Detect Passage of Massive Relativistic Object I came across an interesting article that speculates that we could detect an object of significant mass ("space-ship size") moving through our galactic neighborhood at a substantial fraction of the speed of light with existing technology.

That got me wondering (for a sci-fi novel I'm writing) whether there would be evidence of such passage near the point in space where it occurred years or decades after such passage far from any star.

The CMB interactions mentioned in the article would spread from the point of passage at the speed of light so would have long departed the area. Is it likely that some artifact of passage would still be present, such as a change to the interstellar medium (change to ionization rates, fusion of lighter elements, etc)? How quickly would any such changes disperse? For purposes of the question, assume that scientific observations are possible at the point of suspected passage.

Probably not.

I believe that your question can be stated "Can we detect a lingering effect on the local environment in interstellar space a decade or two after the passage of a relativistic starship?"

The first step is to ask just how relativistic? If it's just "really fast" (say, 0.99c) there would be one set if effects, but if it was much fast -- 0.99999c, for instance -- some new effects would come into play.

The second is to ask about the density of gas in the vicinity. If it's like normal interstellar space (typically a million atoms per cubic meter) its passage will be much harder to detect than if we're in something like a Giant Molecular Cloud where densities can be 100 to a million times greater.

First off, you won't see any lingering electromagnetic radiation: That will necessarily be light years off by a decade later and you specified local effects.  That leaves two things that I can see: (a) Effects on the pre-existing gas, and (b) bits of ship that remain behind.

The interstellar medium is nearly entirely atomic hydrogen, either neutral or ionized. The temperature is typically 100-1000 K. The mean free path of these atoms is very large: 105 miles up to a million times that. The mean speed of the atoms is about 1 kps.

That means that any effect the passage of the ship has on the local interstellar medium will be entirely erased by the diffusion of the interstellar medium (IM) in a short time.  Imagine that the ship blows a tube of vacuum in the IM a million miles across. Neutral hydrogen atoms from outside of this region (just a few mean free paths away) will move in at 1 kps. In a few million seconds, the void created by the ship's passage will be refilled. (A  million seconds is about 10 days, so a single year is plenty of time for the IM to diffuse back to equilibrium.)

If the IM is much denser, the effect may linger as long as a decade, but we're not talking about the IM anywhere near us.

The second potential source is bits of the ship left behind. Hitting the IM at .99c is precisely the same as the IM hitting the ship at .99c, and that's quite hard radiation -- the IM's protons are hitting at 5 GeV. Unless the ship has force shields of some sort, this is likely to cause some spallation -- atoms of the ship's front end eroded into space. This might be detectable, but the problem I see is that those bits are unlikely to be dumped into space with zero transverse velocity and will thus quickly leave the local area and be undetectable a decade later.

The ship it may dump enough energy into the IM to heat it -- creating a long tube of hotter than normal IM along its path. The problem -- again -- is that the mean free path is so long that the hot atoms would immediately decamp for distant places and the colder atoms outside the ships's range would diffuse in, erasing the tracks.

If the speed were insanely high (as if .99c wasn't insane enough), possibly you might see some sort of effect where a bow wave of plasma builds up big enough to clear a really wide tube along the path, long enough that it can't fill in in a decade, or enough energy is dumped into the IM that it's still excited a decade later.  The problem is that that energy -- which is large -- is coming from the ship's kinetic energy through friction with the IM. (a) This slows the ship and (b) it probably does very bad things to the ship's structure.

The last effect is that we might see some side effect of the ship's drive or shields. But since they're basically magic (Clarke's Third Law) I don't see how anyone can give you a scientific answer.


How to calculate the redshift of a line in a bunch of line spectra? If in a spectra, there are many lines, mixed with gravitational redshift and non-redshift lines. For example, in this paper, they identify lines both redshifted and no-redshifted. Then how can i distinguish those redshifted from those do not?

While @planetmaker's comment is true if the lines come from the same source, you can have lines emerging from different physical processes which still appear to come from the same location.
An example is absorption (or more rarely emission) lines from galactic winds, which are typically blueshifted with respect to the "systemic" redshift, i.e. the "average" redshift of the galaxy. You can also have strong emission from an object — e.g. a supernova — that has a large peculiar velocity inside a galaxy.
In general, to distinguish lines with different redshifts, you need to have an idea of where you expect the lines to be (well, the same is true even if you just want to identify spectra where all lines are redshifted uniformly). In this case, the authors know which lines may be expected from the neutron star, namely highly ionized iron and oxygen lines, but the absorption features are then redshifted with respect to the rest of the spectrum.
For instance, the O VII Ly$\alpha$ line has a rest wavelength of 19 Å, but is seen at ~25.6 Å, i.e. a factor 1.35 higher, so its redshift is $z = 0.35$.


When and where are Pluto and Neptune closest to each other? I found two sources sort of answering this question, but conclusions I made of them contradict each other.
https://www.reddit.com/r/askscience/comments/29jrvq/what_is_the_smallest_possible_distance_from_pluto/ - here is stated that Pluto and Neptune never come closer than two billions kilometres from each other.

Yet from picture in here (I am aware it might be not to scale) Can Pluto and Neptune collide anytime in future? it seems that Pluto and Neptune got a lot closer to each other shortly after 1999 than a half of Neptune orbit radius, judging by their speeds and positions in 1999.

So, how is it? How close can they get to each other and when is it going to happen next time?

Tried to use wikipedia, but while article about Pluto contains information at least about when it was in Aphelion, article about Neptune does not any such information (or I could not find it).

Per page 13 http://articles.adsabs.harvard.edu/cgi-bin/nph-iarticle_query?1965AJ.....70...10C&defaultprint=YES&filetype=.pdf (it's Harvard, so relatively reliable):

  the closest approach of Pluto to Neptune is locked in near aphelion
  and the minimum distance between the bodies is approximately 18 a.u.

which works out to roughly 1.7 billion miles or 2.7 billion kilometers.

The image from the other answer:

is deceptive if not outright incorrect (tineye shows 15 results, some of which seem reliable, but I couldn't find the originating source).

Notice how much Pluto moves from 1979 to 1989, compared to how much Neptune moves in the same period. It seems that Pluto moved a lot more, but, since Pluto was about the same distance away as Neptune, it should be moving at nearly the same speed, not faster.

As Stellarium shows, Neptune and Pluto are about 40 degrees apart in 1999 (as viewed from the Sun) significantly more than the diagram in the other question indicates (this might be an artifact of the planes being tilted 17 degrees and an unusual perspective view):

(Pluto is invisible but at the center of the cross-- the bright object nearby is 13 Oph, not Pluto).

A perhaps more interesting question: what is the closest approach of the two orbits, even if the planets in question aren't occupying that given point in the orbit.


Roche limit - calculation with centrifugal forces Let's assume there is a small natural satellite oribiting a massive planet. The satellite rotates synchronously (revolution period = orbital period) on a circular orbit, both bodies are homogenous and rigid. The question is, how to find the Roche limit with regard to centrifugal forces?

With use of Wikipedia I understand that the Roche limit is basically the orbit where difference of gravitational forces between the planet and two points in the satellite body (its centre and a surface facing to the planet) cancels the gravitational pull of the surface of the satellite to its centre. Wikipedia also states that more accurate is to include also the centrifugal force on the surface of the satellite due to its rotation.

Why is not included the difference of the centrifugal forces acting on satellite centre and surface facing to the planet, caused by satellite's orbital movement?

Are there any other forces I need to take into account, for example orbit of the whole sytem around a star?

The roche limit is a calculation of when pieces of a satellite will be pulled off the satellite. Any force that affects the whole satellite equally can be ignored, as this will affect the motion of the satellite as a whole but not (directly) whether pieces are pulled off.

If you consider an object which maintains a constant orientation, but rotates about another body, the paths each part of the object take differ only by a translation in space, and so each part will experience the same centrifugal force. So, the only rotation that needs to be considered is the rotation of the body around its own axis.

In the special case of a tidally locked satellite in a circular orbit, the satellite's rotation around the planet, and the satellite's rotation around its own axis, are the same rotation, just differing in what axis we are considering the rotation to be around.

In the wikipedia article's calculations, the centrifugal force for the rigid body case is considered around the satellite's own axis, but for the fluid body case it is considered as a rotation about the primary. This is an equally valid calculation for the special case of a tidally locked satellite in a circular orbit.


Why don't stars move in the night-sky as the moon does? I want to know why don't we see a change in the position of stars during the night while we do see a change in the position of the moon. I've checked other online sources and some answers say that it is because they are too far away, is this true?

I mean for example if i'm laying down on the earth since sunset looking at the stars, and let's say I'm close to the equator, no matter the distance, I'll be looking at the stars and earth will turn about 180° until sunrise, so my angle of view will be about 180°.

This is referred to as diurnal motion, due to Earth's rotation on its axis, and it affects apparent motion of stars differently depending on their position on the skies relative to the axis of Earth's rotation. For example, on northern hemisphere, the star that appears not to move at all is positioned so that the earth's axis of rotation points directly towards it, and is called a Polaris due to its role as a pole star. Stars further away from true north (not to be confused with magnetic north), will appear to prescribe a circle around it as the Earth completes one rotation on its axis:

                                A long exposure photograph of the night skies, showing their apparent motion

So what you assert, that the stars don't seem to move isn't exactly true. Stars closer to celestial equator will move at equal apparent radial velocity as any object fixed on the night skies would (so at the speed of Earth's rotation), while the stars closer to celestial poles, while maintaining this radial velocity, will prescribe a smaller circle and appear to be more stationary. And the stars exactly aligned with the Earth's axis (like Polaris, but there isn't any such exact counterpart on the southern hemisphere), would appear not to move at all.


Why do massive stars not undergo a helium flash I understand that for low-mass stars the helium flash occurs due to their degenerate helium cores. Thus the answer to this question is probably that more massive stars do not have a degenerate core, but I do not understand why they wouldn't. Due to their increased mass, I'd assume their central pressure is even higher, so surely their cores should also be degenerate?

