Does the sky get darker faster during the winter? When one lives away from the equator, say at around 37° latitude, does the sky get darker faster after the sun sets in the winter as compared to the summer?  Why is this the case?  I believe that if you were to draw the path the sun traverses at different times of the year, the paths are parallel to each other but shifted, such that in summertime, the path is longer with a shorter section below the horizon and in winter it is shorter with a longer section below the horizon. I imagine that the angle of the path relative to the horizon is approximately the complement of your latitude.

The further north you go, the time between sunset and darkness becomes longer, no matter the season. The reason is due to the velocity at that latitude. If there is 10 min of twilight on the equator, then there is $10\sqrt{2}$ min at 45° latitude, 20 min at 60° latitude, ...

Added:
I used the website that @barrycarter listed above and discovered that there are 2 maxima and minima during the year. The solstices have the longest twilight times and the equinoxes have the shortest twilight times.


What’s the object between the Earth and The Sun currently showing in Google maps? If I select satellite imagery in Google maps and zoom out, the Earth is shown in space. It shows light/dark regions correctly updated, but there’s an object between the Earth and the sun. It looks too small to be The Moon.

Is it real? What is it?

That's the Moon alright, and it's definitely real and definitely there. If you go outside and look at the Sun right now, the Moon will be almost but not quite on top of it, though it's impossible to make out due to the Sun's glare. If it were any closer we would have had a solar eclipse around the time of the New Moon that took place 20 minutes ago. Two weeks from now at Full Moon, it'll be in even better alignment, though on the opposite side of the Earth from the Sun this time, which will result in a total lunar eclipse. Mark your calendar!


What will the mass of the new galaxy be? When the Milky Way galaxy collides with the Andromeda galaxy, what will the mass and volume of the new galaxy be?

The mass will be slightly less than the combine masses of the 2 galaxies since some of the stars will be hurled away.

Since the disks of the galaxies are at an angle to each other, the volume would be (roughly) the volume of the 2 galaxies as they collide. Eventually, the volume will decrease (I'm guessing to between 60 to 70%) as the stars adapt to their new environment and the central massive black holes combine.


How can the next supermoon be analytically predicted? The lunar motion can be predicted with basic celestial mechanics, but the perigee and apogee are not always the same, basically because the attraction of the Sun makes some oscilations in the semi-axis of the orbit (https://en.wikipedia.org/wiki/Perturbation_(astronomy)).

My question is: how do they know when the full moon will get as close as it got in the recent supermoon? For example, the Dailypost article Biggest Supermoon in 68 years will light up the November sky claims that the full moon won't come this close to Earth until November 25th, 2034.
How do they kow that? Are they using the perturbation theory equations to predict this? If so, can anybody give me some source or explanation of how it's done analytically?

A slightly more useful answer than my comment above:

Using http://wgc.jpl.nasa.gov:8080/webgeocalc/#NewCalculation "Angular
Separation Finder" feature, we can determine when the angle between
the Sun and the Moon is at a local maximum. It turns out this isn't
quite the definition of Full Moon, but it's very close. I couldn't
find a way to make webgeocalc compute full moon times exactly, though
I'm sure there is a way, and, of course, there are plenty of tables of
full moons online.

Starting at
http://wgc.jpl.nasa.gov:8080/webgeocalc/#AngularSeparationFinder
we fill out the form as so:

and click the "Calculate" button.

This yields a list of all "full" moons. Scroll down and click on the
"Save All Intervals" button:

We now want to find the moon's distance at all of the times we just
saved (ie, the times of the "full" moons). To do this, start at
http://wgc.jpl.nasa.gov:8080/webgeocalc/#StateVector and fill out as
follows:

NOTE: to get the list of intervals above, drag from the saved
intervals result window into the list of intervals window.

The results, which you can also download, tell you the moon's distance from the Earth at the times of the "full" moons we computed earlier:

I downloaded the results in Excel format, imported them into gnumeric
and sorted by radius (distance from Earth to Moon) and did a little
additional cleanup to get:

(the full spreadsheet is available at https://github.com/barrycarter/bcapps/blob/master/STACK/StateVectorResults.xls)

So, yes, tonight's full moon (yellow) is the closest we'll have until
2034-11-25 (red). It's also the 6th closest moon we'll have this
century.

To see other examples of geocalc in action, see my:

What is the format of the data from the JPL's HORIZONS system?
https://space.stackexchange.com/questions/18718/calculating-positions-of-moons-ascending-and-descending-nodes-from-ephemeris-fi

For more about how NASA calculates these values in the first place: Where can I find a set of data of the initial conditions of our solar system? which is also now linked to the general "how to compute positions" wiki answer: Where can I find/visualize planets/stars/moons/etc positions?


Can a planet have more that one ring? Here is a good example picture of what I mean:

If not, why is this so? I've heard often that in the creation of a planet more the rings have formed, they are most likely just all over the place before they settle to the formation that we know, so why not come to a stable position in such a way. Note that the picture is only an example, rings that horizontal to each other are included too.

Yes, you can have more than one ring. Saturn is an example of a planet with multiple rings on the same axis separated by divisions (gaps between each ring).

About rings that are not on the same axis (similar to your example picture) is addressed in this previous question.


Python 2D Jean Instability in Spiral Galaxy Simulation I have ~1000 snapshots of a spiral galaxy simulation (from Molecular cloud perturbations t=0 to t>1Gyr. What I need to do is determine whether any position at any of the snapshot times meets criteria for Jeans instability.

The simulation is 2D, and the snapshot script can read position, velocity and surface density ∑, but not temperature. The simulation also has no gas, instead the simulation was modeled by giving each individual star extra mass which corresponds to gas.
I looked online and several site state the Jeans mass in 2d equal to c$_{s}$$^4$ / G$^2$*∑. But I have no clue how to determine speed of sound c$_{s}$.
Also I am not really interested in Jeans mass, more so whether any position can collapse to make stars, i.e. satisfy Jeans instability.

First a warning, then a hack:

1) If you didn't run this simulation, if you are not a trained theoretical astronomer of undergraduate level or above, and if this simulation doesn't have the kind of data you need to answer the question you are asking you should not trust the results at all.

It takes years of study to understand the weird numerical effects (things that happen in the simulation because of the way it is set up that would never happen in reality), background physics and loads of assumptions and approximations that go into this kind of stuff.

We often say 'garbage in, garbage out'. The answers you get from this will probably not be very illuminating because they say more about the simulation than real space.

2) That said, if you are just doing this for fun or to learn the skills, GREAT! Keep playing! And a very basic assumption you can make in order to get some numbers out (that will be quite wrong) is that all gas in space is ~10K.


How to obtain a luminosity from an SED I have a spectral energy distribution (flux densities measured at several frequencies), and have fitted a 'greybody' model to these points.
I am confused about how I use this information to determine the luminosity of my source.
I think the integral of my fitted function gives me the total flux, but how is this related to the source's luminosity?

The luminosity is $4\pi d^2$ multiplied by the integral under your SED.
i.e. you cannot find the luminosity without knowing the distance $d$ to the source.
NB This assumes no extinction (unless that features in your SED model), that the radiation from the source is isotropic and that the source is not significantly redshifted or moving relativistically.


Is there an astronomy exam I can take? I am currently living in Bangkok and I love astronomy. However, I haven't been able to find an official astronomy exam that I can do to gain college credit. Could someone help me out please?

There is a Dantes Standardized Subjet Test (DSST) in astronomy. I have no idea if and how widely colleges give credit for these exams. Typically, you will start with physics and choose to specialize in astronomy. Therefore, most standardized exams are for physics which may have a handful of astronomy questions.


The destruction of the Universe by a bubble So I was watching a documentary on television about this and a number of theories that are proposed as to what the fate of the Universe will be:

"Big Crunch"
"Big Freeze"
"Big Rip"

I understand these, but there was also another scenario which I don't understand:

A bubble forms in the Universe and expands at the speed of light. It annihilates everything in the Universe that it comes into contact with, until eventually at a very, very distant time the Universe ceases to exist. Does anyone have any clue as to what they were talking about (because I don't)?

I think I know what they mean. They're talking about a false vacuum scenario.
A vacuum state is a state of lowest energy. It's thought that the vacuum of our universe is in a lowest-energy state and is stable, and so nothing special will happen to spacetime. However, if our universe is actually a false vacuum, then it could merely be metastable, and some tiny perturbation could cause it to fall into a lower energy state - a true vacuum, or else a false vacuum of lower energy. The "bubble" that would appear would be a region of this lower-energy vacuum that would expand across the universe without stopping.
Scientists aren't sure if our universe is a false vacuum. Here's a chart of the masses of the Higgs boson and the top quark:

The latest measurements suggest that their masses lie in a metastable region of the graph (which could be bad), although it's towards the stable end of the region (which is good). This vacuum catastrophe could still happen, but the odds aren't in its favor. I also suggest reading some of the excellent papers in the references section of the Wikipedia article. They're quite comprehensive.

Better explanation:
The universe can be thought of as space containing a variety of quantum fields. A quantum field might be best thought of as something that has a value at every point in space. Put together a bunch of these fields and you can describe particles, particle interaction, and all the matter and energy in a given region of space - in fact, in the whole universe!
Picture a region of space with absolutely nothing in it. Nothing. (I'm ignoring vacuum energy, even though I really should discuss it) That's a true vacuum. Remove all the fields, particles, and other interesting stuff from our universe and that's what you'll get. Nothing.
The false vacuum scenario happens if there isn't actually "nothing" but "something". If our "vacuum" has some extra energy that it shouldn't (again, I'm not talking about normal vacuum energy). In other words, there's something where there should be nothing.
In this end-of-the-universe scenario, the universe goes from having this something to having nothing - from a false vacuum to a true vacuum. The region of space actually having nothing starts at a certain region and expands outward at the speed of light. That's what the scientists were talking about.

For anyone who wants a further explanation and/or could tell how many mistakes I made there: Yes, I'm aware that there were mistakes in that incomplete explanation. I didn't properly explain vacuum energy or an energy state, nor did I discuss stability or even properly touch on quantum fields. But quite frankly, I don't think this explanation needs that baggage. Does it add to the richness of the concept? Yes. But is it confusing? Also yes. I don't think working that in would be productive.


