second-story balcony most of the time.
7yo will likely be most interested in objects seen with some degree of detail -- planets, stars, comets, our moon, other brighter objects. We adults would like to glimpse some of the dimmer objects as well, though we know they'll be faint. 7yo will likely be impressed by glimpsing those objects, as well, since he's a bit obsessed with the vastness of the universe at the moment.
We are true beginners. Think of people who have never touched a scope. But we're patient and smart, so we can learn. Which is kind of the point, I suppose.
7yo is not always patient, so an easy, quick set-up would be desirable.
No children will be using the scope unsupervised, but we will be teaching 7yo how to use it while closely supervised.
If giving specific recommendations, we'd like to keep the cost under $500US.

Cheap telescopes are just fine
I won't recommend a specific telescope, but I can tell you the following: I live in the middle of Copenhagen, and with my random 15 year old 4" telescope, which at that time cost \$300-$400, I can see the Moon's craters and mountain ridges, Jupiter's belts and moons, Saturn's ring, Venus' phases, and some nebulae. My telescope is manual, i.e. has no motor to track the sky, but I'd guess that today for the same price, you could get a telescope that does have a motor and a small computer that can find celestial objects, which makes things much easier.
Stay home
In my experience, going outside the city doesn't really add much to the experience (for amateur telescopes). To really see dark skies, you need to be 50-100 km away from the nearest large city, I'd say. Maybe that's not far from you, but when buying a telescope, you should also take into account that larger telescopes are more difficult to transport and set up, so if it's too large, you might end up never pulling yourself together to take it out.
Stay inside
If you take your telescope outside, it takes some time before it has reached the same temperature as the air, and during this time, there can be quite a lot of turbulence, which makes the image flicker. If I quickly want to show my 7 yo something, I simply use it from inside my living room, looking through the window. Two minutes, and he's impressed (until something else catches his attention).
What you see is not spectacular
Also, you should be prepared that there's no way that you can see anything like the colorful pictures that you can find on the internet. You will be able to see that Jupiter and Mars have slightly red colors, but apart from that, most things will look black/white. The reason is that the eye needs much more light than can be collected through a telescope in order to see colors. The largest telescope that I've looked through with the eye was a 31" telescope, and even that didn't reveal much colors.
To see colors, you need to take a picture with a long exposure. Many telescopes have the possibility of attaching a camera to it, but for this you definitely need a telescope with a motor such that it moves along with the sky, or else everything will just leave long white trail across the image.
You should also be prepared that planets and the like will look quite small. You could see something like this (very approximately to scale):

Know what you see
The beauty of what you see in an amateur telescope, even if it's small and dim, lies in knowing what it is that you see. Knowing that some nebula is 1000 lightyears away, and that its ever so slightly reddish color is hydrogen being illuminated by young, massive stars makes it much more fantastic to see.
My telescope is of medium quality. What you get for an expensive one is probably more clear images, but the most important factor is the light-collecting area, i.e. a 5" will show you more than a 4". I'm sure users like Florin Andrei can tell you more here.


Distance of extra-galactic Classical Cepheids There have been many questions and answers about finding the distance of a star from the earth. But as I did some research on the net, I found that we have specific approaches for finding the distances of different types of stars.

So, my question is more specific now.

How to determine how far a Classical Cepheid is?

What if the Cepheid is extra-galactic? Is there a good method to finding the solution?
Can anybody suggest how to make a start at deriving a relation at finding the distance of a Cepheid from us?

There is this Period-Luminosity-Color relation. Can anyone at least enlist the journal articles where I can make a study to get closer to finding the answer to my question?

Since it seems you don't know the equations, I will try to keep it simple.
However, just keep in mind that, in principle, there is no difference between Galactic and extragalactic Cepheids, in this context.

Now, we know that a Period-Luminosity relation holds for the Cepheids:

$P\sim L$

Where $P$ is the pulsation period observed from the Cepheid, and $L$ is the observed luminosity.

We also know that:

$L\sim d$

where $d$ is the distance of the source.

Then it is like to write $P\sim d$.

This means that, if we observe the pulsation period of a Cepheid, we know its distance, and this allows to know the distance of the host galaxy as well.

Some basic sources:

Cepheid variable

Classical Cepheid variable with some numbers

And some diagrams too


Is there any data source for the rotational axes of bodies in the solar system? I have found several right ascension and declination coordinates for the poles of bodies in the solar system, but there seems to be no comprehensive source that is easily uncovered with Google etc. containing any other than those of the planets and a handful of other bodies.

Most importantly i am missing them for the Moon, but i'd like them for all regular satellites and any other bodies if possible.

Collecting and reporting the requested information is the job of the IAU Working Group on Cartographic Coordinates and Rotational Elements. They issue reports every few years; the most recent report was in 2009. The report is currently available at usno.navy.mil and astrogeology.usgs.gov, and elsewhere. Table 1 provides data for the Sun and planets (including Pluto), table 2 for satellites, and table 3 for other solar system bodies.

The question asks specifically about the Moon. The Moon's orientation is hideously complex. I won't even begin to replicate the expression in table 2 of the cited report. The report specifically addresses the Moon's orientation, admitting that the expression in table 2 is approximate. The best source is an ephemeris such as JPL's Development Ephemeris. These data are available in the form of SPICE kernels.

References:

Archinal, Brent Allen, et al. "Report of the IAU Working Group on Cartographic Coordinates and Rotational Elements: 2009." Celestial Mechanics and Dynamical Astronomy 109.2 (2011): 101-135. DOI 10.1007/s10569-010-9320-4


Auroras: is there any bump in frequency at certain times of night, or times of year? I'm trying to find out if there is a range of times at night, and/or days during the year, in which the probability of seeing an aurora is higher than average.

The aurora section of the British Astronomical Association asserts that there's
a correlation with season (higher probability near equinoxes), and a correlation with 'magnetic midnight'.
http://www.britastro.org/aurora/

Are these assertions correct? If so, what is the definition of magnetic midnight, exactly?

Magnetic midnight occurs when the sun, the observer and the magnetic poles are in the same plane. This occurs around local midnight: In the USA it is about 1 hour before local midnight, in the UK it is about 2 hours before local midnight (local time is the time as shown on a sundial, with no adjustments for daylight saving etc)

The aurora season is the winter, mostly because at the high latitudes where aurora are visible, the sky is not fully dark between April and September.

Aurora are most likely during solar maximum, which occurs every 11 years (with a good deal of variability). Solar activity remains high for about 4 years. There have been maxima in 2000 and 2014 (the last cycle has been long and weak) The next is due around 2023-2025.


How will Starlink affect observational astronomy? I have recently found that people around the world have been posting observations of a line of moving objects in the sky which is very likely to be a part of Starlink satellite constellation, such that there is even a meta question here about them: Do we need a notice about billionaire space ventures?.

In 2019, 19 Starlink satellites interrupted the observation of an observatory in Chile by showing up in the DECam exposure. By far, 122 satellites have been deployed, but spacex plans to launch more than 12000 satellite to complete its satellite constellation.