After H burning has finished, the he mass of the He core gradually increases, as does its density and temperature.
Low-mass stars have denser cores when they reach a temperature at which He is ignited. The density is high enough that the electrons in the core are degenerate. In such conditions the heat from the nuclear reactions goes almost exclusively into raising the temperature of the He ions, but almost none goes to the degenerate electrons which have a very low heat capacity, yet dominate the pressure. This leads to a runaway increase in the nuclear fusion rate.
In higher mass stars the core is less dense when it reaches the He ignition temperature. This temperature is almost exactly the same as the ignition temperature for lower mass stars because of the very strong temperature dependence of the triple alpha reaction. The level of degeneracy depends on the ratio of density to temperature (it does not directly depend on the pressure).
At the lower densities in the higher mass stars, the electrons are not degenerate when He ignites and the heat capacity of the electron gas is higher than the ions (because there are more electrons than ions). In this case the heat from the nuclear reactions can effectively raise the pressure, doing work which expands the core and regulates the nuclear reaction rate.
See also https://physics.stackexchange.com/questions/174801/why-is-the-release-of-energy-during-the-he-flash-in-stars-almost-explosive


Celestron C14 collimation and cleaning When checking the collimation of a Celestron 14" SCT I took this image:

My question is: how much will the particles causing the local aberrations affect the image quality? (i.e. do I need to clean the optics?).

Will this affect image quality?  It depends on "where" the dust is located.  If on the telescope, then not very much.  If on the sensor, you'll get the shadows of the dust on your images (use of dark-frames can be used to "subtract" the dust shadows when calibrating the images via software).
Based on the sharpness of dust, I suspect your dust is probably very close to the image sensor.
When light travels through the telescope, you can think of it as taking all possible paths to reach the sensor.   This means that while some photons are blocked by the dust, other photons are taking paths that bypass the dust.  The dust wont show up on the image because the farther away it is much too far away from any focal plane -- so it's completely out of focus to the point of being invisible.  But as some photons are technically blocked... it may fractionally reduce some contrast if the dust gets pretty bad.
The closer the dust is to the image sensor, the easier it is to spot.  Since your dust shows up fairly clearly, my guess is that it's fairly close to your camera sensor.  You mentioned the telescope is a C14, but you did not mention the camera and/or any filters being used.
Cleaning Telescope Optics
After about 4 years without any cleaning, I removed to the dew shield and found this:

I spent a few hours very careful cleaning the corrector plate (which is where I suspect you'll find the bulk of dust) using Dr. Clay's cleaning solution recipe (you can find that here:  http://arksky.org/asoclean.htm ).
The technique has you gently daubing a lint-free cleaning pad with cleaning solution applied to it.  This pulls off the dust but leaves the optics rather messy.  Avoid rubbing action while the optics are still caked in dust as this will just drag the dust around on the surface.  Again, the corrector plate is much more durable than mirror surfaces and I generally never need to clean the mirror surfaces since Schmidt Cassegrain Telescopes aren't "open" to the air.
Once the dust is mostly eliminated, I'll give it a second cleaning pass and this time I use gentle cleaning action (almost no pressure).
Lastly, I'll use a mist of distilled water as a rinse and daub dry with a clean microfiber cleaning cloth.
The result was this:

While this "looks" clean, my advice is ... never get too close to your telescope optics while holding a bright light.  You'll see all kinds of imperfections.  My advice is to leave them be -- getting too aggressive can result in damage.
With that aside, a corrector plate is not the same as a first-surface mirror.  Mirrors have a soft coating and are easily damaged.  Corrector plates have anti-reflective coatings, but they are hard and not so easily scratched as long as you only use very mild cleaning solutions and are gentle.
Cleaning Camera Sensors
I use a solution called "Eclipse" (the manufacturer is "Photographic Solutions, Inc." and it is available from most stores that sell supplies to professional photographers.  Eclipse is nearly pure methanol.  This means it will act as a solvent to remove not only dust, but oil splatter (some DSLRs will occasionally splatter oil if the reflex mirror was generously lubricated).  Since the substance is nearly pure methanol it will evaporate almost completely and leave virtually no trace.
I tend to clean camera sensors in phases of aggressiveness.

Some cameras have a self-cleaning cycle ... a piezoelectric effect which vibrates a filter in an effort to shake the dust loose (helps to point the camera down while doing this.)

Use a soft blower with ordinary air.  Avoid canned compressed air because those usually aren't "air" and the propellant in the can can leave a haze on the camera ... and that has to be wet-cleaned.

Use a very soft and pristinely clean paint brush (not the type you use to paint a house... the type an artist uses to paint a picture.)  There are versions of this that have a grounding wire which can help release static cling.

Wet cleaning.   This is where I use the Eclipse solution.  The same vendor makes lint-free cleaning pads (PEC Pads).  It only requires a few drops and often just one or two wipes to clean the sensor.

After each of these phases, I inspect the sensor with a magnifying loupe and/or take a sample image.  Cover the front of the telescope with a piece of plain white fabric (e.g. white t-shirt is a popular choice) and stretch it so that there are no wrinkles.  Set the camera to a middle exposure so that the image comes out as a middle gray.  The telescope need not be focused.  Inspect the images carefully to for signs of dust shadows.  Keep in mind that the images on the camera is rotated 180°.  E.g. a dust speck appearing in the upper-right corner is really in the lower-left corner on the camera sensor.


How early/late is the human race as intelligent life in the universe/milkyway? In the history of the universe, there has been a time, where intelligent life had no chance to develop (e.g. in the early universe, when no habitable planets have formed) and there is a time today where we know, that at least one intelligent lifeform has evolved in our universe/galaxy. Between these two times many events took place in order to provide the conditions, that (intelligent) life can evolve. So I wonder myself:

Considering the time, when the conditions in the universe became life-friendly, has the human race as intelligent life evolved early, late or something in between?

I think this is a very subjective question depending on your point of view. If you assume that the history of the universe is 13.8 billion years old, then humans have only been around for 200,000 years which makes us very late at evolving. But of course the entire history of the universe hasn't been written yet!

I always find these sort of calendar/24hr clock representations very insightful too. If you look at this image below, modern "intelligent" humans have been around for the last 6 minutes of the year, if the age of the universe was 365 days long, so not very long at all!

That being said, I think there is another answer to your question, but becuase we don't know how long the universe/milky way will last for, its impossible to say or indeed ever know for how long intelligent life could actually evolve for.

I'm afraid i'm going to have to say somewhere in between too, as we simply cant see into the future.


What did I see go by saturn? I was looking at Saturn through a basic reflective telescope. I had it focused and could see the rings. To the left a small dim white dot came across my view passing in about 3 seconds from my north to south( perpendicular to the Rings) no tail or anything.

As mentioned in this comment if you were using low power, say 40x with a 50 degree apparent FOV eyepiece that would be about 0.4 degrees per second which is in the ballpark of the apparent motion of an artificial satellite in low Earth orbit (LEO).

A geostationary satellite (GEO) or even a geosynchronous one would move much more slowly relative to Saturn, and if your telescope wasn't motorized, it would appear in a fixed position even as Saturn slowly moved across your field of view due to Earth's rotation.

In the bottom plot below, taken from this answer it shows that 0.4 degrees per second is quite reasonable for satellites with a wide variety of altitudes. For satellites in LEO between 300 and 1000 km and viewing angles from zenith to near-horizon apparent motion can range from below 0.1 to over 1.4 degrees per second.

As pointed out in this comment it could perhaps be an airplane as well if there was some source of light. It could have had some lights on, or you were watching at dawn/dusk and it received some sunlight, or I suppose it might be possible that it was reflecting Moonlight.

But I think a comet (also mentioned in that comment) or a tiny near-Earth asteroid couldn't possibly appear to move a substantial fraction of a degree per second unless it was closer than the Moon, and we would likely (but not for certain) hear about that, though there are surprises as discussed in the (currently unanswered) question What is the “interesting story on the limitations of (NASA's) current (near Earth object) survey network”?

For further reading on artificial satellites in low Earth orbit, see What actually happened here with a satellite, the ISS and the moon? and also this answer to Why is taking a picture of the moon surface from earth possible but taking one of an artificial satellite is difficult? and while you're at it, answers to Are commercial communications satellites in GEO being constantly monitored by telescopes?


SDSS - getting separate u, g, r, i, z frames from SWARP tool and astropy I am following this paper to extract the u,g,r,i,z image frames from the SDSS image fits files. I have the fits files downloaded (using the bulk data download option from SDSS) and tried to run the SWARP tool to get the resampled image as output, which I will access through astropy. I am still unclear on the exact parameters that will make up the entire SWARP command. So far I have:
swarp frame-i.fits[0] frame-r.fits[0] frame-g.fits[0] frame-u.fits[0] frame-z.fits[0] -SUBTRACT_BACK N -RESAMPLE Y -RESAMPLING_TYPE LANCZOS3 -IMAGE_SIZE 64,64 -IMAGEOUT_NAME attempt.fits

This is written with the description of it in the paper, which is:

With the command that has been run, I am getting quite noisy images as output that do not relay any sort of information. What am I doing wrong in this process?

What they did was the following: for each individual filter, they assembled overlapping frames into a combined, single-filter image. E.g., they combined several g-band images into a single g-band image, combined several r-band images into a single r-band image, etc. (So they ran SWarp five times, once for each filter.) They then put each combined single-filter image into a separate layer to create a five-layer "datacube"; this is their "64x64x5 pixel datacube".
What you are telling SWarp to do is something different: combine five individual-filter images into a single "$u+g+r+i+z$" image. (People do sometimes do this sort of thing, though with SDSS images they would normally just use the g, r, and i images, since the u and z images are so low-S/N that they mainly just add noise to the result. And in any case this isn't what the paper you're interested in was doing.)


Where is Mars in the night sky I live around Houston, Texas and I'm taking advantage of Earth hour to look at the stars. I've found two objects that I think might be Mars. Both of them are in roughly the same plane as Venus and Jupiter. Both are closer to Venus than Jupiter. One of them has a reddish tint and is higher in the sky than Venus, and the other is closer to the horizon than venus. It is 8:50 as I type this. Is one of these Mars?

At that time (20:50), and that location (Houston, Tx), Mars was closer to the horizon.
You can see that in the sky-chart provided by heavens-above.com (attached below).

I cannot tell what's the other object you describe, but if you are not sure whether something is a planet or a star, the planets doesn't twinkle whether the star does.


What is the Lyman Alpha forest Used For? The Lyman-alpha forest is an interesting feature found in the spectra of sufficiently distant objects. This series of absorption lines extends over a range of frequencies, and are a result of the Lyman-alpha electron transition of neutral hydrogen in between the source and the observer.