How does the appearance of the analemma vary with latitude How does the appearance of an analemma vary with latitude?

This what I take to be a definition of an analemma: If the position of the sun is plotted or photographed at noon clock time (say UTC) every day from a position facing due south (in the northern hemisphere) it forms a figure of eight shape as in the picture.

Photographs usually show it with the long axis at an angle. My question is, assuming that the angle is related to the latitude of the camera, how does this angle vary as the camera is moved from the equator to the pole, and why?

Analemma is a diagram showing the deviation of the Sun from its mean motion in the sky, as viewed from a fixed location on the Earth. Note that it says nothing about time- it can be any fixed time. Analemmas created at different times of the day have slightly different shapes. For example, see the solar analemmas taken at the same place at different times.

Solar analemma at 0700 UTC by Anthony Ayiomamitis, from solar-center.stanford.edu

Solar analemma at 1400 UTC by Anthony Ayiomamitis, from solar-center.stanford.edu
The best way to understand the effect of latitude on the shape of the analemma is to consider the effect of earth's axial tilt on the shape of the analemma.

If the Earth’s orbital path was elliptical, but its axis was not tilted, the analemma curve would be oval shaped. At the Equator, this line would be a straight line spanning from left to right or West to East.
The 23.5-degree axial tilt of the Earth affects the Sun’s apparent position in the sky – as the year progresses and the Earth continues to spin at an tilted axis and orbit around the Sun, the Sun seems to move up and down (North-South) in the sky. This has the effect of generating the two loops of the figure 8.

You're correct that the tilt of the analemma varies with the latitude. Observers in the Northern hemisphere will obtain an analemma curve with the broader loop at the bottom. This reverses for observers in the Southern hemisphere, where the broader loop is on the top of the curve. At equator, the anlaemma lies on its side. At north pole, only the top of the analemma would be visible, while it is the opposite at the south pole.


Are there any measurements of other galaxies indicating the existence of exoplanets? Of course scientists assume there would also be exoplanets in other galaxies besides our own Milkyway.
Is it true that our telescopes are not good enough to see planets in other galaxies, nor by brightness changes of the stars nor by inclination changes of stars. If so could the total mass of those galaxies be an indication of the presence of exoplanets, or is dark matter a problem for that?

Yes, it is true that our telescopes are not big/strong enough to detect planets in other galaxies. All we know at the moment are some few planets (not even complete systems!) around some of the stars in our closest galactic neighbourhood.
But let me explain why this is so.

Far away objects are very faint and thus send out few light-particles our way. To compensate this, we must build bigger and bigger telescopes to catch more light-particles from those faint objects. And at the moment we are limited until the 30-40metre class telescopes become active in ~2025.

However there is a class of 'cosmic accidents' that allow telescopes to catch more light-particles. We call those accidents micro-lensing events, as they briefly concentrate the light of a background star around a planet's gravitational sphere of influence.
The result is a spike of light in our telescopes. The shape of this spike informs us about the characteristics of the lensing star and planet.

To put this into context with your question about other galaxies, I've plotted for you how far away we have discovered planets so far:

Here we have on the x-axis the distance to the planet-hosting star, and on the y-axis the declination (one of the celestial coordinates or directions).
In red I've painted the planets that we have discovered with microlensing, in blue are all other methods.
The microlensing discoveries lie all on one line around dec = -30°, because this is the direction of the galactic center, where the polish OGLE project is looking.
The famous US-satellite Kepler that uses the transit method has it's field of view around dec = +45°, and this is also the biggest blue bunch of discoveries.
The small bump in between is the fov of the french CoRoT-satellite that also did some transit-discoveries and pioneered the transit searches from space.

So what do we learn from this graph?
We see that we can discover planets on a massive scale up to ~2000pc away.
Further away it's only possible with luck (while OGLE also looks at quite an impressive number of stars at any given time) and lensing. Now to put this into context, our milky way galaxy has a diameter of ~40.000parsec, and the next galaxy, the Andromeda galaxy is 1 million parsec away.

So that's gonna be hard, even with microlensing, and even harder with the other techniques you've mentioned.
Although there is one event from 2009 claiming to have found something that looks like a planet in their lensing-curve, I don't think this technique is mature enough to announce that extra-galactic planet searches.


"Juxtapositions" of moon,stars and planets in the sky, What are "Juxtapositions" of the Moon, stars and planets called? When the Moon, stars and planets take exactly the same position after few or many years, how to describe the positions and what is called to describe the position?

There are several words to describe two celestial bodies being in the same place on the celestial sphere.

A conjunction is when two bodies seem to pass close to one another, as seen from the Earth.

A transit is when a celestial  body passes between the observer (us) and a bigger body. For instance, Venus can transit in front of the Sun, when it passes between the Earth and the Sun.

An occultation is when a celestial body in the foreground hides a celestial body in the background. For instance, the Moon occults the stars it passes in front of.

An eclipse designates what happens when one body passes in front of the other for an observer, as a result of a transit or an occultation.

A syzygy is simply when three celestial bodies are roughly aligned. One of those three bodies can be the Earth, in which case, the two other bodies will either be in the same place on the celestial sphere (a conjunction) or at the antipodes.


Simple life on asteroids If there were an asteroid orbiting its star within the habitable zone and this asteroid contained liquid water, would it be possible for simple life such as bacteria to form?

If you take a slowly rotating asteroid with a large basin of trapped water, some solved minerals, and may be some gases, close to the asteroid's surface, to get an environment similar to a black smoker by the heat of the sun, I wouldn't call it totally impossible.

The formation of such an asteroid is close to the limits of my imagination, might be as a remnant of a comet with a crust of dust baked together to a water-proof surface layer.

But that's pure speculation at the moment. May be we can learn more this year with Rosetta.

Some biochemistry details: By the rotating asteroid part of the water undergoes a heating/cooling cycle. Provided there are enough prebiotic molecules, by this heating/cooling cycle kind of polymerase chain reaction (PCR) could multiply short occasional RNA or DNA fragments (nucleotide sequences). The difficult part is providing polymerase. It has originally been isolated from a bacterium living in hot springs, not quite dissimilar to black smokers. Information storage and replication (e.g. in RNA/DNA) is one important ingredient for life.

Polymerase is an enzyme, mainly consisting of amino acids. Glycine is the first amino acid found in cometary dust in 2009. That's a first step, but there is a big gap. As an enzyme polymerase is a protein, hence can be encoded by RNA or DNA.

To decode RNA to a proteine, kind of ribosome is needed. A ribosome itself consists of RNA and protein (chain of amino acids). That's the basics of the replication mechanism.

Now we need a containment.
The simplest form of a cell-like structure is a micelle. Organic molecules with a polar group can form micelles.
This paper contains a list of organic molecules found in interstellar medium.
Some of them could be used to synthesize appropriate molecules, but it's also quite a gap in this case.

A long way from what we know of extraterrestrial organic molecules to a bacterium.


Edu-Science telescope I don't know much about astronomy, but my kids are asking for telescopes to view planets, stars etc. Could anyone tell me if this Eduscience telescope is any good?

  This kit contains a 40x with a 41cm body, 20x 30x 40x eyepiece, a diagonal mirror, 30mm diameter objective lens and tripod.

  Size H51, W44cm.

Someone else posted a question about this very scope not long ago. Simply put, no, it's little more than a cheap toy and a waste of money.  The optical quality is going to be extremely poor.  In my opinion, it is more likely to dis-interest someone in astronomy.

You didn't mention how old your kids are.  Not knowing, it's hard to make a clear recommendation.  Still, I wouldn't recommend this.

If your kids were interested in music and wanted to learn to play guitar, would you buy them the cheapest, lowest-quality instrument available, one that has poor sound, is difficult to tune and then doesn't stay in tune, and with such poor intonation that even when the strings are open-tuned, you can't produce proper chords?  You wouldn't necessarily want to rush out and get them a professional-quality recording-ready instrument, but you wouldn't get them a toy either (at least, I hope not).

I'd say the same thing about a telescope.  You want a real telescope, with quality optics.  I'm sure you don't want to pay a lot, but you don't want to get something that will be so frustrating as to make them lose interest.

Of course, like playing a guitar, unless you know what to do with a telescope, it will be frustrating.  Too many people buy telescopes with the belief they'll see things like they see in pictures on the internet, only to be badly disappointed.  I've seen so many telescopes put up for sale on Craigslist and eBay or thrown in a closet or garage and ignored because the user had no idea what they were doing and were completely disappointed.

Depending on your kids age, my first recommendation would be to find and join a local astronomy club. The club I belong to, the Houston Astronomical Society, does regular star parties for schools and children.  Many of our members join because of their kids - because they want to be able to teach their kids.  We also have many adolescent and teen members who are active in our club.  Membership is inexpensive (most clubs are $50 USD or less per year) and you can learn a lot.

As for a telescope itself,  I would ignore ANY AND ALL telescopes under \$1,000 USD that are computerized, what we call GoTo technology.  While some people might disagree, my experience, which is fairly extensive, with such telescopes is that they are poorly manufactured and prone to breaking down - at which point they usually are not usable in a "manual" mode.  You also end up paying a lot for the electronics and mechanical parts, and getting less telescope.  On top of that, the majority of those telescopes still require you know your way around the sky to be able to find key landmarks (usually a number of bright, named stars) to set up the telescope.

My general recommendation is an 8" Dobsonian telescope.  A Dobsonian telescope is a Newtonian OTA (Optical Tube Assembly, the telescope-proper portion of the full instrument) with a simple altitude-azimuth rocker-box mount (meaning the motion is up/down and left/right, a natural motion for most people).  Dobsonians or Dobs, offer the best ratio of price to aperture, and are very easy to use.  However, an 8" Dob may be too much for smaller children, may be more than you're willing to spend, and may be larger than you're comfortable with. However, if you can, an 8" aperture is a very good starting place for an amateur telescope as it offers a good balance between cost, physical size, and optical capability.  Most 8" Dobs weigh a total of about 50 lbs, evenly divided between the OTA and base, making the heaviest single piece around 25 lbs, with an overall total height/length of about 55 inches, give or take a couple.  If you live in a light polluted area, they typically fit comfortably on the back seat of a midsize sedan or larger, and often will fit in a compact (though not something like a Smart car).  Brand new, you can sometimes find them under \$400, but usually you can expect between \$400 and \$500 with a decent starting accessory package.  I personally recommend the Apertura AD8, though the Orion 8" Skyline and XT8 are also decent models.

