Ramifications of black hole stellar system Recently, I got around to seeing the movie Interstellar. In it, the characters of the movie visit a stellar system that appears to be built around a black hole instead of a star. On top of this, their mission is to find a habitable planet in this system. Question: I was wondering how stable the system, as a whole, would be in terms of operation and would any of the planets be habitable at all due to the obvious differences between a star and a black hole.
In the movie the black hole is supposed to be a 'supermassive rotating black hole'. As for being a binary star system, it's never stated as such nor depicted as such. It's depicted visually as the black hole alone being the center. However one of the characters does mention a "Neutron Star" as part of the system so it could possibly be a binary star system. The ambient lighting for the planets is generated by the accretion disk of the black hole. The size and rotation speed of the event horizon are not defined in the movie. At least not that I can remember.
As for the planets, their proximity varies, with the first planet depicted as being so close to the event horizon that it is effected by time dilation. The distances of the other two are not specified directly, but traveling to the second planet out seemingly takes days, while traveling to the third planet out is stated to take months, if I remember correctly. As for their surface gravity, the first planet is depicted as having higher surface gravity than Earth but not so much higher that movement is impossible, just strained. The second planet is depicted as being '80% Earth's gravity', if memory serves me correctly.
To better help define certain variables relevant to the question I found an info-graphic related to the movie that illustrates the size of the black hole and it's rotational speed: http://tinyurl.com/pqph8wl

For those who haven't seen it:

Some human explorers land on a planet orbiting a black hole. The black hole is surrounded by a large accretion disk. The planet orbits at a distance such that going any closer to the black hole will mean that your odds of getting out are slim; it's also composed of water. Finally, time dilation from the black hole means that even though the characters spend about two hours on the planet, a decade or so passes for their colleague on board.

The basic answer is that a planet can orbit a black hole. There are stable circum-black-hole orbits, just as there are stable orbits around just about any celestial body. There's a problem: A black hole typically forms as a result of a supernova. This will eject most nearby planets out of the stellar system. Alternatively, it's unlikely that a planet could be captured by a black hole and be in a stable orbit, so the whole premise - while possible - is highly unlikely. Then again, it's improbable that a planet will be made out of water, a wormhole will open up near Jupiter, or Matthew McConaughey will star in a decent sci-fi movie, so why should anything else in the story be normal?

However, you can't just put a planet anywhere near a black hole, give it a strong enough push, and hope it orbits. The innermost orbit is at the boundary of the photon sphere. On this sphere, only photons can orbit. Inside it, nothing can orbit. However, the only stable orbit is twice as far away, at $2r_p$.
The radius of the sphere is
$$r_p=\frac{3GM}{c^2}$$
We'll assume that the object is not rotating (I don't remember exactly if it is or isn't rotating, but it's simpler in this demo to say it isn't.). The formula for gravitational time dilation is
$$t_0=t_f \sqrt{1-\frac{3r_0}{2r_f}}$$
where
$$r_0=\frac{2GM}{c^2}$$
Assuming that $t_0$ (the time for the observer inside the field) is two hours (7200 seconds) and $t_f$ is ten years (315360000 seconds),
$$\frac{t_0}{t_f}=\frac{1}{43800}=1-\frac{(3)2GM}{(2)r_fc^2}$$
Simplifying, and saying that $\frac{2GM}{c^2}=\frac{2}{3}r_p$,
$$\frac{r_p}{r_f}=\frac{43799}{43800}$$
$$r_f \approx 1.0000228315715 r_p$$
which is outside the photon sphere, but just barely. However, it's well inside $2r_p$, and so most likely instable. The planet is, in short, not going to survive for long. And so giant waves - mini-spoiler - are the least of Matthew McConaughey's problems.

Post-question-edit modifications:
It couldn't have been a supermassive black hole; these form at the center of galaxies. It could have been a stellar-mass black hole, though an intermediate-mass black hole is also likely - if not likelier, if the massiveness is emphasized.
The existence of the neutron star is interesting. If the black hole were intermediate-mass, I would expect that it would have gobbled up the neutron star by now - and the planets, too. So I'd bet the black hole is a slightly-more-massive-than-average stellar-mass black hole. I highly doubt that multiple planets could orbit a black hole - for the reason I gave above; the supernova would have destroyed them or flung them out of the system.
Would any of the planets be habitable? I doubt it. The accretion disk could heat up enough to provide some light, but there probably wouldn't be a lot. I'll write up the calculations either later today or possibly tomorrow, as I'm a bit LaTeXed-out after writing a math-heavy answer on Worldbuilding to find the luminosity, but I suspect it'll be negligible - as will Hawking radiation, in case any smart-aleck was planning on bringing that up.


Why does the author believe that the central mass that gas cloud HCN-0.009-0.044 orbits is smaller than our solar system? 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."

The paper is Indication of Another Intermediate-mass Black Hole in the Galactic Center and is open access.
The quote in the second paragraph is all over the internet. I have found that sciencedaily and iflscience both link back to the same quote on an official NAO page here.

Takekawa explains, “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.”

Questions: How do they know that the central mass is "concentrated in a region much smaller than our Solar System"?
This answer about the paper and subsequent discussions below it don't seem to shed any light on this particular question. The NAO page does not indicate the origin of the quote, I'm wondering if it is a translation of a corresponding Japanese language page on the same site that explains further.

The authors were able to successfully model the motion of the gas streams as Keplerian orbits around an object of $\sim30000M_{\odot}$. In doing so, they derived some key quantities, such as the pericentric distances of these two gas components (the "Balloon" and the "Stream"). One has its closest approach at $\sim0.21$ pc; the other has its closest approach at $\sim0.07$ pc ($14000\text{ AU}$). This places the pericenter well inside the Oort Cloud, and therefore inside the Solar System.

Therefore, given that it's highly unlikely that the Balloon skims the surface of the central object, we can say that the object's radius is certainly less than that of the Oort Cloud, and by extension, the Solar System. This is a key part of the argument that the object is, in fact, an intermediate mass black hole, as they note on Page 5. It's basically the same logic used to constrain the size of Sagittarius A* based on those infrared flares observed last year.


Seeming conflict between most distant objects and age of universe (both estimated) There was a recent article on bbc.com for laypeople like me titled "The mystery of how big our Universe really is", which prompted me to post the following question in their comments section of Facebook, which so far has received no answer, so I am reposting it here in the hope of a better result:
"I have a simple question, which probably has an equally simple answer: if "the furthest visible regions of the Universe are estimated to be around 46 billion light years away" but at the same time "the Universe burst into existence an estimated 13.8 billion years ago", wouldn't the photons (or whatever) that we observe coming from those distant objects have had to have begun their travel towards us long before the universe existed?"

No conflict. The universe is expanding. The furthest/oldest object we can see emitted its photons approximately 13.4 billion years ago, but the universe has expanded since then and that object is now around 32 billion light-years away. See this answer or search the astronomy stack exchange for "age of universe expansion".


Tongue-in-cheek quote on dust extinction I remember hearing a quote or maybe rather an anecdote about a famous astronomer but I can't recall the exact wording and I also forgot who allegedly said it. Unfortunately, that has thwarted all my googling attempts.

It goes like this:

The retired astronomer gets asked by a journalist if he would do it all again and he replies something along the lines of:

  Only if selective to total extinction were constant.

Is this something that really happened? If so, who said it and what is the exact wording?

Please also provide a source if possible.

Well, I managed to contact the person I heard this from.

Apparently it was Walter Baade who was asked:

  If you had your life to live over, would you be an astronomer again?

To which he answered:

  Only if the ratio of total to selective absorption is everywhere the same.

I will not accept this answer as I have not been able to find a source yet. If anyone can find one, feel free to answer and I'll accept.

EDIT: Decided to accept after all, since apparently no one here knows a source. Follow up question regarding the source here.


Eclipse Visible Light Not Dark as Expected I live in a place where we weren't on the path of totality but were in the 90% range. I was expecting it to be a lot darker than it ended up being. What does it mean to be in the 90% range on this map? I was expecting it to block out 90% of the visible light, but while it was noticeably dimmer it wasn't close to what I thought 10% of light would look like.

As seen in this comment, normal daylight is around 100,000 lux, but 10,000 lux is still considered daylight. So, unless you're getting close to 99%, it will still be fairly light outside.


Absolute magnitudes with dust extinction I have a distant $z\sim3$ galaxy with absolute AB magnitudes $M_B$ and $M_I$, where the central wavelengths of the $B$ and $I$ passbands are $\sim0.4\,\mu\text{m}$ and $\sim0.8\,\mu\text{m}$, respectively.

If this galaxy was to experience $1\,\text{mag}$ of $V$-band extinction due to dust within it, where the central wavelength of the $V$ passband is $\sim0.5\,\mu\text{m}$, how do I work out what the absolute magnitudes $M_B$ and $M_I$ would be?

Thanks for any help with this issue.

Check this lecture note. I believe there is everything you need to understand how to proceed. In brief, you should have extra information about the relative absorption, shown in the table on p.18. Given the correction in one band (which you already have $A_V = 1)$, you can find the correction in other bands using the table.

Note that the relative absorption is different for different scenario, e.g., different RADEC, different host, different type of stars.

Also note that there are two types of extinction: galactic and extragalactic.


Do Astronomers really never call Lagrangian points "libration points"? In this answer to Who called the Lagrangian points as “Libration” points and and why was the terminology “Libration” used? I described my feeling that Lagrangian points were also sometimes called "libration points" because for the the linearized equations of motion in the circular restricted three body problem at least some of these were equilibrium points where a small displacement would result in oscillatory motion and some of these are called halo orbits and Lissajous orbits. We know some of these are unstable even in CR3BP but for a few periods at least they are oscillatory.
Then I wrote:
Interestingly former NASA engineer, flight director and Space Shuttle program manager, Wayne Hale's October 2019 blog post Definition of Terms covers this very question, and seems to disagree with this viewpoint!
I recommend reading the full post for sure, but to summarize the post shows some "libration points" of the Moon, where lunar libration is maximum.

