Is it possible to figure out your location in the Milky Way if you are suddenly wormholed to a random, distant location in the galaxy?

Assuming you had access to the relevant astronomical catalogues and data, then yes it would.

Looking at constellations would of course be hopeless and most stellar catalogues only contain star positions for a relatively small volume in our Galaxy. However, there are classes of object inside and outside the Galaxy that would do the job.

A displacement in our own Galaxy would produce a notable displacement in the positions of local group galaxies, like Andromeda and M33, with respect to distant quasars that define the ICRS coordinate reference frame.

Providing one had the means (i.e. telescopes) to make the necessary observations, then it should be relatively straightforward to reconstruct your position in the Galaxy from the new apparent positions of recognisable local group galaxies.

Further possibilities are offered by globular clusters and pulsars.

The colour-magnitude diagrams and chemical compositions of globular clusters are all subtley different, which should enable their identification. They are also spread in a roughly spherical distribution around the Galaxy.

Radio pulsars have unique pulse shapes and timing properties that make them readily identifiable. A possible/probable problem is that many pulsars would be rendered invisible by the change in viewing angle.

In both these cases, a map of the revised spatial locations with respect to the ICRS should enable you to work out where you are compared with the position of the Earth now. I am unsure what precision could be achieved, but then we are unsure of our present location with respect to distance to the Galactic centre by a few percent.


On the lack of observational records by European astronomers of the 1054 guest star

Of the three known medieval supernovae (in 1006, 1054, and 1181 AD), only SN 1006 was definitely recorded by Europeans, and even then, was only discussed in any detail in a single European source (the annals of the monastery of St. Gall in Switzerland). Although there are multiple accounts (some quite detailed) of the supernova of 1006 in Arabic/Islamic records, there is only one known Arabic record of SN 1054, and none for SN 1181.

Since SN 1006 was by far the brightest of the three, the simplest explanation would be that SN 1054 was, like SN 1181, just not bright enough or long-lasting enough to merit sufficent widespread commentary in Europe. (Remember that China in the 11th Century was a much more advanced society than Europe of the same time, with a dedicated tradition of government-supported astronomical/astrological observations.)


The light ray does not bend (in fact it does a little bit, but that's small and doesn't matter now). However, what you see is not a path of a ray of light, but a projection of this path on the celestial sphere around you. This projection is an arc of a great circle with you in the centre.


Found laser ranging data here: http://www.geoazur.fr/astrogeo/?href=observations/donnees/lune/brutes

Here you can search for data for an arbitrary time period.

The data is what they call "MINI" format, which is hard to read, its basically a long string of numbers.

Here is a sample line:

5120160113152419452625024340653926601301910034002705017 087323+04325 5320a0702

Luckily there is a specification for this format here: http://www.geoazur.fr/astrogeo/observations/donnees/lune/mini-format.html

The spec says the time-of-flight for the laser is charcters 24-37 from each line, measured in .1 picoseconds. So for the above line, the laser's round-trip flight time is 24340653926601 (.1 ps).

The data does not contain the distance, so to calculate the distance from the flight time, I do the following:

Divide 24340653926601/2 to get 1-way flight time in .1 ps.

Multiply the result 1.2170327e+13*.1 to get ps.

Multiply the result 1.2170327e+12*1.0e-12 to get seconds.

Multiply the result 1.2170327*299792458(the speed of light) to get the distance in meters: 364857224.599


Shadows of Light = Space or Dark Matter?

You cannot have dark without light.

Not true in the case of dark matter (for the general case, see below). Dark matter is called "dark" because it appears that it doesn't absorb or emit electromagnetic radiation - light! It can interact with light via gravitational lensing, but dark matter particles have no electromagnetic charge (we think) and so are considered "dark".

  The only thing faster than light is dark (i.e. shadows)

This is true, as the video explains.

  Does this make "dark matter" mere shadows of extremely bright objects?

This is the meat of the question, and the answer is a clear "no".  First, any "bright objects", while capable of casting a shadow of something else, will emit electromagnetic radiation.

Okay, so why can't there just be non-extremely luminous objects casting a shadow? One reason is that the dark matter simply isn't where these shadows should be. Many galaxies have a dark matter halo that extends far beyond the galactic plane. Any light sources emitting light (that is then shadowed) would have to be on the opposite side of the galactic disk. This is simply not the case.

The second, more convincing, reason is that dark matter has mass. In fact, that's the reason the idea was first conceived (to account for anomalies in galactic rotation curves)! Shadows don't have mass; therefore, dark matter cannot be explained by shadows.

  Is that also why it has to get dark before we can really see the majority of stars (subjective experience)?

No. I don't know for sure, but I believe that that's because the Sun's light is still blocking out the light of distant stars.[citation needed]

Addendum after edit to question

  Can you provide sources for your claims?

I think that they're pretty well sourced! The first section can easily be checked via Wikipedia (which I assumed you had seen). The rest seems to be well-sourced, too (aside from the last bit, which is, admittedly, unsourced but only tangential).

  Also I didn't say you can't have DARK MATTER without Light , I said you can't have Dark, as in Darkness. I think you're getting hung up on Dark Matter which is a big part of my question, but not the entire question.

Okay, I may have misinterpreted this.

Darkness is, by definition, the absence of light (electromagnetic radiation). So no, you don't have to have light for there to be darkness.

  Also, are you claiming that all darkness in space is dark matter? I don't like to assume but for scientific purposes (not an accredited scientist) would we not agree that the measures made on dark matter were on a sample, and not the entire universe?

I never said that, but you did. Your original question asked if dark matter could be shadows. The answer is no.

  So to further question this, do shadows not exist in space?

They exist. Any object blocking a light source casts a shadow:

This happens to have been used in the video, so it shouldn't be new to you.


How can artificial gravity hold the water in swimming pool?

"Artificial gravity" should act like gravity. So just as gravity has a direction on earth and holds water in a pool, so should  "artificial gravity" hold water in a pool on a space station with  "artificial gravity." Notice that bubbles rise in the space station pool just as they would on earth.

If gravity were loss on a space station with a pool, then the surface tension of the water would want to pull the water into a sphere.

I'll also point out that water is an anti-ferromagnetic material in earth swimming pools, and gravity works just fine. So using magnetic boots on a steel deck is very different that having gravity which would work with an aluminum deck.


How does using Tidal energy affect the moon?

It is minute and inconsequential.  Without tidal energy harvesting, the motion of the water is ultimately converted into heat, it has a very small heating effect (that is swamped by the heating caused by the sun). With tidal energy harvesting, instead of heating the water, a little of the energy is converted to electricity (which is then converted into light or heat,  ultimately the energy becomes heat)
The tidal dams can change local tidal patterns, which could have a minute effect on the moon's motion, but this effect is minuscule.


How can I hear (or at least detect) a pulsar at home?

Interesting questions - I believe I have answers to them.

You mentioned folding analysis - yes, you will need to do folding analysis to find pulsars - the majority of pulsars are found with what are called prepfold plots - here is some software that is very popular, but you can do this yourself. The general principle is that if you separate a signal at a time interval equal to the period of the pulsar and stack the resulting sections of the signal, you're effectively doing the equivalent of "image stacking" as you would do with images you take in a camera, which will boost the signal to noise ratio of the pulsar you're trying to look at.

The best bet for a small setup would be the Vela pulsar - here is a list of significant pulsars. The Vela is the brightest pulsar in the radio part of the EM spectrum (as per the link, which is from Wikipedia, so take its reliability with a grain of salt), so I would assume that's your best bet in trying to detect a pulsar.

As for antennas or dishes, I would think a dish would work best. You're going to want some decently sensitive equipment to detect a pulsar, so I would recommend a proper dish made for collecting radio waves.

And lastly, you would want to consider which frequencies the target pulsar emits at and check which ones your receiver can collect.

I think you've pretty much covered all the other issues; now you just need supplies and money :).


How do GHA of vernal equinox, GHA of astronomical object and RA relate?

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

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

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

GHA = GAST - apparent right ascension (3.14a)

solving for apparent RA,

apparent RA = GAST - GHA (3.14b)

Comparing your equation

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

So your sign seems to be inverted.

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

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

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


Spectra of electromagnetic radiation emitted during solar flare

Flares involve releasing huge amounts of potential energy that are stored in stressed magnetic structures in the solar corona. So-called reconnection events can release this stress but at the same time accelerate charged particles to high energies. These are trapped along the magnetic field lines, spiralling around them, but travelling until they hit the footpoints of the loop that are anchored in the solar chromosphere and photosphere. The much denser plasma here is sufficient to stop the charged particles and this is where they dump all their kinetic energy, some of which is promptly released in the form of non-thermal hard X-rays, which are insufficient to cool the plasma.

The plasma heats up, increases its pressure and it expands to fill the overlying magnetic structures - a.k.a. chromospheric evaporation. This hot plasma radiates and because it is at temperatures between $10^{5}$ and $10^7$ K, most of that radiation is emitted at EUV and X-ray wavelengths in free-free and transitions in highly ionised species.

I guess the root cause as to why the plasma can get up to these temperatures is that the radiative and conductive cooling processes are much slower than the heating processes caused by the accelerated charged particles.


How to understand exactly why gravity darkening happens on rotating stars?

