What effects do CMEs have on planets like Mercury? Some Background?
Coronal Mass Ejections, or CMEs for short are the topic of this question. Anyone new to this topic is sure to find some really interesting facts (and consequences) of CMEs and Solar Flares here.
Some Wiki definitions?

A coronal mass ejection (CME) is a significant release of plasma and accompanying magnetic field from the solar corona. They often follow solar flares and are normally present during a solar prominence eruption. The plasma is released into the solar wind, and can be observed in coronagraph imagery.

So, what's the question exactly?
If you've seen the video linked above, then you know that CMEs release a lot of solar ejecta (plasma). The Earth's Magnetic field takes care of them in a way that we barely knew things like these even existed before some really advanced equipment came along.
So, Question - What effects do CMEs have on planets like Mercury?
I had trouble finding and/or verifying any resource that I found on this question. Can anybody help?
P.S. Any help with the tags will also be appreciated.

Much like Earth, Mercury also has a magnetic field, only much weaker. Because of this, the high velocity charged particles accompanying CMEs distort Mercury's magnetic field at a much greater magnitude than it does earth. Also, because it extends beyond the sun's corona, it has the potential to burn the sun facing side of Mercury's surface even more than usual. Because of the high solar winds accompanying the CMEs it has the potential to actually strip particles off the surface of Mercury to become space dust.
CMEs contain high radiation, large amounts of high velocity charged particles (causing a HUGE magnetic flux/field), and extreme solar winds. (among some other things) so any way that these phenomena will normally affect Mercury will happen during a Coronal Mass Ejection


If the moon still had a magnetic field how long would have been possible to keep an atmosphere? Let's assume the Moon had a considerable magnetic field to prevent the atmosphere dispersion caused by solar wind. Given the low gravity of the Moon, how many years would pass before all the gases (of an Earth like atmosphere) would escape its body? Would its gravity be sufficient to trap heavier gases like CO2?

I'll give a slightly different answer: The escape velocity at the surface of our Moon is about 2.38 km/s. Derived from this paper, as a rule of thumb, an atmosphere can survive 4.5 billion years, if it's average molecule velocity is below 1/6 of the escape velocity of the planet/moon. Applied to our Moon, it's 2380 m/s / 6 = 396.67 m/s.

Carbon dioxide has a molecule mass of 44u.
The according temperature for an average molecule velocity for carbon dioxide is hence
$$\mbox{temperature}=(v_{\mbox{gas}}/157)^2 \cdot \mbox{molecule mass}\mbox=(396.67/157)^2\cdot 44\mbox{ K}=280.87\mbox{ K}.$$
Moon's mean surface temperature at the equator is about 220 K. That's well below the allowed 280.87 K. A carbon dioxide atmosphere could have survived 4.5 billion years according to these oversimplified assumptions.

When looking closer to the atmosphere, it turns out, that the higher layers of the thermosphere and the exosphere can reach more than 1000 K.
Even carbon dioxide can escape from Moon at this temperature over time:
The average molecule velocity for carbon dioxide at 1000 K is
$$157\cdot \sqrt{\frac{1000}{44}}\mbox{ m/s}= 748.5\mbox{ m/s}.$$

This results in a Jeans escape parameter of
$$\lambda_0=\left(\frac{v_{\mbox{esc}}}{v_{\mbox{gas}}}\right)^2 =\left(\frac{2380}{748.5}\right)^2= 10.1,$$
resulting in an escape rate of about $10^{-4}$ relative to a free molecular flow.

From this it isn't yet straightforward to calculate the period of time, Moon would loose a carbon dioxide atmosphere, since it's a function of the height-dependent temperature curve of the atmosphere. But it's at least clear, that Moon wouldn't loose a carbon dioxide atmosphere suddenly, since this is likely just for $\lambda_0 < 3$.


Problem with making telescope I am using a convex lens with $f=5$ cm and $5$ cm aperture, a concavo-convex lens with $f=100$ cm and $5.5$ cm aperture. I am trying to make telescope and it should have magnification of 20 x. But the image coming of eyepiece is very small and cant be seen clearly. Please help with the problem?

A 50mm aperture with 1000mm focal length is an f/20 ratio - which is a very very small aperture for the focal length and will result in a very faint image. The small opening angle will result in vignetting (that's what you see with the 'small image'). With those pre-conditions you won't get a bright image with a wide FOV.


What is the likely closest encounter for a typical star during the Milky Way Andromeda Merger? This is closely related, but distinct question to What are the chances of a star colliding with another during a galactic collision?.

The accepted answer to that question places the odds of the Sun colliding with another star at about $10^{-12}$, what I want to ask is how near a near-miss is expected? In other words, what is the expected closest encounter between the Sun and another star during the collision? A more sophisticated answer would give the expected distribution of close encounters by distance.

A more general version of the question is -- what disruptive events would a typical star such as the Sun expect to experience during the merger?

Suppose the sun moves through a column of Andromeda stars $L$ parsecs long with density $\rho$ and random positions. The probability of encountering at least one star within $r$ of the sun along this column would be $$F(r)=1-e^{-\pi r^2 L \rho}$$ (basically, what is the probability of no "success" when doing a Poisson sample of the column volume times density).

Setting $F(r_{1/2})=1/2$ to find the distance where we should expect a 50% chance of a star more nearby we get $r_{1/2}=\sqrt{\ln(2)/\pi L \rho}$. If we plug in $\rho=0.14 $ per cubic parsec and $L=1000$ parsec (assuming a flat disk-disk collision) I get $r_{1/2}=0.02$ parsec. That is 4125.296 AU, reassuringly far.

To get the mean distance we can use the not very well known but useful formula $E[r]=\int_0^\infty (1-F(r)) dr =\int_0^\infty e^{-\pi r^2 L \rho} dr = 1/2\sqrt{L \rho}\approx 0.042$ parsec.

Depending on your views of disk thickness and angle on merger $L$ and $\rho$ may vary a fair bit. A sideways collision might have $L=67$ kpc reducing the distances by a factor of 8.1. Direct collisions still remain very unlikely, but messed up Oort clouds seem very likely.


Why the earth orbits the sun from X to -X and not the opposite I wonder all the time, why the earth orbits the sun as expected from east to west(X to -X) ?
It could be different, for example from the west towards the east .
What would happen if that were the case?

The thing that keeps the Earth from falling into the Sun is that it orbits the Sun, once a year. We usually don't say that the Sun orbits the Earth.
Orbits have a direction or sense in space. What that direction is called is an issue of history and linguistics,
The Earth rotates around its own axis, this rotation also has a sense, and the same caveats about what this is called apply.
It is the daily rotation of the Earth around its axis and the tilt between this axis and the axis of the Earth's orbit around the Sun that determines for a given spot on the Earth where along the horizon the Sun will rise and set.
If nothing else changed except the Sun suddenly started rising where it normally sets, and setting where it normally rose, that would mean that the Earth started to rotate in the opposite direction around its axis.
We would call that retrograde rotation and it would be a surprise.
It's generally believed that the sense of rotation of most (but not all) planets around their axes is similar to the rotations of the planets around the Sun is evidence of the rotation of all the matter that formed the Sun and the solar system. The Moon also rotates with the same sense around the Earth and the Sun rotates with the same sense around its own axis.

What would happen if that were the case?

This is a great question and is hard to answer. It depends a great deal upon if the Moon was also now in the opposite sense or not.
But I can say one thing that would happen is that people would wonder why the Earth was rotating in the opposite sense of most other things in the solar system!


How do I find out when a constellation reached its zenith? I found this website to look at the sky at the specific time and location: http://www.fourmilab.ch/cgi-bin/Yoursky - you can insert latitude and longitude and time to see how the skies looked at any location/time. What I need to find out is when a particular constellation reached its zenith. I see that if I change the Universal time the location of the constellation in the sky changes, but where is the zenith of a constellation? The constellation in the question is Bootes. How do I know its zenith time let's say for today? preferably using the website above.
Picture example from the website:

The key number you need to find is the "right ascension" This tell you the "longitude" of the constellation, relative to the other stars in the sky.  It is measured in hours and minutes
Bootes has a right ascention of about 14:30  This means that on the day of the spring equinox (March 21) it will culminate at 02:30  (in local solar time, you'll need to make adjustments for daylight saving time or if your local noon isn't at 12:00) Note that 14:30 is 2:30 pm
Every month it will culminate two hours earlier So on April 21 it will culminate at about 00:30  (conveniently there are 12 months and 24 hours)
Working the other way, On January 21st it would culminate four hours later, or at about 06:30, and now, in Early February it culminates at about 06:00
This kind of calculation is good enough for a rough estimate.  More accurate calculations can be done, but are more convenient to get a computer to do them.


From what distance can one object influence gravity of another object? Each object in the universe has its own gravitational influence on all other objects in the universe. What distance do they have to be from each other to create only one gravitational influence? Example - Every human does in my opinion contribute to the earth gravitation thanks to the mass of our bodies. Now would there be a difference if I distanced myself from the surface by any amount?

What id like to know, in what distance do they have to be from each other to create only one gravitational influence.

At whichever point you decide to call them two objects rather than one object. It's a completely arbitrary choice that depends on you rather than gravitational physics. What's going on is that gravity can be described by a mass density distribution, and which part of that distribution corresponds to "one object" and which to "another object" isn't important.

You may be confused by Newton's law of gravity that says the gravitational force force is proportional to the product of the masses and inversely proportional to the distance between them squared. But this law only applies to spherically symmetric objects. It only applies to you exactly if you're a spherical cow.

This kind of arbitrariness actually applies to the Earth as well. Even if the Earth were perfectly spherically symmetric, one could say that Earth's gravity is due to the influence of the northern hemisphere and the influence of the southern hemisphere, etc. Whether you consider the Earth to be one object, or two, or a trillion, depends on you.

The Earth isn't spherically symmetric. It's closer to an oblate spheroid, since it bulges at the equator. But it isn't exactly that either, having mountains or other topographical feature, to say nothing of mass density variations inside the Earth. In principle, one could describe its gravity including people and trees or whatnot. Gravitationally, there is no fundamental difference between you and some rock. That you consider yourself to not be "part of Earth" is a choice you make for other reasons.

If you distance yourself from the Earth, there is a difference because the mass distribution would change. But again, whether you consider this to be "only one gravitational influence" or "the Earth plus you" is up to you, and this holds true regardless of whether you distanced yourself or not.


Temporal path through Hertzsprung-Russell diagram? Is there graphic of typical temporal paths that stars take through the Hertzsprung-Russell diagram?

This is typically a bit tricky to show because (1) some phases (e.g. the main sequence) are much, much longer lived than other phases and (2) a number of phases are spaced very close together in HR-space.

