on/off filter and their photometric calibration

The on- vs. off-band filter is to let you easily measure the strength of an absorption or emission line in the object's spectrum.   "Off-band" doesn't refer to a shift in position, but rather in the wavelength that the filter lets through. It is "off" the wavelength of a particular spectral line.
This illustrates the idea:

(from Michael Richmond's page here, used under CC-BY-NC-SA 2.0 license).
In this case, the light through the filter shown in red would be affected by the H$\alpha$ absorption line in the star's spectrum, while the light through the filter shown in green would not.  So the ratio of the two would give you a very good measurement of how strong the line is, without worrying about how bright the star is overall, i.e. what the overall continuum level is.
In the paper you're reading, they would be looking for emission lines from planetary nebulae, but the idea is the same - the on-band filter measures line+continuum, and the off-band filter only measures continuum, so it lets you correct for the continuum emission and measure only the line strength.
The design of the filters would depend on which lines you're looking for. In this case they are targeting the [O III] line which is normally at a wavelength of 5007 Angstroms, but because they are looking at distant objects, the emission is red-shifted by the expansion of the universe, so they use a custom filter whose wavelength is shifted to match the red-shifted wavelength of that emission line for this particular group of galaxies.


Is there a fundamental difference between a small galaxy, and a large star cluster?

There are a couple of important distinctions between the two types of objects. Galaxies are objects which range in mass from about $10^{9}-10^{12} M_{\odot}$, and contain 'halos' of dark matter which represent the majority of the mass of the object. Now, there are things called dwarf galaxies (which are less massive than regular galaxies; I'd imagine that there really could be a bit of a blurred distinction between these two kinds of objects), however dwarf galaxies still seem to have dark matter components to them.

Globular clusters are tightly bound collections of about $10^{6}$ stars, and do not have significant (if any) dark matter contained within or around them. So for these reasons in practice it is not hard to confuse the two.

Some References:

1) Evidence that dwarf galaxies are dominated by a dark-matter component.

2) Dark matter present in globular clusters are either unlikely, or incredibly highly concentrated and approximately the same mass as the stellar component.


How fast is Neptune getting brighter? When was was this first noticed and reported?

TL;DR: There was apparent 11% increase of Neptune brightness during 1980 and 2000. This could be due to multiple reasons. Recent observation suggested the reason to be change in the amount and brightness of the banded cloud mostly in the planet's southern hemisphere due to seasonal variations and thus affect overall brightness of the planet.

Neptune's visual magnitude stayed relatively constant before 1980 and after 2000, but during this period, there was an unexpected rise in the brightness, about 11%.  Astronomers have proposed various ideas to explain this peculiar variation but none are convincing. But the more promising explanation would likely to be seasonal change due to tilt and cloud formation variation.

A team from Wisconsin made three sets of observations of the full rotation of the planet in 1996, 1998, and 2002. The images showed progressively brighter bands of clouds encircling the planet's southern hemisphere. The findings were consistent with observations made by G.W. Lockwood at the Lowell Observatory, which show that Neptune has been gradually getting brighter since 1980. They believe that cloud bands are getting brighter because the Sun is warming the atmosphere in the south apparently changing the brighness of Neptune.

Neptune's near-infrared brightness is much more sensitive to high altitude clouds than its visible brightness. The recent trend of increasing cloud activity on Neptune has been qualitatively confirmed at near-infrared wavelengths with Keck Telescope observations from July 2000 to June 2001 by H. Hammel and co-workers which could provide an explanation to this brightness change.
Tilt could also be a reason. Hubble images reveal a real increase in Neptune's cloud cover consistent with seasonal change is the apparent absence of change in the planet's low latitudes near its equator. This do has a reason. Since Neptune is inclined at a 29-degree angle, the amount of solar energy received determines the season and this seasonal variation might affect the brightness. When the Sun deposits heat energy into the atmosphere, it forces a response like heating the atmosphere that lead to constant moving and condensing of cloud cover.
Another reason could be internal heat source of Neptune which may contribute to the planet's apparent seasonal variations and ultimately affect the planetary brightness.
Current visual magnitude of Neptune is 7.88 but it could reach 8.0 in the near future.
For more information, refer to this paper, "The Secular and Rotational Brightness Variations of Neptune"
Other references:

https://www.sciencedaily.com/releases/2003/05/030516082529.htm
https://skyandtelescope.org/observing/measuring-planet-magnitudes/
https://www.ssec.wisc.edu/media/Neptune2003.htm


A few sources come to mind to add to MAST and Sloan.
Data releases from the SAMI Galaxy survey, which contain (or will eventually contain) emission line spectra for about 3500 galaxies. The papers for the first two releases can be found [here] and [here].
Another (smaller and older) dataset can be found at the Penn State Center for Astrophysics. It contains spectroscopy of KISS emission line galaxy candidates. The download links to the dataset are on the CASt page [here].
I'll continue to update if I find more of a substantial size, but I think this should expand what you have already quite a bit.


What is the oldest known prediction of constellation positions?

No, there is no possible way that humans from any known civilization in the first century AD would have been able to extrapolate current constellation positions.

Prior to the work of Johannes Kepler and Isaac Newton, humans did not have sufficient knowledge of the basic principles of celestial movement to have any kind of basis for making such calculations.

Second, even if they had the basic principles, they would also have needed much more accurate and detailed information about the makeup of the Milky Way galaxy. The information currently available, that goes into the NASA models, is the product of a great many man-years invested in both observations and analysis of those observations.

Third, even if they had the basic principles and the necessary details with which to create a proper model, they lacked the computing power necessary to actually run the model - computation of that time being limited to pencil and paper (or perhaps an abacus).


Gravitational waves: how can scientists derrive the sizes and spin and distance of two merging black holes?

One element of the answer is that they do it in the other direction. They simulate many different possible black hole mergers and then look for the nearest matches to what they observe. I imagine there is some scope for interpolating between the simulations as well as running bespoke simulations to confirm the match. Spin affects the wave because the shape of space around a spinning black hole is different from that around a non-spinning one, which affects both the motion of the black holes and the processes by which the gravitational waves are radiated.


The magnetic field of the rapidly rotating neutron star interacts with the material coming from the other star in the binary. This results in a transfer of angular momentum, spinning the neutron star down but accelerating the material out of the system. The effect is somewhat like a garden sprinkler seen from above, with material flung out ("propelled") of the system.

I think there needs to be a quite delicate balance between the binary period (or separation) in the system and the strength of the magnetic field and rotation rate of the neutron star to form the propeller effect. It seems to happen most in the early stages of the X-ray binary when the neutron star is rotating rapidly. After the neutron star spins down by the torques exerted on it, the material from the secondary star in the binary forms either an accretion disk or is directly accreted onto the neutron star down the magnetic  field lines.

The effect has also been seen, possibly only the once, in Cataclysmic Variable  (CV) systems, which consist of a low-mass star (typically a K or M dwarf) and a magnetized white dwarf. The "poster child" for this type of propeller system in CVs is AE Aquarii A magnetic propeller in the cataclysmic variable AE Aquarii; Wynn et al 1997

The propeller theory seems to be alive and well as there over 800 citations to the original 1975 paper. A recent highly cited review discussing propellers and accreting neutron stars seems to be Alipar 2001


Can a black hole undergo a Penrose process inside another black hole's ergosphere?

Black holes can orbit each other (we know this from the gravitational waves from their eventual mergers) and there is no particular reason against a small black hole passing through the ergosphere of a large black hole.

The main problem is avoiding a merger. The the two-black hole spacetime has no analytic solution and numeric simulations are messy, but a merger will definitely happen if the event horizons overlap. In any case, a rotating black hole has an ergosphere with an equatorial radius of $2GM/c^2$ and event horizon somewhere between this and the limiting radius $GM/c^2$. If we use the extremal case, then we can presumably just barely fit in nearly an identical mass and rotation black hole in the ergosphere. More likely a much smaller black hole could fit in. The direction of rotation would matter for the dynamics, since counter-rotating black holes would tend to coalesce (see this page for a plot of the relevant radii and where the innermost stable circular orbits are).

A single black hole passing through the ergoregion would not be able to perform the Penrose process, but we can imagine a binary black hole pair that is ripped apart by tidal forces and then presumably the remaining hole would be ejected with a lot of energy. So, yes, we could get fast small black holes this way.

In practice this system would lose a lot of energy through gravitational waves, making it likely that the holes would coalesce anyway. The exact limits require numerical relativity simulations to estimate, and this one would be a tough one.

(Note that the gravitational waves also carry away energy, up to 1/8 of the total mass of two equal-sized holes merging, but this energy is very hard to capture. Penrose did note in his classic paper that Misner was first in proposing black hole mining using a variant of this.)


Is a literal instance of Russell's teapot possible?

The brightness of light received from a light source (or an object that reflects light) is inversely proportional to the square of the distance.
So if an astronomical object A which reflects light from the Sun back to Earth orbits at a distance of two AU from the Sun and an idental astronomic object B orbits at twice that distance, or at 4 AU, how much light will Earth get from the two objects at their oppositions?
Since object B is twice as far away from the Sun as object A, it's surface receives one quarter as much light from the Sun as object A.  Since object B will be three tiems as far from Earth at opposition as object A will be at opposition, its reflected light will be one ninth as bright.  So together, those two factors will make the light that Earth gets from object at its opposition one thirty sixth, or 0.02777, as bright as the light that Earth gets from object A at its opposition.
So the farther away from the Sun an astronomical object is, the less bright its reflected light will be as seen from Earth.
And if an object C has a higher albedo, or is more reflective than object D of the same size, it will seem brighter at the same distance.
The brighter a solar system object is as seen from Earth, the sooner it will be noticed and discovered.  The dimmer a solar system object is as seen from Earth, the longer it can remain undiscovered.
Thus the undiscovered objects in the solar system should be the ones which appear fainter as seen from Earth.
They may be fainter and dimmer as seen from Earth because their albedo is lower than objects at the same distance.  Or maybe because they are much smaller than other objects with the same albedo and distance.  Or maybe they are very large objects with high albedos but are very faint because they are very far away.
Over time, astronomers discover dimmer and dimmer objects in the solar system, objects which have darker surfaces, or are smaller, or are much farther from the Sun.
So the largest solar system objects likely to be discovered in the future are likely to be very far fromt he Sun and the Earth and thus appear many times dimmer than objects of equal size in the inner solar system.
You may have heard of the hypothetical Planet Vulcan, inside the orbit of Mercury, once used to explain certain problems with the orbit of Mercury.  It is no known that Vulcan could not possibly exist.  It would have been discovered long ago if it did exist.