If price and/or Size is an issue, the next step down would be a 6" Dobsonian.  The length/height is about the same, but they're somewhat thinner and a little lighter.  The smaller aperture reduces light gathering ability, which reduces overall capability. Brand new, the price runs under \$300 for a decent instrument, but they can often be found used or refurbished for closer to \$200.

Smaller still, I'd consider the AWB OneSky 130 (known outside of the US as the SkyWatcher Heritage 130p).  This is a 130 mm (5") Newtonian telescope on a simple alt-az mount (often referred to as a Dobsonian, but not a true Dob).  The price is around \$200.  Another option at the same size is the Meade LightBridge Mini 130, which is similar in size and capability.  I've seen this closer to \$175 lately.

Stepping down further, there are a handful of 114 mm (4.5") scopes, including the Meade LightBridge 114 and the Orion StarBlast 4.5"  Again, you are stepping downward in capability, but also price, and these smaller scopes are much lighter and easy to transport, though you Won't see as much with them.

I would not recommend ANY telescopes under this.  There are many available, but most are of poor quality or design.  Under \$150-ish, I'd recommend a decent pair of 7x50 or 10x70 binolculars.

Good luck!


Is there a 'South Star'? The star Polaris is commonly known as the Northern Star because it always points North. This being said is there a South star that always points south?

There is a "South Star" called Sigma Octanis located in the constellation Octans, but it is so dim that virtually nobody calls it the south star.

Just another side note: Polaris doesn't always point North. The reason is that our Earth has a precession effect which make the North pole circulate around the sky. This picture illustrates my point.

EDIT: Just to clarify, the picture is showing the circulation of the south celestial pole around the south ecliptic pole.

EDIT2 : References: http://en.wikipedia.org/wiki/Pole_star#mediaviewer/File:Precession_S.gif
http://en.wikipedia.org/wiki/Pole_star


Would our galaxy appear as a spiral when seen from the Andromeda galaxy? I'm wondering how our galaxy is tilted with respect to the Andromeda galaxy.

Seen from the Andromeda galaxy, what would our galaxy look like?

Would it appear as a spiral, or would it be seen edge-on, and appear long and thin?

The "Galactic coordinates" of the Andromeda galaxy are l=121 degrees, b=-22 degrees.

That means the Andromeda galaxy is seen by us at an angle of 22 degrees to the plane of the Milky Way galaxy. Thus if you were at the Andromeda galaxy looking at the Milky Way, it would be neither face-on or edge on, but a bit closer to the latter than the former. You would still be able to see that the Milky Way had a spiral structure, because the plane of the Milky Way is much thinner than its diameter.

NB. This simple argument works because the Andromeda galaxy is more than ten times further away than the diameter of the Milky Way galaxy.


Are the Venusian "continents" likely to have existed before the global resurfacing event? A topographic map of the surface of Venus shows the large highland areas Ishtar Terra, Aphrodite Terra and Lada Terra that have occasionally been described as "continents", plus various other smaller regions such as Beta Regio. Are these "continents" likely to have existed before the global resurfacing event, or is the topography of the planet likely to have been completely altered?

Ishtar Terra belongs to the tessera type of terrain, "one of the most tectonically deformed types of terrain on Venus", representing ~7.3 % of Venus' surface (Ivanov & Head 2011, Global geological map of Venus). Tessera are "dated" (relatively, from stratigraphic relationships) from the Fortunian, the oldest period:

The most important features of the Fortunian are large tessera-bearing
crustal plateaus (e.g., Fortuna, Ovda, etc.) that form a specific
class of first-order (a few thousands of kilometers across) highs on
Venus [...] The lowest stratigraphic position of tessera and its
relative abundance show that this unit is of extreme importance as a
probable ‘window’ into the geological past of Venus.

Stratigraphy of defined and mapped geologic units on Venus. The
visible geologic history of Venus consists of the Fortunian,
Guineverian, and Atlian Periods. The major portion of the surface of
Venus (about 70%) was resurfaced during the Guineverian Period. See
text for discussion. T is the mean model absolute age of the surface

Further works by Ivanov & Head detail the history of volcanism (2013) and tectonism (2015) on Venus. They consider tessera themselves as a resurfacing process:

The globally observed stratigraphic relationships among the tectonic and volcanic units that make up the absolute majority (~96%) of the surface of Venus divide the observable portion of its geologic history into three different episodes, each with a specific style of resurfacing. These are as follows: (1) Global tectonic regime, when tectonic resurfacing dominated. Exposed occurrences of these units comprise about 20% of the surface of Venus. (2) Global volcanic regime, when volcanism was the most important process of resurfacing and resurfaced about 60% of Venus. (3) Network rifting-volcanism regime, when both tectonic and volcanic activities were about equally important. During this regime, about 14% of the surface of Venus was modified.

So, to summarize, "continent-like" (tessera) terrains like Ishtar Terra have existed for a long (yet unknown) time, and have survived the resurfacing from volcanism that covers most of the surface today. But they are themselves considered to represent a (tectonic) resurfacing process that has affected older (and unknown) pre-Fortunian terrains.

A global correlation chart that shows the three major regimes of
resurfacing on Venus: global tectonic regime, global volcanic regime,
and Network-rifting-volcanism regime.


Do all objects get larger in an "expanding universe"? I'm envisioning an analogy of the universe as a plane with marbles in that plane to represent different planets.

Now, if this plane expands how does this work? I assume that it is not like 'nothing' becomes 'something' then but rather expanding/stretching the universe; but doesn't this imply that everything around us is actually stretching/becoming bigger (like every atom/quarks) becomes a little bit bigger? Or is it just the space between those subatomic particles and combinations of these become bigger?

An example could be the structure of carbon dioxide becoming larger, but only because there is more space between the carbon and the oxygen atoms.

It would, if there were no interactions.

However, everyday matter is held together by other kinds of forces (electromagnetic, mainly). These are quite strong, and any "stretching" of space will be counteracted by these forces "pulling" the atom/molecule/object back to its original size.

So the expansion of the universe only matters on the larger, cosmic scale, between objects like galaxies which are separated by a large enough distance that they don't affect each other too much, gravitationally speaking.


Why doesn't the sun pull the moon away from earth? If the suns gravitational pull is strong enough to hold much larger masses in place (all the planets) and at much greater distances (all planets further away from the sun then earth) why does it not pull the moon away from earth?

Why doesn't the sun pull the moon away from earth?

Short answer: Because the Moon is much closer to the Earth than it is to the Sun. This means the gravitational acceleration of the Earth toward the Sun is almost the same as is the gravitational acceleration of the Moon toward the Sun.

The Moon's acceleration toward the Sun, $-GM_\odot\frac{\boldsymbol R+\boldsymbol r}{||\boldsymbol R+\boldsymbol r||^3}$ is indeed about twice that of the Moon toward the Earth, $-GM_\oplus\frac{\boldsymbol r}{||\boldsymbol r||^3}$. This is irrelevant. What is relevant is the Moon's earthward acceleration due to gravitation compared to the difference between the Moon's and Earth's sunward gravitational acceleration,
$$\boldsymbol a_{\odot,\text{rel}} = -GM_{\odot}\left(\frac{\boldsymbol R + \boldsymbol r}{||\boldsymbol R + \boldsymbol r||^3} - \frac{\boldsymbol R}{||\boldsymbol R||^3}\right)$$
This relative acceleration toward the Sun is a small perturbation (less than 1/87th in magnitude) on the Moon's gravitational acceleration toward the Earth. Given the current circumstances, the Sun can't pull the Moon away from the Earth.

Longer answer:

The gravitational force exerted by the Sun on the Moon is more twice that exerted by the Earth on the Moon. So why do we say the Moon orbits the Earth? This has two answers. One is that "orbit" is not a mutually exclusive term. Just because Moon orbits the Earth (and it does) does not mean that it doesn't also orbit the Sun (or the Milky Way, for that matter). It does.

The other answer is that gravitational force as-is is not a good metric. The gravitational force from the Sun and Earth are equal at a distance of about 260000 km from the Earth. The short-term and long-term behaviors of an object orbiting the Earth at 270000 km are essentially the same as those of an object orbiting the Earth at 250000 km. That 260000 km where the gravitational forces from the Sun and Earth are equal in magnitude is effectively meaningless.

A better metric is the distance at which an orbit remain stable for a long, long, long time. In the two body problem, orbits at any distance are stable so long as the total mechanical energy is negative. This is no longer the case in the multi-body problem. The Hill sphere is a somewhat reasonable metric in the three body problem.

The Hill sphere is an approximation of a much more complex shape, and this complex shape doesn't capture long-term dynamics. An object that is orbiting circularly at (for example) 2/3 of the Hill sphere radius won't remain in a circular orbit for long. Its orbit will instead become rather convoluted, sometimes dipping as close to 1/3 of the Hill sphere radius from the planet, other times moving slightly outside the Hill sphere. The object escapes the gravitational clutches of the planet if one of those excursions beyond the Hill sphere occurs near the L1 or L2 Lagrange point.

In the N-body problem (for example, the Sun plus the Earth plus Venus, Jupiter, and all of the other planets), the Hill sphere remains a reasonably good metric, but it needs to be scaled down a bit. For an object in a prograde orbit such as the Moon, the object's orbit remains stable for a very long period of time so long as the orbital radius is less than 1/2 (and maybe 1/3) of the Hill sphere radius.

The Moon's orbit about the Earth is currently about 1/4 of the Earth's Hill sphere radius. That's well within even the most conservative bound. The Moon has been orbiting the Earth for 4.5 billion years, and will continue to do so for a few more billions of years into the future.


