What is the orbital path of the newly discovered star-less planet PSO J318.5-22?

There is no "orbital path" detected, that's why it is a "free-floating planet". There is no radial velocity mesured, but the informations given by its kinematic location show that it belongs to the beta Pictoris group, that is a stellar group.

For more dirty details, have a look at the submitted paper on PSO J318.5-22:
http://arxiv.org/abs/1310.0457

Comment:
Apart from that (the following reflects my personal opinion) the term "free-floating planet" is ill-chosen; it is clearly a very low-mass object, but since it is not orbiting around another larger object, it seems to me that the term "planet" is not pertinent. I think that it should be consider more than a "very low-mass brown-dwarf". The problem arises from IAU definition of a brown-dwarf and a exoplanet, that is probably not well-fitted for this kind of objects (and that is not really physical). You will notice, by the way, that this object is moving in a stellar group, which kind of reinforce my point.


Does Flat Universe mean even distribution?

No, being flat and being homogeneous is not equivalent$^\dagger\!\!\!$.

Flatness refers to the geometry, which depends on the total energy density $\rho$; if it is above or below a certain critical threshold $\rho_\mathrm{cr}$, we call the Universe "closed" or "open", respectively, while if $\rho$ is exactly equal to $\rho_\mathrm{cr}$, we call it "flat".

Homogeneity refers, as you say, to matter being evenly distributed (on large scales). If the Universe is not only homogeneous, but also isotropic (i.e. looks the same in all directions), then there are three different possible geometries, namely flat, open, or closed. But a homogeneous universe doesn't have to be isotropic, and both open and closed universes can have evenly distributed matter.

Thus,
$$
\mathrm{homogeneity} \nRightarrow \mathrm{flatness}
$$

Whereas the local geometry depends on the local density$^\ddagger$, the global geometry does not depend on how the matter is distributed. For instance, you could in principle have a universe with no upper limit of structure size. In our Universe, we find observationally that the largest structures have sizes of roughly half a gigalightyear. Above this scale, it is homogeneous (although it could be inhomogeneous on scales larger than the particle horizon). But another universe might have structure on all scales and thus not be homogeneous, but still meet the criterion $\rho = \rho_\mathrm{cr}$ and thus be flat.
You could also, as discussed in this answer about homogeneity and isotropy, imagine a universe originating at a central point away from which the density always decreases (i.e. is isotropic around this point), but which still has $\rho = \rho_\mathrm{cr}$.

Thus,
$$
\mathrm{flatness} \nRightarrow \mathrm{homogeneity}
$$

Note that while your statement about the geometry determining the fate of the Universe would be correct if it contained only matter and radiation, it seems that $\sim70$% of the density making it flat has the annoying propety of accelerating the expansion of the Universe. Thus, whereas a flat, matter-dominated universe would expand asymptotically toward a finite size, our Universe is dominated, it seems, by something dubbed dark energy making it expand exponentially despite being flat.

$^\dagger$For a universe, at least. For a roadkill, it might be equivalent.

$^\ddagger$For instance, space "bends" enough around a massive cluster of galaxies to make gravitational lenses.


Name and language of symbol in astronomy book

This is a variant of pi, or "pomega".
In LaTeX, you can get it by using
                                          \varpi

$$\varpi$$
TeX has a question on the 'var' prefix, and Wikibooks has something relevant.

The various forms of pi are present in Unicode as:

U+03A0 Π greek capital letter pi (HTML Π · Π)
U+03C0 π greek small letter pi (HTML π · π)
U+03D6 ϖ greek pi symbol (HTML ϖ · ϖ)
U+220F ∏ n-ary product (HTML ∏ · ∏)
U+1D28 ᴨ greek letter small capital pi (HTML ᴨ)
U+1D70B  mathematical italic small pi (HTML )
U+1D6D1  mathematical bold small pi (HTML ).

It is used to denote the longitude of the periapsis, as in your case.


It's different. Black holes are not objects, like planets or stars. Rather, they are powerful distortions of spacetime, maintained by the concentration of mass/energy therein (which itself is kept there prisoner by the spacetime distortion - a vicious cycle broken only slowly by the Hawking radiation). As such, they cannot be "ripped apart", since there is nothing to be ripped apart there - just a giant warped tangle of spacetime.

Instead, two black holes could merge, if they get close enough and lose enough mutual orbital motion.

merging black holes

The result is a single, larger black hole.


@LocalFluff's comment is more-or-less complete. The process of "swapping" planets between stars could happen during the early lives of stars when most of them are born in relatively dense clusters. You can find some more details and references in my answer to this question, but the outcome is that a captured planet would usually be in a very wide orbit.

Approximately 1% of stars have a "hot Jupiter", but even in dense cluster environments it is thought that only a few percent of stars can capture a planet, and because almost all of these end up in wide orbits there are just too many hot Jupiters to be explained in this way.

There are two classes of theory to explain hot Jupiters. One is the dyanmical explanation offered by LocalFluff. This is itself divided into several different mechanisms: "Secular chaos", (e.g. Wu & Lithwick 2011 - this paper also briefly reviews some of the other mechanisms) which is a gradual change in the planetary system that then suddenly drives a transition to a new configuration. Planet-planet scattering where interaction between planets causes one orbit to become highly eccentric and then tidal circularisation results in a hot Jupiter in a close orbit (e.g.Ford & Rasio 2008). The Kozai mechanism is the interaction beteen two planets that results in an interchange between the inclination and eccentricity of the inner planet. This excites an increase in eccentricity that, when combined with tidal circularisation, can  also result in a close-in planet.

The other class of mechanism involve migration through interaction with a disk. This must necessarily happen early in the star's life whilst surrounded by a disk of circumstellar material. Basically viscous drag in the disk itself causes the planet to move inwards (or sometimes outwards). There are three possible subtypes of disk migration and they are briefly defined here.

It is probable that all of these mechanisms play some role in forming hot Jupiters and one of the goals of exoplanetary study is to figure out which may be more important. Typical parameters of interest are the distribution of planetary orbits and eccentricities and whether there is evidence of previous interactions in the form of misalignments between the rotation axis of the star and the orbital axis of the planet.


Do all orbiting bodies eventually collide?

Yes.

Two bodies in orbit around each other will inevitably collide. The reason for this is that the system will give off energy in the form of gravitational waves. This effect is commonly cited in binary neutron star systems, where the two stars are isolated and close together. One of the most famous of these systems is the Hulse-Taylor binary.

The time it will take for the objects to collide can be calculated:
$$t=\frac{5}{256}\frac{c^5}{G^3}\frac{r^4}{(m_1m_2)(m_1+m_2)}$$
where $r$ is the initial radius, $m_1$ and $m_2$ are the masses of the bodies, and $c$ and $G$ are the familiar constants, the speed of light in a vacuum and Newton's universal gravitational constant.

However, tidal acceleration could offset some of the effects.


The larger an object is and the further away it is from the Sun, the greater its Hill Sphere is. Mercury's is very small (only about $175000km$ in radius from what @userLTK said) and its "moon" would therefore be yanked away by the Sun relatively quickly. See this for more details.  The orbit of this moon would be highly unstable and therefore unpredictable, so it may hit the Sun.


How can the observable universe be so small if there are so many stars in it?

If you know the total number $N$ of galaxies, and you know the size $V$ of the (observable) Universe, then you can calculate the number density $n = N/V$ and mean distance (of the order $1/n^{1/3}$) between galaxies, and convince yourself that it actually works out fine.

However, the reason we know how many galaxies are in the observable Universe in the first place is the other way round:

Observe an average number density of galaxies (which of course varies quite a lot from dense clusters to voids).
Calculate the size of the Universe by integrating the Friedmann equation (using as input observed values of the densities of the constituents of the Universe; dark energy, dark matter, gas, stars, radiation, etc.).
Finally, multiply the two numbers to get the total number.

Note that when observing the number density, we must not look too far away from the local Universe, since this means looking back in time to a period where the number density was different from present-day's value.


Which lens is right to see which planets with an Orion StarSeeker 150 mm?

You SHOULD be able to see Mars, Jupiter, and Saturn easily with your scope with any eyepiece.  In fact, You should be able to see them easily naked-eye.  But, I assume you want to see more detail, not just a bright dot in the sky.

Your telescope is a 150 mm Newtonian reflector with a focal length of 750 mm (giving you a focal ratio of f/5).  It looks like you have six eyepieces, each with different focal lengths, and a 2x Barlow.  Based on that, you should get the following magnifications:

Eyepiece        Mag         With Barlow
----------      -------     ---------------
32 mm           23x         47x
23 mm           33x         65x
12.5 mm         60x         120x
10 mm           75x         150x
6 mm            125x        250x
4 mm            188x        375x

Realistically speaking, anything over 200 or 250x is going to be a waste in this telescope.  The maximum useful magnification of any telescope depends on a fe factors, such as aperture, eyepiece and telescope optics quality, atmospheric steadiness (seeing), humidity, air pollution, etc...  While you may often hear people say the maximum useful magnification is about 50 or 60x per inch of aperture, in reality, on an average night, it's more like 25 to 30x.  Based on this, you shouldn't expect anything over 200x to be useable very often, maybe 150x, unless you live in a place with unusually clear, steady air (like living at high altitude in the mountains).  Actually, I just looked, and Orion says the maximum useful is about 177x, which would be based on 30x per inch of aperture.