None of these claims has ever been substantiated after more than forty years of observation. It has been surmised that some of these objects—and other alleged intra-Mercurial objects—may exist, being nothing more than previously unknown comets or small asteroids. No vulcanoid asteroids have been found, and searches have ruled out any such asteroids larger than about 6 km (3.7 mi).[4] Neither SOHO nor STEREO has detected a planet inside the orbit of Mercury.[4][23]

https://en.wikipedia.org/wiki/Vulcan_(hypothetical_planet)[1]
The largest solar system objects discovered in recent times have all been beyond the orbit of Neptune, and have been rather small.  Only a few of the largest ones have been large enough to classify as dwarf planets, and they are all smaller than Eerth's moon.
It is possible that there may be one or more as yet undiscovered planets in the outer solar system.  But since there are limits on how bright those planets could be and remain undiscovered, there are limits on how large and/or close they could be.

As of 2016 the following observations severely constrain the mass and distance of any possible additional Solar System planet:
An analysis of mid-infrared observations with the WISE telescope have ruled out the possibility of a Saturn-sized object (95 Earth masses) out to 10,000 AU, and a Jupiter-sized or larger object out to 26,000 AU.[6] WISE has continued to take more data since then, and NASA has invited the public to help search this data for evidence of planets beyond these limits, via the Backyard Worlds: Planet 9 citizen science project.[96]
Using modern data on the anomalous precession of the perihelia of Saturn, Earth, and Mars, Lorenzo Iorio concluded that any unknown planet with a mass of 0.7 times that of Earth must be farther than 350–400 AU; one with a mass of 2 times that of Earth, farther than 496–570 AU; and finally one with a mass of 15 times that of Earth, farther than 970–1,111 AU.[97] Moreover, Iorio stated that the modern ephemerides of the Solar System outer planets has provided even tighter constraints: no celestial body with a mass of 15 times that of Earth can exist closer than 1,100–1,300 AU.[98] However, work by another group of astronomers using a more comprehensive model of the Solar System found that Iorio's conclusion was only partially correct. Their analysis of Cassini data on Saturn's orbital residuals found that observations were inconsistent with a planetary body with the orbit and mass similar to those of Batygin and Brown's Planet Nine having a true anomaly of −130° to −110° or −65° to 85°. Furthermore, the analysis found that Saturn's orbit is slightly better explained if such a body is located at a true anomaly of 117.8°+11°
−10°. At this location, it would be approximately 630 AU from the Sun.[99]

https://en.wikipedia.org/wiki/Planets_beyond_Neptune[2]
New moons of the four giant planets are still being discovered, but the recent discoveries are all of tiny objects, less than 10 kilometers in diameter.


What is the current estimation for how much time the Sun will function properly?

Just to add to Undo's answer, after the expansion to a red giant, the sun will become a planetary nebula, where (according to the link) the fusion reactions inside the star are 'overtaken' by the gravitational collapse, in turn causing the inner layers to condense and heat up causing the outer layers to be blown away.  After which, the hot core forms the remnant, a White Dwarf star (NASA source), which is likely to last for several more billions of years.

The image below depicts the current ideas of the expected lifecycle and timeline of the life of the sun:

Image source

How do we know what will happen to the sun?

Currently, the main method to determine the solar lifecycle is described by the article "The Future of the Sun" (Cowing, 2013) is to:

  Studying stars with the same mass and composition as the Sun, the so-called "solar twins," can give us more information about our own Sun; solar twins of various ages offer snapshots of the Sun's evolution at different phases

Where the mass and chemical composition of a star provide the information needed to determine its lifecycle.


Does the summit of a small mountain offer better seeing?

In terms of steadiness of the air and hence stability of the image (ie ability to see fine detail on planets or split close double stars), it could go either way, depending on the atmospheric conditions, including wind speed and direction, and the local topography.

In contrast, valleys can sometimes provide better seeing because they provide some protection from atmospheric conditions.

However, and all other things being equal, you will probably have a higher limiting magnitude (ie be able to see fainter objects) from the top of the mountain.


Best way to simulate star sizes to scale in celestial sphere

Your maths looks ok, bar the fact that $1/$parallax is a biased estimate of the distance (but that can be forgiven so long as you are using data where the parallax uncertainties are much smaller than the parallax).

Your main problem is that stars do indeed have a vast range of sizes. Thus if you really do want to show the relative sizes of the stars you have a dynamic range problem.

The conventional way to deal with this would be to use a logarithmic scale, such that each increment in size corresponds to a multiple of the one below. e.g. Make your plotted radii proportional to $\log_{10} R_{\rm rel}$.

EDIT: In response to altered emphasis of question.

It seems you no longer want to scale the stars according to their actual radius, but rather according to their brightness. The logarithmic scaling is therefore already handled by the apparent magnitude (which is on a logarithmic scale).

Therefore all you need to do is decide what your smallest and largst radius are and compare that with the brightest and faintest magnitudes you want to plot.

e.g. If your biggest star is 5 units and your smallest is 0 units, and your bright and faint limits are $m_{\rm bright}$ and $m_{\rm faint}$ respectively, then the size of an arbitrary star of magnitude $m$ is
$$ r =\frac{5(m -m_{\rm faint})}{m_{\rm bright} -  m_{\rm faint}} \ .$$

And if you want your faintest stars to be size 1 and brightest to be 5, this modifies to
$$ r =\frac{4(m -m_{\rm faint})}{m_{\rm bright} -  m_{\rm faint}} +1 \ .$$

Yet another alternative would be to make the area proportional to magnitude. Again, scaling between radii of 5 and 1:

$$ r^2 =\frac{24(m -m_{\rm faint})}{m_{\rm bright} -  m_{\rm faint}} +1 \ .$$


Strictly speaking the Sun's motion is going to be dominated by the gas giants and the planets and one asteroid is going to have negligible effect on it compared to those.

However if we assume there are no other items we can get an approximate idea of the effect.

You can use the similarity of the orbits.  The Sun ( in this simplified model ) follows a complimentary orbit around the center of mass of the two bodies.

The distances you have for the size of the obit scale like this :

$$r_1 = \frac \mu m_1 r$$

$$r_2 = \frac \mu m_2 r$$

where $\mu$ is the reduced mass :

$$\mu = \frac { m_1 m_2 } { m_1 + m_2 }$$

And as you can see $r = r_1 + r_2$ is the distance between the two bodies.

Now as you can see from those equations, the most massive body moves the least.

The mathematics comes from the way the standard two body problem is solved.

It might be instructive for you to compare the effect of your asteroid to that of Jupiter or Earth.


Is it possible to do planet observation during the day?

Jupiter can be seen during the day.

This image is by Philip Crude. Philip is an experienced astro-photographer. On his webpage http://www.billionplanetsquest.com/p/planets.html he gives details of the equipment and settings used:

This image was captured through my [i.e., Philip Crude's] Celestron CGEM-800 using a ZWO ASI120MC camera at prime focus. Resolution 640 x 480, 35fps best of 900, Brightness @ 1, Gamma @ 48, Gain @ 11. The image was acquistioned with SharpCap and processed with RegiStax and CS6.

Philip also tweeted the image here.
The main difficulty is pointing the telescope in the right direction, as you won't be able to align it against the background of stars, so you need to predict the position and point the telescope accordingly.


How can astronomers determine the difference between "hydrostatic equilibrium" and "just happens to be spherical"?

I think you're asking, "If we know an object's shape, can we determine if it is in in hydrostatic equilibrium?" If so, one might wonder if astronomers classify basketballs or ball bearings as being in hydrostatic equilibrium, since they're so spherical.

Below about 100 km in radius, the answer is in general no. Given some population of randomly lumpy objects (like asteroids), some of them will be close to the shape of a sphere purely on accident. The composition matters too -- an object of this size made of hydrogen gas would assume a spherical shape from hydrostatic equilibrium, but an object made of rock might not (like Mathilde below). We could make better predictions given detailed knowledge of the materials and environment of the object, but this is not always possible, as you mentioned. For small objects, intermolecular and atomic forces dominate gravity.

Once you get to a certain size of object, it becomes much easier to make a prediction about hydrostatic equilibrium. This is still very much context-dependent, and you still get complications from material composition, temperature, etc. However, atomic binding forces have some fixed strength, but gravity scales like mass. Given ordinary astrophysical materials, we can be very sure that a body like Jupiter is under hydrostatic equilibrium.

You can make some order of magnitude estimates by assuming that the atomic interaction energy must be at least as large as the thermal energy (Hughes and Cole 1995). If you check out equation 5 of that paper, you'll see an explicit expression for a radius that divides spherical objects and nonspherical ones. At some mass, the binding atomic energy becomes dwarfed by the gravitational potential, and you always get a spherical object.

tl;dr -- small objects no, big objects yes, medium objects may require detailed modeling.


Astrophotography - Unable to achieve focus with current setup

Yes, the vast majority of small telescopes made for visual observations (like most dobsonians and other newtonian-based architectures) have very limited back focus. In other words, with the focuser retracted all the way in, the focal plane of the primary mirror is sticking out only a short distance. A large back focus is not typically seen, nor is it desirable, in a visual scope. (However, most astrographs have a large back focus.)

This is compounded by the fact that most DSLR have a flange distance (focus-flange distance) which is pretty large - over 45...50 mm in many cases.

Furthermore, an additional distance is wasted by the classic T-ring adapters, which tend to push the camera way back from the focuser's shoulder.

All these factors conspire in your case to prevent you from achieving focus.

Some ideas and solutions:

Adapters

I tell everyone who is trying to do prime focus photography with visual scopes (dobsonians, regular newtonians) to ditch the old T-ring system and just buy dedicated adapters. All you need is a piece of metal that goes in the bayonet mount on the camera on one side, and into the focuser on the other side. No need for T- components.

Some such adapters are made by telescopeadapters.com:

http://www.telescopeadapters.com/

I have two adapters like this, they save a very large amount of back focus, compared to the classic T-ring / T-adapter system. In other words, they allow the camera to get much closer to the focuser, and therefore catch the focal plane on the sensor. You just need to choose the kind of adapter that works with your camera.

Most of these adapters are 2", so in a 1.25" focuser a size adapter is needed - but that would waste some back focus distance.

Camera

Classic DSLR cameras have a very large flange distance, so a large back focus is needed to overcome it.

Other removable-lens, large-sensor cameras have a much smaller flange distance, if they are mirrorless. Examples: Micro Four Thirds cameras (Panasonic G series), Sony NEX, Samsung NX, etc. All these systems have large sensors (comparable to DSLR) which work well for astrophotography, but don't have moving mirrors, so their flange distances are tiny.

https://en.wikipedia.org/wiki/Mirrorless_interchangeable-lens_camera

I have a Panasonic G1, used with the adapter mentioned above. I only need 25 mm of back focus for this to work, which is an extremely small value.

Use a barlow

A barlow would increase the amount of back focus available. Try and see what happens. Keep in mind it cannot work miracles - if you need a very large amount of back focus, it may not work.