The argument goes something like this.
Hydrostatic equilibrium means that the local pressure gradient is proportional to the local density multiplied by a latitude-dependent local gravity. If the pressure just depends on density and temperature, this means that those quantities will also just depend on latitude and therefore will be constant along an equipotential surface. i.e. pressure, temperature and density are functions of the effective gravitational potential $\phi$.
For stars with radiative outer envelopes, the heat flux is proportional to the temperature gradient, multiplied by some stuff (like inverse opacity) that just depends on density and temperature.
But
$$\nabla T(\phi) = \frac{dT}{d\phi}\nabla \phi = f(\phi)g_{\rm eff}$$
If we now say that at the surface $\sigma T_{\rm eff}^4$ equals the radiative flux, then we recover the Von Zeipel gravity darkening law that $T_{\rm eff}$ is proportional to $g_{\rm eff}^{1/4}$.
The missing step in this argument is to show the $f(\phi)$ is constant. Given that the photosphere is defined as where the optical depth is some fixed value (usually 1 or 2/3), and may be assumed to depend only on temperature and density, then this also lies on an equipotential. But $dT/d\phi$ also only depends on $\phi$ and so must also be constant along an equipotential.
For more detail, although missing the last paragraph above(!), see https://www.astro.umd.edu/~jph/Stellar_Rotation.pdf .
The situation is much more complex for stars with convective envelopes or differential rotation and I think can only be tackled through detailed modelling.


What is the unit used for star brightness due to skyglow?

The brightness of a diffuse area is it's luminance, and can be quoted in units of $cd\,m^{-2}$. Measuring it would require knowledge of the properties of the camera: the sensitivity (at various wavelengths) of the ccd, the aperture, length of exposure and so on. This could probably be found using calibration images, in which a known luminance is imaged and then the brightness of the skyglow is compared.

The brightness of a star is usually given as a magnitude which is related to the amount of electromagnetic energy received from the star per second, in a particular range of wavelengths.

It's unclear what you mean by "brightness due to skyglow". The brightness of stars is not due to skyglow, but excessive skyglow can mask dimmer stars and diffuse objects.


Find a local astronomy club; they will help you get started, learn the sky, and show you different types of equipment. Also, read "Turn Left At Orion", H.A. Rey's "A New Way to See Them" and "Find The Constellations" to learn the sky.

And get a pair of 10x50 binoculars. These are the best "telescope" (actually, two telescopes!) to start with: inexpensive, wide-field of view to find objects (anti-soda straw effect), and 100x the light gathering power of the naked eye.


The picture is a mocked-up fake and is not an actual picture of the solar spectrum. You can easily see this because the black "Fraunhofer lines" extend beyond the spectrum and H alpha should have an appreciable width.
The table is massively incomplete. It list only a tiny fraction (the strongest) absorption lines in the solar spectrum. There are thousands of others.
I guess whoever mocked up the spectrum didn't use that table.
A real solar spectrum is shown below. The Fraunhofer lines are not easy to see at all in a spectrum that has low enough resolution to include the whole thing in one linear image. Any lines due to mercury will be weak because mercury is not an abundant element in the solar photosphere.

A better view is offered by an echellogram taken at higher resolution. This begins to demonstrate just how many lines there are in the solar spectrum.


Could Planet Nine be much smaller than thought previously?

It could. As posted by the original authors on their blog pages findplanetnine.com it might be ~5 earth masses

https://arxiv.org/abs/1902.10103 and the answer to this thread


Black holes really look black.

Things look black because they don't reflect or emit light. A black hole is a region of spacetime which light can enter, but can't leave. Black holes don't reflect or emit light, so they look black.

Black holes are not the only things that bend light. Every object with mass bends light. Black holes bend light more than other objects because there is so much mass in such a small region.

Black holes don't have a surface, so they aren't black for the same reason that soot is black.  They are black because they don't reflect or emit light. Space also appears black because empty space does not refect or emit light.

"At the centre" is a difficult idea for black holes, because space and time are bent so much near the black hole that the centre becomes a "when" not a "where".  The laws of physics aren't defied, but we don't know what the laws are at the time in the centre of a black hole.

(Hawking radiation is predicted to be emitted by black holes. This makes black holes emit light, but for a large black hole Hawking radiation is very low energy and very dim.  It is theoretical, it has never been detected, but the theory is very strong and well supported)

(Around the outside of the black hole matter may be falling in. This matter gets very very hot, and so shines very brightly. This means we can see where a black hole is, because we can see the matter that is falling into the black hole)


Who discovered the relation between CMB fluctuations and large-scale structure formation?

I'm not an expert in this, but I'll give it a try.

I think you've got the argument the other way round here. This was never discovered. It was assumed. But why?

We know for the CMB being a redshifted blackbody-spectrum. This means the CMB consists of photons that were emitted in interaction with matter.
The CMB fluctuations inform us about the depths of gravity wells that those photons had to escape from (amongst other things). Thus we get a picture of the baryonic and dark matter density at the time of photon-matter decoupling.
This allows us to set up linear perturbation theory + simulations to look what happens to a uniform gas with the measured fluctuations on top. That's the rough idea.

You could now of course ask "who was the first to implement this?".
The earliest theoretical work that I'm aware of (Peebles 1965) linking "a possible universal blackbody radiation" with the formation of galaxies came in the year of the discovery of background radiation by Penzias & Wilson 1965.


What are the cloud-like blobs in the Martian southern hemisphere?

From the Nature article, the authors propose two alternative explanations:

We used photometric measurements to explore two possible scenarios and investigate their nature.

For particles reflecting solar radiation, clouds of CO2-ice or H2O-ice particles with an effective radius of 0.1 micrometres are favoured over dust.

Alternatively, the plume could arise from auroral emission, of a brightness more than 1,000 times that of the Earth’s aurora, over a region with a strong magnetic anomaly where aurorae have previously been detected

Importantly, both explanations defy our current understanding of Mars’ upper atmosphere.

I think, they are the most trustworthy interpretations that one can get.
Please note that, even if the data are from 2012, the paper is published in 2015, therefore the interpretation is not outdated.


Why isn't the asteroid belt affected by Jupiter's gravitational field?

The asteroid belt is affected by Jupiter's gravity.

There are stable orbits inside of Jupiter's orbit. Jupiter's Hill Sphere has a radius of 53 million km. If you are more than 53 million km from Jupiter, then the Sun's gravity dominates and you can orbit the sun. But Jupiter orbits 780 million km from the Sun, so there is plenty of space between Mars and Jupiter for asteroids to have stable orbits.

As they pass Jupiter on their orbit they will get first a pull forward, and then a pull back but as these pulls happen at different locations on each orbit they tend to cancel out and result in effects like precession of the asteroid's perehelion or periodic variations in eccentricity. These effect act on the Earth and other planets too.

However if the time the asteroid takes to orbit the sun happens to be an exact fraction of the time that Jupiter takes, then it might receive the same pull at the same location. Resonant orbits like this tend to be unstable. The asteroid will be nudged out of the resonant orbit and into one that doesn't match up with Jupiter.  This creates the Kirkwood Gaps in the asteroid belt.

Source


Usage of the standard solar system (astronomical) symbols

As Anders Sandberg wrote, the Sun, Earth, and Jupiter symbols are used for putting things in context, for example M☉ for solar masses or M⊕ for Earth masses. However, the astronomical community has stopped using the other symbols early in the twentieth century. Astrologers still use them, but not astronomers.


Need help with telescope mount; stars have tails

If you can't get any longer than 20 seconds, chances are it's your polar alignment. The mount you have is a motorised GoTo EQ mount (I believe), but if these aren't aligned properly, then they can still have trouble.

Also, depending on your software, try a 2 or 3 star align to get a better accuracy.

Also check the scope is nicely balanced with the camera on the end, and that your mount can take the weight.

EDIT :

If you try all this and you are still getting trailing, it may be worth going to a professional to get the mount looked at. It may be faulty.


Why does the Simbad page "A.A. Michelson's Jovian Galilean-satellite interferometer" show data for Betelgeuse?

The SIMBAD link might be there just because Osterbrock's 2004 AAS presentation about the interferometer mentioned an observation of Betelgeuse.
This would be consistent with the policy stated in Wenger et al. 2000:

No  assessment  is  made  of  the  relevance  of  the  citation in terms of astronomical contents: the paper can be entirely devoted to the object, or simply give a side mention of it - in both cases this gives a reference in SIMBAD.
[...]
SIMBAD approach favours exhaustivity, at the cost of increased information noise.

Other articles about SIMBAD bibliography, from Laloë et al. 1993
to Delacour et al. 2018,
show an evolution of their process from mostly manual to largely automated, but knowledgeable humans are still in the loop.
Sometimes an instrument or project intended primarily to investigate one subject also makes important contributions when reused on another subject.
For example, the Dark Energy Survey of distant galaxies has also contributed to the discovery of several trans-Neptunian objects.
However, that is not exactly what happened here.
Michelson 1920 discusses another interferometer spanning the objective of the 100-inch telescope at Mt. Wilson, noting that atmospheric seeing conditions affected the results less than expected.
Betelgeuse required a longer baseline, so Michelson and Pease 1921 used four 6-inch mirrors mounted on a 20-foot steel beam.


Let's start with a quick clarification: Red Shift is not the same as red light.

Red light is just electromagnetic radiation with a 400–484 THz frequency range, the lowest our eyes can see - highest being violet light, with a 668–789 THz frequency.

Red Shift is an observable effect when you analyze the spectroscopic signature of an electromagnetic radiation source that is so far from us that the expansion of the universe (whose unit of measurement is called the Hubble Constant) is noticeable because of the frequency drop, called Doppler effect.

In short, it works like this:

Pick one of the elements the universe is made of - hydrogen (75%), helium (24%) and trace amounts of everything else (1%). For this example, let's stick with hydrogen.
Determine the spectrum signature of hydrogen by running the light generated via electric discharge on the gas through a spectrometer:

source: flatworldknowledge.com

You'll notice some markings as you examine the light dispersed by the prism. Those markings are called narrow bands, and they appear as a consequence of quantized energy states. Hydrogen narrow bands are well known : 656.2, 486.1, 434.0 and 410.1 nanometers.
Now look at the sky. Pick any bright dot, run its light through a prism, and knowing that it is also made of roughly 75% hydrogen and 24% helium compare its spectroscopic signature to the bands you obtained earlier.