My question is:
What information can be gleaned about the universe from this spectral feature?  In other words, what areas of research use the Lyman-alpha forest as a tool, instead of seeing it as noise?

According to UCLA's web-tutorial Lyman Alpha Forest (Wright, 2004), there are many 'clouds' of gas between us and a distant quasar (as an example) that absorb

  ultraviolet light at the wavelength of the Lyman alpha line of hydrogen at a wavelength of 122 nm.

However, as the clouds of gas have less of a redshift than that of the distant quasar, then their absorption lines are less redshifted than the distant object - a cartoon example is shown below (from the UCLA website):

Now, the significance is of the forest is that it represents clouds that are smaller than the smallest galaxies, thus

  We can only see these very low mass clouds by the absorption they produce in the strongest line of the most abundant element: Lyman alpha. Thus by studying the Lyman alpha forest we can learn about the density fluctuations in the Universe on the smallest observable scales.

A similar explanation of the significance of the forest, is given by Quasars, the Lyman Alpha Forest, and the Reionization of the Universe (Mortlock et al. 2011) is

  quasars like ULAS J1120+0641 are bright and at high redshift, lower redshift intervening material can absorb some of their light, leaving fingerprints on the final spectrum that we observe on Earth. Because hydrogen is the most abundant element in the universe, it should come as no surprise that it leaves the most prominent spectral signature, in the form of a forest of absorption lines.

Specifically, of major significance is that Mortlock et al. (2011) state that

  The Lyman alpha forest can be used to trace this re-ionization of the universe.


I saw a weird thing around Venus I pointed my telescope
(50mm aperture, 500mm focal length, 4mm eyepiece, image taken through a window)
at Venus and saw this:

What is it? I don't think this is Uranus or Neptune as my telescope is very bad. I haven't increased the image size too much and it was 6 A.M. (UTC+3, EEST)

Venus is a very bright "morning star" at the moment, but what your image shows is only a blur. The telescope is very far from focus and you can see no details of the the planet. You may also be getting some reflections from the glass of the window
If Venus was in focus you would see a very small "half moon" (or perhaps "waxing gibbous" shape) at the moment. Try to practice focusing on distant landmarks during the day. That makes it easier to get the focus right at night.


Naming convention for multiple star systems Okay, so I've seen different notations for the stars in β Capricorni. Some sources state that the stars are Aa, Ab1 and Ab2, Ba, and Bb, while others say that they are Aa, Aba and Abb, Ba, and Bb. So, what is the proper way to name stars in a multiple star system, and are there any exceptions to this convention?

The purpose of the notation is to indicate how the system is physically constructed, to indicate which stars orbit which - especially useful in hirachically organized stellar systems. So both ways are correct as their meaning is understood.
The notation I encounter most often is the first one you indicate: indicating with capital letters the systems which orbit the common pericenter of the overall system. With non-capital letters the components of the upper level are distinguished. And keeping it more readable, numbers are used for the 3rd level if further distinction is needed for a 3rd level.
However this nomenclature is not unique - and there is so far no general consensus. See the Washington Multiplicity Catalog for a take on getting some order at a given time and the IAU working group on stellar names.
For a given system the designation might be more traditional and depend on order of discovery and naming by discoverer than following a fixed scheme. See also the discussion of future nomenclature on stellar system components in the corresponding wiki article.


As viewed from Mars, what are Jupiter's and Saturn's maximum brightness in apparent magnitude? From Earth, Jupiter's maximum brightness is -2.94 and Saturn's is -0.24. But what about from Mars? They should be brighter, but by how much?

There are equations at the obvious wiki entry but I'm not sure I understand them. Plus I'm afraid to attempt this calculation myself because I've heard that this sort of thing does not follow the inverse square law. They're reflecting light (not generating it), and I read somewhere that it should follow an inverse 4th power law because of that. I don't see any 4th powers in the apparent magnitude equations.

The inverse-fourth-power law you're referring to is valid for light emitted from a source, reflected non-specularly  — i.e. in all directions — from a reflector, and detected by the original emitter. If the reflector is a mirror, the observed flux just follows the normal inverse-square law with the nominator equal to $(2d)^2$ instead of $d^2$, since the light has to go back and forth. But if the reflector scatters light in all directions — i.e. into a $2\pi$ hemisphere — then the detected flux is $\sim r^2/d^4$, where $r$ is the radius of the reflector (see this answer for a more throrough explanation).

An example of this is a radar. But in our case, it isn't us that emit the light, it's the Sun. The amount of light reflected from Jupiter and Saturn depends on their distance to the Sun, and that distance doesn't change if you move to Mars. The relevant distances (that I got from NASA's Planetary Fact Sheet) are:

Earth semi-major axis    $d_\mathrm{E} = 1.00\,\mathrm{AU}$
Mars aphelion$^\dagger$  $d_\mathrm{M} = 1.64\,\mathrm{AU}$
Jupiter semi-major axis  $d_\mathrm{J} = 5.20\,\mathrm{AU}$
Saturn semi-major axis   $d_\mathrm{S} = 9.58\,\mathrm{AU}$

Now the differences between them:

Earth to Mars    $d_\mathrm{M-E} = 0.64\,\mathrm{AU}$
Earth to Jupiter $d_\mathrm{J-E} = 4.20 \,\mathrm{AU}$
Earth to Saturn  $d_\mathrm{S-E} = 8.58\,\mathrm{AU}$
Mars  to Jupiter $d_\mathrm{J-M} = 3.56 \,\mathrm{AU}$
Mars  to Saturn  $d_\mathrm{S-M} = 7.94 \,\mathrm{AU}$

Hence, from Mars the distance to Jupiter is $d_\mathrm{M-J} = 0.85 d_\mathrm{J-E}$, and the received flux  is thus $1/0.85^2 = 1.4$ times that on Earth. The change in apparent magnitude is then
$$
\Delta m = -2.5 \log\left( \frac{0.85^2}{1}\right) = -0.36,
$$
i.e. Jupiter would be $m = -2.94 - 0.36 = -3.30$ as viewed from Mars (assuming the values you provide are correct; I didn't check this).

Following the same approach for Saturn, I get $m = -0.41$.

$^\dagger$The reason I used Mars' aphelion instead of Mars' semi-major axis is because Mars has a rather eccentric orbit ($e\sim0.09$). Jupiter and Saturn are somewhat closer to circular orbits, though ($e\sim0.05$). This of course is still an approximation. If you wish to take into account the eccentricities of all orbits, you also need to know the angle between their semi-major axes. This also doesn't take into account the inclination of the orbit; however, these are only 1º-2º.
And of course this minimum distance will not occur every Martian year.


How could a neutron star collapse into a black hole? White dwarfs usually do not collapse, as they have electron degeneracy pressure due to the Pauli exclusion principle. However, if one accretes mass beyond the Chandrasekhar limit, it is energetically favorable for the electrons to combine with protons and form neutrons. This gives us a neutron star.

However, neutron stars usually do not collapse into black holes due to neutron degeneracy pressure. How is it possible that beyond the LOV limit, the Pauli exclusion principle no longer prevents the collapse? Shouldn't it still prevent neutrons, which are fermions, from being compressed together any further?

I've seen answers involving quark stars, but those are purely hypothetical. What is the most accepted explanation for this?

The scenario you describe may occur. On the other hand it may actually be that neutronisation in a white dwarf is the trigger for a thermonuclear type Ia supernova.

You may be misunderstanding the Pauli Exclusion Principle (PEP).The PEP states that no two fermions can occupy the same quantum state, not that they cannot occupy the same space or be compressed to whatever density you like. The quantum states here consist of two spin states for every possible momentum state. In a degenerate gas, all these states are filled up to the Fermi energy. All that happens when the neutron star gets smaller (or collapses), is that the Fermi energy just keeps increasing as the neutron density climbs, and the neutron degeneracy pressure just keeps increasing as a consequence.

However, in General Relativity, pressure (like mass/energy) is a source of gravitational curvature and actually increases the required pressure gradient needed to support the star. At a certain threshold radius - a small factor larger than the Schwarzschild radius, a point of instability is reached where increasing the pressure is actually counter-productive.  Beyond this, you can make the pressure as large as you like and it will not prevent the formation of a black hole.

Even inside the BH there is not necessarily a problem with the PEP. You can compress fermions to infinite density so long as they can have infinite momentum.


Watching the Mercury transit with improvised devices I have learned that it is not possible to watch the upcoming Mercury transit with the plain eye (using special filter glasses).

Is it possible to watch the transit with improvised devices (like a camera obscura)?

Mercury's angular diameter on transit day will be 12 arcseconds.
A camera obscura using a 12 mm aperture could resolve it; one lens from +0.75 diopter reading glasses, if you can get them, will project a bright 12 mm image of the Sun at a distance of 1.33 m.
Note that a larger aperture or a shorter focal length will make the Sun image hotter than direct sunlight unless you add a filter.

Test with sunspots before relying on it for Mercury.
If it counts as improvised, projection with 7x35 binoculars easily showed me the 2012 transit of Venus.

Update: I tried both of the above with the Nov 2019 Mercury transit.
With handheld wobbling, neither showed it.
Someone else's 50mm binoculars did, probably because they were mounted on a tripod.


Why are orbits elliptical instead of circular? Why do planets rotate around a star in a specific elliptical orbit with the star at one of it's foci? Why isn't the orbit a circle?

Assume the planet has a negligible mass compared to the star, that both are spherically symmetric (so Newton's law of gravitation holds, but this normally happens to a very good approximation anyway), and that there aren't any forces besides the gravity between them. If the first condition does not hold, then the acceleration of each is going to be towards the barycenter of the system, as if barycenter was attracting them a gravitational force with a certain reduced mass, so the problem is mathematically equivalent.

Take the star to be at the origin. By Newton's law of gravitation, the force is $\mathbf{F} = -\frac{m\mu}{r^3}\mathbf{r}$, where $\mathbf{r}$ is the vector to the planet, $m$ is its mass, and $\mu = GM$ is the standard gravitational parameter of the star.