Do we know of any tidally-locked planets with atmospheres? If not, is there any reason such a planet couldn't exist?

I ask only because that planet would literally have a twilight zone and I want to know that that's a thing somewhere.

As we haven't measured the atmosphere of many rocky planets yet, there aren't many examples. But I found 55 Cancri e:
It is a super-earth with very close tidally locked orbit and an atmosphere that has been measured by Hubble. A year on the exoplanet lasts for only 18 hours and temperatures on the surface are thought to reach around 2000 degrees Celsius. The detected atmosphere seems to consists mainly of hydrogen and helium. So you do have a constant Twilight Zone, just a rather hot one.

Sources:
- Beeing tidally locked
- Detection of the Atmosphere


What's the name of outside the cosmic horizon? The word we use to refer to what's inside the cosmic horizon is the 'universe', so what would you call the empty space outside of the horizon?

The outside is also the universe, the inside is just the observable universe.


Does time also pass more slowly for a galaxy that is traveling at relativistic speeds, where the speed difference is due to Hubble expansion? If time passes more slowly, relative to Earth, for a traveler at relativistic speeds, say at .8c, traveling between stars inside this galaxy, does time also pass more slowly for a distant galaxy that is moving away at relativistic speeds, where the speed difference is due to the Hubble expansion of space?

Yes, but it has little to do with the Special Relativistic time dilation of a moving body - in the sense that recession velocity is due to the expansion of space between galaxies rather than their relative motions.
The cosmological expansion leads to a redshift $z$ and also to time dilation by a factor $(1+z)$. i.e. Those distant clocks appear to run slower to us by a factor $(1+z)$. And vice versa - our clocks would appear to run slow from the perspective of an observer in a distant galaxy.
The time dilation effect has been observed in the light curves of high redshift supernovae (Blondin et al. 2008) and the durations of gamma ray bursts (Zhang et al. 2013) and is consistent with this prediction. This is an important observation since it is very difficult to account for this time dilation effect using a "tired light"  explanation.


Are neutrinos affected by gravity fields? Am asking if a neutrino can be pull by the gravity field of a quasar or a black hole, assuming the neutrino have momentum. It could be possible in the that a neutrino can orbit a black hole at an orbit equivalent to the Schwarzschild radius give or take. If it even possible for a neutrino to go through a black hole?

If you buy into General Relativity (you need to throw out an awful lot of modern physics if you don't), gravity is a manifestation of the shape of spacetime around massive objects. It's not that they are being "pulled," but rather that they are just following the straightest path through spacetime that they can. Neutrinos would be the same as any other particles in that respect, and following anything other than this path would require some extraordinary gymnastics and external forces.


Can Mercury hit Earth or Mars in the next 5 billion years? Mercury is the most unstable planet of the solar system, with an oscillating orbital eccentricity (between 0 to 0.45). It seems there is a 1% probability in the next 5 billion years Jupiter and Mercury might enter in a 1:1 orbital resonance.

The ending variants for this event are :

colliding with Venus
colliding with the Sun
being ejected from solar system.

Can it hit other targets on its way out of the solar system beside those enumerated, like Earth, Mars or the outer planets?

Yes, after a close encounter with other planets, almost anything can happen, including a split of Mercury into smaller bodies by tidal forces, or capturing as one or more moons of one or more planets, or a sequence of captures by, and escapes from planets, conversion to a ring of all or some of the fragments, collisions with other moons, fragments falling into the Sun, others ejected from the solar system.


Are asteroid complexes and types the same? Very novice astronomer here. I keep seeing terms like "S-Complex" thrown around when talking about asteroids, but no matter how much Googling I do I can't figure out what that exactly means (every article just writes about it assuming the reader knows the definition).

Finally I looked up a free textbook and read the section on asteroids. In Composition and Classification it talks about there being three "types" of asteroids (another term I have seen thrown around): C-type, S-type, and M-type. (I've also seen them referred to as Type C, Type S, etc.)

So are the words "type" and "complex" just synonymous words to classify asteroids? If not, what does "_ -Complex" mean when talking about things like S-Complex asteroids?

Thanks!

Yes they are basically the same. The original asteroid taxonomies were defined from photometry (brightness) in a few filters such as the Eight Color Asteroid Survey of Tholen (1984). This resulted in asteroids with similar slopes being grouped together into the main complexes or classes. Once we got more detailed spectra of asteroids subtle differences between sub-members of e.g. the S-complex showed up and so the large complex was broken into more sub-classes such as Sc, Sq etc.


Will the nebula of Betelgeuse be visible to the naked eye? How bright, how large, how soon, for how long? When Betelgeuse goes core collapse supernova it will leave a supernova remnant. Will it become visible to the naked eye? If so, in what time frame will it be visible. Will it be star like from the beginning, or only after decades or centuries of expansion, to then fade away as it thins out?

The Crab Nebula seems to have had a progenitor of similar mass as Betelgeuse, also a core collapse supernova about 800 years ago, and is about ten times further away. Its supernova remnant seems to be about 300x400 arc seconds in size, so about a fifth of the Moon's diameter on the sky. But it is not visible to the naked eye.

If Betelgeuse's remnant becomes a pulsar, would light from its jets hitting gas be visible to the naked eye, and further out earlier than the supernova remnant itself?

There is an article here that describes the visible effects quite well. In essence, within a week or so, it would be comparable in brightness to the moon and therefore visible during the day.

  Betelgeuse would then start a phase of final, rapid dimming and again reach its current brightness level after possibly three years. After six years, it would be too faint to see with the naked eye. This would forever alter the visual appearance of Orion and we might need to think of another object the remaining constellation might represent.

This means that after 3 years, Orion would go back to "normal", and after three more years, we wouldn't see it at all with the naked eye - so no, not centuries or decades.

As for the pulsar question, I highly doubt that the jets would be visible... pulsars are very, very dim optically. Their pulses are best observed in the radio spectrum, so even if the emissions did hit gas clouds, the resulting "light" would likely be in the radio spectrum. It would also likely be spinning incredibly fast, since it would be young, meaning that even if there was some optical light from the jets hitting the gas, it would just appear constant to us given just how fast the pulses are - at least with the naked eye, as you asked.

Now, with other equipment than our eyes, we could likely get some sort of light curve or time vs. intensity profile which would make that pulsation apparent. This would also best be done in radio waves, which are the best bet when dealing with pulsars. (There are a few exceptions, as you mentioned with the Crab pulsar being visible in some other frequencies - see here for more details.)


How to interpret extreme ultraviolet/ x-ray data in the literature? I am interested in a sun-like star 138 light years away that is in the EUVE catalog. A table shows that a photon count of 2.6+/-0.5 photons per kilosecond at 100 angstroms has been detected for this star. I can not determine if this is high or not. I am even having a hard time finding what solar emission is at this wavelength so i can attempt the math to compare with the sun. Can somebody help me here?

The EUVE catalog gives you a count-rate in a certain EUV band. To convert this into a flux, you need a flux conversion factor to go from counts per second to energy flux at the Earth per second. This conversion factor, also often known as an "ECF", depends on the intrinsic spectrum of the source and also how much absorption (known as the hydrogen column density) there is between Earth and the source. So there is no one answer that can be given.
I
Your best port of call is the PIMMS flux conversion tool, which has an EUVE module and can do these flux conversions for you if you know something about the spectrum and the absorption. I'd try it for you but I couldn't immediately see how to do it for EUVE, though it claims the information is there...


Does the Earth have another moon? I was just wondering what are the chances that there is a small object (say less than 1 km but more than few meters) that orbits the Earth but has remained undetected by us? Are we actually scanning the space around the Earth continuously for orbiting bodies?

Not strictly satellites/moons, but certainly companions are 2010 TK7 with a diameter of ~300 m, an Earth trojan at the L4 point, and the ~5 km 3753 Cruithne in a peculiar orbit locked to the Earth's.


What is "g-98" in plots of stellar spectra? I was looking to find some example carbon star spectra and came across this page on the Harvard website which shows several plots of carbon star spectra. The horizontal axis is labelled as wavelength in ångströms, but the vertical axis appears to be labelled with "g-98" which I am not familiar with - any idea what quantity this is?

It looks like the floating "g-98" is just an extension of the header of the plot (made by IRAF), which has wrapped around the image for lack of room on the right-hand side. Looking at the N0 plot, we see that the header should read

NOAO/IRAF V2.10.4EXPORT perry@sparky.sao.arizona.edu Mon 06:04:51 03-Aug-98

which just contains information about the software, user and creation date behind the plot.

(Hat tip to Pela for confirming my guess!)


How to make projection from altitude and azimuth to screen with screen coordinate system? Thanks to @ohoh for helped translate ra and dec into alt and az. Now that I have this data(altitude and azimuth), how can I plot it in something like this, with the coordinate system going like this?

The example sky map appears to be a stereographic projection; circles of right ascension and declination are projected as circular arcs, and their intersections are projected as right angles.
The example has a 195° field of view, and the eastern and western horizons are mislabeled.

Let the celestial sphere be centered at the origin with radius 1, zenith at (0, 0, 1), northern horizon at (0, 1, 0), and eastern horizon at (1, 0, 0).
Assuming azimuth 0° is north:
$$\begin{align}
x &= \sin \mathrm{Az} \cos \mathrm{Alt} \\
y &= \cos \mathrm{Az} \cos \mathrm{Alt} \\
z &= \sin \mathrm{Alt}
\end{align}$$

To project from the nadir (0, 0, -1) onto the plane z = 0:
$$ (x', y') = \left( \frac{x}{z + 1}, \frac{y}{z + 1} \right) $$

To fit this between (0, 0) and (2R, 2R) on a canvas with northern horizon at (R, 0) and eastern horizon at (0, R), scale by -R and translate by (R, R):
$$ (X, Y) = (R(1 - x'), R(1 - y')) $$

For a sky map centered overhead, if north is up, east should be on the left.


Why does a mirror bent 'like a potato chip' allow space telescopes to be smaller and have a wider field of view? I was browsing NASA featured items and came across this - Out With the Old, In With the New: Telescope Mirrors Get New Shape

  Called freeform optics, this emerging mirror technology, brought about
  by advances in computer-controlled fabrication and testing, has
  triggered a sea change in optical engineering... the technology holds
  great promise for scientists who want to develop compact telescopes
  for CubeSat and other small satellites — an increasingly popular and
  cost-effective alternative to more traditional missions that are more
  expensive to build and launch.