Still, with even 177x, you can get some pretty good views of planets.  The 4mm will give you 188x, which should be a nice view, though probably a little fuzzy.  dropping to 6mm will give you 125x, and that shouldn't be too bad.  For Jupiter, you should be able to make out cloud bands and possibly the Great Red Spot, if it's in view.  And, of course, you can see the four Galilean moons.  With Saturn, you'll be able to see the rings, but I wouldn't expect to see the Cassini division - but you might get a hint that it's there.  With Mars you can make out hints of the polar caps, and just maybe a large feature like Syrtis Major.  The filters may help with planetary viewing, helping you bring out the details by increasing contrast.

Here's a couple links to useful discussions of filters for planetary viewing:

Sangamon Astronomical Society Filter Advice
Agena Astro Planetary Filter Info

You will find it easiest to start out with a wider view eyepiece, like your 32mm.  Get the planet centered, then switch out eyepieces. Trying to get right-on a target with a high-power eyepiece is much more difficult, even with GoTo.

For other objects, things are a bit different.  Deep Sky Objects (DSOs: star clusters, galaxies, and nebulae, basically anything outside of our solar system), are many, varied, and can be quite difficult.  Even through larger telescopes, they are often nothing more than fuzzy patches.  They will NOT look like you see in pictures.

Your scope has an alt-az GoTo mount.  This is not always a good thing.  It must be properly aligned, and even then, a lot of alt-az GoTo mounts just aren't all that accurate. When you do your startup routine, it should ask you for at least two alignment stars.  You need to get these aligned as exactly as possible.  If it gives you the option to add more, do so. The more you add, and the more accurately you align them (meaning centered in the field of view, not just somewhere in the eyepiece), the better the GoTo and tracking performance will be.  I've had observing sessions with GoTo scopes where they were off by several degrees.  If you don't know your way around the sky to begin with, you have no idea where you're looking.

For DSO's, aperture is important, but so is getting the right exit pupil.  A guy I know on Reddit wrote up a great discussion of exit pupil you can read here.  Simply put, you should tailor your exit pupil to the need.  This partly will involve measuring your own pupil (this Sky and telescope article includes how to measure your own pupil size).

Finding them, then is the issue.  Even with GoTo, it's not always easy. And just because your GoTo scope points at the object (if it's properly aligned) doesn't mean you can see it.  The description on Orion's Website of your telescope's GoTo capability says: "Huge GoTo database of over 42,000 objects, and handy built-in Tour mode lets you enjoy great views on any clear night."  Unfortunately, a lot of these objects simply aren't going to be viewable in your scope, or from where you're observing.

And observing site is an important consideration.  If you're in an area with a lot of light pollution, the sky background will be so bright that you can't get enough contrast between it and the object you're trying to view to make the object visible.  Still, that said, there are a lot of things you should be able to see from even moderate light pollution.  I would start with the Messier objects, as they are some of the brightest DSOs in the night sky and fairly readily found (a couple are even naked-eye objects).  There should be a menu option in your scope's hand control for these.  If you're properly set up and aligned, it should be able to find them for you.

I mentioned above, if you don't know your way around the sky, it makes it harder.  You should definitely take the time to learn the constellations, at least the more common ones, as well as the names of some of the bright stars.  Most GoTo alignment systems ask you to locate 2 or 3 bright stars by name, and they mostly use the same stars.  There's between 150 and 200 commonly used bright stars for this purpose.  I wouldn't say you have to memorize them all, but if you can find a good chunk of them, say maybe 50-75 of the most common, it makes it easier to find your way around the sky.  If someone says "it's 5 degrees south of Betelgeuse" or "Find Arcturus and move East a few degrees," you'll have a better understanding of what you're looking for.  Also, learning to estimate angular measurements (there's some great tricks out there for estimating degree measurements using fingers on an outstretched hand) can make a difference.

One of the best resources for the beginner is the book Turn Left At Orion by Guy Consolmagno and Dan M. Davis. It's written to help people new to the hobby learn their way around.  You can find it on Amazon, Barnes & Noble (bn.com), or order it from any good bookstore (if they don't stock it already).

Probably the BEST resource, however, is to join an astronomy club or society.  In the US, membership is usually $50 (USD) or less per year.  I suspect they are similarly priced in other countries.  They are a great place to meet people with the same interest.  The best part is, most such groups have people who enjoy teaching newcomers the tricks to finding things and how to really observe them.  They also usually have people who can teach you how to properly collimate, maintain, and even repair your telescope.  And a lot of clubs have their own observing sites, or at least have scouted out good observing sites in the area.  It's really a worth it to join a club.

My club, the Houston Astronomical Society, only costs $36 per year, and we have some great benefits for members (including loaner telescopes, a dark site for observing, and even small observatory).

Good luck and clear skies!


Why was Pluto's resonance not discovered until 1965?

This much, I can answer:

  Was orbital resonance just not a conceived concept until the sixties?

Laplace recognized orbital resonance long before then, and he died in 1827.  I looked, couldn't find a precise year of publication.

https://en.wikipedia.org/wiki/Orbital_resonance#Laplace_resonance

Now, 3/2 resonances were perhaps suggested/discovered quite a bit later, I'm not sure.

Some mathematics on the subject in the following link, a little bit over my skill-set, but it makes the point of stable resonance.

http://www.applet-magic.com/solaresonance.htm

Also, the link points out that the the first resonance discovered, Io, Europa, Ganymede the 1-2-4 resonance, have ratios of 2.0076 and 2.0143, or, 0.4% and 0.7% off resonance.

I'm tempted to say that small variations like that fall within a stable zone, but I'd basically just be guessing.   I can't say for sure.

Pluto was discovered in 1930 and Charon, it's rather large moon, not discovered till 1977.   Perhaps in the 1930s, Pluto was no longer cutting edge physics and all the big brains were looking at Quantum physics, Hubble's discovery and other galaxies, implications from Einstein's discoveries, such as black holes and Nuclear bombs . . . but again, I'm only speculating.   It does seem that a discovery of the 3/2 resonance should have come before 1965 and I think it's a good question.

Not to take this too off subject, but I love the story of how Neptune was discovered.   Short summary here:   http://coolcosmos.ipac.caltech.edu/ask/146--When-was-Neptune-discovered-

Longer one here: https://en.wikipedia.org/wiki/Discovery_of_Neptune

My favorite potential orbital resonance is the Jupiter/Mercury one, which could, eventually, pull Mercury away from the sun.

https://en.wikipedia.org/wiki/Stability_of_the_Solar_System#Mercury.E2.80.93Jupiter_1:1_perihelion-precession_resonance

The Jupiter/Mercury resonance is described as "coincidence" in the article, where as, Pluto's 3/2 with Neptune and Jupiter's 3 inner moons are (I think), probobly not coincidental resonance but some form of stable orbits.

Hope you don't mind my amateur/hobbyist answer.


Has SETI data been used for astrophysics? How, or why not?

Well, that might depend on exactly what definitions you use.  SETI itself is a broad program.  In this answer I'll focus solely on what I consider the most iconic part of modern SETI, which is SETI@home.

We have the following quote from SETI@home's webpage (emphasis their own):

  SETI (Search for Extraterrestrial Intelligence) is a scientific area whose goal is to detect intelligent life outside Earth. One approach, known as radio SETI, uses radio telescopes to listen for narrow-bandwidth radio signals from space. Such signals are not known to occur naturally, so a detection would provide evidence of extraterrestrial technology.

SETI@home's specific purpose deals with the radio SETI data.  So they are implicitly asserting that the data they collect is expected to be useless outside of their namesake: they're specifically looking for things that are not known to occur naturally.

That being said, it turns out that the SETI@home data collection is almost never active (see here on their site).  It is done passively, as a piggyback collection when the Arecibo radio telescope is being used for other science projects.  So if your question is "is any data useful outside of SETI actually collected while SETI(@home) is gathering data?", then the answer is "yes".

Now, we can also note that one of the original goals of the SETI@home project was to prove the viability and practicality of distributed computing for scientific projects.  This we can say was a resounding success, as it spawned BOINC, which serves as a distributed computing platform for a wide range of non-SETI science projects: Folding@home and Rosetta@home deal with protein folding, Einstein@home is for gravity wave detection, the now-defunct Mesenne@home was for finding Mersenne primes, and many others (my copy of BOINC specifically states that there are over 30 currently active).

So if you are willing to consider the advances they spurred in distributed computing in scientific projects, and how this has impacted scientific projects inside astrophysics (or outside), then the answer is again "yes".

EDIT:  Einstein@Home recently made the following announcement about a gamma ray pulsar that was detected via the program; it includes links to a published article on the detection, including an arxiv version link if you find the official publication is behind a paywall.  So that's a concrete, published, and current discovery in astrophysics.  If you're willing to give SETI some of the credit for having spawned the @Home style of distributed computing projects in science, as suggested above, then there you go.


No. Tetanus is caused by a soil bacteria. You cannot get tetanus from rusty iron, if that iron has been sterilised.

There are no Earth bacteria on Mars. As far as we can tell, there is no life of any kind on Mars.


How quickly does a supernova heat up/expand?

Ricky, it is very rapid. The core collapse and initial neutrino burst takes seconds to tens of seconds. We don't normally think about neutrino interactions, but so many are released that even this might be a problem for a nearby habitable planet. It then takes a few hours for the shockwave from the core collapse and bounce to make it out to the surface, accompanied by an intense flash of UV light that would likely sterilise anything in its planetary system. The outer layers of the exploding star are hurled out at around 1000 km/s, so could travel an astronomical unit in a day or so.

The supernova continues to become more luminous for about a week thereafter, increasing in luminosity, from what must have already been enormous (thousands of solar luminosities) by another factor of $10^5$. The equilibrium temperature of any planet scales roughly as $L^{1/4}$, so temperatures would rise by more than an order of magnitude in a week.


What was the field of view of the Ohio State University Radio Observatory of Wow! signal fame?

