September 26 twelve hours sunrise to sunset?

A day would be twelve hours long at the equinoxes if the Earth had no atmosphere and if the Sun was a point rather than a sphere. Sunrise and sunset are defined as the time at which the upper limb of the Sun appears to rise above or set below an ideal oblate spheroid Earth and assuming average atmospheric conditions. While this calculation ignores terrain and variations in atmospheric condition, it does not ignore that the basic facts that the Earth does have an atmosphere and that Sun is a sphere. The standard atmosphere conditions means that sunrise (sunset) occur when the upper limb of the Sun first reaches (first falls below) 34 arc minutes below the horizon. That the Sun is not a point adds another 16 arc minutes to this, making sunrise/sunset occur when the center of the Sun is 50 arc minutes below the horizon.

This means that at the equinoxes, a "day" is at least six minutes longer than the twelve hours we are naively taught (and incorrectly reported twice a year by naive weathermen). As an aside, days are always longer than nights at the equator.


Why does Mars appear to retreat across the sky?

The effect is called apparent retrograde motion.

What happens is that Mars has a 'direction opposite to that of other bodies within its system as observed from a particular vantage point' when this loop occurs.

That's a bunch of words that don't mean a lot to me. A picture is worth a thousand clearer words:

(Imagine this turned sideways and you get the effect in your image)

Basically, because Earth and Mars are orbiting the sun at different rates, our vantage point of Mars changes for each combination of points in the orbit of each planet.

On this scale, the background of stars is pretty much stationary - any apparent movement of the stars due to this effect is going to be negligible. Thus, the stars are our point of reference.

As our vantage point of Mars changes, it appears to shift directions on the stellar background, creating the effect you describe.


What effect does Earth losing a part of its mass have, is there an impact on the Earth's orbit around the Sun?

According to this answer to Is Earth getting heavier or lighter?:

tl;dr: The Earth receives 40,000 tons of dust from space every year, but looses 95,000 tons of Hydrogen and 1,600 tons of Helium every year as well. After all additional effects are balanced, the Earth looses about 50,000 tons a year.

According to this answer to Does launching a device into orbit change earth's orbit?

As long as "what happens in GEO stays in GEO" and there's no mass lost to deep space, the Shuttle and the Earth will mostly orbit around their common center of mass.
When the Shuttle is moving (relative to Earth) in the prograde direction (Earth's motion around the Sun), the Earth will be moving slightly slower than it normally would, but a half-orbit later when the Shuttle is moving retrograde, the Earth will be moving slightly faster than it normally would.
When the Shuttle returns, to 1st order (and maybe 2nd order as well) the Earth will be in the same place it would be if the Shuttle had never taken off.
It's quite a bit like the Earth and Moon orbiting around their common barycenter, and it being pretty much the EM barycenter that moves in a nice elliptical orbit around the Sun, except that as @RussellBorogove points out very effectively, it's a quite a bit smaller effect than that.

So as long as the launched mass stays in orbit around the Earth, the Earth's orbit is essentially unaffected. If the spacecraft goes off into deep space, then there will be an extremely tiny change in the Earth's velocity (rather than due to its mass change) by conservation of momentum.
If the Earth loses mass by natural process isotropically, the orbit will be essentially unaffected. Of course the Earth and Sun orbit around their common center of mass, so if the Earth gets lighter it will have an extremely tiny second-order effect on the orbit, but that's nothing compared to the effect that the outer planets have on the Sun.


What percentage of habitable-zone planets are detectable by transit?

