Who or what will the Arecibo message reach? The Arecibo Message was aimed at the globular star cluster M13, but as far as I read, M13 will move out of the original position and thus not be able to receive it. So, who or what will be able to receive the message?

The Arecibo Message was not broadcast with the aim that it would intercept with a notable astronomical object. Neither do we realistically expect that anyone will ever listen to the Voyager Golden Records.

Such symbolic gestures are simply vectors for public relations and education. We send these "time capsules" into space because we can, as a showcase of our technological achievements, and as a token of hope.

  "As the choice of frequency, duration of message, and distance of the
  target clearly shows, the Arecibo message is very unlikely to produce
  interstellar discourse in the foreseeable future. Rather, it was
  intended as a concrete demonstration that terrestrial radio astronomy
  has now reached a level of advance entirely adequate for interstellar
  radio communication over immense distances."

  -- NAIC staff (1975), http://adsabs.harvard.edu/abs/1975Icar...26..462

  "The spacecraft will be encountered and the record played only if there are advanced spacefaring civilizations in interstellar space. But the launching of this bottle into the cosmic ocean says something very hopeful about life on this planet."

  -- Carl Sagan


How would a planetary defence against comets work? Mappings of practically all potentially dangerous asteroids is well underway. But comets seem almost hopeless to protect against.

How far away are approaching comets detected today? How long time before they would pass Earth orbit?
Is it feasible in our time to map comets in their beyond Neptune orbits, or will we only detect them while approaching?
How much more difficult is it to reach a comet compared to an asteroid? I imagine that the high speed and potentially high inclination of a comet is a challenge for rocketry. Rosetta reaches a comet soon after 10 year trip, but I suppose they picked the easiest to reach comet known.

They used several Mars and earth gravity assists to raise Rosetta's aphelion to around 5 AU. The comet's aphelion is also in that ball park. The comets from Jupiter's neighborhood (about 5 AU) are moving about 40 km/s when they're in our neck of the woods (earth moves about 30 km/s). A comet moving 40 km/s wrt the sun at 1 AU can be moving anywhere from 70 to 10 km/s wrt to the earth, depending on inclination. A head on collision would be 30 + 40. If the comet was moving the same direction as the earth, it'd be 40 - 30.

A comet from the Kuiper Belt would be moving just a hair under escape velocity when in our neighborhood. Escape is sqrt(2) * circular velocity. So a comet from the outer system would be moving 42 km/s wrt to the sun.  The velocities wrt to the earth could range from 72 to 12 km/s. The Kuiper Belt is 30 to 50 AU from the sun.

A comet from the Oort Cloud would also be moving a hair under escape velocity, 42 km/s in our neighborhood. But Aphelion would be much higher, up to 50,000 AU. Presumably many of the non-periodic comets are from the Oort. Click the comet links in this list and you can compare discovery date with last perihelion. For example Comet C/2011 W3 (Lovejoy) was discovered November 27, 2011 and it's .55 AU perihelion occurred December 16, 2011.

In JPL's description of the LINEAR NEO search program, they say "… with most of the efforts going into searching along the ecliptic plane where most NEOs would be expected. " But objects from the Oort don't tend to the ecliptic plane. In my opinion a high inclination comet from the Oort with our name might escape undetected until it was way too late to do anything about it. But an Oort comet with our name on it is very unlikely.


Calculating distance from center of mass of two stars in a binary system I am confused on how to calculate the distances for each star in a binary system to their centre of mass. I am trying to solve it from a book I am reading which shows the solution but I do not understand the solution.
So the info I am given is:
A system is located at a distance D (in parsecs). The two stars orbit around their centre of mass. Star A has an angular separation of a (in arcsecs) and star B has an angular separation of b (arcsecs).
So now I am trying to find the physical separation of the stars to their centre of mass.
The book's solution (once converting to correct units) is:
Distance for star A = D tan(a)
Distance for star B = D tan(b)

I don't fully understand why they use tangent here because when I draw out a diagram of this situation i seem to need to use sine not tangent.
Here is how I drew it:

Why do they use tangent function and not the sine function? Have I drawn my diagram correctly?

It actually doesn't really matter: the angular separation is sufficiently small that you can use the small angle approximations $\sin \theta \approx \theta$ and $\tan \theta \approx \theta$ (where $\theta$ is in radians). In both cases the next term in the Taylor series is $\mathcal{O}(\theta^3)$ and the effect on the answer of dropping this and all subsequent terms of the Taylor series is in most realistic cases going to be utterly dwarfed by the uncertainty in the distance.


How does a Super-Massive Black Hole 'flare'? This week we have read of evidence of the Super Massive Black Hole in the centre of the Galaxy 'flaring'.

My understanding is that light cannot escape from a black hole.

My question is: How does a Super-Massive Black Hole 'flare'?

The flare happens just outside the black hole.  Matter which aproaches a black hole but not perfectly straight on doesn't fall into it immediately. It gets pulled around the black hole, just as a satellite is pulled around the Earth.

If the object is a single small particle, that's almost the end of the story. Gravitational radiation and the precise effects of GR will cause its orbit to decay until it eventually falls in, but if it's small enough not much of anything is emitted in the process.

If there are lots of objects in the neighbourhood, or if your obect is something relatively large and diffuse (like a gas cloud, or a planet or star) on the other hand, different things happen. As gravity pulls on different parts of it with different strengths (tides) the object gets stretched and then pulled to pieces, and multiple objects collide and interact in all sorts of ways. Magnetic fields also start to play a role. Some of that matter escapes completely, but much of it ends up in a disk (a little like Saturn's rings, but less well organised) around the hole. By this point all the stretching and collisions have made it very hot, so it shines.

Now even within the disk there are collisions and friction, and they have the overall effect of heating the matter in the disk and pushing some of it out of orbit, so that it falls into the hole, getting even hotter in the process.

So finally we come to a flare. There are probably many kinds of flare-up in this basically unstable setup, but the most obvious is when a particularly large object or cloud of gas falls into the accretion disk, or when some instability in the disk causes a large amount of matter to fall out of the disk. Either way, the larger amount of hotter matter shines more, and we see a flare.


Simple definition of Bondi Radius Can anyone help me by defining what is exactly Bondi Radius? I have searched far and wide but the results are very complicated and so far I have only rough idea of it but I just cant explain it very well

From the mysterious site Vaporia, which is a fantastic index for astrophysics terms (and other things I'm seeing now):

  The Bondi radius is analogous to the Hill radius, both giving an idea
  of how far from an object, a medium (gas, dust) is likely to be drawn
  in and accreted. The Bondi radius takes into account the (relative)
  speed of the object through the medium, and the medium's density and
  sound-speed. Accretion due to material falling within the Bondi radius
  is called Bondi accretion.

Try to picture a heavy object, perhaps a planet, travelling through a gaseous or dusty medium. Now, some of the gas is eventually going to be the victim of accretion to the planet. The planet is going to pull gas/dust from the medium and the Bondi radius tells you how far from the planet gas/dust is likely to be accreted to the planet.

Hopefully this is more clear!


Is Mars brightest from Earth at opposition, perihelion or somewhere in between? After reading this article, I was inspired to run some simulations at NASA's JPL Solar System Simulator to see Earth's and Mars' orbits from above.

From what I could visually determine (from the JPEG images), Mars is at opposition on July 26, but its perihelion is sometime around September.

When (and where in their orbits) is Mars brightest from Earth this year?

JPL HORIZONS gives these values for Mars as seen from Earth.
r is the Sun-Mars distance in AU, delta is the Earth-Mars distance in AU, and S-O-T is the Sun-Earth-Mars angle in degrees.
I added [] to indicate minimum or maximum values.

Date__(UT)__HR:MN   APmag         r     delta    S-O-T

2018-Jul-23 16:00   -2.75   1.40246   0.38911   172.30
2018-Jul-25 14:00   -2.77   1.40106   0.38730   173.27
2018-Jul-27 12:00  [-2.78]  1.39971   0.38600  [173.51]
2018-Jul-29 10:00  [-2.78]  1.39840   0.38522   172.94
2018-Jul-31 08:00   -2.77   1.39714  [0.38497]  171.73
2018-Aug-02 06:00   -2.75   1.39592   0.38523   170.11

2018-Sep-14 12:00   -1.74   1.38147   0.50640   128.93
2018-Sep-16 12:00   -1.69  [1.38144]  0.51602   127.49
2018-Sep-18 12:00   -1.64   1.38147   0.52588   126.08

Mars is at opposition on Jul 27, closest to Earth on Jul 31, and at perihelion on Sep 16.
Mars is at peak brightness for several days between opposition and closest approach to Earth.
A difference of 0.01 magnitude is imperceptible to the eye and smaller than the uncertainty of most ground-based CCD photometry.
Albedo features and dust storms probably contribute at least that much variation in magnitude anyway.


Why do we expect Type Ia Supernovae to be brighter in a decelerating Universe than what has been observed? This seminal review paper in the PNAS by Robert P. Kirshner (https://www.pnas.org/content/96/8/4224) says the following:

  If the universe had been decelerating—–in the way it would if it contained the closure density of matter, that is, if Ωm = 1—then the light emitted at redshift z = 0.5 by a SN Ia would not have traveled as far, compared with a situation where the universe had been coasting at a constant rate—characteristic of an empty universe, where Ωm = 0. For a universe with Ωm = 1, the flux from the distant supernova therefore would be ≈25% brighter. But the distant supernovae are not brighter than expected in a coasting universe, they are dimmer. For this to happen, the universe must be accelerating while the light from the supernova is in transit to our observatories.

I didn't understand this.

Why do we expect Type Ia Supernovae at a particular distance to be more or less brighter based on the matter density of the Universe?

In other words, why should the brightness of anything at a particular distance  depend on whether the Universe was accelerating or not?

Observed brightness (or magnitude) depends on the distance between the source and the observer. The further the distance, the fainter we will observe given the same source.

If the distance r1 is measured between the source and the observer given the accelerating universe, then the distance r2 measured given the decelerating universe would be r2 < r1. Therefore, this answers your second question. For the first question, the answer is directly followed the current model that predicts less matter density given faster expansion of The Universe.