For a single dish radio telescope we define the primary beam as the telescope's response on the sky as a function of angle. What this means is that half way between the centre and edge of the primary beam, a radio source with a flux of 1 Jy will be observed to have a flux of 0.5 Jy. This response is circular i.e. it is a function of angular separation, and is not dependent on altitude/azimuth. The full-width of this response is typically written as:

where  are the primary beam width in radians, observing wavelength and dish diameter.
As for the big ear telescope, it is obviously not a parabolic reflector or Gregorian offset like the Lovell or Greenbank telescopes (www.naapo.org/W8JK/Images/JDK097l.jpg). Since the secondary reflector is the important one in the determination of resolution, we shall use its dimensions, which are 103m long by 21m high.
The Wow! signal was received at 1420MHz which has a wavelength of 0.21m. Using the above equation, the response would not be circular, but 'fan-shaped'. The response would thereby have approximate dimensions on the sky of 0.12deg by 0.58deg (in azimuth and altitude respectively).
Caveats of this calculation are the non-circular reflectors, which means the actual shape of the telescope's response on the sky would not be Gaussian/symmetrical for that matter. But as for an approximate, angular resolution, 0.12deg by 0.58deg will suffice.
Further to this, if you can find the LST (local sidereal time) of when the Wow! signal was received, and assuming that the big ear telescope is pointing at the zenith (i.e. straight up), the spot the telescope was pointing to would be easy enough to derive based upon telescope latitude and LST. Enjoy!


Measuring star distance by parallax using a small telescope

Short answer - not really, parallax for the closest stars is right on the limits of resolution for a good ground-based amateur equipment.

The nearest star would show a parallax angle of under an arcsecond.  (The Parsec even takes its definition based on one arcsecond of parallax.)

Ground based amateur observations are probably limited to an arcsecond at most, so it's probably not possible to measure them.

A special (and totally fictional) case - if a nearby star like Alpha Centauri were exactly between us and a much more distant star, and it were possible to see them both without Centauri overpowering that distant star, then maybe it would be possible to observe them as a close double.  That's not true in practice however.

Update:

Prompted by Rob's much better answer, the earliest parallax determination I've been able to find was using

  the 6.2-inch (157.5 mm) aperture Fraunhofer heliometer at Königsberg

(Source: Heliometer article on Wikipedia.)

In purely aperture terms, that's an amateur-sized instrument (with some admittedly very precise measuring gear built in.) So my suggestion now is to ignore my initially doubtful response and try measurements, perhaps using a camera at fairly high magnification as Rob suggests...


How often does Earth cross the galaxy regions with higher supernovae activity?

The paper "Frequency of nearby supernovae and climatic and biological catastrophes" by Clark, McCrea, and Stephenson published in Nature estimates (at 50% probability) that the Solar System passes within 10 parsecs of a supernova every 100 million years. This supernova would be part of a 20-parsec strip in which an estimated 50 supernovae occur.

They do speculate that a connection between the Solar System passing through these regions and climatic and biological events (such as ice ages) is possible.

You can read the paper yourself for more details. It's only two pages and you can purchase 48 hour access for $5 from Readcube.

Source:

"Frequency of nearby supernovae and climatic and biological catastrophes" - Clark, McCrea, and Stephenson


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

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

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


Io is not dominated by sulfur. Io is mostly silicate rock and iron. Sulfur is a thin coating on the surface and sulfur dioxide makes up most of the atmosphere, but that's because sulfur dioxide is a volatile. In other words, it has a much lower melting and boiling point than silicate rock. So while the rock remains solid, the sulfur dioxide is gas.

Carbon is not common on Io's crust for the same reason that it is not common on Earth's crust. Carbon may be common in the universe, but it binds to oxygen to make a gas. When the solar system was forming most of the carbon was in the form of CO or CO2 and would not easily have coalesced into a solid body. Silicon is much less common than carbon, but Earth and Io are made of silicon because silicon makes solid stuff at the relevant temperatures.


Does Doppler shift affect apparent pulsar frequency?

Yes, it is correct.

Consider the classic example of a (say) $1000$ Hz siren on an ambulance approaching you.  Once a single wave is produced, the source moves towards you before the next wave is produced.  The wavelength of the sound waves in the air is reduced, so the frequency you hear is increased.  Simple, basic Doppler shift...

But consider the case where the siren is being switched on and off with a five-second overall period.  So you have a $1$ kilohertz signal modulated at $0.2$ Hertz.

But the same argument applies to this modulation frequency.  A chain of $1$ kilohertz waves is generated for $2.5$ seconds and sent on its way.  The ambulance approaches you for the next silent $2.5$ seconds, then starts the next chain on its way, closer to the last chain because of the distance travelled in the silent phase.  The same Doppler shift factor applies to the carrier frequency and the modulation frequency.

The same physics almost derailed the Cassini-Huygens probe. The extreme velocity changes of the probe caused a Doppler shift in both the transmission frequency and the data rate carried by this transmission. The first was handled;  the second was not!

See http://www.thespacereview.com/article/306/1

or

http://spectrum.ieee.org/aerospace/space-flight/titan-calling


What do the acronyms in "AT 2017 gfo" mean?

You are correct. The AT is short for Astronomical Transient. The follows the year and an arbitary sequence of letters to uniquely identify the event, which is assigned by the Transient Name Server

This is the official IAU designation for Supernova Candidates, however true supernovae will be re-designated "SN YYYYX". The neutron star merger was not a supernova, so retains its AT designation

This is described at (source) with more details at Transient Name Server (TNS)


How ready are we to deal with a asteroid/comet/giant-meteor on a collision course with Earth?

The short answer is "not very, but we're getting better".

In the case of particularly large Earth orbit intersecting bodies, kilometers across and bigger, they're fairly well known and tracked. We'd probably get months to years of warning. Technologically, we've probably got the capability of diverting such an object if we have enough warning, but it'd likely be the most complex human endeavor ever. Blowing it up is unlikely to help, we'd need a nice long shove from as far away as possible to divert it.

The bigger problem is smaller bodies in the tens to hundreds of metres. Some of these are known about and tracked, but plenty more aren't. The Chelyabinsk meteor was about 20 metres across and, in terms of size/density/speed/impact angle wasn't that far off being a city killer. We never saw it coming. (Ironically, a very similar but unrelated 30m object was being tracked through a near-miss that same day.)

I'd suggest that a city-killer impact with little or no warning is still quite possible. Something capable of taking out, say, a small country we'd probably have some warning of, but probably not enough. A mass-extinction event we'd probably have lots of warning for, but doing anything about it would be a very big challenge.


How did Kepler "guess" his third law from data?

Kepler's account of how the third law came to be is as follows (Caspar p.286; emphasis mine):

  On the 8th of March of this year 1618, if exact information about the time is desired, it appeared in my head. But I was unlucky when I inserted it into the calculation, and rejected it as false. Finally, on May 15, it came again and with a new onset conquered the darkness of my mind, whereat there followed such an excellent agreement between my seventeen years of work at the Tychonic observations and my present deliberation that I at first believed that I had dreamed and assume the sought for in the supporting proofs. But it is entirely certain and exact that the proportion between the periodic times of any two planets is precisely one and a half times the proportion of the mean distances.

Although Kepler does not actually describe inspiration that led him to believe this, the curious phrasing provides a very strong clue when combined with some background biographical information:

John Napier published Mirifici Logarithmorum Canonis Descripto in 1614, which contained the then-new invention of logarithms. Kepler was aware of Napier's work by 1617 (Caspar p. 308), perhaps earlier.
Joost Bürgi published work on logarithms almost at the same time as Napier, and Kepler was similarly aware of Bürgi, even praising his mathematical abilities as surpassing most professors of mathematics.

Thus, Kepler's statement is is equivalent to saying that the data makes a slope of 1.5 on a log-log graph, a which is a very simple linear relationship on this scale.

References:

Caspar, Max, Kepler, (Dover, New York, 1993).


Are we still going to have rainbows if Sun is replaced by another star?

Rainbows would lack most blue, and some green for red stars.
For a blue star, the blue part of the rainbow would be more intense.

For more complex colors, the rainbow may show some gaps. A rainbow is essentially a spectrum of that star light portion, which is visible to our eyes, and to which the atmosphere is transparent.

Stars vary in brightness. A blue giant would be large and glaring, a red dwarf faint.

Colors in the rainbow would be blurred, hence closer to white, for large stars, and sharper, more distinct, for small stars, according to the angular size of the respective light source.

... This all assumes, that there is still rain. With small, red stars, it would get too cold for rain. With large blue stars, Earth would heat up too much.
To adjust for these effects, the distance to the star would need to be modified.
And of course, the length of a year, and the orbital velocity may change.
This could then cause different tides, changes in volcanism, etc.

Exchanging the star could cause various other effects, too, other polar lights, effects to the ionosphere, the ozone layer, atmospheric erosion, more...


Where is Voyager 1 (or 2) going? In which direction?

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.


Perception of an objects history passing as an observer moves toward it

Yes, it will be 20 Julian years ago.

Yes, it will take 40 years to reach there with speed half the speed of light through vacuum without any obstacles and star must be stationary (though you can't travel with speed of light since physically it's not possible to achieve that speed).

And as you are travelling with speed half to that of the speed of light you will experience a phenomenon of time dilation. Check all the details about time dilation here:

http://en.wikipedia.org/wiki/Time_dilation

You will see time dilated by $\gamma = 1/\sqrt{1-v^2/c^2}$, factor so if you substitute this values it will take you 60 years to reach that star. And it squares all the things.

I hope this explanation helps you. If it is not understandable do let me know; I will explain it in a more mathematical way.


What happens if you fail to reach escape velocity for the galaxy?