Be that as it may: any elementary text book on stellar structure will guide you through the path of a star through the HR diagram, usually accompanied by pictures. The book by Prialnik on this subject matter, or Ostlie & Carroll, are both excellent.

Here is a very crude version of the temporal path through the HR diagram I made some time ago. Numbers show the different phases (2 is MS), colorcoding shows the ages.


can moon rise before sun and set after it in the same day I found it mentioned in ancient Arabic astronomic book Kitāb al-Amwāl by Ibn Qutaybah (889 AD)

Cresent can't be seen in the morning in the east in front of the sun and in the evening i the west behind the sun in the same day

ولا يمكن  أن يرى الهلال بالغداة فى المشرق بين يدى الشمس، وبالعشىّ فى المغرب خلف الشمس فى يوم واحد.

Is this information valid for all locations and dates
Sun and Moon have very near orbits and both daily apparent motion (east-west) but moon has apparent retarding motion (west-east) could this retarding motion make the sun catch the moon in the first day of lunar month?
could this happen in normal (non polar) latitudes?

It's possible, but it's not very impressive.
This month (August 2020 New Moon) from, say central Asia (I used Aral for my simulation on Stellarium) The moon rises at 05:48 local time (UTC+5) and the sun a few minutes later at 06:01
Of course you can't see the moon rise, it is the day of the new moon and it is invisible.
Then in the evening, the sun sets at 19:51 and the moon sets at 20:28.  Again, it is not visible at this time.  (I've not accounted for atmospheric refraction here, which will change the times but not the example.)
This doesn't happen from all locations every month, it just depends on whether the sun passes the moon during the day or during the night, I'd venture a guess that in any particular month, half the world will have the moon rise before, and set after the sun on the date of the new moon.
So this isn't valid for all locations.  If you were in America, then the moon would pass the sun during the night: on one day the moon would rise and set before the sun, the next day it would rise and set after the sun.
Its is also not valid for all dates at a single location.  It is possible that the New moon in September might pass the sun during the day when viewed from America, but not from Asia (though I haven't checked)


What is the shape of the Sun's orbit around the Earth taking into account elliptical orbits? Consider the non-inertial frame that is at rest with respect to the Earth's revolution around the Sun.  (Ignore the Earth's rotation around its own axis.)  My question is, what is the shape of the Sun's orbit around the Earth in this reference frame?

Now if we assume the Earth moves in a circular orbit around the Sun, then the Sun's orbit around the Earth will also be circular. (Just use geometric the definition of a circle - the set of all points a fixed distance from a given point.). Specifically it's the great circle formed by the intersection of the ecliptic plane with the celestial sphere.

But the Earth does not move in a circular orbit - Kepler's first law states that the Earth moves in an ellipse around the Sun, with the Sun at one of the foci of the ellipse.  So if we consider the Earth's elliptical orbit around the Sun, what is the shape of the Sun's orbit around the Earth.  I doubt it's an ellipse, so would be it a more complicated-looking curve.

Note that I'm not interested in gravitational influences from the moon and other planets - this is pretty much a purely mathematical question: if we assume the Earth moves in an ellipse, what would be the shape produced?

The Earth moves in an elliptic orbit around the sun (or around the barycenter). If, in helocentric coordinates the Earth is at position (x,y), then in Geocentric coordinates the position of the sun is in position (-x,-y)

So the locus of the Sun in Geocentric coordinates exactly matches the locus of the Earth in Heliocentric. The path of the sun in geocentric coordinates is an Ellipse.


How long can you be trapped orbiting around a black hole for? I have a story I wanna write but I want to be sure it's not completely scientifically irrelevant.

I know there is  black hole modelisation called the Kerr black holes, in which there is a limit around black holes called the event horizon. It's a spatial limit ; no light ray crossing this limit will ever be able to escape the gravity of the black hole. That's the definition of event horizon.

By extension, I guess any physical object (such as a spaceship) crossing the event horizon will be doomed to orbiting around the black hole until it gets destroyed. My question is about the time you can spend between the moment you cross the event horizon (= you are doomed) and the moment you actually die (for example by the tide effect, which basically destroys bodies because of the difference of attraction force between the feet and the head of a human body)? Could this moment last... 100 years ? 1000 years ?
(for my story the more the better)

I know black holes studies are very theoretical, I'd just like to avoid any huge scientific plot holes.

I am quite unskilled in that domain of astronomy, about time dilatation, etc. So if anyone has any idea of the order of magnitude of the time a object can spend beyond a black hole event horizon before it gets destroyed ?

Thanks for your answers :)

You're right to mention time dilation, because the point of view is important. The standard metaphor is that of an astronaut (A) falling towards the event horizon, wearing a watch, while another astronaut (B) watches from a distance. Astronaut A will simply fall through the horizon and hit the singularity, but B will never actually see A cross the event horizon, instead they will see A's watch run slower and slower.

In terms of survival time, bigger is better. Getting too close to a really small black hole could rip you apart with tidal forces without even having to cross the event horizon (just as moons and planets exert tidal forces on each other) - but with a supermassive one you'd not even notice you'd crossed it.

For a relatively "standard" black hole of about 30 solar masses the infalling astronaut will have around 0.0001s between crossing the EH and hitting the singularity (from their point of view....see above, B will never see them cross the EH).


Is there a name for a planet and its moons/satellites? I'm wondering if there's a name that encapsulates the concept of a planet and the objects that orbit in its gravity. There's a solar system that encapsulates a star and its multiple planets and other objects, but I don't think that "solar system" describes the Earth and its moon, or Jupiter and it's many moons, for example.

This question has been asked before on the Space Exploration page.
In summary, the term used is system, e.g. the 'Jupiter system'.


Seeing Azha (Eta Eridani) in Northern Hemisphere When I use Stellarium, I am unable to see Azha in London or Atlanta, Georgia.  However, it is possible to see Azha in Tokyo and Auckland.

I am not changing the time, only the location.  I presume that my time is converted to their time.  So for example, if my time is 11:30 (London) then when I select Tokyo, I should see what the sky is like at 11:30 Tokyo time.

I can see Azha in Auckland too.  I would've thought the ability to see a star is based on N/S hemisphere not E/W hemisphere.  Is it right I should be able to see Azha in the Eastern Hemisphere but in the West or am I using Stellarium incorrectly?

I am asking because of the Eta Eridanids. They might be an obscure or a disappointing meteor shower. The reason for such differing locations was to see where in the world it's possible to see them.

Eta Eridani has a declination of about -8 degrees, so it is visible from most of the Northern Hemisphere. Eridanus is a constellation that is next to Orion, and like Orion, it is easiest to see during winter.

The Eta Eridanids are a meteor shower that have a peak on about 11th of August. The radiant on that date is just south of Eta Eridani, at about -11 degrees. On 11th of August, Eta Eridani rises at about 3am in London, About an hour before twilight.

Tokyo, which is much further south, and doesn't use Daylight saving, sees Eta Eridani rise at about 1am (local time).

When you change location in Stellarium the time doesn't change. So if you change from London at 3am in August (UTC+1), the time in Tokyo will be 11 am (UTC+9).

The Eta Eridanids are a minor shower with a ZHR of about 6 meteors per hour They will be visible whenever the radiant is above the horizon, but from more northern locations it is only visible for a short time. It also clashes with the Perseids. One may only see a couple of Eta Eridanids during the viewing window, which is why it was only identified in 2005.


Was Newton the first to mention the orbital barycenter? A barycenter is the common center of mass of an orbiting system. Here is a illustrative gif from wikipedia:

The first mention of something like a barycenter that I could find is in a translation of Newton’s Principia, published in 1687, proposition 12, theorem 12:
https://en.wikisource.org/wiki/The_Mathematical_Principles_of_Natural_Philosophy_(1846)/BookIII-Prop2.
This includes the quote that: “the common centre of gravity of Jupiter and the sun will fall upon a point a little without the surface of the sun” which to me demonstrates a clear understanding and articulation of the idea of a barycenter. Does anyone know of an earlier clear reference to a barycenter?
Related: What are Kepler's laws (as he wrote them)?

I'm pretty sure Newton was the first to apply the notion of the barycentre to celestial motion, although the concept of a centre of mass may predate him.
Before Newton, there was a very strong demarcation between celestial motion and terrestrial motion, inherited from the Greek philosophers like Aristotle. (Similar philosophies applied outside the Western world). Celestial bodies had a celestial nature, and moved according to celestial principles, eg, they were perpetually in motion, and that motion was composed of perfect circles. Bodies on the Earth obeyed a different set of laws appropriate to their earthly nature. The essential quality of the earth element is to move towards the centre of the universe and to not otherwise move, unless something else interferes. So terrestrial objects didn't move unless they fell, or were pushed in some way, and would soon stop moving.
Newton did a lot of great scientific and mathematical work, but his outstanding insight was that a single scheme could describe both terrestrial motion and celestial motion. And that is the "core" point of the anecdote about Newton seeing the falling apple. Prior to Newton, the very idea of attempting to apply mundane mechanics to the heavens would have seemed strange to most natural philosophers, if not downright blasphemous.


Terminology question: gas giant vs gas planet While not exactly the most exciting question, I'm wondering: is there any real, semantic difference between a gas planet and gas giant, or are the two terms used interchangeably by most in popular usage (i.e. popular-scientific texts)?

now, a gas planet is obviously the umbrella term for all planets composed mainly of gas, but I often see "gas planet" and "gas giant" used interchangeably without any real change in meaning. I'd class Neptune & Uranus as "ice giants" and Jupiter & Saturnus as "gas giants", but (popular) dictionaries like https://www.thefreedictionary.com/ ; https://en.oxforddictionaries.com/ still seem to group them all into gas giants.

"Gas giant" is the most standard term for planets in the Jupiter/Saturn class, and, as you said, "ice giant" (since the 90's) is the term for planets with compositions like Uranus and Neptune.

"Gas planet", on the other hand, is just not a commonly used term. Judging by Google n-grams, it is less than 1/12 as frequent. Also, if you do a Google book search, of the times it is used, it is very often as a part of the phrase "giant gas planet." I would consider "gas planet" simply to be a description rather than a technical term, at this point.

A lot of likely materials for Earth-sized non-rocky planets would be more likely to be classified as ices in the outer solar system. So planets that deserve to be called "gas planets but not gas giants" may be quite rare, so it's possible a distinction may not be needed.


How much illumination do the background stars provide? I'm trying to get a better understanding of what things look like in "outer space." The main problem I'm trying to solve right now, is how much illumination is provided by the background stars. Is it even perceivable? If we were to ignore the light contributions from our sun and local planets, how difficult would it be to see an object (within arm's reach) that was lit only by the remaining ambient light?