Because of the Doppler effect caused by the expansion of the universe the light waves are stretched, making the frequency drop - shifting the whole spectroscopic signature towards lower frequencies. The further the object is, more noticeable the shift towards red is.

Thus, the term Red Shift.

Now, answering your question:

  How do we know that its not just red light from red stars?

Because even regular light from red stars will shift towards the infrared side of the spectrum as a consequence of the Doppler effect.

Image Sources:
General Chemistry: Principles, Patterns, and Applications, v. 1.0
Starts With A Bang!

More information:
Spectral Lines of chemical elements


Is Ceres in the inner solar system or outer solar system?

Inner. The entire asteroid belt is in the Inner Solar System (now). The definition of "Inner" vs. "Outer" is non-arbitrary, based on the current "frost line", approx. 5 a.u. radius.

https://en.wikipedia.org/wiki/Frost_line_%28astrophysics%29

https://en.wikipedia.org/wiki/Solar_System#Inner_Solar_System


How can we know the nature of the companion in a neutron star binary?

One way to figure out if one (or both) of the objects is a black hole, neutron star, white dwarf, or other compact object would be to try to measure its mass. For example, a neutron star and a white dwarf are both compact stellar remnants. However, there is one decisive factor that determines which type of stellar remnant a progenitor star will become: the remnant's mass.

The Chandrasekhar limit ($\sim1.39M_{\odot}$) is the maximum mass of a white dwarf; the Tolman-Oppenheimer-Volkoff limit ($\sim1.5-3.0M_{\odot}$) is the maximum mass of a neutron star. The large uncertainty in the latter means that it's not particularly useful for identifying compact objects that could be neutron stars or low-mass black holes.

So if you know the mass of one of the objects - and let's assume that it's well within the mass range of one of the classes of stellar remnants - you might be able to figure out which type it is. How would you measure the mass? Well, you could study the orbit of its companion star to try and determine the stellar remnant's effect on its orbit.

Another way would be to study gravitational waves emitted by the system. These waves can only be emitted under certain circumstances - for example, in a system of binary neutron stars - see, for example, the Hulse-Taylor binary system, also known as PSR B1913+16. The power radiated as these waves, as well as the orbital decay, depend on the masses of the objects. While detecting gravitational waves is an incredibly difficult task, there are several detectors planned or already in operation, such as eLISA and LIGO.

You could also attempt to look for other effects that characterize certain stellar remnants. For instance, a certain pattern of radio waves would characterize a pulsar (white dwarfs and black holes are not normally strong radio sources). However, if the objects are not luminous, this may not be effective.


What would happen if a rogue planet hit one of the planets in our Solar System?

It would depend entirely on the size, composition and trajectory of the two objects coming into the collision. They could annihilate each other leaving a mess of debris. Or one or both could survive the collisions in some form. And each of the surviving objects could be flung out of the solar system, flung into a spiral into the sun, flung into an elliptical orbit, or thrown into other objects. The space between planets is big so odds of direct and immediate collision of the "planets like billiard balls" variety you see on some sci-fi aren't likely. But anything the size of a planet would be seen well before any impact.


If I could see from a Gravitational-Time dilated place to somewhere far which is not dilated, would I see fast moving people?

a Gravitational-Time dilated place to somewhere far which is not dilated

"Dilated" or "not dilated" are not absolutes. It's always a relative thing. "Dilated" relative to me, but perhaps "not dilated" relative to you. There is no dilation that happens "in general"; it's always relative to another place/observer/reference.

  Would those people look fast moving?

Yes.

Now, everything coming out of that place would seem like it's "speeding up time" relative to you. If you're looking at them through a telescope, all light would be blue-shifted. If you're receiving a TV transmission, the frequency of the carrier would also be higher.

Keep in mind, this time dilation is not an illusion. It's not just "how you see things". Their flow of time, as measured by you, is indeed faster. This affects everything in that place, as measured by you.

But it's also relative. Some other observer may be in a place where time flows differently from yours - they would look through the telescope and would see a different rate of change. And it's not an illusion for them either.

Time dilation is real. What's not real is a unique time, ticking the same for everyone - that's the actual illusion.

Time is a local thing. (more precisely, the rate of ticking for a clock is a local thing)


What is the difference between equatorial and ecliptical coordinate system?

To locate an object you need a system of coordinates.

You need to define a plane (to define the latitude) and a direction (from which to measure longitude). On Earth, the obvious plane is the Equator. There is no obvious direction from which to measure longitude (which in the past led to different definition of longitude in France and Britain)

There is no single, natural plane in the sky. Several are possible: You can project the equator into the sky. You can use the plane of the Earth's orbit around the sun. You could use the plane of the galactic disc.

Using the Equator is simple, but rather Earth-centric. The Ecliptic system is more natural for solar system objects. The galactic system nice for deep space objects. The lack of a single natural plane in the sky means that several different coordinates systems can co-exist.

As an example Rigel, at RA 5h14'32'', Dec −08° 12′ 05''

Has a Lat-Long of:

-31.11°  76.09°   Ecliptic
-4.40°    219.66°  Galactic


Did nobody in the Astronomy community think 12,000 new satellites in LEO might be a problem?

“No one thought of this,” she said. “We didn’t think of it. The astronomy community didn’t think of it.”

What utter nonsense. SpaceX are either lying or they didn't bother to go looking for any other opinions.

Astrophotographers have been complaining about satellites messing up their photos for years. When Starlink was announced astronomers complained even louder.

Go to any popular astronomy forum and search for "starlink" and see for yourself #1 #2.


How does an eclipse differ from an occultation?

There are a number of terms for these things:

conjunction:
When two objects have either the same right ascension or the same ecliptic longitude (usually as observed from Earth). That is, they lie in (almost) a straight line from the point of view of the observer. In the case of the Solar System, such as when the Sun and Venus line up from the point of view of an observer on Earth, this is a conjunction. (It is also a conjunction when Earth lines up between e.g. Mars and the Sun - it is an inferior conjunction from the point of view of Mars (but we can't observe from Mars)).

Furthermore, when Venus lines up between the Earth and the Sun this is an inferior conjunction, and when Venus lines up with the Sun but on the opposite side of the Sun then this is a superior conjunction. Thus the inner planets (Mercury, Venus) can have both inferior conjunctions and superior conjunctions with Earth while the outer planets (Mars, Jupiter, etc.) can only have superior conjunctions with Earth.

The "almost" in "almost a straight line" is important because the definition is by longitude. Although the planets are said to lie in the ecliptic plane, this is not completely flat - they do not all have the exact same orbit inclinations. Thus not every conjunction is a transit, occultation or eclipse - they do not pass directly in front or behind.

transit:
When an object passes in front of a larger object*, this is a transit. These are rarer than conjunctions because the objects need to be directly in line so that one can be seen crossing the other.

In the case of recent transit of Venus, when Venus passed in front of the Sun, the sunlight passing through Venus' atmosphere gave us the opportunity to collect data on its composition. When one of the moons of Jupiter pass in front of Jupiter itself* this is also a transit.

occultation:
When an object passes in front of a smaller object*, this is an occultation. The farther object is hidden by the nearer object. These are also rare** because, like transits, they also need to be directly in line so that the nearer object can be seen crossing in front of, and blocking out, the farther object.

Asteroids occulting stars are studied because while we can calculate the asteroid's orbit and its distance from us using its speed, knowing how gravity and orbits work, etc., we can't see clearly enough to determine the size (think few kilometres in diameter but millions of kilometres away). But when an asteroid passes in front of (occults) a distant star, the star is effectively a point source (looks tiny), from the amount of time the occultation lasts and the asteroid's speed we can calculate its size***.

eclipse:
The Moon passing in front of the Sun is, depending on the conditions, a partial eclipse, total eclipse or annular eclipse.

Clearly, by the definition, when the Moon fully blocks the Sun it is an occultation and when it only partially blocks the Sun it is a transit, but
this is a special case and is called a solar eclipse rather than a transit or occultation. (Similarly when the Moon passes through the Earth's shadow - when the Earth is directly between the Sun and the Moon - it is called a lunar eclipse.)

The word eclipse comes to us via Latin and French from ancient Greek ekleipsis, from ekleipein 'fail to appear, be eclipsed', from ek 'out' + leipein 'to leave'. (see eclipse on Wikipedia)

This alignment is also known as a syzygy****.

* Smaller or larger as they appear to the observer, an asteroid is most definitely smaller than a star but because it is so much nearer it looks bigger.
** A lookup of occultations can return a long list so that it may not seem rare, but when you compare how many asteroids there are, and how many stars, in relative terms it actually does not happen very often.
*** We can set limits on its size, given that the "average" asteroid looks like a lumpy potato rather than being some nice, regular shape that would make it easy to work these things out.
**** We could keep adding terms and googling them all day... ;-)
***** There is no footnote with five *


Super-Earths and Mini-Neptunes are the "in-between" types of exoplanets you're looking for. A sweeping generalization would put most in the range of $\sim1$-$10M_{\oplus}$ (Earth masses), with some outliers a bit above that. They may have significant quantities of hydrogen and helium in their atmospheres, as well as water, in liquid or vapor form. The latter should also have what are referred to as volatiles (colloquially called "ices") - compounds such as ammonia and water that are present in the atmospheres of proper ice giants like Uranus and Neptune.

One reason we don't have any of these in the Solar System is simply that no worlds with the proper mass formed. I explained here that there is a transition region of sorts, but most super-Earths require minimum masses of $1.5$-$2M_{\oplus}$ to keep their large hydrogen/helium envelopes (which terrestrial planets may accrete early in their lives but later lose). There would have to be a planet just in that mass range and at the right distance from the Sun (see D'Angelo & Bodenheimer (2016) for general constraints on semi-major axes). The environment in the protoplanetary disk may simply not have been suitable to produce the right body at the right time.