Conservation Laws

Because the force is purely radial $(\mathbf{F}\parallel\mathbf{r})$, angular momentum $\mathbf{L} = \mathbf{r}\times\mathbf{p}$ is conserved:
$$\dot{\mathbf{L}} = \frac{\mathrm{d}}{\mathrm{d}t}\left(\mathbf{r}\times\mathbf{p}\right) = m(\dot{\mathbf{r}}\times \dot{\mathbf{r}}) + \mathbf{r}\times\mathbf{F} = \mathbf{0}\text{.}$$
If the initial velocity is nonzero and the star is at the origin, then in terms of the initial position and velocity, the orbit must be confined to the plane of all points with vectors $\mathbf{x}$ from the origin that satisify $\mathbf{L}\cdot\mathbf{x} = 0$. If the initial velocity is zero, then the motion is purely radial, and we can take any one of infinitely many planes that contain the barycenter and initial position.

The total orbital energy is given by
$$\mathcal{E} = \frac{p^2}{2m} - \frac{m\mu}{r}\text{,}$$
where the first term part is the kinetic energy and the second term is the gravitational potential energy of the planet. Its conservation, as well as the fact that it invokes the correct potential energy, can be proven by the fundamental theorem of calculus for line integrals.

Define the Laplace-Runge-Lenz vector to be
$$\mathbf{A} = \mathbf{p}\times\mathbf{L} - \frac{m^2\mu}{r}\mathbf{r}\text{.}$$
It is also conserved:
$$\begin{eqnarray*}
\dot{\mathbf{A}} &=& \mathbf{F}\times\mathbf{L} + \mathbf{p}\times\dot{\mathbf{L}} - \frac{m\mu}{r}\mathbf{p} + \frac{m\mu}{r^3}(\mathbf{p}\cdot\mathbf{r})\mathbf{r}\\
&=& -\frac{m\mu}{r^3}\underbrace{\left(\mathbf{r}\times(\mathbf{r}\times\mathbf{p})\right)}_{(\mathbf{r}\cdot\mathbf{p})\mathbf{r} - r^2\mathbf{p}} - \frac{m\mu}{r}\mathbf{p} + \frac{m\mu}{r^3}(\mathbf{p}\cdot\mathbf{r})\mathbf{r}\\
&=& \mathbf{0}\text{.}
\end{eqnarray*}$$

Finally, let's also take $\mathbf{f} = \mathbf{A}/(m\mathcal{E})$, which has the same units as $\mathbf{r}$, and since $\mathbf{L}\cdot\mathbf{f} = 0$, it lies along the orbital plane. As it's a conserved vector scaled by a conserved scalar, it's easy to show that $\mathbf{f}$ is conserved as well, as long as $\mathcal{E}\neq 0$.

Simplifying

By employing the vector triple product, we can write
$$\begin{eqnarray*}
\frac{1}{m}\mathbf{A} &=& \frac{1}{m}\left[p^2\mathbf{r}-(\mathbf{p}\cdot\mathbf{r})\mathbf{p}\right] -\frac{m\mu}{r}\mathbf{r}\\
&=& \left(\mathcal{E}+\frac{p^2}{2m}\right)\mathbf{r} - \frac{1}{m}\left(\mathbf{p}\cdot\mathbf{r}\right)\mathbf{p}\\
\mathcal{E}(\mathbf{f}-\mathbf{r}) &=& \left(\frac{p^2}{2m}\right)\mathbf{r} - \frac{1}{m}\left(\mathbf{p}\cdot\mathbf{r}\right)\mathbf{p}\text{,}
\end{eqnarray*}$$
the norm-squared of which is easy to crank out:
$$\mathcal{E}^2|\mathbf{f}-\mathbf{r}|^2 = \left(\mathcal{E} + \frac{m\mu}{r}\right)^2r^2\text{,}$$
where $\mathcal{E}$ was used throughout to switch between kinetic and potential terms.

Why Ellipses?

Since $\mathcal{E}$ is energy relative to infinity, to have a bound orbit we need $\mathcal{E}<0$. Thus, from the previous section, $|\mathbf{f}-\mathbf{r}| = -\mathcal{E}^{-1}\left(\mathcal{E}r + m\mu\right)$ and therefore
$$|\mathbf{f}-\mathbf{r}| + |\mathbf{r}| = -\frac{m\mu}{\mathcal{E}}\text{,}$$
which defines an ellipse with foci $\mathbf{0},\,\mathbf{f}$ and major axis $2a=-m\mu/\mathcal{E}$.

Why Not Circles?

The circle is a special case where the foci are the same point, $\mathbf{f} = \mathbf{0}$, which can be restated as
$$\mathcal{E} = -\frac{1}{2}\frac{m\mu}{r} = -\frac{p^2}{2m}\text{.}$$
In other words, circular orbits require the orbital energy to be the negative of the kinetic energy. This is possible, but almost certain not to hold exactly. Since any values of $\mathcal{E}<0$ are allowed for bound orbits, there are many more ways to have elliptic orbits. (Although some of them would actually crash because the star and planet have positive size.)

Note that hyperbolic orbits have $\mathcal{E}>0$, and we can still find the foci using the above method, though being careful with the signs. For $\mathcal{E}=0$, the second focus $\mathbf{f}$ is undefined because this is a parabolic orbit, and parabolas only have one focus within a finite distance from the center.

Additionally, the eccentricity vector $\mathbf{e} = \mathbf{A}/(m^2\mu)$ is an alternative choice for the LRL vector; as the name suggests, its magnitude is the orbital eccentricity.


Which Lowell Observatory telescope is this? (used to draw Moon maps for the Apollo program) The Fox News article Arizona city played critical role in moon exploration history covers several interesting activities that took place in Arizona in preparation for the Apollo Moon landings.

It shows the telescope (shown below) and mentions that it was used to draw lunar maps.

Question: Is it possible to identify which telescope this is exactly? Is there some place where one can read more about it?

click images for full size viewing

  above: Lowell Observatory telescope used by scientists who collaborated with artists to map out the moon for Apollo astronauts (Fox News)

below: Screenshot from July 9, 2019 Fox News video NASA lunar legacy in the Arizona desert

That's the historic 24-inch Alvan Clark refractor.
It was installed in 1896 and restored in 2014-15.
Besides the moon mapping project, Percival Lowell used it to observe Mars, and Vesto Slipher used it for some of the first measurements of galaxy redshifts.
Its current mission is education; in mild weather the observatory lets visitors look through it.

Roger Vine's Scope Views has more photos and a detailed description.


Why nobody cares to give the nearest stars better names? The star occupying 6th place on Wikipedia's list of nearest stars and brown dwarfs can be called either Lalande 21185, BD+36 2147, Gliese 411 or HD 95735. That seems to me like a fine, quadruple choice of equally bad names.

I mean, I know these come from positions held in various star catalogues and maybe aren't that bad compared to monstrosities like WISE J085510.83−071442.5, but can't we keep those for stars like 50 light yrs away and invent something better for our nearest neighbours?

This year we saw a batch of confirmations for planets orbiting in habitable zones of many of these stars. In not so distant future we may find ourselves discussing possibility of sending probes to some of these systems. Aren't better names long overdue?

There are a thousand or so stars within 50 light years, most are very dim red dwarfs, and most are unexceptional. If we invent "names" then we would have to learn the names of all the stars.

If we want to name all the stars that can be seen with a telescope, we would have to come up with a billion or so names.  The idea of naming stars rapidly becomes impractical - nobody is ever going to learn thousands of names which were chosen on whim.

So what could be a "better name" than just making up a name based on one's heroes, friends, gods, or jokes?

Instead we can make a list of stars, and name them systematically. We could use the part of the sky that the star appears in, and some letter or  number. Or we could use the coordinates that the star appears at to label it. Or we could just give each star a number. This way we don't need to invent names, we just use our system. When we do this we get "names" like "omicron ceti" (using the letter "omicron" and the part of the sky "ceti")
or "HD 128621" (star number 128621 in the list prepared by Henry Draper). or WISE 0410+1502 (a brown dwarf discovered by the Wide-field Infrared Survey Explorer, and located at 04hr10min, 15deg02min in the sky)

Cataloguing is simply a better way of naming stars as there are so many of them.

When a star is exceptional in some way, then there may be a reason to give it a name - Barnards Star (moves very fast), Sirius (very bright). But most stars are unexceptional, and there are so many of them.


Do the earth and the moon reflect the different fractions of total solar radiation proportionally? Reading on this website that: "All of the energy from the Sun that reaches the Earth arrives as solar radiation". "Solar radiation includes visible light, ultraviolet light, infrared, radio waves, X-rays, and gamma rays".
Here Albedo is explained to refering to the entire spectrum of solar radiation.
The earth and the moon obviously reflect different amounts of total solar radiation, but question is if they reflect the different solar radiation fractions proportionally?
(Proportional reflection is when all the different light spectrum fractions increases, or decreases, with the same percentages.)
Or do earth and/or moon reflect light dis-proportionally?
Rephrasing the question:
We know how much the difference in visible geometric albedo between earth and the moon
is, but is that figure the same when comparing  reflection of infrared, or any of the other parts of the light spectrum?

The obvious answer is: "of course". Different colour simply means different wavelength-dependent reflection; different brightness means different wavelength-independent reflection. Earth and moon have different colours when viewed from space.
The real question is by how much does it differ as a function of wavelength. There are already here nice comparisons of the solar spectrum and the Moon's spectrum.
There's a relative measurement of Earth's reflectance by Björn et al (2009). Similarily Turnbull et al (2006) discuss the reflectance of Earth compared to the Moon. Reality is complicated as exact reflectance depends on the observation angle and angle of incidence - and the exact materials looked at which varies slightly for the Moon and can vary a lot for Earth, especially when considering cloud and ocean coverage. Earth might be slightly redder due to increased thermal radiation, especially in the absorption bands of water in the atmosphere.


From our world's perspective, does the full moon look larger in one city than it does in another? When I was a very little girl, my dad talked about the HUGE full moon he saw while in the navy. He made the comment during a full moon we were having in Western NYS. He said, this (harvest) moon is nothing compared to the one he saw.

Now years later, I was with my grown son, who's a geologist and who has been all over the world but not during a full moon. He told me my dad was wrong. My son said that the full moon is always about the size of a penny when you hold the penny up to compare the size of the moon to it.

But I have seen pictures of HUGE full moons over cities my whole life and I assumed my dad was correct. Is there a place where I can travel to see a HUGE full moon that would be much bigger than the full moon I see from the shores of Lake Erie? I understand that the horizon makes the moon bigger but the pictures I have seen were not moons on the horizon.

I've searched Stacker's data base but I didn't see this question.

Your son is more or less correct. The moon's size doesn't change appreciably. At best you only see the size vary by about 13% (in angular diameter).