Another way this might be thought of is: how dark is the dark side of Pluto? If a person were able to stand on the surface of Pluto, would a moonless/Charonless night be totally pitch black, or would there be enough ambient light to see? How bright might that be?

EDIT: The example about Pluto is a little deceptive because Pluto will block a large portion of starlight. I'm thinking of the starlight that is neither blocked by nor supplemented by any of the bodies in our solar system.

Although sources go way back, the current best estimate seems to be that of Abdul Ahad from March 2004:

  For an observer located anywhere within the Solar System - excluding the contribution of light from
  the Sun (Solar constant) - provisional integrations using stellar magnitude data sourced from astronomical catalogs places an an approximation for this constant in the range:  -6.0 <= mAhad <= -7.0
  having a mid-point value of -6.5 magnitudes as a consensus within the astronomical community. This
  is the total integrated brightness of the night sky that we visually experience from Earth.

Other sources for same value:

  -6.5. Brighter than Venus, but dimmer than the brightest Iridium satellite flares.


Do/Can Ringed Stars Exist? The other night, while playing Elite: Dangerous, I came across a rather strange celestial body - one I never imagined possible.

It was a Brown Dwarf star with a very large ring.

Is something like this even remotely possible? Have such objects actually been seen?

It was also quite interesting that this star wasn't even the primary star in the system. It was orbiting a larger star (blue-white, I think), along with several planets.

In a sense, that is what a circumstellar disk is.

Source: European Southern Observatory (ESO)

These are usually most noticeable around young stars as protoplanetary disks, disks that form planets. In the picture above, the gaps in the "rings" likely represent forming planets, similar to how the gaps in Saturn's rings indicate the presence of moons.

Our own star has a few "rings", though likely not as visually impressive: the asteroid belt, the Kupier belt, the scattered disk, and the Hills cloud.

That said, there are differences between these structures and the ring systems we are familiar with from planets. For example, rings are made of solid particles whereas protoplanetary disks contain a lot of gases. Rings may exist around "mature" planets, whereas protoplanetary disks represent an immature phase of planetary systems. Protoplanetary disks are not visible to the naked eye, whereas ring systems may be. In technical terms, a circumstellar disk is not a ring system.


Clarification on calculating IR Dust Extinction $A_K$/$A_V$ So I'm calculating V_Alf Ori's Luminosity (or Betelgeuse) using Davies et al. (2013) SED fitted formula from Near-Infrared/NIR K-Band magnitudes and it requires the subtraction of 'extinction'. However, the conventional prescriptions for extinction in the NIR K-regime is a little way off for a massive star with a very dusty environment (likely thick as well due to stellar mass loss and winds enriching it). So instead of using those, I may need to do a quick approximation of it.
This maybe a rookie stuff but as I scour in, which needs some quick clarification or validation, I kept seeing this "$A_K$/$A_V$" which I presume came from this one?
$\frac{A_λ}{A_V}$ = $\frac{R_λE_{(λ-V)}}{R_VE_{(B-V)}}$
where;

$R_V$ is the canonical Cardelli redenning value of 3.1.
$A_V$ is optical extinction.
E(B-V) is conventional color excess.
While the remaining upper variables are 'almost' the same depending on the bandpass of choice λ.

Edit (1): Is that 'correct'/applicable or there's something else on that like applying power law etc.?
Edit (2): Already got the idea of $\frac{A_λ}{A_V}$. Then how about $A_K$?how can I get that?

The extinction $A$ of a medium (e.g. dust) is defined as the number of magnitudes an object is increased (because higher magnitude means less light), as the light from that object travels through the medium. It is usually (but not always) a function of wavelength $\lambda$, and is thus written $A_\lambda$ (less often $A(\lambda)$).
The wavelength can be a single wavelength such as the Lyman $\alpha$ line at $\lambda=1216\,\mathrm{Å}$, in which case we may write that exact wavelength as a subscript, $A_{1216}$. Alternatively it can be the total, average extinction in a wavelength band such as the $K$ band, i.e. the near-infrared wavelength region roughly between $\lambda = 2\,\mu\mathrm{m}$ and $\lambda = 2.5\,\mu\mathrm{m}$ (depending on the manufacturer of the filter).
In this case we write $A_K$.
I can't seem to find the paper you're referring to, but you ask "How can I get that?", and the answer is that you measure it. To measure an extinction, you must know how much light you'd expect in the case that some object would not have been obscured (it doesn't necessarily have to be at the particular wavelength you're interested in; it could also just be at a set of other wavelengths, and then you fit an extinction law to these points, under certain assumptions of the properties of the medium). Or you could simply not measure anything, and the assume that the extinction is "similar" to some "standard" extinction law, such as the Cardelli (1989) law, or the Calzetti (2000) law.
Relative extinction
Because extinction depends not only on the dust properties, but also one the distance traveled, it makes sense to measured it relative to a fixed wavelength, e.g. the optical band $V$ centered on $\lambda \sim 5500\,\mathrm{Å}$, thus factoring out distance. Depending on available data, sometimes the $B$ band at $\lambda \sim 4400\,\mathrm{Å}$ is used as reference, but since $A_B$ is usually larger than $A_V$, the flux is larger in $V$ and hence more precise. From the definition of the color excess
$$
E(B-V) \equiv A_B - A_V
$$
and the absolute-to-selective extinction
$$
R_V \equiv \frac{A_V}{E(B-V)},
$$
the relation between the two ratios is
$$
\frac{A_\lambda}{A_B} = \left(\!\frac{R_V}{1+R_V}\!\right) \frac{A_\lambda}{A_V}
$$
You could call the function $A_\lambda$ the extinction law, but I think it's more common to use the term for the quantity
$$
k(\lambda) = \frac{A_\lambda}{E(B-V)}.
$$


What causes the bright white area in this photo of the earth? Regarding the photo of the earth described here,

I know that the green tinge on the left hand side is to do with the aurora borealis. What's causing the bright white light on the right hand side?

Taken on March 28 2012 by André Kuipers, the version here is labeled

  Amazing aurora lights and atmospheric glare with the countries Ireland, Netherlands, Scotland, England and Wales in the foreground.

Off to the north and east like that, it's likely refracted early sunrise. That's a lot of streetlights for that time of the day; likely a longish exposure to capture aurora. That'd increase sunrise glare as well.


How big can a rocky planet get? If I get the numbers right, at about 17 Jupiter masses a mainly gaseous planet would ignite into some type of star. What about rocky planets? Is there a theoretical limit to their size, so that they can grow up to size x? (I'm imagining a "liquid" planet would congeal into rock after certain gravity value, and would somehow fit into the "rocky class" as well).

Giant planets have an abundance of H and He. Hydrogen burning occurs at relatively low temperatures. Rocky planets, on the other hand, don't have much hydrogen and even less helium. For them, the first relevant process would be C burning, which occurs at about 8 solar masses.

But this is valid only as a thought experiment. Rocky planets loose or don't accrete H/He mainly because of insufficient mass. If they were massive, they would consist of H/He and therefore "ignite" at 17 Jupiter masses.


Solar Eclipse: How it occurs In relation to the image above, although the topic relates to math, how can I visualise the situation described in the final equation presented, i.e., the declinations, ascensions, solar and lunar radii. A diagram would be of great help.

For a small region not far from the equator,
an equirectangular projection will do.
If you want celestial north to be up,
declination should increase from bottom to top,
and right ascension should increase from right to left.
To correct the aspect ratio, scale the RA axis by cos(dec).
Then the sun and moon are circles of the respective angular radii, without the RA scale factor.

In the equation for $z$, $\Delta$ is the declination of the sun, $\delta$ is the declination of the moon, and $a$ is the difference in their right ascensions.


Can we see evidence of sunspots on other stars Searching for exoplanets by measuring the change of a stars brightness when an exoplanet transits the star's disk also has the potential to detect exo-sunspots. Have they ever been detected?

Can we see evidence of sunspots on other stars?

Strictly speaking, no. Since there is one and only one Sun in the entire universe, there is one and only one star in the universe that has sunspots. That's rather pedantic. Going beyond that, the answer is almost certainly an emphatic yes. The generic term is starspots rather than sunspots, and they have been observed on other stars. For a very nice review, see the open source, peer reviewed article on starspots by K.G. Strassmeier.

The key issue with observing starspots on other stars is that even the largest sunspot observed on our Sun would be far too small / far too dim to resolve with even the largest of telescopes. But some stars apparently do have starspots much larger and much cooler than the largest / coolest sunspot on our Sun. Other than being larger and cooler, these very large starspots on other stars exhibit behaviors analogous to the sunspots observed on our Sun.

Starspots the size of the sunspots observed on our Sun almost certainly exist, but they currently can't be detected given the limitations of current technology.


Is it possible to create an artificial Black Hole? I am quite curious about black holes. Once I just thought whether it is possible to create an artificial Black Hole, i.e. an human made black hole. Also explain there formation if it is possible.

There has been a lot of speculation, whether it's possible with particle accelerators to create micro black holes. This is only feasible, if space is higher dimensional on very short distances. This kind of black holes would be very short-lived due to decay by Hawking radiation. This kind of black holes could be detected by their decay products.

There seems to be no evidence, that black holes have formed in a particle accelerator.

In standard 3d space the smallest black hole needs the energy-equivalent of more than $10^{19}\mbox{GeV}$ of the Planck mass of about $20\mu\mbox{g}$ to form. That's far beyond the energy a particle collider can focus onto a single particle collision.

Even the most energetic cosmic rays are several orders of magnitude below the needed energy.

LHC provided up to 8 TeV (8,000 GeV) for proton-proton collisions, scheduled are 13 TeV for the upgrade. That's 15 orders of magnitude below a smallest 3d black hole, but might allow short-lived black holes for higher-dimensional space. This way LHC can constrain the structure of space on short distances.


Why do the planets in our solar system orbit in the same plane? (Yes I'm excluding Pluto from this the same way it was excluded for not being a planet)

Observing the planets orbit of the Sun they all seem relatively planar and roughly all orbit along the same plane.

Is this due to the way that our Solar system was formed or is this a physical phenomena observed in other systems?

In the protostar stage of the Sun, it was surrounded by a (spinning) gas cloud. This cloud behaved like a fluid (well, a gas is a fluid), so it flattened out into an accretion disk due to conservation of angular momentum. The planets eventually formed from the dust/gas in the disk from compression of the dust in the disk. This process won't end up moving the dust out of the plane (all the vertical force of gravity is toward the disk), so the final planet is in the plane too.