The spot on the edge of the moon that is tilted the most toward an earthly observer is called ‘the libration point’.
Have you heard that term before?  I bet you have but in a different context.
Joseph-Louis Lagrange (1736-1813) was an Italian mathematician who played a large part in the development of the metric measurement system (SI) in post-revolutionary France.  He also studied orbital mechanics involving three bodies (e.g. sun/earth/moon) and mathematically proved there are locations around such an orbit which are gravitationally stable.  These points are called Lagrange points in his honor.  There are typically 5 such points and I will leave it to the student to research their locations.
As you can see Lagrange points and Libration points are quite different and literally have nothing to do with each other.
But if you read any number of popular media stories – and even several NASA technical papers – there appears to be confusion and the terms are used interchangeably.  This is so widespread that some dictionaries have started changing the definitions to keep up with what appears to be popular usage.
STOP THAT!
Unfortunately, the curmudgeon in me realizes that this erroneous usage has become so common that it will be hard to change usage in popular literature.
But at least you now know the difference.  And you, like me, will stop when you hear some ‘expert’ (never an astronomer) mixes the terms and think about how much ignorance is being displayed.

Question: Do astronomers really never refer to any Lagrangian points as "libration points" as Wayne Hale's blogpost asserts?

A full-text search using the astronomical-literature collection on the Astrophysics Data System (more complete and more astronomy-specific than Google Scholar) yields 10,845 hits for "Lagrange point" and 2,302 for "libration point" (not sure how long that URL will be valid, but it's easy enough to redo the search). So it's clear the former is more common than the latter.

Nonetheless, my impression from looking at article titles is that use of the "Lagrangian" sense of "libration point" significantly outnumbers the "lunar" sense that Hale prefers.

And it's clear that actual astronomers and physicists have been using "libration point" to refer "Lagrange point" since at least the beginning of the 20th Century. E.g., in Trunpler (1923), "Search for Small Planets at the Triangle Points of Mercury and the Sun", the Lagrange points are referred to interchangeably as "triangle points", "libration points", and "Lagrange's libration points." And Plummer (1903), "On Oscillating Satellites", refers to "points of libration" when discussing what are clearly the Lagrange points in the 3-body problem (e.g., "the three collinear points of libration").

So, yes, astronomers often refer to Lagrange points as "libration points", and have for at least a century; Hale is clearly wrong.


Why do they think that WDJ0551+4135 is "snowman shaped" i.e. a contact binary? The BBC's Huge 'space snowman' is two merging stars says

Researchers have discovered a huge snowman-shaped star with an atmospheric composition never seen before.
It is more massive than our Sun but only two-thirds the Earth's diameter.
The object is thought to have resulted from the merger of two so-called white dwarf stars that often explode as powerful supernovas.
Dr Mark Hollands, of Warwick University, said the team's discovery could help scientists better understand how this process occurs.

and later

But this star, named WDJ0551+4135, has an atmosphere unusually rich in carbon. Dr Hollands said that initially these observations "didn't make any sense. The only way you can explain it is if it was formed through a merger of two white dwarfs".

I assume this means that it is suspected of being a contact binary
Question: Is WDJ0551+4135 thought to be a contact binary only based on chemistry and evolution arguments? Wouldn't a light curve of this object alone be enough, and wouldn't the period be extremely short making this somewhat easier to do?
update: C|NET's Astronomers find a bizarre, dying star that 'didn't make any sense' suggests that the argument is indeed based on chemistry, and that asteroseismology may be required to back it up. It also links to Nature Astronomy An ultra-massive white dwarf with a mixed hydrogen–carbon atmosphere as a likely merger remnant.

It depends who the "they" is in your question.

If by "they", you mean the astronomers who discovered it, they don't. The paper (which is available on arXiv) clearly states it formed from a merger, and doesn't mention the term "contact binary" at all. The merger has already happened long ago, what we're seeing is the end result of it.

If by "they" you mean the BBC, it looks like their correspondent has gone and read far too much into the artist's impression attached to the press release. The artwork depicts the final stage of a merging white dwarf, rather than the object as it appears today. It's probably understandable since a contact binary does look quite bizarre compared to the usual spherical star, and the title was "Astronomers find a bizarre, dying star that 'didn't make any sense'".

There's definitely an argument to be had about whether artists' impressions are misleading when attached to this kind of science communication, but that's something for another website...


Is it fair to describe a solar flare as a big lightning strike on the sun? Solar flares and lightning strikes are both kinds of electric discharges.
A lightning strike is a sudden discharge of electrical energy between an electrically charged cloud and another object, like the ground or another cloud.
A solar flare is a sudden discharge of electrical energy between two electrically charged regions, or "clouds" in the atmosphere of the sun.
Duration
A lightning strike lasts about 0.2 seconds, and it is made up of a series of short strokes of about 60 to 70 microseconds each. It produces an electromagnetic pulse (EMP) that consists of electromagnetic radiation across the spectrum from radio to gamma waves.
A solar flare lasts on the order of minutes to tens of minutes, and is also comprised of series of short bursts. This video of the Cinco de Mayo solar flare shows the way that the solar flare is comprised of multiple short discharges of electromagnetic radiation, which can be seen in the visual spectrum and also detected in other wavelengths like radio, ultraviolet, and x-ray, and gamma.
Amount of energy

Large lightning strike: 109 joules
Large solar flare: 1025 joules

Some other gifs of solar flares

Berkely.edu
Spaceweatherarchive.com
Cosmosmagazine.com

Relevant links

Lightning - Wikipedia
Solar flare - Wikipedia
Full-spectrum EMP

Related questions

What triggers solar flares?
What are the main differences between solar flares and coronal mass ejections
Can lightning occur in stars like the sun?
Typical wavelength of solar flare

My question
I'm also wondering if a solar flare is an electrostatic discharge. My main question is:
If we were to make an oversimplification about solar flares, would we say that they are big lightning strikes on the sun?

ProfRob gives a more detailed explanation of what a solar flare is in his answer to "What triggers a solar flare?" It helps a lot in distinguishing between CME's, eruptions, prominences and solar flares. There is a lot of confusion, even among space enthusiasts, between these phenomena. This is largely because images of prominences and CMEs always seem to be labeled as solar flares. This has been confusing to me, too. A solar flare might be associated with a large eruption of matter, but that is secondary, and the primary feature of a solar flare is plasma being suddenly moved from one location to another by magnetic reconnection, causing it to collide at relativistic speeds with the chromosphere and photosphere, producing the intense light and x-rays seen from earth. Unlike lightning, a solar flare is not an electrostatic discharge. A solar flare is like lightning in the sense that there is a near light-speed flow of charged particles that result in the emission of bright electromagnetic energy, spanning the entire spectrum from radio to gamma waves.


Considering our methods of exploration, how likely is it that there are unfound planets (not dwarf planets) in our solar system? I think it's probably unlikely that there are more planets between Mercury and Mars, but out from Jupiter, there's lots of empty space between the planets. Could there be some small planet hidden out there?

There are no undiscovered planets between the sun and Neptune.
Objects closer (to Sun) than Neptune that are large enough to be considered planets (and not dwarf planets) can't remain 'hidden'. If it's there, the light from the sun will bounce off it and we will see it. As it moves in its orbit, we will notice the position in the sky change, so we will know it isn't a star.
I would like to give a more broad answer to this question though:
What is the maximum size of an undiscovered solar system object and how does it change as you get further from the sun?
The further out a solar system object is, the harder it is to detect. The rate at which it gets harder is severe; the light we receive from an object scales roughly as $1/r^4 $.
($1/r^2$ for the light travelling from the Sun to the object, and again, $1/r^2$ for the light travelling from the object to us on earth).
We are able to detect some very small earth-crossing asteroids. Some as small as ~50 metres across. At a guess (and this is just based on my intuition, not any calculations), there are probably no undiscovered objects larger than 1 km close to earth.
As you travel to the outer solar system (Jupiter to Neptune), the number of bodies increases dramatically. There are currently ~700,000 known solar system bodies and most of them occur in this area. It is believed that all asteroids larger than 10 km have been found.
In the area immediately beyond Neptune (30 AU to 100 AU), we have been finding many pluto-like objects in the past two decades, objects with diameters of 500 km or more. For reference, Pluto is about 2200 km across. It is entirely plausible that there are some, if not many, similar objects at this distance range that have not yet been found. Some of these would classify as dwarf planets, in that they are round, orbit the sun, but have not cleared 95% of their orbit of other matter.
And finally - and this is where it gets exciting - where it gets really far out, there may be an undiscovered large planet, out at ~700 AU. Referred to as Planet Nine, this is a hypothetical object that some (noteworthy) astronomers believe exist because of patterns they see in the orbits of other distant dwarf planets. From calculations, they estimate that it would be as heavy as 10 earths and travel between 200 and 1200 AU. However, these distances are so large and the sunlight out there is so dim, that even with the best telescopes and two years of looking, they haven't found it.
Finally, I'd like to share two graphs. The first one is of distance (of closest approach, or perihelion) vs diameter of various outer solar system bodies. In the bottom right corner there is a noticeable lack of dots, indicating roughly where we don't yet have the capability of seeing objects that small and distant.
The second is an expanded but slightly less accurate diagram of the same.

Image 1 from Johnston's trans-Neptunian objects page. Image 2 from xkcd.


Finding the lens diameter of a telescope by magnitude We have a telescope that can see stars with about +14 magnitude. how to find its lens (or mirror) diameter? (I mean the $D$ in formula $\theta = 1.22 \lambda / D $)

I don't know whether I can assume that the temperature of star is equal to temperature of the sun or not. but if we think that they are equal we get this:

$m_{\mathrm{sun}} - m_{\mathrm{star}} = -2.5 \log_{10} (\frac{b_{\mathrm{sun}}}{b_{\mathrm{star}}}) = -2.5 \log (\frac{\sigma t^4 \times 4 \pi R_{\mathrm{sun}}^2 /(4 \pi r^2) }{\sigma t^4 \times 4\pi R_{\mathrm{star}}^2 / (4\pi r_{\mathrm{star}}^2)})$ if we solve this with what we know about magnitude of sun (-26.83 or something close to it between -25 to -27) and another things like radii we get:

($R/r = 3.16605 \times 10^{-11} \mathrm{rad} = \tan \theta/2 = \theta/2$ and from that and formula
$\theta = 1.22 \lambda / D $ and $\lambda = 550 nm$ we get something like D = $10596.8 \mathrm{m}$. which is very big for a telescope. what's the problem? how to solve this? Is this answer correct?