If you fail to reach escape velocity from the earth, you won't break free of its gravitational well, in other words, you will keep orbiting the center of mass of the earth. If the path of your orbit intersects the surface of the earth, then you will, you know, crash.

If you fail to reach escape velocity for the galaxy, you'll keep orbiting the center of mass of the galaxy, just like the other couple hundred billions of stars in the galaxy. The galaxy has no hard surface to crash into, so you'll keep orbiting until you come close enough to another orbiting body to perturb your orbit (which could, in the right circumstances, leave you with escape velocity and kick you out of the galaxy.


What is the maximum planet mass for human habitation?

When astronomers discuss the "habitable zone" around a star they're really concerned with someone likely to provide liquid water and the right amount of radiation for basic life (cells) as we understand them.  But this doesn't mean they're suitable for humans.

Maximum mass is not an issue, it's gravity that matters.

We need a few things from gravity to be human habitable.

Firstly we're "designed" (evolved) to operate in a one-gee environment.  As far as I know there have been no studies on long term high gee effects on humans (low gee, yes, but not high gee).  However, considering the "plumbing" and structure involved in a human, it seems unlikely to me we'd adapt long term to more than a small percentage larger than one-gee.  Much more on a sustained basis would seem likely to increase physical stress on the bones, muscles, tendons and so on and lead to more rapid development of related illnesses and disabilities.  The plumbing (fluids are being moved all over your body) can cope with quite high gees for short periods, but it's simply not designed to cope with sustained high gee.  The strain on the circulatory system strikes me as the greatest vulnerability.

Movement in high gee is also an issue.  I'd be surprised if people could adapt to long term requirements without serious health side effects.  We'd also be more vulnerable to injury from falls and slips.

Consider how long you could walk in 1.5g.  It would be the equivalent of walking while carrying something weighing half as much as you do yourself. If you were 100 kilos, how long could you walk while carrying 50 kilos ?   If you had to do that all day, everyday, how long before the strain caused serious injuries and other health problems ?  Not long, I think.

So just on this basis I'd suggest a maximum sustained gee of about 1.1 is ballpark.  Walking around with ten kilos when you are 100 kilos sounds like a lot all day everyday, but it's probably doable.

But gravity does more than provide a way to nail us to the ground.  It provides a way to nail the atmosphere to ground at an atmospheric pressure high enough for us to use, and not too high for us.

Now it would appear from human habitations in the deep sea, that we can sustain quite high atmospheric pressures, so that would not seem likely to be a problem.  What tends to hurt humans is shifting quickly from high to low pressures.  As I've argued that we can't sustain high gees for long and need something close to one gee for health reasons, the atmosphere doesn't seem to be a problem in that case.

So what does 1.1g mean in terms of planet size and density.

The gravitational pull is proportional to planet average density and radius.

So we need a planet that's around the 1.1g mark meaning :

$$1.1 \rho_{earth} r_{earth} \approx \rho_{new-earth} r_{new-earth}$$

where $\rho$ is density and $r$ is radius and this represents the limit.

So this may make it easier to find suitable planets.  We aren't restricted to specific mass ranges, but can look for planets that are less dense and larger or more dense and smaller.

But this does make me wonder if the notion of human colonization of planets ever made any sense.  Building our own large habitats in space is starting to make a lot more sense to me as it restricts us less.


It seems to me that you're mostly asking about star naming conventions, which is unfortunately a difficult thing to master because there are many many conventions. What makes this process difficult is that the conventions you seem to be referencing don't come from Hipparcos at all. Likely you're seeing the names of these stars as they're referred to by some other, more popular convention. I'll try to go through the things you have listed and describe how those conventions work.

Hipparcos Names

A star in the Hipparcos catalogue is given a number. That number is preceded by HIP and you cite a good example of HIP 36623. The number itself is more or less meaningless for the star. Some catalogues provide numbers to stars based on their RA and/or Dec, but for the Hipparcos catalogue the number is the same as the that in the Hipparcos Input Catalog (all HIP stars were pre-chosen for observation and defined in an Input Catalog). The star entries are, with a few exceptions, ordered by increasing HIP number, which basically follows the order of the object's right ascension (Equinox J2000) independent of declination.

You can find a pretty good list of Hipparcos variable conventions on this page.

Bayer Convention

You also cite the names Delta Orionis and Omicron Puppis. These names come from the Bayer naming convention. The general scheme for this convention is to group stars by the constellation they're in. Then you rank all of them by brightness. You give each star a name which includes a Greek letter then the Latin name of the constellation it exists in. The brightest star is alpha (equivalent to A), the next brightest beta (equivalent to B), then gamma (equivalent to C), and so on. There is a convention about what to do once you've run out of Greek letters, but honestly, I've never seen anyone use the Bayer convention for such stars. It's generally only used for the brightest stars in a constellation.

From the examples you gave, Delta Orionis is the 4th brightest star in the Orion Constellation (since delta is the fourth letter in the Greek alphabet). For Omicron Puppis, that's the 15th brightest star in the Puppis constellation. In this convention, you can sometimes see the greek letter written as a Greek letter, rather than spelled out, and occasionally the constellation name is shortened. Thus, for your examples, you might see them written as $\mathrm{\delta}$ Ori and $\mathrm{\omicron}$ Pup. Note that lower-case Greek letters are used (first), not upper case.

Variable Star Type Designations

Variable stars are complex and come in many many different flavors. As such, there's a whole slew of different names and schemes to be used when referring to a variable star. A good reference site is the International Variable Star Index page. You can refer here to answer most of your questions, but I'll post the descriptions of the non-alphabetic characters.

  A pipe character ( $|$ ) between two different types signifies a logical OR; the classification is uncertain and all possible types are indicated. An example of this is ELL|DSCT, where the star may be an ellipsoidal binary system or a DSCT-type pulsating variable with half the given period.

  A plus character ( $+$ ) signifies a logical AND; two different variability types are seen in the same star or system. An example of this would be ELL+DSCT, where one of the components of an ellipsoidal binary system is a DSCT-type pulsating variable.

  A slash character ( $/$ ) indicates a subtype. In the case of binary systems (eclipsing, ellipsoidal or reflection variables) it is used to help describe either the physical properties of the system (E/PN or EA/RS), the luminosity class of the components (EA/DM), or the degree of filling of their inner Roche lobes (EA/SD). This is the GCVS classification system. In cataclysmic variables, slash characters are used to indicate some properties of the system, as in the degree of polarization (NA/DQ) or the nature of their components (UG/IBWD).

This site doesn't specifically indicate the meaning of the colon, but research elsewhere indicates that the colon is used to "indicate the uncertain nature of the classification". It looks to me that within variable star designations, and particularly by the General Catalogue of Variable Stars, a colon used anywhere (not just on the variable type designation) implies uncertainty.


How did the authors of Surya Siddhanta find the diameters of other planets in the solar system?

The authors assume a geocentric universe (first thing that is wrong). They then assume that the planet Mars has the same apparent diameter as a globe 30 yojana in diameter (about 150 miles) in the same orbit as the moon, from a perspective at the centre of the Earth. This is just stated, and appears to be supposition. It is an incorrect figure. It is conceivable that some kind of system for sighting through a pinhole could be used for this estimate. This method tends to over estimate the apparent size of planets.
They next claim that the other planets have apparent sizes that would be the same as globes with diameter 37.5 (Saturn), 45 (Mercury) 52.5 (Jupiter) and 60 yojana (Venus). These form an arithmetic sequence with a common difference of 7.5, but apart from that, these values seem to be arbitrary (and are wrong). They might have been based on some kind of pinhole observations, but rounded to a simple sequence for the sake of the poetry or for easier memorisation.
There is uncertainty about the ancient value of the yojana. A different conversion to miles will give different values throughout.
Using this assumption and the distance already calculated to the moon they calculate the apparent angular diameter of each planet. The actual values they get are wrong, In fact, the angular diameter varies as the planet's distance varies.

Planet
calculated value
modern observed value

Mars
2 arcmin
(actually 0.06 - 0.39)

Saturn
2.5 arcmin
(0.25 - 0.34)

Mercury
3 arcmin
(0.08 - 0.17)

Jupiter
3.5 arcmin
(0.51 - 0.83)

Venus
4 arcmin
(0.16 - 1.05)

As you see the calculated values are much too large (ie wrong). They are comparable with values supposed to have been found by Hipparchus, but without a known observational basis. Tycho Brahe also gave similar values, found by sighting through a pinhole. This simple method doesn't give a good estimate of the angular size. The simple pattern in the calculated values is due to the original assumption that the globes were in an arithmetic sequence.
There is then a calculation of the geocentric orbital radii of the planets. They assume that Venus, Mercury and the Sun all orbit at a distance of 3.4 million miles.  This is a significant underestimate, especially for the sun (this is wrong). The distances to Mars, Jupiter and Saturn were also underestimated.
One can then combine the apparent angular diameter (which is too large) with the calculated distance (which was too small) to obtain an estimate of the planetary diameter. When this is done, it gives exceedingly good estimates of the diameter of Mercury and Saturn (within 1%). A good estimate for the diameter of Mars (about 10% out), and a poor estimate for Venus and Jupiter (about 50% of the accepted modern values).
It is clear that there is much that is wrong. A combination of overestimates of some values and underestimates of others cancels out to give some impressive looking figures for two out of five planets. But do note that the estimates for Jupiter and Venus were a long way out, and all the numbers which the calculations were based on were wrong. In total there are 15 values calcuated here 5 apparent diameters, 5 geocentric distances and 5 actual diameters. Of these only 2 values are impressively accurate.
They did better than other pre-telescopic observers, but it isn't clear that they obtained values by anything other than luck.
Source


Is the SETI's `needle-sharp at 982.002 megahertz` signal accessible?