Telescopes for cubesats?!

The article had this illustration, and I'm trying to figure out how that is converted into a clear image without distortions, and why they say it allows the telescope to be much more compact.

It was a short article that didn't even try to explain this, and when I went looking for other articles on the topic for non-experts I found zip. Perhaps this is very challenging to explain, but I thought I'd ask anyhow. I told an enthusiastic youth just a few weeks ago that it wasn't possible to launch a telescope capable of anything useful on a cubesat. Oops...

Freeform optics are a response to the specific challenge of cramming a telescope in a very limited space. A traditional instrument would have all optics symmetrical and aligned on the same axis. It would waste a lot of space within the cubesat. Also, traditional designs tend to be much longer than they are wider; they don't fit well in a cube; it is very hard to make classic instruments that are as short as they are wide.

But with freeform optics you could bounce light in a few directions within the cube. You'd still achieve a decent focal length, and you would use all the volume available to you.

Since light is reflected from mirrors at angles different from normal, you cannot use the traditional symmetric shapes such as parabolic, spherical, etc. You need to basically take a paraboloid and squish it in one direction so that it works about the same like a parabolic mirror (I'm simplifying), but at an angle of reflection of, say, 45 degrees.

In such an instrument you could have multiple "potato chip" mirrors, as in the diagram above. You have to design the instrument as a whole; computer simulations will adjust the shape of each mirror until the performance of the whole instrument is close to a classic straight design.

As far as I can tell, the manufacturing precision is such that freeform optics are only usable at long wavelengths such as infrared, where less precise optics can be used. But technology improves all the time. It also depends on how much aberration you can tolerate in your image.

For usage from ground level this is less useful, unless you absolutely need a telescope in a very small form factor for some reason. Classic optics are still preferred when space and shape are not restricted.


How do we know so many details about Big Bang, but we do not know if Universe is finite/infinite? I understand that this question may be too vague, but I am confused about the following:

On one hand we know a lot of "details" about Big Bang, like really precise details, evolution of universe, age of universe...

On the other hand we do not know size of Universe, or even if it is finite or not.

This seems quite weird to me, since even if we may never interact with Universe that is not observable those parts still interact with parts of our Observable Universe.

Some examples:

Age of universe: 13.787±0.020 billion years (±0.15%) determined several different ways
Hubble constant: roughly 69 to 74 km/s/Mpc (±1.6%) different measurements disagree with each other
Size of universe: unknown

By observing the observable universe we can gather a lot of data about the constituents of the universe now and in previous epochs, right back to the cosmic microwave background. This data places quite narrow constraints on viable models of the Big Bang. This is why cosmologists are confident about the details of the Big Bang, certainly from electroweak symmetry breaking onwards.

However, it is much more difficult to make an observation of the observable universe that could distinguish between a very, very large but finite universe and an infinite universe. That is why this is still an open question.


When do radio and tv signals become indistinguishable from background noise of the universe? I understand the rate an EM signal broadcast uniformly from the Earth will decrease in its power is governed by the inverse square law. How far from Earth will radio and tv signals become indistinguishable from background noise from varied sources in the galaxy, even CMB.

I know nothing about the type of equipment that would be used to gather and analyze such a signal.  I don't know what the geometry of the observation means.  Assuming the optimal equipment and conditions we currently have and no other line-of-sight sources.

If an extraterrestrial civilization has a SETI project similar to our own, could they detect signals from Earth?
  In general, no. Most earthly transmissions are too weak to be found by equipment similar to ours at the distance of even the nearest star. But there are some important exceptions. High-powered radars and the Arecibo broadcast of 1974 (which lasted for only three minutes) could be detected at distances of tens to hundreds of light-years with a setup similar to our best SETI experiments.

SETI FAQ


"Ideal" primary thickness in a Dobson I plan to buy a 12" or 14" dobson, the choice depending on my final budget and on the total weight of the instrument. I have several options for the primary mirror. I tried to find a formula that would give the acceptable range for mirror thickness depending on the diameter, maybe also on the material (BK7, pyrex, ....). Is there such a formula, or a back-of-the-enveloppe calculation, which could help me? If not, what do people go with usually (for these diameters, I mean)?

Thank you for your help.

If you're buying the instrument, as opposed to making it from scratch, my advice is to worry about the overall quality, as opposed to a single parameter such as mirror thickness. It boils down to whether you trust the vendor or not.

In any case, there is no simple formula. You could run the PLOP simulator and try a few different scenarios, and see what the overall RMS error is in each case. But the values will vary very significantly. The main factors are:

the diameter / thickness ratio
the number of support points in the mirror cell
the relative positions of the support points

All of the above assumes a traditional mirror with a flat back, and a classic mirror cell with multiple-of-three support points. But the analysis gets even more complex if it's a cell mirror (the back is not flat), or a honeycomb structure (the back is flat but there are empty spaces in the volume of the mirror). And then there are conical mirrors, etc.

It's difficult to second-guess the vendor, especially if you've never built an instrument. Just read the reviews, ask other owners, try and gauge the overall build quality.


Kepler and the Sun position When Kepler went to Tycho Brahe, he learned that Tycho Brahe was using the mean position of the sun to note down the position of Mars. Well, Kepler needed to convert these datas of Brahe to do his own calculations for the orbit of Mars.

In this link, the difference between the real and mean position is shown.
http://www.keplersdiscovery.com/MeanSun.html
http://www.keplersdiscovery.com/CorrectedTable.html

But how did Kepler calculate the real position of the Sun? I don't get it. And due to my first language is not English it might be possible that I didn't understand some important aspects of the given links.

It depends what you mean by "real position".    Also, just to clarify something that should be gotten out of the way, the Earth rotates, obviously, and the night sky appears to rotate around the Earth a complete 360 degrees in a 24 hours.  That's why the geocentric model lasted as long as it did.  That's what it looks like, and that's what everyone believed prior to Copernicus, but that's not really relevant to your question.   I mention it, because what is relevant is that the stars are so far away that they appear fixed.   They rotate together, they don't move past each other so the stars operate as a fixed background against which the sun, moon and planets move.   So, when we talk about "movement" in the sky or position, we're talking about against the fixed stars in the background, not the rotation of the Earth.

After the fixed stars, the Sun is the simplest object in the sky, it moves a little bit against the background every day and it returns to the same place every year, the Sun's position against the fixed background stars was known long before Kepler, so Kepler always knew the position or direction of the Sun, even at night because the Sun held it's position against the background stars, moving only a little bit each day.   He didn't know the distance to the sun (more on that later), but he knew the position.

Kepler also knew that Mars' orbit around the sun was 687 days.   How did he know that?   Well (these words aren't my own, but it's discussed here)

  It's a simple matter to measure that the synodic period of Mars is
  about 780 days. This is the time between successive oppositions of
  Mars, meaning that Mars is due South at midnight. Given the
  heliocentric model, it is easy to see, as shown in the Wikipedia
  article, that 1/Synodic Period = 1/Earth's orbital period - 1/Mars
  orbital period. So, once the heliocentric model was understood, it all
  becomes easy.

Kepler was a brilliant Mathematician.   Brahe wasn't so much.   Now, I'm not saying Brahe was dumb, cause that's not accurate either, but he wasn't skilled in math the way Kepler was.   Brahe was very patient and dedicated and he made the most accurate observations to date, some 10 times more accurate than anyone before him.   He had the best equipment for making accurate observations in history, so to a person like Kepler, Brahe's notes were pure gold, containing decades of great accuracy.    There was even a rumor that Kepler killed Brahe to get access to his observatory and his work, but most people think that's just a rumor, that it didn't actually happen, but there was enough uncertainty that Brahe's body was exhumed and tested for poison.  (none was found).    But I digress.

A 3rd bit of information is that by using parallax and the maximum angle that Mars varies in the sky, you can get a rough estimate of the distance to the planets relative to each other.   Copernicus did that so Kepler didn't have to.   A bit more on that here.  and here.

So Kepler had a lot of information at his fingertips when he got access to Brahe's notes.   He knew that Mars was about 1.5 times as far from the Sun as the Earth (from Copernicus) and he knew that Mars orbit was 687 days (not difficult given the heliocentric model), which means, every year and 322 days, Mars is in the same spot relative to the Sun, but the Earth has moved. - So you have 2 still objects and one moving object that's some "43 days" earlier in it's orbit each time Mars is in the same place and by taking that measurement every 687 days, you can learn a lot about the relative position of Mars, Earth and the Sun.   And that's what Kepler did.

His two major breakthroughs (in my opinion) were observing that Mars wasn't on the same plane as the Earth (He did that observing Copernicus' observations, not Brahe's - I read that, I can't find the notation right now).    And his other breakthrough was to figure out that the Sun wasn't at the center of the Earth's orbit.   He did that by measuring the angle between the Earth and Sun and Earth and Mars every 687 days.   More on that here.

His measurements telling him that the sun wasn't at the center of the Earth's orbit led him to his ellipse model.   After that it was all clarification and more detailed measurement that led him to his 3 laws, but I think, recognizing that the Sun wasn't at the center of the Earth's orbit was his big "aha" moment that and his discovery of Mars being on a different orbital plane.   Kepler's measurement was quite accurate.   Mars' inclination to Earth is 1.87 degrees.   Kepler had it measured at 1 degree, 50 minutes, or 1.83 degrees.


Is there any definitive proof of stars being "born" in nebula formations? Is there any absolute evidence of star formation inside of nebulae (AKA stellar nurseries) as is suggested and even stated as fact by many textbooks?

In science there is no such thing as an irrefutable fact, there is just the best theory we currently have that explains all the data and makes sense. That is very different from an assumption.

A theory is something that you can test - make models and predictions and check that they match up to real life. An assumption is something that you don't have evidence for but you think it should make sense, so you choose to believe it.