The unaided eye can typically see mag 6 objects. With your telescope, you can see an additional 8 magnitudes. This requires a factor of $100^{8/5}$ of additional light gathering power (since 5 magnitudes equals 100 times the brightness). I.e. about 1580 times larger aperture than the pupil of the eye.  Assuming the pupil has a diameter of 7mm, then the diameter of your telescope would have to be $7\times 1580^{1/2}$, which is about 280 mm.

This is fairly close I think - perhaps a bit on the high side but not much.


The tidal locking problem concerning Earth sized planets in habitable zone around Red Dwarfs Would twin planets orbiting around each other present a solution to tidal locking of habitable planets in red dwarf systems? It's an awfully easy equation to set up. I'm fibbing; I can't do it. I was hoping someone might show me when they get a moment.

There is quite simple formula that will give you the tidal-locking half-time

$$T = C\, \frac{a^6 R \mu}{  M_s M_p^2}$$

$a$ - semi-major axis, or simply the radius of the planet circular orbital trajectory in meters
$R$ - planet radius in meters
$M_s$ - planet mass in kg
$M_p$ - parent planet/star mass in kg
$\mu$ - rigidity, approximately $3×10^{10}$ for rocky objects and $4×10^9$ for icy ones.
$C$ - a prefactor dependent on units used. On Wikipedia, they write $C = 6 × 10^{10}$ if the time is to be in years, but this number is probably wrong. (See the discussion.)

Solving the problem of interaction between the parent star and between the planets will be difficult. Generally, if tidal-locking half-time between the planets is much shorter than the tidal-locking half-time between the star and the planet, the planets will tidal-lock to each other rather than to the star. So we ask, whether

$$ \frac{A^6}{  M_s^3} \ll \frac{a^6}{  M_s M_p^2}\;,$$

where $A$ is the semi-major axis of the orbit of the double-planet. For typical planet in the habitable zone of the red dwarf star, $a$ = 9,000,000,000 m (0.06 AU), while the distance of the double-planet could be for example $A$ = 30,000,000 m. $M_s = 6\times 10^{24}\;\mathrm{kg}$ and $M_p = 2.9\times 10^{29}\;\mathrm{kg}$. We get

$$2\cdot 10^{-41}\ll 6.3\cdot 10^{-36}\;,$$

which is well-satisfied and the planets will tidal lock to each other rather than to the parent star. Were the distance between the planets $A$ = 300,000,000 m, the inequality would yield

$$2\cdot 10^{-35}\ll 6.3\cdot 10^{-36}\;,$$

which is not satisfied and the tidal forces of the star would probably disrupt the double-planet. (The planets would not even be in the Hill sphere, which approximately gives the stable orbits.)

So the answer is: a very close double planets can tidal-lock to each other rather than to the parent red-dwarf star, but they would have to be much closer than the Earth to the Moon.


Precessing of the Earth It's been known that the Earth's axial tilt varies from 22.5 to 24.5. Does this result in a wobbling while the value of the tilt varies about these two limits? If so, how does it affect the mechanics of the Earth's movement (its rotation and revolution about the Sun)?

The Wikipedia Precession article explains this.

Precession of the Earth's axis is caused mainly by the Sun's gravity acting on the oblate spheriod of the Earth. The Earth bulges at the equator and the force of the Sun's gravity on the bulge makes the axis wobble like a child's spinning top - the axis traces out a cone shape in space. It has no effect on the Earth's orbit.

Perihelion precession is caused by the gravity of Jupiter and the other outer planets.


Will NEOWISE be known as the Great Comet of 2020? Many or all comets visible to the naked eye are referred to as Great Comets of their year(s). Might NEOWISE too be called like this? Why or why not? Is the IAU responsible for declaration of Great Comets or another organization or no such body?

I'm inclined to say a hard no using the casual definition of a "great comet"

A great comet is a comet that becomes exceptionally bright. There is
no official definition; often the term is attached to comets such as
Halley's Comet, which during certain appearances are bright enough to
be noticed by casual observers who are not looking for them, and
become well known outside the astronomical community.

So, there's a dual definition of visible brightness and/or recognition outside the astronomical community.   With that in mind, I think it's pretty clear that Comet NEOWISE doesn't make the cut due to it not being easily visible   Granted, it peaks in 2 days from now (7-23-20) but so far it's hard to see by the naked eye and that's not likely to change significantly as it peaks.

Although it’s brighter than all but around 20 stars, its extended,
diffuse nature makes it a challenge for human eyes.

NEOWISE is the brightest comet since comet McNaught, but McNaught, with a peak mangitude of -5.5, was about 500 times brighter. (based on the estimate that Neowise is about as bright as the 20th brightest star, roughly 1.25 magnitude).  McNaught was actually brighter than peak Venus, though it probably didn't look that bright to the naked eye because it was more spread out.
Comet Lovejoy was a bit unusual.  It's "greatness" had nothing to do with it's visibility, which was unspectacular.   Lovejoy passed too close to the Sun and it lost too much material in passing, so it was much anticipated and I remember articles saying it was potentially a super-comet, but it lost everything on the other side of the sun and we never got a good view from Earth.   Perhaps if we'd been on Mercury or Earth's L3 trojan point, we'd have seen a rare and magnificent show.
Wikipedia doesn't list Comet Lovejoy in it's list of great comets.


What are the most recent minerals modified by water found to date on Mars? There likely was quite a bit of liquid water on the surface of Mars less than 4 billion years ago. We know this because we found rocks that old that have been modified by the presence of water. For example :

If the martian blueberries, or hematite spherules found by Opportunity formed by slow evaporation in mineral-rich liquid water, they likely formed 3 billion years ago.
Serpentines have been detected by CRISM.

What is the most recent rock we know of on Mars that was modified by the presence of liquid water on the surface?

If you have access, I would recommend you to read Carter et al. (2013). It is a compilation of all occurrences of hydrous minerals that have been detected on Mars by spectrometers OMEGA (Mars Express/ESA) and CRISM (Mars Reconnaissance Orbiter/NASA). This data was used to investigate the spatial distribution, composition, and age of hydrous minerals on Mars.
Regarding age, the most recent hydrous minerals are opal and zeolites of Amazonian age (< 3 Ga):

The exposure with the youngest unit age is a series of small cones and
dykes or fractures which have been identified thanks to a dust‐free
window in the Utopia Rupes region, in a unit of Amazonian age. Both
opaline silica and zeolites (or alternatively sulfates) are found
there as shown in Figure 7. Zeolites are found mainly on the butte
that could constitute a small volcano, as suggested by the presence of
radial patterns resembling lava flows, to the southeast especially.
Silica is found on an elongated structure which is either a dyke or a
structural horst. Of interest is the presence of two small cones
exposed in the NE corner displaying opaline silica. Such small cones
on Mars have been interpreted either as volcanic vents or mud
volcanoes [e.g., Allen, 1979; Farrand et al., 2005]. In both cases,
the observation of opaline silica pleads for a local fluid circulation
inside a hydrothermal context consistent with volcano‐ice interactions
as predicted by previous studies [Allen, 1979; McGill, 2002; Farrand
et al., 2005].

However, it does not really answer your question as their presence is likely not due to "liquid water on the surface", but rather to impact-driven hydrothermal systems inside the crust. Exposures of their "horizontal deposits" (i.e. sedimentary) and "alluvial fans and deltas" categories, most likely linked to surface liquid water, are of Hesperian and Noachian age, although I couldn't find which is the youngest.
A limitation to keep in mind: some exposures of hydrous minerals may be older than they seem, as the terrain could have been excavated by an impact long after the alteration process (Noachian terrain exposed by Hesperian or Amazonian impacts).


How could original astronomers have used the clouds to find distances? I was looking at a very bright moon tonight through light cloud cover and saw a halo of light far bigger than the moon imposed on the clouds. Almost in a conic shape by comparison. I was wondering if early astronomers could use this difference to calculate the distance of the moon given the height of the cloud and the dimensions of earth.

No, because the size of the halo is determined by the properties of ice crystals, the distance to the Moon is irrelevant. So you get a 22° halo
 with both the Sun & the Moon, even though the Sun's much further away.


Binary stars system ultimate fate I read an article mentioning that when two stars orbiting each other they'll produce gravitational waves that can carry away energy into space. These phenomenon increases the speed of the two orbiting stars and draws them closer to each other until they finally collided, is it true that gravitational waves can carry away energy into space?

Well, yes apparently that is how it works and a Nobel prize was won in 1993 by Hulse and Taylor for demonstrating this phenomena in the object PSR B1913+16.

PSR B1916+16 is a pulsar (a rapidly rotating neutron star with a strong magnetic field that emits beamed radiation along its magnetic axis). The pulsar provides an extremely precise ticking clock with which to study the motion of stars in a binary system. The pulsar is in orbit with a companion that is also a neutron star (though not an observable pulsar). The orbit is elliptical with a period of 7.75 hours.

The period of the orbits is decreasing. That is, the two stars are gradually getting closer together. This effect is a consequence of general relativity, that predicts that such massive objects that are so close to each other will act as a source of gravitational waves. The gravitational waves have not (yet) been directly observed, but GR makes a prediction for how much energy they carry away and hence at what rate the orbital period changes.

The rate of decrease in the distance between two bodies is given by
$$\frac{dr}{dt} = -\frac{64G^3}{5c^5}\frac{(m_1 m_2)(m_1+m_2)}{r^3}$$
and leads to a merger between the two objects in a time of
$$t = \frac{5c^5}{256G^3}\frac{r^4}{(m_1 m_2)(m_1+m_2)}, $$
where $m_1$ and $m_2$ are the masses of the two orbiting bodies, $G$ is the gravitational constant and $c$ is the speed of light.

In slightly more friendly units
$$ \frac{dr}{dt} = 7.8\times10^{-19} \frac{(M_1 M_2)(M_1+M_2)}{r^3}\ au/yr,$$
where $r$ is in astronomical units and the masses are in solar masses, and
$$ t = 3.2\times10^{17} \frac{R^4}{(M_1 M_2)(M_1+M_2)}\ yr. $$

Thus you can see that the effect gets much stronger (and the merger times shorter) if you have two high mass objects (it depends on the product of the masses) with a small orbital separations.