The variation in the apparent size of the moon comes from the fact that the moon's orbit around us is not perfectly spherical and sometimes it is closer and sometimes farther. The moon will appear largest when a full moon coincides with the moon being at its closest point in its orbit. This is colloquially referred to as a "Super Moon". The opposite, being at its farthest point in its orbit during a full moon, is known as a "Micro Moon" or "Mini Moon". These are really the two extremes, but most often the moon is a more average size that doesn't appear to change much. Shown below is an image of both a Super Moon and a Micro Moon. You can see the size difference, which is noticeable, but not hugely so. If you want to see a large moon, find the time when the next Super Moon occurs (a good upcoming Super Moon is on November 14, 2016).

You also do mention the Moon Illusion, where a Moon on the horizon appears larger than it really is. The cause for this is debatable, but it is a known optical illusion. I believe the reason for this effect is that near the horizon, you have points of reference that allow you to judge the size of the moon different than when it is high in the sky, far from other comparison to other objects. You can attempt to view the Super Moon as it is rising so as to take full effect of this illusion, but try to do so in an area that isn't flat and devoid of structures (e.g., don't watch the Moonrise over a body of water).

As for your father, possibly he saw a Super Moon near the horizon which maximized both the optical illusion and the moon's natural size variability. Possibly he his memory also became exaggerated over the years so he remembered a Moon which was larger than it really was.


Predicting accurate time of local sunrise in mountainous area? Is there any software which will tell me accurately at what moment any part of the solar disk will become from a specific point on the ground taking account of terrain. I feel like Google maps should have the terrain data I need, and the rest is geometry, but it looks like quite a bit of work, so I wondered if it already existed.

Specifically, I am interested in when a viewer 1.5m above local ground level at latitude 46.497300, longitude 7.714651 will first see direct sunlight on the morning of 1 August 2023, taking account of the surrounding mountains. Ideally I'd like to know how that varies with height above the ground (1-2m) and for locations within 10-20m around that location (in other words, how and when the shadow of the mountains moves off that area).

The PeakFinder app1 can do most of what you're looking for.  It's available for Android and IOS as well as through a web browser.  The app shows the paths across the sky for both the sun and moon2 and displays rise/set times taking into account the local topography and viewpoint elevation3.  At least the Android version works without a network connection (nice for hiking).  IIRC, it uses SRTM data4 for elevations so it's not super-accurate but does a good job over all.

I plugged in the location, elevation (ground + 1m), and date you gave and got sunrise at 05:09 local time, reaching the ridgeline at 08:37 local time.  Here's a link to the page; screenshot is below.

1 I'm not affiliated with PeakFinder, just a satisfied user.

2 If using a web browser, click on the "sun" and "moon" icons on the page's toolbar.

3 It also shows what the surrounding terrain looks like and the names of visible peaks.

4 As does OpenStreetMap, IIRC.

Source: Cropped screenshot of PeakFinder site


Could a habitable satellite of a gas giant have a stable subsatellite? I have set a science fiction story on a moon, orbiting a gas giant (which orbits its star at approximately the same orbit as the Earth around the Sun), and given this moon its own satellites.  The moon is Earth-sized, and the moon’s satellites are on the order of a large asteroid each (perhaps up to half the size of the Moon … Luna if there was any confusion with the number of times I said ‘moon’ in this post.)

I am a computer nerd, not an astronomer, so I’m wondering if this is something that is at least possible?  Could the orbits of the gas giant around the star, the moon around the gas giant, and the subsatellites around the moon remain stable for the requisite billion or so years for life to develop on the moon?  Could the moon be habitable?

This is a fun question, so I spend some time thinking about it. This is what I came up with.

Is this possible?

Yes, this is certainly possible. Just consider the Solar System itself. We have a massive central body, the Sun, and several tiny subsystems (planets with their moons) orbiting it.

Is it stable?

Again, from the simple observation that the Solar System exists in its current form, we know that such a system can remain more or less stable over a significant portion of the stellar lifetime.

A more difficult problem, I think, is if can it be formed in the first place.

Could it be habitable?

This is a tricky question as it critically depends distances between the several orbiting bodies. Lets consider the orbital mechanism and derive from what we know of the Solar System.

In the following I've assumed the gas planet is Jupiter and the fictional planetary system consists of a Sun', Jupiter', Earth' and Moon' such that the apostrophe denotes the fake object. These fake objects are identical to the real equivalents, except for their orbital configuration and relative distances. If you want a different system you can follow the same line of reasoning, but plug in different numbers for the fake objects.

Now, we know that the Earth-Moon system exists in a stable orbit around the Sun, which means that at 1 AU the gravitational force of the Sun is not strong enough to destabilize the Moon orbit. From this we can find a lower limit on the gravitational force that Jupiter' may exert on the Moon'
\begin{aligned}
  F_{J'M'} &= F_{SM} \\
  \frac{ M_{J'} M_{M'}}{r_{J'M'}^2} &= \frac{ M_{S} M_{M}}{r_{SM}^2}
\end{aligned}
Solving this equation we get
\begin{aligned}
 r_{J'M'} &= r_{SM} \left( \frac{ M_{J'} M_{M'} }{ M_{S} M_{M} } \right)^{1/2} \\
   &= r_{SM} \left( \frac{ M_{J'} }{ M_{S} } \right)^{1/2}
\end{aligned}
which evaluates to about 0.03 AU. So the Earth'-Moon' system should have an orbital radius around Jupiter' of at least 0.03 AU.

Now the question is if the Earth-Moon system can remain bound in its orbit around Jupiter. To answer this we can safely ignore the Moon' and assume it remains tightly bound to the Earth'. The potentially destabilizing force on the Jupiter'-Earth' system is again the gravitational pull of the Sun', but now on the Earth'. We can do a similar trick as before, but now we need to account for a different mass ratio. So lets require that the fraction of forces be equal in both cases:
$$
  \frac{F_{SM}}{F_{EM}} = \frac{F_{S'E'}}{F_{J'E'}}
$$
with some algebra this gives
$$
 r_{S'J'} = r_{J'E'} \frac{r_{SM}}{r_{EM}} \left(\frac{M_E}{M_J}\right)^{1/2}
$$
which for $r_{J'E'} = $0.03 AU evaluates to a minimum orbital radius of about 0.7 AU.

This greatly surprised me as it means you can comfortably fit the entire Jupiter'-Earth'-Moon' system inside the habitable zone, meaning that in principle life should be possible.

Of course the seasons on such a planet would be rather extreme.


What is the last nuclear reaction in a binary system before supernova? I understand that binary star systems may cause type Ia supernovae.

I have a question regarding supernova explosion in a binary system. What is the last nuclear reaction before massive explosion?

I am thinking of two 2 close stars of 3M and 6M.

I guess it depends on the type of Supernovae.

Type 1a supernovae are produced by a white dwarf star orbiting a companion close enough that matter transfers from the companion to the surface of the white dwarf.

Assuming only your star system of a 3M star and 6M star, the 6M star would become a white dwarf star first.  At the end of it's life, it would have been furiously converting hydrogen to helium, and helium to carbon via the C-N-O cycle and the triple alpha process (C-N-O being dominant in stars above 1.3M sol).  The highest reaction in this stars main sequence lifetime would have been the creation of oxygen, but oxygen is a catalyst in the C-N-O cycle, and therefore the last probable reaction in the star would be that of Helium in some form of shell burning in the outer core (more on that next).

The white dwarf, while orbting the companion, may siphon off hydrogen from the surface of the donor star.  This hydrogen will accumulate evenly over the surface of the dwarf star, slowly growing the mass of the dwarf.  Also, the hydrogen will be shell-burned into helium during this time increasing the density.

If sufficient mass can be accumulated, then the carbon within the white dwarf can ignite in a runaway reaction - literally blowing the entire star apart.  These supernovae produce every element up to Nickel (Ni-56).


How long is the moon full Is the moon the most full only for one moment

As stated  here and here

Or does if stay the same full for a longer period of time (maybe because, the sun is grater then it, and light spreads out)

Full Moon Position of Sun

tl;dr: Perfectly full moon lasts, never more than, approximately 2 minutes.

Estimating an answer from the point of view of Plain Geometry:

If we approximate, that the sun is at an infinite distance, then exactly half of the moon is illuminated.

Next, how much of the Moon do I see when I look at it from Earth? (This has nothing to do with illumination of the Moon.) Estimate: Radius of moon (Rm), 1740km. Distance to the moon (Dm), 384,000km. Doodle this into a diagram that is VERY much not to scale...

arctan(Rm / Dm) gives us 0.260 degrees. So as long as the observer is within that angle from the Moon-Sun-centerline, the portion of the Moon which is visible, would be a circular disc fully illuminated.

Here, I suspect that the variation in the Moon's distance is the biggest thing I've estimated away.

The question now is: combining the Moon's motion of revolution (aka orbit about the Earth) with the Earth's rotation, how long does it take an observer to perceive the moon moving half a degree (1/4 degree either side of the Moon-Sun-centerline)? Rotation of the Earth gives 15 degrees-per-hour, minus the Moon's 1/2 degree per hour... 14.5 degrees per hour apparent motion.

I get approximately 2 minutes duration when the Moon's shifting of apparent position would only hide/show illuminated portions of it's disc.

But the lunar orbit is inclined

So it's only 2 minutes if the Moon is very close to the plane of the Equator. As the Moon's orbit carries it out of the plane, the time would be reduced. Beyond half a degree out of the plane, you would actually never (pure geometry) seen a fully illuminated disc.

So I think the best answer is probably, "Never more than 2 minutes."


What kind of collision of celestial bodies would be visible to the naked eye? When asteroids crash into the Moon they give of huge amounts of energy resulting in a "bright" flash, as can be seen in this video. At the end of the video its also says that anyone looking at the Moon at that moment would have been able to observe it without a telescope.

But what kind of collision would be required to be able to be seen from Earth when it would involve more distant bodies? Would one of Jupiter's retrograde moons colliding with one of its prograde moons be visible?

Edit:
A real example of a high energy collision would be the collisions of the fragments of comet Shoemaker-Levy-9 with Jupiter. However according to this video from ESA it happened on the opposite side of Jupiter, so not in our view.