How to reduce magnification and increase FOV of a camera attached to a telescope?

You are using the webcam as an electronic eyepiece. People have been using various types of cameras as eyepiece for decades.
The reason for the small field of view is the small (1/4") sensor of the camera. Only a part of the actual field of view strikes the sensor. The field of view of your telescope will likely cover a 2" eyepiece. You are putting a 1/4" sensor into the middle of a 2" circle. Tne sensor only sees the center portion of the total field of view.
Most use focal reducers in front of their cameras to increase the field of view of their electronic eyepieces.
I started electronic viewing twenty years ago with webcams and CCTV security cameras. I now use several Mallincam Astro cameras.
Google "electronic eyepiece for telescope" or "electronic assisted astronomy". You will useful information once you get past all the advertising.


Black hole collision animation: What are these extrusions?

The shape of the surface shown in the video is a depiction of the spacial curvature of the spacetime. (The relationship with time are depicted seperately by the arrows and the colors.) More particularly, the shape is depicting the curvature of equatorial plane of the binary. The depicted surface has been embedded in a (fictional) 3D space in such away that the curvature of the surface is equal to the intrinsic curvature of the equatorial plane.

Lets try to unpack what this means for the interpretation of the extrusions. First note, that there is no physical meaning to whether something is show as an extrusion or a depression, this does not actually effect the curvature. The video makers could have also chosen to depict the depressions around the black hole as extrusions instead --- without changing the meaning.

What is relevant, however, is that some regions are shown as depressions while others are shown as extrusions. This means that somewhere in between, there must be a saddle point in the depicted surface. Saddle points correspond to regions with negative (spatial) curvature (i.e. an area where if you would draw a triangle its angles would sum up to less than 180 degrees). The extrusions themselves quite clear have positive curvature (a triangle would have more than 180 degrees).

Note that the sign of the spatial curvature has little to do with whether gravity is attractive or repulsive. If you want an indication of direction in which gravity is working, the arrows give a better idea (although that interpretation should also be taken with a pinch of relativistic salt).

EDIT:
Clarifying the last point a bit. The animation depicts three aspects of the spacetime curvature: The rate at which time flows (the lapse) as a color map, the rate at which space is dragged (the shift) as gray/silver arrows, and the spatial curvature as the curvature of the surface. Together these three completely characterize the curvature of spacetime. Consequently, they dictate how a test object would move in the spacetime, i.e. "how gravity acts". Although all three elements are important for the motion of particles, some give a better qualitative indication of the behavior of test particles than others. In this respect, the color map and the arrows are more important than the spatial curvature. The typically a particle will want to move with the arrows, and along the gradient of color towards the redder regions (in both cases this generally means toward the black holes). The spatial curvature plays a somewhat secondary role, and matters mostly for particles moving at high velocities. Hence my comment that the sign of the spatial curvature is not a good indicator for whether gravity is attractive or repulsive at some point.


Does 2 merging black holes necessarily make a quasar?

Basically no.  While the merging of 2 black holes is a very interesting event, a quasar is what you get when 1 very large black hole eats a whole bunch of matter and the light from the quasar comes from the intense heat and interactions from that tightly bound, rapidly spiraling and very excited matter.

Quasar's were probably most common when galaxies are young but there are a few more recent ones (see examples in comments).  Quasar

Related Stack Exchange Question

  If the size is smaller than what would be considered as a supermassive
  black hole would it not be a quasar?

Almost all large galaxies have a super-massive black hole in their center.  Source.  The sizes vary with the size of the galaxy.   Quasars are much more rare, only in a few galaxies.

As for the specific size of black hole that can form a quasar, credit to Rob Jeffries below.


Why hasn't the Earth fallen into the sun yet?

No not nearly enough drag.  There just isn't much dust in space, and the Earth has so much momentum that the tiny amount of interplanetary gas has a tiny effect on the Earth's orbit.

There's only about 5 particles of mostly hydrogen gas per cubic cm in the region around the Earth. It has a drag on the earth of a few thousand Newtons. But the mass of the Earth is $5\times10^{24}$ kg, so the deceleration is of the order $10^{-20}$ m/s2.

At this rate it would take longer than the age of the universe for drag to make a significant change to the orbit of the Earth.

Things were different when the planets orbited in an accretion disc. At this time, large planets could make excursions from the outer part of a stellar system to the inner, due to friction from the dust that they encounter.

The Earth does gain a little mass from dust, rocks and asteroids that continuously hit it, but it also loses a little mass to atmospheric lose. The total is slightly negative (overall Earth loses a little mass) but this doesn't affect the orbit, as orbital velocity is independent of mass.

Large bodies do change the orbit very very slightly, but comet and asteroid impacts happen from all directions, and there is little cumulative effect.

The gravity of other planets does perturb the orbit of the Earth, but not to make it spiral into the sun, but to cause the elliptical orbit to become first more, then less eccentric, and change the orientation of the ellipse.


If it only takes 8 minutes for that light to reach the Earth, what took up most of the time?

TL;DR: Because light doesn't travel at an infinite speed.
Light travels at exactly 299,792,458 meters per second, and one AU is equal to 149,597,870,700 meters. Therefore, it takes about 500 seconds, or 8 minutes and 20 seconds, for light to travel from the Sun to Earth.


Number density of stars on the Galactic plane

The number density of stars is not well constrained, because it hinges critically on the behaviour of the stellar mass function at low stellar masses. Low-mass stars tend to be dim and hard to detect. What is reasonably well constrained, instead is the mass density of stars, where low-mass objects contribute little, even if their number is large.


Blowing up an asteroid/comet really potentially worse?

Well there are some things to consider. Initially if you could make sure that after you blow up an asteroid you will end up with numerous but small enough pieces so that they will either: one, burn up in the atmosphere or two, be headed away from Earth (and not hitting us five years later) then we are OK, and blowing up the asteroid with a missile would be a fair solution.

The problem here lies in the fact that we know little about the internal composition of asteroids in general, and presumably even less about a particular one, so it is very hard to predict exactly where the pieces of the asteroid generated by an impact are or aren't going to end up or, be headed towards or even its size.

Another scenario could be that if you effectively smashed the asteroid into small pieces that could then burn into the atmosphere, and if those pieces were coincidentally to end up being consumed by Earth's atmosphere, it would heat up provoking presumably an unpleasant day on Earth of course depending on the mass of the object.

But there is a much better solution than that Armageddon-Hollywood inspired one. It is call gravitational tethering. There is something we know, and we know very well about asteroids, and that would be their trajectory paths or orbits. Even when a new asteroid is discovered, its orbit can be computed pretty quick and with great accuracy (because we know the solar system's gravity very well). So if an asteroid is to impact Earth, it is likely that we will know with years, probably decades in advance. And so we can just send a space vehicle (called gravity tractor), with enough mass and time in advance, and place it just beside the asteroid, hence allowing us to tilt its orbit by just a tiny amount, due to the gravitational pull between the two objects. Now when you consider the effect of that tiny amount in the long run, it effectively deflects the path of the given asteroid from that of the Earth so that it won't hit us 20 or 30 years later.

And this is something we have control over, and something we can predict with great accuracy. It is the (safe) way to go.

If you are still not happy with my answer, you can listen to Neil de Grasse Tyson himself explaining it in this 5 min video.

Also check out this talk from the American Museum of Natural History on "Defending Earth from Asteroids" LINK

Further reference here.


Is there a common metric to measure telescope power?... Is there something like an APR for telescopes?... Is there any objective rating system?

If there was one single metric, it would be aperture, but there really isn't because there are several other aspects that can make one telescope much better suited for viewing than the other.

The diameter is most of the power of a telescope, dim stars always appear brighter, doubling the diameter of the objective results in 4x the area to collect light, so the apparent magnitude of a point object like a star goes up by 1.5 or so.  Also a greater diameter increases the detail visible to the point where you can use more magnification, or at least make the same view much sharper.
The ratio of the focal length to the diameter is the "focal ratio", for instance a short focus reflector with an 8" mirror and a 32" focal length has an f4 ratio, and a refractor with a 3" lens and a 36" focal length has a f12 ratio.
Shorter focal ratios are generally better for observing extended objects such as galaxies and nebulae since the available light is concentrated into a smaller angular size for the image, whereas long focal ratios are better for observing detail such as planetary observations and splitting close double stars because a given eyepiece will result in greater magnification. I take it you're a novice at observing astronomical objects, so I'd recommend dipping your toe in the water by buying a relatively small middle range focal length scope for general observing without breaking the bank.  Oh, and avoid those department store telescopes which are shaky and don't have good optics. Claims of 700x magnification don't do anything but make the wind blown blur more apparent.


Thought experiment about the outskirts of the universe

I will answer all of your points with zero tact and/or precision in order to fit them all in.

1) It will take you about a "split second" given you stated that it only takes your brother a split second to travel from Earth to the point 13Bly away and back again. He will be exactly said split second older and no more, because as he is traveling faster than the speed of light, the rules governing his and your aging no longer apply, meaning you have created a scenario where both his and your time's are connected and simultaneous.

2) At the point when you meet up with your brother a split second later, the universe will be a split second older... At the time you've reached your destination the universe will be [ > 13By + (the age of the universe when you left) ]. The fact that your ship takes one year to accelerate to and from the top speed is irrelevant given the duration of the travel time.

3) .999c ... Why not, assuming that you have enough "fuel" your imaginary ship should be more than capable than making a single one way trip. There given the distance and time it would take you to get there, there is close to zero reason to accelerate either sooner or later, although the whole near speed of light travel is sort of the point of your question, it would kind of be necessary.