For "normal" stars you can never get them closer than their stellar radii and generally means that the effect is too small to be important. However, compact objects (white dwarfs and neutron stars) can be brought much closer together and this is why the effect has been seen there. The Hulse-Taylor binary components
have masses of $\sim 1.4 M_{\odot}$ and are separated by an average of about 2 million km (0.013 au). Inserting this into the formula above give a merger time of 1.7 billion years. A more accurate calculation that takes account of the elliptical orbit gives 300 million years.

However, if you made the neutron stars orbit with a separation of 0.1 au (more typical for a normal stellar binary system), the merger timescale would increases to well beyond the current age of the universe.


Can we actually predict an asteroid impact? I have been seeing all these "hoaxes" on the internet about Planet X and how it would destroy Earth, on September 23rd 2017. It didn't happen.

I started to wonder if it's possible - with our current technology - to map the path of an asteroid with calculations of how an asteroid will pass near or collide with Earth. I've heard one is going to "safely" pass Earth on January 13th 2019.

There's a wide variety of answers for how much warning.   Most of what the Near Earth Asteroid Program or focuses on is Earth Crossing Objects or Near Earth Objects.   Those pose the greatest risk of impact and they are also, somewhat conveniently, the easiest to track.

The hardest to track is a high eccentricity long period orbit that approaches from the sun side of the planet.   Chelyabinsk was a sun side asteroid (an asteroid that hits during the day generally approaches from the sun side.   Those are the hardest to see, and even though Chelyabinsk was too small to worry about (20 meters diameter estimate).  It's possible that it would have been observed by the luck of having a telescope pointed in the right direction had it approached from the night side of the planet.   Still, an object that size, it's unlikely that they'd have had more than a few hours warming.

The larger the object, the easier it is to see and once an object is observed for some time and it's trajectory has been studied, it's path can be predicted years, even decades ahead of time.   The warning time drops sharply for an object not yet observed.

Potentially Hazardous Objects are at least 100 meters across, sometimes 150 meters is used as a minimum.   25-50 times the reflective surface of Chelyabinsk, so an object that size could be seen a fair bit further ahead of time.

Finally, Planet X doesn't exist and if it did exist, it would have cleared it's orbital path by now and it wouldn't be sending asteroids and/or comets towards Earth once every orbit and predicting an exact date (Sept 23rd) from an unobserved planet X. That's just silly.

There is stuff out there in our solar system that we haven't seen.   Lots of stuff, maybe even a hundred or more unidentified dwarf planets, possibly another planet or two.  (Click here for cool diagram)

The thing is, orbits tend to clear out over time and our solar-system is over 4 billion years old.   Most of the stuff that had an orbit that was going to crash into the Earth did so long ago and Planet Nine, assuming it exists, has cleared out it's orbit and it rarely sends stuff flying towards Earth.

Space is also so big that Earth is, by comparison, a very tiny target, so an asteroid coming from the outer reaches of the solar system and hitting Earth is like shooting a hole in one from the space shutting.   Stuff is far more likely to miss than hit the Earth.

But to answer your question on an object coming from a distant orbit, sent our way by a theoretical planet, it's hard to say when we'd see it in that scenario if it's a brand new observation and it would also depend on whether it came towards the Earth on the sun side or the night side.  (Objects can hit the Earth from any direction), but sun side observations are the hardest to see.

The good news is, high eccentricity impacts of objects large enough to be considered potentially hazardous are rare enough that you are unlikely to be affected by an Asteroid Impact in your lifetime.   It's somewhere around 140 on the list of things you should worry about.   And low eccentricity orbits that could hit Earth, or NEOs, those are being carefully tracked.

The other good news is that Planet X (if it even exists) doesn't threaten Earth.  No object in distant orbit around our sun would be a threat.  In fact, having a planet out there would actually send more asteroids away than it would cause to fly in towards the inner solar system.   They would reduce, not increase the number of impacts, similar to Jupiter.

But a large object passing through might post an asteroid/comet impact threat. Gilese 710 could do that, but it won't pass close enough to have that effect for about 1.36 million years.

I don't want to make light of this.  Asteroid impacts do happen and Earth may have been hit by a very destructive asteroid or comet about 12,900 years ago (the Younger Dryas impact hypothesis) and a global changing event like that might happen every, 50,000 or 100,000 years or so, which means it's 100%, absolutely worth looking out and seeing as much as possible, because Earth will get hit by a large object again.    But such impacts are very rare on a human lifetime scale.

But I'm not sure I ever answered your question.  How much warning for a large object, say 150-250 meters?   If the trajectory makes it tough to see, maybe only a few hours.   If it approaches from the night side, maybe a few days or a few weeks.   Longer if someone gets lucky and has a big telescope pointed in the right place, but I'm kinda guessing.    As our technology improves, the warning period will increase.  (I invite correction, especially on this last paragraph).


Does the age of the universe take into account General Relativity / Special Relativity? It is generally accepted that the age of the universe is approximately 12-15 billion years old based on the speed of the expansion of the universe. Since everything is moving very fast away from us, doesn't relativistic effects from high speeds come into play here? If something is moving quite fast, time slows down, affecting how "old" it is. Is all this taken into account when calculating the age of the universe? I don't know if I'm making any sense to anyone else or not.

This is a confusing question - your title mentions GR, but of course the age of the universe is entirely derived as a result of using GR to solve for the dynamics of the expanding universe.

The text of your question is talking about time dilation and the effects of special relativity (a subset of GR). Here there is a point to be addressed. It is not a good way to think about the explansion of the universe in terms of things receding from each other at ever increasing velocities. It is better to think about space expanding. It is also not possible to apply special relativity except in local inertial frames.

It can and has been shown that the redshift-distance relationship cannot be simply interpreted as the doppler shift due to things moving away from us at ever increasing speeds - since this of course predicts a maximum recession velocity that approaches the speed of light. In fact, beyond redshifts of $z=1.5$, the "recession velocity" in all viable models exceeds the speed of light. What actually happens is that light emitted at some earlier epoch travels across expanding space to get to us and this expansion increases the wavelength of the light.

Essential reading: Davis & Lineweaver (2003).


What is meant by "there is no refraction at the zenith"? Wikipedia states:

  Atmospheric refraction of the light from a star is zero in the zenith

and other sources seem to agree.

I don't see why this should be so. Consider a situation in which air gets warmer from left to right. Then light should curve towards the warm air, and a star at the zenith would appear refracted.

So why do so many sources claim there is no refraction at the zenith? What do they mean by this?

To predict the position of a star one has to account for atmospheric refraction. In a simple model, the atmosphere is treated as being composed of layers of gas with variation in density and temperature due to height, but no variation with latitude or longitude and no turbulence. In such a model, one can predict the position of a star to a great deal of accuracy (so it is a useful model), and in this model there is no refraction at the zenith (as the light doesn't pass at an angle through a medium of varying optical density).

In such a model there would be no twinkling, however modelling twinkling is not needed for positional astronomy. Large variation in optical density that could cause stars to appear significantly displaced at the zenith doesn't occur. The variation in optical density with height is much more significant for positional astronomy.


What is the average color of the universe? Our Earth, being a beautiful blue orb, has an overall perceived wavelength of somewhere near 470 nm, that being the wavelength of light blue, because the oceans play an enormous role in the weighted average of all of the Earth's colors.

But what about the whole universe? The Earth is nothing in comparison to the host of multicolored stars, ranging from deep red to fiery blue. Taking into consideration the number of stars and each separate color's weighted average,  could we calculate the average color of the universe?

According to astronomers at Johns Hopkins University, it's beige. Note, they previously announced it was turquoise, but they identified a bug in their calculations, redid the numbers, and now it was determined the average color is beige. Kind of boring, yes, but kind of what you'd expect.

In short, they collected "detailed light measurements from more than 200,000 galaxies. They then constructed a "cosmic spectrum," which represents all the energy in the local universe emitted at different optical wavelengths of light" and averaged it. More details at link below.

NPR : The Color of the Universe Is...


What are the prospects for follow-up observations of phosphine on Venus? Today, it was officially announced that astronomers have detected phosphine on Venus via the $\text{PH}_3(0\to1)$ transition (Greaves et al 2020). While the line was found by both the James Clerk Maxwell Telescope and ALMA, and while the team is fairly confident that the detection is robust, follow-up observations would nevertheless be nice, particularly in other bands. Sousa-Silva et al 2020 note that phosphine has strong features in the 2.7-3.6, 4.0-4.8, and 7.8-11.5 micron bands, and while they're pessimistic about detecting phosphine in $\text{CO}_2$-dominated atmospheres around Sun-like stars using less than ~200 hours of observing time, those numbers are for exoplanets, where we'd expect substantially lower fluxes than we'd receive from Venus.
With all that in mind, what are the most promising bands to search for phosphine on Venus, in addition to to the $1\to0$ transition? Are they the three infrared bands discussed by Sousa-Silva et al, or could other trace components of the Venusian atmosphere block the signal at some wavelengths? I see that $\text{SO}_2$ was the only remotely feasible possibility for a source mimicking the observed phosphine transition, but that would require temperatures twice as high as observed.

As a side note, it's been announced that BepiColombo will use an onboard spectrometer to try to detect phosphine on Venus during two flybys of the planet en route to Mercury. The first will be on October 15, 2020, and the second will be on August 10, 2021. I haven't been able to find out more details on the planned observations, but the spectrometer (MERTIS) operates in the 7-14 micron band.

This figure is from the paper "Phosphine as a bio signature gas in exoplanet atmospheres".  It shows the absorption cross section of Phosphine compared to other molecules. We can see that Phosphine has a distinct enough profile from the others molecules in the 7.8-11.5 microns range, with the exception of NH3.  Probing from 2-11.5 microns should then allow to distinguish Phosphine from other molecules. Simulations seem to indicate that current state of the art
nIR ground-based spectrometers could detect phosphine to a minimum of a few ppb, so expect a lot of follow-up observations from the ground in the coming months.


Are black holes really singularities? Can't black holes just be super dense objects? They could still be black (having the color of black never really required special physics, after all) and have a really strong gravitational field. If we suspect that it actually absorbs light due to its gravity then it is possible that there is gravity strong enough to capture light, not letting any radiate out when matter falls into this object.