What are the cloud-like blobs in the Martian southern hemisphere? In a very recent edition of New Scientist, an article Mystery cloud-like blobs over Mars baffle astronomers (16 Feb, 2015), detailed that

  on 12 March 2012, amateur astronomers around the world noticed a strange blob rising out of the planet's southern hemisphere, soaring to 250 kilometres above the surface.

According to the article, this anomaly grew to be about 1000km across and even had 'fingers' stretching to space.

The article, details several theories (some being a bit wild) including atmospheric processes, aurora and stretching reality quite a bit - aliens (from the article, not me).

An image of the plume is shown below (a still image from this YouTube clip, which states that a 2nd plume was observed some months later:

The plumes appear to have formed in a region called Terra Cimmeria.

What are scientific theories, models etc for the formation of the cloud-like blobs in the Martian southern hemisphere?

From the Nature article, the authors propose two alternative explanations:

We used photometric measurements to explore two possible scenarios and investigate their nature.

For particles reflecting solar radiation, clouds of CO2-ice or H2O-ice particles with an effective radius of 0.1 micrometres are favoured over dust.

Alternatively, the plume could arise from auroral emission, of a brightness more than 1,000 times that of the Earth’s aurora, over a region with a strong magnetic anomaly where aurorae have previously been detected

Importantly, both explanations defy our current understanding of Mars’ upper atmosphere.

I think, they are the most trustworthy interpretations that one can get.
Please note that, even if the data are from 2012, the paper is published in 2015, therefore the interpretation is not outdated.


Why aren't planet orbits circular? Please forgive the simplicity of my question.  I have only had basic science courses.  Common sense seems to indicate that planetary orbits should be very close to circular, any perturbations having been eliminated over the ages.  Could it be that the composition of the planet be influencing the pull of gravity to produce a different gravitational attraction due to its mass?  I mean that if a planet were composed all of one substance, would the planetary orbit be perfectly circular?  I think about earth having constantly shifting water and gasses and even solids that affect its own rotation.  Might these shifts also affect the orbit of the planet around the sun, or does the total mass of the earth and moon together determine the orbit?

They are close to circular, although we might quibble about the meaning of "close". Except for Mercury, all planetary orbits have eccentricities below 0.1. That's close enough to a circle to actually pass for a circle at a superficial examination.

Mercury 0.2056
Venus   0.0068
Earth   0.0167
Mars    0.0934
Jupiter 0.0484
Saturn  0.0542
Uranus  0.0472
Neptune 0.0086

Check out these ellipses. The first one has an ecc. of 0.1. Looks like a circle, doesn't it? Even the second one (ecc = 0.2) kind of looks like a circle, too, if you don't examine it too long.

As for why they are not exactly circular, down to the last digit? There will always be perturbations that will squish perfect circles a little. Planets interact with each other, and pull each other off perfect circular orbits. It's very unlikely that in a complex system like ours you'll find highly circular planetary orbits.

But, again, the differences between the current planetary orbits and perfect circles are for the most part quite tiny, and you should think of them as "basically circles", unless you're an astronomer or a NASA engineer.

Comets, OTOH, those do tend to move on narrow elliptic orbits.


Original Hubble diagram units incorrect? Figure 1 from Hubble 1929 (original paper with open access pdf) has velocity with units of km. Is this just a typo or was there a reason the plot is printed this way?

It is in fact supposed to be in $\mathrm{\frac{km}{s}}$ (KILOMETRES per second).

I think that whoever authored the paper, probably took it as a random shortcut and assumed anyone reading it would either have read the paper and understand it is supposed to be $\mathrm{\frac{km}{s}}$, or just assume it from the fact it says velocity.


If the Earth was in Venus's orbit, would it be habitable? If the Earth as-is were moved to Venus's orbit (and Venus removed), would the Earth remain habitable, would it eventually turn into what Venus currently is, or would it turn into something else?

I'm mainly wondering which characteristics of Venus (hot, no magnetic shielding, etc) are the result of how it is composed compared to it's proximity to the sun.

Similar question: If Venus and Mars changed places, would we then have 3 habitable planets?

A planet's environment is a complex interplay of factors.  If you dramatically change the insolation of a planet, it becomes very difficult to predict how the other factors which influence the planet's environment will respond.

Venus, for example, seems locked into its high-CO2 environment (which makes its surface hot) because its surface is too hot to allow for the formation of carbonates.  Even if it had water (also needed for the formation of carbonates), it would be continuously boiled away from the surface.

All planets - Earth, Mars, and Venus included - have their temperatures influenced by their albedo.  Albedos of planets are often strongly correlated with cloud cover.  Cloud cover generally increases as a wet planet warms, reflecting more light.  But atmospheric composition can overcome this balance, as in the case with Venus, which has a very high albedo but is still swelteringly hot on the surface.

Even if you get rid of the feedback factors like water and carbon dioxide, the planets have very different nitrogen levels - Venus at about 2 1/2 times that of Earth (enough to cause mild nitrogen narcosis on its own), and Mars almost nothing.  Really, Mars in Venus's position would be in a terrible state - it's been so stripped of its atmosphere even as far from the sun as it is, it would be all the more stripped being much closer to the sun.  Over geological timescales it'd become highly water stripped as well, as it'd be much easier for water to enter a gas phase, and the solar wind can very effectively strip water planets even as large as Venus, let alone Mars.

Venus in Mars' position would probably fare much better.  But it's difficult to say exactly how much.  Given that we can't say with great confidence exactly what set in motion the path that led to its current state today (thicker, more viscous lithosphere preventing crustal overturn? cause or consequence?), it's hard to say whether that would have happened had it been in Mars' orbit.  One curiosity if you took today's Venus and moved it to Mars' position: you may well cause an atmospheric collapse to a supercritical state. It's been theorized that even in its current orbit that's happened before (temperature runs counter to the formation of supercritical fluids).


Am I using my Celestron 8 Schmidt-Cassegrain telescope correctly? I became interested in astronomy a few years ago and started with a Celestron AstroMaster 70AZ refractor telescope. That seemed nice, and I wanted to upgrade to a go-to telescope that was also more much more powerful, so I bought the 8 inch Celestron Schmidt-Cassegrain.

It's great, but honestly I'm a bit disappointed because the magnification doesn't seem all that much stronger than my 70AZ. I haven't done a side by side comparison. But for example when looking at Saturn, I was expecting to see more than just a white ball with a ring around it - I was able to see that with my 70AZ. I thought I'd see colors, and maybe be able to differentiate between some of the bigger ring bands.

I haven't tried any deep sky objects. I'm using various lenses starting from 30mm down to 6mm.

Am I doing something wrong?

Keep in mind that many of the photos you see with colors are composite images.  To differentiate colors try to use some various color filters.

To see bands on Saturn, try using a yellow.  This will bring out different details in what you are seeing.

This link provides some more details for using various filters.
http://www.telescopes.com/telescopes/howdifferentfilterscanbetteryourviewarticle.cfm


How long would it take for a rogue planet to evaporate in the late stages of the Universe? I've read once a popular science book, in which the author calculated how long would it take for the most massive black holes to evaporate because of Hawking radiation. He claimed that after that time the Universe will be the soup filled only with elementary particles.

However, what would happen with celestial bodies that aren't black holes? For example, a rogue planet of the size of Earth, drifting somewhere in deep space, in an extremely cold Universe? What known effects would cause that planet to evaporate? How long would it take?

Before the evaporation of black holes that you mention, our universe will enter what is known as the Black Hole Era. About $10^{40}$ years from now, all the protons and neutrons in the universe will have decayed into positrons and neutrinos. This is how the last of the planets will pass away.

The Black Hole Era is so named because at this point only black holes and elementary particles will remain.

It should be noted that there is currently no evidence for proton decay, and so this phase of the Universe is strictly theoretical.

Source:

The Decay of Heaven - DeepAstronomy.com


What is the difference between a dwarf spheroidal galaxy and a globular cluster? This Wikipedia article states that Segue 1 is either a dwarf spheroidal galaxy or globular cluster associated with our own Milky Way.  I am wondering what the difference is between dwarf spheroidal galaxies and globular clusters, since at first glance is seems that the morphological properties and stellar ages are similar between the two classes of objects.

There are many differences starting from size to the dark matter content... Please check the following link  it has given much information :
 http://www.answers.com/topic/dwarf-spheroidal-galaxy


How to calculate when the moon is highest in the sky to the earthbound observer? Like when is the new moon highest in the sky? Or third quarter? How do I know this for any general moon phase? Thanks.

The Moon moves round the ecliptic. At full Moon it is opposite the Sun. Hence when the Sun is furthest South (in the Northern Winter), that full Moon is that which will be furthest North in that year and so on. But plainly, the Moon moves round the ecliptic every 28 days, so every 28 days or so it is furthest North etc.

In other words there is no answer based on phase or time of year alone.


The physical processes of emission lines in cosmic nebula I think I understand how absorption lines in cosmic bodies occur. But after reading about the emission lines in quasars I am wondering more and more about the physical processes causing the emission lines in cosmic nebula. I hope you can explain these processes to me, more precisely from where to where energy needs to transition and if there are temperature requirements on the involved objects. Below are some of the statements that I encountered and I don't understand:

Emission clouds/nebula need to be ionized - But they need to be only partially ionized, right? If they are fully ionized energy transitions of the electrons wouldn't be possible, so emission lines wouldn't be possible either and there would be only scattering.

Only hot gases create emission lines - Is this really true? Wouldn't it be possible, that a cosmic cloud much cooler than the object illuminating it would also cause emission lines (see drawing below)?

Regarding emission lines from stars I read that they occur due to recombination - is this true? But why only discrete initial energy of the electrons would allow them to recombine with a (partially) ionized atom?

So can you explain the physical processes of nebula emission lines in a bit more detail than the links do? Tnx

Just a short answer, and likely others will fill in more details.

If there is ionization of some atoms, then generally there is recombination as well - you will have both processes going on, roughly in balance with each other.  Typically when an electron recombines with an atom, it does so into some excited state.  Then as it drops from that excited state into lower states, it emits one or more photons on the way down.  This is how ionization gives you emission lines - it’s the inverse process that matters for the emission.

To have emission lines, the cloud only needs to be hotter than the background behind it.  So in your sketch, an observer at the bottom would see an emission spectrum (as you’ve drawn), but an observer to the right would see an absorption spectrum (with the same lines) - from that perspective, more light is lost relative to the (hot) background source than is re-emitted in that direction by the cloud.