Our current understanding of star formation and stellar nurseries is an excellent theory - it has been around for a long time, built up of many parts that have each been tested and confirmed, it matches observations in many ways. It's not perfect and there are still things we don't know (otherwise scientists would have no jobs) but in high school level textbooks we can reasonably present this as a 'fact' because we are pretty certain it won't change any time in the next 50 years.

An assumption would be like me assuming you are reading a high school textbook. It fits the facts (the level of detail in your question) but I haven't tested it (asked you about the book).


Is blue light not traveling faster then red light in space? I just can't understand why light and other massless particles are not traveling at different speeds?

Gravity effects anything with a mass but light has no mass but a black hole, and weak lensing changes the trajectory of light to create more red shifted light more than blue shifted light. That tells me light has mass. Can someone clarify this to me in Laymons terminology?
Pictures would be helpful.

Photons are massless. This doesn't depend on their energy, so doesn't depend on their frequency or wavelength.

Massless particles travel at the speed of light.

Even if we abandon particles and look at classical electrodynamics, we find that the speed of an electromagnetic wave (in vacuum) has a fixed value. It doesn't depend on wavelength.

Gravitational lensing affects light of all wavelengths equally. Lensing is caused by light travelling on null geodesics - basically the "straightest line" it can follow in space-time that has been curved by the presence of mass.

EDIT (Not required for any understanding of my answer, which stands as is, but to address some comments below).

The gravitational bending of light must take into account the curvature of space-time. If that is not done, then one gets the "Newtonian" formula which is half the deflection of a correct GR treatment. For particles with mass (which isn't what this question is about), the appropriate formula for their deflection will depend on their speed. At low speeds, the Newtonian formula will work just fine. At high speeds (approaching $c$) the deflection asymptotically approaches that for a light beam (i.e. twice the Newtonian value and the same as light). A mathematical proof of the latter in the Schwarzschild and Kerr metrics is provided by Barrabes & Hogan 2004 (see their eqn 4.5). The reason, as discussed by Misner, Thorne & Wheeler (in "Gravitation", 2017 Princeton University Press, chapter 1, section 1.6 - freely available from the linked site) is that gravitation is not just due to the curvature of space, but the curvature of space-time. The tracks of light and (non-relativistic) particles follow distinctly different paths, with different curvatures, through space; but their tracks through space-time have the same curvature.


Sources of Turbulence in the ISM What sources of turbulence exist within the Interstellar medium (ISM)? Which ones are physically the most important for newly forming stellar systems?

Turbulence sources:

There are numerous sources of turbulence in the interstellar medium, at all scales:

at large scales, there is the shear from galactic rotation. One way to sustain turbulence and to couple large and small scales would be the magnetorotational instability (MRI).
at large scales, gravitational instabilities can also play a significant role, through spiral structures.
outflows and jets from forming stars play an important role, releasing a lot of energy in the ISM.
in star forming regions, massive stars are also important. Radiation and stellar winds from massive stars are an important input of energy in the ISM. And eventually, the most massive ones will blow up in supernovæ, releasing even more energy.

Therefore, one could then regards separatly three processes related to massive stars:

stellar winds
ionizing radiation
supernova explosion

Importance for star formation:

They are all relevant to star formation, one way or another. One key property of turbulence is to cascade from large to small scales; therefore, even if you inject turbulence at large scales (galactic scale) you'll get turbulent motions down to the scale of a molecular cloud.

A nice illustration of the tubulent cascade is the Larson's relation (Larson 1981):

The Larson's relation shows the evolution of the velocity dispersion with the size of the structure you're looking at. Velocity dispersion is an indicator of turbulence. Indeed, these dispersions are non-thermal: knowing the typical temperature of the MIS (about 10 K), one can estimate the thermal velocity of, for example, the CO molecule ($v_th = \sqrt{2kT/\mu m_H}$, with $k$ the Boltzman constant, $T$ the temperature, $\mu$ the mean molecular wright and $m_H$ the hydrodgen atom mass) that is about 0.07 km s$^{-1}$. Measured velocity dispersions are of the order of 1 to 10 km s$^{-1}$, and these is interpreted as a turbulence signature (and estimate).

Details:

Energy rates: values are given (roughly) for the Milky Way

MRI: $ \small \dot e = 3 \times 10^{-29} {\rm erg\ cm}^{-3}\ {\rm s}^{-1} $;
Gravitational instabilities: $\small \dot e = 4 \times 10^{-29} {\rm erg\ cm}^{-3}\ {\rm s}^{-1}$;
Outflows: $\small \dot e = 2 \times 10^{-28} {\rm erg\ cm}^{-3}\ {\rm s}^{-1}$;
Ionizing radiation: $\small \dot e = 5 \times 10^{-29} {\rm erg\ cm}^{-3}\ {\rm s}^{-1}$
Supernovæ explosions: $\small \dot e = 3 \times 10^{-26} {\rm erg\ cm}^{-3}\ {\rm s}^{-1}$
Stellar winds: it strongly depends on the type of star: it varies as the power of -6 of the star's luminosity. It therefore ranges from an energy comparable to a supernova explosion (or even more for Wolf-Rayet stars) to almost nothing.

Sources:

Mac Low & Klessen 2004


Percent-illumination of crescent moon and its naked-eye visibility? What level of percent-illumination of moon (waxing crescent), given by Stellarium, is enough to make it visible with naked eye, in clear sky?

thanks

Any percentage at all would be visible, if the sky is dark enough. The surface on the illumined part is as bright as the street in front of your house during the day. So even a very tiny sliver would be clearly visible.

But you're running into a different problem here. As the illumined portion decreases, the apparent position of the Moon in the sky gets closer to the Sun. At some point it will be lost in the glare. It's a contrast issue.

Your best bet is a coincidence: if the Moon's apparent position is very close to the Sun's, while the Sun is setting, and you have clear visibility to a flat western horizon. In that case, the Moon's crescent could become very thin indeed, and still remain visible.

I've seen it like that, once. The Moon was very close to the Sun, several lunar diameters away. The Sun had just descended under the horizon. The Moon was very low in the sky, extremely thin, almost too thin to estimate the width of the illumined portion. The expression "razor edge" comes to mind. The thin curved line made less than half a circle (maybe 1/3?), with the two ends gradually tapering off in terms of brightness.

Due to the unusual shape, it took me a second or two to realize that that was actually the Moon. I showed it to my kids and their reaction was: "That's the Moon? NO! Wait... Really?" It was the Moon, really, confirmed with SkySafari on my phone.

It was not drawing attention to itself by any means, being almost lost in the sunset glare - not bright, just a thin, discreet, whitish curve in a yellow-red sky. But once you saw it, it was clearly there. It would have been invisible if the Sun was not below horizon, I think, or maybe not. Definitely invisible at mid-day, without any doubt.

I wish I had access to a telescope during that rare sight. :( Although, with the Moon at such low elevation, seeing must have been atrocious.


Radiation from Neutron Star impact? As I understand it, objects falling into an old/inactive/dead/non-pulsar neutron star would be ripped apart, form an accretion disc and eventually emit a burst of radiation as they fall onto the surface of the star.

If this happened to:

a comet
a dwarf planet
an ice giant

...roughly how much energy/radiation would each produce? Also, if possible, over what kind of time frame? I'm only looking for ballpark figures in all cases.

Back of the envelope? The energy released is about $GM/R$ per unit mass. For a 1.5 solar mass neutron star of radius 10 km, this amounts to $2\times 10^{16}$ J/kg, or approximately 2/9 of the rest mass energy of the accreted material.

Of order half of this would be radiated and half would go into heating the neutron star.

Timescales are more difficult. Starting when? Once the material gets to the innermost stable circular orbit at around 15 km, then the final 5km to the surface takes a tiny fraction of a second. But from where an object is shredded (much further out), much would depend on its angular momentum.


Is it possible to observe strong gravitational lensing with amateur telescopes? Strong gravitational lens systems like the Cosmic Horseshoe have been imaged by scientific space telescopes, but have any amateur astronomers accomplished this?
Or are amateur/small ground based telescopes not suited for this?

The cosmic horseshoe is beyond amateur instruments. It is a magnitude 20 object. In a large (2.5 m) professional telescope it looks like:

This image taken from the SDSS III data.  It is small (10'', half the size of Mars at opposition) but that is not insurmountable for amateur equipment. But is it very dim.  If an object like this was visible in moderate telescopes, it would have been found long ago.  It is really only the sky surveys, and computerised searches of the survey data that allows for such objects to be found.
Processsing the SDSS data can bring out more detail, as can be found on the discovery paper https://arxiv.org/pdf/0706.2326.pdf
While this isn't visible, another example of gravitational lensing can be: the Einstein cross is a gravitationally lensed quasar, and it is at the limits of amateur equipment.  It is even smaller, but as it is composed of four point-sources it can be seen. Skyhound recommends at least 18 in telescope, but practically 24'' to see the four components, exceptional dark skies, fully adapted eyes and perfect conditions (especially perfect seeing: this object is small). Your reward for this is "four smudges of light embedded in the galaxy" The SDSS view gives an indication of how hard this object is.


If Jupiter would stop rotating, would its atmosphere become homogeneous and single coloured? Looking at pictures of Jupiter, there are many different colours and patterns. I imagine that the rotation has much to do with this, since there are horizontal "bands" across the planet along the direction of rotation.

If the rotation would slow down and eventually stop, would the atmosphere get mixed and even out over time until we could only see a fuzzy brown (?) color but no color shifts or other patterns on the surface?

They're often called jet streams, but the visible bands you see in Jupiter are probably more like Hadley cells.  Convection currents that move heat from lower in the planet's surface to higher before cooling and circling back downward.   The darker bands are rising warmer atmosphere, the lighter bands, made lighter in color by more ice, are falling.

https://en.wikipedia.org/wiki/Hadley_cell#/media/File:Earth_Global_Circulation_-_en.svg

https://en.wikipedia.org/wiki/Atmosphere_of_Jupiter#Zones,_belts_and_jets

Jupiter's fast rotation gives it several Hadley cells per hemisphere.   By comparison, Earth has 3 and Venus has 1.   Faster rotation and perhaps Jupiter's larger size as well creates more Hadley cells.