I'm just having a hard time accepting that anything can exist with an infinitely large property as it would lead to infinitely large mass leading to infinitely large forces... that would just destroy the universe infinitely fast, wouldn't it? So I think a supermassive object can be a badass just as well without having to be a singularity.

What is a singularity?

A singularity is a point in the universe where matter is infinitely dense. The singularity is at the centre of the black hole, and is often hiding behind an event horizon. So basically, a singularity is a point in space where a lot of matter is squeezed together in a very small space. When you have a singularity matter does not go to infinity. Density is what goes to infinity. Density and matter are different things.

A singularity is often created in a supernova explosion. A supernova is where a star is caused when there is so much matter in the centre of a star (at the end of its life) that the star cannot support its own gravitational force, and it collapses in on itself and explodes.

The second way a supernova can happen is in a binary star system - where if there is a white dwarf, it steals matter from the other star and eventually it accumulates so much matter that it explodes.

A singularity is caused by the big stars, not small ones. Small stars that explode in a supernova create something called neutron stars.

So what is a black hole?

A black hole is a area in space with a gravitational field so strong not even light can escape. Black holes don't have the colour black, they don't have a colour because we can't see any light coming back from them. So no, light is not absorbed. Black holes can also eventually dissipate, due to a mechanism called Hawking Radiation

There are multiple types of black holes including:

Supermassive black holes (often found at the centre of galaxies)
Stellar black holes
Primordial black holes
Rotating (Kerr) black hole

Finally, there are 3 main areas inside a black hole.

This diagram can show us many things, for example, how black holes don't have a gravitational pull which goes across the whole universe.

On a final note, we can work out the area where even light cannot escape by using the Schwarzschild radius. Seen as you said this answer shouldn't include relativity, I won't include any maths either. The Schwarzschild radius, in summary, is where the event horizon is in a black hole.

Summary

Sorry for providing a awkwardly long answer, I was having trouble understanding what you didn't understand, and I thought that if I explained a singularity - then I'd surely help you, sorry if I've just said things you already know. I've provided some Wikipedia links if you want to have a look at some of the terms I left in italics. You might also want to look at black hole thermodynamics

Schwarzschild radius
Ergosphere
Hawking radiation
White dwarf
Black hole thermodynamics


"Chain reaction" supernovae? I know not all stars can "go nova", and that of those that do there are different types of novae, among which are some that aren't even explosions (at least not as we think of them), and that even the ones that are explosions are really not like what we do with TNT or C4 anyway.

Nonetheless, I can't shake the comparison from my head, and that got me thinking: Are there any circumstances under which one star going nova can trigger novae in other nearby stars, creating a chain reaction?

Be careful when saying "nearby stars". The scales involved in "nearby" are beyond what most people imagine. If our Sun was the size of a soccer ball located in California, the nearest star would be another soccer ball located in Greenland.

Let's say the average distance between "nearby" stars is a few light years. Let's say an average supernova is like the recent one in the M82 galaxy. That supernova, at, say, 10 light years, would be less bright and less hot than the Sun as seen from Earth. To make that supernova appear as bright and as hot as the mid-day Sun, you'd have to be at 1/3 light years from it (the supernova would have to be in the middle of the Oort cloud).

Interstellar distances are stupendous. Even a supernova cannot do much to compensate for distance.

But what if the stars are very close to each other? They would pretty much have to orbit each other, forming a double star. Moreover, both would have to be on the verge of going supernova. That's a pretty unlikely situation.

Even then, there are many kinds of supernova. There's type Ia, where a dwarf orbiting a red giant leeches too much material from the giant and eventually explodes:

Then there's the core collapse supernova, where a star is simply too bloated and too old to sustain enough pressure in the core; the core collapses into itself due to gravity and re-ignites fusion, effectively exploding like a bomb and ejecting the outer layers:

Is a nearby stellar explosion going to increase instability in its about-to-blow-up companion, or is it going to reduce its chances of exploding? The answer is not trivial.

Bottom line: I wouldn't say it's mathematically impossible, but it's a combination of quite unlikely scenarios. In a "normal" galactic neighborhood like ours, it's never going to happen.


How to explain that Polaris is a circumpolar star in the sky of any North America city? Circumpolar star definition: the stars that remain perpetually above horizon.

My explanation: Latitude affects the constellation we see and Polaris(the North star) has altitude of approximately 89.4 degree. For example, in Miami(25 degree latitude) the Polaris is 90-25=65 degree above horizon. This means that for any city above equator, people are able to see the Polaris.

My explanation is verbose and probably incorrect. Could someone provide a valid one?

In Miami, Polaris will be approx 25 degrees above the horizon, not 65. Height above the horizon is equal to the latitude of the observer.  At the North Pole, Polaris is very nearly directly overhead, all the time, because the Earth's axis of rotation points very nearly towards it.

Due to precession, Polaris will not always be so close to where the Earth's axis points.


Can a black hole formation be stopped or interrupted My question is a hypothetical: if I were to create a black hole the size of a family house, in New York, is there any way that it can be stopped, or interrupted? If a black hole of this size is formed, will everyone die?

Your black hole is quite massive. It has a mass that is much greater than the mass of the Earth, (about 1000 times the mass of the Earth. And the rest of the Earth isn't moving very fast relative to the black hole.  So in a short amount of time everything falls into the black hole. This will be a quite energetic process with a lot of high intensity radiation being given off. But in short "everybody dies" and quite quickly too.


What is the difference between azimuth angle and hour angle? What is the difference between an hour angle and an azimuth angle? When I see illustrations of the two, the angles look very similar.

Azimuth is measured along the local horizon: 0° is geographic north, 90° is east, and so on.
Lines of equal azimuth converge at the zenith and nadir.
Hour angle is measured along the celestial equator: 0h is the local meridian, 6h is in the western sky, etc.
Lines of equal hour angle converge at the celestial poles.
In other words, the two angles are measured around different axes.
The angle between those axes is the complement of the observer's latitude.


Would there be any difference for our solarsystem to be not in the Milky Way? Guessing, the solar system would be taken out of our galaxy of any reason, would there be any difference to the live on earth or even to the mechanics in within our solar system?

No it wouldn't make a difference unless you moved it near another star or body. If it was moved such that something was close enough to exert significant gravitational attraction it would have an effect on our solar system.


Would it have been possible to send a radio signal towards ʻOumuamua? As we could not get any radio signals from asteroid ʻOumuamua, couldn't we have sent a powerful radio signal to it and then check if we can get any radio signal in response?
This way we can at least check if it is just a rock or any alien spaceship. If it were an alien spacecraft, then they should definitely give a radio signal in response I believe.

Answer posted under the wire during closure. Others may be able to write an additional and/or different answer as soon as two more reopen votes are cast.

Would it have been possible to send a radio signal towards ʻOumuamua?

Yes. It is certainly true that at any time, a large radio telescope or even a small transmitter can always sent a signal toward something. New Horizons can receive our signals now even though it's in the Kuiper belt because it knows when and where to listen and at what frequency.

As we could not get any radio signals from asteroid ʻOumuamua, couldn't we have sent a powerful radio signal to it and then check if we can get any radio signal in response?

The asteroid was quite close to us in the 2nd half of 2017, but now it's much farther than New Horizons. However if it had advanced technology it would know we were very active in the electromagnetic radio spectrum and it might be listening. If so, we certainly could have sent it a signal that it could receive.
Checking for a response is also hard because we'd need to know what frequency and when. We'd have to allocate resources, etc.
For the frequency, see also

https://scifi.stackexchange.com/q/197905/51174
https://scifi.stackexchange.com/q/16348/51174

Below: Plots of the distance of ʻOumuamua (bumpy, blue) and the New Horizons spacecraft (straight, green) from Earth as a function of years since 2017-01-01.


Stars in multiple constellations Are there examples of single stars that are a part of multiple constellations? What about stars that are part of one Western constellation and one non-Western?

Thanks in advance!

In traditional constellation shapes / outlines, yes. But the International Astronomical Union (IAU) standardised on a set of 88 constellations in 1922, and from 1924 to 1930 formalised the constellation boundaries, splitting the sky into separarate areas (that cover the complete sky) so there's no ambiguity now.

For example, before the constellations were standardised in 1930, the same star (on the boundary of Auriga and Taurus) was known both as Gamma Aurigae and Beta Tauri.

And the traditional constellations of Ophiuchus (The serpent bearer)  and Serpens (the snake) (Where Serpens ran across Ophiucus) was handled by splitting them into Ophiucus and Serpens Caput (snakes head) and Serpens Cauda (snakes tail).

see http://www.ianridpath.com/boundaries.htm for more details.

Different cultures have different traditional constellations. If you have a look at the free Stellarium program (see http://stellarium.org/), that has a set of overlays showing constellations from a number of different cultures. (I don't remember if that's part of the standard installation, or an optional extra you can download).


What would the CMB look like somewhere else in the universe? An answer to a question on Worldbuilding.SE opined that if you were suddenly transported an unknown distance and direction away by magic you could use the Cosmic Microwave Background to determine your position relative to the Earth.

This lead me to wonder, what does the CMB look like at different positions in the universe? Can changes in how the CMB looks be used to determine your position, direction, or velocity relative to Earth?

The CMB is nearly featureless, but the operative word is nearly. The tiny temperature fluctuations that we've measured should look about the same anywhere within tens to hundreds of millions of light years of Earth or so. This would allow you to get an absolute orientation relative to the CMB and thus relative to Earth.

It's fairly easy to measure velocity relative to the CMB, so if you measure your velocity relative to the CMB somewhere else (within that 10-100 million lightyears of Earth, anyway) and if you know your orientation, you can compute your velocity relative to Earth.

But I don't see any way to easily measure the distance from Earth except very approximately. (The CMB that you see from any point is different because it comes from a different sphere. If you move far enough from Earth, the temperature fluctuations you see are no longer the ones we see from Earth because you're sampling a different part of the 300-year-old universe.)

Later:
I've been thinking about the position problem and there's a solution, but it's complicated.  The CMB samples the universe of 13.7 billion years ago and from any point in it see the distribution of matter in a shell that old. If we move to another point, in effect, the shell shifts also and samples different parts of the gas. The key point is that the faint ripples in the gas's temperature that we see as mottling in the CMB are three-dimensional bubbles.