This link may help with some additional explanation of these ideas.


Can two detectors for gravitational waves be combined into a very long baseline interferometer? While detectors for gravitational waves are interferometers themselves, the question here is about an increase in resolution resulting from two or more detectors far away from each other.

They do use two or more detectors to triangulate the origin of a source. Most recently the Advanced Virgo detector in Italy was used in conjunction with the two LIGO detectors in the USA. This has been published in Physical Review Letters (Phys. Rev. Lett. 119, 141101 – Published 6 October 2017).

With two detectors the sky area for "pinpointing" the source is quite a large area, but with three detectors the area is reduced by a factor of about 20.


Why aren't there galaxy-sized balls of iron out there? Since we know that stars burn out, collapse, or blow up when too many of their atoms fuse into iron, then how come, with all the ancient galaxies we've been observing, there isn't one galaxy whose stars have become giant balls of iron, which would eventually collide into one galaxy-sized iron globule in space?

In short, you can't make a galaxy-mass ball of iron because there's no way to support the ball against the inward crush of its own gravity. First, it would collapse into a neutron star, but even neutron stars can't hold themselves up beyond about 2-3 solar masses.  Your galaxy-sized object would collapse into a black hole.  But we do see these!  All massive galaxies are believed to have supermassive black holes at their centres. These are up to 10 billion solar masses in size, which is comparable to smaller galaxies.

In any star, be it a main-sequence star like the Sun, a giant, or even a white dwarf or neutron star, the inward force of gravity must be balanced by some outward force or pressure.  (In fact, we basically assume this when we construct stellar models, although it's a very good assumption.) In main-sequence stars, this pressure is provided by the nuclear reactions happening inside them but as a star evolves, it gets a bit more complicated. In some situations, the nuclei get so close together that they their electrons begin to share quantum states. And because no two electrons can be in the same state, they have to occupy higher energy states. This in effect exerts a kind of pressure (called degeneracy pressure), because the electrons are forced to move around faster than they otherwise would. This is what supports white dwarfs and the cores of some evolved stars. e.g. low-mass red giants.

Degeneracy pressure is fine up to a limit known as the Chandrasekhar limit, which is about 1.44 solar masses (depending on the composition of the object in question). Anything above this would collapse. First, this collapse would yield a neutron star, where things are a bit different. But a similar principle holds, and there's a degeneracy pressure because the neutrons themselves also occupy one quantum state each.  The details are much more difficult here, but the overall consensus is that neutron degeneracy pressure can support objects up to about 2-3 solar masses. After that, there's nothing left to balance gravity, and the star collapses into a black hole.


Why doesn’t the Sun fill the sky on Mercury? I’ve seen a lot of photos showing Mercury in front of the Sun when it passes by and you can see just how tiny it is in comparison.

Here’s a great example:

So I’d expect if I was travelling towards Mercury in that photo that as the planet got bigger in my view so would the Sun and eventually I wouldn’t be able to see the edges of the sun because it’s so big at this distance... makes sense that an object gets bigger the closer you get to it!

However what confuses me is that if you were on the surface of Mercury... this is the view you’d get where it’s about a 6 times larger than when viewed from Earth.

So how is this so small when it’s taking the whole view up when we’re viewing it further away from mercury in the first photo? But we’re closer in the second photo but now the sun is smaller... How is the first photo possible (it’s an actual photo taken by NASA) why isn’t it taking up the whole sky in the second photo when you’re actually on the planet?

When looking at the first image, you need to take into account perspective distortion in images with a narrow field-of-view. This is the same effect that makes people look closer together than they actually are when photographed through zoom lenses (which is useful to bear in mind when reading the news these days!). Mercury is not as close to the Sun as the first image makes it appear.

You also need to bear in mind that objects in space tend to lack "depth cues": in the atmosphere, objects further away will become blurrier and more "washed out" due to scattering. In space, this does not apply.


What is the physical interpretation of the derivative of the emitted and observed wavelength? The flux F of an event at redshift $z$ is related to its luminosity L as
$$F=\dfrac{L}{4\pi d_L^2}\,,$$
where $d_L\equiv d_L(z)$ is the luminosity distance between us (the observer) and the event (the emitter).

We also know that a photon emitted at a wavelength $\lambda_{em}$ is observed at
$$\lambda_{obs}=(1+z)\lambda_{em}\,.$$

We might want to know how does the observed spectra of the event $f_{\lambda}$ is related to the spectral energy distribution (SED) of the source $l_{\lambda}$. We can then deduce that
$$\dfrac{dF}{d\lambda_{obs}}=\dfrac{1}{4\pi d_L^2}\dfrac{dL}{d\lambda_{obs}}=\dfrac{1}{4\pi d_L^2}\dfrac{dL}{d\lambda_{em}}\dfrac{d\lambda_{em}}{d\lambda_{obs}}$$
$$f_{\lambda}(\lambda)=\dfrac{1}{4\pi d_L^2}l_{\lambda}\left(\dfrac{\lambda}{1+z}\right)\dfrac{1}{1+z}$$
$$f_{\lambda}(\lambda)=\dfrac{1}{4\pi d_L^2 (1+z)}l_{\lambda}\left(\dfrac{\lambda}{1+z}\right)$$

Although I understand that the factor $(1+z)^{-1}$ comes from the change of variable between emitted and observed wavelength, what is the physical meaning of it?
As an example, the $d_L^{-2}\propto (1+z)^{-2}$ and those two factors arises from the photons losing energy because of its wavelength being stretched and arriving further apart because the space between them grew.

I think you can understand that factor as follows:

When photons lose energy, they're spread out over a larger wavelength range. Since there is a fixed number of photons, that means that the number of photons per observed wavelength bin decreases by a factor of (1+z), and hence the flux density, which is really what $f_\lambda$ is, decreases by this factor.


What is the density profile within the Sun's photosphere? Which one of these is wrong? The Sun's photosphere contains the Sun's surface as defined by opacity = 2/3 point. I'd like to see the profile of mass density from bottom to top of the photosphere. I did a quick search and got confused.

The image below is found on the Wikipedia Photosphere page. If I trace the dotted line labeled "Density" to the photosphere layer and read the density axis at the top, I read something like $8 \text{ to } 3 \times 10^{-7} \text{ g/cm}^{-3}$, which you could call $1 \times 10^{-6} \text{ g/cm}^{-3}$. However, the Sun section of Wikipedia page where this image is shown says:

  The Sun's photosphere has a temperature between 4,500 and 6,000 K (4,230 and 5,730 °C) (with an effective temperature of 5,777 K (5,504 °C)) and a density of about 1×10−6 kg/m3; increasing with depth into the sun.

and links to the solar-center.stanford.edu page The Sun's Vital Statistics for the $1 \times 10^{-6} \text{ kg/m}^{-3}.$ Converting the units, this is only $1 \times 10^{-9} \text{ g/cm}^{-3}.$

Question: Is it possible to clear up this disparity, and to see a plot of the density versus depth from the bottom to the top of the Sun's photosphere, which would likely contain both positive and negative heights above the Sun's surface?

Source: File:Sun Atmosphere Temperature and Density SkyLab.jpg
Original source: SP-402 A New Sun: The Solar Results From Skylab

I usually don't answer my own questions, but sometimes when the question itself is called into question I make an exception.
The density of the photosphere at $\tau_{5000}=1$ is predicted to be $3 \times 10^{-7} \text{g/cm}^3$ in the Holweger-Müller Model Atmosphere7.
As pointed out in comments there is a spread in values here.
The 1E-06 g/cm^3 density value (plot) is more consistent with  = 1 or "bottom" of the photosphere, while the density in the quote is more consistent with the cooler "top" of the photosphere (circa 4300 K).
From Chapter 2: The Photosphere of Timo Nieminen's thesis Solar Line Asymmetries: Modelling the Effect of Granulation on the Solar Spectrum

Figure 2-3: The Holweger-Müller Model Atmosphere
7 Holweger, H. and Müller, E. A. “The Photospheric Barium Spectrum: Solar Abundance and Collision Broadening of Ba II Lines by Hydrogen”, Solar Physics 39, pg 19-30 (1974). Extra points have been cubic spline interpolated by J. E. Ross. The optical properties (such as the optical depth and the opacity) of a model atmosphere are, obviously, very important, and will be considered later. See table C-4 for complete details of the Holweger-Müller model atmosphere including all depth points used.
8The height scale is not arbitrary. The base of the photosphere (height = 0 km) is chosen to be at standard optical depth of one (i.e.  5000Å = 1 ).


Correlation between stellar mass and galactocentric distance As we move away from the centre of, say, a spiral galaxy, is there a relation between the stellar mass and distance?
I know the how the stellar abundance and density varies with the distance but does stellar mass also follow a pattern?

In terms of the distribution of stellar masses when stars form -- the Initial Mass Function (IMF) -- there doesn't seem to be much variation, at least for spirals. (There is evidence that massive elliptical galaxies may have different IMFs as a function of radius, with relatively more low-mass stars formed near their centers. The cause of this is unclear; it might have to do with different conditions for star formation, with very dense, high-pressure gas in what would become the centers of massive elliptical galaxies possibly producing different distributions of stellar masses.)

The current distribution of stellar masses at a given radius in a spiral galaxy will depend almost entirely on the local star-formation history. Since stellar lifetime depends on stellar mass, regions with recent star formation will have more stars that are massive; regions where star formation ceased a long time ago will be dominated by low-mass stars, since all the higher-mass stars there have died.

In most spiral galaxies, the mean age of stars tends to decrease as you move further out in radius (as indicated by optical colors: reddish central regions are mostly old populations dominated by red giants, while the disk further out is blue because there are younger populations dominated by hot, massive stars). This is an example of what's called "inside-out" star formation.

However, a lot of lower-mass spirals seem to show a reversal of this trend at large radii, with redder (older, mostly lower-mass) stars beyond a certain radius. Since this is the same radius at which the density of stars in the disk stars falling off more rapidly (the "break" or "truncation" radius), it's usually thought this is the result of a combination of two effects: 1) A radial threshold/cutoff in star formation, with very few stars being formed beyond a certain radius (possibly because the gas density becomes too low); and 2) Radial scattering of stars over time by spiral arms, which means that older stars tend to be scattered to larger (or smaller) radii simply because they've had more encounters with spiral arms. In this case, the relative fraction of more massive stars would increase with radius out to the truncation radius, and then decrease beyond.