Use eyepiece projection

With some eyepieces, you can thread adapters on the user side, and then put the camera body (without lens) on the adapter. Move the eyepiece WAY out of focus, and it will generate a real image which can be projected on the sensor. This is not prime focus photography anymore, but it works and it's better than nothing.

However, such adapters are hard to find, and many eyepieces won't support them.

Use a dedicated USB "eyepiece camera"

You can find relatively cheap USB cameras made for imaging, some are less than $100. With one of those, you could achieve focus relatively easy. Example:

http://www.telescope.com/Astrophotography/Astrophotography-Cameras/Orion-StarShoot-USB-Eyepiece-Camera-II/pc/-1/c/4/sc/58/p/102083.uts

Would not offer the same performance like a DSLR, but it's better than nothing.

Use a telescope made for imaging

Visual scopes, like a small newtonian, are not really that great for photography, for a variety of reasons. A dedicated astrograph would always work better. There are many Cassegrain systems out there made specifically for imaging, some are relatively inexpensive (not much more expensive than a DSLR).

In many cases, you need very long exposure, and you need to track the sky while the shutter is open. Many such scopes come with a tracking mount, or are made to be installed on a tracking mount. That would make it very easy to do tracking well.

It would be very hard to do tracking with other kinds of scopes (tracking platforms for dobsonians do exist, but are not very common).

Move the primary mirror up

Think about it - if you move the primary mirror up in its cell, the focal plane will stick out. This is a way to gain back focus only if you know what you're doing. Additional spacers in the primary mirror cell could do it. Modifying the cell is also doable.

Beware, you could ruin the scope if you make a mistake. Also, if the secondary mirror is too small to begin with, you'll increase the amount of clipping.

Afocal photography

Put the eyepiece in the scope, just like for a normal visual observation. Now put the camera at the eyepiece and move it around until you capture the image. Click the shutter. This is called afocal photography.

You'll need a camera mount, that will attach to the focuser, holding the camera at the eyepiece. Something like this:

http://www.telescope.com/Astrophotography/Astrophotography-Accessories/Orion-SteadyPix-Deluxe-Camera-Mount/pc/-1/c/4/sc/61/p/5338.uts

Keep in mind that your camera is very likely too heavy for this trick to work. Also, afocal is usually the lowest quality of all techniques, and rarely works well enough.

(There are special adapters that go on the eyepiece on one end, and thread into the camera lens like a filter on the other end. These would work better, but are hard to find, only a few eyepieces support them, and your camera is too heavy anyway.)

Keep in mind that your scope has a 1.25" focuser. Even if you do get the sensor in focus, the small size of the focuser will clip the image; you will only be able to use the central part of the sensor. It is recommended to use 2" focusers for prime focus imaging with large sensors.

Also, your f/4 scope would exhibit strong aberrations at the edge anyway, which would blur the image. For prime focus imaging, longer focal ratios are usually desirable.

All in all, your scope and your camera are mismatched. You have a pretty good camera, but a very small scope that is not made for imaging. Either use a dedicated astrograph with that camera, or keep the scope and use a cheap USB "eyepiece camera".


What is the horizontal distance needed to observe an object just as badly as if it where in space at the zenith?

If the atmosphere could be redistributed so that it has sea level density $D$ up to height $H$ and $0$  density above that, then we can calculate
$$H = \frac{P}{gD} = \frac{100,000}{9.8 \times 1.27} \approxeq 8,000$$
(SI units), where $P$ is sea level pressure.

So it is the same as looking through $8$ km of air horizontally.


Is it possible to use the vacuum of space to help with pollution

No.
It is hard to get stuff into space.  If you built a pipe that went to space, it wouldn't suck stuff up for the same reason that the atmosphere doesn't get sucked off the Earth: Gravity pulls things down.  Instead you would have to pump stuff up the pipe, you would need lots of power to do this.  And when it reaches the top it would just fall down, unless you make your pipe 36000km long, so that the rotation of the Earth becomes strong enough to push it away
What you have essentially built is a space elevator. There are huge engineering challenges in making a space elevator.  It is honestly the very hardest way to get rid of pollution.


When will the next transit of Earth be visible from Mars? Was the last one really on May 11, 1984?

Yes. The Clarke story is quite correct. Earth (and Moon) would transit the sun from the viewpoint of Mars on 8 May 1905, 11 May 1984 and again on 10 Nov 2084.

full list over a period of about 1000 years, from Wikipedia Transit of Earth from Mars

November 10, 1595   May 5, 1621     May 8, 1700     November  9, 1800
November 12, 1879   May 8, 1905     May 11, 1984    November 10, 2084
November 15, 2163   May 10, 2189    May 13, 2268    November 13, 2368   May 10, 2394
November 17, 2447   May 13, 2473    May 16, 2552    November 15, 2652   May 13, 2678

They follow a pattern of occurring after 26, 79 then 100 years.


If you want to trigger outflows, there are two main processes that can help:

magnetic field;
radiation pressure.

Magnetic fields is well-suited to launch such outflows, through magneto-centrifugal effects. Basically, if there is magnetic fields tighed to the disk surrounding the protostar and the angle between the field lines and the disk is right, a particle from the disk that follows the magnetic field lines can be ejected. This is the well-known Blandford & Payne (1982) mechanism.

Note that, at earlier stages of star formation, magnetic field could also trigger outflows by a "magnetic field tower" growth. Basically, during the collapse of the prestellar core into a protostar, the field lines are dragged and twisted (due to the rotation of the core), the twist propagating from the inner to the outer region of the core, that triggers outflows. This kind of thing is discussedd in Hennebelle & Fromang (2008).

By the way, these two mechanisms are complementary, because slow outflows (of the order of 10 km s$^{-1}$) are observed around young protostar, whereas "fast outflows" -- jets, of the order of 100 km s$^{-1}$ -- are observed around late protostars, and these two mechanisms explained well the former and the later.

For more massive stars, radiation pressure will plays a role in outflows triggering. The protostar radiates away a part of its accreting energy, and strong a radiation pressure can trigger the growth of a kind of "bubble", that is an outflow (you can see Hennebelle & Commerçon (2012) for a brief review of the question).

Regarding your comment on your own question, it is mostly radiation pressure that will clear up the gas around the new-born star, that is clearly not an outflow or anything alike.


does a minority of all planetary systems show transits?

Consider a circular orbit and approximate the planet as a point.

The distance from Earth to a star is so large compared to the diameter of the star, which we'll call d, that we approximate the lines joining the two poles of the star with the observer as parallel to each other and to the x-axis that we choose. Thanks to symmetry, we choose our coordinates so that the plane of the planet's orbit passes through the tangent plane of the celestial sphere at the star in two places on the y-axis. By symmetry again we ignore half the cases for rotation and also ignore the direction of the orbit and only have to consider how the orbit is inclined at between 0 and pi/2 radians from the x-axis (Edit for clarity: in the XZ plane). Call that angle a.

When the radius r of the planet's orbit is such that d/r = sin(a), the planet will just barely touch the limb of the star when it is closest to the Earth. If sin(a) > d/r it will never happen and if sin(a) < d/r there will be a transit of finite length. Since there is no preference for the value of a, we conclude that the probability of the planet transiting must be $$P(sin(a) < d/r) = P{ \left( a < {arcsin \left( { d \over r } \right)  } \right) } = { 2 arcsin \left( { d \over r } \right) \over \pi } $$

You will notice that for a star-grazing planet, the probability of a transit approaches certainty. Overall probability will depend on the probability distribution of semi-major axes and possibly eccentricity which I ignored here.


Which observatories help NASA find 5 new asteroids per day?

Given that the article seems to be referring to NASA's Near Earth Object Program, it appears that there are five subprograms scanning the skies:

Catalina Sky Survey (CSS) (four telescopes: 0.50 m, 0.68 m, 1.00 m, 1.50 m)
Pan-STARRS (one telescope: 1.80 m)
LINEAR (two telescopes: 1.0 m (retired), 3.5 m)
Spacewatch  (two telescopes: 0.9 m, 1.8 m)
NEOWISE (one telescope - WISE: 0.4 m)

According to JPL, CSS and Pan-STARRS are responsible for 90% of new Near Earth Object (NEO) discoveries. LINEAR is responsible for detecting objects one kilometer in diameter or greater, while NEOWISE provides additional measurements from previous WISE targets, so it isn't making new detections, merely follow-up observations.

NASA's statistics show that over 15,000 Near-Earth Asteroids (NEAs) have been discovered so far, from all programs scanning the sky. In each of the past three years, about 1,500 NEAs have been discovered, most by the five programs listed above. That's roughly consistent with the five-per-day claim. However, it's not yet near 2,000 per year.


How much damage would collision with a similar mass primordial black hole do to earth?

Consider an initially stationary particle of matter and suppose a 1 Earth mass black hole flies past it at speed $v$ on a trajectory that passes the initial position of the particle at distance $r$. The particle will be mainly affected by the gravity during a time period of roughly $r/v$ (up to some "geometric" constants), during which time it will accelerate at roughly $GM/r^2$. This means that it will move a distance about $GM/r^2 \times (r/v)^2 = GM/v^2$. So if this is small compared to $r$ it will be left "more or less" where it was, rather than being ripped away. So we can expect a hole of diameter a few times $GM/v^2$.  $GM$ is about $4\times 10^{14}$ so to keep the hole diameter to say 1m you could need $v$ about $2\times 10^7 m/s$ which is about $0.07c$.
We can also estimate the typical velocity change achieved by our test particle from this interaction, which is $GM/r^2 \times r/v = GM/rv$, giving a KE per unit mass of about $$G^2M^2/r^2v^2$$.
So, suppose the Earth has density $\rho$  and radius $R$ the cylindrical shell of thickness $dr$ will mass $R\rho rdr$ (still ignoring "small" constants like $2\pi$). That shell will acquire kinetic energy $G^2M^2R\rho dr/rv^2$ from the interaction. Integrating from $GM/v^2$ to $R$ we get a total energy deposited in the parts of the  Earth which are not "ripped away". $$\frac{G^2M^2R\rho}{v^2}\log\frac{Rv^2}{GM}$$
Using $R = 6\times 10^6$, $GM = 4\times 10^{14}$, $\rho = 5000$, and $v = 2\times 10^7$  (all SI units) we get about $10^{25}J$, not enough to actual destroy the Earth (Earth's gravitational binding energy is about $10^{32}J$) but about $10^{10}$ megatons or an earthquake of about 13 on the Richter scale.
A black hole moving 1000 times slower (typical solar system velocity) would essentially destroy the Earth.


Why 2014-2015-2016 were the years when the number of discovered exoplanets increased the most?