Why does an accretion disk have to be flat? Well, firstly, let's imagine the protostar and gas cloud before the accretion disk formed. Usually such a setup will have particles spinning in mostly one direction. The ones spinning in retrograde orbits will end up reversing themselves due to collisions.

In this gas sphere, there will be an equal number of particles with positive and negative vertical velocities (at a given point in time; due to rotation the velocity signs will flip). From collisions, eventually these will all become zero.

A particle revolving around a planet will always revolve such that the projection on the planet is a great circle. Thus we cannot have a particle with vertical velocity zero but with vertical position nonzero (as that would imply an orbit that isn't a great circle). So as the vertical velocity decreases, the orbit inclination decreases too. Eventually leading to an accretion disk with very little vertical spread.


How is the diameter of a gas giant calculated? As we know atmospheres of celestial bodies don't just stop at a given distance. They gradually become less dense as you move away from the center.

I understand that the diameter of stars is typically given to be that of their photosphere (i.e. the boundary where a laser shined from behind the star toward an observer would just barely be visable).

I assumed that a similar method using optical depth also applied to gas giants. However wikipedia currently has the somewhat confusing explanation:

  As Jupiter has no surface, the base of its atmosphere is usually
  considered to be the point at which atmospheric pressure is equal to 1
  MPa (10 bar), or ten times surface pressure on Earth.

Which isn't quite the same thing as diameter, but it got me thinking: When wikipedia or a textbook or whatever lists a diameter for a gas giant how is that figure arrived at?

I won't argue with the wikipedia definition (although the NASA Jupiter fact sheet lists it as the radius at 1 bar), but just to point out that the scale height of the atmosphere of Jupiter is given by, $h \sim kT/m g$, where $T$ is the temperature, $m$ is the mean molecular mass and $g$ is local gravity. Putting in some numbers: $ g \simeq 24.8 m/s^2$, $m \simeq 2.2 m_u$, $T \simeq 165K$, gives $h \simeq 25\ km$. Thus the pressure changes extremely rapidly compared with the actual radius of a gas giant,and is very small compared to the difference for instance between the polar and equatorial radius for a gas giant (like Jupiter) with significant rotation.

So, whether you give a planet's radius at 1 bar or 10 bar isn't going to make a lot of difference $<100\ km$.

However your comment about whether the definition is analogous to the photosphere of a star does raise an interesting question. The effective opacity radius of a gas giant can vary with the wavelength of light. It is this property that can be used in conjunction with transiting exoplanetary systems to learn something about their atmospheres. For instance, it requires less physical depth of a gas to block wavelengths corresponding to the strong 589nm sodium doublet and thus the exoplanet is opaque at larger radii at this wavelength and thus has a (slightly) deeper transit. This provides the opportunity for transit transmission spectroscopy. e.g. http://adsabs.harvard.edu/abs/2012MNRAS.426.1663S

Another example is that at UV and X-ray wavelengths there have been observations suggesting that hot Jupiters have very extended (and opaque to high energy photons) atmospheres and may be being photoevaporated. e.g. http://adsabs.harvard.edu/abs/2013ApJ...773...62P

EDIT: So this raises an important distinction. A definition based on the radius at a particular pressure is mostly used by theorists and solar system scientists. However, we still don't know what the atmospheric structures of exoplanets are and can't measure pressures. The radii quoted for exolanets are more akin to how the radius of a star is measured - based on its opacity at the wavelengths being observed.


How long do the phases of sunset take? How long does it take from the moment the Sun star 'touches' the horizon, to the time it totally disappear? (I'm not talking about the radiation (rays), but just about the appearance of the star by itself. I don't care of the rays). If I were near a beach or plane I'd check it, but I'm in crowded urban region so it's difficult to check it I believe).

@Calc-You-Later's answer is correct in places where the path of the Sun is perpendicular to the horizon (at the equator during the equinoxes, somewhere between the tropics the rest of the time). There, the sunset lasts 2 minutes. However, at other latitudes, the sunset may last a bit longer.

Let $\alpha$ be the angle the path of the Sun makes with the horizon. This angle depends on your latitude and the date.

The speed of the Sun along its path is $\vec{v_s} = $ 0.25°/min. This speed can be projected along the vertical and horizontal axes. The vertical speed is now $\vec{v_s} . \sin (\alpha)$. The sunset length is still the apparent angle of the Sun divided by the vertical speed of the Sun.

So at higher latitudes, if the Sun sets, for instance, at a 45° angle, then the sunset will last 2 min / sin(45°) = 2.85 minutes = 2 minutes 51 seconds. If the Sun sets with a 60° angle, the sunset will last 4 minutes.

To get the value of $\alpha$, you can add your latitude and the solar declination:

Determine how many days have passed since December 21st (i.e since the winter solstice). Let's call this number $x$
Multiply this number by the amount of turning the Earth does in one day
$ y = x * 360/365.25$
Get the cosine of this number and multiply it by the tilt of the Earth $z = -23.5 * cos(y)$. This is the solar declination
Add the latitude $\theta$ of your location $\alpha = z + \theta $

This does not account for atmospheric effects. The atmosphere will bend the rays of light, and can make it possible to see the Sun even though it is under the horizon.

The atmospheric effects, however, depend on the atmospheric conditions (temperature gradient,...), and are more difficult to quantify.


The relation between the energy of star and heating effect Because a star's temperature is an indication of the energy passing through each unit of its surface, it follows that for cool red and hot blue stars of identical luminosities, the cooler red star must be considerably smaller so that the energy escapes through a reduced surface area and has a far greater heating effect. Hence, knowledge of a star's colour and luminosity can reveal its size.

In particular, I can't understand the relation between energy and the heating effect. Of course, I searched for the word "the heating effect" in my own language. Also, I Googled in English.

In this context, what is the heating effect? Could you explain to me? If you give me any help, it will be very useful to me.

I'm not sure what you mean by "heating effect". The amount another body is heated by a star with luminosity $L$ is quantified by its effective temperature:
$$T_{eff}=\left(\frac{L(1-a)}{16\pi\sigma D^2}\right)^{1/4}$$
where $a$ is the albedo, $D$ is the distance to the other body, and $\sigma$ is the Stefan-Boltzmann constant. If the luminosity is given, then there is no direct dependence on the size of the star.

The only thing that matters is the luminosity. You stated yourself that the luminosities are the same, which means that the red star in question is likely a red supergiant - a member of a group containing many of the largest stars in the universe.

Why, mathematically, is this the case? Well, stellar models can tell us that this is the case, but we can also figure this out via the Stefan-Boltzmann law by assuming that the stars are black bodies. For an object of radius $R$ and temperature $T$, the luminosity is approximately
$$L=4\pi\sigma R^2T^4$$
If star 1 is a blue supergiant and star 2 is a red supergiant, then, setting the luminosities equal, we have
$$L_1=L_2\to R_1^2T_1^4=R_2^2T_2^4$$
We know that $T_2$ is much less than $T_1$ - possibly by an order of magnitude - so $R_2$ must be much larger for the luminosities to be the same. This matches what we observe and what models predict.


Milky way: How do we know its appearance? How can we know how does the Milky Way look like if we are in it?
Sorry if the answer is evident, I am not an expert.

There are lots of reasons. First of all because we can see that a typical galaxy is a spiral galaxy and we seem to live somewhere pretty typical.

Secondly because we can see the arms of the spiral through close examination of the Milky Way (ie the band in the sky) - through stellar distances and measurements of hydrogen cloud red shifts and so on.

But you your question is a perfectly good one because, ultimately, we are guessing a bit as we cannot see the galaxy from anywhere other than one vantage point.


Using emission lines to determine redshift of a quasar I am attempting this past paper question and am not sure how to tackle this, this is not homework! :)

  The spectrum of a distant quasar shows two broad emission lines with observed
  wavelengths of 317.7nm and 404.7nm. The strongest lines are most likely to be either Lyman-α with a rest wavelength 121.6nm, CIV with a rest wavelength of 154.9nm or MgII with a rest wavelength of 280.0nm.

  (a) What is the redshift of the quasar? Please show your working.

  (b) What two lines have been detected?

  (c) The two lines have widths of 10nm and 20nm respectively. If I have measured
  the mass of the black hole in the quasar to be 10^9Msun, how far from the centre of the quasar are the regions of ionised gas responsible for the emission lines?
  Please specify any assumptions that you make to obtain your result.

I am assuming I must find z to be the ratio of the change in wavelength to the rest wavelength. However I'm unsure of how to identify these in the question.

So the two observed lines must be two of the three suggested lines, red-shifted by the same amount. That means the ratio of their wavelengths will be unchanged, so we need two of the suggested lines whose wavelength is close to the ratio $404.7/317.6$.  It's easy to check that only the first two are close to that ratio, so they must be those two. So the redshift should be equal to $317.7/121.6 = 2.61$  and to $404.7/154.9$ which is also $2.61$ (such close agreement in real data would be a bit unlikely, and lead one to check the authenticity of the data.


Why do J2000 coordinates change with time in Stellarium? I'm new to the field so I'm sure this is a basic misunderstanding.
As I currently understand it, J2000 equatorial coordinates describe the position of objects by referencing a specific point in time (12:00 Jan 1st 2000) to account for the effect of the precessional cycle in the reference point for RA. Presumably this means that the same object measured on different dates should still have the same J2000 coordinates?

In Stellarium, if I look at the RA/Dec (J2000) of a star on one date, then compare to the previous year the J2000 coordinates don't match:

I also notice that if I set the current date/time to 12:00 1/1/2000 the RA/Dec (J2000) and RA/Dec (on date) values are different when I would have expected them to be the same:

Where am I going wrong here?

Apparently, Stellarium's "J2000.0" reports the coordinates of stars in J2000 frame, but at the epoch of the date you specify, instead of reporting the coordinates of the star at epoch 2000.0, which can be misleading to say the least. So the difference in J2000 coordinates that you noticed corresponds to the proper motion of the star. And the star you picked, 61 Cygni, actually has one of the largest proper motion in all of the catalogued stars, with about 4.1"/year in RA and 3.2"/year in declination. Looking at the differences that Stellarium reports in your example, we can notice that it matches perfectly with this proper motion : 31.3" - 28.2" = 3.1" ± 0.1" in declination, for RA: (55.27s - 54.92s) x 15 x cos(declination) = 4.1" ± 0.1".

If you pick a star with a smaller proper motion in Stellarium, like Deneb (only 2 mas/year for each axis), the J2000 coordinates change much more slowly.