Can I look at the sky and find the day of the week? Suppose I wake up from a coma on a desert island in the 19th century (i.e. we already use the Gregorian calendar but have no satellites yet). I have a clear view of the sky and a couple of days to make observations and calculations. Can I determine the current day of the week? Also, can I find the answer instantly?

On a related note, is it possible to define days of the week in a purely astronomical (e.g. planet-agnostic) manner such that for the Earth we get the traditional meaning?

It is possible to determine the day of the year using the distance of the sun from the vernal equinox (you have to account for the equation of time etc, but it is possible). It is also possible to determine the year if your measurements are precise enough, using the current position of the vernal equinox (which changes due to precession). The ambiguity of the factor of 26000 years can be removed by noting the positions of other stars (for example, Barnard's star). Knowing these two, you should be able to find the day of the week.

About your second question, Astronomers like to deal with Julian date, which is a continuous day counting system, and you can determine the day of the week from the Julian date, though only if given a reference, for example Jan 1, 2000, was a Saturday. As such, a week and days of a week are man made (https://en.wikipedia.org/wiki/Week).


How Soon Could a Waxing Crescent Moon Be Seen? Question:

Apart from a Solar Eclipse, How much time is needed until a Waxing Crescent Moon be seen following a New Moon?

Would the time of year be significant? The Vernal Equinox + 1 Month.
Could it be seen the same night that it occurs?
Could it be seen in the day when it occurs?
Would this be affected if the viewer was in a rural location, without lights? Or would city lights help?

Context:
Ancient Lunar calendars depended on observation of the New Moon, (like political / religious obligations).
What was the potential margin of error? Hours? a Day? (Not considering bad-weather.)
From Wikipedia - A New Moon Cannot be Seen:

In astronomy, new moon is the first phase of the Moon, when it orbits as seen from the Earth, the moment when the Moon and the Sun have the same ecliptical longitude. The Moon is not visible at this time except when it is seen in silhouette during a solar eclipse when it is illuminated by earthshine.

Related:

Is it possible for a person to not see the new moon at different places on earth?
From which country or area is the new moon visible first?

From the US Naval Observatory: Crescent Moon Visibility:

  Naked-eye sightings as early as 15.5 hours after New Moon have been reliably reported while observers with telescopes have made reliable reports as early as 12.1 hours after New Moon.

USNO is usually a very reliable site for info like this.


Where can I look up the 3D positions of the closest stars? The problem:

I want to simulate a travel from Earth to the closest 15-25 stars. Which is the fastest way to accomplish this?

What i have:

I currently have the distance between Earth and the nearby stars, according to this link: http://www.space.com/18964-the-nearest-stars-to-earth-infographic.html. But from it I can't know if Alpha Centauri (4.3 light-years from Earth) is closest to Sirius (8.58 light-years from Earth) or to Wolf 359 (7.8 light-years from Earth).

What I need:

I need the distance between any two stars, at least the closest ones to Earth. Then it would be possible to run a Traveling Salesman problem, in order to visit all the nearby stars (or a subset of 15-25 of them) in the fastest time.

Where can I get this information?

You can determine the 3D position for a star, $p$, by using the distance from Earth, $R$, as well as the right ascension, $\Omega$, and declination, $\delta$, within the following formula.

$p = R \begin{bmatrix} \cos \Omega \cos \delta \\ \sin \Omega \cos \delta \\ \sin \delta \\ \end{bmatrix}$

Note this will yield positions within the Earth-centered inertial reference frame.


Gift for 16 year old Aspiring astrophysicist,,,, I'm an outsider here, so please bear with me.
I have a 16-year-old nephew interested in the field of astrophysics and am looking for a birthday gift.  I had thought about a book on astronomy, but don't see that field listed as a core pre-requisite for that career.  Is knowledge of astronomy a core pre-requisite/useful for this profession?
Thank you

there is a series of educational Japanese manga comics books called "The Manga Guides" they explain a particular subject of science and maths (like Physics, Relativity, Regression Analysis, Statistics, etc) and they are fun to read, I gifted the one on Physics to my 14-year old nephew and he liked it. Also the book has actual exercises between chapters and explained examples, and each book are independent, so you are not obligated to read all of them to finish the story.
https://en.wikipedia.org/wiki/The_Manga_Guides#The_Manga_Guide_to_Physics
There is one on the Universe and General Relativity too. I read the one on Relativity out of curiosity and it was very entertaining


Help identify a bright and round object photographed through a telescope It was a brightest object in the sky. Attached is the picture taken on 25th January around 6pm in Riga / Latvia, towards the south direction:

Apologies for a quality of the picture, the telescope was inside the flat and the picture was taken through the window.

The object was visible for few hours while it was traveling from the south-east towards south-west. I observed this bright object previously on couple other days, it's the first time I pointed my tellescope in its direction.

View of the sky: https://imgur.com/iwRfmKa

It was a brightest object in the sky.

This already makes one think "Venus?"

At about 6 PM on 25-Jan-2019 from Riga Latvia the planet Venus was about 17 degrees above the horizon in the Southwest.

Venus is always strikingly bright when it's visible, it's apparent magnitude stays between about -3.8 and -4.8.

From in-the-sky.org's planetarium mode

This matches nicely to your photo out the window:

But what about the donut-shaped image?

Your image is way, way out of focus. The the outside of the annulus represents the diameter of the telescope's primary and the inside represents the diameter of the blockage by the secondary mirror holder. The blockage is about 45% of the radius or about 20% of the area, which seems reasonable for some telescopes. Here's a random example of a telescope with a similar blockage.

Here are some other celestial donuts to enjoy! In this case they are caused by a tiny spot of dust which is out of focus, and one is dark rather than bright, but they are other examples of how point-like objects that are out of focus show you your aperture.

above: from this answer, below: from this answer


Where exactly is the modification that first corrected the spherical aberration in Hubble's primary mirror? The April 2015 CBS News article How NASA fixed Hubble's flawed vision - and reputation describes the modification taken to the Hubble Space Telescopes by the Space Shuttle that was used to correct for the strong spherical aberration of the primary mirror due a highly accurate but incorrect figure.

above: "Due to an oversight during fabrication, the concave shape of Hubble's primary mirror was too shallow toward its outer edges by 2 microns, a tiny fraction of the width of a human hair. As a result, starlight was not brought to a focus at the same point, resulting in blurry images. These three images show the same star as viewed from a ground-based telescope, left, and from Hubble's uncorrected mirror, center. The corrected image, after a space shuttle repair mission in 1993, is seen on the right" from here. credit: NASA
From the article:

But the mirror's biggest impact was on the Wide Field Planetary Camera, the instrument expected to provide the dazzling visible and near-infrared images most easily understandable by the public and of tremendous value to astronomers.
"The real killer was Jim Westphal (the WFPC principal investigator at the Jet Propulsion Laboratory in Pasadena, California), who tended to be very negative," Weiler recalled. "Basically, he said he couldn't do anything. So that's what I was preparing to walk into the press conference with."
But as it turned out, Weiler had two aces up his sleeve. One he knew about, one he didn't.
The ace he knew about was drawn seven years earlier, when Weiler kicked off work to build a backup camera, the Wide Field Planetary Camera 2, a carbon copy of the original. John Trauger was the principal investigator, the man in charge of the instrument at JPL.
The ace Weiler didn't yet realize was built into the design of Trauger's camera.
Shortly after the first light image came down, "one of the grand old men of optics here at JPL, Aden Meinel, he and his wife Marjorie wanted to see what the image looked like," Trauger said in an interview. "We brought it up on the screen. ... And he looked at that for five minutes and he said 'that looks like spherical aberration.'
"That was the first time I heard spherical aberration," Trauger recalled. "The thing was, he also said 'well, if it is spherical aberration, you could fix it with Wide Field Planetary Camera 2.' He knew that we had these nickel-sized mirrors in our optical system, which received a sharp image of the primary mirror itself. ... The very last mirror before the image is created on our CCD was an opportunity to straighten the wavefront out. He recognized that, too."
By giving that mirror a prescription that exactly countered the spherical aberration in the primary mirror, the WFPC 2 would be able to achieve a perfect focus and do all of the science that was intended.

Question: Where exactly is the modification that first corrected the spherical aberration in Hubble's primary mirror? Can it be identified within the image below, or is it located elsewhere? Was the modification only a change in the figure of one of the small mirrors, or were other optical elements modified as well as part of the correction?

above: "The business end of COSTAR: Small mirrors on motorized arms directed corrected light into the Hubble Space Telescope's spectrographs and faint object camera. COSTAR and the Wide Field Planetary Camera 2 were installed during a 1993 shuttle mission, correcting Hubble's flawed vision." From here. Eric Long/Smithsonian Institution

The image provided is of COSTAR which replaced the High Speed Photometer and fed corrected light to the Faint Object Camera (FOC), Faint Object Spectrograph (FOS) and the Goddard High Resolution Spectrograph (GHRS). Since the WFPC2 was already going to replace WFPC1, the correction for the primary mirror's spherical aberration could be designed in (as it has been for all subsequent instruments).

The overall layout of the HST optics (the Optical Tube Assembly (OTA)) and the WFPC camera optics is shown below in a figure taken from the Rodgers and Vaughan 1993 SPIE paper:

Light enters from the top left and is focused as normal by the HST primary and secondary mirrors and then part of the beam is picked-off into the WFPC2 camera.

The WFPC2 optical configuration is shown in more detail below (taken from the instrument description section of the WFPC2 Instrument Handbook) . The text just above that states:

  The aberrated HST wavefront is corrected by introducing an equal but
  opposite error in each of the four Cassegrain relays. An image of the HST
  primary mirror is formed on the secondary mirrors in the Cassegrain relays.
  (The fold mirror in the PC channel has a small curvature to ensure this.)
  The spherical aberration from the telescope's primary mirror is corrected
  on these secondary mirrors, which are extremely aspheric.

indicating that the correction for HST's primary is done on the part labeled 'Secondary Mirror' in the diagram. Note that is the secondary mirror of the mini-Cassegrain telescope that is inside each of the 3 widefield and 1 planetary camera channels inside WFPC2 itself and not the main telescope secondary mirror which is off the diagram and part of the 'OTA' (Optical Telescope Assembly)


What telescope might this be a model of, with Sir Fred Hoyle? The BBC article Archaeopteryx: The day the fossil feathers flew has nothing to do with Astronomy but it shows a file photo of the famous astronomer Sir Fred Hoyle with a model of what looks like some kind of telescope.