4) The N times scenario is redundant and a little weirder than your other questions, as the speed of light is the same in all reference frames, you would never breach that boundary, however to each previous ship the rate of travel would appear exponential even though there would only be infinitesimal gains. Would you ever reach free space... no, without faster than light travel it is not possible. I would venture a guess that space would look somewhat similar to the here and now regardless where you ended up within your lifetime. a) by tinker, you could assume that you were able to travel as fast as your twin in part (1). b) Think Star Trek, impulse power was sub light speed travel, all warp speeds are some large multiple of c.

Look into relative simultaneity and space-time diagrams for reference on the whole twin paradox issue. Also you could read something like ... the physics of star trek for faster than light travel scenarios with actual physics explanations or problems.


What is the real position of the tidal bulge?

There is no great contradiction. The offset of the tidal bulge is about 3 degrees. It is exaggerated in diagrams for clarity. The diagram is correct, but not to scale.

This causes the tides to be slightly late. Imagine a person standing on the Earth of the diagram, with the moon directly overhead. The tidal bulge is on their left. The rotation of the Earth will take them towards the left (the moon is also orbiting but its motion is much slower), so a little later (12 min later) they will reach the maximum of the tide. The maximum is delayed by about 12 min.

Actual flows of water around the coast are driven by this tidal bulge, but are complex effects of local topography. The actual flows of water are highly non-linear, including multiple locations at which there is no tide.


Why do we presume that the universe has always expanded?

We can, after making some approximations, trace the movements and locations of the matter in the universe back in time, using the laws of physics (mainly general relativity).

At the moment we see a certain density of matter which is, on a large enough scale, expanding uniformly -- as if space were being blown up so that every bit of it doubles in size every seven billion or so years.

When we extrapolate that back we find that long ago the matter density was much higher, but it was still expanding in a similar kind of way, and this would seem to hold true back to a state of very high density indeed, at which point our existing understanding of the physical laws stops giving us an extrapolation that makes any kind of sense and we have to stop.  Up to that point though, the extrapolation gives us a picture which is pretty consistent with all the various ways we have of observing it.

So, as someone has said, the universe might have been created moments ago with everything, including our memories set up just so, but this doesn't seem a very useful line of speculation. Instead we say that, based on the physics we understand, and everything we can see and measure in the universe, it looks like its been expanding ever since it was incredibly hot and dense about 13.8 billion years ago.


Would an ultra bright moon indicate that the world is ending?

I remember that well. I presume you're referring to an episode of "The Outer Limits," which was based on a short story by Larry Niven. Having looked into the idea back when I read the story in the 1990s, I can say it's plausible in the sense that it's not strictly prohibited by any law of physics.

That said, a solar flare would have to be of precisely the right size, coming in at precisely the right trajectory for precisely the right duration, and with the moon in precisely the right spot for the scenario in the story to play out. More likely, the entire planet would be completely roasted, and all higher life forms extinguished.

Of course, most solar flares are not nearly that powerful. Typically, we experience transient disturbances in global telecommunications, spectacular aurorae, and little else when a flare hits. In the late 1800s a more powerful flare hit, and some people speculate that a flare that size could affect not only telecommunications but also power grids. However, the planet would not even be singed.

There are in fact several ways to burn the Earth, as you put it. Astronomer Phil Platt describes these in wonderful detail in his book Death from the Skies. In addition to a solar flare, a gamma ray burst from a nearby supernova (< 20 ly) would do it. I won't mention the others since I don't want to spoil the fun of the book.


Did Edmund Scientific 4¼ inch Newtonians have 90° prisms as secondary mirrors in the 1970's?

My Edmund F/10 4 1/4" was bought in 1975, and had a rectangular front-surface mirror as the diagonal, with a single arm support attached to the tube near the eyepiece holder.

I believe the Astroscan 2001 (or at least some models of it) used a prism for the diagonal, supported by the flat glass plate at the front of the tube; haven't found a reference though.

The advantage of a prism is that it is more maintenance-free then a mirror which will eventually need to be re-aluminized.  The disadvantages are that there are two optical surfaces instead of one in the light path that could reflect or scatter light, you can get color dispersion, and a rectangular prism will block more light than an elliptical mirror (especially important in a fast scope like the AstroScan).  The star diagonal Wikipedia page has a pretty good discussion of the trade-offs.


How distances are measured in the Solar System, in the Galaxy, and in the Universe?

Stellar Parallax
Stellar parallax uses differences in perspective to determine the distance from an object. When the earth goes around the sun, our perspective of the star, galaxy etc. changes and so the angle from us to the object changes. Because we know how the earth moves around the sun, we know the distance between the points that we take the measurements. From this we can use the Sine Rule to determine the distance from the object.
For Example, if the distance between the two points of measurement were 2 units, and the two inside angles (A and B) were (A)80° and (B)75° (and therefore the other angle (C) is 25°) we can say that:
$a = \frac{2\sin80}{\sin25}\ $
where "a" is the opposite side to angle "A". This means that the distance between the object and the second (B) point of measurement is approximately 4.66 units.
Diagram here.

Standard Candles
 A standard candle is an astronomical object that has a known absolute magnitude. (source)
By using the known absolute and the measured apparent magnitude (what is seen by instruments etc.) it is possible to determine the distance of the object by using the formula:

m-M = 5 log d – 5
where m is the apparent magnitude of the object, M is the absolute magnitude of the object, and d is the distance to the object in parsecs.(same source)

One of the most widely used examples of these standard candles is the supernovae type Ia, which was used to discover that the universe is accelerating in its expansion. This led to the awarding of the 2011 Nobel Prize in Physics to Saul Perlmutter, Brian Schmidt , and Adam Riess.
Additionally, RR Lyrae are another form of standard candle used to measure galactic distances (SN Ia are much more luminous and therefore can be used to much higher distances). RR Lyrae are a type of variable star with a relationship between pulsation period and absolute magnitude, which is what makes them good for use as standard candles.
Other information here.

Redshift
Redshift is the effect where light from an object is "stretched" out so that the wavelength is moved towards the "red" end of the spectrum. Distances are calculated by using Hubble's Law, but I'll let others explain it: here, here and here.


Is it possible that the Sun has a binary partner (the Nemesis Theory) that has eluded detection?

The first part of your question has been asked before:  Is Sun a part of a binary system? and the current (lack of) evidence for such a companion is discussed on the relevant wikipedia page about "Nemesis".

To summarise: if it were a small companion star, or even a brown dwarf that had been cooling for 4.5 billion years and it had a 26 million year orbit, then Kepler's third law tells us how far away that object would be. It turns out that this is not far enough away that the object would have eluded detection in all-sky surveys. The recent WISE survey in the near-infrared should have been capable of detecting even a very cool brown dwarf (and indeed it has detected some very cool brown dwarfs, just not that close to the Sun - e.g. a 250K brown dwarf only 6 light years away Luhman et al. 2014) that was close enough to the Sun to be a Nemesis candidate.

Since the details do not appear to be on the wikipedia page, I'll fill some in. If we take a very low mass brown dwarf - say 20 Jupiter masses - in a 26 million year orbit, then Kepler's 3rd law tells us this will be about 90,000 au (1.4 light years) from the Sun (assuming a circular orbit). According to the evolutionary models of Saumon & Marley (2008), such an object has an intrinsic luminosity of $10^{-7}$ times (1 ten millionth) that of the Sun and a temperature of 400 Kelvin and would appear to have a spectral type of late T or early Y.

From the calibration of absolute magnitudes versus spectral type for cool brown dwarfs in Marsh et al. (2013) we know that at 90,000 au, such a brown dwarf would have magnitudes of $H=14$ and $W2=8$. The former is bright enough to have been seen in the 2MASS all-sky survey and the latter is easily detected by WISE. The combination of data would also easily have revealed the large parallax of such an object. We can conclude that an object would have to have much lower mass than this to remain undetected.

Now the middle part of your question: A star will "wobble" in response to its companion with exactly the same period as the orbit. So, if you are prepared to wait for an appreciable fraction of 26 million years, then yes, the presence of the binary companion might be revealed in a net small oscillation of average proper motions over the whole sky, with a period of 26 million years! Otherwise not. Measuring the "wobble" by doppler methods is currently capable of detecting Jupiter-mass objects in orbital periods of 10-20 years. Measuring the wobble "astrometrically"  - that is measuring the position displacement of the star due to an unseen companion is more sensitive to distant companions, but still, one is limited by the fact that the wobble period will be the same as the orbital period of the companion. An assessment of the performance of the Gaia astrometry satellite by Perryman et al. (2014) suggests detection of planets with orbital periods out to 10 years might be possible.

For the final part of the question - yes, the motion/wobble of a star can be decomposed (using Fourier techniques) into components that are due to each planet. There are numerous examples of multiple planetary systems that have been discovered by the radial velocity technique in this manner. Similar techniques can and will be employed to analyse any astrometric wobble. An interesting plot is to show the motion of the Sun in the plane of our solar system, in the rest frame of the solar system centre of mass. The Sun executes a complicated trajectory in this plot due primarily to the influence of Jupiter (on an 11 year orbit), but with superimposed "epicycles" due to the influence of the smaller planets on different orbital periods. See below.

Motion of the Sun around the solar system centre of mass (from http://commons.wikimedia.org/wiki/File:Solar_system_barycenter.svg )


In what parts of the world is it impossible to view the night sky due to light pollution?

Googling for 'light pollution map' gives a pretty good looking result. Those things usually are created using satellite measurments that are susceptible to scattered light (by measuring the polarity of light) in order to separate direct illumination from actual light pollution.

I also suggest you visit a city and then compare the light pollution index on this site to what you know as home, in order to get a feeling for the severity of the numbers.

I cannot answer your question about how many people have never seen our galaxy, as this is a function of travel willingness of each individual living in a city and therefore maybe even impossible to answer for our intelligence agencies.