There have also been a few specific questions about constellations of satellites and astronomical instruments here and they reference Starlink.

However, the following question has not yet been asked point-blank: How Will Starlink Affect Observational Astronomy? Is there any citable research or science-based speculation as to the extent that Starlink will influence observational astronomy? Is it possible to address it here now?

According to Did nobody in the Astronomy community think 12,000 new satellites in LEO might be a problem? SpaceX chief operating officer Gwynne Shotwell has said

  “No one thought of this... We didn’t think of it. The astronomy community didn’t think of it.”

Now that some time has passed and people have definitely thought of it, is the likely impact on observational astronomy better understood?

tldr; some modelling has been done with a lot more to do, but generally the impact of these constellations is fairly negative, but potentially manageable.

Ok, there's a lot to unpack here. First things first, while people have been aware of Starlink and have thought of it, modelling the impacts to observatories hasn't happened as of yet in quite a few places.

The main observatory which has had modelling done, LSST, released this statement;

  LSST is particularly sensitive to bright trails of these satellites, due to LSST’s unprecedented product of field-of-view and light collecting area.  Most LSST images will contain such a trail if plans for multiple LEOsat constellations materialize. The first group of Starlink satellites are sufficiently bright during dawn and dusk (when LSST would be surveying) that the trail would exceed sensor saturation, generating uncorrectable artifacts in the data
  LSST Statement

AURA (the owners of LSST) provided some additional information;

  The LSST Project Science Team has been simulating the potential impacts to LSST observations. Their latest update of preliminary results from November 2019 indicates that (assuming the full deployment of planned satellites) nearly every exposure within two hours of sunset or sunrise would have a satellite streak. During summer months there could be a 40% impact on twilight observing time (less in winter) and saturation of sensors by the satellites can continue well past astronomical twilight. Because of scattered light in the optics by the bright satellites, the scientific usefulness of an entire exposure can sometimes be negated. Detection of near-Earth asteroids, normally surveyed for during twilight, would be particularly impacted. Dark energy surveys are also sensitive to the satellites because of streaks caused in the images. Avoiding saturation of streaks is vital. The LSST team notes that “For the full Starlink constellation, at any one time during the night over 200 Low Earth Orbit (LEO) satellites will be visible. This will increase by the mid-2020s with additional planned constellations of LEO satellites.”
  AURA Statement

I've found some other astronomy authorities that have released statements relating to the problem at hand, but they don't have that much information as to potential impacts of these constellations.

As of June 2019, the IAU stated that

  We do not yet understand the impact of thousands of these visible satellites scattered across the night sky
  IAU Statement

In regards to this lack of information the AAS created this survey is to gather as much information as possible about the impacts to various observatories so that they can be better informed:
AAS Survey
From what I can tell, the results of the survey aren't out yet and of course when released it will provide a lot of insight into the potential impacts and how to manage them.

These are the properties which the AAS expect of a constellation of 1,584 Starlink satellites.

Consider satellites to be visible only when altitude ≥ 30 degrees and in direct
Sunlight or Earth's penumbra
magnitude 5
satellites cross sky in 4 minutes
6-9 satellites visible at any time during 1-hour starting (ending) at 12-degree -evening (morning) twilight
spatial density on sky ~ 7E-4 deg-2
angular speed across sky ~ 0.5 - 1 degree / sec

In regards to radio astronomy, the NRAO were happy enough with SpaceX Starlink, though there might be concerns with other constellations such as Oneweb.
NRAO Statement
Oneweb concerns

In reference to the previous point, all of this comes with the caveat that Starlink is potentially only one of perhaps 4 or 5 LEO constellations. Even if SpaceX acts responsibly you still have to convince all of the others to do so as well, which might be difficult.

SpaceX and AAS are working together on ways of reducing the impact of Starlink to astronomy. AAS on mitigating Satellites Constellations impact


How can 'HD 140283' be older than the universe? Scientists have known about the star HD 140283, informally nicknamed the Methuselah star, for more than 100 years, since it cruises across the sky at a relatively rapid clip. The star moves at about 800,000 mph (1.3 million km/h) and travels the width of the full moon in the sky every 1,500 years or so, researchers said. Its apparent magnitude is 7.223.

Previous research had estimated that the Milky Way galaxy's so-called
"Methuselah star" is up to 16 billion years old. That's a problem,
since most researchers agree that the Big Bang that created the
universe occurred about 13.8 billion years ago.

Later estimates show that the star could be as old as 14.5 billion years (± 0.8 billion years), which is still it older than the universe's calculated age of about 13.8 billion years. This is an obvious dilemma.

How can a star be older than the universe itself?

It's not - the current age is an estimate. The discrepancy comes from astronomer's attempts to fully and correctly measure aspects of the star, particularly its distance.  From the NASA page:

  The new Hubble age estimates reduce the range of measurement uncertainty, so that the star's age overlaps with the universe's age — as independently determined by the rate of expansion of space, an analysis of the microwave background from the big bang, and measurements of radioactive decay.

Currently, if you take the uncertainty into account, the current estimate for the star's age is 14.5 +/- 0.8 billion years  or between 15.3 and 13.7 billion years - the latter end of the estimate overlaps with the current model of the age of the Universe - so a more correct way of describing this age is that HD 140283 is at least 13.7 billion years old by this estimation.

Another paper based on this star in Nature - "Nearby star is almost as old as the Universe", a team has narrowed down the age to 13.9 +/- 0.7 billion years, stating that:

  Taking into account that experimental error, the age does not conflict with the age of the Universe, 13.77 billion years.

more concisely stating that

  The star's age is therefore at least 13.2 billion years

No doubt further work is being performed to further refine the age of the star.


Betelgeuse in false colors, is this common? First, take the original Betelgeuse image from ALMA and remove red and green channels. Then boost the blue channel luminosity by x10 to x20 and a lot of detail pops out.

This is the resulting image that can be reproduced from the original.

If you zoom into the picture you'll notice tiny concentric "ripples". Blue shifted light denotes high energy particles (as opposed to red shift).

My hypothesis (I'm not an astonomer!) is that we are seeing background radiation, while the star itself obstructs the background. The ripples would then indicate gravitational bending as those heavy particles cross the super-extended corona of the star.

Is there an accepted hypothesis about these artifacts?

EDIT:

The inner ring of pixel-mess around the black spot is out of gamut and thus "invalid data". The ripples are inside the corona and, finally, the outer parts of the image are relatively uninfluenced background radiation from other objects. There are NO ARTIFACTS on the R and G channels.

This image shows what I mean by "ripples":

As a counter-argument to myself I show this more recent picture taken by SPHERE with similar artifacts. Maybe I should conclude that the high-level structure is real while the low-level structures are artifacts caused by the measuring instrument itself hitting its sensibility levels.

PS: I posted this question also on reddit.

EDIT: blue channel histogram

This is how to see the blue channel color information. I use Krita which is by no means a scientific tool but its immediate usage helps.

