How close would merging black holes have to be to feel gravitational waves?

It's a bit hard to answer this question as there are some ambiguities and certain parameters of the setup aren't defined, but I can try to give you a general idea.

  If we were to feel such gravitational waves, what would they feel like?

I'll start here as we need to have some well established meaning for how we can feel a wave before calculating what conditions produce such a wave. However, this is the ambiguous part. A gravitational wave passing over us would effectively just be felt as a differential gravitational field, which can induce spaghettificiation as Andy points out. One side of your body might feel a stronger pull than the other being that one side is experiencing slightly different gravity than the other. But I have to say that you'd need extreme waves to have this effect. The frequency of the waves would have to be extremely high for the wavelength to be on the order of the human body's size. If the wavelengths are much longer than your body, the change, even from a powerful gravitational wave, would probably be too slow for you to detect.

I'll note that the frequency of the waves detected by LIGO were about $250\:Hz$ for the first detection (at the peak) and $450\:Hz$ for the second detection (again at the peak). That implies a wavelength of $\sim600\:km$ at the shortest. At these extreme wavelengths, the gravity differential is negligible. You'd need wavelengths on the order of a meter which implies frequencies of $10^9-10^{10}\:Hz$ which is right near the peak of astrophysical possibilities. So already your proposal of Earth-sized black holes is out. There's no way they can produce these extreme waves.

But now we run into another problem. In order to make the wavelengths on the order of the human body such that you can actually feel the differential, you've made the frequency extremely high. Could your body even notice gravitational changes happening in $10^{-10}\:sec$? I doubt it. That'd be like me blasting you with hot air, then cold, but flipping back and forth between the two at a rate of 10 billion times a second. I have a feeling everything would just cancel out and you wouldn't really feel anything, but it's hard to say.

  How close would a black hole collision have to be for us to feel the gravitational waves pass through Earth?

Let's ignore the issue of whether you'd ever feel a wave and look at the main question. I'll just consider raw energy in the wave and hope that somehow equates to feeling it. The power output of two orbiting bodies of mass $M$ (assuming the same mass for simplicity), and orbiting at a radius of $r$ is given by

$$P_{grav} = \frac{32}{5}\frac{G^4}{c^5}\frac{2M^5}{r^5}$$

Note that $G=6.67\times10^{-11}\:m^3kg^{-1}s^{-2}$ is the gravitational constant and $c=3\times10^8\:m/s$ is the speed of light. Now $P_{grav}$ is the total power radiated out in energy per second in the form of gravitational waves. You of course would not be experiencing this entire power though because this is radiated in all directions. The power is not radiated spherically symmetrically, but for simplicity's sake, let's just assume it is. That means we can say the power you experience given your area $A$ and distance $d$ from the system is given by

$$P_{exp} = P_{grav}\frac{A}{4\pi d^2} = \frac{32}{5}\frac{G^4}{c^5}\frac{2M^5}{r^5}\frac{A}{4\pi d^2}$$

Play with some numbers for this and see what is necessary to have an appreciable power.

I will point out though that at the distance you'd have to be where this power became appreciable, it would be overwhelmed by the tidal forces involved.

So in short, no I don't think you'll ever be able to feel gravitational waves, no matter the circumstance.


strange moonset around 10 pm (3.2 2014, NZ) - explanation?

February 3, 2014 has been short (three days) after new moon. The moon is then about 40 degrees or about 3 hours behind the sun. Hence, what you observed is quite normal, in Europe, too.

The only difference is, that the moon looks to the other side when seen from the southern hemisphere. Here, for 2014, a moon phase calendar of the northern hemisphere.


What's that very bright star (south-oriented, appearing at 5 p.m.)?

That star is probably the planet Venus.

It is always close to the Sun, and Stellarium shows it to be in the south and well above the horizon at around 5pm local time.

To check, right above it should be Mars (also a planet).

In general, if you see a very bright star, it is very likely that it is a planet, and most likely to be Venus, Mars, or Jupiter. The latter two can appear at any time during the night, but Venus is always close to the Sun, so it can be seen just after sunset or right before sunrise. All planets appear close to the track of the Sun across the sky.


Spacing (2422 days) between solar eclipses - unrelated saros numbers

Since a solar eclipse is a coincidence of a new moon and
an ascending or descending lunar node,
the interval between solar eclipses must be a common multiple of
a whole synodic month
(or lunation, 29.53 days between same lunar phases)
and half a draconic month
(27.21 days between same lunar nodes).
These line up around eclipse seasons,
173.3 days apart on average.

The saros cycle of 18.03 years is 223 synodic months, 242 draconic months,
or 239 anomalistic months
(27.55 days between lunar perigees or apogees).
This multiple coincidence of phase, node, anomaly, and time of year makes the saros especially useful for prediction.
However, there are other relationships between eclipses.

Between consecutive saros series there is another interval called
the inex:
358 synodic months, 388.5 draconic months, or about 28.94 years.
George van den Bergh showed in 1955 that the interval T between any two eclipses can be expressed as T = aI + bS, where I and S are the inex and saros periods, and a and b are integers.
Some such combinations have names:

                   Synodic  Draconic  Anomalistic  Eclipse
Name      Formula  Months    Months     Months     Seasons

Semester  5I - 8S      6      6.511       6.43       1.02
Hepton    5S - 3I     41     44.493      43.94       6.99
Octon     2I - 3S     47     51.004      50.37       8.01
Tritos     I - S     135    146.501     144.68      23.00
Saros      S         223    241.999     238.99      38.00
Inex       I         358    388.500     383.67      61.00

Source: F. Verleben

Between the eclipse pairs in question, 2422 days = 82 lunations = 10S - 6I or 2 heptons.
The two pairs also happen to be 3 heptons apart.
The same hepton family includes southern hemisphere eclipses in Dec 2020, Nov 2030, and Jul 2037.
Tabulated in a
saros-inex panorama,
all seven eclipses would fit on a single diagonal line of slope 5/3.

            Greatest eclipse  Sun      Moon
   Date      UT     Lon Lat   Decl  EcLat Node  Saros  Inex

2017-08-21  18:26   88W 37N   +12   +0.43 asc    145    39
2020-12-14  16:13   68W 40S   -23   -0.29 desc   142    44
2024-04-08  18:17  104W 25N    +7   +0.35 asc    139    49
2027-08-02  10:06   33E 25N   +18   +0.14 desc   136    54
2030-11-25  06:50   71E 44S   -21   -0.39 asc    133    59
2034-03-20  10:17   22E 16N     0   +0.29 desc   130    64
2037-07-13  02:39  139E 25S   +22   -0.71 asc    127    69

Sources: F. Espenak,
JPL HORIZONS

From a Moon-centric point of view, the umbra of a solar eclipse moves straight along the ecliptic, but several factors shape its path on the Earth's surface.
The terrestrial longitude of a path is directly related to the time of day (UT).
A path's terrestrial latitude is very sensitive to the Moon's
ecliptic latitude
and is also affected by the Sun's declination.
Due to the tilt of the Earth's axis, an eclipse path
slants southwest-northeast near the March equinox,
bends southward near the June solstice,
slants northwest-southeast near the September equinox,
or bends northward near the December solstice.
The Moon's motion in ecliptic latitude also contributes a slight slant,
northward at ascending node or southward at descending node,
as the umbra passes west to east.

Both the 2017-2024 and 2027-2034 eclipse pairs occur near opposite equinoxes and at similar times of day, so their paths naturally have opposite slants and similar ranges of longitude.
The 2030 and 2037 eclipse paths have similar ranges of latitude but barely cross because they occur 4 hours apart in the day.
Eclipses in the same saros series, if not widely separated in longitude, occur at the same time of year and have nearly parallel paths with few opportunities to cross.


Is it safe to view moon at daytime through telescope

As long as you don't directly look at the sun through your telescope, then there is no problem. But don't try to see the moon during solar eclipse without special filters (easily available for your telescope's eyepieces)

In  my telescope, I have seen the moon day time when there was very light cloud cover. It was awesome scene. You can feel that the moon is spherical by observing its terminator and the lighted edge. You can easily make out the dark side edge.


What exactly is interplanetary scintillation; what was the Interplanetary Scintillation Array looking for? Did it successfully observe any?

To fully answer your questions, let me introduce scintillation before interplanetary scintillation.
Atmospheric scintillation
The imaging of an astronomical source is affected by a collection of effects that goes under the name of astronomical seeing, the main ones being smearing, motion and scintillation of the image. All these effects are caused by the deformation of the light wavefront due to random inhomogeneities in the refractive index of the atmosphere.
Let's visualize the basic idea behind the seeing. Suppose to have a light source located at infinite distance, such that it's ideally a point source and its light reaches Earth in plane waves. When entering the atmosphere these waves face changes in the refractive index, and in geometrical optics approximation the wavefronts are deformed according to Snell's law. In the following image you have two simplified cases: on the left, it is shown that a vertical gradient in the refractive index produces a simple tilt of the wavefront; on the right, it is shown that a horizontal gradient produces a deformation of the wavefront. Here $\textrm{WF}_1$ is the incoming unperturbed wavefront, $\textrm{WF}_2$ is the same wavefront after entering the atmosphere and $n_i$ are the refractive indices.
$\hskip2in$
Let's now approach the real case. It is known that variations in the refractive index are related to variations in density (e.g. via Gladstone-Dale relation). Since our atmosphere has an overall more-or-less stationary distribution of temperatures, densities, and pressures, the most dramatic variations of the refractive index are found only in the turbulent layers. Here each turbulent eddy can have different density, temperature and so on, causing local variations in the refractive index. These layers are located in the first km of atmosphere (the planetary boundary layer) and at $\sim$10 km (near the tropopause). In the following image, it's depicted how a plane wavefront is deformed after crossing a turbulent layer and all of its eddies with different $n_i$. The lenght $r_0$ is the Fried parameter, which can be interpreted as the characteristic lenght of the turbulent eddies.
$\hskip2in$
What is atmospheric scintillation then? Well, as you see from the previous image, the curvature of the wavefront leads to a convergence (or a divergence) of the light rays (i.e. the lines perpendicular to the wavefront). This means that when you take an image by collecting photons with your sensor (i.e. when you collect a portion of the wavefront) the image brightness can rise or decrease depending on whether light rays are converging or diverging. Furthermore, this brightness variation is time-dependent, since the eddies evolve with time and the turbulent layer has an horizontal drift velocity. This is scintillation.
To see scintillation you need that (i) the source has small angular dimension, (ii) the turbulent layer is far from the telescope, (iii) the characteristic length of the collected wavefront's portion (e.g. the diameter of the telescope) is comparable to $r_0$, (iv) the exposure time is less than the lifetime of the deformation. If (i) and (ii) are not satisfied you might see the motion of the source instead of scintillation, while if (iii) and (iv) are not satisfied you will see speckles or a smeared image.
Scintillation in the radio domain
In the radio domain, scintillation is not limited to atmospheric scintillation.
For wavelengths between millimeters and centimeters we still have atmospheric effects, mainly due to the water vapor vertical gradient near the ground. Indeed, for this wavelengths the refractive index of air deviates from unity less than few houndreds of ppm. Since the turbulent layers are closer to the ground and usually the radiotelescope apertures are bigger than the characteristic length of the turbulence, seeing is not dominated by scintillation.
For wavelengths beyonds centimeters, the frequency of the waves gets closer to the plasma frequency in the ionosphere, hence the wavefront deformation provoked by the passage through the electron clouds in the ionosphere becomes relevant. Since their great distance from the ground, this will result in scintillation.
The timescale of ionospheric scintillation is between minutes and tens of minutes, but Hewish (1955) started to notice that there was another scintillation with a timescale of few seconds, and that its intensity was greater for sources near the Sun.  Following this hint Hewish et al. (1964) have shown that in fact this faster scintillation was produced by plasma clouds in the interplanetary medium, delivered by the solar wind.
Going further, Sieber (1982) have shown that there's also a scintillation with timescales between days and months due to plasma clouds in the interstellar medium.
Conclusions