I don't know if the "dish" is supposed to be fixed metal to reflect radio waves, or a rotating dish of mercury to reflect light, but I'm guessing the former is more likely. It looks like an off-axis, tilted secondary mirror reflects radiation back to ground level, but it's just a model so I can't be sure.

Are there any ideas what real instrument (if any) or at least what design that this fixed-"dish" model might be a model of?

What's scary is that at the bottom right it looks like a building several stories tall, which means that the telescope is huge, and the size of Sir Hoyle is on par with Godzilla. The time frame seems right for that too! ;-)

The photo looks like it could be from the late 1950s/early 1960s. Jodrell Bank's Lovell telescope had been commissioned and people were looking for the next big thing in radio astronomy.

That next big thing was the fixed Arecibo dish, with a diameter of 305m, commissioned in 1963.

It is clear though that the picture is not a (very accurate) model of Arecibo and the location is wrong. I would guess instead that this picture purports to be the design for a radio telescope on the Moon or is an upscaled model of the Dover Heights "sandpit" telescope (see below) built in the 1950s in Australia.


How can I identify a star? I'm obviously a novice. I see a star in the eastern sky every night. It is quite large and impressive. How can I identify it. Thank you.

If an astronomical object appears to be very bright, and does not appear to 'twinkle' very much, then it is probably a planet, rather than a star. Jupiter is almost on the opposite side of the Earth to the Sun right now, so it will rise in the East when the Sun is setting in the West. Since Jupiter does look like a very bright star to the naked eye, and since it is in the part of the sky that you describe, then the object that you are seeing is almost certainly Jupiter.

As for recognising actual stars, you really need to learn the layout of the constellations to some extent. This will allow you to find your way around the night sky. A useful tool for this is a program like Stellarium. This will show you the positions of stars and other astronomical objects, including the planets.


Are there any simulations of the Grand tack? The BBC program with Brian Cox "The Planets" involved some description of the Grand Tack hypothesis to explain the current state of the solar system.

This theory has Jupiter wandering through the solar system causing various orbital disruptions. Are there any computer simulations/CGI videos available to show how Jupiter may have achieved this? On a sightly unrelated note: I recall reading Velikovsky's "Worlds in collision" many years ago. He had a similar idea but attempted to attribute these changes to biblical events / comets. He was many orders of magnitudes out in terms of timing and the size of the disruption, but I suppose he could be credited with hinting that Geological uniformitarianism might not be applicable to the solar system.

Kevin Walsh, lead author of the original Grand Tack paper has a page on his website discussing the Grand Tack model and subsequent work. At the bottom of this page there is a movie of the evolution taken from the numerical simulation that was the basis of the paper and the Grand Tack model. Sean Raymond (one of the other authors on the Walsh et al. 2011 paper) also have some movies on his research webpage on the Grand Tack model. There is also a presentation from the 2019 Sagan Workshop which discusses terrestrial planet formation in more detail and presenting the 2 other possible solutions, along with the Grand Tack model, to the early Solar System formation problems such as the so-called "small Mars" problem.


Which kind of properties can we get for cosmic ray particles hitting on an optical ccd? It is very common that we meet cosmic ray particles in optical images recorded by CCDs.
You can see the "snowflakes" in the hubble images below:

Generally we should remove them in order to get photometry or spectroscopic info.

However, there may be useful info we can derive, for example, can we get their directions or energies statistically before they arrive at a CCD?

There is some correlation between the ADU value of a cosmic ray point and its real energy?

Cosmic ray "hits" are artifacts caused when high energy particles (cosmic rays, often muons) slam into atoms in the CCD itself and liberate large numbers of electrons, which then show up as bright spots and streaks when the CCD data is read out. They can also occur as a result of radioactive decay processes in the material in the detector and instrument itself and also the immediate surroundings.

They are a tremendous nuisance that can be removed to a certain extent by median stacking images - though this also fails if you get into a situation where cosmic ray hits are found in the same pixels of $>1$ images in your stack. This limits the lengths of exposures that can safely be made with astronomical CCDs. Depending on exact usage, this limit is usually between 30 minutes and an hour.

The flux of muons and the distribution of angles at which they reach the detector is actually very well known already. It is largely unaffected by local shielding (unless your telescope is buried far underground!)

It is possible to distinguish different types of events (muons, locally produced gamma rays etc.) from looking at the pattern made on the detector (e.g. Groom 2004), but I don't think you can tell much about their energies, unless it could be by looking at the flux as a function of how much shielding (lead?) or how far underground the detector was, since often only a small fraction of the cosmic ray energy ends up deposited in the CCD.


Does the new moon appear at the same time around the world or in one region last My query is basically, say for example if the new moon is seen in the Middle east region, does it mean that it was the day of new moon in sub continental country as well ? You can't have a new moon in India for example the next day scientifically, right ? Please help me understand this with diagrams so I can explain it to others.

To me it doesn't make sense that we will have two consecutive new moons in the same region of the planet. New moon in middle east but new moon in India is on the next day.

I am aware since it's a broad topic so timezone can change the answer so in this question. I am restricting my question to the sighting of the new moon in the same region as in say Moon is sighted in Dubai, doesn't it make sense that India will also have the first day of moon on the same day given that it all falls in the same plane.

If the new moon is measured by moonrise then it's possible that it could land on a different day in different longitudes.   Or, if it's not measured by day, but by astronomical position, it's still possible that it would fall on a different day in different timezones.    180 degrees of longitude should equal roughly a 50% chance of landing on one day before or after (2 days apart isn't possible).

30 degrees of longitude should be roughly 1 chance in 12 of landing on a different day.    More detail is in the possible duplicate question.

Map of timezones.  While the lines aren't precisely straight, each line should correspond to about 15 degrees longitude.


Problems with computing X,Y,Z positions of planets at a certain time in Unity3D (C#) I'm trying to create a little program in Unity3D that positions the planets in 3D space at a given time. I've followed this document: http://stjarnhimlen.se/comp/ppcomp.html.

Using the data from section 4 and the code from section 6 and 7 I should have the correct algorithm right? I did skip on the changes in the orbital attributes (eg + 4.70935E-5 * d) because I just need to have them move in a set orbit. I don;t really need them to change over time. For now I'm just increasing the mean anomaly over time without taking the period in account.

The problem is that the planets do not seem to get the correct offset to the ecliptic, and their movement is really weird: they speed up, then slow down and reverse in their paths. Here is a video of their movement: https://www.youtube.com/watch?v=zHOBChFpan0

It looks like the eccentricity is correct on their ellipses and the distances to the sun are also correct, but the rest is just wrong. I'm not that great at math or astronomy to fully understand what's going on in this algorithm, but I'm hoping someone here might spot the error.

using UnityEngine;
using System;

public class Orbit : MonoBehaviour
{
public float longitudeOfTheAscendingNode;
public float inclination;
public float argumentOfPerihelion;
public float semimajorAxis;
public float eccentricity;

public float meanAnomalySpeed = 0.1f;

float M = 0; //mean anomaly

void Update()
{
    SetPosition();
}

void SetPosition()
{
    float N = longitudeOfTheAscendingNode;
    float i = inclination;
    float w = argumentOfPerihelion;
    float a = semimajorAxis;
    float e = eccentricity;

    float E = M + e * (180 / Mathf.PI) * Mathf.Sin(M) * (1 + e * Mathf.Cos(M));

    float xv = a * (Mathf.Cos(E) - e);
    float yv = a * (Mathf.Sqrt(1 - e * e) * Mathf.Sin(E));

    float v = Mathf.Atan2(yv, xv);
    float r = Mathf.Sqrt(xv * xv + yv * yv);

    float xx = r * (Mathf.Cos(N) * Mathf.Cos(v + w) - Mathf.Sin(N) * Mathf.Sin(v + w) * Mathf.Cos(i));
    float yy = r * (Mathf.Sin(N) * Mathf.Cos(v + w) + Mathf.Cos(N) * Mathf.Sin(v + w) * Mathf.Cos(i));
    float zz = r * (Mathf.Sin(v + w) * Mathf.Sin(i));

    transform.position = new Vector3(xx, yy, zz);

    M += meanAnomalySpeed * Time.deltaTime;
    if (M > 360)
        M -= 360;
}

}

I think E and M are in degrees in your code, but the trig functions in the math library in C# expect an argument in radians.

One way to handle this is to define constants  radians = Pi/180 and degrees = 180/pi then always do Mathf.Sin(M*radians). You will also need to do this for the lines float xx=.

Alternatively, you could work entirely in radians. Convert all angles to radians at the start and only use radians in the code. You would need to convert n, i, w and M to radians and remember that e and v will also be in radians.


Is there a database of past sunspots? I would like to plot a butterfly diagram. I've been looking through the SOHO archive for a table that would indicate the date and position of each sunspot, but to no avail. The closest I got was to use the archived images to create the table myself, but that would be extremely fastidious.

Is there a place where I might find this kind of data?

The National Solar Observatory historical archive lists at least one source of sunspot data. Under 'Sacramento Peak Telescope'->'Evans Solar Facility' there is a Dropbox link to Sunspot numbers data which seems to have spot numbers against time in ASCII format; this may be a suitable resource to get you going.


What are the small things, look like stars, moving in the sky? I have always been observing something that looks like a star but not as bright as a star moving from south to north at the same time every night.

What is it? It cannot be a star because I know stars do not move.

A starlike point visibly moving across the sky is a satellite. The biggest and brightest is the international space station, but there are many others.

They can be seen any night that is clear, usually shortly after dusk, or before dawn. http://www.heavens-above.com/ allows you to see predictions of when satellites will be overhead.


What is the exact meaning of vicinity of Black hole or a disk? the same as inside? What is the exact meaning of vicinity of Blackhole or a disk? Is this the same as inside the horizon for black holes? If so, why people don't say inside of BH, simply. What is the difference?