So if you shift a short distance in space, the CMB will shift the most in the direction of motion (both in front and behind) and the least at right angles. Furthermore, the amount of shift will depend on the angular size of the fluctuation you're looking at. Low-angular-frequency fluctuations come from physically large ripples int he gas and you need to move a long distance to move the shell you are sampling out of the bubble. High-angular-frequencies ripples come from smaller bubbles and shorter shifts in position will take the shell you observe out of one bubble and into another.

Waving my hands just a tiny bit, it should be possible to use statistical methods to estimate the amount and direction of movement by comparing two detailed CMB maps. A major limitation of the method is that the finer the angular resolution, the sooner you lose all correlation between the two views fo the CMB, but the coarser the angular resolution, the coarser your measurement will be.

Numbers? As long as we're not talking about jaunteing many billions of light years, we can ignore general relativistic issues. A fluctuation with an angular size of 1 degree corresponds to a CMB fluctuation which appears to us right now to be around 250 million light years across and consequently ought have a good chance to be visible for moves within roughly that distance. This suggests that the CMB method ought to be able to give you a rough position up to maybe a half-billion or even a billion light years, but will drop off rapidly in accuracy after that.

Note that for shorter hops, it ought to be possible to actually recognize galaxy super-clusters and get a position that way.


How is the height of the Great Red Spot measured? This answer states that the Great Red Spot is getting taller. This means that we are able the determine the spot's height. How do we do this?

The Great red spot looks oval from above but is actually saucer shaped as the storm curves towards the center and the deepest point lies at the center. It is 16350 km wide, so the height is negligible compared to the width. The storm has its roots beneath the surface of the atmosphere. NASA’s Juno spacecraft calculated that the storm penetrate about 300 kilometers into the planet’s atmosphere.
Scientist also calculated the "thickness" of the storm which I believe the portion of storm that is above the atmosphere. The calculated thickness value agreed well with measurements at the cloud level since the Voyager mission in 1979 and was same during the Juno space mission indicating the thickness is more or less constant and is not affected from the apparent shrinking of the spot. The scientists calculated that the Great Red Spot is about 105 miles (170 kilometers) thick.
The cloud cover around the center of the spot vary by about 30 km and at the extreme end, the clouds hovers at a higher position which is about 8 km above the surrounding clouds. The difference is apparent in the false color Galileo image:

The deepest clouds are dark blue, higher clouds are light blue, high thin hazes are pink and high thick clouds are white.
Scientists were able to calculate the dimensions and dynamics of the Jupiter spot by performing a simple experiment. They investigated the dynamics of large Jupiter vortices by performing numerical simulations from a simple laboratory experiment involving a salt-water-filled Plexiglas tank that measured 20 inches by 20 inches by 28 inches (50 by 50 by 70 centimeters). You can find more information in ref. 4.
References:

https://earthsky.org/space/juno-probes-depths-of-jupiters-great-red-spot
https://www.space.com/jupiter-great-red-spot-shrinking-thickness-steady.html
https://www.sciencedaily.com/terms/great_red_spot.htm
Lemasquerier, D., Facchini, G., Favier, B. et al. Remote determination of the shape of Jupiter’s vortices from laboratory experiments. Nat. Phys. 16, 695–700 (2020). https://doi.org/10.1038/s41567-020-0833-9
Space: From Earth to the Edge of the Universe, Carole Stott, Robert Dinwiddie, Giles Sparrow, Dorling Kindersley Ltd, 2010


How can I make Stellarium match the magnitude of stars visible in London, UK? Can someone recommend some Stellarium settings to mimic the amount of light pollution, and magnitude of stars that should be visible on a clear night in London, UK?

Ok go on Stellarium then hover over the left side of the screen afterwards click on the location and select London, England (City Of London is the original walled city), then go and hover over the left side of the screen and select sky and viewing options and in the sky section and select the air pollution level from location database then it will automatically set the air pollution depending on where you are on Earth, cool right?


Help understanding this unsettling image of Titan, Epimetheus, and Saturn's rings? The NY Times article Saturn’s Rings Are Sculpted by a Crew of Mini-Moons is really interesting and links to the recent paywalled paper in Science Close Cassini flybys of Saturn’s ring moons Pan, Daphnis, Atlas, Pandora, and Epimetheus

But I absolutely can't understand one of the photos in the NY Times article, shown below. Titan seems to be...

behind Saturn's rings, and yet it is
huge relative the spacing of the rings, and yet it
appears to be out of focus while the rings and Epimetheus are in focus.

Can someone help me understand how all of these can be true at the same time?

  In the foreground, the moon Epimetheus appearing to hover above Saturn's rings. Epimetheus is dwarfed by Titan in the background.CreditNASA/JPL/Space Science Institute

This NASA page says this photo was taken on April 28 2006.

Using Celestia, I managed to find the picture from Cassini that best lines up with the photo. It doesn't match up precisely, but that's to be expected as the calculated orbital elements of all these moons (and cassini) in the software won't necessarily match up to reality precisely.

Below is the zoomed out version of this shot. You can see Titan in the centre and Epimetheus as a dot on top.

And here is the top down shot from Cassini to the moons. Circled are Epimetheus and Titan.

So to answer your question: Titan is really big compared to epimetheus (about 50x), Titan has an atmosphere and so appears fuzzy (It is actually in focus, everything in space is very far away and so is effectively at infinity for focusing purposes), and the rings are highly oblique, so you only see a small slice of them.


What is the exact position of the Large Magellanic Cloud? I am not able to visualize the position of the LMC and have not found a satisfactory explanation...
Can someone please answer it this way?

Relative to a ray joining the Earth to the Galactic-Center(Milky Way's), please describe the position in the plane of the ray as- to the South/East/West or their combinations. And will the galaxy be below/above the plane of the ray?

Note: Please consider the direction of the ray(Earth->Center) as the axis of a compass that points towards the north.
E.g://Answer format that i was hoping for:
   To the South-West of the Segment, and below its plane.

First of all: your coordinate system (as I understand it) is not "well defined". You only provide a line (Earth --> Milky way centre) and define that as the North-South axis. You don't give a reference for the east-west axis or the up-down axis. So, I'll assume that the East-West axis lies along the plane of the Milky way and up is where the Boötes Dwarf system is. This way, our coordinate system is equal to this image from the LMC Wikipedia site.

The problem with this picture is that the position of the Earth is only implicitly given. It is "behind" the center of the Milky Way. So from that position, the LMC is almost to our right (east) and a bit below the plane of the Milky way; about 30 to 40 degrees down.

This is also nicely validated by this image, acquired here (note that the grid is in 30 degree angles).

So, finally we can answer your question in the way you wanted: in the north-east to east segment, and below its plane.


Why do the magnetic field lines of the earth go from south pole to north pole? I saw these figures and could not get why the magnetic field lines are not pointing from the north pole towards the south pole but they do the opposite.

Or also this picture which is about magnetic characteristics of the sun says the same thing:

That was a curiosity that I also had once.
The reason for that is that the magnetic pole near earth's geographic north pole is actually the south magnetic pole.
You can find a more detailed answer on https://wtamu.edu/~cbaird/sq/2013/11/15/why-does-a-magnetic-compass-point-to-the-geographic-north-pole/

Happy Thanksgiving!


TASCO 49060700 60x700mm Spacestation Refractor So, my grandfather so graciously gave me his TASCO 49060700 60x700mm Spacestation Refractor Telescope, which is extremely wondrous. Any way, I found this answer to my 3x barlow lens, which I believe is saying multiply the magnification on the lens to the barlow. I've got three separate lenses; 25mm, 10mm, and a 4mm. My question is, do I multiply 25 by three(25 being the 25mm) and I get 75x magnification?

A telescope's magnification is its objective focal length divided by the eyepiece focal length:
$$ \frac{700~\text{mm}}{25~\text{mm}} = 28 $$
A Barlow lens multiplies the effective focal length of the objective:
$$ \frac{3 \times 700~\text{mm}}{25~\text{mm}} = 84 $$
The 10 mm eyepiece gives 70x magnification without the Barlow.
Other configurations exceed the maximum useful magnification of 2x per mm of objective diameter (60 mm → 120x).


During an eclipse, does the size of the moon and sun match perfectly? During a lunar or solar eclipse, from our perspective when we look towards the moon and the sun is behind the moon, does it fit exactly?

Has anyone calculated the sizes of each in a relative context?  Obviously the sun is thousands of times larger, but I am talking about the size of the sun/moon when looking from our perspective.

I hope my question makes sense.

To expand on what taupunkt commented, on average, according to the article Solar Eclipse: What is a Total Solar Eclipse & When is the Next One? (Rao, 2014), it is a unique quirk of nature that

  The sun's 864,000-mile diameter is fully 400 times greater than that of our puny moon, which measures just 2,160 miles.  But the moon also happens to be about 400 times closer to the sun than the Earth

This is often stated in middle and high school science textbooks.

However, the orbits of the Earth around the sun and the moon around the Earth are elliptical, as illustrated in the diagram below:

Image source: NOAA

This results in the exact ratio mentioned above not occurring resulting in an annular eclipse, where the moon is too 'small' to fully cover the solar disc.

A comparison between a total and annular eclipse is shown below:

Image source: ScienceJedi

NASA's Eclipse data page, in particular, their Eclipses and moon orbit information page discuss in detail (with considerable data), the

  interaction and harmonics of the synodic, anomalistic, and draconic months not only determine how frequently eclipses occur, but they also control the geometric characteristics and classification of each eclipse.

Within the NASA link, data about how much of the sun's disk that would be obscured by the moon has been calculated (from observations and predicted). For example, using the data for solar eclipses from 2011 to 2020, The average fraction of the sun's disk that is and is expected to be obscured (this is referred to as 'Eclipse magnitude') in an annular eclipse varies from about 0.95-0.99 (95% to 99%) for that time period.

A catalog of eclipses up to the year 3000 is provide on this NASA page.

The greatest difference would occur if the moon is at its furthest from the Earth (apogee) and the Earth closest to the sun (perihelion).


How can a planet have a 90° inclination? Planetary systems as we understand it exist from a disk of mass when the parent star is young.  That's why all our planets are in the same plane, or close to it.  See also Why is the solar system often shown as a 2D plane?,  Inclination of planets, and answers there.