Download the image from the ESO source link above (it's the "Full size original" 91.6MB TIFF image).

Then, in Krita, go to "Filter > Adjust > Color Adjustment curves...". On the channel drop-down select "B" and observe the following picture:

I have highlighted with RED the interesting zone. On the background you'll notice there is a histogram and that color information is only present to the far left between the bars. All other intensities (maybe byte value 10 and above?) do not occur in the image.

  To the far right there is a barely visible population of "byte value 255 pixels", those are the out-of-gamut pixels.

To see those pixels drag the upper-right corner knob to the left so that the curve is steep from 0 to 255 between the two red bars I've shown. This will magnify intensity by a factor of 10 or 20 and map very dim pixels to brighter blue.

The uploaded image (second link above) is the resulting image of this process. There is NO EXACT SPOT I'm talking about. It is an aura all around the black blob in the middle.

This is a partial and perhaps temporary answer only, posted in order to get to the bottom of this. I think that the comments are correct, you're probably seeing an artifact of image manipulation and the original data is just a 2D intensity map.

I downloaded the original image at https://cdn.eso.org/images/screen/potw1726a.jpg, converted back to .png and analyzed in Python using  matplotlib and numpy. Right now I don't see these mysterious rings, but if you can provide pixel coordinates from the original image and explain more how you made the rings visible, that would be great!

Image breakdown into three color channels:

Line scans in x and y through center and maximum:

Same but multiplied by 256 (8 bits per channel) and plotted on a log scale:

Scatter plot of G and B versus R show that these are correlated. We're probably just looking at artificial coloring from a table, not different channels of data. It could be a similar thing going on here as well.

import numpy as np
import matplotlib.pyplot as plt

fname = 'Betelgeuse in false colors.png'

img = plt.imread(fname)
print(img.shape)

if True:
    plt.figure()
    plt.subplot(2, 2, 1)
    plt.imshow(img)
    plt.title('RGB')
    for i , char in enumerate('RGB'):
        plt.subplot(2, 2, i+2)
        plt.imshow(img[..., i], cmap='gray', vmin=0, vmax=1)
        plt.title(char + ' channel', fontsize=16)
    plt.show()

rgbx_cen = img[640]
rgbx_570 = img[570]

rgby_cen = img[:, 640]
rgby_570 = img[:, 520]

things = rgbx_cen, rgbx_570, rgby_cen, rgby_570
names  = 'rgbx_cen', 'rgbx_570', 'rgby_cen', 'rgby_570'

if True:
    plt.figure()
    for i, (thing, name) in enumerate(zip(things, names)):
        plt.subplot(2, 2, i+1)
        r, g, b = thing.T[:3]
        print(r.shape, g.shape, b.shape)
        plt.plot(r, '-r')
        plt.plot(g, '-g')
        plt.plot(b, '-b')
        plt.title(name, fontsize=16)
    plt.show()

if True:
    plt.figure()
    for i, (thing, name) in enumerate(zip(things, names)):
        plt.subplot(2, 2, i+1)
        r, g, b = thing.T[:3]
        print(r.shape, g.shape, b.shape)
        plt.plot(255*r, '-r')
        plt.plot(255*g, '-g')
        plt.plot(255*b, '-b')
        plt.yscale('log')
        plt.title(name, fontsize=16)
    plt.suptitle('scaled 0 to 255', fontsize=16)
    plt.show()

if True:
    r, g, b = [x.flatten() for x in np.moveaxis(img, 2, 0)[:3, ::8, ::8].copy()]
    plt.figure()
    plt.subplot(1, 2, 1)
    plt.plot(r, g)
    plt.title('green vs. red')
    plt.subplot(1, 2, 2)
    plt.plot(r, b)
    plt.title('blue vs. red')
    plt.show()


Significance of astronomical twilight What is the significance of astronomical twilight? Astronomical twilight is when "the geometric center of the Sun's disk is between 12 and 18 degrees below the horizon" , but what, if any, is the significance of that event?
Here's a description of what astronomical twilight is, but I fail to see any mention of the event's significance.

After the sun sets the sky remains light for a while. The significance is that observations during astronomical twilight may be impaired by the continuing glow of the sky. After astronomical twilight, the sky is as dark as it can be, and you can start your long exposures of faint objects.
There are various definitions of "twilight": When it is "not completely dark".  For civil purposes, it is "dark" when the sun is 6 degrees below the horizon, for ships it is dark enough to see navigation stars when the sun is between 6 and 12 degrees below the horizon.  But for astronomers, who need the sky to be completely dark, twilight doesn't end until the sun is a full 18 degrees below the horizon
I see you've already found TimeandDate's series of articles on the different types of twilight but I'll link it here for convenience.


Size and mass of comet ATLAS (C/2019 Y4) I read some news about this new comet and some unreliable sources said it is twice as big as Jupiter. I tried to find a reliable estimation of its size or mass, but I couldn't find anything. Would you please provide me a general way to distinguish reliable facts from unreliable facts about comets? Like their size and mass range, how much mass they lose, etc. Thank you in advance.

The confusion comes from the difference between the nucleus and the coma.

The nucleus is a small icy body, only a few km across.

The coma is the cloud of gas and dust released from the nucleus as it warms up.  With not much gravity, the coma spreads out into space, and it can be hard to say exactly where the edge of the coma lies, however, a coma "the size of Jupiter" would not be unusual. Active comets can have comas that are larger than the sun.

But the mass of the coma is tiny. It is made of a very low-pressure gas. It is nearly all empty space, most of the mass of the comet remains locked in the nucleus.


Why does the concentration-mass relation of galaxy clusters have a negative slope? As reported in some recent observational works works (ref. Merten et al. 2015) the concentration-mass relation of galaxy clusters has a negative slope. This means that at a fixed redshift, haloes with a higher mass have a lower concentration.

Some old works say this is a consequence of the fact that the concentration depends only on the density at the assembly time. For instance, Bhattacharya et al. 2013  says:

  The correlation of halo concentration with mass is based on the idea—as first explicated by NFW—that the concentration is determined by the mean density of the universe when the halo is assembled, with higher concentrations corresponding to higher densities. Thus cluster mass halos, which are assembling today, should have a lower concentration than halos of lower mass that were built up at an earlier epoch, where the mean density was higher. Furthermore, one may expect this trend to flatten out (sufficiently) beyond the nonlinear mass scale $M_∗$, and therefore, since $M_∗$ falls rapidly with redshift, flatten out over an extended range in mass as redshift increases.

I cannot understand how this can lead to a negative slope between the concentration and the mass

The hierarchical picture of cosmology posits that the smallest structures formed first, early in the evolution of the Universe, and that more massive systems assemble from these initial structures.  The smallest structures therefore have the highest concentrations according to the picture you present, as the Universe had a higher mean density at the time. As the Universe evolves, the average density decreases, and the largest systems form from the smaller structures. Therefore, the higher the mass of the system, the smaller the concentration, whereas the smallest structures keep their initial concentration from their formation time.