Stabilizers/dampers for refractor telescope

It depend where vibration are. Often (if you have a good mount for the telescope) all the structure vibrate and there are some good anti-vibration feet that can be useful. If the tripod isn't the problem and the vibration are only near the telescope, probably the problem is the mount that is under-sized for you telescope. In this case the only good way is change the mount with a sturdier one.


How to compute the perihelion precession of Earth's orbit?

For simulation only, you should be able to show it by simple code as following:
$$
\vec{a}_t = - \frac{GM}{r_t^2} \hat{r} \\
\vec{v}_{t+1} = \vec{v}_t + \vec{a}_t \Delta t \\
\vec{r}_{t+1} = \vec{r}_t + \vec{v}_t \Delta t
$$
where $t$ is time step, $\vec{a}$ is acceleration, $\vec{v}$ is velocity, $\vec{r}$ is position, $\hat{r}$ is a unit vector in radial direction from the origin (set the Sun at the origin), $G$ is gravitational constant, $M$ is Sun mass in this case, and $\Delta t$ is size of time step. You will have to play around with the initial values and step size to get certain orbit representing the case. To make it more real, you derive the initial values from the real parameters, if known.


Yes, for a given galaxy type…
The term "a distribution of number of galaxies per redshift interval" does make sense, but usally we restrict ourselves to a certain type of galaxies, e.g. Lyman $\alpha$ emitters, Lyman break galaxies, sub-mm galaxies, etc. The reason is that galaxies are detected in a number of different ways, and no single selection technique captures all types of galaxies.
…and comoving volume…
Because the Universe is expanding with time, the "true" (physical) number density of galaxies — and everything else (except dark energy) — decreases with time. To be able to compare densities at different times, it is useful to measure them in comoving coordinates, which is the coordinates that expand along with the Universe. In these coordinates, densities only change if some physical process happen. For instance, the  physical energy density of photons decreases both because the volume of the Universe increases and because the photons redshift. In comoving coordinates it also decreases, but only due to the redshifting since the volume change is factored out.
…and luminosity.
Moreover, since galaxies have no well-defined lower size/mass/luminosity threshold, we have to define a threshold above which we count them.
For this reason, rather than talking about "the number density of galaxies per redshift", we usually use the very useful observable luminosity function (LF) which measures the number of (a certain type of) galaxies per luminosity bin, and per comoving volume, at a given redshift.
LFs have been probed out to very high redshifts, at the times when galaxies were just beginning to form at $\sim$half a gigayear after Big Bang (e.g. Schenker et al. 2013).
You end your question by asking about the number of galaxies at $z\le0.01$, which would be considered very low redshift in the context of galaxies. You can find the answer by considering all kinds of selection criteria, checking for overlaps between surveys (most local galaxies will be detected by multiple techniques), and integrating over volume the LFs down to your chosen threshold (e.g. Small Magellanic Cloud-sized). Doable, but not trivial. An alternative is to integrate numerically calculated dark matter halo mass functions, using e.g. this online tool, and assuming the every halo hosts a galaxy. I get a number density of $n\sim0.4$ halos per $h^{-3}\,\mathrm{Mpc}^{3}$, or $1.2\,\mathrm{Mpc}^{-1}$. The comoving volume out to $z=0.01$ is $V=3.3\times10^5$ Mpc, so the total number of (super-SMC-sized) galaxies is $N = n V \simeq 3.8\times10^5$.


How do isolated binary compact systems survive their formational supernovae?

OK, here's what I've been able to piece together with a little research (not really my field, so I may get a few things wrong).

The short answer is a combination of: A) It's pretty hard to actually disrupt a massive star via a neighboring supernova, and the two stars aren't intially that close to each other; and B) Lots of complicated effects due to the binary interaction.

I'll focus on what seems to be the best current model for forming a compact double neutron-star (DNS) system -- one with a small enough separation in the resulting neutron stars so that gravitational-wave emission can cause them to merge within a few billion years -- which is discussed in this 2017 article by Tauris et al.. It looks a bit like this:

Massive binary star system forms, with moderate separation (not as close as the final DNS binary will be).
Primary star (initially more massive) evolves faster, to the point where it expands and fill its Roche lobe, leading to mass transfer to the secondary.
This means the primary loses mass while the secondary gains mass. So when the primary reaches the supernova stage, it has less mass in its envelope to eject -- and the companion star may now be the more massive of the two.
Primary's core collapses, producing first SN. This probably won't destroy the companion; it might disrupt the system -- not so much because the companion is kicked a lot but because the NS remnant is lower mass than the pre-SN primary star, and it probably gets a kick from explosion asymmetries. But recall that the companion star may now be more massive, so the gravitational disruption is less.
Companion star evolves, expanding to overfill its Roche lobe, leading to a "common envelope" phase where the NS is actually inside the companion star's outer envelope.
Common envelope phase leads to

A. Significant shrinking of the orbit as the NS experiences dynamical friction;

B. Ejection of much of the companion star's outer envelope.
System is now much more compact, with a NS and a stripped helium-star companion.
Possible further mass transfer from companion onto neutron star, leading to even less of an envelope around the companion and an even lower mass.
Companion's core collapses, causing second SN. This has little effect on the first NS (i.e., the former primary star), since the NS is a very small target with extreme density and surface gravity: very hard to disrupt. In addition, the companion has very little envelope left to eject at this point.

Some key points:

The binary starts off with a significantly wider separation than the final DNS will have, so the effects of the first SN on the companion will be less extreme than you might imagine;
Mass transfer (and ejection) during the Roche overflow and common envelope stages removes mass from the stars, which means there's less envelope to eject when a SN kicks off (so it does less damage to the other star/NS) and also that the SN ejection of a star's envelope doesn't have a strong effect on the gravitational binding energy, since there's less of an envelope to eject.

This paper by Liu et al. (2015) presents some modern simulations of the effect of a SN blast wave on a nearby companion star. It turns out to be pretty hard to disrupt the companion: even for relatively small separations, the companion only loses about 10% of its mass, and this goes down to a few percent or less as the separation increases. (Remember that the companion is itself a massive star most likely still on the main sequence, and so relatively compact, with high surface gravity.)

Can the explosion push the companion star away? Probably not very much: again, the companion star is itself quite massive -- maybe more massive than the original primary because of the mass transfer in stage 2 (above) -- so it's hard to push it around.


How much illumination do the background stars provide?

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

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

Other sources for same value:

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


How to calculate what is the best date and latitude to observe a certain object in space?

It's a very general question, as a rule of thumb the higher an object is on the sky the light is going to be less dispersed as the atmosphere is thinner. So, the best point and moment to observe an object is when the object is in zenith (directly over your head). Usually it is not possible to travel around the globe to make your observations, so we are going to see better the objects that happen to be near zenith, or they that are crossing the meridian line (with circumpolar objects, it's better wait until the object the meridian line nearest to the zenith)

I don't have here my books so I can't give you a proper formula, but if you want, I can try to give you that formulae.


Is there a name for the maximum in local solar elevation angle?

I think this is an indirect answer, but the name for this event is solstice. In the strictest definition, the solstice is

  either of the two times in the year, the summer solstice and the winter solstice, when the sun reaches its highest or lowest point in the sky at noon, marked by the longest and shortest days.

By definition, the solstice is the exact point in time when the Earth's axial tilt is pointed as close to (or as far from) the Sun as possible and by extension, the Sun is at its highest (or lowest) point in the sky during the year. If you're in the northern hemisphere, this point is the summer solstice, whereas it is the winter solstice for the southern hemisphere.


Why when we look through a telescope in space, do the billions of stars not block our view from seeing further?

OK, Imagine the stars at distance x block an area of the sky. At a distance of 2x there should be four times as many stars, but they would seem four times smaller in terms of area covered. Thus, the blocking grows linearly until a significant part of the sky is blocked. The stars within 20 light years (excluding the Sun), blocks approximately $4.3 \cdot 10^{-16}$ of the sky. Even if we scale that up to the radius of the universe, 45 billion light years, only about $2 \cdot 10^{-6}$ of the sky is covered.

Even that tiny number is an overestimate, as we live in an extremely star-dense part of the universe, our galaxy.

Space is simply too big.


How to improve the quality of an image of the planets when taken through the eyepiece of a telescope with a dlsr camera

If you were using both a camera lens and an eyepiece (afocal method), try eyepiece projection.
In this method, the eyepiece is the only optical element between the prime focus and the detector.
You would need an eyepiece projection adapter tube to connect the camera body to the telescope's focuser.
A longer tube will increase magnification but also increase exposure time.
Any tube length which reasonably balances those concerns should be OK.

It's also possible to use a Barlow lens and no eyepiece to extend the effective focal length of the telescope.
If your adapter tube is wide enough to accommodate the Barlow, you may be able to use it this way too.

Astronomy Source has an article with more details about eyepiece projection.
Tele Vue has an overview of that and other methods, plugging their own products of course.


Question about tidal forces and the Roche limit

When we do mathematical calculations, sometimes terms cancel and we get to draw a line through them.

But in the real world cancellations and "total cancellations" don't really happen.

Forces just add. Often there's a big force and one or more little forces, and as long as the big force substantially dominates the little forces, we tend to notice only the big force. But it doesn't cancel, it just covers up the effects of the others to the casual observer.

Let's think about gravity. We are so used to moving around in 1 g of gravity that we don't notice it. You could say that our brain "cancels" out 1 g from our conscious awareness and we compensate for it automatically. We don't feel and notice the 1 g when we stand or sit or lay down, even though we make use of it to stay put.

However, get in an elevator or an amusement ride and add a little acceleration, and we notice it right away! That sinking feeling when a fast elevator starts accelerating downward for example watch (but don't listen to) the video The Equivalence Principle

The purpose of the above discussion is to discount the idea of "total cancellations". A bit of rock or soil on the surface of a body feels the gravitation of the rest of the body pulling down, the "centrifugal force pulling" it up if the body is rotating, and the gravitation of nearby bodies (e.g. Sun or Moon or planet) pulling up or down or sideways depending on orientation. It feels all of these all the time.