If those data were available to the public, the Breakthrough Listen Open Data Archive would most likely have them.
Queries there currently return no 982 MHz data for any sources, and "PROXCEN" data only at 3009 and 3094 MHz from 2017-18.
This Scientific American article mentions work in progress by Sheikh et al.; they may have dibs on the 982 MHz data from 2019 until they publish their paper.


Equation to find distance between objective and eyepiece

1 A.U. is same as infinity. The difference in terms of eyepiece position is infinitesimal, you can't measure it. Anything beyond a few kilometers away is pretty much "at infinity".

Regardless of that - from the practice of designing and building telescopes, calculations only offer you a starting point. You do the math, and the distance is 105 cm. But in practice lenses will deviate from the ideal focal length. Even if they didn't deviate at some temperature, put them in a cold environment, and the focal length will change a fraction of mm.

So take the calculations as a starting point, and build the instrument in such a way as to allow fine adjustments of the position of the eyepiece. There's a device called focuser that allows such fine adjustments. Or simply rely on friction to move the eyepiece back and forth until the image looks best, and hold it there.

When using the instrument in practice, you'll forget the ideal distance. What you will do is adjust the position of the eyepiece until the image looks best. You will do that every time you observe, and often multiple times during the same observation.

If you want some math, take a look at the thin lens equation, and apply it to the objective lens.

f = focal length of the lens

o = distance from lens to object

i = distance from lens to image

Then the thin lens equation is:

1/f = 1/i + 1/o

i = 1/(1/f - 1/o)

If o = infinity, then i = f.

But what happens if o = 1 A.U.?

i = 1/(1/1 - 1/(1.5 * 10^11)) = 1.0000000000067 meters

The difference is something like 6.7 * 10^-12 meters. It's smaller than an atom.


Stellar core mass as a function of metallicity

I feel silly, feel free to laugh at this, but I found my mistake. In the Hurley paper, he uses "log" for the logarithm of base 10. I had been using the natural logarithm. With the fix, the graph looks more realistic:


Blackbody curve in RGB for objects less than 1500 K?

So I use the following logic:

Define 'red', 'green' and 'blue' light in terms of the wavelength ranges those colours encompass (from here)
Calculate the average spectral radiance across each wavelength range, for a defined temperature
Define the RGB value to be the relative ratios of each colour's spectral radiance with 255 defined as the colour with the largest spectral radiance

I made a Python script which you can find here to calculate, display and show the appearance of the corresponding RGB value. Within the script, just change the value of the stellar_temperature variable to see the RGB values of stars of different temperatures.
The script is then finished off with a simple plot to show the appearance of that RGB value.
Now there are a couple of assumptions of course:

RGB values can be defined in terms of radiance/luminosity/flux ratios at each wavelength... This obviously breaks down if you have a detector which gets saturated at certain flux levels! I presume the human eye might fit that description.
Stars can be modelled by perfect blackbody spectra. I believe this to be a pretty valid assumption when the accuracy level needed is differentiation of different RGB tones...
No foreground extinction/absorption by atmosphere/dust etc.


Why is the Sun's brightness and radius increasing, but not its temperature?

The effective temperature $T_\mathrm{eff}$ of a star, which is presumably what's been plotted, is defined through its relationship with the star's radius $R$ and luminosity $L$ by

$$L=4\pi R^2\sigma T_\mathrm{eff}^4$$

This comes from the assumption that the star radiates like a black body at the photosphere. While this isn't strictly true, it's quite accurate, and regardless, that's how we define the effective temperature. The actual surface temperature will be slightly different but also behave roughly as plotted.

So, even if $T_\mathrm{eff}$ is constant, the star expands if it grows brighter.  Also, you can see that the sensitivity to temperature is steeper than radius, so a moderate change in luminosity can be absorbed by a relatively small change in effective temperature.

While the luminosity is basically determined by the simple behaviour of the nuclear reactions in the core (in terms of temperature and density), the surface properties depend on how energy is being transported near the surface. For radiation, you have to consider what the opacity of the material is, which itself depends on ionization states and whatnot. It's easy enough to see why the luminosity grows (the core gets denser and also hotter, producing energy faster) but the determination of the surface properties is more complicated. For the Sun, it turns out the way shown in the plot after you solve all the equations with the relevant opacities.

Also, as an extreme counterexample to "brighter means hotter and smaller", remember that red giants are much brighter but also much cooler!

PS: I'm not sure the source of the data but I would guess the wiggle at the start is because of the star finishing its contraction onto the main sequence.  That is, before the first minimum, energy is being released by gravitational contraction.  After it, the energy from nuclear reactions starts to dominate.


Any gas or particles you see in the video are not matter of outer space, but part of the propulsion system or the spacecraft. The high velocity of the gas is a consequence of the propulsion. The exhausted gas (water vapor, aluminium, carbon dioxide, soot, many other compounds, depending on the propellant) continues to move roughly with constant velocity relative to the rocket as soon as it has left the thrusters (Newton's first law); the rocket accelerates as long as the thrusters are active (Newton's third law), but that's not noticible in the video relative to the exhaust fumes. While the rocket is moving through Earth's atmosphere there may be some interaction of the exhaust fumes with Earth's atmosphere.

There is no interaction of gas or fragments with matter of outer space visible  in the video.

Although outer space isn't quite empty. It contains e.g.

interplanetary and interstellar dust, mainly consisting of silicates, but also of other compounds like oxides, water ice, traces of organic compounds,
hot plasma, mainly consisting of protons and electrons emitted by the sun (solar wind),
galactic cosmic rays, mainly consisting of fast-moving atomic nuclei,
electromagnetic radiation across the electromagnetic spectrum, from radio waves to gamma rays,
magnetic fields.


Converting FITS (NuSTAR) coordinates to WCS

You can upload your FITS files to astrometry.net (or use an API) and get the coordinates. You can optionally get new FITS files back with the coordinates included in the metadata. From there you will be able to proceed with your processing.

And, instead of uploading everything, you can also install a copy of the astrometry.net package locally if you're running Linux.


The answer is that they don't come very close. As Wikipedia notes,

  It is rare for a comet to pass within 0.1 AU (15,000,000 km; 9,300,000 mi) of Earth.

Even better, though, is an actual list of some of the closest approaches of comets to Earth. The closest one listed, Comet Lexell in 1770, came 0.0151 AU away from Earth. The list only shows 20 comets have come with 0.1 AU of Earth. So Earth has not had a close encounter with a comet (like Mars is having) in a long time, if ever.

This site says that C/1491 came within an incredible 0.0094 AU in 1491, but I'll take the IAU site over it. And it does admit that that comet has an uncertain orbit, so that figure could be wrong.

But when comets come even a tiny bit near to Earth, there are some articles written about them. See here and here.

As for the distances to famous comets:

Closest Ever

Halley's - 0.0334 AU
Great Comet of 1760 - 0.0682 AU
Tempel-Tuttle - 0.0229 AU

Also, some near-Earth asteroids used to be comets. Extinct comets have lost a lot (if not all) of their ice and so may be classified as asteroids. This blurs the lines between the two groups. Some extinct comets have become near-Earth asteroids. This explores the issue pretty well. So if you include extinct comets as "comets," you may have comets that have come even closer to Earth.


What is the apparent rotation time for a planet relative to an other planet?

Note, an observer on Planet A can't see a rotation of Planet B. He can see a the Sun orbiting around him, and Planet B orbiting around the Sun. Thus

If Planet B has an orbit of smaller radius than Planet A, then the observer can see Planet B sometimes before, sometimes after its Sun.
If Planet B has an orbit of larger radius than Planet A, then the observer can see Planet B to cycle on the sky, roughly on the plane of the Ecliptics,  but a little bit faster as the Sun.

The calculation is this:

Planet A takes $\frac{1}{m}$ full circle in a day.
Planet B takes $\frac{1}{n}$ full circle in a day.
Thus, the observer on Planet A can see an $|\frac{1}{m}-\frac{1}{n}|$ full circle of rotation of Planet B in a day.
Thus, he can see a period of Planet B of $\frac{1}{|\frac{1}{m}-\frac{1}{n}|}$ Earth days, which is in simplified form

$$\underline{\underline{|\frac{mn}{m-n}|}}.$$


If the whole observable Universe is as big as planet Earth, how big would Earth be?

The radius of the observable Universe is 41.5billion light years (ref).
A light year is $9.5\cdot10^{12}$ km.
The radius of the Earth is 6371 km.

So the answer is $6371 km \cdot \frac{6371 km}{41.5\cdot10^9\cdot9.5\cdot10^{12}km} = 1.03\cdot10^{-16}km = \underline{\underline{1.03\cdot10^{-13}m}}$.

The size of the atoms are $\approx 10^{-10}m$.


How are the masses of celestial bodies determined?

Newton's analysis of Kepler's laws lets you determine the ratios of the masses of everything in the solar system with quite high precision, but it doesn't let you determine the actual mass.  The actual masses were one of the large unsolved problems of physics for almost a century, until 1798 when Cavendish's work with a torsion balance (described by him as "weighing the world") determined the density of Earth, from which the masses of everything could be computed.


Why don't globular clusters flatten with a galactic disc?

Globular clusters formed whilst the gas of the proto Milky Way was still approximately spherically distributed.

The gas forms a dissipative system that loses energy and collapses (within the first billion years) to a disk whilst conserving angular momentum.

Formed stars and clusters are essentially collisionless so the halo stars continue to have a spherical distribution, whilst more recently formed stars formed from gas already in a disk and so continue to be in a disk.