The big increase is due to the Kepler spacecraft. Kepler is responsible for the majority of the discovered exoplanets (https://exoplanets.nasa.gov/keplerscience/), so its data releases will have a big effect on the planet’s discovered by year. It’s easy to see in the plot attached (From: https://link.springer.com/article/10.1007/s12145-020-00464-7 ) how the years you mentioned match the announcement of discovered exoplanets using Kepler’s and K2 (same spacecraft).
On why it is ~2014 and not 2010/2011/2012, from my understanding that's when the campaigns' data started being released (https://keplerscience.arc.nasa.gov/k2-data-release-notes.html#k2-campaign-1).


What are the effects of using non-spherical lenses in refracting telescopes?

Short answer: they exist, but they aren't cheap (unless made of molded plastic).
Long answer:
Telescopes use parabolic, hyperbolic, elliptical, and less frequently, non-conic aspheres to achieve optimal performance for their design intent.  A reflective telescope using 1 or 2 mirrors simply cannot achieve acceptable performance with spherical mirrors.
Also worth noting, the traditional hand-grinding method of shaping a telescope mirror tends to form a parabolic shape anyway, so it just works out.
Transmissive optics (i.e. lenses) however, have the ability to put many elements/surfaces in a single axis, and thus can achieve better performance quality.  Having said that, designs often limit the number of lenses for various reasons: size, weight, total transmission (since each lens absorbs some light), simplicity of assembly, and sometimes cost.
Thus, engineers will use aspheric surfaces to achieve these performance goals. In mass production, plastics and some glasses can be molded into aspheres with relatively low cost. Some can be formed by polishing a flat or spherical shape while the glass is under stress, and then when the stress is removed, it relaxes into an asphere. A lot of military/aerospace equipment use aspheres even if they are not low-cost, because it makes the equipment meet the size- and weight-to-performance goals.
There is even a common telescope design that implements an aspheric lens.  The common Schmidt-Cassegrain telescope (SCT) has a front glass element which looks flat, but is actually a very slight 4th-order asphere ("w" shaped).  This is manufactured commercially using the stressing method mentioned before.


Cosmological parameters are measured in a variety of ways, and their values will depend on which measurements you trust the most. The paper you link to (Planck Collaboration et al. 2016) with the 2015 results from the Planck observations of the cosmic microwave background is probably the one that most people will accept, but even in that paper you will find different values, depending on which observables you combine.

You will find the values in their Table 4. I think that most people use the values in the column called "TT+lowP+lensing" (e.g. Geil et al. 2016, Ricotti et al. 2016, and Liu et a. 2016), which is the "conservative" choice. However, you'll also find some (e.g. Chevallard & Charlot 2016 and Silk 2016) who use the values in the last column, called "TT,TE,EE+lowP+lensing+ext". These values takes into account external data (baryonic acoustic oscillations and supernovae data), which reduce the uncertainties, arguably to unnaturally small values. The TT, TE, TT, and lowP refer to the polarization maps used, and "lensing" to the weak gravitational lensing measurements by Planck.

Standard cosmological parameters

The table below is a modified version from the Planck paper where I show only the most used parameters:

Here, $n_s$ is the slope of the primordial power spectrum, $H_0$ is the Hubble constant in km s–1 Mpc–1, $\Omega_\Lambda$ and $\Omega_m$ are the density parameters of dark energy and total (dark+baryonic) matter, $\Omega_\mathrm{b}h^2$ and $\Omega_\mathrm{c}h^2$ are the density parameters of baryonic and dark matter, multiplied by the factor $h \equiv H_0/100$ (squared), $\sigma_8$ is the matter density fluctuations on scales of 8 (comoving) Mpc, $z_\mathrm{re}$ is the redshift at which the Universe was reionized (assuming instant reionization), and the last row shows the inferred age of the Universe in billion years.

Curvature density

The constraints on the curvature parameter $\Omega_K$ is given in Table 5, which has somewhat different combinations of data. All in all, Planck constrains the curvature to $|\Omega_K| < 0.005$, but you will rarely offend anyone by simply setting the curvature to zero.

Temperature and radiation density

The radiation density is a bit more convoluted. It has a contribution from both photons and neutrinos, and their densities are related as
$$
\rho_\nu = N_\mathrm{eff} \frac{7}{8} \left(\frac{4}{11}\right)^{4/3}\rho_\gamma,
$$
where $N_\mathrm{eff} = 3.046$ is the effective number of neutrino species. Following the procedure by Pulsar in this answer, but with updated parameters (i.e. the $N_\mathrm{eff}$ given above, and the average CMB temperature of $T_0 = 2.722\pm0.027$ (Eq. 83a)), I get that
$$
\begin{array}{rcl}
\Omega_\mathrm{rad}h^2 & = & \Omega_\nu h^2+ \Omega_\gamma h^2 \\
                       & = & (1.7018 + 4.6213) \times 10^{-5} \\
                       & = & 4.1620\times10^{-5},
\end{array}
$$
that is, with $h = 0.6781$,
$$
\Omega_\mathrm{rad} = 9.0513\times10^{-5}.
$$

Recap

So, answering your question is a little difficult, as there's no single answer, and since the curvature is given with 95% confidence ("2$\sigma$"), rather than 68% ("1$\sigma$"). For the Hubble constant and the matter and dark energy, I'd recommend $H_0 = 67.81\pm0.92 \,\mathrm{km}\,\mathrm{s}^{-1}\,\mathrm{Mpc}^{-1}$, $\Omega_\mathrm{m} = 0.308\pm0.012$ and $\Omega_\Lambda = 0.692\pm0.012$.

For curvature, I would use 0 (especially because your calculations probably would need to switch between ordinary and hyperbolic trigonometry depending on the sign),  but if you do want to include uncertainty, you can say $\Omega_K=0\pm0.005$ (95%). Or, you could simply use $\Omega_K = 1 - \Omega_\mathrm{m} - \Omega_\Lambda - \Omega_\mathrm{rad}$ and propagate the errors to get $\Omega_K = 0\pm0.017$, which will give you a more conservative value.

For radiation, to propagate uncertainties you would have to know the covariance matrix of the input parameters, but since the total error is dominated by that of the Hubble constant, standard error propagation $(\sigma_{\Omega_\mathrm{rad}}/\Omega_\mathrm{rad} \simeq 2 \sigma_h/h)$ yields a value of $\Omega_\mathrm{rad} = (9.0513\pm0.2456)\times10^{-5}$.

So, to be explicit my recommendation is:
$$
\{\Omega_\mathrm{m}, \Omega_\Lambda, \Omega_\mathrm{rad}, \Omega_K\}
= \{ 0.308, 0.692, 9.05\times10^{-5},0\}
\pm \{0.012, 0.012, 2.46\times10^{-6}, 0 \}.
$$
But I think the most important thing is to state where you take the parameters from. People rarely state why the choose a particular set of parameters, and although Planck gives very small error bars, other probes give error bars so small that they're basically mutually incompatible. That's why you can still easily get away with $\{\Omega_\mathrm{m}, \Omega_\Lambda, \Omega_\mathrm{rad}, \Omega_K\} = \{ 0.3,0.7,0,0\} \pm \{0,0,0,0\}$.


Is the list of the most abundant elements in the universe a fact?

Neil de Grasse Tyson is extrapolating.
Chemical abundances can and have been measured in a huge numbers of stars in our own Galaxy and considerably fewer in local galaxies. The chemical abundances of glowing gas clouds can also be measured and this technique has a much greater reach. In addition, there are ways in which the summed light from all the stars in a galaxy can be used to give crude information about their average chemistry, and again, this can be used out to large distances.
From these measurements we have a pretty good idea of the chemistry of our local part of the universe. One can then construct an inventory to make a table like the one in your question. This is dominated by stars and gas - planets cannot be measured but are a negligible fraction of mass.
Now there is a complication. The chemistry of the universe changes with time, because hydrogen and helium are gradually being turned into heavier elements inside stars, and then much of the products are distributed into the interstellar and intergalactic medium when stars die. Thus to get an "up to date" inventory one should exclude the older stars and perhaps focus more on the gas, which gives an idea of current chemistry in the interstellar medium.
Having done all this, you get the table in your question - which applies to the local universe.
It is then a fundamental assumption in cosmology that the universe is homogeneous on large scales. There is thus no reason or evidence to suppose that things are different elsewhere. Indeed, given that we now pretty much understand why the chemical league table looks the way it is - a simple consequence of the physics of star formation, nuclear fusion and mass loss in stars - then it is difficult to imagine any scenario in which it could be very different anywhere else.


what kind of telescope eyepieces is best for viewing planets up close

To get good views of Jupiter and Saturn you need quite high magnification - probably 100 times or more. The XT8 has a focal length of 1200 mm and so to get that magnification will need an eyepiece of 12 mm focal length. I believe the scope comes with a 10mm eyepiece which will give 120 times magnification, which should give decent views.

You could go for higher magnification with shorter focal length eyepieces, but if you do you will find that the object you are observing scoots across the field of view rather quickly and you will spend more time moving the telescope than actually observing.  As you get more experienced you may find you get more adept at moving the scope, and so could use more magnification.

It’s important that the mirrors of the scope are properly aligned (collimated) so as to get the sharpest views. The manual should tell you how to do that.


What is the resolution in megapixels of the Hubble Telescope?

The size of the cameras that took the Pluto pictures is easy enough to find: the ACS Wide Field Camera has two 2K by 4K (hence 8 Mega pixel) CCDs, with a field of view of 202×202 arcsec. The high resolution HRC had a square 1 Megapixel camera with a 26×29 arcsec field of view.

The reason that the images of Pluto are relatively poor, is that Pluto is small! The size of the camera is irrelevant; it is the size of the pixels and angular resolution that are vital.

The pixel size of the HRC (which I think is what the pictures are mainly due to) is 0.025 arcsec, which matches the angular resolution of 0.05 arcsec. The angular diameter of Pluto is about 0.1 arcsec. Hence you effectively have an image that is only made of 4×4 pixels!  I'm surprised it looks as good as it does - but that is because these are not the raw images (see below).

Edit: In response to comments I had a look at some details from the original paper describing these images, by Buie et al. (2010).  http://iopscience.iop.org/article/10.1088/0004-6256/139/3/1128/meta  It seems that what is shown is not an actual image from the ACS, but is a best-fitting map of what Pluto looks like, obtained by modelling all the images available.

The model image is described by a much finer pixel scale than the original images, but then blurred with a Gaussian filter. However, to quote from the paper, "The width of the Gaussian filter was chosen to impose a limit on the spatial frequencies in the output image comparable to the diffraction limit of HST. "

Thus the angular resolution of the final "reconstructed images" is still similar to and limited by the angular resolution of the HST.