If the net force, the sum of all the forces, pulls up, then the body starts to disintegrate. If the net force is still down, it doesn't.

Think of it as a threshold, rather than a cancellation.

  ...if that's so, ¿why does the body deforms even before crossing the Roche limit?

As long as the body is made of material capable of deforming or flowing (which rock certainly is, especially when hot) then it will continue to move or flow until the sum of all the forces reaches zero. Gravity usually dominates, but it's the force of the compressed layer below pushing up that counters the sum of all the forces.

If the body is rotating, then the net force is weaker on the equator than at the poles, and so less pressure is needed to compensate. That means the higher pressure at the poles will cause some of the material to move to the equator.

To read more about several of the competing forces pulling on us all the time as we stand on the Earth, see:

@DavidHammen's cool table
my (sadly down-voted) calculation
all of the answers to Did the Mars rovers actually confirm the gravity of Mars?

Once you are comfortable with the idea of equilibrium, multiple forces adding and pressure, then have a look at all the interesting answers to these particular Roche limit-related questions:

Is there any orbit at which the Roche limit can be “felt”?
Roche limit - calculation with centrifugal forces
How Wide is Earth's Roche Limit?
Filling the Roche Lobe


Why was the Vera C. Rubin Observatory LSST Camera tested with a broccoli and a pinhole?

From the first article you linked:

Taking the first 3,200-megapixel images of a variety of objects, including a head of Romanesco – a type of broccoli – that was chosen for its very detailed surface structure, was one of these tests. To do so without a fully assembled camera, the SLAC team used a 150-micron pinhole to project images onto the focal plane.

Which implies:

the advantage of using a pinhole is that it allows the sensors to be tested without requiring the full camera to be assembled yet.
the Romanesco was chosen to test the sensors' ability to image an object with detailed structure.


Why are radar images of comets shaded only on one side?

There is a very good blog article here that explains this effect but basically the Doppler shifting of the radar return means that certain parts of the asteroid light up more than others in the observing wavelengths from Earth. Its better explained in the article>>

Taken from the article "How radio telescopes get "images" of asteroids"
Posted by Emily Lakdawalla:

  As an asteroid rotates, some parts of it are moving toward us, while other parts are moving away. As the broadcast radio wavefronts hit the part of the asteroid that is moving toward us, the asteroid smacks into each wavefront faster than it would if it were not rotating. The speed of the wavefronts does not change, because the speed of light is constant, so the wavefronts end up being packed closer together. This is a Doppler shift. The asteroid has taken the broadcast wavelength and reflected it at a shorter wavelength from the parts of the asteroid that are rotating toward us. On the other side of the asteroid, which is rotating away, the opposite thing happens; each arriving wavefront smacks into the asteroid a little later than it would if the asteroid were not rotating, so the reflected waves are spread farther apart.

Source: http://www.planetary.org/blogs/emily-lakdawalla/2011/3248.html?referrer=https://en.wikipedia.org/


Can the Moon provide momentum to an object in Earth's orbit? Gravity Assisted Boost

The Moon has a "Roche lobe", where the Moon's gravity dominates, and all you need for the situation you describe is that the Roche lobe lies physically outside the Moon.  That will be true, because the Lagrange points are on the outside edge of that Roche lobe, and it's outside the Moon.  The Moon's Roche lobe is shown here: http://hyperphysics.phy-astr.gsu.edu/hbase/Mechanics/lagpt.html


What's the difference between the apparent horizon and event horizon of a black hole?

The black hole region of a spacetime is defined as a region where nothing can escape to infinity and an event horizon at a given time is the boundary of a connected region of space which is part of the black hole region. As you're after a simple answer I won't give a formal definition of a black hole or an event horizon, but they can be found in Wald.

The problem though is to know whether anything in a region of space can escape to infinity depends on precise knowledge of the future, also not all spacetimes have a suitable notion of infinity. Yet clearly even when we don't know the entire history of the spacetime or when there isn't a suitable notion of infinity, we can still identify objects that are functionally equivalent to black holes. The apparent horizon if you like is the spatial boundary of what we might consider to functionally be a black hole.

As nothing can escape a black hole, even light directed away from it is pulled back towards the singularity so we know that a black hole can cause even light directed outwards, relative to a point in space, to move inwards to that point. The apparent horizon is the boundary between where outwardly directed light moves outwards and where it moves inwards.

In Schwarzschild spacetime, or more generally Kerr-Newman spacetime (in standard coordinates) the apparent horizon and the event horizon coincide. However, more generally, the location of the apparent horizon depends on how you 'slice' spacetime (it is observer-dependent if you like - unlike the event horizon). Apparent horizons needn't be associated with (formally-defined) black holes, however as long as the spacetime has certain properties, they will indeed be associated with black holes and will lie at or inside the event horizon.


The USNO page refers to USNO Circular 163 which in turn refers to a paper by Aoki et al. and one by Moyer. These papers explain how the expressions created by Simon Newcomb at the end of the 19th century were updated to work with the new system of constants introduced in 1984. If you want to change the epoch, you could redo these calculations with the epoch of your choice.

Further changes to celestial mechanics calculations have occurred since then, but apparently the USNO has decided that for the accuracy stated in the USNO web page, 0.1 second, it is not necessary to overhaul the calculation to account for all the changes since 1984.


Does the KBO 2014-MU69 have two numbers and entries in asteroid databases? How did it get promoted to Major Body designation?

The 3713011 object (that is the internal SPICE/SPK id) is a special trajectory computed by the New Horizons team at SwRI which included data which is not in the public domain. The other one for (486958) 2014 MU69 (its official asteroid number and temporary discovery designation) is for the publicly available data.

Generating an ephemeris in HORIZONS for the former produces:

 Revised: Dec 29, 2018            486958 (2014 MU69)                    2486958
 *SPECIAL*

 This pre-computed trajectory is consistent with the New Horizons spacecraft
 Kuiper-Belt extended mission, with the planned 3500 km flyby of Ultima Thule
 on 2019-Jan-1 @ 05:33 UTC.

 Trajectories were provided by the New Horizons mission planning team at SWRI
 and are fits to internal flight-project data that has not been made available
 outside the flight project.

 To use this official mission trajectory to ...

  A) Generate ephemerides with respect TO the mission target:
      Specify the coordinate center in Horizons as "@2486958"

  B) Generate ephemerides OF the mission target
      Specify the target in Horizons as "2486958".

 To instead use the latest JPL ground-based orbit solution (based on the
 publicly available data) and numerically integrate over an arbitrary
 time-span with statistical uncertainties of that public dataset, do a
 look-up in Horizons with syntax "2014 MU69;" or "486958;" or "DES= 2486958;"
 or "DES= 2014 MU69" or "2486958;" (i.e., with a semi-colon).

 Note: the object previously had a provisional SPK ID of 3713011.  It was
 then assigned an IAU number of 486958 such that the primary SPK ID became
 the present 2486958 (2000000 + IAU number).

 Trajectory name                            Start (TDB)         Stop (TDB)
 ---------------------------------------  -----------------  -----------------
 20180601_20250101_od141_tcm30_burn_V0.1  1994-Jan-08 00:00  2033-Dec-21 00:00

(486958) is 2014 MU69's official number and is used at e.g the MPC. 3713011 and 2486958 are the JPL SPK id's for the kernels with and without the extra trajectory information respectively; they are not used outside JPL's software such as HORIZONS, Small Body Browser or the SPICE toolkit. Tutorial 18 of the SPICE Tutorial covers SPK files if you want more details.

The JPL software system handles multiple versions of the trajectory and/or orbital elements by creating multiple "bodies" for the same object. This is most commonly seen with comets which have 1 or more orbital elements for each perihelion passage e.g. this screenshot for comet 46P (Wirtanen):


Why do we have so few neutrinos to study?

It's not that the neutrino detectors don't have a chance of detecting a neutrino, but that the neutrinos do not interact with matter (well hardly ever).

If there are $6.5\times 10^{10}$ neutrinos per second per $cm^2$ and if the detector has an area of $10^{6} cm^2$, and it detects 1 neutrino per year, then 1 out of $2.05\times 10^{24}$ neutrinos interacts with the detector. $2.05\times 10^{24}$ pass through the detector unnoticed.


Troop of moving stars through a straight line

That was likely the latest StarLink satellite constellation.

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


How active was the Milky Way in its first 5 billion years?

Although the question was asked a while ago, it still sounds interesting, so I did a bit re-search and I found a Nature news-feature by Adam Mann: Hidden history of the Milky Way revealed by extensive star maps

The team had spotted a set of 30,000 renegade stars. Unlike other objects in the main body of the Milky Way, which orbit in a relatively flat disk shape, these non-conformists were moving backwards, in orbits that were carrying them out of the Galactic plane.
Within weeks, the team had worked out that the luminous horde pointed to a long-hidden and especially tumultuous chapter in the Milky Way’s history: a smash-up between the young Galaxy and a colossal companion. That beast once circled the Milky Way like a planet around a star, but some 8 billion to 11 billion years ago, the two collided, massively altering the Galactic disk and scattering stars far and wide. It is the last-known major crash the Galaxy experienced before it assumed the familiar spiral shape seen today.

The highlighting in the quote is my attempt to answer your question - it is not talking about the Milky Way's first 5 billion years, but it somehow implies that its early history might have been indeed turbulent.
References
The original article is by Amina Helmi et al. is called The merger that led to the formation of the Milky Way’s inner stellar halo and thick disk


Are black holes orbiting around each other?

A black hole (briefly), is nothing more than a dead star whose mass was more than 3 solar masses. When that star's life ended it collapsed under its own gravity and went through a supernova, losing a portion of its original gas in the process, thus losing part of its mass, and thus its total gravitational pull is diminished compared to the original star (contrary to common believe for a black hole having the "ultimate gravity"). So first of all, black holes vary in size and are not all the same (thing).