The key here is that smaller structures are older than larger structures, therefore have a higher concentration.


I made a graph trying to show the Sun's movement relative to the Galactic Plane. It doesn't seem to be correct I made a Desmos Graph following the most voted answer of this question, but I'm not sure if it is correct, because it suggest that the formation of the Sun happened shortly after it would have crossed the galactic plane, but the Sun did not exist before that. Is that possible?

Unfortunately, what you are trying to do is not possible. We do not know the details of the Sun's motion or the Galactic potential precisely enough to make a prediction of exactly where the Sun was 4.5 billion years ago.
The amplitude and period of the oscillatory motion that I specified in my answer to the previous question is known now to of order 10%. If you extrapolate backwards for more than a few cycles then the phase could be anything.
In addition, the Sun has likely changed its radial position in the Galaxy during its life. If so, then the amplitude and period of the vertical oscillatory motion will also have evolved.


How much did we know about asteroids at the beginning of the 20th century? When was the nature and size of asteroids discovered? I was under the impression that it was not until the 1980s or 1990s with deeper space probes and Hubble imagery that we learned the nature of asteroids. However, today the wonderful fortune UNIX script gave this quote from a man who died in 1910:

  Noise proves nothing. Often a hen who has merely laid an egg cackles
  as if she laid an asteroid.  -- Mark Twain

How would Samuel Langhorne Clemens have known enough about asteroids to make such a statement in the early 20th century?

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

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

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

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

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

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

and

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

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

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


Appearance of the Cresent Does the latitude of your position on earth change the appearance of the cresent like it changes the position of other objects visible in the sky? Does it rotate it?

Yes it does. If you are at a high latitude the crescent Moon will set looking like a reversed C orientation but in the tropics it will set more like a U orientation. Though the time of the year will also have some effect due to the angle of the ecliptic to the horizon.


How can a very old star be formed among Population III stars? After looking at some articles about old stars, (this, that, and this Wikipedia article), I thought about how such an old and small star could have formed. As the early universe had very massive, early-type stars and (no? almost none?) smaller, late-type stars. So how could a ~$1M_\odot$ star have formed in the early universe, where massive Population III stars dominated?
Here are my theories, if you may find them helpful.
Mass loss via a long-dead companion
Star Foo and Star Bar are both $300 M_\odot$ stars closely orbiting each other. Star Bar is slightly more massive than Star Foo, causing Star Foo to lose mass to Star Bar. Now, Star Bar is $580 M_\odot$ and Star Foo is now drained to $20 M_\odot$. As Star Bar ages and becomes a supergiant, Star Foo gets engulfed by Star Bar, reducing Foo's mass to $\text{1-2} M_\odot$. Star Bar then goes supernova, ejecting Star Foo. 13 billion years later, Star Foo ages and becomes a subgiant, what we see today.
Formed in a massive star's disk
When the universe was young, there were plenty of materials to create stars. One massive star, star Foo, formed. It had a vast disk of gaseous material, weighing over $2 M_\odot$. In the disk, perturbations caused a massive clump of gas to form together, causing the ignition of hydrogen and creating a small, yellow dwarf of mass $0.8 M_\odot$, called star Bar. When Foo went supernova, Bar was ejected. 13 billion years later, we see an ultra-metal poor, very old, late-G type borderline-subgiant.
Split off from a massive star's envelope
A massive star forming region creates a star, Foo. Foo is very massive and large, and begins losing mass through the stellar wind. In a massive eruption, over $1 M_\odot$ is ejected from the star's envelope. This gas clump collapses under its own weight to form a small, Sun-like star. As the universe ages, we see a metal-poor, very old red giant.

Many astronomers and cosmologists conjecture that Population III stars were very massive, perhaps even extremely massive. While this opinion is not universal, it does appear to be dominant. If this is the case, at least some of the very first stars died shortly after they formed, within a million years or less. This in turn would have meant the seeds for the first Population II stars would have been present in some places shortly after the very first stars formed and then died.
The Population III stars did not all form (and die) at the very same time, and those seeds needed to create the first Population II stars would not have been dispersed uniformly. So at least for some period of time in the early history of star formation, there would have been a mix of Population III stars in some places, Population II stars in others.


Please Guide me to buy my first Telescope I am a beginner and I've planned to buy a telescope(Newtonian reflector)for viewing planets and deep sky objects. As far as I know the most important factor to get a brighter view is that the telescope need to have a good aperture. I need a telescope that would be good both for viewing planets and deep sky objects.So my question is here as a beginner how much aperture is necessary to get amazed with the space. And I don't know much regarding telescope so please let me know the other factors for getting good view of celestial objects.

It's a broad question, but I'll take a stab trying to provide the essentials for a good start. Also see this post which contains important additional information:

Best telescope for the viewing of Nebulae, Stars and Planets

There are many, many factors involved in choosing a telescope. You seem settled on a dobsonian. That's not the only possible choice, but a dob is a good scope to start with.

You are correct that aperture (diameter of the primary lens or mirror) is the primary driver of telescope performance. Bigger aperture = better performance. However, performance is not the only thing you should worry about. Think of a car - do you only care about speed when buying a new car?

Any size scope will show you something. I've a 50 mm (2") achromat refractor that does double duty as finderscope. I can push it up to 100x magnification, and it will show lots of craters and mountains on the Moon, will show Jupiter as a disk with two equatorial belts, will show the rings of Saturn, will show the M13 globular cluster (barely) and the M31 Andromeda galaxy (again, barely). Not bad for a little scope.

A small 150 mm ... 250 mm (6" ... 10") dob will show you more. When seeing is good, you see several belts on Jupiter, and even individual curls on the big belts. Saturn's rings start to show the divisions, such as Cassini. These days Mars is at opposition - such a scope will show you the polar ice cap and the major land features. M13 fills the eyepiece with a dust of stars. The Orion Nebula looks big and clear even in the city. The Ring Nebula is visible, etc. Under a very dark sky, you can see (barely) the 3C 273 quasar, at 2.4 billion light years.

A mid-size dob (12" ... 18") will show you even more. More details on the planets, when seeing permits. More faint fuzzies (galaxies, nebulae, star clusters) especially out and far away from the city.

A large dob (20" and over) is an amazing instrument. On those very rare occasions when seeing cooperates, Jupiter at 1000x in such a scope is unforgettable. Also, under a very dark sky out in the desert, point it anywhere and the sky is full of faint fuzzies.

So, which size should you choose? It depends on price, maintenance, and ease of use (size, etc). Let's talk about maintenance.

As a dob owner, there are two things you must never forget: collimation, and cooling.

Collimation

It means "keeping all the optics aligned and centered". The mirrors will move around a bit, and in time will lose collimation. As a dob owner, you have to collimate the scope once in a while (and every time before you observe high-resolution targets such as planets). Once you learn how to do it, it takes about 2 minutes, it's super-quick and easy.