Just to hammer home the message. It really doesn't matter here how many pixels the HST cameras have, so long as they have enough to sample the angular resolution properly. At a 0.1 arcsec angular diameter as seen from Earth, Pluto has an angular size equivalent to trying to take a photo of a quarter from 50 km. It really isn't going to help you if you have the latest camera phone with a gazillion pixels unless you also connect it to a 2.4m telescope like HST with an angular resolution better than this.


Is it safe to view a solar eclipse with eyeglasses along with eclipse glasses?

Yes, it's safe :).  Your eyeglasses are just correcting your vision. The net effect is just to give the equivalent of normal vision - and in any case, you'd normally be holding the eclipse glasses in front of your eyeglasses, in the same way that front-of-scope solar filters work to reduce the incoming light to a safe level.

Even if you're longsighted (which uses converging (positive) lenses to correct your vision, the glasses are close enough to your eye that you're nowhere near the focal point - so the light doesn't get concentrated much (and probably gets unconcentrated again by your eye, anyway).

If you're shortsighted, like me, your glasses will be diverging (negative) lenses which spread the light out rather than concentrating it (and again, are close enough to your eye that's there's not enough distance for it to affect things much.


Exploiting symmetry in the axisymmetric Jeans equations

Background: If $f(\mathbf{x},\mathbf{v},t)$ is the distribution function of the stars in phase space and $n = \int f\,\mathrm{d}^3\mathbf{v}$ is the star density, then in the cylindrical coordinates $(R,\varphi,z)$, the $z$-component of the second of the Jeans equations applied to an axisymmetric system is
$$\renewcommand{\expv}[1]{\langle{#1}\rangle}\partial_t(n\expv{v_z}) + \partial_R(n\expv{v_Rv_z}) + \partial_z(n\expv{v_z^2}) + \frac{n\expv{v_Rv_z}}{R} + n\partial_z\Phi = 0\text{.}$$
In a steady state, the first term vanishes, so if the positive and negative terms of the $z$-component of velocity terms balance, then so do the second and fourth, in which case:
$$\partial_z(n\expv{v_z^2}) = -n\partial_z\Phi\text{,}$$
which we can then integrate.

If the the star and mass distributions are axisymmetric and symmetric about the equatorial ($z = 0$) plane, then the potential has even symmetry in $z$, $\Phi(R,z) = \Phi(R,-z)$, and so does $n$. Thus their derivatives with respect to $z$ are odd functions, and $\Phi_{,z}(R,z) = -\Phi_{,z}(R,-z)$ implies
$$\frac{1}{n}\int_{-z}^z n\Phi_{,z'}\,\mathrm{d}z' = 0\text{,}$$
since the integrand has odd symmetry. Hence,
$$\frac{1}{n}\int_z^\infty n\Phi_{,z'}\,\mathrm{d}z' = \frac{1}{n}\int_{|z|}^\infty n\Phi_{,z'}\,\mathrm{d}z'\text{.}$$


Which wavelength is the most quiet, for a ground-based radio telescope?

You may be looking for the "water hole". See http://www.setileague.org/general/waterhol.htm.


There are many, many ways of doing this.
Gravitational lensing
This is by far the most well known. It has been mentioned by the others, but I'll touch on it.

Light coming from distant bodies can be bent by gravity, creating a lens-like effect. This can lead to multiple or distorted images of the object (Multiple images give rise to Einstein rings and crosses).
So, if we observe a lensing effect in a region where there isn't any visible massive body, there's probably a black hole there. The alternative is that we are peering through the dark matter 'halo' which surrounds (and extends passed) the luminous components of every galaxy and galaxy cluster (See: Bullet Cluster). On small enough scales (i.e. - the central regions of galaxies), this is not really an issue.

(This is an artist's impression of a galaxy passing behind a BH)
Gravitational waves
Spinning black holes and other dynamical systems involving black holes emit gravitational waves. Projects like LIGO (and eventually, LISA) are able to detect these waves. One major candidate of interest for LIGO/VIRGO/LISA is the eventual collision of a binary black hole system.
Redshift
Sometimes we have a black hole in a binary system with a star. In such a case, the star will orbit the common barycenter.
If we observe the star carefully, its light will be redshifted when it is moving away from us, and blueshifted when it is coming towards us. The variation in redshift suggests rotation, and in the absence of a visible second body, we can usually conclude that there's a black hole or neutron star there.
Salpeter-Zel'dovitch / Zel'dovitch-Novikov proposals
Going in to a bit of history here, Salpeter and Zel'dovitch independently proposed that we can identify black holes from shock waves in gas clouds. If a black hole passes a gas cloud, the gases in the cloud will be forced to accelerate. This will emit radiation (X-rays, mostly), which we can measure.
An improvement on this is the Zel'dovitch-Novikov proposal, which looks at black holes in a binary system with a star. Part of the solar winds from the star will be sucked in to the black hole. This abnormal acceleration of the winds will, again, lead to X-ray shock waves.

This method (more or less) led to the discovery of Cyg X-1
Cosmic gyroscopes
Cyg A is an example of this. Spinning black holes act like cosmic gyroscopes — they do not easily change their orientation.
In the following radio image of Cyg A, we see these faint gas jets emanating from the central spot:

These jets are hundreds of thousands of light years long — yet they are very straight. Discontinuous, but straight. Whatever object lies at the center, it must be able to maintain its orientation for very long.
That object is a spinning black hole.
Quasars
Most quasars are thought to be powered by black holes. Many (if not all) of the candidate explanations for their behavior involve black holes with accretion disks, e.g. the Blandford-Znajek process.


The vast majority of the particles in Saturn's rings are small, on the order of $\sim10^{-1}$ m or lower. The columnar number density, according to data from Voyager 1 and Earth-based observations, can be approximated as a function of particle radius by a power law for all particle radii $a$ in meters such that $0<a<1$, as can be seen on this log-log plot (Fig. 15.5, Cuzzi et al. (2009); taken from Fig. 8 of Marouf et al. (1982)):

Though it is only acknowledged in the original paper, the vertical axes for the three different ring regions have been shifted upward different amounts to fit all three on the same graph.

After $a=1$, there's a deviation from the law, and then a steep dropoff at about $a=3$. Obviously, particles larger than this exist, and they certainly play an important role in ring structure, but they're relatively rare.

Obviously, the trends show that smaller particles are much more common, and thus while there are indeed particles larger than pebbles - some as big as boulders, perhaps, or bigger - they are certainly few in number. Most particles are extremely small, smaller than pebbles.

This data covers observations from ring semi-major axes of $\sim$ 75,000 km to $\sim$ 135,000 km - a fairly big spread, covering most of the rings and ending near the Roche Division. The paper doesn't have one single graph of particle number density of a given size at a given semi-major axis, but it does have several subdivided plots (Fig. 15.1 and 15.2) of optical depth as a function of distance from the center of Saturn, which should give you some helpful data on total number density, if you want to make some basic assumptions about mean particle radius. This data is a bit newer, from Cassini, but the Voyager 1 data is just as helpful.


Depending on the mass of the universe, the universe has three different possible shapes.
http://en.wikipedia.org/wiki/Shape_of_the_universe

It can be finite or infinite depending on the shape it has, and there are some ways we can discard possible shapes (if it started to shrink instead of expand).

Hope this helps


How to create synthetic test images of stars?

This doesn't need programming but the SkyMaker package will generate very realistic astronomical images from catalogs produced by Stuff. It includes realistic models of stars and galaxies and the point spread function including e.g. atmospheric blurring and noise effects in the CCD (blooming, saturation etc)


Apsidal Precession: What Am I Getting Wrong?

The value obtained is in degrees per day (which I didn’t know); this is simply multiplied by 3600 to convert to seconds per day, then by 36525 to convert to seconds per century.


Can a telescope be built to focus on a point source star?

A smaller field of view would not necessarily help. The data transmitted back to Earth by Kepler was already limited by telemetry constraints and the portions of the CCD that were read out were already focused on small windows around (many) selected stars. Having a field of view that is at least 10 arcsecond across would be sufficient. What is more important is to make sure your star images are well-sampled. This is a bit of a problem for Kepler. The CCD pixels are quite big (4 arcseconds) to give a big field of view, but this is similar to the angular resolution of the telescope. This means all the light can fall in basically one pixel, which is not ideal for precise photometry. If you were focusing on one star you would make the field of view smaller but have lots of CCD pixels across your star image. This improves the precision of the photometry considerably.

If you were to construct a similar telescope designed to look at one star at a time, then in principle you could have a much higher telemetry rate for that star. Hence, if you had enough photons and a low CCD readout noise, you could take much shorter exposures and see finer detail in the light curve transits - maybe moons, rings, starspots etc. The standard Kepler cadence was 30 minutes for most stars and one minute for some of the brightest stars - see this link. A more rapid cadence would allow a much better characterisation of the host star through asteroseismology (e.g. mass and radius), which in turn leads to better characterisation of its planets.

To take advantage of the higher telemetry rates, it would be advantageous to have a larger aperture telescope, simply to collect enough photons. The angular resolution of the telescope is not that important if investigating obviously single , bright sources. You would need an absolutely enormous telescope (100m+) to resolve the disks or planetary systems of even nearby stars, so this is not really feasible.

A larger telescope would also allow you to split the light by wavelength and investigate transits in a wavelength dependent way. This is really important for investigating atmospheric composition and chemistry. This is perhaps the only area where focusing on a single star might be useful - because you could design a slit spectrograph to monitor the transit spectrum. On the other hand, these days it is possible to design multi-object spectrographs using fibres or slit masks, so overall I can't see any real advantage to having a telescope that is capable of gathering data for only one star at a time.


Is it possible to achieve a stable "lunarstationary" orbit around the moon?

First off, such an orbit wouldn't be a geostationary orbit since geo- refers to the Earth. A more appropriate name would be lunarstationary or selenostationary. I'm not sure if there is an officially accepted term since you rarely hear people talk about such an orbit.

You can calculate the orbital distance of a selenostationary orbit using Kepler's law:

$$a = \left(\frac{P^2GM_{\text{Moon}}}{4\pi^2}\right)^{1/3}$$

In this case, $a$ is your orbital distance of interest, $P$ is the orbital period (which we know to be 27.321 days or 2360534 seconds), $G$ is just the gravitational constant, and hopefully it is obvious that $M_{\text{Moon}}$ is the mass of the Moon. All we have to do is plug in numbers. I find that

$$a = 88,417\:\mathrm{km}=0.23\:\mathrm{Earth\mathit{-}Moon\:Distance}$$

So I at least match your calculation pretty well. I think you were just relying on Wolfram Alpha a bit too much to get the units right. The units do work out right though.