How many days a month can you see a moonrise during daylight?

It's too hard for me to come up with reliable figures for the amount of visible moonrises/moonsets during daylight, but I can answer

what elongation of the moon is needed for the moon to be visible?

It will depend on the conditions, but according to the Wikipedia article about the Islamic calendar, the Moon may already be visible 17 hours after the conjunction with the Sun:

But, the lunar crescent becomes visible only some 17 hours after the conjunction, and only subject to the existence of a number of favourable conditions relative to weather, time, geographic location, as well as various astronomical parameters.

The synodic month is about 708.734 hours, so a period of 17 hours corresponds to an elongation of $360^{\circ} \cdot \frac{17}{708.734} = 8.6^{\circ}$.


Can Earth be ejected out of the solar system this way?

It looks like the model is a simulation of what happens if a massive object flies by a very narrow ring system type of thing with a circular orbit.   The ring system setup seems improbable to me, and the massive object doesn't appear to fly in a straight line, so it's not very accurate for two reasons, but the gravitational effect seems accurate.    If a massive body was to fly into the solar system and pass near Earth it could send the Earth out of the solar system or in towards the sun.

Fortunately, bodies that massive almost never fly inside the solar system, so that scenario is unlikely to happen over the lifetime of the sun.


What was the orbit of the meteor that impacted the Moon on 2013 September 11?

Out of curiosity I looked into this a bit.    North/South on the moon is pretty irrelevant for the objects orbit, but it struck at East 339 degrees (or West 21 degrees) near the Lubiniezky craters in the northwest part of Mare Nubium - which can be seen here - towards the center but a little bit in the south eastern part of the map.

and you can see the phase of the moon here. It changes slowly enough that a 24 hour range doesn't make a huge difference.

http://lunaf.com/lunar-calendar/2013/09/11/

but apart from the crater looking roughly straight on, it's probably impossible to say the orbit it came from because the angle of approach is too big an unknown.   It hit from kind of both the earth side and sun side direction, so it could have been a temporarily captured asteroid around the earth or a small comet or asteroid coming back from around the sun.  Either could have made that impact I would think.


It is the semi major axis of the relative orbit i.e. the ellipse whose major axis is the sum of the two major axes and whose focus is in the barycentre. Essentially, this is like fixing the reference frame at the barycentre and looking at how the stars' position vectors change. It will turn out (try looking up two body problem) that this is equivalent to an object of reduced mass $\mu = \frac{m_1*m_2}{m_1+m_2}$ orbiting the one with total mass.As uhoh pointed out, this image illustrates semi major axes : https://i.stack.imgur.com/298tF.png .


Do NEA (Near Earth Asteroids) have minable water ice?

No and yes.

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

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


Are there pictures from a solar eclipse of the earth taken from a moon landing unit or orbiter?

Such images are most readily available from full-time Earth-observing satellites.

The DSCOVR climate observatory sees the Earth from Lagrange point L1, 3.9 lunar distances toward the Sun.
Its EPIC camera captured sequences of images half an hour apart for the solar eclipses of 2016-03-09 and 2017-08-21.

Geostationary satellites orbit 0.11 lunar distance from the center of the Earth.
The CIMSS group at Wisconsin has a collection of various weather satellites' images of several eclipses.

A lunar point of view would fall between these two examples, qualitatively more like the former than the latter.


As you said, the mass range of the object indicates that it could be a neutron star, rather than a black hole.
But also the density of the object is a clue. The density of a neutron star is much higher than that of a stable white dwarf.
Extremely regular electromagnetic pulses have to be observed. Known pulsars have a pulse period between 1.4 ms and 8.5 seconds.
The wavelength of the pulses is in the radio range.
Pulsars slow down gradually, at a known rate, so you can measure how the pulse period evolves over time, and check that is compatible with the slow-down rate of a pulsar.

Adapted from this page


How to correctly create planetary orbits from kepler elements

If you're running your own N-body dynamics,
you can get initial Cartesian positions and velocities for each body in your system (including the Sun) by asking
JPL HORIZONS
for Ephemeris Type: Vectors.
I would choose Coordinate Origin: Solar System Barycenter so that the Sun does not drift away from the origin,
and Target Body: [Planet] Barycenter (or Sol) unless you are modeling the Earth and Moon as separate bodies.

If you still want to compute initial Cartesian positions and velocities from Keplerian elements, first do a sanity check: Earth should cross the X axis at the equinoxes and the Y axis at the solstices.
Second, those approximate elements for the major planets may not be precise enough for your purpose.
Instead you might ask HORIZONS for Ephemeris Type: Elements, and Time Span beginning with the epoch of your elements for Apophis.
You could also get elements for Apophis in a more convenient epoch this way.

Note that Keplerian elements are subject to dynamic evolution and are only valid for a limited time near their epoch.
If you ask HORIZONS for Apophis's orbital elements one week before and after its close approach to Earth, you can see that they are quite different. Of course the latter are also highly uncertain.

For the gory post-Newtonian details of the dynamical model the Apophis prediction article discussed extending, see JPL's
notes on DE430 and DE431, part III.


I ended up using PARAM - http://stev.oapd.inaf.it/cgi-bin/param

Cheers,


It will always expand (probably).

The mass and energy in the universe curves space, and it seems that there is exactly the right amount of mass and energy to keep the universe expanding forever.

Had you asked this question 20 years ago the answer would have been "we don't know". Recent discoveries of "Dark Energy" suggest that the expansion of the universe is not slowing down, but speeding up. This would mean that the universe is destined to expand forever, at a faster and faster rate.

The notion that there will be a "Big Crunch" followed by another "Big Bang" was seriously considered by multiple scientists: among them Willem de Sitter and George Gamow. There are problems with this hypothesis: It is not well supported by observations of the rate of expansion. Moreover, there would need to be way to reduce the entropy of the big crush, as entropy can only increase. And moreover any "pre-big bang state" may be hidden behind the singularity at the start of the (current) universe and so fundamentally hidden from us.


Why is the shadow of the earth not visible during sunsets?

To have a sharp shadow, you need a point source of light.  But the Sun is extended enough that it does not set all at once, so there is not a sharp shadow.  Also, there is refraction of the redder light, so instead of seeing a shadow, you'd see reddish clouds-- which I wager you see quite often (though much of that effect is from the way the atmosphere scatters away the blue light-- those effects are hard to disentangle but you know refraction matters because you see how the Sun's shape is distorted upon setting).


How many astronomers are there in the world today?

"The worldwide community of professional astronomers is only about 10,000; most are located in the us (with about 1,000 in the UK and 250 in Australia)." From So You Want to Be an Astronomer by Duncan Forbes.

Another source stated that the number of professional astronomers is about the size of a small town.


Is the rotation of the Sun and the rotation/orbit of the Moon around the Earth a coincidence?

This is purely coincidental.

Why? Because the moon used to rotate faster and is continually slowing down. Long story short, you just happen to be at a point in history where they two numbers roughly coincide. A few million years ago it wasn't true, and a few million years from now it wont be either.

And just to cover all my bases, there's nothing to suggest that the slowing orbit of the moon can have any reciprocal effect on the sun's rotational period.


Why do the planets orbit in the same direction?

The same reason (almost) all of them rotate in the same direction: because of the conservation of angular momentum.

Before a star and its planets exist, there’s just a cloud of disorganized gas and small molecules. The Solar System formed from such a cloud around 4.6 billion years ago.

On that scale, there is some small amount of rotation within the cloud. It could be caused by the gravity of nearby stellar objects, local differences in mass as the cloud churns, or even the impact of a distant supernova. The point is, all molecular clouds have at least a little rotation.

In a large system like a molecular cloud, each particle has some angular momentum, and it all adds together across a very wide area. That’s a lot of momentum, and it is conserved as the cloud continues to collapse under its own gravity. That angular momentum also flattens the cloud, which is the reason why the Solar System is near-planar.

When the cloud finally collapses, it forms a star and shortly after planets. However, angular momentum is always conserved. That's why planets all follow the same orbit, and why almost all of them rotate in the same direction. There's nothing to turn them the other direction, so they will continue spinning in the same direction as the original gas cloud.

There are a few exceptions, though. Whenever objects formed in such a way that sent them orbiting the opposite direction, they usually collided with objects going in the same direction as the original cloud. This destroyed any outlying objects or sent them in the same direction as the original cloud.

Still, two huge exceptions are planets Venus and Uranus. Uranus spins on an axis of almost 90-degrees (on its side). Venus meanwhile spins the opposite direction as Earth and the other planets.

In both cases there is strong evidence that these planets were struck by large objects at some point in the distant past. The impacts were large enough to overcome the angular momentum of the bodies, and give them a different spin. There are also a range of other theories; for example, some astronomers think that Venus may have been flipped upside-down. Point is, there were irregular events that happened to both of these planets.


The effect of gravitational wave is like tidal forces?

It depends what you mean by "very strong".

The announced first detection (14th September 2015) was considered a very strong signal and that still only stretched the 4 km detector arms by less than the width of a proton (a tiny particle inside the nucleus of an atom). This wave would have no effect on a person.

You could hypothesis a stronger wave that would stretch and squish a person, I'm not sure what effect this would have but there are no objects in nature capable of producing waves this strong, even something as close as our sun becoming a supernova wouldn't be that strong, so it will never happen in real life.


Can the Earth be used as a gravitational lens?