What exactly is interplanetary scintillation?

Scintillation is a time-depending change in the intensity of a light signal, and it is caused by deformations in the wavefront due to random variations of the refractive index. In the case of interplanetary scintillation, the refractive index variations are caused by plasma clouds traveling in the interplanetary medium and delivered by the solar wind.

What was the Interplanetary Scintillation Array looking for?

Precisely interplanetary scintillation. From it you can study the properties of both the interplanetary plasma and the scintillating radio sources.

Did it successfully observe any?

I guess yes!


Why is the ring of light around the M87 black hole bigger than the photon sphere?

The apparent radius of something residing in a Schwarzschild metric, when viewed from infinity is given by
$$ R_{\rm obs} = R \left(1 - \frac{R_s}{R}\right)^{-1/2}\ ,$$
where $R_s$ is the Schwarzschild radius $2GM/c^2$.

This enlargement is due to gravitational lensing and the formula is correct down to the "photon sphere" at $R =1.5 R_s$.

Most of the light in the EHT image comes from the photon sphere. It is therefore observed to come from a radius
$$ R_{\rm obs} =\frac{3R_s}{2}\left(1 - \frac{2}{3}\right)^{-1/2} = \frac{\sqrt{27}}{2}R_s\ .$$

There are small (<10%) corrections to this for a spinning black hole governed by the Kerr metric.

A sketch which illustrates why gravitational lensing leads to an enlarged image can be found in Pela's answer to a related question.


Trajectory: What would happen if the sun disappeared one day?

On the first of January, the direction of the sun from Earth is towards Sagittarius, with an ecliptic longitude of 280.  The velocity of the Earth around the sun is perpendicular that that, and towards Virgo, in particular, it is close to the direction of Gamma Virginis (Porrima), with an ecliptic longitude of 280-90 = 190.  The Earth will move in a straight line in that direction.  (Conveniently Jan 1st is very close to perihelion, so elliptic corrections are needed even less than normal)
It won't actually get near to Gamma Virginis because that star is 38 light years away and at the Earth's speed of 30km/s it would take 400,000 years to get close and by then the star will have moved away from its current location.
It wouldn't come close to a gas giant (none are in a position in which they would come anywhere near the trajectory of the Earth, even without checking, they are all on the "wrong" side of the sun)
It would be very unlikely for the Earth to come close to a gas giant.  If it did then the gravity of the giant planet would change the direction of the Earth, it could go in almost any direction.
For another date, the equation is simple enough: find the ecliptic longitude of the sun on that date and subtract 90.  That gives the ecliptic longitude of the Earth's velocity.
You can now use a star map to find the direction that the Earth will travel in.  On January 1st the Earth will travel in a direction toward an ecliptic longitude of 190, but remain in the plane of the ecliptic.  On the map below, the ecliptic is the line through the middle of the map and you can find where 190 is by looking at the numbers at the top and bottom.
Because in the absence of the sun, there would be nothing to cause the Earth to orbit, it would travel in a straight line towards that point.


Relationship between Hubble paramater and absolute magnitude

Maybe. To calculate the Hubble constant (H=v/d), you need the know the velocity (v≈c*z￼￼) and distance of a distant object (or rather as many objects as possible). The velocity of these objects must be dominated by the expansion of the Universe. So stars in our local group that are still bound to us via gravity wont work. So as long as the stars you use are not really bright or even supernovae in distant galaxies there is no way to get the Hubble constant.
If your stars are far enough away, however, you still have get their distance. This you can do with the distance modulus, i.e., difference between absolute and apparent magnitude:

m-M = 5 log(d)-5

So assuming the stars are distant stars for which you also know their apparent magnitude (e.g. Cepheid variables), than yes you can calculate H.
If they are just random stars, than not.


Well, if you buy a car and then you go "this thing doesn't drive well, where do I find a better engine" - then folks are going to ask you questions such as: have you driven stick shift before, do you understand how the clutch operates, etc.

There are many reasons why a telescope does not provide a satisfactory image.

Perhaps the expectations were too high. An amateur telescope can provide quite amazing views, but it's never going to be equal to a large professional telescope in orbit around Earth.

Also in line with the expectations thing, all else being equal, a large telescope will always perform better than a small one. A 6" scope and a 24" scope will provide very different levels of performance. If you have the 6" scope, but you expect 24"-level performance, that's never going to happen.

Perhaps seeing (atmospheric turbulence) was bad during the time when you tested the instrument. For objects in the solar system, seeing is one of the major impediments - it reduces resolution and muddies up the image. It changes in time; sometimes it's good, other times it's bad.

Perhaps light pollution was the problem. DSOs (deep space objects, which have low brightness) will easily get masked by the regular light pollution in cities. It changes with the place; it's bad in cities, it's good far from cities.

Was the telescope collimated? Some telescopes, such as dobsonians (and most scopes with a newtonian OTA) require regular collimation. Many SCTs do too. Performing collimation on such a scope is like changing the oil on your car - you just have to do it periodically.

Was the telescope stored in a warm place (inside the house), then suddenly taken outside in the cold night? The temperature differential will be similar to bad seeing for up to one hour, until the scope and the air take the same temperature.

Was the scope properly focused? In many cases, finding perfect focus is a little tricky. A small turn of the knob could make a big difference.

Can you even reach focus? It's surprising how often people have weird setups where the eyepiece cannot even reach the focal plane of the primary mirror. In that case, no amount of twiddling the focuser will provide a good image.

Is the eyepiece bad? Sometimes mass-produced scopes come with eyepieces that are just not very good. It happens. Trying a known good eyepiece instead should provide proof.

Are the optics very dirty? A layer of dust on the primary will not affect performance in any way that you can perceive, so leave the dust alone. However, if there's a big fat oily smudge on the back side of the eyepiece, then of course performance will be degraded.

Are the optics seated properly in their mirror cells, etc? A primary mirror not sitting flat in its cell will mean the scope will never perform well; this could happen during transport with some cell designs.

Is the frame or the OTA intact? Has it been bumped or otherwise deformed?

Finally, yes, in some cases mass-produced scopes have bad optics. In that case there's nothing you can do except replace something - either the defective mirror or the whole instrument. Keep in mind that a custom mirror will cost more than your whole instrument, if it's high quality.

TLDR

You're not providing any details, so it's hard to say what's going on. Based on experience, I'd say the most likely candidates are:

If your scope is a dobsonian or an SCT, I would start with collimation. Learn how to collimate the scope, and do it regularly. It has a very, very large impact on observations of the Moon and the planets. Do this before you try anything else.

If you're observing DSOs (galaxies, nebulae), maybe light pollution is the reason. Load up the scope in your car and drive about 1 hour away from the city, and try again.

Learn how to focus the instrument. It could be trickier than it seems.

Finally, go to a local astronomy club meetup and try and use an instrument similar to yours. Maybe your expectations are not calibrated properly.


Is any consensus forming on the solution to the "Lithium Problem"?

No. There is no consensus.

The discrepancy between the predicted big bang nucleosynthetic abundance of Lithium 7 and the measured value can be summarised as follows.

If we take what we know about the the baryonic mass density of the universe and the Hubble constant, we get a self-consistent picture between the cosmic microwave background, observations of galaxy recession etc. and the estimated primordial abundances of Helium and Deuterium.

The problem arises because these same cosmological parameters predict a primordial lithium abundance of $3\times10^{-10}$, when expressed as a ratio to the hydrogen abundance.

On the other hand, measurements of the Li abundance present in the photospheres of the oldest stars (a.k.a. "halo stars") in our Galaxy suggest that the primordial abundance was about $1.2\times10^{-10}$.

The factor of 2-3 difference between these numbers is about 4-5 times the measurement precision. This is the so-called "Lithium problem".

The potential solutions are reviewed by Fields (2012). They fall into the following categories.

Astrophysical solutions - that we don't understand our measurements of the Li abundances because of an imperfect understanding of the atmospheres of low metallicity stars; or that we don't understand interior mixing mechanisms that mean at the photosphere, we see material that has been mixed upwards from the interior where the Li has been depleted in nuclear reactions.
Nuclear physics - maybe the details of the reaction rates and cross-sections in the big bang model are awry? There are still some sizeable uncertainties here which remain to be nailed down, but are seen as rather unlikely solutions.
Additions to the standard big bang model. This includes things like inhomogeneous nucleosynthesis in the early universe - i.e. that it was clumpy even at this early stage. Other possibilities include the possibility that the equilibrium reactions in big bang nucleosynthesis were upset by the decay of massive dark matter particles.

Thus there are lots of ideas to solve this problem, and other ideas which suggest it is not so much a problem, but that we can't do the measurements properly.

EDIT: Update 18/11/19 from the "Lithium in the Universe" conference in Frascati, Rome.

Having sat through some review talks on the problem, I can summarise progress as:

Explanation 2 (Nuclear Physics) is dead. All avenues appar to be explored and the likely remaining uncertainties are at the level of 10%.