I use this phrase actually! and it may seem obvious but not to me, please somebody make it clear. I searched but I couldn't find an exact answer.

Edited: Until which radius from Blackhole I can use this word meaningfully? It's not clear to me its astronomical dimension. "near" is not enough to me. If it could be inside a horizon and also outside of it, it makes a huge difference from point of view of the observer at infinity, as you know, how I can call them by the same name and sometimes I mean one sometimes the other one.

From the Oxford English Dictionary:

  "vicinity" just means "the area near to or surrounding a place"

so "in the vicinity of a black hole" just means "near to a black hole."

Other dictionaries say "immediately next to" so in response to your comment about vagueness, treat vicinity as the immediate neighbourhood of the black hole. You will not find an exact meaning - it is used as it is in English. There is no specific distance measurement. You don't have a distance from radius which would stop being in the vicinity.


How did Astronomers mostly(?) agree to publish arXiv preprints along with peer-reviewed Journals? Was there pushback? This excellent, thorough and well-sourced answer to Has a gravitational microlensing event ever been predicted? If so, has it been observed? includes four links to papers on adsabs.harvard.edu pages, and all of these have arXiv preprints.
In fact a large number of posts link to the arXiv preprint version of astronomical papers and I suspect most astronomy papers have arXiv preprints.
To my understanding, this allows other astronomers earlier access to and awareness of ongoing research results which may allow the field to progress faster, and also allows everyone access to the paper without penetrating a paywall. This may have implications for the folks who collect the money which is paid at the wall.
Question: How did astronomers manage generally agree to do this? Was there much pushback by astronomical journals and/or by some authors in the astronomical community? If so, how was that resolved?
note: This may also be on-topic in Academia SE, but I think that both the pools of potential answerers and of interested readers of this bit of history in the field of Astronomy here is substantial.

Obligatory XKCD, hat tip to @JamesK

This is an interesting question, and I think it can productively be split into a couple of distinct parts, separating the publisher side from the author side:

Was/is there any pushback from journals about posting pre-publication versions of papers on the web?

Why do research article authors in astronomy post papers on arxiv at a relatively high rate?

Part 1 is easier to answer, since there are fewer individual actors to consider.  As far as I can tell, the answer to this is a flat "no".  As @ProfRob notes in a comment, there was a long tradition of dissemination of preprints on paper before the web existed, so it likely would have been difficult for publishers to forbid that practice once electronic dissemination became a possibility.  Many journal publishers are directly connected with professional societies, and even for-profit journals need to show support for wide dissemination of ideas. I don't know of any journal that publishes astronomy research that doesn't allow posting of preprints.
Contrary to what one often hears, the higher-profile journals like Science and Nature are on board with this idea as well; both of them are very explicit in stating policies that allow both pre-publication posting on sites like arxiv, and free discussion of results at meetings. (current Nature policy, current Science policy).  What they don't allow is authors discussing results with the press before publication. Nature in particular seems to emphasize the importance of preprints; they published their first editorial on it in 1997, and they seem to publish articles every few years that say, in effect, "no really, it's ok."
Part 2 is harder to answer, as is any question about why a practice evolves in some particular setting that involves individual decisions by many different people to adhere to some unstated norm.
First I thought I'd check to see whether it's really true that astronomy is different from other physical sciences in its use of preprints.  The article linked in the answer from @Thomas suggests that astro is on the high side (at least as of 2011) in use of arxiv.  I did my own spot check with a few ADS queries.  To keep it simple, I just picked a few journals. In the years 2015-2019, this ADS query shows that there were a combined 24,634 refereed articles published in The Astrophysical Journal and Astronomy & Astrophysics (to pick a major American journal and a major European journal). Of these, 21,568 have preprints available on arxiv. (You get this by adding the qualification "property:eprint_openaccess" to the query.)  So that's almost 88% of papers, which seems high indeed.  To compare to another sub-field, I chose the journal Physical Review B (which publishes papers on condensed matter physics, since I know there's a cond-mat section of arxiv), yielding 25,952 articles, of which 16,936 have arxiv preprints.  So that's 65%, still pretty high, but a fair bit lower than astronomy.
So, why might this be?  My guess is that part of the answer might lie in the creation of ADS, and in particular the ADS project to scan all of the previous astronomical research literature and make it freely available online. That made it possible to sit at your desk and do all of your bibliographic research, rather than going to the library.  Once people got used to doing that, it was natural to want to do it for the newer literature, too.  Of course those electronically published articles were likely available to many researchers directly from the publisher, through their home institutions, but that early digital access was pretty heterogeneous by journal and sometimes not so smooth.  That might have contributed to people wanting to put papers on arxiv to make them more available.
It's certainly more complicated than this, but I do think the existence of ADS (including its indexing of arxiv) may have made a difference in earlier / more complete adoption of electronic dissemination in astronomy compared to some other fields.
A related thing that I think is interesting is that the use of arxiv in astronomy has now become so ubiquitous that some researchers consider posting on arxiv to be important not just for early visibility, but for visibility at all.  By that I mean that the daily arxiv postings have become, for many, the way that they learn about new literature, supplanting the role of journal table-of-contents services.  I don't know how widespread the practice is (and I'm sure there are some generational differences), but I have certainly heard people say that arxiv is the only resource they use for regularly keeping up on new papers that are published.  In other words, there's a sense in which simply publishing your paper in a refereed journal is not enough to make it fully visible to the community, at least not in the short term (though people will likely find it in literature searches later).
To check this idea, I went back to the ADS astronomy searches referenced above and looked at citation numbers.  The 21,568 papers with arxiv preprints have 532,493 citations total, while the 24,634 papers overall have 561,738 citations.  Subtracting, that means that the 3,066 papers without arxiv preprints have 29,245 citations. Looking at average citations per paper, that means that the arxiv preprint papers have 24.6 citations per paper on average, while the non-arxiv-preprint papers have 9.5 citations per paper on average.  That is such a huge difference that it actually makes me wonder if there is some confounding factor I'm not considering - I encourage others to check the numbers and/or to try other searches.  But if that's true, it's a huge effect, and bears out the idea that arxiv plays an important role in the community in research visibility overall, not just in early access.
Finally, I'll note that the importance of arxiv is now so complete that you even see an open letter from the AAS journal editors to arxiv, asking them to include the Research Notes of the AAS (a new publication of short, unrefereed papers) in the arxiv.  There has been an ongoing discussion of this, since arxiv moderators started (mostly) rejecting arxiv postings of RNAAS papers.


What is the faintest magnitude a telescope can see? This website claims:

  The space telescope can detect objects as faint as 31st magnitude...

It's referring to Hubble, but does not cite any source or math. How do you calculate the faintest (apparent) magnitude that a space telescope can detect?

This is a term known at the limiting magnitude. This term refers simply to the faintest apparent magnitude your instrument can detect. Wikipedia has an article describing limiting magnitude.  I pulled the info from there.

The wiki article begins off explaining the basic way of finding limiting magnitude.

$$5 \cdot \log_{10}\left(\frac{D_1}{D_0}\right) $$

Now, if we simply plug in some values, 254 mm telescope and assuming the eye is 6 mm, we get an answer of 8.13.  This can be rounded to nearest base and we can say that it increases the magnitude by 8.

The article goes into more description about the contrast of the given medium. With Earth, we have the atmosphere, lights, and pollution to worry about. The more you are able to reduce these effects, the better. That's why for telescopes at observatories, we place them in high places, usually away from populated areas to help reduce these worrisome effects. This helps increase the contrast between the object you are looking at and its medium (in astronomy the night sky).

In the article it explains this with the formula below:

$$m \cdot v = m_{nakedeye} - 2 + 2.5 \cdot \log_{10}(D \cdot P \cdot t)$$

$D$ = objective or main mirror diameter in mm
$P$ = power or magnification
$t$ = transmission factor, usually 0.85–0.9

Simply putting in our values from above and estimating a 10" telescope has a 250x magnification with a transmission factor of 0.85, we get how the wiki article states that the answer of a 10" telescope easily getting a magnitude 15 is possible.

Now for observational laboratories and something like the Hubble telescope, they use additional factors, including image processing mechanisms that help reduce noise in the image and integration time of the object you are viewing.  This means using a sensor that can capture the image at low light levels AND while it moves.

If you really want to know more, go here. This link actually explains in very technical detail the technology behind the imaging discovery efficiency.  Basically, it helps boost the limiting magnitude to detect extremely faint objects.


Timescale of ignition of a protostar? What is the timescale of start of nuclear fusion as T Tauri type star transforms into a Main Sequence star?

Wikipedia article on T Tauri type stars mentions:

  Their central temperatures are too low for hydrogen fusion. Instead, they are powered by gravitational energy released as the stars contract, while moving towards the main sequence, which they reach after about 100 million years.

The mentioned 100 million years is the period of the star being in its stable (well, as stable as the turbulent T Tauri type stars get) state without nuclear fusion. Then, once the fusion starts, we get between 3 million and hundreds of billions years of main sequence, depending on mass of the resulting star.

What I'm interested is how long is the period of transition between the two - ignition of the nuclear reaction - time between "all energy produced by gravitational contraction" and "most of energy produced by nuclear fusion".

I imagine this period could be quite short, and the effect quite rapid and turbulent as the initial fusion increases local temperature (and as result pressure) drastically, leading to conditions conductive for fusion spreading rapidly over the volume which is already on brink of entering the fusion everywhere within the protostar, essentially a nuclear wildfire encompassing the gathered gas, a chain reaction starting.

Am I right in my guess this process is rather rapid? Was it ever observed? Or contrary, does the intensity of the fusion reaction rise gradually and slowly from zero over many millions years of star formation?

I have mulled over this a couple of times (it's a really interesting question!), and hopefully come up with a somewhat enlightening answer. I haven't been able to find a good, modern reference for these details (perhaps I just suck at literature searches...) so there's a little mucking around in the history books

The total timescale of evolution onto the main sequence for a protostar in the T Tauri mass range ( < 3 solar masses) is on the order (of magnitude) of several tens of millions of years. The ignition of fusion is not precisely a "runaway" reaction: however it occurs relatively quickly and once it starts, gravitational contraction quickly ceases.