That said, if Jupiter didn't rotate, it would still have heat from below, rising and escaping.   It would still shed temperature from below, because of internal heat.

What it might look like is a complicated question but I think a uniform brown is unlikely.   It would still have regions of darker, warmer rising air and lighter more ice filled, colder falling air.   Jupiter's bands aren't because it's atmosphere doesn't mix, it's because warm Jupiter air is a different color than cold more icy Jupiter air.

Also, no rotation effectively gives the planet a sun side and a night side.    That would vary some depending on no rotation at all, vs a slow tidally locked rotation, but even so, the Sun side would be warmer.    Jupiter's upper atmosphere is more driven by heat from below than heat from the Sun, but the Sun would still have an effect, especially if it only shone on one side of the planet.

I don't think we have a good sense of what non-rotating Jupiter type planet might look like.  The upper atmosphere temperature would be a primary factor, but with no rotation, I think it would probably require some high level mathematical modeling to work out an answer.

Some kind of grand current might be possible, perhaps with patterns perhaps similar to the "Y" shape observable in UV in Venus' atmosphere, or perhaps an ocean current like pattern, but that's just speculation.

https://www.space.com/28835-venus-y-shape-mystery-solved.html


Can we discern black holes with visible light? Sorry for my dumb terms. I'm interrested in astronomy but I have next to zero background.

Given that a black hole absorbs everything, including light, could we ever discern one with visible light (and live to pass the information on)?

This question contains one generated picture of a black hole and Google image shows visual representations that all look similar (black circle in the centre, light that's distorted around it and the further we go away from the center the background is less and less distorted), but are those actually representative? Is there really a black spot in our vision field at the place where the black hole is? Both the answer to the question and this page discards such images as "fantasy", but they don't really answer the questions I'm asking, which are:

Will we ever be able to discern a black hole with visible light and live?
If we can, what is a currently acceptable description of what we'll see with visible light? I don't especially expect a picture, so a textual description is also good, but keep in mind it's in regards to visible light. Other questions on Astronomy SE explain what we can "detect" using radio waves or other means of observation.
In 2014-2015 CERN said they might be creating micropscopic short-lived (10-27 seconds) black holes with the LHC (and this lead to a global nonsensical fear that it'd destroy the planet). Did they create any? If yes, was there any opportunity to picture them despite their tiny size and short life?

There are two "usual" ways to detect a black hole:

If it is part of a binary system (two stars closely rotating around each other) and dragging matter from the other star into an accretion disk, that disk will heat up. Energy distribution will tell us whether the accreting object is indeed a black hole or a neutron star instead. This is primarily visible in X-ray, so not an answer to your question.

If the black hole has a mass of 100.000 to millions of times that of the sun (supermassive black hole), it will sit in the centre of a galaxy, and sometimes accrete matter. This is mostly visible in Radio, and again in X-ray. Still not an answer.

So, how can we detect a black hole in the optical? Best chance is an indirect detection, like in the images you linked above. The one with the distorted stars in the background is not "just fantasy", but a simulation based on our understanding of gravity. The problem when we try to see this is scale, and hence resolution. The event horizon of a stellar mass black hole is on the order of a few kilometers.

Let's replace our sun by a black hole. At a distance of $1.5\cdot10^8$km, with a black hole diameter of 6km, we'd see it occupying an angle of
\begin{equation}
 \alpha = \tan{\left(\frac{6}{1.5\cdot10^8}\right)} = \frac{6}{1.5\cdot10^8} = 2.3^{\circ}\cdot10^{-6} = 8''.3 \cdot 10^{-3}\,.
\end{equation}
So a solar mass black hole at the position of the sun would be 8 milli-arcseconds (mas) in size. A single 8m Telescope at the  VLT has a resolution of 50mas. If we consider that the gravitational lensing effect takes place on scales larger than the Schwarzschild radius, it could just manage to see this effect. (The theoretical resolution limit for an 8m telescope at 500nm wavelength is $1.22 \lambda / d = 15$mas.)

For real black holes at distances of 10s or 100s of lightyears, you would need a truly gigantic telescope.

Next option, again with gravitational lensing: The microlensing effect causes a background star to be temporarily magnified - it shines brighter for short period of time. A black hole passing in the foreground can cause this, and it is an optical observation, but it is very difficult to know that the lens was indeed a black hole and not some star or other compact object.

Last option, Hawking radiation: Hawking postulated that black holes might emit radiation inversely proportional to their size. For small enough black holes, this is a runaway process - they emit more as they get smaller, and because the emission causes them to shrink, they emit yet more again. Since Hawking radiation has a black body spectrum, the maximum of this emission will lie in the visible range for a brief period of time before going into UV, X-ray, and gamma radiation. The problem here is that no one has observed it yet, and you'd need a fairly small (but not too small) black hole close by. Also, you might get killed as the radiation gets harder.

To your questions:

Possible, but probably not. You would need to get very close, but not close enough to get killed by tidal forces. Or you'd need to be able to identify a microlens as a black hole by some means other than lensing.
Either lensing effects (see above), or a brief flash going from red to blue (probably too short to observe).
These black holes would be mostly visible by the absence of other particles. You'd get less elementary particles than you would expect for the reaction,some energy and some momentum would be missing, and there would be characteristic gamma radiation. However, for CERN to announce this, this would have to have happened often enough to not be a statistical fluke (this turned out to not be a detection after all.)


Where is Voyager 1 (or 2) going? In which direction? Knowing it has exited the solar system, is it traveling in the direction of the center of the galaxy, away from it, orbiting it, or maybe even going perpendicular from the disk-shaped galaxy?

Stars and gases at a wide range of distances from the Galactic center orbit at approximately 220 kilometers per second.

From Wikipedia

That's much faster than the Voyager probes relative to the Sun (Voyager 1: 17 km/s). Hence the probes will orbit the galactic center roughly the same way as our Solar System, even after occasional hyperbolic encounters with other stars.

Voyager 1 is travalling in the direction of Ophiuchus, seen from Earth. She's at or beyond the border of the heliosphere.


Why do radio telescopes, but not shorter wavelengths, have this Big Data problem? Only with new radio telescope arrays like ALMA and SKA, have I heard of this "problem" with too much data come up. Too big to store. That one cannot archive it all for future studies, but that one has to select beforehand what fraction of the data to analyse and erase the rest, and the most, of it.

Shouldn't it be the other way around, that shorter wavelength telescopes like those in IR and the visible and UV generate more data because they have higher resolution per unit of aperture?

Shouldn't it be the other way around, that shorter wavelength telescopes like those in IR and the visible and UV generate more data because they have higher resolution per unit of aperture?

This is a very interesting question!

My answer is no, not currently, but someday it may be the case.

Because we have big fast computer and fiber optic technology, we can generate images from arrays of antennas computationally. We can form images and even implement adaptive optics for wavefront correction at radio frequencies in software (rather than mechanical actuators correcting wavefronts by brute-force bending mirrors).

An order-of-magnitude description of ALMA is 100 dishes each one delivering 1 GHz of bandwidth, and SKA arrays will be far larger. While the Event Horizon Telescope literally records all raw data on boxes of hard drives that's not currently possible for ALMA and larger arrays, so only much smaller, heavily processed intermediate results files can be saved for later analysis and fine-tuning of calibrations and parameters.

For "shorter wavelength telescopes" in this context it's better to think of them as "higher frequency telescopes". Even a 100 nm wide band at say 800 nm (near IR where wavefront correction is working well represents 47,000 GHz! Even with conversion to a baseband frequency we don't have the ability to digitize signals at that rate, much less perform correlations of multiple signals.

So instead we let the light do the interferometry for us!

An optical telescope is just a big, simple, one-trick-pony optical computer. We design the surfaces to perform an integral of the electric field at the aperture. The center pixel receives the integral of all electric fields incident on the telescopes clear aperture without any phase bias. A pixel at one edge of the sensor receives the integral of the same electric fields times a phase function that increases linearly from one side to the other.

One day if and/or when detection and computing speeds and parallelisms are big enough, I am sure an electronic optical computer will probably be built.

For more on optical telescopes versus computational interferometry see answers to

Would “layered” radio interferometry work?
Could mirrors be replaced with CCDs?
How would a Fast Fourier Transform Telescope work without a mirror, dish or lenses?


What's the color of Upsilon Scorpii? I was looking to Scorpio the other day, and noticed that the last star in the tail (the telson's tip) -- known as Upsilon Scorpii, υ Scorpii, or Lesath -- seemed yet more red than the "heart" of the scorpion -- Antares or Alpha Scorpii. Was it that I was drunk, or some kind of sky pollution? (Actually, a friend confirmed it was red, and she seemed sober than me.) Or is it, indeed, red? I couldn't find any information on-line. I saw this database, but don't know how to find a star by its name, or if they have information about the color.

I don't know what could cause Upsilon Scorpii to appear red as it is not noted as being a highly variable star. It's spectral type is B2, which implies it would be blue/white in colour. If you know the spectral type of a star, which many stellar databases will note, you can work out its colour from this.


Has anyone ever tried to make a simple telescope using ice? I grew up with long cold winters, and saw a lot of remarkably transparent ice formed by refreezing meltwater, both in puddles and ponds, and in large icicles. I'd always thought about making optical elements but never tried, as I knew getting a good surface figure would be quite a lot of work.

After clicking through links in this answer to the question Are there any natural materials that can be used as a magnifying lens (or to craft one)? I came upon the images below, which have rekindled my interest.

However I'm currently near the Tropic of Cancer and far from any "Winter Wondernalds".

So I'd like to ask if anyone has ever tried to build a simple, low magnification telescope from ice or if there are any references to such an attempt.

I don't care if it's a front-surface reflector for the Sun or a refractor for astronomical or terrestrial viewing. I'd just enjoy seeing any, even slightly successful attempt at getting some optical magnification from surfaces produced in ice. Conventional eyepieces would be fine, as would projection on a piece of paper.