This is why collimation is important:

http://legault.perso.sfr.fr/collim.html

This is an introduction to dobsonian collimation:

http://www.cloudynights.com/documents/primer.pdf

Here are some simpler collimation techniques, which you can use until you acquire all the tools:

http://www.etnaastros.com/documents/No_Tools_Collimation.pdf

http://www.skyandtelescope.com/astronomy-resources/how-to-collimate-your-newtonian-reflector/

Also see the manual that comes with the scope.

Cooling

If you keep the scope in the house or garage, and then take it out for observation, the primary mirror will be much hotter than the night air. As a result, air convection will appear on the mirror and distort the image - just like the shimmering you see in the summer along the hot walls of a house under sunlight. The mirror is too "hot" for the ambient.

This is what happens:

http://www.garyseronik.com/?q=node/55

http://www.garyseronik.com/?q=node/69

To combat convection, take the scope outside at least 1 hour prior to observation, and let it cool off. Very small dobs (up to 6" or so) will probably work fine just with passive cooling. Small-medium and above (10" or above) basically require a fan on the primary mirror (forced cooling).

Always collimate and cool off the scope before you do an observation requiring high resolution (when observing planets, the Moon, double stars). If you don't need a lot of high resolution (like when observing galaxies, nebulae, star clusters) then collimation and cooling are not so important.

If you keep your dob collimated and cooled, it will provide great views of the planets. There is this myth out there that dobs "are not great for planets", and refractors are somehow magically "so much better for planets". This is not true. The myth exists only because of the large number of people out there who keep their dobs in terrible shape; those poor scopes underperform by a very large margin. Don't be one of those folks.

Let me just draw a line in the sand for you: An 8" f/6 dob is a good start overall. Why is that?

At 8", it is big enough that it will be seeing-limited most of the time. In other words, it is so big, atmospheric turbulence will limit the resolution more often than the size of the scope.

It is big enough to show you some faint fuzzies even from the city. All the Messier objects should be visible no matter where you are.

It is not too big to fit on the back seat of a sedan car, if you want to drive to a place with a darker sky. It is not too big to move around in the backyard easily.

The mirror is not so big and thick as to mandate active cooling (fan on the mirror); you can pretty much get by with passive cooling only - take it out at least 1 hour before observation. However, a fan may improve things a bit anyway (you need to experiment with that, but it's not a priority at this size).

At f/6 it's not too hard on the eyepieces, and it's not too hard to collimate. A shorter scope (f/5 and below) produces a steeper light cone; you need higher quality (and price) eyepieces to use it, and collimation requires much better precision.

Finally, the price is right - a few hundred dollars, which is affordable for quite a few people. Take care of it and it will last you a lifetime.

Now, if you can afford it and think you can deal with the steeper learning curve, could you buy a bigger instrument? Sure. 16", 18" - the wallet is the limit. Such big dobs are made differently - they are called "truss dobs". They don't have a big tube where everything is contained. Instead, the tube is replaced by a truss made of a bunch of poles - looks like the arm of a crane machine.

When you're done observing, you disassemble it. It all fits in a sedan car. It's not too heavy for transport, because you handle each part separately.

However, such an instrument is usually f/5 or less. Collimation is much more difficult. You need a coma corrector (like a Paracorr) to fix coma at the edge, and that's additional money. You need better eyepieces. Cooling doesn't work unless you use active cooling (fan). And it's just more expensive overall.

Can you start on a large dob like that? It's not impossible, but it requires more money and a much more significant commitment. This is not a toy that you can discard after two week-ends.


Where to find spectra from Las Campanas Redshift Survey? I am interested in the optical spectrum of a galaxy. In SIMBAD the reference for the redshift of my galaxy is the Las Campanas Redshift Survey. Hence I am looking for the optical spectrum from which the redshift was estimated.

LCRS catalog data should be accessible at VizieR:VII/203.
The associated paper is Schectman et al. 1996.


On Augustus' Actual Prescription to Restore the Julian Calendar to Accuracy I hope that I may ask this question here as I have seen some favorably received questions related to the Julian calendar on this site.
From James Evans' book, ``The History and Practice of Ancient Astronomy,'' is the following snippet (pg. 165):

In praticular, it indicates that ``to bring the [Julian] calendar back into step with the original plan, Augustus decreed in 8 B.C. that all intercalations be omitted until A.D. 8.''
I know the Romans were ``inclusive'' counters---so may I interpret this as meaning that leap years did not resume until 12 A.D.?
Another source: https://www.tondering.dk/claus/cal/julian.php specifically states: ``There were no leap years between 9 BC and AD 8 (or, according to some authorities, between 12 BC and Ad 4).''
QUESTION: Does anyone know for certain if A.D. 8 was a leap year or not; and would you please provide an authoritative reference if you know of one?
Thank you.
I have also read elsewhere: ``

8 AD was definitely a leap year.
It was believed since Scaliger that the leap year sequence was (45), 42, 39 ... 12, 9 BC, AD 8, 12 ... This was based strictly on sources in literature. Chris Bennett claims that an astronomical papyrus published in 1999 (pOxy 61.4175) which gives lunar ephemeris in late 24 BC and implies that Scaliger was definitely wrong, and the correct sequence is actually 44, 41, ... 11, 8 BC, AD 4, 8, 12 ...
The major source for Roman chronology questions is A. K. Michels, The Calendar of the Roman Republic (Princeton, 1967).
If you just want the answers, Bennett's page has clear tables. It also has a very good list of primary sources.


Could there be a M-type asteroid beyond Jupiter? Is there any reason to think an M-type asteroid could not have migrated beyond Jupiter and entered an orbit similar to the Centaurs?

According to the Grand Tack Hypothesis, in the early solar system, Jupiter and Saturn migrated inwards and then outwards, scattering asteroids as they did so.

In this hypothesis it seems inevitable that some m-type asteroids will have ended up in orbits with large semi-major axes.


How do sunspots move seen from the side? Sunspots rise vertically? Do they float on the surface like oil? Do they flow up or down vertically from the Sun's core like a lava lamp?

The sunspots are actually depressions and low thermal pressure zones.

Incredible magnetic fields are piercing the surface of the sun and the magnetic force pacifies the movement and the convection at the zone of the sunspot. that's why the surface becomes darker clouds near those magnetic fields.

Elsewhere, the sun's surface has convection wells called solar granules a bit like a lava lamp, which are about 1000km diameter. You can check images of "solar convection" to study maps of convection.

At a sunspot the movement of colder plasma flows downwards, and there is a hot spot trapped under the sunpot where the thermal energy escapes radially away from the sunspot.

This is a video attempting to illustrate the view under the sun:

https://www.youtube.com/watch?v=zU9O8SmCHFQ

This is a virtual simulation of them. The versions are wildly different:

https://www.youtube.com/watch?v=W5dldC5dPMk

https://www.youtube.com/watch?v=UN_9HMVXjY0

https://www.youtube.com/watch?v=ARIJx0geaOE

here is a speeded up video of some sunspots:

https://youtu.be/BejRxZAa66E

https://youtu.be/l1VpMRMg4P4

https://youtu.be/T5MnZUINzjY


Possibility of certain conditions on a planet I want to know if there can be any planet that can fulfill the following conditions and if any planet has been observed with such characteristics:

The planet must not be tidally locked.
The day-night cycle must be similar to our planet or smaller.
The planet must be so close to the star that the light emitted from the star 'burns' (can cause third degree burns to human skin within seconds after exposure).
Temperatures in the shadow cannot exceed 40-50°C (whether prolonged shadow during daytime or conditions during nighttime).
Atmosphere is not necessary.