However, the following possibilities were also discussed.

Perhaps the fundamental constants have changed with time resulting in somewhat different binding energy differences between different nuclei? Scherrer & Scherrer (2017) for example discuss a scenario where the mass difference between 2 He nuclei and a 8Be nucleus changes with time in such a way that lots of 8Be is produced during big bang nucleosynthesis, but decays back into He later on. This does not alter the primordial He abundance inferred from present-day observations, but the removal of He at early epochs results in a lower production of 7Li.
Inhomogeneous magnetic fields, contributing significant energy density in the early universe may have led to temperature fluctuations at the epoch of nucleosynthesis. Such fluctuations could reduce the predicted Li abundances a bit, but they also change the predicted deuterium abundance and this wouldn't fit with the now quite precise measurements of the primordial D abundance and the Planck value of $\Omega_b h^2$.
It has been hypothesised that magnetic fields can play a role in filtering how much primordial Li makes it into the structures that form galaxies. i.e. There is a sort of chemical differentiation caused when there is a gradient of the magnetic field strength at right angles to the magnetic field. This accelerates ions down the gradient and possibly away from forming structures. Since Li is easily ionised in the early universe (compared with H and He), this might lead to the gas that formed the first stars being Li-poor.

At the end of the meeting, participants were asked (for fun) to vote on what they thought to most likely solution was. #1 Was first, with #3 a distant second.


What was the length of year 1 million years back?

The Hubble expansion has no bearing whatsoever on the length of the year. This is because the whole Milky Way galaxy (and in fact most galaxies, if not all, and even local groups) has decoupled from the Hubble flow long ago. In fact, it could only form after it decoupled. Note that M31, our sister galaxy, is in fact falling onto the Milky Way rather than receding (as the Hubble flow would imply), demonstrating that the whole of the Local Group (of galaxies) is decoupled from the Hubble flow.

What happens is that any over-density expands at less than the Hubble rate and thereby grows. Galaxies (and larger structures) form from small relative over-densities that eventually grow large enough to withstand the overall expansion and instead collapse under their own gravity to form bound objects, such as galaxy clusters, galaxies, star clusters, and stars. This implies that the Hubble flow has no bearing on the inner dynamics of such systems.

Of course, the number of days in a year was higher in the past than today, but that is only because the Earth is spinning down (due to tidal friction with the Moon), so that the days become longer.

If anything has had an effect on the semi-major axis of the Earth orbit (and hence on its period), then that is gravitational interactions with the other planets. However, weak interactions (secular perturbations) can only alter the orbital eccentricity and leave the semi-major axis unaltered.

Finally, there is a tiny effect from the Sun loosing mass (to the Solar wind). The period of any orbiting bodies is proportional to $M_\odot^{-1/2}$.


Why aren't stars like Eternally Collapsing Objects?

We know that stellar radiation pressure balances the gravitational compressive forced of a star.

We do not know that. Degeneracy pressure, thermal pressure, and radiation pressure are what collectively balance gravitation in a star. Degeneracy pressure dominates over the other two in white dwarfs and neutron stars. Thermal pressure dominates in low mass stars such as the Sun. It's only in very massive stars where radiation pressure dominates.

  Shouldn't the process of shrinking be a quasistatic process making stars (Eternally Collapsing Objects)?

You are forgetting about fusion. A star is more or less in equilibrium thanks to a number of negative feedbacks. (Negative feedback in a control system is good. Positive feedback is very bad. The "screech!" sound you occasionally hear from microphones is positive feedback.)

Suppose the fusion rate increases slightly in a region near the center of a star. This increases the temperature and thereby the pressure, which in turn causes the core to expand, which in turn causes the density and temperature in the core to fall, which in turn decreases the fusion rate. The reverse applies for a slight decrease in the fusion rate. Similar negative feedbacks apply for changes in temperature, pressure, and density. A main sequence star is self regulating.

Things get a bit trickier when a star leaves the main sequence, but that is a different question.


It was predicted by some, before the images arrived, that Pluto's craters would be much shallower than perhaps on other rocky objects/satellites. The reason is the icy composition and the relatively low impact velocities in the outer solar system (Bray & Schenk 2015).

There are indeed reported quotes in recent days that there are no impact craters on Pluto (e.g. here or here), based on the highest resolution images.

However, I'm with you - the things you have marked look like very large impact craters, or the remains of them, and so what we are seeing is that there are no small ones.

The gist of what I have heard/read is that this indicates geological activity on Pluto, which is effectively removing the evidence of relatively small impacts and that the surface we see is mostly younger than 100 million years old.
Watch this space, a lot of people are just flapping their lips at the moment (including me).

Another possibility - raised by Wayfaring Stranger - could be that the expected number of impacts was (hugely) overestimated. This is discussed by Singer & Stern (2015). There are indeed uncertainties, and it is hoped that cratering on Pluto & Charon would help constrain the number densities and size distribution of small impactors. My impression is that the uncertainties could be at levels of up to factors of 10 - and this seems to be discussed quite carefully in Greenstreet et al. (2015) (behind a paywall, sorry), who also comment in the abstract that surface ages judged from cratering are " entirely dependent on the extrapolation to small sizes [of the impactor distribution]".

Further Edit: LDC3 suggests all the craters could be too small to be seen. The high resolution images that have been released could resolve craters of a diameter of a couple of km or more and cover a few percent of the surface. According to Singer & Stern (2015), the number of craters at this size and above on Pluto they expected in such an area is $\sim 1000$. So I think that either the planet has been resurfaced recently, the craters have been eroded in some other way or their estimates for the number of impacts/size of the impactors is wrong.


How do we estimate the dimensions and mass distribution of a galaxy's dark matter halo, beyond the effective limits of its visible matter?

A rotation curve cannot be measured beyond the point at which no visible matter can be seen.
Rotation velocity measurements can come from clouds of hydrogen gas, distant planetary nebulae or globular clusters and satellite galaxies. The latter are numerous and extend to about 400 kpc from the centre of the Milky Way. The exponential scale length for the Milky Way's luminosity is a few kpc, so indeed observing these objects does explore the Galactic potential, and hence the matter distribution, to many times the size of the visible matter disc.


Normalizing the supernova spectrum by photometry

I think you understood correctly.

Just to be clear. Your photometric data have enough images including bias, flat, and standard stars, and are reduced + calibrated with standard stars. And, your issue is that you can reduce spectrographs (with bias, flat, and arc lamp) but you don't have standard stars for the calibration. Hence, you want to calibrate your spectra with the photometric data. Is this correct?

The following is for more detailed example how to proceed this.

First, note that a photometric data point is the accumulation of photons within a bandpass. See the response curve shown here for example. In the example, B band has its effective wavelength 4353 A, FWHM 781 A, and with its own shape of the response function.

For simplicity, the following demonstration will assume that the response function of the B band is delta function, i.e., zero elsewhere except at 4353 A. We will extend this example to other response function later. Suppose that you have a spectrum containing 4353 A, and you want to calibrate it with the known B photometry. So you can find a ratio = flux(B; photometry) / flux(B; spectrum), and apply this ratio elsewhere in your spectrum.

Note that the spectrum and photometry must be from the same epoch ideally. Otherwise, use the nearest available epoch as an approximation.

If you have more than one photometric data points to calibrate the one spectrum, ideally the ratios elsewhere will be the same. Realistically, they are not. Hence, you average the ratios by somehow.

Next, if you would like to consider the dispersion of the response function, instead of using the flux at a point, you accumulate the fluxes by considering the shape of the response function.

For simplicity, let's say the B band response function is a Heaviside step function = 1 within [4353-781, 4353+781] A and = 0 elsewhere. So, you accumulate the fluxes weighted by the response function from your spectrum. Then find the ratio, and follow the rest of the process.


What makes some meteor showers continue for days, while the "Unicorn shower" can be shorter than one hour?

It sounds like this effect:

  ... in addition to the usual shower on August 12th, there might be an extra surge of meteors on August 11th caused by a filament of dust newly drifting across Earth's orbit.
  The filament, like all the rest of the dust in the Perseid cloud, comes from Comet Swift-Tuttle. The difference is, the filament is relatively young. It boiled off the comet during the Civil War, in 1862. Other dust in the cloud is older (perhaps thousands of years old), more dispersed, and responsible for the month-long shower that peaks on August 12th. The filament will eventually disperse, too, but for now it retains some of its original ribbon-shape.

The 2004 Perseid Meteor Shower NASA


How can we see stars if they are hundred years of light far from the Earth?

You are right, there is no way to know, what is the current state of these objects "now" - if you think on the Newtonian mechanics.

For example, the 1054 supernova, what created the Crabs Nebula, happened actually not in 1054, but around in 6000 B.C. It is because it is $\approx$ 7000 ly away from us.

In the Special Relativity, "now" is relative. You can't say, that the 1054 Supernova happened in fact 7000 year ago: if you would see the whole story (1. the star explodes 2. the astronomers of a planet far, far away, some millenia later, detect it) from a spaceship going, for example, with half the light speed, you would get totally different numbers. There is no "objective now" in the special relativity.

For us, "now" is what we see in the telescopes. For us, the crab nebula explosion happened in 1054. We have no way to visit a star, for example, 50 light year away, now - we would need at least 50 years to visit it. (It is the theoreitcal limit - in practice, there is no technology to send anything even to the nearest star, 4.3 light years away. If there would be, it would likely require at least centuries to get there.)

However, the overwhelming majority of the stellar evolution processes is much longer. Most stars exists between 10million and 10billion years long. Of course we have no way to watch a star billions years long with the telescopes. However, we can see billions of stars, typically some decade (centuries) long, and even this is enough to understand a lot from them.

We have a strong reason to not expect a huge difference from the "current" state of most stars/extraplanets from what we can see today on them - simply because such events (change of planetary orbits, average lifetime of most stars) last million times longer.


Why was there a gap in the number of asteroid detections between 1807 and 1845?

Ceres, Pallas, Juno and Vesta were all discovered between 1801 and 1807. After that, astronomers looked in vain for 38 years until the 5th, Astraea was spotted on December 8, 1845 by German amateur astronomer Karl L. Hencke by accident. He stumbled on Astraea while looking for Vesta one night.
So, what took so long to discover Astrea? There is a paper1 which discusses 4 possibilities:

Astronomers believed that there should be only four asteroids and thus didn't bother to look anymore. This conception came from Biblical theories and ideas.
Astronomers decided to postpone search for asteroids because of inadequacy of good charts and instruments.
Asteroids that were discovered later were not that bright as compared to the four asteroids that were discovered earlier. So, they had to wait for good telescopes to be able to see them.
Political unrest in Europe during that time. So, no one was that invested in astronomy.