The evolution of a 1 solar mass protostar follows these basic steps. Things are a little different for different masses - too complicated to explain here but the references should provide ample further reading!

A Jeans-unstable cloud of gas and dust begins to contract, exchanging gravitational potential energy for kinetic energy, and thus heat. The luminosity of the protostellar cloud increases as it collapses. It takes around 100,000 years for the initial period of rapid collapse to finish, at this point the cloud is very luminous (perhaps 20 solar luminosities and 8000K).
Over the next 1 million years, the protostellar cloud slowly contracts and cools to around 4500K. The protostar then travels down the Hayashi track, contracting further but changing little in temperature - its luminosity continues to fall. This is the stage where T Tauri stars are at. Most T Tauri stars are younger than 3 million years old.
The star then follows the Henyey track, where the luminosity begins to slowly increase again as a radiative zone develops in the star's core and it continues to slowly contract. This can take some tens of millions of years.
Finally, the conditions in the core are extreme enough for fusion to begin. The timescale from all the energy being provided by gravitational contraction to all the energy being provided by fusion, is on the order of 1 million years. The star's luminosity (counter-intuitively) decreases again when this happens, as the energy from fusion doesn't quite offset that from gravitational contraction, which ceases when fusion begins.

  Figure: the The Lg/L curve describes the amount of energy gained from gravitational contraction over the total luminosity of the star. The logarithmic time axis is in seconds (reproduced from Iben (1965), Figure 3).

  References:

I. Iben (1965), Stellar Evolution. I. The Approach to the Main Sequence
R. Larson (1972), The evolution of spherical protostars with masses 0.25 M_solar to 10 M_solar
L. Henyey et al (1955), The Early Phases of Stellar Evolution

  Interesting reading I came across for somewhat higher mass protostellar formation:

F. Palla and S. Stahler (1991), The evolution of intermediate-mass protostars. I - Basic results


Long-lived supernova remnants? We cannot see the supernova remnant of the star that triggered the formation of the Sun and the Solar System, as the remnant dispersed and became mixed into the interstellar medium. So I am wondering, what is the maximum size/age of a supernova remnant/planetary nebula can reach before it can no longer be distinguished?

Of the order 10000–100000 years.  As an example of an old Supernova remnant, you can see the Vela SNR, estimated at 11,000–12,300 years old.
Nasa describes the evolution of a remnant
Initially, the shock wave expands freely, As it grows it sweeps up material from the interstellar medium which slows its expansion. Expansion causes adiabatic cooling (or the Sedov-Taylor phase) until the point that electrons start to recombine with nuclei. This then allows for more effective thermal emission of UV radiation, which cools the remnant further, and it "disperses into the surrounding medium over the course of the next 10000 years."
A paper (mentioned by Rob in comments) describes this in greater detail, and offers this graph of the radio luminosity of a supernova remnant.

 Illustration of a model radio light curve for Type Ia and CC
SNR, evolving through the ejecta-dominated and Sedov-Taylor phases. The
dashed line marks the transition to the radiative phase, where we assume the
synchrotron emission becomes inefficient,


Dynamical modelling of galaxies with counter-rotating components using the Schwarzschild method I am currently trying to understand dynamical modelling using Schwarzschild's method. I have been doing some reading the past days, but I suspect that I am getting something really wrong. According to my current understanding, the Schwarzschild method for dynamical modelling using spatially-resolved kinematic data works like this:

Fit one or more 2D profiles (e.g., Gaussian) to the image or zeroth moment of the observation.
De-project the above fitted model to 3D space and create a 3D luminosity distribution.
Convert the 3D luminosity distribution to mass using a bunch of mass particles.
Create a series of possible orbits for the above mass particles and calculate the resulting velocity vectors.
From the above velocity vectors create velocity and velocity dispersion maps and calculate the chi-squared between them and the observed velocity and velocity dispersion maps.

The last step is what troubles me. To my understanding, the orbits of the particles in a Schwarzschild model can be counter-rotating. Right? In a galaxy observation, when we have two collisionless components (e.g., stellar components) that counter-rotate, it is possible that we will observe double-peaked spectral lines. Right? This information is impossible to be stored using a velocity and a velocity dispersion map, but it can be stored in a spectral cube. Yet, all papers on dynamical modelling using the Schwarzschild method fit models to velocity and velocity dispersion maps instead of a spectral cube.

So, from the above I have concluded that my understanding of the Schwarzschild method is not correct. I would really appreciate if someone could help me to understand what I am getting wrong.

Thank you in advance.

Convert the 3D luminosity distribution to mass using a bunch of mass particles.

Well, you probably could do that, but I suspect it would be inaccurate and extremely time-consuming. Most Schwarzschild modeling approaches parameterize the 3D luminosity-density distribution using functions from which the gravitational potential can be relatively easily calculated.

For example, one popular approach is to approximate the observed 2D light distribution using the sum of multiple Gaussians (as you mention), which can be deprojected under certain assumptions into the sum of multiple 3D Gaussians to get the 3D luminosity-density distribution. Then you multiply this distribution by a mass-to-light ratio (usually one of the parameters of the model you're ultimately trying to fit to the data) and derive the gravitational potential from that. No need to approximate things with a bunch of particles.

This also allows you to easily add extra hypothetical "dark" components, like dark-matter halos or central supermassive black holes, which have simple analytical descriptions and don't need to modeled with particles.

  To my understanding, the orbits of the particles in a Schwarzschild model can be counter-rotating.

Correct. In practice, if the system is (assumed to be) axisymmetric, you can calculate a set of orbits, and then easily generate their counter-rotating counterparts by just reversing the velocity vectors.

  Yet, all papers on dynamical modelling using the Schwarzschild method fit models to velocity and velocity dispersion maps instead of a spectral cube.

You're clearly not reading the right papers ;-).
Let me divide this into two parts:

1. Converting spectroscopic data to stellar-kinematic data

Before any modeling takes place, the observed spectroscopic data (long-slit and/or integral-field-unit "spectral cubes") must be analyzed to extract the stellar-kinematic data. This is done by dividing up the data into different spatial bins on the sky (bins along the slit, or 2D bins for a spectral cube) and summing the spectra within each bin, to improve the S/N.

Then you fit each spectrum with a weighted combination of stellar spectra (observed or from models), broadened by a model1 line-of-sight velocity distribution (LOSVD). In very simple cases, the LOSVD is assumed to be just a Gaussian with a specific central velocity $V$ and velocity dispersion $\sigma$. The more common present-day approach is to model the LOSVD with a Gaussian which is multiplied by third- and fourth-order Gauss-Hermite polynomials. The strength of the 3rd-order polynomial is parameterized by $h_{3}$ and measures skew; the 4th-order polynomial is parameterized by $h_{4}$ and measures kurtosis. For galaxies with counter-rotating components, $h_{3}$ and $h_{4}$ provide some information about the non-single-Gaussian nature of the LOSVD, although it's rather fuzzy and poorly constrained.

A more sophisticated approach (requiring high-S/N data) is to "non-parametrically" model the LOSVD with a histogram having twenty, thirty, or more velocity bins. To show how the non-parametric approach works, consider the following stellar-kinematic data for the S0 galaxy NGC 4191:

The first plot, from Fig. B1 of Krajnovic et al. 2011 (MNRAS, 414, 2923), shows $V$ and $\sigma$ maps (from the SAURON IFU). In the velocity field, you can see that the velocity reverses between $r \sim 5$ and 10 arc seconds; at roughly the same radius, $\sigma$ peaks. The paired off-center peaks in $\sigma$ are suggestive of a region where the two counter-rotating components are about equal (as argued by Krajnovic et al. in their paper).

The second plot, based on Fig. 2 of Coccato et al. 2015 (A&A, 581, A65), shows the $\sigma$ map (colored panel, rotated 90 degrees with respect to the figure above) and non-parametric LOSVDs (31 velocity bins) for three of the bins (histograms) from VIRUS-W IFU observations of NGC 4191. You can see how, in the second histogram, the double peaks due to the overlapping, counter-rotating stellar components are clearly represented.

2. Comparing the Schwarzschild model with the kinematic data

After computing the library of stellar orbits for a given galaxy potential and assigning a set of weights to the individual orbits, you "observe" the model by orienting the galaxy to match the observed orientation, dividing it up into projected 2D spatial bins that match the 2D binning used for determining real galaxy's stellar kinematics. Then (for each bin) you compute the weighted LOSVD by integrating along the line of sight through that bin, adding up the radial-velocity contributions from each orbit that intersects the line of sight. If your data consists of $V$, $\sigma$, $h_3$, and $h_4$, then you fit the resulting model LOSVD for that bin with the Gauss-Hermite function to get the model bin's $V$, $\sigma$, $h_3$, and $h_4$. If your data are non-parametric LOSVDs, then you can simply bin the model LOSVD using the same velocity-binning scheme you used for the data.

This set of predicted LOSVD values (either $V$, $\sigma$, $h_3$, and $h_4$ or the non-parametric binned velocity histogram) is what you compare with data's LOSVD values for that spatial bin, repeating for all the spatial bins with kinematic data.

1 "Model" in this case having nothing to do with the Schwarzschild modeling.


I can't see any detail through my telescope I have a AILINK 114mm diameter, 900mm focal length, reflector telescope. I'm able to see the moon very well however, I can't see Planets very clearly

The first two images are what I can see and the second two is what I want to see
Is that possible for my telescope also, how can I take better pictures like those examples.

Note* it was pre windy and the image was a little better then in the picture

The planets appears to be smaller than the Moon, so it is more difficult to see the details on them. These are some of the factors that determine how good the image will be.

Steadiness of the scope. You mentioned that it was windy, so I assume that the scope was vibrating a lot. To get a good view, you need a scope that does not vibrate when you do not touch it. (A rule of thumb is that the vibration should dampen out in 1 or 2 seconds when you are done focusing or moving the scope to track an object.)

Steadiness of the air. If you look over a paved road on a hot day, you can see how the air is shimmering. Now imagine magnifying that shimmering air by 50x or 100x, and that is what you are dealing with when using an astronomical telescope. The steadiness of the air will limit the size of details that can be seen. (Astronomers refer to this as the "seeing". The shimmering of the air is generally not as bad as my example of the road, but looking through 20 miles of atmosphere and magnifying it make it seem as bad.) The telescope can also create air currents if the scope is warmer than the air. Waiting 15 to 60 minutes (depending on the design and size of the scope) to let the telescope cool down to the air temperature can help with the air currents inside the tube.