Back of the envelope time.
First, we have to assume perfect data, so the only factor at play here is whether there is a geometric eclipse or not. Of course if you have poorer data then you will miss some planets because they are too small. i.e. We are looking for the fraction that can be detected in principle.
Let's assume the planets are small enough that their size does not really influence the transit probability, which is given by $\sim R_*/a$, where $a$ is the semi-major axis.
Let's assume circular orbits.
Let's assume a bare planet with no albedo so that the equilibrium temperature is given by
$$ T_{\rm eq} = T_* \sqrt{\frac{R_*}{2a}}\ .$$
Let's assume that a typical star in the Galaxy is an M-dwarf with a temperature $T_* \simeq 3500$ K and a radius of $R_* = 0.5 R_{\odot}$, and let's assume that the likelihood of planet occurrence is independent of stellar mass, so that the properties of an M-dwarf can be assumed for statistical purposes (in practice the answer will depend on what kind of star you are considering).
Let's assume that a habitable planet needs the equilibrium temperatures to be between 273K and 350K (arbitrary I know, and ignores the issue of atmospheres). The range of $a$ for this temperature range, around our fiducial M-dwarf, is between $50R_{\odot}$ and $82R_{\odot}$, with a probability of detecting a transit of between 0.6-1.0%.
So that is my answer 0.6-1%
The probability for higher mass stars is smaller and there are fewer of them. That is because although they are larger, the planets have to be much further away to be in the habitable zone (e.g. the transit probability for Earth is 0.4%). The main uncertainty is the probability of close-in planet occurrence for very low-mass stars where the transit probability can be much higher even though there are fewer host objects.


What is the formulat to compute King tracking rate for a given set of topocentric coordinates?

Too long for a comment. Possibly helpful resources:

http://www.bbastrodesigns.com/lib/
http://www.bbastrodesigns.com/lib/coordLib.js (file with actual formulas)
http://canburytech.net/DriftAlign/Equations.html
http://canburytech.net/GoQat/index.html (free software that implements this function)
"A manual of celestial photography: Principles and practice for those interested in photographing the heavens Hardcover – 1931" and it's 1988 reprint (ISBN 0933346468 for the latter, both available on Amazon), the original source.


From where so much of energy would have arrived as was required for such a giant universe to have come into existence?

I think it'd be helpful to explain what you mean by "have gone into the evolution of the Universe". Also, please stay assured that a lot of attention is going towards the shortcomings of big bang theory and other cosmological theories already :)

If by evolution of the universe you mean its expansion and accelerated expansion, then it does not require the energy equivalent of the galaxies. The reason is that it is already happening, without using their energy equivalent.

Where the energy came from: Probably one of the biggest physics and philosophy open questions, if not the biggest one. So, no answer can be given right now.

Last but not least, the energy equivalent of the galaxies is currently considered to be only a fraction of the total mass/energy in the universe. The rest being the unknown factor named dark matter and mainly the unknown factor called dark energy. The latter is actually related to the evolution of the universe (after the big bang of course). Its nature not known yet, but it clearly is not in the form of matter, unlike galaxies.


Could asteroid mining affect gravitational balance?

Something to remember is that the total mass of asteroids in the asteroid belt is tiny compared to the masses of the planets. Ceres, the largest body in the asteroid belt by a wide margin, is only 0.00015 times the mass of Earth, Ceres itself accounts for about a third the belt's mass:

Asteroid belt facts from ScienceDaily

I haven't been able to find any papers or articles that suggest the residents of the asteroid belt have any measurable perturbation on the orbits of Jupiter and Mars (the other way around is a different story!). If the asteroids that spend most of their time near Mars and Jupiter were to disappear, I doubt there would be any observable effect on the orbits of the planets.

Consider that comets and asteroids commonly pass close to Earth and the Moon and their influence on Earth's orbit is negligible.

As for what we would do with what we don't need from an asteroid, there are a few options I could think of based on the size of the target.
With smaller asteroids, if there is a risk that any changes to its orbit (such as an impact or accidental explosion) that would send it close to Earth, a complete controlled demolition of it might be warranted, or simply attach large chemical rockets to it and push it away. In the near-future, it is likely that even if we do start mining asteroids, provided we haven't dangerously altered its orbit, they will simply be abandoned afterwards.


Can there be a 2nd possibility deduced from the fact that galaxies are moving away from us?

Galaxies are moving away from us in every direction.

If it were like your race, galaxies at our same distance from the central point will be slowly approximating. That is, we will note that galaxies nearer to the center than us will move away from us (since they are faster), galaxies in the opposite side of the sky will move away from us (since they are slower), and galaxies in the plane that lies between these two point will be slowly approximating to us (since we converge with them to the same point at the same speed).

So the answer is no, that is not a possibility.