The paper concluded that points (1) and (2) are more important and relevant and points (3) and (4) were met with speculation/contradictions. Refer to the paper for more details.
Reference:

Only the first four asteroids, Hughes, D. W.,1997, Journal of the British Astronomical Association, vol.107, no.4, p.211- 213. (link)


Calculating Postition of ISS / Hubble in space

The orbit of the ISS does vary. Drag from the very thin atmosphere slowly causes the orbit to lower, and the speed to increase. So the ISS needs to be boosted about a couple of times a month.

Its position can be predicted, at least in the short term, and websites such as Heavens Above publish current position, predictions, and orbital information.


Does this photo show the "Little Dipper" and "Big Dipper"?

As mentioned, these are the Pleiades, and the belt of Orion.  These are visible in the South at this time of year. The Big and Little Dippers are in the North, so turn around.  The best way to find them is a map:

I've marked the approximate edge of your photo, with Orion, the V of the Hyades and the small cluster called the Pleiades.  The big dipper is much bigger. It is usually quite easy to find, in the North East,  In this map it is labeled "Ursa Major" which means "great bear".
The little dipper is a rather faint constellation. Even its brightest star, Polaris, is only second magnitude.  You can find it by tracing from the pointers in Ursa Major (I've coloured them green). It is the only moderately bright star in that region of the sky. Once you have found the Pole star, it is possible to make out the faint stars that make up the rest of the Ursa Minor, the little dipper.


Shouldn't we not be able to see some black holes?

Well it doesn't exactly work the way you describe it. Matter doesn't just fall into a black hole.

A black hole still has a finite mass, which means other matter orbits its just like it would a comparable object of the same mass, like a star. In fact the gravitational disturbance it causes in this way is usually how we determine that there are, for example, supermassive blackholes in the center of galaxies.

Now the light you speak of is an accretion disk. When a star orbits a black hole, the star can get torn apart due to gravitational variation in the star as a consequence of the nearby black hole. Matter from the star starts orbiting the black hole, but it cant just do that in any direction. Due to conservation of angular momentum (the star itself was rotating) it is forced to rotate in one specific circular path, thats what we call the accretion disk and is the reason it doesn't become a huge sphere of light. As for light itself, it can't orbit, it either gets deflected or it gets trapped inside.


How is the magnitude of a star measured?

A device called a photomultiplier is attached to a telescope.

The telescope is pointed at the target star and an aperture is
closed down to avoid as much spurious light as possible, and a
reading is taken. The photomultiplier literally count the photons
hitting it.
Next the telescope is pointed at a well-known reference star in the
same part of the sky, the aperture is closed, and a reading is
taken.
Finally, the telescope is pointed at a dark part of the sky, the
aperture is again closed, and a reading taken.

Computer software compares the data from the target star, the reference star, and the dark sky to remove as much noise from light scattered by the atmosphere as possible. It's important to know how much of the measured light is coming from the sky and subtract it from the light measured from the target.

Often, this same procedure is repeated with a series of filters to get the luminosity of the target star at many wavelengths. Searching for a variable star's period requires switching from target star to reference star to dark sky time after time, hour after hour, night after night.

I paid my way through college doing this.


How do you draw the ecliptic on a celestial sphere sketch?

The ecliptic is the plane of the Earth's orbit around the Sun.
From the Earth's point of view, the Sun appears to migrate along the ecliptic, eastward one cycle per year.
The ecliptic and equator intersect at the equinoxes.
The Sun appears to cross the equator northward at the vernal equinox (♈) around March 20 and southward at the autumnal equinox (♎) around September 22.

Equatorial coordinates are linked to the equinoxes; ♈ is at right ascension 0h, and ♎ is at RA 12h.
Where these points appear in the sky depends on the local sidereal time:

♈ rises in the east at 18h, crosses the meridian at 0h, and sets in the west at 6h.
♎ rises in the east at 6h, crosses the meridian at 12h, and sets in the west at 18h.
At sidereal times between the above, apply a polar rotation of 15° per hour to interpolate.

Once you've determined where on the celestial equator to mark ♈ and ♎, then you can draw the ecliptic through them as another great circle 23.4° oblique to the equator.
The ecliptic runs north of the equator east of ♈, and south of the equator east of ♎, in the sense that eastward is counterclockwise as seen from the north celestial pole.


What are the Similarities and Differences between ALMA and FAST?

This is quite a broad question, since there are far more differences than similarities between the two. This answer will focus on the first part of the question and be about the technical aspects of the two telescopes.

Basics: FAST is a "classic" filled aperture telescope. This means it consists of a single large dish that focuses incoming radio waves on a single receiver (this is simplified). In contrast, ALMA is an interferometer: Many (> 60) dishes each focus incoming radiation on their own receiver. The relative positions of the individual telescopes is known to great precsion, and both amplitude and phase of the incoming radiation measured, which allows astronomers to combine several or all of the telescopes for a single observation. The maximum achievable resolution is then not determined by the diameter of a telescope but by the distance between those furthest apart. (Again, this is greatly simplified; interferometry is complicated.) In principle, FAST could also become part of a huge interferometer called VLBA.
Wavelength: FAST is a "classic" radio telescope, observing in the centimeter to meter wavelength range. ALMA is a sub-mm telescope operating at smaller wavelengths of 0.3 mm to about 1 cm.
Resolution: since the resolution is $
\propto \lambda / d$, with $\lambda$ the wavelength and $d$ the diameter of the telescope, ALMA has a much higher resolution - smaller $\lambda$ and greater $d$. ALMAs resolution is given as 10 milliarcseconds, while in the ideal case FASTs is $1.22 \cdot 0.1 / 500 \cdot( 180 / \pi \cdot 3600) = 50$ arcseconds at 10 cm wavelength. For comparison, according to wikipedia the unaided human eye has a resolution of about 60 arcseconds.
Sensitivity:

Atmospheric transmission: FAST observes in a wavelength regime in which our atmosphere is quite transparent. In contrast, the ALMA dishes are located in one of the driest deserts of the world at an altitude of over 5000 m, in order to minimize the effects of water vapour in the atmosphere.
Collecting area: FAST has a much larger collecting area. Also (I'm uncertain in this point, please correct this if I'm wrong) you lose some sensitivity when doing interferometry compared to just pooling the collecting areas of the single telescopes in ALMA.

Point Spread Function: The PSF of a single telescope, even a radio telescope, is much simpler than that of an interferometer. So the data analysis required to turn observations into an image should be much more straightforward for FAST.

There are some differences in the required surface accuracy (imperfections significantly smaller in scale than the observed wavelength do not matter) and surface composition - the ALMA dishes are solid metal, whereas the FAST dish is perforated. This works because as long as the holes are smaller than the wavelength, the incoming radio waves see it as solid.

The receivers used to convert incoming photons to electricity for both telescopes work on the same principles. Practically, it is much more difficult for ALMA due to the high frequencies at the edge of the far-infrared spectrum. ALMAs shortest wavelengths are just a little too long for receivers used in infrared or optical telescopes to work, and almost too short for normal radio receivers to work.


Is the SOI a spherical region or a oblate-spheroid-shaped region?

The sphere of influence is neither a sphere nor an oblate spheroid. It is a surface with no name. An approximation of this surface is

$$\left(\frac r R\right)^{10}(1+3\cos^2\theta) = \left( \frac m M \right)^4$$

This is neither a sphere nor an oblate spheroid, and this is but an approximation. Thefull expression is an absolute mess. Dropping the factor of $(1+3\cos^2\theta)^{1/10}$ (which is close to one) yields

$$r = \left( \frac m M \right)^{2/5} R$$

Tada! The equation of a sphere!

The true surface is defined in terms of two ratios. Consider two gravitating bodies, call them body A and body B. From the perspective of an inertial frame, the acceleration of a tiny test mass toward these two bodies is given by Newton's law of gravitation. These two bodies accelerate toward one another as well, so a frame based at the origin of either body is non-inertial.

From the perspective of a frame at the center of body A, the acceleration of the test mass is the inertial frame acceleration of the test mass toward body A plus the inertial frame acceleration of the test mass toward body B less the acceleration of body A toward body B. Denote the acceleration of the test mass toward body A as the primary acceleration and the difference between the inertial frame accelerations of the test mass and body A toward body B as the disturbing acceleration. Finally, define $Q_A$ as the ratio of these two. Now do the same for a frame with origin at the center of body B. The sphere of influence is that surface where $Q_A = Q_B$.


Why did the distribution of asteroids discovered in 2010 have a radial modulation?

I'm pretty sure that the radial pattern found in the data is a result of WISE's approximately 90 minute sampling cadence (dictated by the satellite's orbit), astrometric precision (about 0.2 arcseconds in the stacked images around launch, see Wright et al. 2010), and the number of free parameters in fitting the asteroid orbits based on that data. See, in the actual images the asteroids appear as points of light that shift appreciably between frames. IIRC, they expected 7 to 12 observations per asteroid. So you have 10-ish observations spanning 15 hours or so to fix the asteroid's orbital parameters around the sun. As you can imagine, there will be more parameters than can be perfectly fit with a single pass in this data set alone.

At a guess: it's related to quantization in uncertainty estimates and how that feeds forward into the orbit fitting algorithm.

I don't know the details behind the striping, but I'd bet it's related to the numerical precision used in the early processing of the data. They've either refined the orbits since using observations from passes separated by ~6 months, or modified how they're handling the numerical precision in their astrometric measurements since. More likely the former, but I'm sure if you asked Amy Mainzer (PI of NEOWISE, and head of the asteroid hunting part of the mission), Roc Cutri (head of the database creation and data processing part of the team), or any of the people on Mainzer's team, they could tell you more.

Relevant background: I was Ned Wright's grad student (original PI of WISE), and he had me design and test an asteroid hunting algorithm in the run-up to launch (we ultimately didn't use it - it scaled like $N^2$, IIRC, and existing efforts in the literature scaled like $\log(N)$ or $N\log(N)$). I ended up working with the extragalctic part of the team, hence my uncertainty on the precise details of what the solar system team did. I think I asked them about the striping during a presentation, but I don't recall the answer, so I'm pretty sure the answer was mundane and unavoidable.