OK, so... you said the collision with Shoemaker-Levy 9 and Jupiter happened on the non-visible side. What if we'd seen it? The energy released by the largest fragment collision was estimated at 6 million megatons of TNT, which is 2.52e22 joules. The Sun's luminosity is 3.846e26 watts, by comparison. So obviously the collision won't rival the Sun. Using the luminosity-distance relation and taking the 6 million megatons as pure luminosity (an overestimate, but we're not going for precision here) gives the result of 0.0033 watts/square metre.

Relating flux to apparent magnitude, we get an apparent magnitude of the collision of... approximately -12. That's about the brightness of a full moon. Bear in mind this is a very rough calculation and it makes the assumption that all the energy was released over a second and all of it was light, so probably the real value would be a few magnitudes down But, assuming I've made no terrible calculation errors, it would definitely have been visible.

Given that was a relatively tiny comet, your example of two of Jupiter's moons colliding would release FAR more energy, therefore would be clearly visible, I'd imagine. All depends on how much energy gets emitted as light.


Can we violate law of physics in any other planet? Can't there be any way to violates any law of physics? If in our earth we can't violate any law of physics can't we violate in any other planet?

If there is a way to violate a law of nature, it will become a law of nature as soon as it is discovered, studied and formalized into a scientific theory. Therefore, many new discoveries in science do violate (then current) laws of nature, but will not do so for very long. Often there are all sorts of medals and prizes involved too.


Luminosity value to use for the calculation of the habitable zone? I'm currently studying the star system of Gliese 570  and I´m trying to calculate the habitable zone of the component A from the luminosity value in relation of the Sun´s. I´ve read that the needed value of luminosity is the value of the bolometric luminosity (here 26%), but the calculations of the HZ that I found is about 0.40 AU, which seems to be calculated from the visual luminosity (about 16%).

So, the questions are: Which of these 2 values do I have to use to calculate the habitable zone? Does the stellar class of the star affect the selection of it?

The number you need is the bolometric luminosity. This tells you how many Watts per square metre are incident upon the planet. That is true for the Sun and it is true for GL 570.

Beyond that, you will need to make assumptions about an atmosphere and an albedo if you are trying to calculate a realistic habitable zone. If you just assume the planet is like the Earth then using the same albedo value will probably do. But, because Gl 570 has a spectrum that peaks at 50% longer wavelengths than the Sun, then the albedo values at longer wavelengths are more important for a planet orbiting Gl 570. Whether that is bigger or smaller than the Earth would depend on exactly what sort of atmosphere/clouds/terrain were on the planet.


In a finite universe, what happens when light reaches the boundary? If a universe is finite and is not expanding at a speed equal to or greater than c, what happens when light or another form of electromagnetic radiation traveling reaches the boundary?

One important concept that has not been mentioned yet is the "cosmological principle."  This is the key simplifying idea for our cosmological models, it says that the universe is the same everywhere on the largest scales at a given age, so alien astronomers 50 billion light years from us that also conclude the universe is 13.8 billion years old will be observing pretty much all the same things we are.  This principle is not proven by data, but the data is consistent with it, most notably the Hubble law and the homogeneity of the cosmic microwave background.  It allows us to understand the past of our own part of the universe by seeing the past of distant parts (a constraint we are stuck with given the speed of light).

So that's why a universe model that is finite in size must curve back on itself-- nothing else would satisfy the cosmological principle.  This doesn't mean the principle is true, it means we are not going to part with it unless we have to.


is pan-starrs' data available to public users? Anybody knows that?

The data products are all images or star catalogs?

This is the link:
http://pan-starrs.ifa.hawaii.edu/public/

Looks like you've to be member of the PS1 Science Consortium to be able to log in.
And you'll need some database query knowledge, at least by using the PSPS Query Builder, or more advanced programming capabilities (XML, SOAP, Python, Perl). A short documentation of the database tables can be found here.
If you like to dig deeper into the database, this FAQ page may be a starting point.

The image processing pipeline (IPP) is described in this paper.


Where are we in an approximate timeline of the possibly habitable universe? Our universe is supposedly 13+ billion years old and our Sun is a third generation star. It seems to me that we are now in a relatively young stage of the universe. How many generations of stars will there be before the end of the universe? When is 'now', relative to the all possible timepoints in the universe, that life could appear?

Here's a very roughly calculated partial answer.

A first generation star and solar-system would obviously not be a good candidate for life or even planets cause you can't do much with mostly hydrogen and helium.   You can't even build an ocean with mostly those 2 elements.  Stars do just fine, but planets don't.  Gas Giants and Suns only.

After the big bang: 75%-76% Hydrogen by mass, 24-25% Helium (with a teeny-tiny trace amount of Lithium).   I'm going to assume 76% / 24% for rough calculations.

Using our solar-system and the Milky-way as a model, elements by mass:

Milky-Way

73.9% Hydrogen

24.0% Helium

1.04% Oxygen

0.46% Carbon

0.13% Neon

0.11% Iron

0.10% Nitrogen

0.07% Silicon

0.06% Magnesium

0.04% Sulfur

Smaller amounts of Argon, Calcium, Nickel, Sodium, Phosphorus, Chlorine, etc.

and composition of the Sun by mass (pretty similar except for more Helium)

71.0% Hydrogen

27.1% Helium

0.97% Oxygen

0.40% Carbon

0.06% Neon

0.14% Iron

0.10% Nitrogen

0.10% Silicon

0.08% Magnesium

0.04% Sulfur

Using these numbers as a rough guideline, if 1.9% to 2.1% of the baryonic mass of the milky-way converts from Hydrogen to Carbon or heavier elements after (about) 2 generations, and we extrapolate a 1% increase in all Carbon or heavier elements to each stellar generation or every 6 billion years, then we can begin to make an estimate.

Helium percentage could stay somewhat constant at about 24% cause it's both consumed and created stars.

Your original question, how many generations, if we figure 1% increase in heavier elements (Carbon or higher) per stellar generation, 10 generations we're still looking at, very roughly, (24% helium), 66% hydrogen, 5% oxygen, 2%-3% carbon, 2%-3% other elements.  It still looks like a workable scenario to me though the stars, being made of some denser elements might have denser cores and as a result, could burn hydrogen a bit faster.    Past a certain point the ratios probably stop being ideal but I see no reason why quite a few generations couldn't still work create life-friendly solar-systems.

Our sun is expected to cast off about half it's matter into a planetary nebula before it settles down into a white dwarf and most of that cast off matter will be hydrogen and some helium.   Some time later if our sun as a white dwarf accretes enough to meet the Chandrasekhar mass and goes Type-1 Nova, it will again cast off a significant percentage of it's matter, so, to put it simply, Stars recycle a pretty good percentage of their hydrogen and helium, which is kinda cool.   If you see Rob Jeffries' brilliant answer here, stars of 7-9 solar masses will eject an even higher percentage of their elements before they go white dwarf, as much as 85% and most of what gets ejected is the lighter elements, hydrogen and helium from the outer layers.   Even smaller red-dwarf stars expel a percentage of their elements due to more active solar flares and lower gravity throughout their lives.

Add to this, the possibility of continued merger of the galaxies and dwarf galaxies in our local group, and life capable solar-systems could continue to form for quite some time, perhaps several tens of billions of years.

This answer isn't meant to be definitive, but only a rough estimate.  Corrections welcome.


Why doesn't the moon always follow the same path? Or why aren't the moon's apoapsis and periapsis fixed? I have seen many times that the moon doesn't follow the same path. Sometimes the moon gets very close to earth which is shown in the news. If the periapsis of the moon is soo close then why doesn't a super moon occur every 27 days? And if the moon is inclined at 5.1° so how can more than 2 eclipses occur in one year?

The Moon's orbit about the Earth is only approximately elliptical. The Moon's orbit precesses both axially and nodally, and the eccentricity of the Moon's orbit varies.

That the Moon's orbit precesses axially means that perigee sometimes occurs when the Moon is close to new, sometimes when it is close to half-full, and sometimes when it is close to full. Supermoons occur when perigee occurs close to when the Moon is full. The variation in the eccentricity means that apogee distance and perigee distance vary considerably.  Super-supermoons (my term) occur when perigee occurs close to when the Moon is full and when perigee distance is much closer than nominal. While supermoons occur about three of four times per year, super supermoons occur only once or twice a year.

With regard to eclipses, suppose that at the time of a full Moon that the Moon's orbit about the Earth is close to being aligned but not exactly aligned with the Earth's orbit about the Sun. This means that a month later, the Moon's orbit may still be close to being aligned with the Earth's orbit about the Sun. The large size of the Earth's shadow makes it possible for lunar eclipses to occur in two consecutive months.

Eclipse seasons, the times during which lunar eclipses and solar eclipses are possible, are about 34 days long, a bit longer than one month. This means that if a partial lunar eclipse occurs just as an eclipse season starts, a nice solar eclipse will occur half a month later, and a partial lunar eclipse will occur another half a month later. Similarly, if a partial solar eclipse occurs just as an eclipse season starts, a nice lunar eclipse will occur half a month later, and a partial solar eclipse will occur another half a month later.


Are exo-meteoroids possible? Is it possible that meteoroids, comets, asteroids or other small bodies from around a nearby star (such as Alpha Centauri) could get into the solar system?

How low is that possibility?

Edit:

Since the Oort Cloud is not being gravitationally attracted enough to the Sun to form a disk, I assume this is the same for Alpha Centauri's Oort cloud. Therefore, my question is: can the Sun gravitationally pull a comet from Alpha Centauri's Oort cloud?

Most, and perhaps all, stars form in stellar clusters.  Most likely, the sun formed in a open cluster.  In clusters the density of stars is quite high, therefore close encounters are common.  So, I think it would be quite likely for such close encounters to disturb Oort clouds enough to send comets out of their host systems.  Some small fraction of these comets would be recaptured by other stars within the open cluster.

In addition, it is thought that in the early period of planetary disk formation, there were many more planets than today, but they often collided with each other, often sending pieces out of the stellar system. Moreove, many small planets and asteroids were thrown out by gravitational interactions with the large planets (see Tremaine).  These 'debris' would be roaming interstellar space.  Occasionally, one could fall into any random stellar system.

The odds that we have any asteroids or comets from any particular nearby star is low, unless it happened to have been from the Sun's birthplace.  We do not have the ability to determine which of the nearby stars are from the Sun's open cluster, but the Gaia Satellite may make that possible soon.   And, if we explore the solar system long enough, we may find exo-solar asteroids among the hordes of asteroids.