As for the difference between J2000 RA/Dec and apparent (on date) RA/Dec, this is caused by the fact that the J2000 frame's equator/equinox was fixed at the mean position of the equator/equinox in 2000 (mean position means that the small periodic effects of nutation are averaged out). So the apparent coordinates on Jan 1 2000 are slightly different because apparent coordinates take into account nutation N of the rotational axis R around the mean P (precession):


Do all stars have an Oort cloud or is it a rare occurence? Do all stars have an Oort cloud like ours that will be filled with comets and other objects? If not, why are they not around every star?

Awesome question, especially since we know so little of the answer.

Nobody knows for sure how the Oort Cloud formed - I'll put that out there right now - but the current hypothesis is that it was originally part of the Sun's protoplanetary disk. All of the ice and rock coalesced into small bodies - proto-comets, if you will. While these bodies were much closer in to the Sun than they are today, they were tossed far out by gravitational interactions with the gas giants. Other interstellar comets could also have been captured by the Sun, adding to the population.

So why is the Oort Cloud spherical? After all, the protoplanetary disk was just a flat disk. Why were the orbits of the objects perturbed? Well, the Oort Cloud objects are only loosely bound to the Sun - relatively, that is. They can be influenced by passing stars or other objects. It appears that galactic-scale tidal forces, combined with the influence of passing stars, molded the Cloud into its current spherical shape.

So what does this all tell us? Well, we know other stars have protoplanetary disks, right? Some also have exoplanets - gas giants like Jupiter. They are also subject to tidal forces and the passing of nearby stars. So, theoretically, there's no reason why other stars shouldn't have Oort Clouds.

So can we find them? The answer is, most likely, no. Here's why. According to Wikipedia,

  The outer Oort cloud may have trillions of objects larger than 1 km (0.62 mi), and billions with solar system absolute magnitudes brighter than 11

An absolute magnitude of a solar system object of 11 is very dim. Now, the object's apparent magnitude is how it would look from a given distance; the absolute magnitude is how it looks from a distance of 1 AU (in the case of Solar System objects, this quantity is denoted $H$). Oort-cloud objects are 2,000 - 50,000 (or more) AU away-- so these objects, to us in the same solar system, have an apparent magnitude much fainter than 11.

The point of that poorly-explained interlude is that these objects are faint. Very faint. And objects in Oort Clouds around other stars would appear even fainter. Using the distance modulus, we can calculate the apparent magnitude of an object if the distance to that object and its absolute magnitude are known:

$$m-M=5(\log_{10}d-1)$$

(from here)

where $m$ is apparent magnitude, $M$ is a scaling of $H$ normally used for stars, and $d$ is the distance in AU.

Given an Oort Cloud object $x$ light-years away, you can figure out how bright (or dim) it would appear, given that 1 light-year is 63241 AU. Try this with the distances of nearby stars, and you'll realize how dim objects in these stars' Oort Clouds would be.

As a final note: We don't know for sure if other Oort Clouds exist. From what I've been able to find, we don't have sufficiently powerful telescopes to observe these hypothetical Clouds, and so we don't (and may never) know if they exist.

I hope this helps.

This paper was instrumental in this answer. Start at page 38 for the relevant information. This page, too, has some good information.

As I found from a link from an answer to this question on Physics, we've found Kuiper-Belt-like disks around other stars. This means it is certainly plausible for these stars to have Oort Clouds, too. And exocomets have been detected, which is another good sign.


What is gravity really? What is gravity?  I want to know more than it being simply the "mysterious force" that attracts things to earth.  Is it a particle, a wave, or something else entirely?

I can attempt to address the second part of your initial question (*"Is it a particle, a wave,...?") Einstein's theory of general relativity states that mass and energy bend space-time. Space-time, in turn, tells matter how to move (John Wheeler put this more elegantly).

This concept is completely different from the theories of the other three fundamental forces (electromagnetism, and the strong and weak nuclear forces). In these quantum theories, forces are mediated by particles called gauge bosons. Electromagnetism is carried by photons, the strong nuclear force by gluons, and the weak nuclear force by W+, W-, and Z bosons. There have been many attempts to find a quantum theory of gravity - that is, to use quantum principles to construct a field theory of gravity. In these theories, gravity would indeed be mediated by a particle, dubbed the graviton. String theory is one example of these theories; the physics community is divided about it.

You probably have heard other terms tossed around that embody interesting concepts. A gravitational wave is essentially a ripple in space-time emitted by an object or system of objects. There are strict restrictions over what sort of objects can emit these waves; binary neutron stars are one consistently cited example. These waves should not be confused with the aforesaid hypothesized gravitons; while gravitational waves carry energy, they do not "mediate" gravity. They should also not be confused with the unrelated gravity waves.

So, basically, the predominant view in the physics community is that general relativity is the best description of gravity; at the moment, gravity is considered to be the bending of space-time, so it is indeed "something else entirely." Many theories of quantum gravity, including string theory, however, attempt to create particles called gravitons as force-carrying bosons. If evidence is found relating to these theories, then we may well learn whether or not gravitons do exist. One more thing about waves: Because of the quantum concept of wave-particle duality, any particle can be described as a wave, which has a wave function. So if gravity is carried by a particle, then it is also due carried by a wave!


What happens to galaxies when they die? Stars explode when they die and blast heavy elements into space.  Do galaxies do the same thing?

Well, it would be useful to define what a 'dead' galaxy is.  Probably the most simple method would be a galaxy that is no longer producing new stars.  We might also consider a galaxy that no longer produces significant light in the visual spectrum, or perhaps EMR across the entire spectrum.

Generally, there's unlikely to be a firm line between living and dead, and not nearly as dramatic as larger stars. More akin to watching a camp fire burn itself out.  Star formation is largely dependent available gases, but as more and more stars fuse those gases into heavier elements, there is less gas available for star formation. For your average sized galaxy, this will eventually result in running out of gas. Eventually the galaxy will dim and go dark, a process purported to begin at the center of the galaxy, where star formation is heaviest according to research based on Hubble images of giant galaxies. (Tacchella, et al.) The matter ought to (mostly) all still be there and still orbiting the (presumed) SMBH, but with no energy coming from fusion, it's going to be a dark, cold, and barren place. Sounds dead to me.

There are some complicating factors.  It's believed that encounters with nearby galaxies can affect available gases. The gravity from a larger galaxy could potentially strip the gases from a smaller one, a fatal blow for the smaller galaxy. Fortunately, it won't suffer much as the death will come (relatively) quickly. This process has been deemed 'strangulation' by a study published in Nature several years months days ago. (Ping, et al.) Note that as the study indicates, the methods of death are proposed solutions - not conclusive understanding of the exact processes that result in a galaxy's death.

S. Tacchella, C. M. Carollo, A. Renzini, N. M. Förster Schreiber, P. Lang, S. Wuyts, G. Cresci, A. Dekel, R. Genzel, S. J. Lilly, C. Mancini, S. Newman, M. Onodera, A. Shapley, L. Tacconi, J. Woo, and G. Zamorani. Evidence for Mature Bulges and an Inside-out Quenching Phase 3 Billion Years After the Big Bang
Science 17 April 2015: 348 (6232), 314-317. [DOI:10.1126/science.1261094]

Y. Peng, R. Maiolino & R. Cochrane. Strangulation as the primary mechanism for shutting down star formation in galaxies Nature 521, 192–195 14 May 2015 [DOI:10.1038/nature14439]

Andrea Cattaneo. Astrophysics: The slow death of red galaxies Nature 521, 164–165 14 May 2015 [DOI:10.1038/521164a]


How Does Earth Move In Relation To The Stars In my simulation, I have stars scattered around Earth, all stars are the same distance from the center of the Earth and positioned based on their right ascension and declination like so:

I am wondering how the movement of Earth can be simulated based on the current time when running the simulation.

I am using the hyg v3 database which can be found here. I think that this dataset has the data necessary to find out the motion of stars, but I may be wrong.

Also, due to the structure of this simulation, with almost 120,000 stars, it is much more efficient to rotate/move the Earth rather than move all of the stars.

How can I show the stars visible over a location at a certain time by simulating the motion of the stars by rotating Earth?

The orientation of Earth wrt the stars repeats each sidereal day (thus ~23:56h). The visible night sky only changes each (synodic 24h) day due to the different direction the sun is at when considering the star's rest frame.

If you have the star's position from like now, you simply rotate Earth for each our by 360/(23.933)° = 15.04°.
If you want to do that for an arbitrary date, take the stars' position: they come with a datum like J2000 which means their position on January 1st 2000 UTC. C calculate the sidereal days passed since then, accounting for the incomplete rotation of the current day which gives the rotation around the Earth's axis as above.

If you want to go into more detail and longer time, and account for centuries and longer, you will need to take into account the nutation and presession of the Earth axis.


Do astronomers, in principle, have more information about older phenomena than younger? The further away we see, the older states of the world we observe. And the volume of the part of space which is observable, increases by the square of the distance to it. So there should basically be a million times more phenomena (data) observable about what happened a billion years ago, than about what happened a million years ago. Given good enough observatories.

This is obviously counter intuitive, and sorting it out for so called redshift effects doesn't help in that respect. Do/can we really (potentially) know more about things/events the older they are? Is there some epistemological problem involved in that?

There are more stars and galaxies but not necessarily more information.

You might choose to measure the information content in terms of photons received, in which case, for a given star/galaxy, this decreases as the inverse square of the distance, so the two effects cancel out.

Astronomy is a constant struggle with the tension between these two competing dependencies. As we look further, not only do we see more of particular types of object, but we also get to see examples of rare objects - i.e. the tails of the distributions. On the other hand as we expand our horizons, so our grasp of what is going on becomes increasingly blurred in terms of spatial, spectral and temporal resolution. It is for that reason that bigger and bigger telescopes are built!

The issue you raise concerning redshift is interesting. The co-moving volume versus look back time in any direction will not be a simple function (i.e. doesn't just go as the square of the lookback time) and depends critically on the adopted cosmological parameters.

There is of course a limit in any case. Once we get back to the microwave background at $z=1100$, then further probing backwards with electromagnetic radiation is stymied by the optically thick nature of the universe at early times.


Why do people choose 0.2 as the value of linking length in the friends-of-friends algorithm? The friends-of-friends algorithm (hereafter FOF) is commonly used to find halos in cosmological simulations. And FOF depends on only one parameter, linking length, $l_\mathrm{link}=b \left(\frac{V_u}{N}\right)^{1/3}$, where $V_u$ is the volume of the simulation box, and $N$ is the number of dark matter particles.
In most literature, the value of $b$ is set at 0.2. What does $b\approx 0.2$ mean physically?