What is the most dense object in the universe?

Let us define this as the largest observable density of a stable object, in order to exclude black holes which may have  a very large (infinite) density at their centers or objects collapsing towards a black hole status.

If we restrict the definition in this way, then the answer should be the core of the most massive neutron star that we know about.

At present there are a couple of neutron stars with mass of about $2M_{\odot}$ (Demorest et al. 2010; Antoniadis et al. 2013. Depending on the exact composition and equation of state at their centres these should have densities of around $2 \times 10^{18}$ kg/m$^{3}$ at their centers and average densities of $\sim 10^{18}$ kg/m$^3$.

Note that these densities are around 3 times the density of a proton or neutron or 5-10 times the density of nuclei at zero pressure.

In principle, the density of a single electron is much higher.


This paper should tell you all you need to know about how LIGO observes and localizes gravitational waves, and in particular its blind spots. The map below is taken from the paper and shows the observability of the system when in the  HLV configuration (Hanford-LIGO, Livingston-LIGO & Virgo). To triangulate a source by definition you need at least 3 measurements so that is why Virgo is included, although the two LIGO detectors can make very broad estimates for source localisation on their own (GW150914 was assigned a 600 square degree error box).

The (x) denotes a blind spot.


The age of the universe as we know t is very hard to determine. Current estimates put down the age f universe to 13.5 billion years +/- 0.8 billion. This is a huge a margin of error when compared to the margin or error you speak of.

Also it will never ever be possible to measure the age of universe because:

The universe started according to the big bang theory and if we take it as true it gives us a near infinite radiation density. We could measure the expansion and try to estimate the age but it is still hazy as we have no way knowing the initial density of the universe.
There is always an uncertainty in any measurement. and measuring something so far away(The edge of the universe) will have a lot of uncertainty.

These measures of today may still have a huge margin of error but they are the best we have till date. And to narrow the error down to about even a few years would take us a lot of time and tremendous development in technologies. So basically we have a very slim chance of determining the precise age of the universe in the foreseeable future


What could be the possible origin of boron found in the Martian soil if not from ancient groundwater?

Research published in the PLoS One paper "Boron Enrichment in Martian Clay" (Stephenson et al. 2013) suggest a volcanic origin for the boron, in the chemical analysis of a Martian meteorite (found in Antarctica), they explain:

  A straightforward geochemical interpretation of our results is that boron, a relatively volatile and soluble element, was first concentrated in the fluid dregs of lava (4–7 ppm boron has been detected in the late stage mesostasis of other nakhlites) and then became further concentrated by any groundwater or hydrothermal fluids that came into contact with the rock.

Further modelling reported in the paper "Partitioning of light lithophile elements during basalt eruptions on Earth and application to Martian shergottites" (Edmonds, 2014), also point towards a volcanic origin. Their modelling is based on analyses of shergottite pyroxenes (Martian meteorites), stating that observed elevated boron alongside other light lithophile elements or LLE are to do with the earlier mantle chemical processes, extrapolated from those observed on Earth:

  The range in LLE and trace elements observed in shergottite pyroxenes are instead consistent with concurrent mixing and fractionation of heterogeneous melts from the mantle.

Also, a volcanic origin is suggested in the Goldschmidt Conference Abstract "Boron isotopes in the nakhlites: Implications for crustal fluids on Mars" (Spivak-Berndorf et al. 2008), also based on analyses of nakhlites:

  The lack of isotopic fractionation between the igneous and
  secondary alteration phases in the nakhlites indicates that (1)
  the B in the fluid that deposited the iddingsite was derived
  from the primary igneous minerals and (2) there was minimal
  B isotope fractionation between the fluid and the iddingsite


What is the closest heavenly body to our solar system

The closest body we currently know to exist is Proxima Centauri which is part of the Alpha Centauri system at about 4.25 light years away.

There may be rogue planets that are closer but we haven't been able to detect any. Objects down to brown dwarf size that are closer than Alpha Centauri have not been found by extensive surveys.

Here is a list of close objects to the solar system: https://en.wikipedia.org/wiki/List_of_nearest_stars_and_brown_dwarfs


Light distorsion/ilusion when looking at bright objects

It's a bunch of different phenomena, somewhat related, but still distinct.

Naked eye

Since you mention naked eye observations - seeing a small amount of short spikes around bright objects is normal. The causes can be multiple: diffraction of light in/around your eyelashes, dirt/secretions on the cornea, just regular aberrations in the eye's optical system, or in extreme cases it might actually signal vision problems.

Your iPhone picture

It appears that you're taking the picture through some fabric or some other kind of rectangular network-shaped material. In this case, the spikes are due simply to reflection of light on the individual threads. If you removed all fabric and kept only one thread without moving it at all, the light would reflect on it exactly in the same place.

Telescopes

Many reflector telescopes show a pattern of cross-shaped spikes around stars. Those are due to diffraction. Those telescopes have a secondary mirror supported by an X-shaped element called the "spider". Light is diffracting on the spider and creates a similar pattern of spikes around stars.


Do celestial bodies actually appear larger along the horizon?

No, it's an illusion. Probably the ancient one. Simple experiment you can do is, set grid on telescope, measure the angle subtended when moon is at horizon and when moon is atop. You will see angle subtended by moon is same, hence the size are same. It's merely an illusion.

For more details:
https://en.m.wikipedia.org/wiki/Moon_illusion


Graph of apparent magnitude of Mars through time from around 2000 to present?

TheSkyLive's Mars information page has graphs of its distance and apparent magnitude from Earth.
If you set the same year range for each, you can see how the quantities are related.
They provide similar pages for all the planets and selected comets and asteroids.


Do liquids exist in the interstellar medium?

Basically, No. for liquid as we commonly know them on earth. If you trow a water bucket in space, it gets quickly vaporized. From this interesting ref:

  having a pressure vacuum will cause the water to boil almost instantly

There is however an exception I've been thinking of. Since glass has an amorphous structure, it can also be regarded as a liquid... And if some advanced form of life or intelligence managed to send some into the interstellar medium, then yes there is.

It's also possible that some stellar explosion send away magma of some form with an amorphous structure similar to that of glass (e.g. obsidian)


The sun "burns" hydrogen and even has "campfires" on it, but has anyone calculated a rate of actual chemical burning on the Sun?

I just became curious enough to search a few keyphrases.
This article talks about hydrogen molecules acting as energy "sink" in the sunspots - much like water phase changes on Earth create climate features. The heat capacity, mass density, adiabatic properties of the molecular hydrogen are pretty much different from the atomic one and these are important for the convection.
Hydrogen changing between atomic and molecular form is clearly a chemical process.
The article talks even about water molecules there. Given the high temperature and low pressure, they are expected to be pretty much short-lived and oxygen constantly moving between water, hydroxyl ion or radical and oxygen atom or ion.
Related matter here

The formation of H2 can significantly alter the thermodynamic properties of the sunspot atmosphere and may play a significant role in sunspot evolution. In addition to the survey observations, we have performed detailed chemical equilibrium calculations with full consideration of radiative transfer effects to establish OH as a proxy for H2, and demonstrate that a significant population of H2 exists in the coolest regions of large sunspots.

Looks like chemical processes not only do happen on the Sun, but are of great importance for the photospheric features we see.
Do these qualify as "burning" is a matter of oppinion.


Is there enough light to grow plants in the ocean of Europa?

The surface of Europa gets some sunlight direct from the sun and some long wave radiation from Jupiter so it's not in darkness.  It's also in Jupiter's radiation belt so it gets hit with high energy charged particles and the surface of Europa, despite being bombarded by particles is still a vacuum.  The surface of Europa is a terrible place to look for life, with the exception being that the liquid ocean of Europa explodes onto it's surface regularly, so evidence of life, if it exists, may well be found on the surface, but it's unlikely that anything can live there for long.  It's very inhospitable.

The inside of Europa is more interesting.  That's why Neil deGrass Tyson says he wants to go "ice fishing on Europa" to see what "licks the camera" and if we find life there we'll "Call them Europeans" . . . funny comments aside, the inside of Europa is probably dark, outside of some infra-red rays due to temperature.   Photosynthesis doesn't work well in darkness and with just long wave light, though maybe it's possible, it wouldn't nearly as much growth due to far lower energy levels.

As noted in the answer to your other question, if there is life in Europa's ocean, it's more likely to be something like chemotrophs.

What makes Europa interesting is the energy it gets from tidal forces being stretched between Jupiter and Ganymede, so it gets a regular influx of heat and perhaps internal volcanism, so it may have undersea vents similar to our oceans.   The energy influx and possibility of underwater volcanoes makes it more interesting than most other moons or dwarf planets.

If life is found in Europa's oceans, even if it's very primitive life by most standards, it would still be a very cool discovery because it could tell us something about how life starts and that's a big unknown right now.  The discovery of life outside the Earth would be a very big story.

We'll probably have to wait till 2030 before we begin to get some answers.  Sorry for the long wait, but it is what it is.


Was the solar system flat before the planets formed? Or does the Kozai-Lidov mechanism flatten orbits after the gas giants form?

The solar system was "flat" before any planets formed.

When a roughly spherical cloud collapses, it conserves angular momentum but loses energy. This means it tends towards a minimum energy configuration for a fixed angular momentum, which is a flat disk. The planets formed in this disk and that is thought to be why their orbits are, by and large, in the same plane and nearly circular. The timescale for forming a disk from a collapsing gas cloud is much less than a million years and is a prerequisite for forming planetesimals, that then go on to form planets on timescales of 10-100 million years.

The Kozai-Lidov mechanism enables cyclic interchange between eccentricity and inclination. It cannot be responsible for making a system flat and stable.


What is the role of the mesh on which dipole elements of the MWA antennas are placed?

The mesh acts as a ground plane.  An antenna design project can be divided into two theoretical regimes: free space, and ground plane.  A free space antenna is an antenna that is located far enough away from anything, that nothing will alter its electrical properties.  This is the simplest way of doing a design, but not applicable to this antenna
This antenna is mounted on the Earth.  The soil under the antenna has electrical properties that can not be precisely known.  The properties may change with the amount of vegetation or rainfall through the year.  To avoid having to deal with the exact properties of the soil, we place a metal shield between the antenna and the soil.  The shield is call the ground plane.  Its purpose is to present known electrical properties to the antenna.  Ideally the ground plane is always at zero Volts relative to the antenna.
The reason for using a mesh as opposed to a solid sheet of metal due to economy of materials facilitated by a neat property of electromagnetic waves: a wave cannot pass through an opening that is smaller than its wavelength.  Therefore instead of spending all the money, and gaining all the weight of a solid sheet, we can adjust the width of the mesh to block (that is, reflect) the frequencies we are interested in.