Do we know at which angle the Event Horizon Telescope will look at the accretion disk of Sagittarius A*?

We don't know the answer to this (or if someone does, I'd really like to hear that answer). Our galaxy's Super-Massive Black Hole (SMBH) is an unusually quiet one, with little to no accretion disk. If one indeed exists, and it almost assuredly exists, even if it is tenuous and small, we have not yet seen it.

Around 2013, there was a gas cloud designated G2 that was expected to be accreted which would have given us a lot of insight into the black hole and its structure. Unfortunately there was little to no activity observed. A Nature article about this gas cloud goes into more detail and talks about the accretion disk, as quoted below.

  The cautiously predicted flare-up would be a valuable probe of the environment immediately surrounding the black hole, about which little is known. What we do know is that the average rate of mass accretion driving the emission from Sgr A* is exceptionally small, thus explaining why our Galactic black hole is extremely dim. That we can detect it at all is due largely to its proximity — it is about 100 times closer to Earth than the nearest supermassive black hole in another galactic nucleus, that of the Andromeda galaxy. We also know that the accreting gas is heated so much that the individual gas particles almost all approach the speed of light, and that the observed emission emanates from a magnetohydrodynamic maelstrom. But the basic parameters of this small-accretion-rate regime are still being sought: does the accreting gas form a well-defined disk or an outflowing jet, as it often does in the more prominent cases observed in many other galaxies? Or is it better described as a more isotropic inflow comprising stochastically distributed density fluctuations?


In addition to the answer above, Pluto is a Neptune crossing trans neptunium object or TNO as it's often abreviated.  Being a TNO doesn't exclude the possibility of an object being a planet but crossing the orbit of Neptune does.   So, even if Pluto cleared out it's entire orbital neighborhood, it would still cross Neptune and naming it a planet would be problematic.

If it's orbit changed to being entire outside of Neptune, it would still face the problem of being one of several minor planets in that general region of similar size.

If you really wanted to push the theoretical to the limit, if Pluto's orbit moved so it stopped crossing Neptune's and if it merged with all the other similarly sized TNOs, then you could make an argument that it stands out as unique for it's region and a strong argument could be made that it should under those conditions, be called a planet, but that's a stack of improbabilities so high that winning next weeks mega millions top prize seems likely by comparison.

Also, if all the large known TNOs were somehow combined, they would still have a mass less than Mercury, though perhaps a slightly larger diameter, but that would be large enough where it just might get planet status.   It won't happen though.  Space is too big and object's orbits are too far from each other.   Planet formation happened within the proto-planetary disk at distances much closer to the Sun.   Uranus and Neptune likely formed much closer to the Sun than they are now.   There's not enough material and too much space in between for a planet to coalesce at Pluto distance from our Sun.


Can a star be massive enough to go 'supernova' yet not be massive enough to leave behind a neutron star? Or, perhaps vice versa?

This diagram from the Wikipedia article on supernovas mostly answers your question I think.

There is in fact an interval that is expected to not produce a remnant, but it is not at the lower end of the mass range, interestingly. Instead it's a result of the particular mechanism that triggers the supernova for these heavy stars, which blows the core apart entirely.


Below is an image of the solar system from that date. You can imagine drawing lines from the Earth to the various planets. You will see that Saturn, Mars, Venus and Mercury are quite closely aligned. And Jupiter is about 25 degrees out of that alignment. This matches with the photo that shows Jupiter further from the other planets.

Note that my image, derived from Nasa shows the orbits of the planets are not equally spaced. Your image does not represent reality.

The photo shows four planets quite close, and Jupiter a little way from them, just as the model of the solar system suggests. Of course the planets are in a straight line in the sky. They always are in a straight line, because they all are on the ecliptic. The alignment was significant because the planets appeared close together.

All the planets move around the sun in (almost) the same plane. The Earth also moves in this plane. The plane that the planets all move it is called the "plane of the ecliptic". Imagine a circle drawn on a piece of paper. If you look at the circle from above you see a circle. If you look at it from an angle you see an ellipse, and if you look at it from the side you only see a line. The Earth is in the same plane as the other planets, so all the planets appear to be moving in a line. This means that the positions of the planets in the sky will "always" be in a line, even when there is no alignment. The plane of the ecliptic looks like a line, because Earth is on the plane.