Yes, it's possible in theory, but beyond current technology to achieve. The focal point of the Earth is 15300 AU away.  By contrast, Neptune is about 30AU.
As a gravitational lens is not like a glass lens, you don't get an image formed and at any distance greater than 15300 AU an Einstein ring would be formed around the Earth. But at that distance, the Earth would be very close to the sun in the sky, and the sun has its own gravity. Moreover the Einstein ring would be have the same radius as the Earth at that distance.  It would be very small. No Einstein ring is formed at less than 15300; the ring that would be formed by a body with Earth's mass is smaller than the diameter of the Earth at distances less than 15300
Larger bodies have a nearer focus, the Sun has a gravitational focus starting at 550 AU, which is perhaps just about within current solar sail technology. It might be possible to make direct observations using the sun as a gravitational lens.  But you don't get a steerable telescope. It would still be a very challenging mission.
See a related question about the mission on space exploration stack and this forum discussion


Delta-v from Mercury surface to Venus surface

Assuming circular, coplanar orbits, Trans Venus Insertion (TVI) is 8 km/s from Mercury's surface. That's with horizontal take off and no gravity loss. That's for the minimum energy Hohmann transfer.

Exiting the Hohmann transfer and soft landing on Venus would take 11.9 km/s if Venus had no atmosphere. But much, maybe all of that 11.9 km/s could be done by aerobraking.

I used my Hohmann spreadsheet to get these numbers.


No. The metrics look very differently.

The metrics of the the Universe is determined by the FLRW metrics. Its spacelike sections are planar.
The spacelike sections of the spacetime around a Schwarzschild (non-rotating) BH is spherical.

More elementarily formulated: the black hole is like a static sphere. The Universe is like an expanding sheet. However, this is what we currently see, matched by our current models. These are not so strong facts, like the density of the iron, what anybody could measure on his desk. I think the correct explanation is roughly this: "it doesn't really looks so, but no one knows".


If there was air between the Sun and Earth, how warm would we get?

This is a really hard question to answer as Carl Witthoft alludes to. The main problem is that there is no way you can finagle putting at atmosphere around the Sun and have it extend to the Earth. It would be wildly unstable and no matter what happens, the general result will be an explosion of energy that kills the Earth. And you can't really say you're going to wave some magic wand and prevent all of that because you're then fundamentally altering physics and you won't end up answering the question you wanted to ask to begin with.

Your only way of answering this question is to look at the current atmosphere of the Sun and try to extrapolate. Take a look at the figure below. This represents both the temperature (solid line) and density (dashed line) of the Sun vs height from the surface (what the "surface" of the Sun means is a whole other question). The temperature at the surface of the Sun is a measly $5800\:\mathrm{Kelvin}$. As you increase in height, this temperature (i.e., the temperature of the plasma atmosphere above the "surface" of the Sun) doesn't change much. This region of the atmosphere is known as the chromosphere. But then, at a particular height known as the transition region, there's a drastic increase in the temperature. It goes from a few thousand degrees to a couple million degrees. This area of the atmosphere is the Corona. This graph cuts off at $10,000\:km$ but the Sun's atmosphere extends far out beyond Pluto.

Hopefully you can fully appreciate the point I'm getting at which is that Earth actually is inside the atmosphere of the Sun and that atmosphere is really hot! The reason we of course aren't boiled to death is that the atmosphere is incredibly sparse. Only a handful of particles per cubic centimeter. So what you're really proposing is to increase the density of this atmosphere by an insane amount. You want to make it approximately $10^{20}$ times more dense! If we just extrapolate properties, that's going to turn the already hot atmosphere of the Sun (around the Earth) into something incredibly hot. And the huge density means we'll really feel the effects. The Earth will likely be vaporized in a flash, before the cataclysmic explosion which I alluded to above can even happen.

So, long answer short: About a billion degrees.


If Mercury and Venus where to collide would it make a habitable planet?

The answer is "Probably, but not very quickly."

The energy of a collision comes from two sources, the kinetic energy of the collision and the gravitational energy. If Mercury were to impact Venus, even if it began by creeping up on it very, very slowly so as to have as little kinetic energy as possible, it would fall into Venus's gravitational well and release gravitational energy enough to melt both bodies.  The resulting mess would take ca. 10-100 million years for a crust to resolidify.

The resulting body could evolve towards habitability.  One theory of planetary development says that active plate tectonics are critical to habitability.  We know that new rock absorbs large amounts of CO2 -- new volcanic rock in Indonesia has been estimated to absorb a quarter of the CO2 released each year.  So the continuous exposure of new rock by plate tectonics may be a major force in preventing the runaway greenhouse effect that has made Venus uninhabitable.

We don't know for sure how plate tectonics starts or ends, but if the resulting planet had a particularly vigorous case of it, this might compensate for its closeness to the Sun.

On the down side, Venus is already short of hydrogen and thus water and a collision with Mercury would eject Venusian volatiles into space and Mercury would not add any new, so the resulting planet might be rather barren.

In any event, whatever happens, it's going to take a very long time for the resulting body to cool enough to be livable.


How to find a Cepheid's pulsation period using its graph?

This graph is not the light curve of a cepheid. It is used to find the average brightness of the star, and from that, estimate its distance.

The rate at which cepheids pulse is related to their average luminosity (and so their average absolute magnitude). This graph relates the period to their average absolute magnitude.

To use the graph, observe a cepheid for long enough to establish a period (say 30 days, for example). Then read the graph (an absolute magnitude of just under -5.5 in the example)

You now know how intrinsically bright the star is on average. By looking at its average apparent brightness, you can calculate an estimate of its distance. See How does the cepheid variables help in finding distance


I think your initial question is a good one, but the text gets a bit more jumbled and covers a few different points.

  Can things move faster than light inside the event horizon of a black
  hole?

Nice question.

  Black Holes are regions of space where things get weird.

I'm 100% OK with this statement.  I think it's a true enough summary and I'm sure I've heard physicists say this too.   Even if "Weird" isn't a clearly defined scientific term, I'm 100% fine with this (even without a citation).

  Past the event horizon of a black hole, any moving particle
  instantaneously experiences a gravitational acceleration towards the
  black hole that will cancel out it's current velocity, even light.
  That means that the gravity well of the black hole must be able to
  accelerate from -C* to 0 instantly✝.

Are you quoting somebody here?   Anyway, this isn't quite true.   Black holes don't accelerate things from -c (which I'm guessing would be a light beam trying to fly away from the singularity but inside the event horizon), to 0 "instantaneously".

Perhaps a better way to look at it is to consider curvature of space, and inside a black hole, space curves so much that all directions point to the singularity.   It's the "all roads go to Rome" scenario, even if you do a complete 180, you're still on a road that leads to the singularity.

I understand the temptation to look at that as deceleration, but I think that's a bad way to think about it.  Light doesn't decelerate, it follows the curvature of space.

  Given that fact, we can assume the gravitational acceleration of black
  holes is C/instant**. Given this, it stands to reason that in
  successive instants, the particle will be moving at speeds greater
  than C, because it is experiencing greater gravitational forces and
  continuous gravitational acceleration.
  Does this actually make sense? Is there something I'm missing here? By
  this logic, it seems like anything inside of the event horizon of a
  black hole could and should move faster than C due to gravitational
  acceleration.

outside of a black hole, continuous acceleration would never lead to a speed greater than C.  You can accelerate for billions and trillions of years and all you'd do is just add more 9s to the right of the decimal point.

You seem to be assuming that inside a black hole this can happen, but I'm not sure why you'd assume that.

"continuous gravitational acceleration" - no matter how strong, is no guarantee for faster than light travel.   That's logically inconsistent with the laws of relativity.

  Edit: I showed this question to a friend and he questioned if the
  hypothetical particles that were radiating from the singularity (The
  photon traveling exactly away from the black hole) might be hawking
  radiation; that is, the gravitation acceleration of a black hole is
  only strong enough to curve the path of light around a non-zero radius
  (thus not actually stopping it, but altering it's course), and not
  powerful enough to decelerate light. Is this actually what hawking
  radiation is, or is he as confused as I am?

I think, a more correct way to look at hawking radiation is to see it as something that forms just outside of the black hole, a particle/anti particle pair and one escapes and the other falls inside, and that's probably not 100% correct either, but the singularity itself doesn't send out particles.   Hawking radiation has to do with quantum properties of space.  It's not a property of black holes.  The black hole just happens to be unique in that it can capture one half of a virtual particle pair and the other half can escape.

This also is a pretty different topic than your original question.

  *Where movement towards the singularity would be considered a positive value, movement away from the singularity is a negative value, that
  is, anything moving at the speed of light away from the singularity
  would be moving with a velocity of -C relative to the singularity.
  ✝If it couldn't accelerate from -C to 0 instantly, any photon
  traveling exactly away from the black hole would be able to escape the
  event horizon.

  **An instant is an arbitrary amount of time, it could be a fraction of a second, a second, a minute....

I think it's a good idea to differentiate mass-less pure energy particles and particles with mass.   You seem to be saying that a ray of light can be traveling away from a black hole at the speed of light, get caught in the gravity, slow down and then fall back into the black hole like a ball that's tossed straight up into the air from the surface of the Earth.   That's probably not what happens.   The ray of light follows the path of space time ahead of it, which happens to be curved so much that it points into the black hole, even if, in the classical sense, the light begins by pointing away.   All space curves into the singularity once you're inside the event horizon, so there is no "away from" anymore.   At least, that's how I think it works.


How does space time differ within a galaxy?

The gravitational redshift factor and also the factor by which time dilation occurs in a gravitational field is approximately $GM/Rc^2$. Here, $R$ would be the radius from which light was emitted and $M$ is the amount of mass within radius $R$ (making the crude assumption of spherical symmetry).

Thus, for an external observer,  light emitted from a radius of 10 kpc, might have $M \sim 10^{11} M_{\odot}$ in the galaxy shown. The redshift/time dilation factor is only $5\times 10^{-7}$ and is basically negligible.