Distinction between metric expansion and objects just moving apart from each other? I see explanations likes this:

Galaxies and other sources, then, are not strictly expanding away from each other but rather are attached to the fixed grid on the expanding fabric of spacetime. Thus, the galaxies give us the impression of moving away from each other. (Scientific American)
[...] space is expanding [...] the most correct way to think about it would be that the grid points are getting farther apart. (Physics.se)

How have we determined that it is some underlying grid or space itself that is expanding rather than the objects just moving apart on a static grid of spacetime?

Time will show

...although that is a very long time.

If the grid of the universe was static, objects moving away from us should continue to move away from us with a continues velocity. That is not the case from observed data: Things farther away from us seems to move away faster. This is known as Hubble's law, and is one of the basic concepts of modern cosmology. The model of an expanding metrics fit that data, in that when there is more space fabric between us and the far object, there are more space that stretches, and therefore the velocity is higher.

That things move away from us, and have an accelerating velocity, should suggest a universe that expands faster and faster. There are however other forces limiting this growth, most notably gravity. The fate of the growth of the universe is determined by that unknown balance, that must also consider dark matter and energy.

The magnitude, or even the validity of those models is the current frontier of astronomy, and cosmology in general. Nothing is settled yet.


Planning observation: determine dates where moon is within 30 degrees of my object? I'm planning an observation for a proposal and I need to determine which dates the moon is within 30 degrees of an object so I can ask to not be scheduled for those dates.  This object is near the ecliptic so it is guaranteed the moon will pass close by at times.

In the past I've used skycalc to manually figure out such dates, but now I need to figure out dates over a several month period, and I was hoping there was some more automated tool I could use for this purpose.

I would use JPL Horizons to calculate the position of the Moon and save the results to a text file. Then use a spreadsheet to calculate the separation between the Moon and your object. $$\cos \theta = \sin \delta_1 \sin \delta_2 + \cos \delta_1 \cos \delta_2 \cos (\alpha_1 - \alpha_2)$$ where $\theta$ is the separation, $\delta$ is the declination, and $\alpha$ is the right ascension.

Since the Moon moves about 1 degree in 2 hours, you can use that to determine what time interval to use in the output, depending on how accurately you want to find the 30 degree separation.


Are the planets Trappist-1 in the solar system? I'm an Astronomy noob, but I'm interested in learning how stars and planets work. Everything started with the Trappist-1. I read about it, and it's fascinating. However, it's impossible for me to localize it with my current state of knowledge. Therefore, I would like to ask some general questions here.

First of all, during high school, I have learned that we are in the solar system and there are some planets orbiting the sun. Easy to remember and imagine, but there are stars, planets, constellations, etc. which I can't manage to relate to this basic system that I learned about.

And what confused me a lot are these two pictures:

and

And this definition of an exoplanet:

  An exoplanet or extrasolar planet[...]

From the pictures, I can see that Trappist is in fact in the solar system. Is this right? And if this is right, why in the world did it take us so many years to find it? We have one Voyager that is almost out from the solar system. Couldn't it find it before? And is it possible to find more planets in our solar system? Is it because Voyager went only in the x-y direction, and not "perpendicular" to the orbits?

I'm sorry for asking these stupid questions, but I simply can't build new knowledge on my current low-level understanding of the solar system. This is my current knowledge:

Can you please help me to answer the above questions?

During high school, I have learned that we are in the solar system and there are some planets orbiting the Sun.

Yes that is correct. According to current definitions of what counts as a planet, there are 8 planets orbiting our Sun $-$ Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, and Neptune. There are also a collection of smaller "Dwarf Planets" which include (in no particular order): Pluto, Ceres, Eris, and Makemake. There's also a lot of other minor solar system bodies out there like asteroids, comets, etc.

  but there are stars, planets, constellations

Yes, there are many other stars out there besides our own (as you can see by simply looking up at the night sky). Each of those stars has the potential for having its own planets, much like ours does. Any planet outside our own solar system is referred to as an exoplanet (exo- meaning outside or external). Within the past 20 or so years we've developed the technology to find exoplanets around other stars and this TRAPPIST-1 discovery was simply the latest and most sensational discovery. To date we've discovered ~3500 exoplanets.

  What confused me a lot are these two pictures. [...] And this definition of an exoplanet.

Those pictures, while seeming to show the the TRAPPIST-1 planets in our solar system, are not indicating that those planets actually reside in our solar system. Instead, what they're trying to do is compare the new TRAPPIST-1 system to our own so you can see the relative sizes. In your first image, you see Jupiter along with the orbits and distances of a few of its Moons. Next to it, you see the TRAPPIST-1 system with the seven planets we know of around it. Next to that is the Sun where you can just begin to see the orbit of Mercury (with the other planets beyond the bounds of the image. The point here is just to compare sizes of orbits. Clearly this TRAPPIST-1 system is much smaller than our own since all 7 planets orbit closer to the main star that even Mercury does for us. However, it is still bigger than Jupiter and its moons. They're not trying to make the point that these systems reside in the same place, just putting them together for comparison.

As for the second image you have, what they're trying to show here (I assume) is the habitability zone. Around every star, depending on the star's size and temperature, there will be a good distance to be from the star that is optimal for life to exist. You don't want your planet to be too hot (by being too close) or too cold (by being too far). This image appears to be showing you the regions where a planet can be at a good distance from TRAPPIST-1 and still be at a good temperature. There is the Inner B line which I take to be the Inner Boundary of this zone. In other words, if a planet is any closer than this, it will be too hot. There's also the Outer B (or Outer Boundary) where beyond that, any planet would be too cold for life. Finally they have the Earth B. I can't be certain but I believe this point is actually the distance for the "optimal" temperature that we have here on Earth. In other words, the best chance a planet has of being at a good temperature for life is to orbit between Inner B and Outer B and ideally be at the distance of Earth B. This image also shows that the orbit of the planet TRAPPIST-1e is close to Earth B and still inside Outer B, meaning it is likely to be at a hospitable temperature for life (which is good news for us!). I won't go into what an AU (aside from mentioning it is a unit of distance) or the frost line, since they're not pertinent to the point of the image. Feel free to ask other questions for that.

Lastly, the definition of an exoplanet you found describes it an "extrasolar planet". That just means it is beyond our solar system $-$ "extra-" meaning "outside" or "beyond" and "solar" referring to our own Sun.


Is there any scientific correlation between cold nights and star-lit skies? I've many times heard expressions such "cold, starry nights". It's apparently poetic to describe star-lit night skies as being "cold".
Is this based on reality in some way? Is there something about cold weather which would make the stars more visible from Earth?
I've been taking many nightly walks and, maybe, I'd say that the stars seem to be more visible when it's cold/winter. But I'm far from sure about that; it's useless as "evidence". I don't really see why this would be. Seems like it would have more to do with how cloudy it is.

Yes. Clear skies allow the Earth to cool more effectively at night. If there are clouds, these re-radiate some portion of the infrared flux from the Earth back towards the ground, keeping it warmer.


Binaries consisting of a black hole and a non-black hole? Are there any known binary systems in which one member is a black hole and the other is not a black hole (main sequence star, giant, neutron star, white dwarf, whatever)?

Googling seems to turn up only discussion of binaries in which both components are black holes. Is there some astrophysical reason why the kind of system I'm talking about cannot form? I think I remember that gravitational wave observatories have talked about black hole-neutron star mergers, but are such systems purely hypothetical so far?

I guess Sagittarius A* is sort of close to being an example, if you count it and the stars orbiting it (or orbiting their common barycenter) as a multiple star system.

Black hole and main sequence star/giant star

We can observe binary systems containing a black hole by looking for emissions from accretion disks which may form when matter is transferred from the companion star. X-ray binaries and microquasars are particularly notable types.

The compact object doesn't have to be a black hole - neutron stars also sometimes show up - but in some cases, measurements indicate that the remnant is a black hole. In other cases, either the mass of the compact object is poorly constrained or could still be a massive neutron star, so for many systems, the nature of this object remains unknown.

Examples:

Cygnus X-1: The companion star here is HDE 226868, a massive O-type star.
V404 Cygni: The companion star is a K-type star (much less massive than HDE 226868); the system forms a microquasar, with occasional high-energy x-ray emission.

Black hole and white dwarf

It seems a logical assumption that black hole/white dwarf systems should be possible, and indeed they are. Mass loss through accretion by a companion object can change a star's evolutionary future (see the Algol paradox for a particularly weird case), but for many systems with a black hole, the "normal" star should still progress through the main sequence and post-main-sequence evolution, possibly becoming a white dwarf.

Matter transfer can still happen at this stage, of course, so emission may continue, although the nature of the radiation could be changed. It's true that there aren't many known systems of this type, but there almost certainly more.

Examples:

X9: The companion star is a white dwarf orbiting extremely close to the black hole. Bahramian et al. are confident that the companion is a white dwarf, and the main object is likely a black hole.

Black hole and neutron star

These type of systems are possibly the most interesting from a gravitational point of view. They should produce gravitational waves, which could be detectable by LIGO.1 LIGO has yet to observe such a system, which seems a bit puzzling, although something like the blue supergiant in Cygnus X-3 could evolve into a neutron star in the future. This is a case where more data and observations are needed.

1 You've probably heard the rumors that LIGO detected a possible neutron star-neutron star merger, which would imply that its sensitivity to neutron star-binary systems could indeed be high enough for detections, as expected. It turned out that it was simply a binary black hole system[1] - not quite as exotic as some people were hoping for.


Parent stars of our Sun - Where are its remains? This is in a way related to this question.

Question is: Our star is third generation star, which is explained by existing Barium. That Barium was created by other stars. Now, those stars must have been Supernova. When they exploded, it was either a neutron star or a black hole.
Where is that Neutron star or black hole?

We have no way of identifying where the Sun was born, what the surrounding environment was, or where the Sun's siblings are right now. This is easy to see from some bare numbers: the Sun's current orbital period around the Milky Way is some 250 million years, and it's been around for some 4.5 billion years, making for some 20 orbits around the galaxy since its birth. There's a lot of stellar-movement dynamics that can happen over those twenty orbits, and any siblings have long parted ways. Similarly, any stellar remnants that are close to us right now are close because we're passing by, not because we were born in their neighbourhood.

As such, whatever the source of the heavy elements in the solar system, it's several billions of years too late to figure out any individual sources.