However when there is an alignment with the Earth (as there was in 2002) the planets will appear to be not only on the line of the ecliptic but close together in the sky.

Of course the planets are not actually close together, and there are no gravitational or other malevolent influence of planets being in a line


What programming languages does SETI (Search for ExtraTerrestrial Intelligences) most commonly use for signal processing?

A significant portion of SETI's work is done through a program known as SETI@home, which takes signals that have been received and processes them for responses, such as recurring patterns. Here is a page on how to download its source code. I grabbed the most recent nightly tarball and it turns out the program is coded in C++. I offer this as evidence that C++ is the most commonly used programming language used to process SETI's workload.


What might some examples be of radio telescope interferometric arrays made from "cheap components relying on massive computing power"?

I'm going to compare the Murchison Widefield Array (MWA) to the Green Banks Telescope (GBT) just to pick two rather big projects.

The first phase of the MWA featured 256 128 tiles and cost A$ 51 million.
About the MWA: http://www.mwatelescope.org/telescope
It saw first light in 2013.

edit: the first phase comprised 128 tiles only but I could not find out if the 51 million were the price for the first or the first and second phase.

The GBT cost US$95 million and saw first light in 2003.

To make it compare better: US$ saw an inflation of 1.27 within 10a making a total of USD 120 million. AUD is 0.68 USD bringing down the cost of the MWA to USD 34.5 million.

both telescopes got a lot of upgrades later.

It is difficult to compare but the GBT has a much higher receiver area while the MWA has a vastly greater baseline. They serve different purposes. From this point of view they don't seem to differ in an order of magnitude in pricing.

But the question was about "cheap" parts. Here the costs of the MWA are broken down (page 8). Only 12% of the total costs are related to the front end (i.e. antenna and beam steering electronics), the rest goes in equal parts to the infrastructure and the back end. Considering that in a design like the MWA a lot of signal processing is already part of the design it might even be cheaper in total when considering costs for later upgrades.


How Does Black Hole Evaporation Look From the Inside?

An infalling observer cannot remain stationary inside the event horizon of a black hole.

According to them, there is a short period of proper time before they meet whatever lies at the centre of the black hole (a singularity in GR).

If the observer looks back, they do not see an infinite amount of time pass by in the outside universe. In fact for the type of black hole where they might (a) survive and (b) have time to make any observations (i.e. a supermassive black hole) then they would see the outside universe only for a relatively short time longer than their experienced proper time and certainly for nowhere near as long as the Hawking radiation evaporation timescale. So the possibility that the black hole is (slowly) evaporating is of no consequence to their final moments.

The necessary material and diagrams to understand the previous paragraph have been covered before on both Physics and Astronomy SE.

https://physics.stackexchange.com/questions/82678/does-someone-falling-into-a-black-hole-see-the-end-of-the-universe

Would time go by infinitely fast when crossing the event horizon of a black hole?


Just how fast is a Fast Radio Burst thought to be?

The publication describing the original detection of the first known FRB (Lorimer, D. R., Bailes, M., McLaughlin, M. A., Narkevic, D. J. & Crawford, F.: A Bright Millisecond Radio Burst of Extragalactic Origin. arXiv:0709.4301) has a plot of the measurement that makes the effects of dispersion on this particular FRB nicely visible. Take a look at Fig. 2 in the paper.

The actual signal is less than 10 ms, while dispersion delays the signal by around 200 ms over the 200 MHz frequency range between 1.3 and 1.5 GHz (note that this relationship is nonlinear).

Your idea about algorithmically removing the effects of dispersion on the signal is regularly done in practice, search for "dedispersion". At our (hobbyist) observatory, we are using D. Lorimer's own sigproc package to do this, and it seems to be in widespread use amongst professional observers as well. The basic idea is to simulate a classical filterbank arrangement and shift each filter channel according to the DM.