This could perhaps gets an order of magnitude bigger as one approaches the denser central bulge, but is still very small.

The only appreciable redshift and time dilation effects will come in the immediate surroundings of any supermassive black hole. By immediate, I mean within about a hundred Schwarzschild radii, where $R_s = 2GM/c^2$. In our Galaxy that would mean within a billion kilometres.


One consisted of two convex lenses of different focal lengths and the
  other one consisted of a concave lens as the objective lens and a
  convex lens as the eye-piece(that's called a Galilean one I guess).

That's the ABC of refractor design - the most basic types of refractor. They are only interesting from a historic perspective.

http://en.wikipedia.org/wiki/Refracting_telescope

You can certainly purchase a large convergent lens (bi-convex, or plan-convex) and use that as an objective. Then, for the eyepiece (ocular), you can use either a small convergent lens (bi-convex, or plan-convex), or small divergent (bi-concave, or plan-concave). Probably only the first one will work well: the image will be inverted, but the field of view will be nice and wide. The second design (divergent lens for eyepiece, Galilean refractor) provides a non-inverted image, but the field of view is extremely narrow and hard to use; avoid it, it's frustrating.

However, this sort of bare-bones design is only interesting if you want to reconstruct the history of astronomy. At least for the eyepiece you could just "steal" any cheap 1-1/4" eyepiece from a telescope or microscope. Even the cheapest one will work much better than your single-lens "eyepiece".

http://www.telescope.com/Accessories/Telescope-Eyepieces/pc/3/47.uts?&refinementValueIds=4469

For the objective, again, I suggest you don't use a single convergent lens. It will work, but it will be full of aberrations (image distortions, color fringes, etc). Instead, go on Surplus Shed, click on the Lens Finder. For lens type, choose Achromat. For diameter, anything over 45 mm. For focal length, anything over 190 mm.

http://www.surplusshed.com/lens.cfm

An achromat is a combination of two lenses, glued together, that are designed to reduce aberrations. The image will be much better.

Their stock varies greatly. Right now, I've found these two:

http://www.surplusshed.com/pages/item/pa1038.html

http://www.surplusshed.com/pages/item/pa1039.html

The 50 mm diameter, 300 mm focal length achromat, combined with a 15 mm focal length eyepiece, will provide 300 / 15 = 20x magnification. With a 5 mm eyepiece instead, it will give 300 / 5 = 60x magnification. Don't go much higher than this (more magnification is not always better).

The focal length of the objective dictates the length of the telescope tube. E.g. with the 300 mm achromat, the telescope length will be 300 mm plus some extra at the back for the eyepiece.

EDIT: Be mindful of the f-ratio. Basically, divide the focal length by the diameter, and that gives you the focal ratio.

A lens with a diameter of 50 mm and focal length of 500 mm has an f-ratio of f/10. A lens with a diameter of 50 mm and focal length of 300 mm has an f-ratio of f/6.

It's best to use an achromat with an f-ratio between f/8 and f/20. If you go below f/8 it will be very hard to collimate (fine tune) the scope. If you go above f/20, it will be too long.

Sometimes you find super-cheap eyepieces there, too, like this one:

http://www.surplusshed.com/pages/item/l2019.html

Also keep digging around that site, maybe you'll find various other parts necessary for building the refractor.

Enormous care must be given to keeping the objective and the eyepiece collimated (aligned and centered). Even a tiny collimation error will greatly affect performance. This is not a trivial engineering problem.

There's also an easier way. You could purchase a complete kit, like the Galileoscope:

http://galileoscope.org/

It's a 50 mm (2") f/10 refractor that has about the same performance like good binoculars. The price is very affordable (and very cheap for what it does). As a telescope and mirror maker, I was impressed by the quality of the achromat doublet objective that they provide (50 mm diameter, 500 mm focal length, f/10). For the price and the size, it is well made and well corrected, and punches well above its price category.

The kit comes with everything you need: all the lenses, the tube, various small parts, etc., including components for the ocular (eyepiece). Another nice thing is that you can swap its eyepiece for a standard 1-1/4" eyepiece used in a regular telescope (like a cheapie on Surplus Shed). Whoever designed this thing put some good thought in it. There are several different instruction documents on the site, that you can choose from.

My 9 and 12 year old sons put it together in an afternoon. It's that easy.

EDIT: What can you do with the Galileoscope?

Obviously, you can watch some of the planets. You'll see the two big equatorial belts on Jupiter, and the 4 galilean moons of Jupiter. Saturn's rings will be quite obvious. Venus will show a crescent - it will look just like the Moon, only much smaller.

You can watch the Moon, too. The big craters are quite visible in an instrument this size.

You can watch some of the deep space objects too. The Orion Nebula is quite visible actually. The Andromeda galaxy is somewhat visible from a place outside the city.

Double stars such as Albireo, which appear as single star to the naked eye, are easily split by a 50 mm aperture. Albireo is very pretty, and easy to see in a Galileoscope - an orange/blue double, easy and wide, very beautiful.

Mizar, one of the main stars in the Big Dipper, is also a double. This one is a lot more tight. I think you should be able to split it in a G-scope anyway.

http://www.synapses.co.uk/astro/bearings.html

For the planets and the Moon, you'll probably want high magnification. 50x or more, up to 100x (the maximum supported by this instrument).

Make sure the instrument is sitting on a stable, firm support. A flimsy, shaky tripod is the enemy of high magnification.

Look on Surplus Shed and other online stores for 1-1/4" eyepieces. A 5 mm eyepiece will give you 100x magnification - the maximum. A 10 mm eyepiece will give 50x. A 20 mm eyepiece will give 25x - nice and wide, good for big objects like the Orion Nebula.

Look for eyepieces with a wide field of view (at least 50 degrees, up to 80 degrees, or more, if you can find them). A wide field makes everything easier.

It is also possible to build everything from scratch, including making the lenses and/or mirrors yourself from a piece of glass. But that is a MUCH bigger commitment, and a much more long-term project (a few months to a year).


Which scenario will be worse for the ultimate fate of the universe?

Well, that's a wee bit opinionated. It depends on your point of view - that is, how you take a liking to the following two scenarios. Everyone has their own opinion; from the details I give below, I'll let you answer that yourself.

To infinity - and . . . well, more infinity: The perpetually expanding universe

There are actually three types of universes that could fit this model: The accelerating expansion one, the constant expansion one, and the decelerating expansion one. At this point in time, we appear to be in a universe undergoing accelerated expansion, thanks to dark energy. However, all three scenarios give us a similar end result: A dead universe.

In the far future, as long as the universe does not curl back into a Big Crunch, slowly all processes as we know them will begin to end. Star formation will eventually cease, and black holes will slowly start to gobble up a lot of matter. In the even farther future, black holes are evaporate to decay via Hawking radiation, leaving the universe a collection of various subatomic particles. (If protons decay, then perhaps there will no longer be any atoms). The temperature of the universe will drop until everything is about thermodynamically equivalent. Not a great place to be (well, there won't be any life, so nobody will be around to see it, but still. . .).

Beyond this. . . well, there are a few different ideas regarding what could happen. The Big Freeze idea states that the universe will eventually just cool off to this low-temperature state. The Big Rip idea says that matter itself could be torn apart bye extreme expansion of space. Cosmological false vacuum theories suggest that quantum tunneling effects could make some interesting expansion possible. However, I emphasize that we do not know whether any of these long-term theories are correct. They are too far in the future, and there is too much uncertainty among the relevant factors.

In the short term, however, it appears our universe will expand forever.

Back to the beginning: The Big Crunch universe and all its friends

The Big Crunch scenario says that if the density of the universe is greater than the critical density, the universe will fall back onto itself into a singularity - just like the Big Bang. This leads to other theories, such as the epykrotic universe, or a Big Bounce, which postulate that our current universe is just undergoing one of infinitely many stages of expansion and collapse.

This isn't exactly a great outcome, either, but again, I'll leave you to judge.


Is blue light not traveling faster then red light in space?

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.


How exactly 'polar aligning' a Telescope with equatorial mount helps?

The Earth's rotation makes the stars and planets appear to move westward across the eyepiece field of a stationary telescope at about 900 arcseconds per minute.
An equatorial mount can counter this by moving only on its polar axis at a constant rate.
An altitude-azimuth mount must move on both axes at variable rates in order to track.

In early October 2018, Mars appeared to move about 1.1 arcseconds per minute relative to the stars.
In an equatorially mounted telescope, properly aligned and tracking at the sidereal rate, Mars would take about 13 minutes to move through its own apparent diameter of 15 arcseconds.
Without a tracking motor, you'd have to move the RA axis by hand, but you could leave the declination axis alone for a while.

If the mount has a tracking motor designed to be usable in either hemisphere, it may have a switch to reverse direction.
If that switch is in the wrong position, stars appear to move westward twice as fast as if the motor were switched off.

If an equatorial mount is misaligned, stars appear to drift in declination, depending on their position in the sky, up to 15 arcseconds per minute per degree of polar alignment error.
For example, if you're in the northern hemisphere and the mount's polar axis is 5 degrees too far to the east, objects near the meridian appear to drift southward at 75 arcseconds per minute, whereas objects 4 hours off the meridian drift only half that fast.
If the polar axis is more than 30 degrees off the celestial pole, it might as well be an alt-az mount.


What are six different methods to measure distances in Astronomy?

Using type Ia supernova light curves.

The RR Lyrae period-luminosity relation.

Moving cluster statistical parallax.

Surface brightness fluctuations.

Redshift-distance relation.

Using merging gravitational wave sources as standard sirens.

These and more can be pursued following this link. https://en.m.wikipedia.org/wiki/Cosmic_distance_ladder