If you want to determine if this orbit can exist however, you need to do a bit more work. As a first step, calculate the Moon's Hill Sphere. This is the radius at which the Moon still maintains control over it's satellite, without the Earth causing problems. The equation for this radius is given by

$$r \approx a_{\text{Moon}}(1-e_{\text{Moon}})\sqrt[3]{\frac{M_{\text{Moon}}}{3M_{\text{Earth}}}}$$

In this equation, $a_{\text{Moon}} = 348,399\:\mathrm{km}$ is the Moon's semi-major axis around the Earth and $e_{\text{Moon}} = 0.0549$ is the Moon's orbital eccentricity. I'm sure you can figure out that the $M$'s are the masses of the respective bodies. Just plug and chug and you get

$$r \approx 52,700\:\mathrm{km}$$

A more careful calculation, including the effects of the Sun is slightly more optimistic and provides a Hill radius of $r = 58,050\:\mathrm{km}$. In either case though, hopefully you can see that the radius for a selenostationary orbit is much farther than the Hill radius, meaning that no stable orbit can be achieved as it would be too much perturbed by the Earth and/or the Sun.

One final, semi-related point. It turns out almost no orbits around the Moon are stable, even if they're inside the Hill radius. This is primarily to do with mass concentrations (or mascons) in the Moon's crust and mantle which make the gravitational field non-uniform and act to degrade orbits. There are only a handful of "stable" orbits and these are only achieved by orbiting in such a way as to miss passing over these mascons.


Stars tend to twinkle for two main reasons: first, stars are very far away (the closest star is about 4 light-year from the Sun) and are therefore seen as point sources. Second, Earth has an atmosphere. Earth's atmosphere is turbulent, and therefore all images view through it tends to "swim". Therefore, sometimes a single point in "object space" is mapped to several points in "image space", and sometimes it is not mapped at all. Since stars are seen as single points, they sometimes seems brighter, sometimes even seems to disapear.

If you look at it in another planet of our solar system, it will depends on the planet's own atmosphere. If you look at stars on Mars, the atmosphere being very thin, the stars won't twinkle that much. Same for Mercury. On Venus, the atmosphere is so thick than you won't see anything apart from the atmosphere itself (if you are not crunched by the atmosphere pressure, by the way...).


Astrometry.net has identified your star field as being part of the Andromeda constellation. The diffuse object in the centre of your image is the Andromeda Galaxy (Messier 31). The bright star to the left of it is Mirach (β Andromedae). Given a reasonably dark sky and averted vision, it is possible to see the core of M31 with the naked eye. Since M31 is approximately 2.5 million light years away from us, it is one of the most distant objects that can be seen with the naked eye.

Diametrically opposite Mirach to M31, and equidistant from it, is the Triangulum Galaxy (M33), although it is only just discernible in your photograph (even here, averted vision helps). If you can see M33 with your naked eye, then you have excellent vision and really dark skies. At a distance of 3 million light years, it is also one of the most distant objects that can be seen with the naked eye.

Astrometry.net

Note that only the core of M31 is visible to the naked eye (as it is in your photograph). If M31 was brighter, the arms of the galaxy would be visible and it would appear to be several times larger than the full Moon, as shown in the composite image below.

NASA


Are brown dwarf masses notoriously difficult to measure?

Yes, they are difficult to measure. The only way to measure a brown dwarf mass is if it is in a binary system. The binary frequency of brown dwarfs is low - perhaps 10-15%.

Having found your binary system, then it has to be wide enough to resolve with your telescope, but not so wide that Kepler's third law means it takes centuries to do an orbit. The low mass of brown dwarfs does not help here. To have an orbit of say a decade or less means they have to orbit close to the other object. This in turn means adaptive optics are required.

Having said all that, the masses of quite a few brown dwarfs have been measured, so it is not extraordinarily difficult.

The object talked about in the article is not in a close binary, so it's mass cannot be measured, it can only be estimated indirectly. But here come the difficulties. To estimate the mass requires accurate calculations of the evolutionary tracks of brown dwarfs. The measurements of brown dwarf masses in binaries indicates that we do not have these. Secondly, to use the evolutionary tracks requires a luminosity (and therefore a distance), but because brown dwarfs are cooling objects, you also need an age. In general, unless the brown dwarf can be associated with a group of other stars with known age then this isn't possible.  Measuring the temperature of a brown dwarf isn't helpful because the cooling tracks are very close together in the Hertzsprung-Russell diagram, and in any case, it is very difficult because brown dwarfs are nowhere near blackbodies and their atmospheres are still poorly understood.

In this case, it seems that the brown dwarf has been newly associated with an (assumed to be) coeval group of stars at 200 Myr. Thus its luminosity can give a (highly) model-dependent mass. I cannot examine the details as the claimed published research cannot be found!


What do "star noise" and "sky noise" refer to for TESS (Transiting Exoplanet Survey Satellite)?

The key formula for signal-to-noise calculations in photometry can be written as something like
$${\rm SNR} = \frac{S_{\rm star}}{\sqrt{S_{\rm star} + S_{\rm sky} + \sigma_R^{2}}},$$
where $S_{\rm star}$ is the photons counted from your target star, $S_{\rm sky}$ is the amount of unrelated photons in your photometry aperture (due to sky or other stars) and $\sigma_R$ is the readout noise in your detector.

The numerator is of course your (interesting) signal and the denominator is the noise. The reason the noise terms are added in the way they are is that each term inside the square root is a variance. i.e. The variance of the detected signal from the star is equal to the number of detected photons (a property of Poissonian counting statistics).

Now for more complexity we can express the signals as being photons per unit time (I'll use a small $s$ for these), such that $S_{\rm star} = s_{\rm star} t$, where $t$ is the integration time.
$${\rm SNR} = \frac{s_{\rm star} t}{\sqrt{s_{\rm star}t + s_{\rm sky}t + \sigma_R^{2}}},$$

Now the thing defined as $\sigma$ in your plots is the inverse of the SNR, plotted on a logarithmic axis.
$$\sigma = (s_{\rm star }t)^{-1}\sqrt{s_{\rm star}t + s_{\rm sky}t + \sigma_R^{2}} $$
$$\log \sigma = -\log s_{\rm star} -\log t +0.5 \log(s_{\rm star}t + s_{\rm sky}t + \sigma_R^{2})\ \ \ (1)$$
This is being plotted against the $I$ magnitude, which is proportional to $-2.5\log s_{\rm star}$.

So if we now look at each of these contributions individually, we see that the noise due to "shot noise" on the star should indeed be a straight line on your plot. From eqn (1), ignoring other sources of noise
$$ \log \sigma = -0.5\log s_{\rm star} -0.5\log t$$
Thus in units of parts per million and scaled for the square root of the exposure time, this will appear as a straight line in a log-log plot with gradient of $-0.5/-2.5=+0.2$.

Now just look at the contribution due to the sky background, from eqn (1), ignoring other sources of noise
$$\log \sigma = -\log s_{\rm star} + 0.5 \log s_{\rm sky} -0.5 \log t$$
So the gradient here is -1 in log space (the sky term is just a constant offset in the log-log plot) and +0.4 in terms of (log) ppm/magnitude. Hence the gradient due to the sky noise term is twice as steep.

Finally, looking at the readout noise contribution
$$\log \sigma = -\log s_{\rm star} + \log \sigma_{R} -\log t$$
Therefore this has the same gradient as the sky contribution (as you can see in the plot), but the offset in the log-log plot should depend on the exposure time. This makes sense because the brightness at which readout noise will dominate does depend on how long the exposure is. So there is some unspoken (at least in the caption you show) assumption about how long the exposure time is in order to place the contribution from readout noise on the same graph, although the gradient is understood.


Gravitational wave detection time difference between LIGO Livingston and LIGO Hanford

Updated to use reported timing confidence intervals instead of trying
  to infer those from reported sky location resolution uncertainties.
  The latter approach was misleading because the sky location has been further
  resolved using more information than just the arrival time difference.

Beware that I am neither an astronomer nor involved with aLIGO.
I just use classic euclidean geometry here,
combined with occasional transformations between cartesian
and spherical coordinates.

We have two observatories, $L$ (Livingston) and $H$ (Hanford)
and a far-away point $E$ emitting a (gravitational wave) signal.
Let us assume that the signal propagates at constant speed $c$
through low-density matter, including planet Earth.
The straight-line distance between $L$ and $H$ (passing through Earth)
is $|HL| \approx c\cdot 10.0\cdot10^{-3}\,\mathrm{s}$.

$L$ receives the signal from $E$ about
$t_{ELH}\approx 7\cdot10^{-3}\,\mathrm{s}$ before $H$.
This implies that $E$ is a distance
$d = c\,t_{ELH}$ closer to $L$ than to $H$.

Now assuming large-scale euclidean geometry and looking at the
plane containing $E,L,H$, the set of all candidate points $E$
where $|EH|-|EL| = d$ is one branch of a hyperbola, namely
the branch closer to $L$:

In fact, setting
$$\begin{align}
    r &= \frac{d}{2}    & r_1 &= \frac{|HL|}{2} & R &= |LE|
\\  L &= (r_1,0)        & H &= (-r_1,0)         & E &= (x,y)
\end{align}$$
we get the cartesian hyperbola equation
$$\frac{x^2}{r^2} - \frac{y^2}{r_1^2-r^2} = 1$$
and in polar coordinates, viewed from $L$:
$$R = \frac{r_1^2 - r^2}{r - r_1\cos\alpha}$$
For very large $R/r_1$, $|\alpha|$ must be very close to its asympotic limit:
$$\begin{align}
    \cos\alpha &\approx \frac{r}{r_1} = \frac{d}{|HL|} = \frac{t_{ELH}}{t_{HL}}
\\  \text{where}\quad
    t_{HL} &= \frac{|HL|}{c}
    \approx 10.0\cdot 10^{-3}\,\mathrm{s}
    \qquad\text{(time directly from $H$ to $L$)}
\end{align}$$
In 3D this means that $E$ lies on one of the two sheets of a hyperboloid
of revolution whose axis of revolution is the line $HL$.
In particular, in $L$'s view of the sky, partially hidden under the horizon,
there should be a fairly thin circular band that covers all projected
candidate points for $E$.