That said, it is important to emphasize that there is no such unique source to begin with. As explained in the answer you linked to, there are plenty of such predecessors:

  There are grains of material trapped inside meteorites that consist of solids that were already present in the pre-solar material. These are important because these grains were thought to have formed in individual stellar events and their isotopic compositions can be studied. These tell us that the Sun formed from material that has been inside many different stars of different types.

Moreover, if you want to place those predecessors, you are fighting against the internal mixing of the interstellar gas on a galactic scale:

  Mixing in the interstellar medium is reasonably effective. The material spewed out from supernovae and stellar winds 5-12 billion years ago has had plenty of time to mix throughout the Galaxy before the Sun's birth. Turbulence and shear instabilities should distribute material on galactic length scales in a billion years or less.

This means that the Sun was not born out of the ashes of its neighbours. Instead, it was born out of the ashes of stars that might have died on the opposite side of the galaxy, several billion years before the Sun's birth, whose ashes then got thoroughly distributed by the mixing of the interstellar medium.


Can A Black Hole Exist? In 1939, Robert Oppenheimer and other concluded that a certain neutron stars could collapse to black holes and no known law of physics would intervene.  As far as I can tell, there is no observational data that shows that an event horizon of a black hole exists.

Assuming that a black hole with an event horizon exists, then there are all kinds of mathematics that explain what is observed externally, approaches the event horizon.  It can be explained what an infalling observer sees as they approach the event horizon.  However, what happens when an infalling observer passes the event horizon, the result is a mathematical absurdity, in that the observer would observe the universe outside the event horizon as beyond "the end of time."

It seems that passing through an event horizon is as mathematically absurd as an object with non-zero mass reaching and surpassing the speed of light, as it would require more than infinite amount of energy in the latter, or more than an infinite amount of time in former.

So, can an event horizon even "exist" in the first place, if "existing" means within the bounds of any moment after the big bang and before an infinite amount of time from any point within the universe.

It seems that the existence of black holes seems to be well accepted, however faster than than light travel does not under the current understanding of the physics of relativity.

There is evidence that both black holes and their event horizons exist.

The primary evidence for stellar-mass black holes arises from observations of the dynamics of binary systems. What has been found, for at least 20 binary systems, is that the optically visible star has a dark companion, that is usually more massive, and more massive than can possibly be supported as a neutron star or any other degenerate configuration ($>3M_{\odot}$ and mostly $>5M_{\odot}$ - see Narayan & McClintock (2014) for an excellent, accessible review).

Supermassive black holes at the centres of galaxies can also be "weighed" using the dynamics of gas, or in the case of our Galaxy, actually watching the stars orbiting something that must have a mass of $4.4\times10^{6}M_{\odot}$, yet is remarkably compact and emits little or no radiation.

The evidence for event horizons is more circumstantial, but not absent. In any case, absence of evidence would not be evidence of absence - event horizons are difficult to prove; they are small and far away, with elusive observational signatures. The lack of an event horizon could only be accommodated by inventing something even more bizarre than a black hole.

The most compelling argument for an event horizon is found in sources that accrete from their surroundings, either in a binary system or at the centres of galaxies (see Narayan & McClintock 2008; read particularly section 4). In some circumstances, the accretion flow can become radiatively inefficient (dark), in which case one expects that if the accretion flow reaches a "surface", it will radiate its kinetic energy strongly as it thermalises on impact. On the other hand, a black hole event horizon can swallow such flows without trace. The prediction is, that in their low-states, transient black hole accretors will be much less luminous than their neutron star counterparts, and this is what is found (by a factor of 100).

Other evidence is that quiescent neutron star binaries show hot, thermal emission, presumed to be from the neutron star surface. None has been seen from the black hole binaries. Type I X-ray bursts are seen in neutron star binaries, caused by an accumulation of matter and subsequent nuclear ignition on their surface. The black hole candidates do not show these bursts. The accretion rate onto the supermassive black hole in the centre of the Galaxy should produce thermal radiation at any "surface" that would be quite visible in the infrared part of the spectrum. No such detection has been made.

All of these observations are best explained if the black hole has no surface and can simply make the accreted energy "disappear" inside its event horizon.
Narayan & McClintock conclude that these lines of evidence are "impervious to counterarguments that invoke strong gravity or exotic stars".


Why are Mars northern/southern hemisphere altitudes so different? I'm looking at the Mars topography map:

original link

...and I can't shake the impression that someone just messed up the placement of the center point of the planet, shifting it about 8km north from where it should be. If the center point was ~8km further south, the north and south altitudes would be roughly the same. Moreover, it's even hard to argue about placement of equator, as the northern pole is about 16km closer to the equatorial plane than the southern - due to the altitude difference. Move the equator and it all evens out.

It's fairly hard to believe such a mistake to have persisted so long, so... is there any deeper reason behind this strong asymmetry of the planet, other than just a topographical snafu?

The placement of the center of the map is messed up, but it's nobody's mistake. From chapter 4 of The Surface of Mars:

The elevation difference between the two hemispheres
offsets the planet’s center of figure from its center of
mass by 2.986 km as measured along the polar axis

As explained in another answer, the altitude is defined based on Mars' center of mass.
Regarding the question that is in the title ("Why are the hemispheres so different?"), this sharp contrast between the Northern and Southern hemispheres is known as the Martian dichotomy.
The elevation on Mars follows a bimodal distribution. That is most of the land is either 4 km under the mean altitude or 1.5 km above the mean altitude.
But not only is the altitude different between the two hemispheres, but the crustal thickness and the number of impact craters are also very different.
There are several competing hypotheses to explain the dichotomy. Here are a few.
Geological phenomena
Part of the crust was overturned
Very early in the history of Mars, there was a magma ocean that crystallized. Solomon et al. (2005) suggest that the top the mantle ended up denser than what was underneath, and it flipped.
There used to be active margins
By analogy with the dichotomy seen on Earth between oceans and continents, Sleep (1994) suggests that the northern hemisphere may be the result of new crust being generated and spreading from the center. The limit between the two areas would have been an active margin, similar to the Andes where the oceanic crust drops under the continental crust.

However, this would result in the formation of magnetic stripes that are observed in the southern rather than in the northern hemisphere.
Thermal convection in the mantle
As suggested by Zhong and Zuber (2001), there might have been strong convection in the mantle with a single convection cell.
These models succeed in reproducing some differences between the two hemispheres.
Impacts
A giant impact in the northern hemisphere
Wilhelms and Squyres (1984) suggest that a giant impact very early in the history of Mars might have caused the northern depression. They also indicate the location of the impact.
However, no thinning of the crust has been found there, like what is observed for younger and smaller craters. The ring deposits that such an impact would have created are missing. The edge of the northern plains are very irregular, when impact craters tend to be somewhat circular.
A giant impact in the southern hemisphere
Leone et al. (2016) propose that the dichotomy is the result of an impact in the southern hemisphere and that the result of the impact in more crust, rather than missing crust.
A few not quite as giant impacts
Explaining the irregular boundary between the two areas is rather difficult with a single impact, so Frey and Schultz (1988) suggest that a few big impacts might have caused the peculiar geography of Mars.
However, this hypothesis doesn't completely explain why the impact basin isn't circular.

So the question of the martian dichotomy is still open and doesn't have a definitive answer yet.
Further reading:

Mysteries of Mars by De Blasio, chapter 2
The Surface of Mars by Carr, chapter 4
Origin of the Martian crustal dichotomy, George E.McGill and Steven W.Squyres


Does a photon honor the causality principle? We know that the faster you move, the slower your time is perceived from an external observer and the space along the direction of the movement shrinks.

Photons, having no mass, move at the speed of light c (in the vacuum) and for them the space has no meaning because it's collapsed into one single point (edit: better say into a single plane, because only the direction along the path collapses). Actually, they don't "move", just "are" at the same time in all places along their path.

In the same way, for them time has no sense. Because they don't "move" they take 0 time to reach their destination.

This seems to be odds we the causality principle, but of course I'm wrong.
I try to explain it with a simple example. Let say a photon is emitted from a very far star millions of years ago and I receive it in my eye now. It means its path was started on that star and ended in my retina.

It took millions of years, for us, and 0 seconds from its point of view.
So, I apologize for the stupid question, but how it was able to "know" that my retina will be there? If I was few cm apart, perhaps it would have continued for other million of years until it will eventually reach something other. But for it, the time is still 0.

How this could be explained?

And, we know the speed of light is less than c in other media. But it's not clear to me if from the point of view of a photon changes something. For it, the space and time are still 0? As an external observer I can say it still moves as fast as it can. From its point of view it's always in all points of its path.

Is it correct?

Instead of saying "space has no meaning because it's collapsed into one single point" it's a lot easier if you say that a photon is not at rest in any frame. Then all of the paradoxes you mention go away.

It's easy to use the Lorentz transformations (the mathematical core of special relativity) to show that it's impossible to transform a frame moving at $v<c$ to $c$, or vice versa. (Actually, the Lorentz transformations were developed / discovered a decade or so before Einstein proposed the theory of special relativity. His great insight was to realise that those transformations made more sense if you abolish the hypothesis of the luminiferous aether, and treat space and time as a united geometric structure).

In relativity theory, it simply doesn't make sense to talk about "the point of view of a photon". A photon doesn't have a point of view because it's not geometrically possible for an inertial reference frame to have a velocity of $c$ relative to any other inertial reference frame.

The geometrical structure of spacetime proposed by relativity theory has important implications for causality because no causal influence can travel faster than $c$. Two key concepts here are light cones, and the relativity of simultaneity. There are plenty of questions on this topic on the Physics stack, so I won't go into the details here, but very briefly, if a photon leaves spacetime event A and is absorbed at spacetime event B, then all observers will agree that A occurred before B, although different observers may measure different spatial distances between A & B and different time durations between A & B, but they will all agree that the (local) speed of the photon was $c$. We say that the spacetime interval between A & B is light-like (or null).

If spacetime events A & B have greater space separation than time separation, eg in my frame, B happens 1 second after A, but the distance between them is 2 light-seconds, then we say that the spacetime interval between them is space-like. No causal influence can travel between them, and different observers will disagree on whether A happened before B, or vice versa, and for some observers A & B occur simultaneously. Here's a nice diagram from the Wikipedia article on the relativity of simultaneity:

  Events A, B, and C occur in different order depending on the motion of the observer. The white line represents a plane of simultaneity being moved from the past to the future.