If we say the original star was orbiting around another star (A Binary system), (putting other factors aside) the effect on the other star from the black hole's gravitational forces would be less compared to its original star.

However that gravitational pull increases drastically as you go closer to the blackhole, since all the remaining matter is concentrated into a single point (singularity). Gravity becomes too severe, close to a  blackhole that even light can not escape, an area called "the event horizon" of the black hole. But that's too close to the black hole. Nothing is special about its gravitational effect on the further objects as I pointed out with the orbiting star.

Regarding your statement: "Is it true also for objects like black holes? Are they even able to be moved by another force?" I don't know what you mean to be moved by another force, but black holes do move as any other object in space, they are nothing more than the remainder of their original star, and they inherited its motion relative to the other objects around. If say the original star was rotating around the center of it's galaxy, it will continue to do so.

Your statement: "bodies with low gravity force orbit around another one with higher gravity force" is not very accurate, since (bodies with higher gravity force - the more massive object), and  (bodies with low gravity force - lesser mass) both orbit one another.

For instance, as the Moon orbits the Earth, the Earth also orbits the Moon, they both orbit around a point between the two of them called the "Barycenter". This point is always closer to the more massive object.

In the case of the Earth-Moon system it's inside the Earth's volume. Therefore it's common believe that only the Moon's orbiting the Earth, without the other way around as well.

Topics you must check out besides the link provided in the previous answer:

Orbital Mechanics
Newton's Universal Law of Gravitation
Black Holes


Exoplanet radius density relation according to NASA exoplanet archive

What you are seeing is a combination of several things as you move across the diagram from left to right with increasing planet radius. As this diagram  from a talk by Cayman Unterborn at the recent "Know Thy Star - Know Thy Planet" conference shows, is a transition from rocky (like Mars, Venus, Earth; the 'Ma', 'V' and 'E' on the diagram) exoplanets with increasing density as ~1.5 Earth radii is reached to "Neptune-like" gas giants with increasing amount of a gas envelope, possibly with a denser core. (The size and type of Jupiter's core is not a totally settled question for example and so the wider range of exoplanets could have a wide variety of core properties). There is a lot more information in this blog post on 'Which small exoplanets are rocky' which talks about what we can learn about planets in the 0.5 to 4 Earth radius regime.

At higher radii (approximately 11 Earth radii/1 Jupiter radius) what you are seeing is the effect of electron-degeneracy pressure. This causes the radii of exoplanets with a mass of about Jupiter, brown dwarfs up to 80 times the mass of Jupiter and low mass M dwarf stars (with a mass of hundreds of Jupiters) all to have very similar radii but with widely varying densities. This causes the near-vertical line on the right hand side of your plot. This is also what makes it very difficult to know whether you have a star, brown dwarf or a planet if you only a measure of a transiting object's radius (from e.g. Kepler) and don't have a mass measurement (usually from radial velocities) to go with it.


Which data should I use to take the fast Fourier transform (FFT) to find time period for eclipsing binaries?

For this, you should create a lightcurve, a graph of brightness over time, to view the data. For Kepler data, the bjd(date) column is the time in BJD. The dtr_flux stands for detrended flux, meaning that it should contain "cleaned" data.

First, I would plot the dtr_flux column over bjd(time) to obtain a lightcurve. Depending on the type of eclipsing binary, you should see a minimum and maximum dip occurring periodically. The data also contains a "phase" column, meaning the data has been "folded" to allow you to plot each cycle of the eclipsing binary on top of each other. See http://www.southampton.ac.uk/~sdc1g08/BinningFolding.html

After that, you can take the FFT if you wish. Remember that since this is an eclipsing binary with both maximum and minumum brightness dips, the FFT will have a spike both at P and P/2.


According to wikipedia it is 10 degrees.  Wikipedia cites Peißker et al,
which uses the formula:
$$\Delta\phi = \frac{6\pi G}{c^2}\frac{M}{a(1-e^2)}$$ to derive a relativistic periapse shift of 9.9 degrees per orbit.


Could there be Earth-like planets in binary or trinary systems?

Firstly, a few planets have been discovered around binary or trinary systems, mostly Neptune sized or larger.  An example is are the two planets described in the NASA article "Kepler-47: Our First Binary Star 2-Planet System", where they are described as:

  Now Kepler mission has discovered Kepler-47b and 47c, the first transiting circumbinary system — multiple planets orbiting two suns. To compound the excitement of the discovery, one of those planets is in the binary system's habitable zone (where liquid water may exist)

(A representation of the Kepler 47 system is pictured below)

The second planet discovered in the system, Kepler 47c, is within the binary stars' habitable zone, but is slightly larger than Neptune and is likely to be a gas giant - but that raises the potential for terrestrial satellites.

Another example is the Saturn sized Kepler 16b reported in the article "Weird Exoplanet Discovered Orbiting Two Stars" (Klotz, 2011), which orbits 2 stars that are smaller than our sun, thus has habitable zones closer to their parent stars, in this case, the equivalent of the Venus' orbit.

In terms of the habitable zones around multiple star systems, the article "Habitable Binary Star Systems" suggests a major advantage of a binary (indeed a trinary) system is that the combined energy between the stars could indeed extend the habitable zones, particularly,

  Low-mass twins could make the best hosts, because their combined energy extends the habitable region farther away than would exist around a single star.

However, temperatures would vary significantly and the orbits would be irregular, and in simulations reported in "Double-Star Systems Can Be Dangerous for Exoplanets" (Wall, 2013) results in such disruptions that one or more of the planets therein could be ejected and sent hurtling into interstellar space.  But such perturbations are likely to take a considerable amount of time (millions of years) to fully occur.

This is also discussed in "Exoplanets Bouncing Between Binary Stars" (Moeckel and Veras, 2012), where the planet in such as system is described as 'bouncing' between the binary stars, then resuming a relatively regular orbit.

The recent discovery of a terrestrial-like planet in the Alpha Centauri system, as reported in "Discovery! Earth-Size Alien Planet at Alpha Centauri Is Closest Ever Seen" (Wall, 2012) - despite this planet not being in the habitable zones of either star, this does raise the hopes for the possibility of Earth-like terrestrial planets existing in the habitable zones of other multiple star systems.

There is also the possibility that an Earth sized moon could orbit the larger planets such as those in the Kepler systems mentioned earlier.


What kind of things I could "see" with an amateur radio telescope?

Rather than looking for ready-made systems, take a look at projects. Right now, plenty of amateurs are using software defined radio coupled to various antennas for astronomy. Start here:
http://www.rtl-sdr.com/rtl-sdr-for-budget-radio-astronomy/
And while it has nothing to do with imaging, there's plenty of radio astronomy that amateurs can do using simple (albeit sometimes large) antennas:

http://radiojove.gsfc.nasa.gov/
http://radio-astronomy.org/pdf/qex/radio-jove-proof.pdf
http://www.radiosky.com/project.html


What did the night sky look like 100,000 years ago?

(Here I assume by "constellation" you mean "pattern of stars", rather than "region of the sky". To be a pedant one should use "asterism" for a pattern of stars.)

Some asterisms will be recognisable, but there can be some significant proper motion over that period.

Some stars will move more than others. Generally, closer stars will appear to have moved further than more distant stars. So the stars that make up the asterism of "the plough" (which are fairly close) will have moved more, but will still be in the same general region of the sky. On the other hand the bright stars of Orion are more distant. In 100000 years ago the general appearance of Orion would already have formed.

Videos showing
The past and future of the Plough https://www.youtube.com/watch?v=txJH8RlIoXQ

The future of Orion https://www.youtube.com/watch?v=5KFQD6Bpnm0


If the Earth is rotating counter clockwise, how does the sun rise from east?

Try this non-Gedanken-but-real experiment:

Hold an apple in your hand so the Sun (or a light) shines on it. Nerds may use a proper globe.
Along the apple equator draw arrows to the right with an E for East: E→E→E→E, .... Nerds be sure to use a non-permanent marker.
Spin it counterclockwise (when looked at from above, i.e. the North pole). You're doing it right when all the Es move in the direction of the arrows. West to East.
The Sun rises on all points on the apple that go from dark to bright; identify that line. It should be half a circle.
Imagine a near indestructible humanoid killer machine from the future standing on that line (called terminator, the line, that is). What direction does the machine look when it glances at the Sun the moment he (the Sun, that is) rises?


How should exoplanet discoveries affect SETI?

Well you are correct and of course that kind of research has been going on. The SETI programme targeted a set of Kepler stars with evidence for a planet (or planet candidate) in the "habitable zone". The initial searches for narrow-band radio emission (at 1.1-1.9 Hz) appears to have concentrated on those systems containing multiple planets.

The (null) results for $\sim 100$ system were published by Siemion et al. (2013). The observations were not sensitive enough to detect the sort of "radio chatter" emitted from our planet, but might have detected some sort of deliberate signalling beacon.

A slightly more imaginative approach, which I'm struggling to find the details of, is that one can observe Kepler multiple planet systems when one calculates that two planets are in conjunction as seen from the Earth. There is then the chance of seeing communications from one planet to the other. (e.g. http://adsabs.harvard.edu/abs/2014ebi..conf..5.2K ).

In a more general sense, the exoplanetery discoveries have provided great impetus to the SETI programme. It is now clear that planetary systems are common and present around most stars. It is also now apparent that small planets are common. New possibilities are also being considered. Most exoplanets in habitable zones are orbiting stars smaller and less luminous than the Sun. No exact Earth analogues have yet been found. This has led to an explosion of activity investigating planetary atmospheres and habitability around K- and M-dwarf stars.


Milky way: How do we know its appearance?

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

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

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