Right ascension, Equinox, Positions Help

For an object at about RA 6h and close to the celestial equator, viewed from 37 degrees N, it becomes visible in the early morning on about July 20th, and it is is visible at some point during the night until about May 20th, when it sets shortly after the sun.  Between May 20th and July 20th it is not visible. (Determined Stellarium)
An example of such an object is Betelgeuse.
An object that is at 6h of right ascension will culminate at midnight (ie reach its highest point at midnight about on the winter solstice on Dec 21st. The sun has a right ascension of 6 hrs on the summer solstice on June 21st and a RA of 18 hours on the winter solstice.  So the best time to observe the object would be on the winter solstice.


What is the difference between emission line and recombination line?

A recombination line is a special case of an emission line.

Emission lines

An emission line is any spectral feature that rises above the continuum — i.e. the average amplitude of the spectrum (in some wavelength region) — and is due to atomic transitions (where "atomic" include atoms in molecules and dust grains, and "transitions" may be electronic, vibrational, or rotational).

These transitions may arise from absorption of light, or collisions of an atom with another particle. In astrophysics, this other particle will usually be a free electron, since these are much faster than atoms. The absorption of energy puts the atom in an excited state, and when the atom de-excites, a photon is emitted.

Recombination lines

Recombination is the process of a free electron being "caught" by an atom missing an electron, i.e. an ion. The ion, in turn, has been produced previously by being ionized by some high-energy process, e.g. an ionizing photon or a violent collision. The difference between the kinetic energy of the electron before the encounter, and the energy of the state it goes to, is emitted as a photon.

The electron may go directly to the ground state, in which case another iozing photon is just produced. But it may also go to an intermediate state, emitting a lower-energy photon. The electron is then in an excited state, from which it will soon de-excite, possibly through several levels until it reaches the ground state. This process is called cascading.

The photons that are emitted during this cascade are called recombination lines.

Recombination lines from gas around stars

In the vicinity of very hot stars (O and B stars) which produce many ionizing photons (the Lyman continuum; LyC), you will have a region of neutral hydrogen gas (HI) being ionized. Timescales for recombination in these regions are quite small, so the LyC is "almost immediately" reprocessed into recombination lines.

Quantum mechanics tells you the probabilities of ending up in the various states, and it turns out that, for instance, for each LyC photon you get roughly 0.68 Lyman $\alpha$ photons (with a small dependence on temperature), i.e. photons emitted when the electron false form the first excited to the ground state. In terms of energy, this translates into ~1/3 of the total power.

This is actually quite amazing: It means that 1/3 of the total continuum of photons more energetic than the 13.6 eV needed to ionize hydrogen is converted into a single emission line! For this reason, galaxies — in particular young galaxies that still host many hot stars — are often quite luminous in Lyman $\alpha$; sometimes even only in Lyman $\alpha$.

Absorption lines

Emission lines are related to absorption lines: In the first case, you have a continuum of light (e.g. from thermal processes) with some physical process "adding" some extra light to the total spectrum. In the latter case, you have the same physical process removing light from a continuum.

For instance, if you observe a (non-ionizing) source through an HI cloud, you'll see absorption lines at the wavelengths corresponding to the various transitions of neutral hydrogen. If you were able to observe the same cloud from another angle, you'd see emission lines at the same wavelengths.


Is the nose of the solar system and the solar apex the same thing?

No, they're roughly 60 degrees apart.
The Solar Apex.
The solar apex does indeed indicate a component of direction of travel within our corner of the galaxy:

The coordinates as obtained by visual observation of the
apparent motion is right ascension 18h 28m 0s and declination
of 30° North in galactic coordinates: 56.24° longitude, 22.54°
latitude. The radioastronomical position is 18h 03m 50.2s and dec
30° 00′ 16.8″ galactic coordinates: 58.87° longitude, 17.72°
latitude.

There are two different precise measurements here which conflict, I won't speculate as to why Wikipedia is unreferenced on the subject. The galaxy is a place containing a great deal of complex movement and the Local Standard of Rest is moving relative to the galactic plane, the centre of the galaxy and the other local stars:

In astronomy, the local standard of rest or LSR follows the mean
motion of material in the Milky Way in the neighborhood of the Sun.
The path of this material is not precisely circular. The Sun follows
the solar circle (eccentricity e < 0.1 ) at a speed of about 255 km/s
in a clockwise direction when viewed from the galactic north pole at a
radius of ≈ 8.34 kpc about the center of the galaxy, and
has only a slight motion, towards the solar apex, relative to the LSR.

We're traveling relative to an ever shifting cloud of stars towards Hercules (also shifting), southwest of the star Vega.

The sun's motion in the Milky Way is not confined to the galactic
plane; it also shifts ("bobs") up and down with respect to the plane
over millions of years.

So you see, there are many components of direction of travel to consider.
There's an unquotable abstract from a Harvard article here, which may give you an idea of how much needs to be taken into account in determining the relative vectors.
The Nose.
The ""Nose" of the solar system is sometimes used to refer to the direction of travel of the local galactic wind. Just as a comet's tail points away from the sun, the tail of the Heliopause points away from the wind, the nose points in the direction it's coming from.

The nose is as you have stated is pointing towards Scorpius.


You can, in fact, differentiate between the crossings of the instability strip in certain types of variable stars. There have been quite a few evolutionary models devoted to this, and many, as well as scores of observations, show that the pulsational period $P$ of a Cepheid variable changes during each crossing. Look at, for instance, the period changes of 60 galactic Cepheids, shown in Figure 2 of Pietrukowicz:

The first and third crossings occur when the star is evolving redwards (towards the right of the HR diagram), and show positive $\dot{P}$. The second crossing occurs when the stars are evolving bluewards (towards the left of the HR diagram), and show negative $\dot{P}$.

Models and observations by Neilson et al. (2014) show the same thing (see Figure 1):

The initial large period change indicates the first crossing, the smaller negative period change indicates the second, and the third less drastic increase indicates the third.

Similar indicators should be present in further crossings (fourth and fifth crossings in the later stages of core and shell helium burnings), and while the amplitude of $\dot{P}/P^2$ would likely be different, the same general positive/negative trends should be present. For more information, see Turner & Berdnikov (2004).

There doesn't seem to be as much information on period changes in non-Cepheid variable stars, although Smith (2013) states that most Mira variables show only minute period changes, if any, and RR Lyrae variables may show period increases as they evolve redwards across the instability strip toward the end of their lives, as is the case with Cepheids. This, by the way, does not fully agree with evolutionary theory.


Do interstellar asteroids decelerate and eventually stop?

An object moving through the interstellar medium will experience a weak drag force. If the drag force is $f(v)$ the velocity will decline as $dv/dt=-f(v)/M$ where $M$ is the mass. In fluids $f(v)\propto v^2$; this appears to work even for low density gas. Note that this differential equation leads to a velocity that declines forever, slowly approaching zero but never reaching it.

We can make a simple ballpark estimate of the timescale of slowdown by estimating how long it takes for it to run into its own mass of interstellar material. If it has radius $r$ it will encounter $\pi r^2 v \rho$ kg of ISM per second. So for mass $M$ the timescale of slowdown is $\tau \sim M/\pi r^2 v \rho$.

If we take a $r=1$ km spherical asteroid of mass $M= 8.8802\cdot 10^{12}$ kg (assuming density 2.12 g/cm$^3$) moving at 26.33 km/s and use $\rho=10^{-20}$ kg/m$^3$ (this varies a lot) then the timescale is about $10^{22}$ seconds, or 340 trillion years. So after maybe a quadrillion years the asteroid would be nearly at rest relative to the local gas if nothing disturbed its trajectory...

However, it is fairly likely for the asteroid to encounter a star during this time. Stellar density is about $\rho_*=0.14$ per cubic parsec, and the timescale for getting within 100 AU from a star is $\sim 1/\pi (100 AU)^2 v \rho_* \approx 10^{11}$years. This is likely to act as a gravity assist giving it some of the star's relative velocity (on the order of kilometers per second). So the true answer is that the asteroid will never settle down as long as there are stars in the galaxy.


How can I collimate a dobsonian telescope with a laser collimator?

Note to future answerers: We have been working in the Astronomy.SE main chat room with OP to solve this problem and established it's actually a matter of focusing that resulted in problems described (telescope was either too intrafocal or extrafocal), not collimation itself. This is a video on YouTube that OP said was exactly how the problems looked like. Granted, focusing can be technically called collimation too, and indeed that demonstration video calls it as such (albeit misspelled), since it's still about adjusting positions of primary and secondary mirror in dobsonian or newtonian telescopes (dobsonian is a type of newtonian), but the former is a lot simpler to solve by turning focusing knobs and achieve optimal illumination.

Now, let's answer the actual question.

Preparing our laser collimation setup

First thing to know is that any laser collimator will likely have the beam strong enough to cause damage to your eyes if pointed directly in them, or damage strong light sensitive equipment, like for example digital cameras, so be careful with it and take precautions. Carefully read instructions on how to handle and use it, before you do so. This is how a laser collimator might look like:

Next thing to do is make it functional by inserting batteries or plugging its power adapter and testing it actually works by turning it on with the switch, if it has it (probably does). Once you'll establish your laser collimator is turned on and functioning by observing (usually ruby red) light beam spot it makes on any neutral surface, turn it back off and attach any adapters on it you might need to help you attach it to your telescope's focuser. This is how your telescope's OTA (Optical Tube Assembly) might look like:

Now remove the barlov and eyepiece from your telescope's focuser if you're using a direct focuser adapter for your telescope model, tighten the laser collimator's compression ring by rotating its rubber ring and attach the collimator into the focuser. Rotate collimator's compression rings to tighten it in place. If you're not using an adapter, your laser collimator might attach directly to your eyepiece. Consult your collimator's user manual, if unsure.

Rotate your collimator's targeting faceplate towards your telescope's primary mirror (on the telescope's bottom) and turn on the laser.