The friends-of-friends (FOF) algorithm (Huchra & Geller 1982; Press & Davis 1982, Davis et al. 1985) for finding groups of particles can be used for various numerical problems, not just cosmology, and linking lengths may vary between these problems. The larger the values you use for $b$ (the linking length in terms of the mean interparticle distance), the more particles you will include in your group of particles. Since the density falls off with distance from the center of the group, this means that the average density will be smaller for larger values of $b$.
In cosmological applications of the FOF algorithm, particle groups are called halos, and experience shows that using $b = 0.2$ result in an average enclosed overdensity of close to $\Delta = 180$ times the average "global" density. This value of 180 equals the virial overdensity predicted by the spherical collapse model in the Einstein-De Sitter cosmology (see e.g. Mo et al. 2010). Also, the value of 180, or perhaps 200, is also typically the overdensity used for another popular halo finder, the so-called spherical overdensity algorithm.
That $b=0.2$ results in $\Delta \simeq 180$ is actually only true for an isothermal density profile; the exact value will depend on the actual profile and halo concentration $c$. Surhud et al. (2011) offer an analytical solution to the overdensity as a function of $b$ and $c$.


Would a Phobos full impact make Mars temporary habitable? Due to tidal deceleration, Phobos is predicted to be destroyed by Mars in the next 30-50 million years either by collision or by breaking up in a planetary ring. If Phobos falls on Mars in one piece could the impact could be comparable with Chicxulub meteorite? Since Mars is smaller that Earth, would the effects of the impact cause a global restart of volcanic activity and temporary warm the planet? Would this be sufficient to make Mars more friendly to life for a few thousand years?

The impact of Phobos, even in one piece (less likely), would be different from Chicxulub. Phobos is probably a little larger than the Chicxulub impactor, but much slower, and comes in almost perfectly tangential. The impact energy would be less than a 10th of the Chicxulub impact, and the energy would be distributed over a large region around the Martian equator. It wouldn't start volcanism, and it wouldn't warm Mars in a relevant way. It wouldn't make Mars more friendly to life by warming. The only noteworthy contribution to habitability would be additional organic matter which Phobos is presumed to contain.

Phobos will (probably) desintegrate due to tidal forces, before its fragments will collide with Mars. Any loose rocks, dust, regolith on the surface will separate from Phobos below the Roche limit of
$$d=1.26~R_M\cdot\left(\frac{\rho_M}{\rho_m}\right)^\frac{1}{3},$$
with $R_M=3389.5\mbox{ km}$ the radius of Mars, $\rho_M=3.9335 \mbox{ g/cm}^3$ the mean density of Mars, and $\rho_m=1.887\mbox{ g/cm}^3$ the mean density of Phobos. This results in a Roche limit of
$$d=1.26\cdot 3389.5\mbox{ km}\cdot\left(\frac{3.9335 \mbox{ g/cm}^3}{1.887\mbox{ g/cm}^3}\right)^\frac{1}{3}=5455.6\mbox{ km}.$$
That's for Phobos' center about 2059 km above the equatorial surface of Mars (taking 3396.2 km as equatorial radius.)
Therefore desintegration will start long before the impact. Loose rocks will collide with Phobos and unlock more material. Whether there will be a considerable solid remnant to finally cause a big (tangential) impact is difficult to predict.

The orbital speed of the fragments at the time of collision will be about
$$v_0=\frac{v_e}{\sqrt 2},$$
with $v_e=5.03\mbox{ km/s}$ the escape velocity of Mars.
Phobos' inclination of 1.093° (relative to Mars's equator) is almost neglectable for the calculation of the impact velocity.
Mars' equatorial rotation velocity is about 241.17 m/s.

The impact velocity will hence be about
$$\frac{5030\mbox{ m/s}}{\sqrt 2}-241.17 \mbox{ m/s}=3316\mbox{ m/s}.$$
Phobos' mass is about $1.0659\cdot 10^{16}\mbox{ kg}.$
This results in a kinetic impact energy of
$$\frac{1}{2}mv^2=\frac{1}{2}\cdot 1.0659\cdot 10^{16}\mbox{ kg}\cdot (3316\mbox{ m/s})^2=5.86\cdot 10^{22}\mbox{ J}.$$

The heat of fusion of water is about 334 kJ/kg, the heat of vaporization of water at 0°C is about 2504 kJ/kg, together 2838 kJ/kg are needed to evaporate water ice of 0°C.
Hence the impact energy of Phobos would be sufficient to evaporate
$$\frac{5.86\cdot 10^{22}\mbox{ J}}{2.838 \cdot 10^{6}\mbox{ J/kg}}=2.06\cdot 10^{16}\mbox{ kg}$$ of water ice of 0°C, corresponding to $2.06\cdot 10^{13}\mbox{ m}^3$ of liquid water.

Mars' surface area is about 144,798,500 km², or $1.447985\cdot 10^{14}\mbox{ m}^2$.
Equally distributed, we could get a layer of water ice corresponding to
$$\frac{2.06\cdot 10^{13}\mbox{ m}^3}{1.447985\cdot 10^{14}\mbox{ m}^2}=0.142\mbox{ m},$$
(correspoding to 14.2 kg/m² of liquid water) evaporated by the impact energy.

With a surface gravity of 3.711 m/s², this would correspond to
a pressure of $14.2 \mbox{ kg/m}^2\cdot 3.711 \mbox{ m/s}^2= 52.8 \mbox{ N/m}^2=0.528\mbox{ hPa}$.

The current atmospheric pressure on Mars is about 6.3 hPa, with diurnal, seasonal and regional changes.
Thus we could get a maximum of 10% short-term pressure increase in the Martian atmosphere by water vapor, in an ideal case.

At the end, an impact of Phobos, or its fragments, would result in a regional reworking of the Martian surface in the equatorial region, but not in a long-term global change.

Just melting water ice would lead to a global ocean with a depth of about 1.2 m. But this ocean would boil due to the low atmospheric pressure, and freeze out again. Same with local lakes due to the uneven topography.

Model calculatios with a release of carbon dioxide would cause some temporary greenhouse effect, but in the equatorial region a release of carbon dioxide is even more unlikely than a release of water vapor.


Does the Sun impose back it's tidal forces onto the Earth (such as the Earth's to the Moon)? It is known that each year the Moon is moving away from the Earth due to tidal forces.

What about the Earth relative to the Sun?

Yes - the earth and sun do have tidal forces like the moon and earth. There are two main reasons this is happening.

The sun is always losing mass due to nuclear reactions, the sun is always converting a tiny amount of its mass into energy. This means the pull on the earth is weakened.
The tidal forces that happen between the earth and moon also happen between the sun and earth, except on a much smaller scale.

The sun is also losing its angular momentum at a rate of 0.00003 seconds per year


What did I see streaking across the sky last night? I was enjoying a casual naked eye view of the summer sky last night around midnight in a new jersey beach town. The sky was pretty clear for around here.The Milkyway was just visible and there was no could cover or buildings higher than me so I saw from Cassiopeia to Sagittarius.

I was hoping for a meteor and I got one I have never seen before. I will try to describe it.

Appearance: .It appeared out of no where like any other meteor but it had no streak AT ALL. As it traveled it shimmered. It was blue colored and not particularly bright perhaps fuzzy. The shimmering was the best way I can describe it.

Trajectory: It began high in the sky slightly to the east around the "southern" part of the constellation Aquila. It traveled west. It was as fast as any other meteor but as I said had no tail. It lasted very long... say 4 or 5 seconds. It didn't have an arching trajectory like something lower. It traveled straight and as it disappeared, it seemed to travel off into vs down to the horizon.

I was so astonished that I practiced the speed it went with my finger after it finished. It moved from say Vega to Altair in less around a second.

It couldn't have been any kind of craft since it appeared and disappeared and traveled so quickly. I have seen 100's of meteors of many kinds but not any like this. What did i see?

I tried to be as detailed as possible. If I missed anything, I'll add it in the comments.

For the record, I am not a believer in visiting aliens or supernatural events. BUT you will laugh at this. Years ago,I had read about an Ozark superstition that if you say "Money" three times when a meteor appears, you will get rich. I liked that one and since I stargaze a lot, I figured I'd get the opportunity. Well I had my chance and then some but I was so astonished, I forgot to say it. Looks like I  will have to keep working for a living :)

It could have been an Iridium Satellite.  Their solar panels are at an angle such that the sun reflects off them causing them to look like a bright flash that crosses the sky for a few seconds.  I use a site called Heavens Above that allows you to enter your exact location so you can see what objects appear above, at what time and in what direction.  Check that site out and see if any Iridium pass over there.


Are the distances between two celestial bodies the distance of their centers? For example, the average Sun-Earth distance is 1 AU. Is this

the distance between the center of the Sun and the center of the Earth
the distance bewteen the surface of the Sun and the surface of the Earth,
or something else?

Distance is an ambiguous term that can mean both centre-centre and surface-surface depending on the context it is used.

In cases where the size of the objects is trivial compared to the distance, 'distance' is almost exclusively used for the "distance between centres" of the two objects. This is the case for Sun-Earth, Earth-Moon and almost any other pair of significant solar system object. The alternative meaning must be specified if intended, like: "the distance from the surface of the Sun to Earth"
It may however also mean the "distance between surfaces", especially when the two objects are very close to each other. This is often the case with binary asteroids.

For  an orbital spacecraft, the word 'distance' is mostly not used when referring to the central object. The terms 'altitude' or 'orbital radius' are preferred. Altitude is the distance to the surface (or rather, to a theoretical sealevel. It may also refer to the actual surface though, so be careful). Orbital radius is the distance to the centre of the Earth.

As a side note "average distance" in an orbit is referring to the semi-major axis as that is the number used for calculating the orbital period and different velocities.


What can I see using a celestron nexstar 127SLT? I am planning to buy celestron nexstar 127SLT and I want to know what can I see using my telescope.

First of all, the high end of magnification is generally not as important as several other factors. However, considering that no additional information was given in the question beyond "reflecting", I will make some assumptions and only give you a very general estimate.

The book "Fundamental Astronomy: H. Karttunen et al.", page 51, formula (3.5) gives a relationship between the diameter of a telescope's aperture and it's useful magnification.This relation is as follows ($e$ is the resolving capacity of the human eye, $D$ is the aperture diameter, $ω$ is the maximum magnification and $λ$ is the wavelength of visible light, used in the formula for the telescope's resolving capacity:

$$θ ≈ 1.22×λ/D ≈ λ/D,$$

$$ω_{max} = \frac{e}{θ} ≈ \frac{eD}{λ} = \frac{(5.8 × 10^{-4} × D)}{(5.5 × 10^{-7} m)} ≈ \frac{D}{1 mm}.$$

That's a key concept for you, because any additional magnification above that can be thought of as "empty magnification". It exists mathematically, but the image is so blurry that you will not see any additional detail, and probably be quite disappointed in your purchase if you paid a lot of money for that magnification!