From Lorimer et al. (cited above):

  Figure 2: Frequency evolution and integrated pulse shape of the radio burst. The survey data, collected on 2001 August 24, are shown here as a two-dimensional ‘waterfall plot’ of intensity as a function of radio frequency versus time. The dispersion is clearly seen as a quadratic sweep across the frequency band, with broadening towards lower frequencies. From a measurement of the pulse delay across the receiver band using standard pulsar timing techniques, we determine the DM to be 375±1 cm−3 pc. The two white lines separated by 15 ms that bound the pulse show the expected behavior for the cold-plasma dispersion law assuming a DM of 375 cm−3 pc. The horizontal line at ∼ 1.34 GHz is an artifact in the data caused by a malfunctioning frequency channel. This plot is for one of the offset beams in which the digitizers were not saturated. By splitting the data into four frequency sub-bands we have measured both the half-power pulse width and flux density spectrum over the observing bandwidth. Accounting for pulse broadening due to known instrumental effects, we determine a frequency scaling relationship for the observed width W = 4.6 ms (f/1.4 GHz)−4.8±0.4 , where f is the observing frequency.
  A power-law fit to the mean flux densities obtained in each sub-band yields a spectral index of −4 ± 1. Inset: the total-power signal after a dispersive delay correction assuming a DM of 375 cm−3 pc and a reference frequency of 1.5165 GHz. The time axis on the inner figure also spans the range 0–500 ms.


Has radio astronomy ever been done on objects that appear very close to the Moon? Is this avoided?

Yes, and lunar occultations have proved useful in several cases.

Hazard et al. 1963 used a lunar occultation to produce a high-resolution brightness profile of the now well-studied radio quasar 3C 273. Scheuer 1965 goes into a little bit of detail on general computations.

A slightly different tack was taken by Vedantham et al. 2015. They were attempting to create interferometric maps of the cosmic 21 cm line signal using LOFAR. Unfortunately, they needed a calibration source to be able to measure this "global signal". The Moon, occulting portions of the field of view, provided a calibration source as they observed that patch of sky.


Why do planets and satellites in the Solar system look so wildly different if they came from more or less the same matter?

This questions can be split in two; for planets and satellites.
The diversity of planets reflects in part the diversity in terms of chemical composition of the protoplanetary disk. We know that UV radiation from the sun can dissociate complex molecules or even very simple ones; for example, when UV rays split water molecules the result is free hydrogen and oxygen atoms. Since hydrogen is extremely lightweight they can be transported easily on the flux of stellar winds. So water, to keep with that example, if close to the sun could end up been dissociated and depleted from the region of the disk, but above the so called "snow line" UV radiation from the Sun was so weak that this couldn't happen as often and thus water molecules (which are very heavy in comparison to single hydrogen atoms), stayed there. That only explains the dichotomy between interior and exterior planets in terms of water content, and even then, some processes (like the late heavy bombardment) might add some water to the interior (as it happened on Earth). But this reasoning is not only for water, carbon dioxide, ammonia, methane and hundredths of different molecules have their own "frost lines". Closer to the sun the carbon can't be methane is a volatile gas that gets quickly pushed outwards, but at some tenths of AUs methane can remain in stable conditions and can even condensate into liquid droplets.
All of this just to say that the protoplanetary disk was NOT homogenous in terms of chemical composition, and was not homogeneous in terms of density or pressure. The thermal and chemical gradient across the nebula ensures some diversity and complexity for the entire planetary system.
Here you have a beautiful diagram showing how different chemical compounds could condensate in different temperatures and pressures on the protoplanetary disk.