Angular alignment of the secondary mirror

By looking into the telescope top down from the secondary (top) mirror into the telescope, we will adjust the angular alignment of the secondary mirror by referencing the laser dot projecting on the primary (bottom) mirror. The secondary mirror's support has a center ring, the collimating cap, with three bolts on the top of the secondary mirror holder. Use Allen wrench or Phillip's screwdriver to rotate them individually to bring the laser dot into the center of the doughnut ring of the primary mirror, like so:

Try to bring the dot more directly into the doughnut center than on the photograph above, though. I intentionally didn't select a photograph from when it was in its dead center, because of the light reflection covering the doughnut ring completely, rendering it nearly invisible.

Primary mirror collimation

Unlock the primary mirror plate locking screws, position yourself towards the panel side of the primary mirror tube so you can observe the laser's center spot on the targeting faceplate on the collimator and start adjusting collimation screws by turning them gently until the laser dot is in the center of the targeting faceplate:

Another way to check your collimation progress when adjusting collimation screws is by checking the projection laser dots on the secondary (top) mirror and making sure both laser dots are overlapping into one single spot, but that might make your adjustments of the primary mirror somewhat harder, so using the targeting faceplate of the laser collimator would be my suggestion and perhaps check alignment of the secondary mirrors laser dots projection for the last finishing touches to collimate your mirrors as precise as possible.

Now you're done collimating your dobsonian or newtonian telescope, so it's time to remove the laser collimator by first switching it off, then untightening its compression ring and taking it out of the focuser. Once you've removed the laser collimator, insert in the focuser barlov and the eyepiece, and tighten them back in place.

Additionally, and as requested, here is the Youtube video of the Newtonian Collimation using SCA Laser Collimator that are also the source of the last two example photographs and inspiration to write the collimation procedure, courtesy of HoTechUSA YouTube channel. Make sure to browse through other similar videos on telescope collimation in this same channel.

Collimation of a dobsonian without a laser collimator

Standard collimation technique for the exact dobsonian telescope make OP mentioned has can be read in this Instruction Manual for Sky-Watcher dobsonians (PDF):

  Collimation is the process of aligning the mirrors of your  telescope
  so that they work in concert with each other to  deliver properly
  focused light to your eyepiece. By observing  out-of-focus star
  images, you can test whether your  telescope's optics are aligned.
  Place a star in the centre of  the field of view and move the focuser
  so that the image is  slightly out of focus. If the seeing conditions
  are good, you  will see a central circle of light (the Airy disc)
  surrounded by a  number of diffraction rings. If the rings are
  symmetrical about  the Airy disc, the telescope's optics are correctly
  collimated  (Fig.g).

  If you do not have a collimating tool, we suggest that you  make a
  "collimating cap" out of a plastic 35mm film canister  (black with
  gray lid). Drill or punch a small pinhole in the exact  center of the
  lid and cut off the bottom of the canister. This  device will keep
  your eye centered of the focuser tube. Insert  the collimating cap
  into the focuser in place of a regular  eyepiece.

  Collimation is a painless process and works like this:

  Pull off the lens cap which covers the front of the telescope  and
  look down the optical tube. At the bottom you will see the  primary
  mirror held in place by three clips 120º apart, and at  the top the
  small oval secondary mirror held in a support and  tilted 45º toward
  the focuser outside the tube wall (Fig.h). The secondary mirror is
  aligned by adjusting the central bolt  behind it, (which moves the
  mirror up and down the tube),  and the three smaller screws
  surrounding the bolt, (which  adjust the angle of the mirror). The
  primary mirror is adjusted  by the three adjusting screws at the back
  of your scope. The  three locking screws beside them serve to hold the
  mirror in  place after collimation. (Fig.i)

  Aligning the Secondary Mirror

  Point the telescope at a lit wall and insert the  collimating cap into
  the focuser in place of a regular  eyepiece. Look into the focuser
  through your collimating  cap. You may have to twist the focus knob a
  few turns  until the reflected image of the focuser is out of your
  view. Note: keep your eye against the back of the focus  tube if
  collimating without a collimating cap. Ignore the  reflected image of
  the collimating cap or your eye for  now, instead look for the three
  clips holding the primary  mirror in place. If you can't see them
  (Fig.j), it means  that you will have to adjust the three bolts on the
  top of  the secondary mirror holder, with possibly an Allen  wrench or
  Phillip's screwdriver. You will have to  alternately loosen one and
  then compensate for the  slack by tightening the other two. Stop when
  you see all  three mirror clips (Fig.k). Make sure that all three
  small  alignment screws are tightened to secure the  secondary mirror
  in place.

  Aligning the Primary Mirror

  There are 3 hex bolts and 3 Phillip's head screws  at the back of your
  telescope, the hex bolts are the  locking screws and the
  Phillip's-head screws are  the adjusting screws (Fig.l). Use an Allen
  wrench  to loosen the hex bolts by a few turns. Now run  your hand
  around the front of your telescope  keeping your eye to the focuser,
  you will see the  reflected image of your hand. The idea here being
  to see which way the primary mirror is defected,  you do this by
  stopping at the point where the  reflected image of the secondary
  mirror is closest  to the primary mirrors' edge (Fig.m).  When you get
  to that point, stop and keep your  hand there while looking at the
  back end of your  telescope, is there a adjusting screw there? If
  there  is you will want to loosen it (turn the screw to the  left) to
  bring the mirror away from that point. If  there isn't a adjusting
  screw there, then go across  to the other side and tighten the
  adjusting screw on  the other side. This will gradually bring the
  mirror  into line until it looks like Fig.n. (It helps to have a
  friend to help for primary mirror collimation. Have  your partner
  adjust the adjusting screws according  to your directions while you
  look in the focuser.) After dark go out and point your telescope at
  Polaris, the North Star. With an eyepiece in the  focuser, take the
  image out of focus. You will see  the same image only now, it will be
  illuminated by  starlight. If necessary, repeat the collimating
  process only keep the star centered while tweaking  the mirror.


Video astronomy requires much more equipment than traditional eyepiece viewing. It helps to understand what is going on between the camera lens and the video screen.

Most important is the conversion of the cameras analog signal to the computer digital format. This is usually done with a frame grabber and its associated drivers. The cameras typically have 700 ish lines of resolution which are usually translated into 640 by 480 digital signal. The camera analog signal is 60 cycle interlaced. The computer display frequency is whatever the PC is set to. I'll let the Math people tell us how the drivers fit all those numbers together and make a nice picture. I think the answer is PI or at least within a decimal place or two. The numbers get rounded and every once in a while the computer screen hiccups. National Enquirer sort of news.

The frame grabbers come in a wide range of cost and reliability. There's no rhyme or reason to their reliability. They depend on the computers operating system which is not identical on every computer. The frame grabber and driver will work well on one PC but not another similarly equipped PC.

Typically the astro video cameras two output connectors, a coax and an S-Video. The person making the video could have proven the "Waves" were and artifact of the PC by simply connecting one of the cameras outputs to an analog video screen, a television with a composite input for example and the other output to the computers frame grabber. Provided the camera and analog video are functioning properly he could have shown us a rock steady image on the analog screen and the waves on the Computer screen. With a little skill he could have shown us an alien or two surfing on the waves.

Smoke, mirrors and snake oil. Suckers are said to be born at the rate of one per minute. I have a feeling the con artist birth rate is up there as well.

It is a bit of a shame, Video or near live viewing as many of us call it, is gaining in popularity. Having someone use what is not unusual for a Video astronomer to see just for a bit of sensationalism serves no useful purpose.

Another point to edit in. The uTube is a video of a video. The image we see is taken by a second video cameras pointed at a video image being displayed on a computer screen. Crunch the computer screen frequency and resolution with the frequency and resolution of the second camera. You end up with images of airplane propellers that appear to stand still or move very slowly. I wonder how much adverting revenue that sort of video would bring into uTube.


What lens should I use for my Raspberry Pi?

Thought I would revisit this and answer my own question...no lens is needed at all :)

The telescope itself is the lens, so once that finally made sense to me I then took the lens off the Raspberry Pi camera and I can now use the camera with the telescope


I will make a small calculation here, but please proceed to the results if you may like to.

Calculation

Stars are spherical and static, so metric near their surface (photosphere) and outside on is Schwarzschild. Hence time-time metric component on the surface is:

$$g_{44}=1-\dfrac{R_{grav,*}}{R_*}$$,

where $R_*$ is the radius of the star and $R_{grav,*}$ is its gravitational radius.

Then, if the velocity of the star is much smaller than the speed of light, gravitational redshift in the lowest order does not depend on this velocity. Therefore, the emitting star can be assumed at rest.

The light from the star propagates along isotropic geodesic in Schwarzschild metric. The geodesic is described by lagrangian:
$$ \mathcal{L} = \dfrac{1}{2}g_{\mu\nu}k^{\mu}k^{\nu}$$,
where $k^\mu=(\vec{k},\omega/c)$ is the 4-vector of the light wave and $\omega$ is the frequency of light. Since the metric is static $\dfrac{d\mathcal{L}}{dk^4}=g_{\mu 4} k^4=g_{44}k^4=\textrm{const}$. Therefore:

$$(1-\dfrac{R_{grav,*}}{R_*})\omega = \textrm{const}$$

for the light as it travels towards us. So:

$$\omega_{obs}=\omega_{emitted}(1-\dfrac{R_{grav,*}}{R_*})\Longleftrightarrow \lambda_{obs}=\dfrac{\lambda_{emitted}}{(1-\dfrac{R_{grav,*}}{R_*})}$$,
where $\lambda$ is the wavelength.

Redshift is simply $z=\dfrac{\lambda_{obs}-\lambda_{emitted}}{\lambda_{emitted}}$. Assuming $z\ll 1$ one has a simple formula:
$$
z_0=\dfrac{R_{grav,*}}{R_*}
$$

If $z_0$ turns out to be comparable to unity, one should compute
$$
z=\dfrac{1}{1-z_0}-1,
$$
which then gives the correct value of redshift. Note that redshift does not depend on $\lambda$.

Nice numerical forms for this would come from $R_{grav,*}=2.95\textrm{km}\dfrac{M_*}{M_\odot}$:

$$z_0=0.295\dfrac{10\textrm{km}}{R_*}\dfrac{M_*}{M_\odot}\Longleftrightarrow z_0=4.24\cdot 10^{-6}\dfrac{R_\odot}{R_*}\dfrac{M_*}{M_\odot}$$

It is also nice to express the redshift in $\textrm{km}/\textrm{s}$:

$$
z_0=8.84\cdot 10^4 \dfrac{10\textrm{km}}{R_*}\dfrac{M_*}{M_\odot} \textrm{km/s}\Longleftrightarrow z_0=1.27 \dfrac{R_\odot}{R_*}\dfrac{M_*}{M_\odot}\textrm{km/s}
$$

Summary and discussion

In summary, when redshift $z$ is small, it is approximated by $z_0$, the numerical expressions for which are given just above. If $z_0$ turns out to be not small, one can calculate $z=\dfrac{1}{1-z_0}-1$, which then gives correct redshift.