The basic condition is you'd need is a more massive, hotter star than our sun.  Not necessarily huge, but 1.5 - 2.5 times the mass of the sun should be enough because that corresponds to considerable extra luminosity and UV rays, so you could move the planet far enough away than Earth is from the sun and avoid any risk of tidal locking or slowing, so the planet could retain fast rotation period.

And you'd need a very thin atmosphere so the planet doesn't retain much heat under shadow.

If the star is bright enough in the sky that you'd get a burn in seconds under direct light, then shade probably wouldn't be cool unless it was permanent shade, as you noted, perhaps a cave.

If there's no atmosphere, you'd probably get colder temperatures than what you mention in shade, so some atmosphere, but a thin one.   Maybe somewhere between 1/10th of an ATM give or take and very low greenhouse gas, so basically very little water, CO2 or CH4.    1/10th of an ATM is just a very ballpark guess.

The only part I don't see happening is 3rd degree burns in seconds.   3rd degree burns by photon radiation from a star would probably require longer than that, unless the planet's surface was hot enough to melt rock.   Bad sunburn in minutes, sure.   3rd degree burns in seconds with your other criteria sounds unlikely to me.


Energy Source for Gamma Rays from Merging Neutron Stars When the two neutron stars merged in the latest event, large amounts of gamma rays were emitted.  Given that their deorbitting is a result of gravitational wave production, where does the energy for the gamma rays come from?

The only sources I can think of are the rotational kinetic energies of the merging stars or the potential energy released by the final collapse into a single spherical mass.

Not Nuclear fusion;
There is also a huge amount of nuclear fusion that creates heavy elements from all the iron there is in neutron star: they in fact are not only made of neutrons. But as said in comment, building of heavier elements rather consumes energy than it generates some.

Not sure scientists  has a clear agreement on how this works.

Gravitational collapse

For example I found this (rather old) abstract which states it could be a result of gravitational collapse, without singularity (since here the result of the merger might not be a black-hole, or the lightest ever: around 2.7 solar masses!)

  theoretical developments (string theory, quantum gravity, critical collapse), which suggest that complete gravitational collapse can occur without singularities or event horizons

Thermonuclear event again: fission

From Space.com, they say Kilonova energy could be explained by rapid atomic decay of heavy elements

  The mergers of dense cosmic bodies that are thought to cause short gamma-ray bursts can also blast out neutron-rich gas that rapidly generates heavy elements such as gold and platinum, scientists say. These "r-process" elements can undergo radioactive decay and release an enormous amount of energy — 1,000 times or so that given off by stellar explosions such as novas. These powerful events are thus known as "kilonovas"


Did Mercury clear its neighborhood? For a body to qualify as a planet according to the IAU definition it must have "cleared its neighborhood". What evidence is there Mercury indeed cleared its neighborhood? Perhaps it migrated there afterwards, when the neighborhood had already been cleared. Does the Grand Tack hypothesis impact our definition of the inner planets as planets?
https://en.m.wikipedia.org/wiki/IAU_definition_of_planet
See Grand Tack Hypothesis.
The present definition of a planet is vulnerable as it seems connected to a model of formation of the solar system. The answer below states that in practice an operational definition used that I believe is adequate.

One calls Mercury a planet because it doesn't share its orbit with any other bodies of comparative size. The fact that it is in its orbit means that that orbit hasn't been cleared by another body (for if it had been cleared by another planet, then Mercury would have been cleared too.)
In practice one doesn't look at "How" an orbit is cleared, only if there is one body that dominates the orbit, or if there are many with no one body dominating.  So Mercury is clearly much larger than any other bodies in roughly 88 day orbits.  But, (for example) Ceres does not dominate its orbit, as it shares its orbit with Vesta, Pallas and Hygeia and many other asteroids.


Catalogue of black holes Where can one find a reliable list of black holes that have been discovered/observed so far, directly or indirectly, along with their properties, like mass, spin, etc.?

The Wikipedia list given by @NeutronCat in a comment contains this external link. This compilation by Wm. Robert Johnston has not been updated since 2004, but contains both stellar and supermassive black holes, along with some parameters (coordinates, distance, mass...).


Evidence for homogeneity of the universe Cosmologists assume that the universe is homogeneous on large scales and, in particular, that the unobservable parts are similar to the observable universe. As I understand it, one important reason for this assumption is simply something along the lines of “hey, it would be really strange and surprising if our corner of the universe turned out to be special.” While this does seem very reasonable to me, I was wondering what other reasons (if any) we have for the assumption. Is it, for instance, predicted by the big bang model?

In the standard big bang model, homogeneity is an assumption - therefore it cannot predict homogeneity. Homogeneity is not a requirement of the model and indeed you can build "inhomogeneous big bang models" - though they must still be consistent with the observational evidence of a high degree of homogeneity on large scales within the observable universe.
The obvious observational evidence for homogeneity on large scales (and of course, in the observable universe) is the smoothness of the cosmic microwave background (CMB). The CMB is very, very uniform - with perturbations (bar the large-scale dipole) at the level of 1 part in $10^5$.
We can say very little about the unobservable universe, since it is... unobservable. And because it is unobservable, it has no influence on us and cannot affect the utility of the models we use to explain the observable universe.
NB: the big bang is just a model of how the universe works, with adjustable parameters including dark matter, vacuum energy density and homogeneity, not a theory.


Google Sky Rectangle with No Stars I've been viewing Google Sky, which is quite amazing, and came across a rather expansive rectangle where no stars are visible at 22h13m04.44s   4°59'29.49" (see image below). My only thought is that there's probably a satellite or something up there that Google is required to hide. Any thoughts on what this is?

This is a known issue affecting multiple locations in Google Sky, including a region in Orion and this area in Pegasus. If you look closely at the interface, you can see that Google Sky gets its data from preexisting surveys, including the Digitized Sky Survey and the Sloan Digital Sky Survey - Google doesn't operate its own fleet of telescopes (at least not yet), so it uses existing archival data. You can query the DSS online.

Here's what I see when I look at the region in question on Google Sky:

Now here's what I see when I look at the same region with DSS-1 data:

The blank spot is now filled with the field of stars, as expected.

If you want, you can do the same thing with the other DSS and SDSS datasets. I wasn't able to find any blank spots in the raw images, meaning that this is presumably an issue on Google's end, not an indicator of the presence of some object that needed to be hidden.


What discoveries have come from the DASCH project (Digital Access to a Sky Century @ Harvard) @astrosnapper's answer to the question Could stellar occultations be used to survey Kuiper belt objects? mentions the TAOS-II project and asteroid occultation measurements (which occur on the timescale of seconds) and that got me thinking about the term Time Domain Astronomy.