Yet the newly announced Kepler-452b apparently has an inclination of close to 90° (89.806+0.134−0.049°).  Do we have any model of solar system formation that explains how this can form?

As Wikipedia notes,

  Since the word 'inclination' is used in exoplanet studies for this line-of-sight inclination then the angle between the planet's orbit and the star's rotation must use a different word and is termed the spin-orbit angle or spin-orbit alignment. In most cases the orientation of the star's rotational axis is unknown.

As of now (7/26/2015), no other planets, smaller objects, or debris disks have been found around Kepler-452. This means that the situation is not as unusual as might be thought. If there were other planets in the system with orbital inclinations near 0°, then things would appear exceedingly odd. Why would this one planet be different than the others?

One possible scenario is that the star's rotation axis has been changed. Think about how Uranus's axial tilt is skewed. Nobody knows for sure what happened, though one popular theory is that it was hit at some point in time by another object. Granted, it would take a rather large object to have any effect whatsoever on the axial tilt of Kepler-452, but it is still possible.

The point of this idea is that perhaps it is the star, not the planet, which is unusual.

Several papers have been written by those much more knowledgeable than I, notably Crida & Batygin (2014) and Xue et al. (2014). Both refer to Hot Jupiter models, but the ideas behind them are still possible for other planets.

The former suggest

  Contemporary observational surveys suggest that a considerable fraction of solar-type stars are born as binary or multiple systems (Ghez et al. 1993; Kraus et al. 2011). Moreover, most stars form in embedded cluster environments (Lada & Lada 2003) where dynamical evolution can lead to the acquisition of transient companions (Malmberg et al. 2007). Recently, Batygin (2012) showed that the presence of a companion to a young star can force the proto-planetary disk to precess around the binary’s axis, so that the plane in which planets eventually form and the equatorial plane of the central star can differ.

They also suggest that encounters with other planets could be responsible for some cases of this (although an angle of 90° makes this seem rather improbable); a related idea that has been applied to Kuiper Belt Objects is the Kozai mechanism. The companion to this theory is that the star was perturbed by an encounter, possibly with a companion (as I discussed earlier).

Xue et al. discuss similar scenarios, focusing in part on "dynamic" interactions between planets. Direct tidal interactions between the star and the planet are also a possible scenario.


Could an orphan/rogue planet have a moon? Is there some reason why an orphan or rogue planet wouldn't have a moon? Let's say it started out in a normal system and then just got flung out by an unstable orbit.

Could an orphan planet have a moon?

Yes.

s there some reason why an orphan or rogue planet wouldn't have a moon?

Yes, this answer lists some possibilities.
What you didn't ask was:

Is there some reason why no orphan nor rogue planet could ever have a moon?

and based on the previous two, No.
Any non-contact gravitational acceleration that could pull a planet away from its star has some chance of accelerating its moons sufficiently to stay with the planet.
If the perturber were a few times the Moon's orbit farther from the planet, staying with the planet might even be likely.
Looking at this obliquely from a delta-v perspective, Jupiter orbits the Sun at 13.1 km/s, and its inner Galilean moon Io orbits Jupiter at 17.3 km/s.
Without doing the math, I think it's possible that you could give Jupiter a 13.1 km/s kick impulsively in an optimal direction and Io would stay bound to it, but now in a much more elliptical orbit.


planets are point like Need Help. I have a problem with my 50 mm objective with 500mm focal length refractor telescope. I can see moon clearly. I have 18 mm and 9 mm eyepieces.

I see Jupiter and Venus as a bright point when focused at 30x or 60x and I see white disk when out of focus. While I have successfully resolved double stars with it and I saw Pleiades star cluster clearly with it, I cant see the moons of Jupiter.

Should I buy 3x Barlow to see the bigger image? is low magnification is my problem?
Or there is another solution to this problem.?

You should definitely be able to see the 4 main moons of Jupiter. You won’t need additional magnification - they can be seen through 10 x binoculars. It’s possible that when you looked the moons were close to or behind Jupiter, so worth trying again. Or it may be that James is right and you were looking at a star.
As for the bands on Jupiter, these are quite subtle but at 60x magnification, 2 dark bands should be visible. Don’t expect to see the Great Red Spot, at least initially.


Are the stars seen at the equator different throughout the year? If you live on the equator and look up at the night's sky at, say, midnight, would you see a completely different sky on the 1st of January than you would on the 1st of July?

I assume you would as the night side would be facing away from the sun by 180 degrees and necessarily looking outward from the solar system 180 degrees from each of the two points at those times.

Is that correct?
Why?

Yes, the constellations that are visible at midnight on Jan 1 are different than those visible at midnight on July 1 (6 months later). The website Stelvision has a star chart that you can customize for date, time, and location. This is due to the Earth's revolution around the Sun and the direction that an observer is facing at midnight.

From a given location, a given star or constellation rises, transits (when an object is due south), and sets 2 hours earlier each month.

In generally, the above is true for all latitudes. Of course, stars/constellations that are circumpolar never rise and set, and from the Earth's pole all objects are circumpolar. But even in this case, if you would pick a direction to look at midnight on Jan 1, those constellations would be behind you (180 degrees away) at midnight on July 1 (assuming you could see them in the daylight on one of those dates ;-).


"Pluto Time" for any solar system body A while back NASA made a site called Pluto Time, where you can enter your address and find that "...moment near dawn and dusk each day, where the illumination on Earth matches that of high noon on Pluto."

Is there a generic formula where we can calculate this for any body in the solar system?

How about any body around any star?

To really get exact times, you would have to take into account the opacity of the atmosphere, but I will be ignoring this here, and consider the light flux outside of the atmosphere of any planetary body.

There is no [planet] time for planets closer to the Sun than Earth, or for exoplanets orbiting brighter stars than Earth at the same distance. This  because they receive more light flux than we do. On a side note, there is an Earth time on Mercury.

For planets in further orbits than Earth around the Sun, [planet] time can be determined by comparing the solar flux at any place on Earth to the flux received by the planet at midday.

Flux received by the planet at midday

The average light flux on a planet can be determined based on the fact that the  flux varies inversely proportional to the square of the distance.
Outside Earth's atmosphere, the Solar flux is $S_0 = 1370$ Watts per square meter, for a surface placed perpendicular to the Sun rays.

Since Earth is 1 AU from the Sun, the received flux for a planet at a distance of $x$ AU from the Sun is then given by

$
F_{planet} = S_0 * \Big(\frac{1}{x}\Big)^2
$

Solar intensity on Earth

Based on this article, the flux received from the Sun at a given location at latitude $\phi$ on Earth, and when the Sun is at declination $\delta$ in the sky can be determined to be

$
F_{location} = \frac{S_0}{\pi} \Big(H \sin \phi \sin \delta  + sign(H) \cos \phi \cos \delta \Big)
$

where $H = \arccos \big( -\tan (\phi ) * \tan (\delta ) \big)$, so $H/\pi$ is the length of a day.

Putting it all together

So given your latitude $\phi $ on Earth, you want to find the declination $\delta$ of the Sun by saying that $F_{location} = F_{planet}$.

This gives you a trigonometric equation that you can solve numerically with a tool such as this, or any other method you fancy.


Can an asteroid have rings? I know an asteroid can have moons, but how about rings? Does it also get affected by the Roche limit?

The asteroid Florence has two moons.

The simple answer is yes, an asteroid can have rings. The current known example is 10199 Chariklo, whose rings were discovered in 2014 (see Braga-Ribas et al. (2014)). It actually has two rings, at distances of 391 km and 405 km, 7 km and 3km in width. As the discovery paper notes, they may be the (young) remnants of a debris disk around Chariklo, and have some moderately-sized bodies embedded in them. The observations indicated that water ice is likely present. Another small body, 2060 Chiron, may also have rings (see Ortiz et al. (2015)). Again, water ice is thought to be a major component of the putative rings.

Both ring systems were hypothesized in part due to evidence from stellar occultations, i.e. cases where the minor body passed in front of a star. Stellar occultations have proved useful in many cases involving small bodies, such as Kuiper Belt Objects.

As Ortiz et al. noted, Chariklo and Chiron are the two most massive members of a class of minor planets called centaurs. They orbit (unstable) in between the giant planets of the Solar System. Insights into ring formation around centaurs could perhaps provide information about centaur formation and evolution. It seems likely that the rings form from either

The impact of a smaller body on the surface of the centaur
The breakup of a satellite orbiting the centaur (as it crosses the Roche limit, as you speculated)

Now, rings have not been observed around Trans-Neptunian Objects in similar stellar occultation data, leading the authors to hypothesize that centaurs are a separate population. More observations could shed some light on this hypothesis.

A final interesting point I'd like to make is that many moons around minor planets have been found, but only one or two ring systems are known.


Roll, Pitch and Yaw of Orbital Planes I have been reading about celestial mechanics and particularly about planetary orbits. I understand that a planet's orbit can be tilted (pitched) with respect to the Earth's ecliptic and that it might precess (yaw) over time. But I have not come across any indication that another planet's orbital plane might be rolled with respect to the Earth's.

I have also found references to Keplerian orbital elements, but I don't see that any of them indicate roll. Am I misunderstanding the situation? If not, would someone point me at an elementary resource where I can learn about this?

You're using the wrong terms. Engineers use yaw, pitch, and roll to describe the orientation of a vehicle. Some erroneously call these rotations Euler angles. Astronomers and physicists use true Euler angles, a rotation about the Z axis of some reference plane, followed by a second rotation about the once-rotated X axis, followed by a third rotation about the twice-rotated Z axis. Note that the Tait-Bryan angles (aka Cardano angles) use a sequence of rotations about three distinct axes. Euler angles use only two axes.

The first rotation is the planet's axial precession angle. The second rotation is the planet's axial tilt, or obliquity. The third rotation represents the planet's daily rotation. The rates at which the precession and obliquity change are much smaller than the quickly-changing third angle.


What is the longitude of the winter solstice for the various planets? I am looking for data showing the longitude at which the various planets  in the solar system reach their winter solstice.  Where could I find this data?

You might find in-the-sky.org's Custom astronomical graph plotter by Dominic Ford helpful. For instance, for Jupiter, a start date of today, you can obtain the declination graph of the next 10 years for a given location (which you can change as well):

This does not solve your problem that you want to determine the longitude of maximal declination, but still.