Stars

One can see that:

For normal stars like the Sun ($R_*\sim R_\odot, M_*\sim M_\odot$) the redshift is of order $1\textrm{km/s}$. It is almost important since stars in solar neighbourhood move normaly at the velocity of a few tens of $\textrm{km/s}$
For white dwarfs ($R_*\sim 10^4 \textrm{km}, M_*\sim M_\odot$) redshift is a few times $100 \textrm{km/s}$ and gets very important when doing proper spectroscopy. Therefore one typically accounts for it.
For neutron stars ($R_*\sim 10 \textrm{km}, M_*\sim M_\odot$) redshift is very important $z\sim 0.4$, but neutron stars are general relativistic objects anyway, so it would be expected beforehand.

So, in summary, when measuring light from individual stars one does have to account for gravitational redshifts to get accurate results, and particularly so when studying white dwarfs.

Groups of objects

Now, the same formulae are correct to an order of magnitude when applied for larger volumes of space, with $R_*$ and $M_*$ now meaning the size of the volume and the mass inside it. However, as typical interstellar distances are of order of parsec and $\textrm{pc}=3\cdot10^{13}\textrm{km}$, resulting $z$ will be very small even for such dense groups as globular clusters ($z$ is of order $10^{-8}$ in this case). So, groups of objects do not affect the redshift.

Cosmological overdensities

Nevertheless, cosmological scale underdensities of order few $100\textrm{Mpc}$ in size may affect the apparent redshift of distant objects, as we would be inside the underdensity. However, such an underdensity would have to be significantly symmetric around us in order to explain the lack of corresponding anisotropy in cosmic microwave background. Therefore, it is considered unlikely.


What was the process of calculating how much darkmatter/matter/dark energy the observable universe consists of and how did it come at this conclusion

We are not actually measuring the $\Omega_r$,$\Omega_m$,$\Omega_{\Lambda}$.
At first, we measure the distance modulus to certain celestial objects by using the standart candles in the astronomy.
This distance can be calculated from,
$$d_L=\frac {c} {H_0}z(1+\frac {1-q_0} {2}z)\,\,\, (Eqn.1)$$
Here $q_0$ represents the deceleration parameter. Now I am not going to go in mathematical details but you should now that, using the acceleration equation we can write $q_0$ in terms of $\Omega_r$,$\Omega_m$,$\Omega_{\Lambda}$.
$$q_0=1/2 \sum \Omega_{w,0}(1+3w)$$  and for matter ,$w=0$, for radiation $w=1/3$ and for dark energt $w=-1$ we get,
$$q_0=\Omega_r+1/2\Omega_m - \Omega_{\Lambda}\,\,\, (Eqn.2)$$
And the relationship between distance modulus $(m-M)$ and $d_L$ can be written as,
$$m-M=5log_{10}(d_L/1Mpc)+25\,\,\, (Eqn.3)$$
If we combine Eqn.1 and Eqn.3 we get a relationship between distance modulus and z.
We plot the $m-M$ for distant objects with respect to $z$. From the slope of the graph we can determine the $H_0$ (for $z\rightarrow 0$) And for large z,  deviation of the plot from the straight line gives us to determine the $q_0$.
Now we have experimental data. Then we put different values for $q_0$ or equally $\Omega_m$ and $\Omega_{\Lambda}$ (with respect to Eqn.2) to determine the best fit for the graph (since we can take $\Omega_r=0$)

Edit: I clearly missed the "The Planck data" in the OP's question but its the same general idea. For different density parameters, the CMBR would look different, like the graphs that I shared are different for different density parameters. So by looking at the CMBR, we can determine the density parameters and which one fits well to the CMBR.

Further Note: I want to also suggest that the $\Omega_b$ can be calculated from the early universe by measuring the amount of deuterium and other elements present in primordial gas clouds.
And observering the rotational curves on the galaxies we can have approximate values on the $\Omega_{dm}$. (finding the dark matter density)
We cannot deduce the $\Omega_m$ and $\Omega_{\Lambda}$ just by looking at the Supernova data, for example the supernova data is also consistent with $\Omega_m=0$ and $\Omega_{\Lambda}=0.4$ or with $\Omega_m=1$ and $\Omega_{\Lambda}=1.7$


Can a liquid be used to de orbit debris?

Temperature is not the only factor for matter phases, pressure does make a difference too.

As space is practically a vacuum, thus pressure is zero. As far as I know, there can be no liquids in the vacuum of space. No element nor compound is liquid in zero pressure no matter what the temperature might be. The best candidate for this would be Helium, but even in this case the vacuum of space cannot keep equilibrium between liquid and gas phases, and thus all liquid helium will evaporate into gas.

Liquid could exist in space, if it would be able to collapse on its own gravity and maintain gaseous atmosphere bound to the liquid body. This would require way more matter than is feasible for your purpose, however.

EDIT: Also adding mass to an orbiting object does not in general cause its orbit to decay, as mass does not significant role in orbit equation. Gravitational force IS proportional to product of masses, but then again acceleration is force divided by mass - ie it gets cancelled out; applies only if adding mass does not significantly change the barycenter of said two objects (ignoring relativistic effects and gravitational waves since obviously far outside the scope of this question)

Adding surface area does not affect orbit in vacuum since there is no drag nor friction. Also shape of the object does not matter for the very same reason.

EDIT 2: As noted in comments, liquids may remain liquids in vacuum long enough for this scenario. I couldn’t find very good references how this might look like in the vacuum of space, but seemingly water can exist as a liquid for extended periods of time in very low pressures, see eg https://www.hou.usra.edu/meetings/lpsc2014/pdf/2060.pdf which states that 250ml of liquid water did fully evaporate in about 2.5 hours in 0.003 arm (Personally I find this very surprising and contradicting to my prior knowledge but I have been wrong many times before)


Question about the relevant physical parameters to build a backyard observatory

Many astronomers are using electronic eyepieces to overcome light pollution. You may want to see what these devices can do in your area'

I use several of the Mallincam cameras in my observatory.

https://www.mallincam.net/

https://groups.io/g/MallinCam/topics

Some photos of my observatory and assistants

https://www.flickr.com/photos/142615862@N04/sets/72157672437929982

More home observatory sites

http://obs.nineplanets.org/obs/obslist.html


Is there a chance that UFOs could be messages from other planets?

No.

The purpose of sending a message is to have someone else read it. It defies logic to send a message over light years of space and then have it do anything but broadcast its presence very clearly to anyone who might possibly want to investigate it. If the goal of these UFOs was to tell us something, they wouldn't be such elusive specimens that only appear to specific people at specific times and places, and then disappear without a trace. If UFOs are in fact meant to communicate with humanity, they are indeed rather poorly designed, because they haven't accomplished that goal very well at all!


Of the planets, two have ice caps: Earth and Mars.  The Earth's ice caps are water ice, and Mars has ice caps of Carbon dioxide, and some Water ice. The CO2 on Mars would give a weakly acidic solution if melted but it would not be a caustic acid.

Titan has hydrocarbon lakes at its poles, and other planets are too hot, or are uniformly frozen.


Can I use an array of inexpensive cameras as an alternative to a telescope?

Yes it would.  However, it also depends on your definition of "Detail".  For some, it's a tight shot where the whole frame is only a few dozen arcseconds wide, and others think of detail as being wide field, but also "deep" - or having a very high signal to noise ratio.

If you want wide field, and a high signal to noise ratio, then yes, a multi-camera array will work well, and you'd be able to accumulate a large number of high integration time images in a relatively short amount of time.

So here's a breakdown of what you and others who may not know need to know about deep sky astrophotography:

The main goal for all photos is to produce as visually pleasing image as possible.  A single image taken of any target will accumulate light from many sources.  The first, and obvious source is the light from whatever target you're trying to photograph.  The second, less obvious source is light pollution from man-made sources, the moon, and natural sky glow.  The third source is the electronic and thermal noise of the camera itself.

Technically, all of these sources are "signal" - however two of the three are unwanted, so we will call the light pollution signal and the camera signal 'noise'.  Longer exposures produce more signal from what you want, but also more noise from what you don't want.

Luckilly, noise can be reduced by using several techniques in conjunction with each other.  First, you need to take long photos.  The signal will increase linearly with time.  A 60s photo with a signal of "x" will have a tenth of the signal as a 600s photo.  Noise, however, increases with the square root of the exposure.  Using the same example, the 600s photo will have a signal to noise ratio that is about 3.2 (sqrt of 10) times better than the single 60s photo.

Increasing the Signal to noise ratio by taking longer exposures is the first step to producing 'detailed' wide field images.  Camera noise can be further reduced by taking many images.  An array of cameras makes this trivial.  Two cameras will produce twice the images as one camera, and 10 cameras will produce 10 times the images as a single camera.  That's obvious.  Why does it matter?

When we stack images, the more we have, the better the stacking algorithms can preserve the signal we want to keep, and reduce the camera noise we don't.  Once again, the signal to noise ratio mathematics work out the same as above.  1 image will have a certain signal to noise ratio, but 10 images will have a signal to noise ratio that can be about 3.2 times higher - because the total integration time has increased ten fold.

Of course, if you're stacking photos, you should be calibrating them as well (dark frames, bias frames, etc) - which really reduce the amount of noise per sub-frame.  Because of this, your actual SNR boost by stacking should in theory be a lot higher than 3.2.

If photographed a target across 60 degrees of the sky (30 before meridian and 30 after) then you'd be able to acquire photos for 4 hours.  One camera taking 10 minute photos would get 24 photos in the 4 hour period of time.  If you had 5 cameras, you'd get 96 photos.  72 additional photos in the same amount of time is a lot of extra data to help reduce the noise, and really let the faint details stand out above the background - especially if all of your photos were well calibrated.


How difficult is it for an amateur to discover a new exoplanet or astronomical object?

Zooniverse has a citizen science project called Exoplanet Explorers that used volunteers to examine data on exoplanet candidates.  It was so successful that it has now run out of data, but there are other space projects that are looking for contributors