A quick search led me to an Astrobites post with a catchy title A Big Step Backward for Time Domain Astronomy that (among other things) describes the DASCH project (Digital Access to a Sky Century @ Harvard).

The post is dated 2012 and says:

  The project is working to (1) digitally scan all of the plates, (2) detect sources on the plates, (3) assign physical coordinates to the plates, and (4) measure/calibrate the magnitudes of all sources.  The end product will be a publicly available database which can be queried to produce the 100 year light curve for individual objects.  As a part of the project the team had to develop a specialized scanner, as commercially available units would have been too slow to finish the project in anything resembling a reasonable amount of time. As it is, when operating at full efficiency, the project can now scan and analyze 400 plates a day.  As of the writing of this article, 38,346 plates have been scanned (~8% completion; see Figure 2 for a map of their current coverage) and over 2.7 billion magnitudes have already been measured.

The article ends with:

  ...Given that we’re only 8% of the way through analyzing the plates, we can look forward to further interesting discoveries from this survey in the next few years.

It is now six years later. Presumably the scan rate has accelerated a bit and a much larger fraction of the plates have been scanned, and techniques for reducing and analyzing the huge amounts of data have improved as well.

Since 2012, what new discoveries have been made from this effort? There may be too many to list, so a few highlights and a link to a summary might be the best way to answer.

The most notable one I'm aware of is the detection of the secular dimming of Boyajian's Star (KIC 8462852), as described in Schaefer (2016). Follow-up by Schaefer et al. (2018) suggest that the ratios of the dimming across the BVRI bands are consistent with dust, though the reason for the lack of detectable infrared excess remains unknown.

DASCH data has also been used to constrain the orbital period of the ringed object transiting J1407, see Mentel et al. (2018). The DASCH plates did not show any previous transit in the data, ruling out a large fraction of orbital periods and opening the possibility that the ringed object J1407b may not be gravitationally bound to J1407.


How would one estimate the rotation period of a star from its spectrum? The figure is shown; the measurements were taken on two consecutive observing nights. The Ordinate is the flux normalized to continuum and the abscissa is the wavelength scale. You can see the "bumps" indicated by the arrows referring to some Starspot as the spot moves on the profile; assuming a single time-stable position-stable spot.
The "bumps" slightly shifts, as indicated by the arrows in the top line profile compared to the bottom profile, as the spot "moves across" the surface as the star as it rotates; i'm just not sure how to get an estimate of the rotational period from this? Given just the wavelength for the abscissa.
Figure reference : Gray, D. (2005). The Observation and Analysis of Stellar Photospheres (3rd ed.), page 498. Cambridge: Cambridge University Press. doi:10.1017/CBO9781316036570

It takes a spot "bump" half the rotation period to traverse from the blue side of the line profile to the red.
You can see that the bump has maybe moved across 20% of the profile in one night, so the rotation period would be estimated as $\sim 10$ days.
Looking up the period of Sigma Gem - it is 19.6 days. So my method is not very accurate by eye, but you would need to simulate the star plus spot spectrum more accurately to take account of limb darkening and the spectral resolution of the spectrograph.


Movement of the satellites of the planets . . . Is the movement of the satellites (moons) of a planet coplanar, like the planets being coplanar around the local Sun?

Inner moon orbits tend to low inclinations with regard to planet's equatorial plane. Charon is an exception to this general rule. Nineplanets.org has a nice data base.


Which eyepieces I can use it for best viewing experience using my existing telescope? Telescope: Orion 09843 SpaceProbe 3,3 inch aperture(76mm), 700mm focal length, eyepieces - 25mm and 10mm focal lengths
Hi, I am very happy and excited to   see moon craters, Jupiter and Saturn. And planets like Mars, Venus, and Uranus are just looks
like a brighter star but not able to see its surface color. Similarly Jupiter looks like white small ball
instead of an orange surface. I am very much excited and  want to go bit closer. So please guide me list of eyepieces that I can use with this scope at least to see Jupiter  surface and big moon craters.

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

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

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


How did astronomers determine the path of 'Oumuamua so quickly? The interstellar asteroid 'Oumuamua was discovered on October 19, 2017. Once discovered, less than a month's worth of precovery observations were found on various sky surveys. By mid-November, scientists had worked out its orbital eccentricity, and soon after its probable origin and destination.

With such a quick visit and so little data, how did scientists figure out so quickly where it came from and where it was going? Did using space probes for observations impart enough parallax to easily determine that, or did Earth-based observation have enough parallax themselves?

Three accurate observations are sufficient to fix a Keplerian orbit (ie an elliptical or hyperbolic orbit with the sun at the focus)

In practice, observations are not perfectly accurate due to limitations of the equipment and observations over a short time are particularly prone to observational error being magnified. Moreover the orbit will be perturbed by the gravity of the planets, so won't be perfectly Keplerian. For this reason pre-discovery images will help fix the exact orbit more accurately.
As the time difference between first and last observation is critical in the quality of the orbital determination, we talk about the length (in days) of the observation arc as a measure of how well defined the orbit is.

Multiple observations can reduce error by an averaging effect (the Gaussian curve is so named from Gauss's use of it in orbital determination)

However once you have three or more observations of a body you can determine its orbit almost immediately, using a computer to do the calculations for you. It would have been immediately obvious that the object was on a very unusual trajectory.

There is no need to use space probes for the observations. The relative motion of the Earth and the Object provide sufficient parallax to determine distance, position and velocity.


Image processing techniques to distinguish between stars and galaxies in an image? Say I have an arbitary image of stars and galaxies without any extra information (no other spectra info other than image itself). Can you help me distinguish them?

Are there any specific features / properties I could look for? Say for eg the R G or B value or maybe values in some other color space (YUV etc.) be more or less for stars than galaxies, etc.

EDIT:

I dont have any "special" data (no sensor or spectral data), I only have data from the internet (common images). The point is to detect stars and galaxies from images commonly found on the internet.

One of them is this: https://cdn.spacetelescope.org/archives/images/large/heic0916a.jpg

If you have the CCD data and instrument profile, then a star may be treated as a "point source" and a galaxy as an "extended source". This means that, under ideal conditions a star will have it's light spread through the "point spread function" of the telescope, usually these objects have a Gaussian(ish) intensity profile with one pixel of the CCD claiming most of the light and adjacent pixels falling off in intensity, while a galaxy will have an extended, flattened profile with 2 or 3 or 20 pixels claiming most of the light and then falling off in adjacent pixels.

If you have the full CCD information, this is the method of choice. This breaks down when galaxies are so far away that they are resolved smaller than the "point source" stars (larger telescopes can resolve farther away galaxies as extended than smaller, back yard telescopes).