Some related imagery to "get the ball rolling":

From http://wildwoodsurvival.com/survival/fire/ice/ist.html also see Fire from Ice

From http://wildwoodsurvival.com/survival/fire/ice/istmake.html

From http://www.primitiveways.com/fire_from_ice.html

I'm not sure it counts as "simple" but there is the ice cube neutrino observatory whose detector consists of a cubic kilometer of very clear ice a mile or so down in the antarctic icecap.


Is it possible for a moon to continuously have a side facing its star whilst orbiting a planet? For example; is there any possibility for the moon to always have a side facing the Sun whilst orbiting Earth? And if so then what would the day cycle be like?

I worked out the math, it's pretty straight forward.   If we take the Hill sphere, which is an estimate for the furthest possible orbit, and I'm going to just run the math on circular orbits.  Eliptical orbits are harder to put in resonance anyway.

the simple Hill Sphere formula.  $3\frac{r^3}{a^3} = \frac{m}{M}$, where $a$ is the semi major axis planet to sun and $r$ is the distance moon to the planet, little $m$, mass of the planet, big $M$, mass of the star.

To calculate tidal forces, the force changes with the 3rd power of the radius, so for equal tidal force on a moon from both the star it's planet orbits and the planet it orbits, $\frac{r^3}{R^3} = \frac{m}{M}$, and that rather neatly puts the equal tidal forces outside the outer edge of the Hill Sphere by a factor of the cube root of 3, or about 1.44

The outer half of the Hill Sphere isn't long term stable anyway in real systems, so for the Moon to experience a stronger tidal force from the star it's planet orbits than it experiences from it's planet, it would need to be well outside the stable orbital region around the planet, by at least a multiple of 2.88.

There's no way for any moon in any system to be tidally locked to it's star and not it's planet.


How can the IAU 2000A vs IAU 2000B nutation comparison be reproduced? I am attempting to plot how the IAU 2000A nutation model degrades as its terms are omitted. As a spot-check, I decided to compare it to IAU 2000B, which includes only the 77 most important lunisolar terms. It is widely claimed that IAU 2000B only sacrifices 1 mas of accuracy. For example, in this PDF presentation from SOFA:

  The IAU 2000B nutation series is almost as accurate (1 mas) as the full
  IAU 2000A series

http://www.iausofa.org/publications/aas04.pdf

And in this published paper:

  IAU 2000B is more than an order of magnitude smaller than IAU 2000A but achieves 1 mas accuracy throughout 1995-2050.

https://www.aanda.org/articles/aa/full/2008/04/aa8811-07/aa8811-07.right.html

But I have not yet succeeded in comparing them in such a way as to produce a difference that small. When I compare the angles they return each day over the two decades 2000–2020, I see a difference in the first angle — delta psi — of >2 mas. Using the USNO NOVAS implementation of the two series, because it was easier to install (pip install novas) than pysofa, I get:

from math import tau
from novas.compat.nutation import iau2000a, iau2000b
T0 = 2451545.0  # Year 2000.0

dpsi_differences = []
deps_differences = []

for day in range(0, 366 * 20):  # Years 2000.0 through ~2020
    dpsi_a, deps_a = iau2000a(T0, day)
    dpsi_b, deps_b = iau2000b(T0, day)
    dpsi_differences.append(abs(dpsi_a - dpsi_b))
    deps_differences.append(abs(deps_a - deps_b))

def report_difference(name, differences):
    radians = max(differences)
    days = differences.index(radians)
    degrees = radians / tau * 360.0
    arcminutes = degrees * 60.0
    arcseconds = arcminutes * 60.0
    mas = arcseconds * 1000.0
    print('Maximum difference for {}: {:.4f} mas at T0 + {} days'
          .format(name, mas, days))

report_difference('delta psi', dpsi_differences)
report_difference('delta epsilon', deps_differences)

Result:

Maximum difference for delta psi: 2.1867 mas at T0 + 1396 days
Maximum difference for delta epsilon: 0.8631 mas at T0 + 7017 days

Am I misinterpreting the output of the NOVAS routines? Or, alternately, am I misunderstanding the meaning of the two angles? I understand the angles as being a pair of rotations which each, in the worse case — that of a point on the great circle of the rotation — will move a coordinate through the same angle as the rotation itself. So I understand a 2.1867 mas difference in Δpsi, for example, as changing sky coordinates by a maximum of that same 2.1867 mas when the nutation matrix is used to translate coordinates into or out of the equinox-of-date.

My next step would be trying to get the sofa library installed locally and then running a similar routine against it in case the NOVAS implementation is simply broken, but before trying to install a library by hand, I wanted to double check in case my understanding of the angles was itself faulty.

Thanks for any snags that can be identified in my reasoning!

I suspect that the problem is in your assumption that changes in $\Delta \epsilon$ or $\Delta \psi$ lead to coordinate changes of the same size. In fact, the position of the equator and celestial pole are complex functions of those angles, notably with both positive and negative terms.  If you look at Equations (8) and (9) (for the coordinates $X$ and $Y$ of the pole) in the 2008 A&A paper you linked to, you can see this dependence.  For example, in the $X$ equation, there is a $\Delta \psi \sin \epsilon_0$ term, but later there is also a $- (\psi^2_A / 2) \Delta \psi \sin \epsilon_0)$ term that could at least partially cancel it for a given value of $\Delta \psi$.  There are also cross terms between $\Delta \psi$ and $\Delta \epsilon$.

While those equations are only for an approximation to IAU 2000B, looking at Capitaine & Wallace 2006 for the exact equation (Eq. 36) shows the same behavior, i.e. that there are both positive and negative terms, as well as cross terms, involving those quantities.

So to do a comparison of accuracy, I think you would need to calculate the $X, Y$ coordinates of the pole and the position $s$ of the origin with both models, and compare those.


Is it possible that Venus was once a rogue planet? As Venus rotates in clockwise direction, while other planets rotate in anti clockwise direction:

Is it possible that Venus was once a rogue planet?

This is considered very unlikely. It is actually very difficult for things to be captured into orbits. They pick up speed as they fall in towards the larger object, and that's automatically enough speed to carry them back out. Capture either requires dust and gas to slow the body down, or very lucky gravitational interaction with a third object.

This has happened with moons, but the solar system is much more crowded than the galaxy. For something the size of Venus to pass through the solar system at all would be exceedingly rare.

Finally, it is easy to explain the retrograde rotation by other means. This is where the other question that James K mentioned in the comments comes in.


Do some galaxies collide faster than the speed of light? If some galaxies expand away from each other faster than the speed of light then do they collide faster than the speed of light?
Update 1:

I guess my question is what's the max speed/velocity at which galaxies can move towards each other?
How long does it take for galaxies to collide?

If you could use this example that would be awesome.
Galaxy A: diameter of 3,000 light years
Galaxy B: diameter of 300,000 light years
If Galaxy A collide with Galaxy B , what would be the velocity at the collision?
I have no idea about what the time and distance would be.

If some galaxies expand away from each other faster than the speed of light then do they collide faster than the speed of light?

The short answer is "no".
The expansion of the distant universe at apparent speeds faster than light is due to the expansion of space.  But the velocity of that expansion depends on how far apart the observer and object are.  If there are two objects A and B which we see apparently moving away from us at faster than the speed of light due to the expansion of spacetime then that is not how the objects will see each other.  They will be close to each other so from their reference frames the expansion of spacetime is negligible (just as it is for us and our "local" galaxies).
At some future time we think the universe may expand even faster at "small" ranges but in that case the two objects could not collide.  Once they are able to collide, they are close enough so that the effects of the expansion of the universe are not going to make them move apart faster than light.  Once objects move apart faster than light they cannot collide (unless the entire universe starts to contract).
If they are expanding away from each other faster than light (due to the expansion of space-time) then there is no way for them to collide.  No attraction could overcome the expansion of spacetime if it has already reached the point where they are moving apart faster than light.  It would require the entire universe to start contracting to do that.
So if it's going to collide with you, an object cannot be moving faster than light relative to you.

How long does it take for galaxies to collide?

This depends on how fast they are approaching each other.  In theory they cannot approach faster than light, so the upper limit is (simplistically) the distance divided by the speed of light.
A more realistic example would be Andromeda and the Milky Way which are approaching each other at a relative speed of 300 km/s.   Andromeda is about 2.5 million light years away and current estimates are for a collision in about 4.5 billion years, so that's moving much, much slower than the speed of light.


Calculating the sun’s position in ECI I’m trying to calculate visible ISS passes and I need to calculate if the ISS is behind the earth by “drawing” a line between the ISS and the sun and seeing if it crosses earth. I have the ECI position of the ISS, but I don’t have the position of the sun. How do I calculate the suns ECI position given a date and time?

Edit: I need to use c# for this so I can’t use python libraries

If you know how to use python, you can do something like :

import astropy as ap
ap.coordinates.get_sun(time)

It will return  a position in geocentric coordinates for the given time.


Do the orbits of planets change sometimes? Do planets sometimes wobble and get off their paths? What if an asteroid were to hit it?

The orbits of the planets and other bodies are changing all the time, because of the influence of other bodies.

Yes, planets can change orbit significantly because of the influence of other planets. There is a theory that has Neptune and Uranus swapping orbital positions a long time ago.


How long before Phobos reaches the Roche limit of Mars? Phobos orbits really close to Mars, so close that one Phobos period is shorter than one sol. Due to tidal friction, it is also getting closer and closer to Mars. How much time will it take before Phobos reaches the Roche limit of Mars?

Sometime before now and "tens of millions of years".
Phobos currently orbits at a radius of about 9400km
The theoretical Roche limit is different for rigid and fluid bodies.  If Phobos were a fluid body, it would already have passed the Roche limit at 10500km.  This is because fluid bodies will deform into an ellipse pointing towards the planet, and this amplifies the tidal forces and so causes break up a larger radius.
If Phobos is fully rigid body, then the Roche limit is half that at about 5400km. In fact a rubble-pile-with-crust like Phobos is neither fully fluid, nor rigid. Wikipedia gives an estimate of "30-50 million years" before it is fully disrupted.
So Phobos is already a body under stress, and (as anybody who knows a little about earthquakes knows) rock under stress is often unpredictable in how it responds: It can creep, fracture, or hold rigid while stress increases and then fail catastrophically.