Delta Velorum - non-observed variable star I am a beginner variable star observer from the Southern Hemisphere (but with experience in astronomy) and, when I was looking for VS to observe, I saw that Del Vel has almost no observations. Thus, there is no graph for when the eclipses occur.
VSX data: mag range of 1.95 - 2.43 V, period  of 45.15 days and 1.3% (0.587 days) eclipse duration
It doesn't have data about how the magnitude varies in the visual spectrum, althought I know it is a bright star (already saw it at naked eye).
Does it vary in the visual spectum, or there simply is no data enough to determine this? Could I collect this information visually, or is this waste of time?

Yes it varies in the visible spectrum.
The paper  The nearby eclipsing stellar system δ Velorum describes why this is difficult target.  Surprisingly it is because it is so bright.
The absolute brightness of a star will vary significantly due to absorbtion in the atmosphere.  If the sky is slightly hazy, then the stars will be dimmer.  To measure the magnitude of a star, then, you don't directly measure how much light is coming from it, but compare it to nearby stars.  But δ Velorum is one of the brighter stars in the sky, so there are fewer nearby comparison stars.  This, combined with the long period of the orbit makes observations harder. This probably explains the shortage of observations.


Relation between absolute magnitude of UV and star formation rate I surfed the internet and searched a lot of sites,
so far I got this

but unable to find the formula which relates only Muv and the Ψ as I don't know the value of parameters Auv and μ(z).
If there exists any such formula, then do help!

The UV luminosity of a galaxy can be calculated, given a stellar population. This population, in turn, can be calculated given an initial mass function (IMF), i.e. the distribution function of stellar masses. In this case, the UV luminosity should be linearly proportional to the star formation rate (SFR), sometimes written $\Psi$.
The UV is primarily emitted by young stars with lifetimes of $\lesssim10^8\,\mathrm{yr}$. Assuming that the SFR is constant over longer time scales, Kennicutt (1998) calculates in his seminal paper (with ~5000 citations) for a Salpeter (1995) IMF (equally seminal; ~6700 citations) with mass limits 0.1 and 100 $M_\odot\,\mathrm{yr}^{-1}$ that
$$
\frac{\Psi_\mathrm{Sal}}{M_\odot\,\mathrm{yr}^{-1}} = 1.4\times10^{-28}\frac{L_\nu}{\mathrm{erg}\,\mathrm{s}^{-1}\,\mathrm{Hz}^{-1}}.
$$
were $L_\nu$ is the emitted UV luminosity density, i.e. luminosity per frequency bin. At a (luminosity) distance $d_L$ from the galaxy, the observed flux density $f_\nu$ is then
$$
f_\nu = \frac{L_\nu}{4\pi d_L^2}.
$$
The flux density is related to the apparent (observed) UV magnitude $m_\mathrm{UV}$ as (Oke & Gunn 1983; ~1900 citations despite their +/- error in their most important equation)
$$
m_\mathrm{UV} = -2.5\log \frac{f_\nu}{\mathrm{erg}\,\mathrm{s}^{-1}\,\mathrm{cm}^{-1}\,\mathrm{Hz}^{-1}} - 48.6.
$$
Distance modulus and dust extinction
The apparent magnitude is related to the absolute magnitude $M$ through the distance modulus
$$
\mu \equiv m - M.
$$
Furthermore, the above relations don't take into account the fact that some of the emitted light is absorbed by dust. Dust extinction increases the observed magnitude (because magnitudes are backwards) by a number $A_\mathrm{UV}$.
That is, the apparent UV magnitude is
$$
m_\mathrm{UV} = M_\mathrm{UV} + \mu + A_\mathrm{UV}.
$$
Typically, the extinction $A_\mathrm{UV}$ isn't measured explicitly (because UV measurements are difficult); rather the extinction is measured in two optical bands, e.g. $B$ and $V$. The extinction $A_\lambda$ at a given wavelength $\lambda$ can then be estimated as
$$
A_\lambda = k_\lambda E(B-V),
$$
where $E(B-V)\equiv A_B - A_V$ is the color excess, and $k_\lambda$ is given by the assumed extinction law, e.g. a Cardelli (1989) law or a Calzetti (2000) law.
The role of the initial mass function
The Salpeter IMF is old, and at that time we didn't have too good data for the low-mass stars (which are faint), so he just fitted a single power-law. With better data, it was eventually found that there were less low-mass stars than previously thought, i.e. it takes less stellar mass to produce a given UV luminosity. A popular choice today is the Chabrier (2003) IMF (~5100 citations), which gives roughly a factor of 1.8 more UV than the Salpeter IMF per star formed, on average. That is, the conversion factor is 1.8 times smaller.
With a Chabrier IMF, you hence get that the relation between absolute magnitude of UV and star formation rate (which is what you're asking) is
$$
\frac{\Psi_\mathrm{Cha}}{M_\odot\,\mathrm{yr}^{-1}} = 7.8\times10^{-29}\,\times\,4\pi d_L^2 \times10^{-(M_\mathrm{UV} + \mu + A_\mathrm{UV}+48.6)/2.5},
$$
where $d_L$ is measured in cm.
Simpler relation
The above formula resembles the one you give, but is a bit strange in that it both contains $\mu$ and $d_L$, which are just two different measures of the same thing. Using the definition
$$
\mu = 5\log\left(\frac{d_L}{10\,\mathrm{pc}}\right)
$$
we can — after a bit of algebra — express the definition of the Oke & Gunn (AB) magnitude as
$$
M_\mathrm{UV} = -2.5\log L_\nu + 51.6.
$$
With this, we can can write the relation as
$$
\frac{\Psi_\mathrm{Cha}}{M_\odot\,\mathrm{yr}^{-1}} = 7.8\times10^{-29}
\,\times
10^{-(M_\mathrm{UV}+A_\mathrm{UV}-51.6)/2.5},
$$
or even
$$
\boxed{
\frac{\Psi_\mathrm{Cha}}{M_\odot\,\mathrm{yr}^{-1}} = 3.4\times10^{-8}
\,\times
10^{-(M_\mathrm{UV}+A_\mathrm{UV})/2.5}.}
$$


Circumpolar Equation Derivation Edit: Geometric answer from Math.stackexchange

I am trying to understand where the circumpolar equation comes from.

$\delta \geq \frac{\pi}{2} - l$

I can understand that the sort of "rotation" of the reference frame that one experiences in horizon coordinates. I'm not sure where to go from the fact that a star's lowest altitude must be greater than 0.

It is pretty simple to show that $a = l$ and $c = \frac{\pi}{2} - l$. $a$ in this case is the altitude of the north celestial pole. What I need to do is show that the angle of the cone made by a circumpolar star with $\delta + l \geq \frac{\pi}{2}$.

You can deduce that $a_{min} \geq 0$ for a star to be considered circumpolar. I just need to do this geometrically in terms of its declination from the celestial equator.

Edit: I think my latest attempt has found a solution.

For a star to be circumpolar, the star must, at the very least, come above the horizon.

$0 \leq a \leq \frac{\pi}{2}$

In the celestial sphere coordinate system, $c$ can be found to be

$c = \frac{\pi}{2} - \delta$

In the horizon coordinate system, $c$ can be found to be

$c = l - a_{min}$

Therefore

$l - a_{min} = \frac{\pi}{2} - \delta$
$\\$
$l + \delta = \frac{\pi}{2} + a_{min}$

$a_{min}$ can be set to 0 because it is the lower bound for circumpolarity.

$l + \delta = \frac{\pi}{2}$

$\delta$ must be in the cone made by the star's rotation about its declination angle. Therefore

$\frac{\pi}{2} - l \leq \delta$

There is nothing wrong with your explanation.  You could add some explanatory notes like $a_{min} = l-c$ and $a_{max}=l+c$. Also, I suggest that you put the equation $l+\delta = \frac{\pi}2-a_{min}$ on a separate line.

It is very good to figure out these equations on your own.

$ $

Here is an alternative explanation with less algebra:

If you walk outside at night, you will notice that the north star has an elevation of $l$ degrees if your latitude is $l$.
The north star does not move.
The angular distance between stars does not change  (in the short term!).
In order for the a star to be circumpolar, it needs to be within $l$ degrees of the north star (otherwise it will dip below the horizon when its hour angle is 12 hours.)
So, the angle $\theta$ between the north star to a circumpolar star needs to be less than $l$.  That angle is $\theta=90^\circ-\delta = \pi/2 -\delta$ where $\delta$ is the declination of the star.
Finally,
$$
\begin{align} \theta &\leq l\\
\pi/2 -\delta &\leq l \\
\pi/2  &\leq  \delta + l \\
\pi/2  - l &\leq \delta.
\end{align}
$$


What is the Maximum Speed that can be acheived Because of Acceleration Due to Gravity? Planets and other bodies in the universe have different ranges of gravitational fields. What is the maximum speed that can be achieved because of acceleration due to gravitational field.

In other words which is the strongest and largest gravitational field discovered in the universe till date. What is the maximum speed An object accelerating in these fields can gain?

It all depends on your frame of reference, but the biggest speeds caused by gravity will involve black holes or perhaps neutron stars.

From the point of view of a far external observer, an infalling object (from infinity) reaches a maximum inward speed of $0.385c$ (where $c$ is the speed of light) towards the black hole at 3 times the Schwarzschild radius. Thereafter, the object will appear to slow down and tend towards being stationary at the event horizon.

From the point of view of a "shell observer", that is somehow held stationary in the gravitational field of the black hole, the infalling object would have a speed that tended towards $c$ as the shell observer tended towards the event horizon at the Schwarzschild radius. It is meaningless to try to extend this inside the event horizon since (stationary) shell observers cannot exist inside the event horizon.


Relationship between Color Index and Temperature Is there an exact relationship between color indices and temperature?
I have seen some numerical formulas relating B - V color index to temperature, of the form $B - V = a_1 + \frac{a_2}{T} $ .

I am wondering is there any way to derive these formulas or are they completely empirical?

They are mostly empirical. Found by measuring the $B-V$ for stars of known $T_{\rm eff}$ (which are in turn measured by knowing the luminosity and radius of a star, and this is only known for a small number of stars). The relationships also depend on stellar surface gravity and composition.

An alternative approach is to derive "synthetic" relationships by folding spectra calculated from model atmospheres through the appropriate filter responses.