If you just have catalog data, such as colors, then you want to look for items falling off of the "stellar color locus." Stars, in color space tend to sit on a "ribbon" in color-color space (see this from Yanny+2000)

Objects with colors which do not lie on this "ribbon" are special objects. For example the box labeled "Q" is Quasars, which is a type of galaxy. Not shown in this image is a "cloud" of objects which are galaxies (in this image, the cloud will start around that quasar box and extend much redder in both color axes). If you take all of your objects from a telescope and plot them in any(ish) color-color space you will find a tight band, which are stars, and a cloud near this band (which are galaxies). This is because stars operate as a blackbody, while galaxies operate as a dual blackbody (with a lot of light emitted by the stars, following the same blackbody radiation profile, and a lot of gas absorbing this light and re-emitting it in cooler, infrared blackbody colors).

This method is not perfect, this is a statistical method for selecting say 90% galaxies.

Note: Rob is absolutely correct that you cannot do any of these with jpeg images. A lot of jpegs, especially promotional images have had their colors doctored, or even include false colors (for example to represent ultraviolet or infrared light; this is also why you can't do this with random jpegs, each camera will have it's own way of representing color, so my polaroid camera picture will give different colors than a modern random DLSR camera... you can only do this if all the images are collected from identical cameras). jpegs also do not include the "full CCD information" ie. the intensity of light falling onto each pixel of the CCD.

Note 2: Yes you can estimate the PSF for a telescope with an image, but only a CCD image which shows the photon count in each pixel. A CCD image is a 3d image, which shows how many photons fall into each bin, a jpeg does not have this information, so it is... impossible I think, unless the jpeg is produced to show the photon count and not the color.


Exploiting symmetry in the axisymmetric Jeans equations I'm solving the axisymmetric Jeans equations to determine the initial conditions for an exponential disk inside an NFW halo. The density profiles of the two components are
\begin{equation}
\rho_d(R,z)=\rho_{0,d}\exp\left(-\frac{R}{R_0}\right)\exp\left(-\frac{|z|}{z_0}\right)
\end{equation}

\begin{equation}
\rho_h(r)=\frac{\rho_{0,h}}{r/r_s(1+r/r_s)^2}
\end{equation}

Due to the axial symmetry, for the meridional velocity dispersion we have
\begin{equation}
\sigma^2(R,z)=\frac{1}{\rho}\int_z^\infty \rho \frac{\partial \Phi}{\partial z'}dz'
\end{equation}

As the galaxy is symmetric about the disk plane ($z=0$), we would expect the distribution function at a distance $h$ above the disk plane to be identical to that at $z=-h$; in other words, if we flip the galaxy, the only difference will be a change in direction of particle motion, but not speed or dispersion.

Therefore, can't we replace the $z$ in the limit with $|z|$?

Background: If $f(\mathbf{x},\mathbf{v},t)$ is the distribution function of the stars in phase space and $n = \int f\,\mathrm{d}^3\mathbf{v}$ is the star density, then in the cylindrical coordinates $(R,\varphi,z)$, the $z$-component of the second of the Jeans equations applied to an axisymmetric system is
$$\renewcommand{\expv}[1]{\langle{#1}\rangle}\partial_t(n\expv{v_z}) + \partial_R(n\expv{v_Rv_z}) + \partial_z(n\expv{v_z^2}) + \frac{n\expv{v_Rv_z}}{R} + n\partial_z\Phi = 0\text{.}$$
In a steady state, the first term vanishes, so if the positive and negative terms of the $z$-component of velocity terms balance, then so do the second and fourth, in which case:
$$\partial_z(n\expv{v_z^2}) = -n\partial_z\Phi\text{,}$$
which we can then integrate.

If the the star and mass distributions are axisymmetric and symmetric about the equatorial ($z = 0$) plane, then the potential has even symmetry in $z$, $\Phi(R,z) = \Phi(R,-z)$, and so does $n$. Thus their derivatives with respect to $z$ are odd functions, and $\Phi_{,z}(R,z) = -\Phi_{,z}(R,-z)$ implies
$$\frac{1}{n}\int_{-z}^z n\Phi_{,z'}\,\mathrm{d}z' = 0\text{,}$$
since the integrand has odd symmetry. Hence,
$$\frac{1}{n}\int_z^\infty n\Phi_{,z'}\,\mathrm{d}z' = \frac{1}{n}\int_{|z|}^\infty n\Phi_{,z'}\,\mathrm{d}z'\text{.}$$


Whats the name of this eye-shaped galaxy? I would like to know the name of this eye-shaped galaxy:

Strange eye-shaped galaxy has black-hole iris, Betsy Mason, 23.07.2009, Wired.com:

That is galaxy NGC 1097, a barred spiral galaxy in the constellation Fornax. It has been distorted by interactions with neighbouring galaxies and has a weakly active galactic nucleus, with jets coming from the nucleus. The jets contain stars from a galaxy that has recently (in astronomical terms) been cannibalized. The ring in the centre is a region of star formation, surrounding the nucleus of the galaxy at about 5000 light years.


Formulas to determine the illuminated phase and orientation of the Moon The phase of the Moon (the bright limb) looks different depending on where on Earth the observer is located and the time of the observation. I've located this video and the accompanying spreadsheet that supposedly handles this sort of calculation, however I am somewhat confused on what formulas are actually used. I would greatly appreciate if someone with a bit more experience could decipher the spreadsheet a bit and perhaps write down or link me to the formulas being used.

I'm most interested in determining the illuminated phase (i.e the size of the bright limb) and the apparent orientation of the illuminated portion relative to the horizon. My goal is to understand how this type of calculation is done, in order to re-implement it in code for an application to use - displaying the typical crescent moon icon for night-time, but at the correct angle for a specific location and time.

PS! First post here, so I hope I have used the correct terminology for things.

EDIT: Eventhough the author of the spreadsheet left an answer to the topic, it doesn't feel like it really answers the question. Not to sound clueless and needy, but the question talks about formulas, which I know can be derived from the spreadsheet, however as I have no idea how to approach this, then the answer I was hoping for would help a bit more.

I created the spreadsheet.  It's a bit complicated, to say the least.  Initially I was inputting the formulas by hand from the book I cited in the video, but the process was rather error-prone so I ditched those tables and simply scanned the pages electronically and used the numbers from the scans to construct the tables of the Fourier series used to compute the moon's position.  Some of the tables are therefore redundant and aren't used.  You may be able to use a much simpler method to compute the position of the moon though; for my purposes I needed arcsecond accuracy.  If all you're interested in is computing the approximate phase angle and apparent orientation you can get away with a much more simplified calculation for the moon's position that is accurate to within a few tenths of a degree.  In that case the main thing you need from my spreadsheet are the formulas for computing the phase angle and apparent orientation.  You'll need to calculate the approximate location of the sun as well in order to determine these things.  The spreadsheet does this as well, using a simplified method described by Jean Meeus in "Astronomical Formulae for Calculators."  He also gives a simplified method for computing the position of the moon.  Either "Astronomical Formulae for Calculators" or "Astronomical Algorithms" (both by Jean Meeus) would be excellent references for your project.  I can walk you through the formulas you need if you need help with that part.