The solid angle of view enclosed by the circular band is
$$  A_{\text{ste}} = 2\pi(1 - \cos\alpha)
\approx 2\pi\left(1 - \frac{t_{ELH}}{t_{HL}}\right)$$
Uncertainties in $t_{ELH}$ yield uncertainties in $\cos\alpha$
which widen the circular band.
Thus the circle becomes an annulus.
The solid angle of view covered by that annulus measures
$$\begin{align}
    A_{\text{ste.max}} - A_{\text{ste.min}}
    = 2\pi(\cos\alpha_{\text{min}} &{}- \cos\alpha_{\text{max}})
    \approx 2\pi\frac{t_{ELH\text{.max}} - t_{ELH\text{.min}}}{t_{HL}}
\\  \text{briefly:}\quad
    \Delta A_{\text{ste}} &\approx 2\pi\frac{\Delta t_{ELH}}{t_{HL}}
\end{align}$$
Substituting reported data for $t_{ELH}$, with $\text{min}$ and $\text{max}$
referring to 90% probability intervals, we have:
$$\begin{array}{rcrrr}
    \text{entity} & \text{unit} & \text{nom} & \text{min} & \text{max} & \Delta
\\\hline
    t_{ELH} & 10^{-3}\,\mathrm{s}   & 6.9   & 6.5   & 7.4   & 0.9
\\  \cos\alpha & 1                  & 0.69  & 0.65  & 0.74  & 0.09
\\  90^\circ - \alpha & {}^\circ    & 44    & 41    & 48    & 7
\end{array}$$
And thus the annulus, when projected to a unit (celestial) sphere, covers
$$\Delta A_{\text{ste}} \approx 0.18\pi \approx (1.9\cdot10^3)^{\circ\circ}$$
(The ${}^{\circ\circ}$ means
square degrees.)
In contrast, the reported value of the credible region varies between
$140^{\circ\circ}$ (50% probability) and $590^{\circ\circ}$ (90% probablity).
This is because large parts of the annulus could be ruled out based on
additional information such as signal strength ratios.
But those refinements are not the topic of this thread.

For easier visualization, let us move to a point $B$ on Earth's surface
where the center of the annulus is in the zenith,
and the annulus itself is at constant altitude $90^\circ - \alpha$,
as given in the above table.
To that end, let $O$ be the location of Earth's center and require
that the 3D ray $OB$ points in the same direction as the ray $HL$.
This determines the latitude and longitude of $B$.
Note that the parallel shift from $HL$ to $OB$ incurs a parallax,
but that is negligible because the distance to $E$ is so large.

Idealizing our planet to a sphere and doing the calculations
(transforming the locations of $H$ and $L$ from spherical coordinates to
cartesian coordinates, computing the difference vector, normalizing it,
and transforming back to spherical coordinates), I find $B$ to be at
$$B\colon\ 27^\circ 21'54.22''\mathrm{S}\ 38^\circ 36'33.45''\mathrm{W}
= 27.365061^\circ\mathrm{S}\ 38.609293^\circ\mathrm{W}$$
which is some 700km southeast of Rio de Janeiro in the Atlantic.
If you had been there at 2015-09-14 09:50:45 UTC
and had spun around yourself, drawing a band at constant altitude
$90^\circ - \alpha \approx 44^\circ$ and width $7^\circ$ in the sky,
that would have marked all the directions
where the signal source $E$ could have been.

To get the sky view simulated, I have written a tiny script gw150914.sts for
Nightshade:

# Display the sky at the time of the aLIGO GW150914 observation
# from an Earth-based position chosen such that the set of possible sky
# positions of the source form a circular annulus with azimuth-independent
# altitude = asin(delta_t/10ms), about 44 deg.
clear
timerate rate 0
date utc 2015-09-14T09:50:45
flag show_tui_datetime on
set home_planet "Earth"
moveto lat -27.365061 lon -38.609293 alt 0 heading 0 duration 2
wait duration 2
flag azimuthal_grid on
zoom fov 120 duration 2

This gives the view below. I have added dashed green circles to delimit
the annulus.

But do not expect to find likely candidate spots within that annulus.
The annulus hides the radial depth of the projected space volume.
A range for that radial depth is given by the estimated bounds
for the distance $R=|LE|$. The corresponding flat space volume is
$$\Delta V = \frac{1}{3}\Delta A_{\text{ste}}
\left(R_{\text{max}}^3 - R_{\text{min}}^3\right)$$
Plugging in the reported values, again from 90% probability intervals,
$$\begin{align}
\Delta A_{\text{ste}} &\approx 590^{\circ\circ} \approx 0.0572\pi
&   R_{\text{min}} &\approx 0.23\cdot10^{9}\,\mathrm{pc}
&   R_{\text{max}} &\approx 0.57\cdot10^{9}\,\mathrm{pc}
\end{align}$$
yields
$$\Delta V \approx 10^{25}\,\mathrm{pc}^3$$
To get a very rough estimate of the number of galaxies in that volume,
let us multiply $\Delta V$ with a rough galaxy density estimate of $10^{-20}\,\mathrm{pc}^{-3}$,
which results in about $10^5$ galaxies in that volume.
My conclusion from this back-of-the-envelope calculation is
that we cannot hope to localize $E$ further, unless given more information.

Things would have been a bit different if we had had a third aLIGO observatory
distant from the line $HL$ operating and capable of detecting the signal
from $E$.
Then we could draw three annuli, take intersections and thus
determine the direction of the signal within one reasonably small spot
on the celestial sphere.

Further reading

As this LIGO post tells
us, parts of the annulus could be ruled out by considering additional features
of the data such as varying signal strength.
A more detailed description of the probability distributions for the results
is given in this LIGO paper.
More information about the methods used to further narrow the credible
sky regions of the source location is given
in this paper.

Addendum

An equivalent script gw150914.ssc
for Stellarium:

// Display the sky at the time of the aLIGO GW150914 observation
// from an Earth-based position chosen such that the set of possible sky
// positions of the source form a circular annulus with azimuth-independent
// altitude = asin(delta_t/10ms), about 44 deg.

var obs_lat = -27.365061;
var obs_long = -38.609293;
var obs_alt = 0;
var obs_locname = "Atlantic, 700km SE from Rio de Janeiro";
var obs_planet = "Earth";
var obs_date = "2015-09-14T09:50:45";
var duration = 2;

core.clear("deepspace");
core.setTimeRate(0);
core.setDate(obs_date, "utc");
core.setObserverLocation(obs_long, obs_lat, obs_alt, duration,
                         obs_locname, obs_planet);
core.wait(duration);
core.moveToAltAzi("90", "0", duration);
core.wait(duration);
GridLinesMgr.setFlagAzimuthalGrid(true);
StelMovementMgr.zoomTo(120, duration);

And yes, core.setObserverLocation expects longitude before latitude.


Does wavelength affect redshift caused by the metric expansion of space?

Standard cosmological models predicts that the cosmological redshift and the speed of light are wavelength-independent. This result is confirmed observationally e.g. by Ferreras & Trujillo (2016), who used $\sim500\,000$ SDSS galaxy spectra down to a precision of $\Delta z \sim 10^{-6}$ and $\Delta z \sim 10^{-5}$ for galaxies at $z<0.1$ and $z>0.1$, respectively.


Do far away planets reflect light in the same way our moon/earth reflects light?

The short answer is yes, the outer planets of the Solar System reflect light in much the same way that the Earth does, but exoplanets are too far away for us to see clearly.

If you looked at Jupiter, you would be able to see the same colors that show up in Juno's photographs.  Even though the sun is less bright out there, it is still bright enough to illuminate the surface.  At Earth, the sun's irradiance (its intensity) is ~1361 W/m$^2$.  At Jupiter, the solar irradiance is ~50.26 W/m$^2$.  This is about the intensity of a common lightbulb.  Going further out, the solar irradiance at Pluto is ~0.873 W/m$^2$.

Now, this is not a very high intensity.  When you move away from the sun, things get darker and darker.  However, we have special instruments to counteract that.  We use telescopes to gather up large amounts of light and focus them, so that we are able to get a clear picture.  That is how telescopes work.  We have pictures of Jupiter from the Earth, most notably from the Hubble Space Telescope, and you can see that they are, indeed in color. (The black dot in the middle of the Great Red Spot is the shadow of Ganymede, one of Jupiter's moons).

There are plenty of resources online to learn how telescopes work.

As for planets outside our Solar System, we can't actually get pictures of them.  Any images you've seen of exoplanets are actually drawings made by artists.  They're so far away that we just don't have the resolving power to see them clearly.


Why is constant velocity drift not an accepted explanation of Hubble expansion?

The simplest answer is that Milne's explanation requires a non-uniform cosmos, with a formerly crowded "starting area" that expands into empty space.  Observations indicate that the cosmos is nearly uniform at very large scales.  With Milne's model of a non-expanding space and uniform cosmos, galaxies that started near to us with high velocities that ended up far away would be replaced by galaxies that started far away and ended up near, so there would be no observable relationship between distance and velocity.

There is a slew of other observational evidence against the naivete' of Milne's model, of course.  The cosmic microwave background shows that space use to be dense and hot; galaxies evolve, interact, and combine over time; there are intergalactic clouds that influence galaxies' trajectories.  Only when you can account for all that does incorporating elements like dark energy into your model make sense.


What I mean is, even if it is capable of expanding, if everything originated at the big bang how can that space turn to be infinite in a finite time - the age of the universe.

In the standard ΛCDM model of the Big Bang, the universe is infinite and has always been such. The Big Bang singularity happened everywhere, in the sense that far back enough in time, the density diverges to infinity at every place.

But this is just a particular model--it assumes that the universe if spatially flat and is globally homogeneous and isotropic. There are extended models in which it is not exactly flat, and so could be finite even if it is still homogeneous and isotropic (if the curvature is even slightly positive). And of course we don't actually know whether it is homogeneous and isotropic at scales much larger than we actually see. Some inflationary models imply that it isn't.

To clarify: the ΛCDM model uses the assumes a spatially flat FLRW solution of general relativity, in which space is the Euclidean $3$-space The Euclidean $3$-space is the only flat homogeneous and isotropic $3$-manifold, so there no way to make it finite without violating at least one of those modeling assumption (e.g., a flat torus could have the same form for the metric, but would not be globally isotropic).


Since this appears to be a screenshot of Eyes on Exoplanets, I searched for "exoplanet 485" and found Kepler-485 b.
With the colors exaggerated, the label area could be "Kepler-485" in yellow overwritten with "Sun" in orange.
If this is correct, it's not a cluster, just a star with a planet, plus some artifacts from the instrument which took the image.


Is cosmic shear generally agreed to have been observed?

Yes it has been observed. There are numerous papers covering this concept. Before I get into weak-lensing (because that is what you reference in your question) I'll note that cosmic shear is simply a specific type of distortion of an object by a lensing event. It is not limited to the weak lensing regime and cases of strong lensing exhibit cosmic shear as well. So images like this one show cosmic shear.

As far as cosmic shear by weak lensing goes, the first published and recorded case of a weak lensing measurment was in 1996 and published by Brainerd, Blandford, & Smail (1996). The cosmic shear you speak of is referred to as polarization in that paper, but they're essentially trying to measure cosmic shear. Note also, while this is weak lensing, most of that lensing occurs by one galaxy lensing another. As such it is also termed galaxy-galaxy lensing.

Since the first detection, there's been steady, but limited work on galaxy-galaxy lensing and shear measurments by papers such as Mandelbaum et al. (2006), van Uitert et al. (2011), and Velander et al. (2013). I think the Velander paper is the latest and greatest measurements you can get and they have a lot of good material in there. Because you requested an image, I'll pull out one from this paper.