Viewing an object when it is high in the sky will help with the seeing because you are looking through less atmosphere (and less distortion). Viewing an object when low in the sky will usually be disappointing. (Move to a location where the object is higher in the sky! :-)

Quality of the mirror A telescope mirror needs to be ground and polished to a high degree of accuracy. Without someone performing a test on your scope, it is hard to know if the figure is excellent, acceptable, or poor. (I have not ground a mirror myself, but I have read that it is not hard to make a good mirror. Or course, it is even easier to make a poor mirror!)

Quality of the eyepieces The eyepieces are just as important as everything else. Some eyepieces provided with small telescopes are of low quality. If the diameter of the eyepiece barrel is about 1 inch (25 mm) or has the letter "H" in front of the eyepiece size (like H20mm), the quality is likely to be poor. 1.25 inch (32 mm) diameter eyepieces are generally better quality.

Magnification A rule of thumb is that the maximum useful magnification is 50 times the diameter of the optics in inches (or 2 times the diameter of the optics in mm). For 114 mm, that gives a useful magnification of 220x. You will find that lower magnifications are used more often because the image gets worse at higher magnification, especially when the seeing is poor. 50x to 150x should be enough to give good views of Jupiter and Saturn.

Collimation A Newtonian reflector has two mirrors. If the mirrors are not aligned properly, the image quality will deteriorate. (The alignment is known as collimation.) The way to check the collimation is to look at a bright star high in the sky. When in focus, the star should be a sharp point surrounded closely by a faint ring. When you un-focus the image slightly, the star will be a larger dot, and the faint rings (diffraction rings) will be a little larger and easier to see. The rings should look like a bulls eye around the dot of the star. The image should be the same on either side of the focus.

If the image is not round, or if the image changes from inside focus to outside focus, either the quality of the mirrors is poor, or the mirror is warped in the mount (too much force in the holding mechanism is bending the mirror!) or the collimation is off. The warping of the mirror can be corrected by not applying as much force on the clips that hold it in the cell. The collimation can be adjusted to correct for problems with the alignment. See How to Align Your Newtonian Reflector Telescope on the Sky & Telescope website for example. A poor quality mirror cannot be easily corrected unless you know someone with the skill to re-polish it.


Would a satellite in geosynchronous orbit between the earth and moon track across the sky together? On Nov. 25 at 2:30a.m. from southern California while viewing the moon with a 70mm D x 400mm F telescope a point of light was seen at about the 7 o'clock position below the moon. As viewed the point of light was clear and seemed to be tracking with the moon. As I tracked the moon the point of light stayed within the same distance and in view. If it was a planet or star wouldn't they track slightly different than the moon ?

The answer to the headline question is clearly: No.

The full moon has an apparent diameter of about 0.5°. Its apparent motion is roughly 360°/24h, neglecting the orbital motion of Earth and Moon. Hence it apparently moves about 24h*0.5°/360° = 0.033h = 2 min.
Hence the moon moves one moon diameter in about 2 min.
(The neglected orbital motion induces an error of about 3.65% (0.997d/27.3d), divide the siderial Earth day by a siderial moon orbit, hence an error of just a few seconds for the apparent motion of 0.5°.)
In 35 minutes, Moon's apparent motion would be about 17 moon diameters.

The geosynchronous satellite, in contrast, remains staying at the same apparent position, by definition.

Planets or stars move apparently about the above error estimate different from the moon, hence roughly 27.3-times slower relative to the apparent moon than a geosynchronous satellite.


Can black holes be used as an energy source I’ve heard that black hole will be rotating continuously which stores a huge amount of energy in its rotation. Will this energy be accessible, since the rotation is imposed on the space outside the hole.

What you are thinking of is called the Penrose process - a means of extracting rotational energy from a black hole.

The basic idea is that you allow a composite body to fall towards a rotating black hole. Once it reaches the ergosphere - that region, outside the event horizon, where the "frame dragging" due to the rotating spacetime is so severe that no body can remain stationary from the point of view of a distant observer - the body should split. One half falls into the black hole and the other orbits and flies out again. If things are arranged correctly then the returning part of the body can have a kinetic energy that exceeds the rest mass energy of the half that has fallen into the black hole. The energy comes at the expense of slowing the rotation of the black hole.

The Penrose Process (from Brito et al. 2015)

If the rotating black hole has an accretion disc (as many do if they are in binary systems - that's how they are identified) then another method is the Blandford-Znajek process. This may actually be the way that quasars and many of the energetic phenomena in gamma ray bursts and black hole binary systems are powered. In simple terms, a magnetic field that threads the accretion disc can be twisted up by the ergosphere. Material can then be propelled in jets along the rotation axis, but is given a spin by the field lines, which extracts angular momentum (and rotational energy) from the black hole. This energy can then be released when these jets run into other things.


Conversion from Equatorial Coordinate to Horizon Coordinates I'm developing my own Planetarium software using C++ following the book Practical Astronomy with your Calculator or Spreadsheet 4th Edition.

I don't know if I my formulas are correct or not, but when I check my results (Azimuth and Altitude) with Stellarium, they are different.

I have also checked my values with this page, but they are different. By the way, the azimuth and altitude given in that page are also different from the values given in Stellarium.

Do you know where can I find the formulas to covert celestial coordinate to horizontal coordinates?

It depends from which celestial coordinates you are converting. I suppose you are working in equatorial coordinates. Then, you can convert as following:

$$\tan A = {\sin h \over \cos h \sin\phi_o - \tan\delta \cos\phi_o} $$ $$ \begin{cases}
 \cos a \sin A = \cos\delta \sin h \\
 \cos a \cos A =  \cos\delta \cos h \sin\phi_o - \sin\delta \cos\phi_o
\end{cases}$$ $$\sin a = \sin\phi_o \sin\delta + \cos\phi_o \cos\delta \cos h$$

Where $A$ is azimuth, $a$ altitude, $\alpha$ right ascension, $\delta$ declination, $h$ hour angle and $\phi_0$ the observer's latitude.

Source: Wikipedia


Do NEA (Near Earth Asteroids) have minable water ice? Water is very useful for rocket fuel and life support in space. But does water, in abundance and form which conceivably could be usefully mined this century, really exist in NEA's? Or do we have to go all the way out to the frost line, the Asteroid Belt, to find asteroids with available water? (Lunar polar water ice is not topic here, I think, but low delta-v asteroidal water).

No and yes.

I'll address the "no" answer first. The answer is no if you truly mean "water ice". Ceres is close to the limit of where water can exist as water, as opposed to in the form of hydrated rock. Those intriguing bright spots on Ceres may well be water ice that has been exposed to vacuum (but they might just be salts). Ceres most likely is inside the frost line, but just slightly. That's part of what makes Ceres so intriguing. Ceres orbits at 2.7+ AU. It is not a Near Earth Asteroid. Contrary to mainstream thinking, perhaps the current frost line is at 2.7 AU (as opposed to 5 AU). There's no way that it is at 1 AU.

The answer is yes if you look beyond ice as a source of water. Some carbonaceous asteroids contain water in the form of hydrated minerals. Some of them contain quite a bit of water, up to 22%. We'll find out more when OSIRIS-REx launches next year and returns a sample from 101955 Bennu in 2023.


Is parallax a "technique to measure the star distances" or a "tiny shift in star's position"? I am not completely sure whether the Parallax is a "technique to measure the star distances" or a "tiny shift in star's position"?

The a book says two points about Parallax:

Astronomers have developed a variety of clever techniques for measuring the vast distances to stars, one is called Parallax.
Astronomers can measure a star's position once and then again 6 months later and calculate the apparent change in position. This tiny shift in stars position is called its parallax.

Apart from the book source, whatis.techtarget.com defines Parallax as the way an object’s position or direction seems to change depending on viewing angle.

It is both - a small shift of the position of a star on the sky as we see it, and a means of estimating the distance to the star.

The apparent position (with respect to very distant objects like quasars) changes because our viewing point changes as the Earth moves around the Sun in its orbit. The amount by which the position changes is inversely proportional to is distance.


How do you calculate length of day in a multiple-star system? How can you calculate the length of day for a planet (given a particular latitude and longitude on that planet) orbiting multiple stars?

What about a planet that is orbiting one star, but that star has another star orbiting it?

Are you asking how to measure a day for an exoplanet?  Because that is pretty hard given that they are far away and hard to image their surfaces.  But, if you are asking how to define the day, say for simulations, then that is pretty easy, actually.
From Wikipedi's entry on Day.

For a given planet, there are three types of day defined in astronomy:
stellar day - an entire rotation of a planet with respect to the distant stars
sidereal day - a single rotation of a planet with respect to the vernal equinox
mean solar day - average time of a single rotation of a planet with respect to the sun as the central star

The first two have to do with the spin of the planet with respect to the remote stars and therefore the complexity of the stellar system does not enter, to first approximation.  The 'stellar day' depends on how much time it takes for some remote galaxy, for instance, to come back to the same point in the sky as seen by an observer on the planet.  The 'sidereal day' depends on how much time it takes for the direction that is at the extension of the equator of the planet with the plane of the orbit (pick 1 of the 2 points) to come back to the same point in the sky. So presuming that the spin axis and orbital plane are not changing much (and conservation of angular momentum law helps here), this is all well defined.  If the planet's orbital plane is changing rapidly then there simply is no well defined sidereal day, but still there would be a valid stellar day.
The 'mean solar day' is the average time it takes for the nearest star to return to the same spot in the sky.  Generally, even in a binary or triple system, the planet is close enough to one that this is well defined.  This is because, if  not, the planet is probably not in a stable orbit and will quickly leave the system.  However, if the stars are in a tight configuration  and the planet's orbit is wide and far enough away from them, it could be stable.  Then the simplest definition for 'mean solar day' would use the center of mass of the stellar system instead of the nearest star (which would change over time).
There are a few stable complicated orbits, like a figure 8 orbit, where the planet can switch back and forth between two stars.  Unlikely, but theoretically possible.  In such a case, the mean solar day is ill-defined, but the stellar and sidereal day are well defined.