Let's try backtracking from 500x.

If this is the true maximum useful magnification this telescope can achieve, then its aperture diameter would be about 50cm. This aperture allows for $D^2/d^2$ brighter stars to be observed ($d$ is the diameter of the human pupil). When this is calculated and converted to magnitudes, it turns out you should be able to see stars up to magnitude $9+6=15$.

However this is a very rough estimation and you certainly will not be able to see magnitude 15 stars due to atmospheric extinction, seeing and other factors ... my estimate is you'll see stars about the 12th magnitude on a good night.

If the 500x telescope you are looking at has a much smaller aperture than 50 cm (and probably that's true) then that 500x is not going to be so useful in reality.

Take some time and read further about telescopes. This site has many existing questions and answers on the subject.


Is Earth unique in its fairly clear atmosphere? So, we have surface pictures from two alien planets, Venus (captured by the Venera 13), and Mars (captured by the rover).

Both of these pictures appear to be very dusty. For Venus we see strong storms; that makes sense. However, the Martian air also appears very dusty. Is Earth relatively unique in its clear atmosphere?

Mars

Venus

Titan (Thanks LocalFluff!)

Our atmosphere is only transparent to visible light, In most other wavelengths, some or all of the light is absorbed

Image from Wikipedia, adapted from image by NASA

Our eyes have evolved to take advantage of the transparency at these wavelengths. If we had evolved in an atmosphere with a very different mix of gases. One in which visible light was absorbed, we would have evolved eyes that see different wavelengths.

There is a notable "window" at about $10\mu m$ in the diagram above. And you might wonder if any animal has evolved to see in this window. However, our own bodies emit thermal radiation at about $10\mu m$, eyes wouldn't work as they would be swamped by their own glow. However this window is used by thermal imaging devices.


Troop of moving stars through a straight line Last night,I and my brother suddenly saw that some(50-60) dots of light( exactly like stars)  were moving through a straight line(from NW to SE) above us  with same velocity and towards exactly same directions.These moving stars were seen for just a few seconds.Then they didn't disappear, they just crossed us. We became confused. My brother is highly interested in astronomy. But he was unable to detect the phenomenon. I think this case is rare.  What may they be?

That was likely the latest StarLink satellite constellation.

Note that usually it is important to give an exact time and place to answer questions like this, but in this case the answer looks fairly obvious. It really does look odd to see the constellation fly overhead (later it will disperse into separate orbits and not be very conspicuous).


A plot of the Moon-rise/sets on the Horizon Location We have been trying to see where the moon rises/sets on the horizon.   Unfortunately, the moon is not cooperating.   Even when we know the moon rise/set times (timeanddate.com), we often cannot find the moon on the horizon.
We are wondering if the moon wobbles back and forth along the horizon the same way the sun wobbles.  We cannot find any website that talks about where the moon is on the horizon.

We are wondering if the moon wobbles back and forth along the horizon the same way the sun wobbles.

Yes, but over a period of 27 days compared to a full year for the Sun.
If you visualize a compass, the angle that a star rises and sets is the same angle measured from due south (for those in the northern hemisphere). This figure shows a star rising 60 degrees from due south (at an azimuth of 120 degrees). The same star sets 60 degrees from due south (at an azimuth of 240 degrees).

The rise angle and set angle can be calculated from
$$cos(angle) = -\frac{sin(dec)}{cos(lat)}$$
where $dec$ is the declination of the star and $lat$ is the observer's latitude.
The same calculation applies reasonably well to the Sun and Moon. (The apparent diameters makes a small difference, as does atmospheric refraction. Let's ignore those effects for now.) Due to the motion of the Moon during the day (and to a smaller extent the Sun), the change in declination changes the rise and set angles by a small amount - a few degrees at most. For the following graphs, I have ignored this motion during the day.
For the year 2021, you can see how the rise/set angle and azimuth of the Moon changes from "north of east/west" to "south of east/west" during one month (approximately). Of course, the Sun requires an entire year to go through the same cycle.

The azimuth changes because the declination changes, and the declination changes because the Earth's axis is tilted 23.5 degrees from perpendicular to the orbital plane. The Moon's orbit is inclined another 5 to 6 degrees from that. As Uhoh has mentioned, the inclination of the Moon's orbit sometimes adds to the declination and sometimes subtracts from it. This longer cycle takes 18.6 years to complete. The rise/set angle for a full cycle looks as follows:

Here is another factor that may be affecting your observations. Between New Moon and Full, the Moon rises during the day. It may be difficult to see the Moonrise until it gets close to full. During this time span, the Moonset occurs at night. (So "measure" the Moonset and calculate the set angle from due south.) The opposite occurs from Full Moon to New: it rises during the night and sets during the day.


Protocol for establishing longitudinal meridians on other heavenly bodies When we first observe a new heavenly body (it could be a new star, asteroid, etc.,or a minor planet in our own solar system), are there any procedures set in place for establishing a system of longitudinal meridians?

Being that a prime meridian is an arbitrary concept that you can pick and establish anywhere, is a location decided based on a physical feature, or perhaps from the first points of data we gathered when making detailed observations of the heavenly bodies for the first time?

What about heavenly bodies with no easily discernible or non-stationary features (Gas Giants)? Also does the IAU regulate this process?

I have always wondered if the USSR and the US shared common reference points for locations such as on the moon.

As an example consider the prime meridian for Vesta.

Based on Hubble images, the IAU had established a prime meridian based on the observation of the Obler's regio (a dark region). The prime meridian was defined as passes through the middle of this region.

When the Dawn mission arrived at Vesta, it was discovered that the asteroid's pole was not in the same place that it had been assumed to be (based on Hubble data) and that Obler's regio was not a convenient basis for a prime meridian, as it was not even discernable from up close. Instead they chose a small crater, Claudia, that was close to the equator, and defined the prime meridian based on that (the prime meridian is defined as a passing 4 degrees from the centre of Claudia) This is now the generally accepted coordinate system, although it has not been "officially" adopted by the IAU.

In general, if you are the first to map a new world, you get to choose where do draw your meridian.


How did the authors determine both the spatial size of gas cloud HCN-0.009-0.044 and its central mass at the same time? Phys.org's Hiding black hole found says:

A research team led by Shunya Takekawa at the National Astronomical Observatory of Japan noticed HCN-0.009-0.044, a gas cloud moving strangely near the center of the galaxy 25,000 light-years away from Earth in the constellation Sagittarius. They used ALMA (Atacama Large Millimeter/submillimeter Array) to perform high-resolution observations of the cloud and found that it is swirling around a massive invisible object.
Takekawa says, "Detailed kinematic analyses revealed that an enormous mass, 30,000 times that of the sun, was concentrated in a region much smaller than our solar system. This and the lack of any observed object at that location strongly suggests an intermediate-mass black hole. By analyzing other anomalous clouds, we hope to expose other quiet black holes."
Tomoharu Oka, a professor at Keio University and coleader of the team, adds, "It is significant that this intermediate mass black hole was found only 20 light-years from the supermassive black hole at the galactic center. In the future, it will fall into the supermassive black hole, much like gas is currently falling into it. This supports the merger model of black hole growth."
These results were published as Takekawa et al. "Indication of Another Intermediate-mass Black Hole in the Galactic Center" in The Astrophysical Journal Letters on January 20, 2019.

The paper Indication of Another Intermediate-mass Black Hole in the Galactic Center (open access) is pretty hard to read as it details the careful analysis of ALMA data reduction and analysis.
Question: How did the authors determine both the spatial size of gas cloud HCN-0.009-0.044 and the mass of the central object at the same time?

The gas clouds are much larger than you think: From Fig. 1 of the paper (Takakawa 2019), the largest cloud (which they call the "Balloon") is roughly 10 arcsec across. Now they can't actually measure its exact distance, but its projected distance from the Galactic center is only 7 pc, and the authors argue that it's at least 5 kpc away due to absorption from molecular gas which is known at this distance (Sofue 2006), and since such hot and high density ($\sim 10^7\,\mathrm{cm}^{-3}$) molecular gas is abundant in the Galactic center, it is probably located there, i.e. at a distance of $R=8\,\mathrm{kpc}$.

The cloud's absolute size is thus
$$
D = 10'' \times 8\,\mathrm{kpc} = 80\,000\,\mathrm{AU} \simeq 0.4\,\mathrm{pc}.
$$

The mass is measured from dynamics, i.e. they measure the line-of-sight velocities of several clouds. These velocities depend both on the mass of the dynamical center (i.e. whatever object the clouds orbit), and on the total velocity of this object wrt. us. Since they only know the line-of-sight velocities and not the full, 3D motions, they cannot know with certainty the mass, but performing a chi-square fit of 3D models that fit the 2D observations, they determine the best-fit value of the mass to be $(3.2\pm0.6)\times10^4 M_\odot$.

Their Fig. 4 shows the model:

3D model of the motion of the clouds. The red cloud is the Balloon. Other observed clouds are shown in green, blue, and orange colors. The black dot indicates the position of the black hole. The actual observation is the 2D projection, seen in the bottom plane.

The model of the velocities also gives the actual trajectories of the clouds around the central mass. Because the radius of the Balloon's orbit is $\sim0.07\,\mathrm{pc}$, it means that this much mass must be concentrated within a region of maximum that size, and probably significantly smaller. This implies a density of $>2\times10^8 M_\odot\,\mathrm{pc}^{-3}$, at least an order of magnitude larger than even the most densely packed globular clusters, and since no luminous counterpart is observed at that location, the authors consider an intermediate-mass black hole as the most promising candidate for the gravitational source.


How does 67P rotate? What can be said about 67P/Churyumov-Gerasimenko axial tilt?

Is it fixed (if so, at what value), only changing with precession (at what rate?) or does 67P "tumble" with axial tilt changing over time? (and if so, at what rate?)

I'm interested in estimating the 'polar night' area of the comet, and how it changes over time - but without knowing the rotation properties this is quite hard to do.

Looking at this site, there is a table, which has some information.

  Rotation period: 12.4043 ± 0.0007 * hours

It also rotates around a fixed axis

  recent observations suggest that the axis is tilted about 40 degrees

There are actually only two objects known to rotate chaotically in the solar system - one is Saturn's moon Hyperion, and the other is a small asteroid called 4179 Toutatis (source: this link from Guinness World Records)


Why does the Gaia color-magnitude diagram have this shape? I was looking at the visualization tools of Gaia, and tried plotting a color-magnitude diagram. On the right hand side of the image, photGMeanMag (mag) (the amount of light) is plotted as a function of BpRp (mag) (the color).