  Caption: Galaxy-galaxy lensing signal around red lenses which have been split into luminosity bins according to Table 1, and modeled using the halo model described in Section 3.2. The black dots are the measured differential surface density, $\Delta \Sigma$ , and the black line is the best-fit halo model with the separate components displayed using the same convention as  in Figure 3. Grey triangles represent negative points that are included  unaltered in the model fitting procedure, but that have here been moved up to positive value s as a reference. The dotted error bars are the unaltered error bars belonging to the negative points. The grey squares represent distance bins containing no objects.

Note that in this paper, they're ultimately trying to measure the tangential component of the cosmic shear (you can't measure the radial component), denoted by $\gamma_t$ and by equation (1) in their paper, they relate this to the surface mass density $\Delta \Sigma$. So in their plot above, when they say they're measuring $\Delta \Sigma$, they're really measuring $\gamma_t$ and converting through a multiplicative constant.

The limiting factor in progressing further in this field is the lack of detailed observations required to do a true cosmic shear measurement. Since this measurement can only be observed statistically and not directly, you need observations of the sky which are both deep and wide. It is hoped that the WFIRST satellite will provide these measurements and thereby expand ten-fold the weak lensing work astronomers can do which in turn may some day allow us to make important statements about the content, distribution, and properties of Dark Matter.


What will happen to life on Earth when the Andromeda and Milky Way galaxies collide?

What will happen to life on earth or human beings on earth?

Assuming that human beings, or life, still exists on Earth at that time, they will have survived so much due to the ongoing death of the sun, that the gravitational pertubations due to the galactic collision will be nothing.

Keep in mind that in about 1-2 billion years, the sun will be so hot and large that all the water will have boiled off the earth into space.  About 3 billion years from now, the surface of the Earth will be so hot that metals will be melting.

Any life that has survived those events and still lives on Earth will surely take a galactic collision in stride.

I imagine, though, that most humans will have fled Earth - if not for distant star systems, then at least for planets in our own system that are going to be warming up enough for human habitation.

  If we are about to collide in next 4 billion year then how long before we should take action for inter-stellar voyage?

As soon as possible. When interstellar voyage becomes possible, we should start sending out ships to colonize other planets and star systems.  This will likely take a long time, but if we are to survive more than a billion years, it is necessary. Keep in mind that the Sun and Andromeda are events we can predict.  We don't know, and can't predict, the next cataclysmic asteroid strike, which is likely to happen in a shorter time than a billion years.  There are lots of reasons to exit the planet, we should be worried about the ones we can't predict or see, not the ones we can predict.

  Are scientists working on such projects for inter-stellar voyage?

Yes, but in small steps.  Manned missions to space, to the moon, and living aboard the ISS have provided significantly valuable information that will be used in such interstellar missions.  As we continue to push the boundaries of our ability to survive in space we eventually will be able to live in space, perhaps whole lifetimes will be spent in space.  As engine technology progresses beyond simply lifting people out of Earth's gravitational well, we will eventually be sending people on long voyages outside our solar system.

It's a very, very long way off, but each advance takes us closer to that eventual goal.


The first accurate determination of the rotation period and the first hints that the rotation period was different from the orbital period (as originally expected) was in 1965 from 430 MHz radar echo measurements from the now-collapsed Arecibo Observatory (RIP) in Puerto Rico. These were reported in a very short paper (1/2 page) in Nature by Pettengill and Dyce 1965. This was followed by a slightly longer paper by Peale and Gold (Peale and Gold, 1965, Nature) giving the theoretical context which says (after discussing tidal friction expected from the Sun on Mercury):

For a planet on a circular orbit the final condition would then be one of synchronous rotation like the motion of the Moon with respect to the Earth. Mercury's motion around the Sun takes 88 days and for synchronous rotation the sidereal period would thus be 88 days also. The observed value of 59 ± 5 days differs markedly from this (see preceding communication).

(the preceding communication is the radar measurement of Pettengill and Dyce)
As the OP says, the later November 1965 Nature paper by Giuseppe Colombo seems to be the first to raise the possibility of the rotation period being 2/3rds of the orbital period and therefore it being in a 3:2 spin-orbit resonance:

IN a recent communication by S. J. Peale and T. Gold the rotational period of Mercury, determined from radar Doppler-spread measurements to be 59 +/- 5 days ....snip...
A very nearly uniform rotational motion of 58.65 sidereal-day period, that is 2/3 of the orbital period, may indeed be a stable periodic solution.

This is borne out by the 1967 paper by Dyce, Pettengill & Shapiro 1967AJ..72.351 (free to view) which discusses the determination of the rotation periods of both Venus and Mercury via radar and discusses what lead to the misleading earlier optical results for the rotation period of 88 days. It also states (page 358):

In fact Colombo (1965) pointed out the possibility that Mercury's rotation period might be 58.65 days i.e. exactly two-thirds of its rotation period. This value of the rotation period would apparently make Mercury unique in the solar system in having its axial motion locked to its orbital motion in such a manner.


How useful are filters for spotting nebulae?

With an 8" scope, a filter will very likely give you better results than observing without a filter. Although a filter does block light, the crucial aspect is that a filter increases contrast (by blocking light pollution and extraneous wavelengths of light more than the nebula), thereby allowing you to spot low contrast diffuse nebulae (like IC59 and IC1318) much more easily. This is in fact more critical for visual observation than for photography because it is possible to increase contract in post-production with photography. You will find that visual astronomers go to great lengths to increase contrast--baffling, flocking, premium mirrors, etc.

Light pollution filters are broadband, meaning they allow all light except light emitted by streetlights. From darker skies (like bortle 4.5), this will not give dramatic results. I would recommend using an OIII filter to start with. The OIII filter is a narrowband filter, meaning it cuts all light except at a very narrow wavelength range from ionized oxygen. Very often OIII filters make the difference between being able to see an object and not being able to see it, even from very dark skies. As an excellent test of the OIII filter, check out the Veil Nebula too; from your skies, this object would be rather difficult without an OIII but rather stunning with an OIII filter.


Do radio telescopes see other stars better at night?

The Sun doesn't substantially impact radio observations during the day, because radio telescopes operate at long wavelengths. In general, light at longer wavelengths scatters less than light at shorter wavelengths, and so visible light from the Sun scatters much more than radio waves from the Sun.$^{\dagger}$ The former effectively fills the daytime sky, while the latter does not, leaving the radio sky dark even during the day. Couple this with the fact that the Sun - though bright in radio waves - is not extremely bright compared to other radio sources,$^{\ddagger}$ and you find that daytime radio observing isn't significantly different than nighttime observing.
Strictly speaking, you could argue that there may be indirect effects from the Sun. Water vapor is an important source of opacity, causing several extremely opaque bands, and atmospheric emission can add noise to the system temperature of your instrument, causing complications to observations. Humid summer days, for example, can be problematic for observing astronomical water vapor emission. As the Sun does impact weather and the climate, there are minute differences between day and night, but I don't believe those are significant when compared to how much the Sun affects optical observers.
As a final note: radio telescopes don't commonly observe stars, which aren't usually strong radio sources. There are occasionally highly-publicized stellar bursts or flares - BLC1 is an example, assuming it's natural - and star-exoplanet interactions can lead to radio emission, as in the case GJ 1151. But radio astronomers rarely look for radio emissions from stars, in comparison to other objects like radio galaxies, pulsars, and clouds of neutral hydrogen.

$^{\dagger}$The same is true for light traveling through interstellar space, which is why infrared and radio emission can travel through dust while visible and ultraviolet light can't; you need wavelengths much larger than the sizes of dust grains for minimal extinction.
$^{\ddagger}$A commonly quoted value is that around $\sim1\text{ GHz}$, the Sun has a flux density of $S_{\nu}\sim10^6\text{ Jy}$ when it's quiet, with an increase of 1-2 orders of magnitude during active periods. That's only a couple of orders of magnitude brighter than the next brightest sources in the sky, and the difference shrinks substantially in the $\sim100\text{ MHz}$ regime.


Does Hawking radiation in fact bring mass into the universe?

The black hole loses mass in order to generate the Hawking radiation. The energy carried by the radiation is exactly proportional to the mass lost by the black hole according to Einstein's formula E = mc^2

No new mass is brought into the Universe this way.

EDIT:

(Keep in mind the explanation below is not rigorous. As Stephen Hawking himself has said, this is more like a mental model, good enough to visualize things.)

Virtual particles are called virtual because they don't really "exist" in the common sense. They have mass and they have energy. Due to Heisenberg's uncertainty principle, it's okay if particles with a total energy E almost "exist" for a very short while, as long as they return to nothingness afterwards.

The production / destruction of virtual particles could happen all the time, anywhere, not just at the edge of a black hole. This is fine, as long as each particle only seems to "exist" a short time. Conservation of energy is not violated.

However, at the edge of a black hole, something different happens. You have a pair of virtual particles popping up into existence. Soon after, one of the particles is captured by the black hole, while the other particle moves away from it. Since the captured particle is virtual, it doesn't really have mass and/or energy. No new mass is added to the black hole.

Now the two particles cannot re-annihilate each other. So you have one lone particle moving away from the black hole. It's not a virtual particle anymore, it's a real particle. But it has energy and perhaps it has mass also. Where does the energy come from?

The energy required to make the new particle real comes from the gravitational field of the black hole. The black hole is losing mass in order to power the creation of real particles moving away from it.

Again, don't get stuck in details. This is not what really happens. This is more like a metaphor.


Why does fusing iron in a stellar core use more energy than it releases

An atomic nucleus contains protons and neutrons. There is a balance of forces: the (residual) strong nuclear force attracts protons and neutrons together. The electro-magnetic force pushes them apart.

When nuclei fuse, energy is required to push them together, as you have to push against the electromagnetic repulsion. But when the nuclei come close enough, the nuclei are attracted by the strong force, and energy is released as the nuclei fall into each other. For small nuclei, the energy released in fusion is greater than the energy needed to overcome electromagnetic repulsion. For large nuclei (which have a larger positive charge) the reverse is true, more energy is required to push the nuclei together than is released.

Iron is at the balance point, at which the energy required to overcome electomagnetic repulsion just exceeds what is given out on fusion. So iron is the smallest nuclei to absorb energy when it engages in fusion. And so when substantial amounts of iron form in the core of a star, the core will no longer be able to heat up to maintain the balance with gravity, the core collapses and the star explodes. See Why is iron responsible for causing a supernova?

Iron is therefore one of the most stable nuclei (not the most stable: that is Nickle-62, but for various reasons this is not likely for form inside stars) Iron acts like the "ash" of nuclear fusion.

A more detailed answer would require understanding the quantum mechanics of how binding energy in nuclei works, which can explain why it is Iron that has close to maximum binding energy.