I was surprised by the distribution of the points in the plot, since I was expecting to see something like the main sequence of the HR diagram, like in this plot:

Why are the points distributed in a sort of wedge, rather than along a diagonal line?

Try plotting absolute G magnitude (i.e. corrected for the fact tthat all the stars are at different distances) on the y-axis.


Could an object, assuming possible, escape black hole by travelling above constant speed? I am always hearing that after travelling inside a given BH's event horizon, the escape velocity is greater than c, which makes it impossible to escape.

Suppose an ship is inside a BH. Assume it is possible to travel faster than c. If the ship applied the faster-than-light travel, would it be able to escape a black hole?

Furthermore, it seems to me that above event horizon, the escape velocity is less than c, while inside event horizon, the escape velocity is greater than c. This leads to a further question: Does the escape velocity increase the closer you get to a BH's singularity?

Assume it is possible to travel faster than c

We normally cannot use physics to answer questions about scenarios that break the laws of physics. But for this scenario, we can say that if the ship can travel backwards through time, then it can escape from the event horizon. Just about every FTL scheme can be used for backwards time travel, and as far as we know, backwards time travel is physically impossible.

Rather than thinking about this problem in terms of escape velocity, its better to think in terms of worldlines. The velocity of a body depends on the observer measuring it, and which coordinates they're using, and dealing with that stuff near a black hole gets tricky. The usual Schwarzschild coordinates break down at the EH, and different observers measure quite different things, depending on their location and motion.

Inside the event horizon of a Schwarzschild black hole, all worldlines travel towards the singularity, without exception. As you move towards the future, you move towards the singularity. So the reason you can't escape from inside the EH is that there are no paths that go outside the EH. Unless you can magically travel backwards in time, so you can follow an in-falling worldline in reverse.


How do GHA of vernal equinox, GHA of astronomical object and RA relate? I have zero clue in astronomy and I need astronomical data to write a program. What I learned so far:

celestial coordinates are usually measured relatively to vernal equinox aka meridian of First point of Aries (which is not located in Aries anymore)
the GHA of first point of Aries changes constantly

I was searching for vernal equinox GHA data and it seems to me that this personal, generated Almanac version lists it as "Aries" (it was misleading me).

I expected that
( GHA(object) - GHA(v. equinox) ) mod 24h = RA (object)
as per this article. However I have found a program which calculates declination and right ascension and I have found out that the relation is sort of inverse. For example:

https://sv-inua.net/wp-content/uploads/2016/12/almanac2020.pdf

And the program output:

               The Moon

  JD 2458849.50,  2019 December 31 Tuesday 23h 58m 49.174s  UT 2020
  January 1 Wednesday  0h 00m 00.000s  TDT nutation dRA -1.043s dDec
  -6.12" Geometric lon -13.866 deg, lat -4.894 deg, rad 2.6996e-003 au   Apparent geocentric longitude 346.129 deg   latitude -4.894 deg
  Distance 63.319 Earth-radii Horizontal parallax    0d 54' 17.67"
  Semidiameter    0d 14' 47.60" Elongation from sun 66.20 deg,
  Illuminated fraction 0.30 Phase 2.0 days before First Quarter
      Apparent:  R.A. 23h 16m 38.283s  Declination -  9d 58' 41.62"
  Local apparent sidereal time   1h 54m 46.005s diurnal aberration dRA
  0.012s dDec -0.03" diurnal parallax dRA -104.977s dDec -2485.74"      atmospheric refraction 0.033 deg  dRA 4.296s dDec 102.06" Topocentric:
  Altitude 25.678 deg, Azimuth 224.447 deg Topocentric: R.A. 23h 14m
  57.615s   Dec. - 10d 38' 25.32" local meridian transit 2019 December 31 Tuesday 21h 16m 02.396s  UT rises 2019 December 31 Tuesday 15h 49m
  26.475s  UT sets 2020 January 1 Wednesday  2h 50m 11.075s  UT Visible hours 11.0124

So:
110 deg 56,9 sec - 100 deg 7,1 sec = 10 deg 49,8 sec
which is inverse of R.A. which program lists.

Do I have a misconception?

I'm using the Explanatory Supplement to the Astronomical Almanac 3rd ed. as a reference. The glossary tells us that Greenwich Apparent Sidereal Time (GAST) is the GHA of the true equinox of date, that is, "GHA(v. equinox)" in your equation. On page 80 equation 3.14 states

local hour angle = local apparent sidereal time - apparent right ascension (3.14)

If one is located on the Greenwich meridian, then that equation becomes

GHA = GAST - apparent right ascension (3.14a)

solving for apparent RA,

apparent RA = GAST - GHA (3.14b)

Comparing your equation

( GHA(object) - GHA(v. equinox) ) mod 24h = RA (object)

So your sign seems to be inverted.

Looking up values using UT1 and the Multiyear Computer Interactive Almanac, rounding to nearest second,

apparent RA 23 h 16 m 38 s
GAST         6 h 40 m 29 s

Substituting into eq. 3.14a, GHA = 7 h 23 m 51 s


Why was the size of the solar system not defined by Mercury Transits? The first measurements of the (absolute) size of the solar system was made using the Transit of Venus, an event that arguably will only happen twice in a lifetime.

Transits of Mercury occur far more often. Why have astronomers had to wait for Venus Transits to make historical measurements of the size of the solar system?

It is indeed possible to measure the AU using transits of Mercury, and Edmund Halley tried to do just that in 1677.  However, there are two advantages to a transit of Venus.  The first is that during the transit, Venus is only 0.28 AU from Earth, whereas Mercury is about 0.7 AU away.  This makes the parallactic effect twice as big.  The second difficulty is that you need to measure the time between second contact and third contact (i.e., the first and last time that the planet is entirely in front of the Sun).  Mercury is so small that it is hard to determine when second and third contact are.  (In fact the limiting effect for the measurement of the AU during the transit of Venus was the so-called "black drop effect", which made it impossible to measure the times of second and third contact to a precision less than about ten seconds.)

A former professor of mine from when I was in graduate school has a really excellent writeup of the history of these measurements:

http://www.astronomy.ohio-state.edu/~pogge/Ast161/Unit4/venussun.html


how to calculate distance between galaxies in the distant past for example, assume there is a high redshift galaxy called "A" with z=6, so how to calculate the distance between us and another galaxy B when the light left the galaxy A?

For example, say B has a redshift of 0.5, and let's assume when the light emitted from A, us and B both exist.

I'm assuming you're talking about physical distances (as opposed to any of the other distance measures in cosmology).

The comoving distance to a galaxy at redshift $z$ is
$$
d_C(z) = \frac{c}{H_0}\int_0^z
         \frac{dz}{\sqrt{
         \Omega_r(1+z)^4 +
         \Omega_m(1+z)^3 +
         \Omega_k(1+z)^2 +
         \Omega_\Lambda
         }},
$$
where $c$ and $H_0$ are the speed of light and the Hubble parameter, and
$\{\Omega_r,\Omega_m,\Omega_k,\Omega_\Lambda\}$ $\simeq$ $\{\lt 10^{-4},0.3,0,0.7\}$
are the radiation, matter, curvature, and dark energy density parameters, respectively.

By definition, the comoving and the physical distance $d_P$ coincide today, but at the time a galaxy at $z$ emitted the light we see today, it was at
$$
d_P = \frac{d_C}{1+z}.
$$

In your example, $z_A = 6$, and $z_B = 0.5$. The comoving distance to galaxy $B$ is, from the equation above and with a Planck Collaboration et al. (2016) cosmology, $1946\,\mathrm{Mpc}$ away. Hence, when $A$ emitted the light we see today, its physical distance was a factor $(1+6)$ smaller, i.e. $278\,\mathrm{Mpc}$.

In Python, with astropy:

from astropy.cosmology import Planck15
from astropy import units as u
zA,zB      = 6,.5
dC_B       = Planck15.comoving_distance(zB)
dP_B_at_zA = dC_B / (1+zA)
print(dC_B)
   -> 1945.5612969107992 Mpc
print(dP_B_at_zA)
   -> 277.93732813011417 Mpc
print(dP_B_at_zA.to(u.Mlyr)) # convert to light-years
   -> 906.5103217447773 Mlyr


What is the escape velocity of our Galaxy relative to? I've been thinking that everything has a relative escape velocity, if it has Mass, seemingly. You can escape orbit around the earth by accelerating enough. You can escape the orbit of the sun by accelerating enough. Heck, this is kind of a stretch, but even electrons can escape an atom relative to how much electric charge the nucleus holds. What would it be like to escape the orbital velocity of our Galaxy-- and what exactly would it be relative to-- a black hole at the center of the Galaxy?

The quote was:

  Solar system's orbital velocity is estimated at roughly 220 km/s, and galactic escape velocity for our vicinity at about 537 km/s. So in the direction of Solar system's velocity vector, velocity required to escape Milky Way is ~ 317 km/s. And much more, if this Solar system's own orbital momentum cannot be used to full extent and a launch in other directions is required. This is of course assuming you can launch on a trajectory that avoids getting too close to gravitational influence of other solar systems.

How in the world do you calculate the escape velocity of other galaxies let alone our own-- do we know what the solar systems in the Milky Way orbit and how they orbit it?

One way to think of the Escape velocity is to imagine it backwards.

Instead of a rocket being fired into space, think of the same rocket, starting at rest relative to the Earth at a great distance (for this thought experiment pretend the sun, doesn't exist and the rest of the universe is empty), then let it fall to the ground. No matter how far away you start, the speed when the rocket hits the Earth will always be 11km/s or less.

So if you start with 11km/s or more at the Earth's surface, you have enough energy to coast forever into space. The escape velocity at the surface is 11km/s, the escape velocity would be different, and lower, if you were already in a high orbit.

We can do the same thought experiment with a galaxy. We know, roughly the distribution of mass in the galaxy (most is in a large blob of dark matter  with the visible disc inside it) Now take a rocket ship a very great distance from the galaxy, and allow it to fall.  As it passes the sun, the rocket will be moving at 537km/s relative to the centre of the galaxy. (there is a black hole there, but it is tiny compared to the galaxy, so has very little effect on the gravity of the galaxy)

If you know how the mass of a galaxy is distributed we can get a good idea of how fast something would be travelling at any point, if allowed to fall. And by the same argument as above, that gives the escape velocity at that point.

The escape velocity around the sun is about 1/600 of the speed of light. The journey to leave the Milky way is at least 30000 light years long, This means it is going to take about 20 million years to reach the edge of the galaxy, since you will be getting slower as the galaxy's gravity pulls you back. Space (as has been observed) is big.