Also accretion of planetesimals is more energetic closer to the Sun (meaning that breakups can occur more frequently and is difficult for a planet to grow large), while in the outer regions planets can increase in mass with regularity since collisions with other planetesimals are performed at lower relative speeds (because of how two similar orbits have a difference in periods that gets larger when you get closer to the Sun and thus larger relative velocities). This coupled with the gravitational interactions of the protoplanets and the early disk (see planetary migration and nice model etc...) allow for different accretion rates and the accretion of materials of different composition of what was found in the original place of formation of a particular planetesimal. This also helps to keep a wide variety on planetary masses.
A wide variety in planetary masses is the starting point for a larger variation as the planets evolve in time and diverge from their initial conditions. A rocky small planet (Mercury) might have less heat trapped inside than a larger one (Earth), due to the smaller energy released by smaller accretion rates. Thus it might get cold rapidly and a magnetosphere due to a melted interior can't happen. The absence of a magnetosphere allows for solar wind charged particles to erode your atmosphere by sputtering. Instead on a planet like Earth, the larger mass has led to a melted interior that in turn generated a magnetosphere that lasted for billions of years, in Mars it lasted some time but now is almost gone so the atmosphere has also been almost destroyed. On Earth the presence of an atmosphere leds to all sorts of chemical erosions, and phenomena. Also, it's melted interior coupled with the specifics of its chemical composition and thickness of the crust allow for a mechanism called plate tectonics. Tectonics can't happen on Venus because the crust is not as thick (due to different composition) and thus it doesn't break apart in plates but just deforms and folds in a complex behaviour that is unique to Venus.
Also collisions with planetesimals can alter the future evolution of similar planets. Venus was probably very similar to Earth (similar mass, very similar composition and not so dissimilar temperatures as one might think) but their paths diverged completely as tectonics on Earth recycled the lithosphere and on Venus the carbon dioxide got more trapped in a greenhouse effect, and because Earth had a collision with another planet that have us our Moon, which is a mechanical stabilizer while a random collision with Venus (with different impact parameters) led to an extremely slow rotation and long days (but no moons). Longer days mean different insulation, and that changes drastically the climate of a planet. On Mars days are similar to Earth's but since it is smaller and the atmosphere has gone many things are very dissimilar from Earth. Also, Mars has no tectonics (the crust is thicker than Earth's and moves like a monolithic object) so there are few volcanoes and they grow huge (while on earth a single hot-spot generated by a mantle plume drills several holes on the crust as the plate moves on what is called a volcano chain) and there are stress faults like Valles Marineris (unique in the Solar System) that on Earth would have been relaxed by tectonic movements.
To see how different the evolution of two planetary objects can be just by making them different mass take a look at our Moon. It has the same chemical composition (it is a chunk from Earth in fact), it is basically at the same distance to the Sun as Earth, it lives in the same interplanetary environment (same solar radiation, solar wind, impact rates etc...), and still it is completely different. This is all due to mass! The moon can't retain a large atmosphere as Earth because it has less gravitational pull. The same temperature for our atmosphere there means that the particles reach escape velocity easily and start to escape from the gravity well. With no atmosphere, not internal heat, the moon lacks almost any type of erosion in across billions of years of evolution. Processes of erosion on Earth have made the diversity of geological formations to explode in comparison to those found on the Moon. Even then the moon has its own peculiarities and dynamical features unique to it.
Now we are getting closer to the satellites question. They in fact should look almost the same, since they are formed from very very similar material in extremely similar conditions. And indeed we believe that moons originally were very similar (for example the 4 Galilean moons). But Io is to close to Jupiter and the other moons interact with it in such a way that the geological processes are completely different. Water and volatiles evaporated quickly as it heated up by tidal forces from Jupiter. These tidal forces were not as strong in Europa since it is farther away, thus it only melted part of the icy crust creating an ice analog of plate tectonics which generated a plethora of diverse formations. Satellites evolve. Enceladus shoots jets because of tidal interactions and orbital resonances with other moons. Some moons like Japeto have a dual colored surface because of the material sprayed by Enceladus landing on on one of its sides. Some moons like Triton have nothing to do with the other because they formed in another region of the solar System and later got trapped by the gravitational pull of a planet (Neptune in this case).
As I mentioned before. Atmospheres (density, composition and pressure) are largely dependent on the mass of the planet or moon. Look at this graph:

It shows the velocity of gas molecules in relation to temperature of the gas. For larger temperatures gas molecules move faster. In a planet with low mass, the escape velocity is lower than one with a larger mass. Thus a planet closer to the sun (at higher temperature) needs to be larger in size if it wants to preserve the same gas molecules in its atmosphere as a planet that is farther away (colder). You can see why Earth's atmosphere could trap and retain water, oxygen, carbon dioxide, ammonia, methane nitrogen and other gases while it is unable to trap hydrogen and helium (because they are lighter and thus for the same temperature they can move as fast as needed to escape Earth). Meanwhile, the Moon, which has the same heat coming from the Sun as Earth, since it is less massive it can't retain almost any gases (maybe a bit of Xenon). Titan, is a huge moon thus it can retain many gaseous molecules like Nitrogen and Oxygen (those in turn make the pressure high enough to retain also volatiles like methane in a liquid form on the surface). But why doesn't Ganymede has the same atmosphere as Titan if they are basically the same size? Because Ganymede is closer to the Sun, larger temperature means that the molecules move faster and thus they escape its attraction easily.
As you can see the complex processes of atmospheres of a moon or a planet change everything (erosion, recycling processes, chemical corrosion, etc...) and in turn that diversity in atmospheres comes from a diversity of masses and distances to the Sun.
I think the Solar System is a chaotic system, dynamically, geologically, chemically etc... Chaos means that for an small difference in initial conditions the system will evolve in exponentially diverging different states. Planets and moons might have started as similar objects but history and the chaotic dynamics of the system have evolved into completely different environments. Not only that, but the truth is that planets didn't started as equals but were very different from the beginning, so imagine how far away is Venus to become a Titan, or an Io to become an Earth.
Also there are processes and conditions that are specially well suited for divergence. For example: Earth is very dynamic while Mars, Venus, Mercury, the Moon and others are totally not. Why? because on Earth water can exist in 3 different states of matter. We can find liquid water, water vapor and ice in different regions and seasons. And that is because Earth is at an average temperature and it's atmosphere has just the right pressure to allow this. Earth's conditions are very close to the triple point of water (where all three states of matter coexist), that's why we have a water cycle on Earth, with rivers and glaciers eroding the landscape and clouds regulating the climate.

Mars, Venus, Mercury, all have temperatures and pressures were this can't happen, not only on water but on many compounds present there. You know where this can happen? On Pluto! This was very surprising, Pluto shows a variety of terrains and geological features that exceeds all expectations. Now we know this is because Pluto is extremely dynamic (as Earth) and a lot of erosion and geo-chemical processes can occur, but this is not because of water (since Pluto has low pressure and low temperatures), but because of Nitrogen and Neon! Both elements have their triple point inside to Pluto's range of conditions and thus Neon rivers, Nitrogen glaciers and hazes are expected on this dwarf planet.
It is indeed an interesting question. How incredible are the laws of nature that allow for extreme variety even between brothers. I wonder how a planet around any other star might be, our simplistic categories of hot Jupiters, mini-Neptunes, super-Terras, etc... are just so primitive and restrictive. What wonders await us in this complex and diverse cosmos is beyond our comprehension.


Is there or will there be any search after a primordial black hole 500 au from the Sun?

Maybe, Harvard scientists have proposed a way to determine, once and for all, whether Planet Nine actually could be a black hole. Specifically, the new method would scour the outer solar system for evidence of telltale flares that are emitted when a black hole devours a comet or other distant object. Such flares, they say, should be detectable by the upcoming Vera C. Rubin Observatory in Chile, which is expected to begin a 10-year survey of the southern sky within the next few years.
I suppose if funding is available for this it will be done when the telescope comes on-line. This may solve the problem of if it's a black hole but a large planet could still be possible but unknown.
Ref: https://astronomy.com/news/2020/07/is-planet-nine-a-black-hole-or-a-planet-harvard-scientists-suggest-a-way-to-find-out


Second Telescope: Maksutov Cassegrain vs Refractor

I have viewed through many telescopes over the years and in doing so have met many people like yourself. To suggest a F/12.5 Maksutov will satisfy you would be the equivalent of telling someone to dive into the deep end of a swimming pool without knowing if that person can swim.

If it's bang for buck that motivated your question I think you would get the most if you just explore the capabilities of the 3 inch refractor for a while. There is a lot to be seen with that telescope and you have barely scratched the surface.

Some prefer electronic eyepieces. Have you seen what they are capable of doing? I viewed with eyepieces for fifty years before trying a Mallincam ten years ago. I haven't used an eyepieces for serious viewing since.

Take a look at the NightSkiesNetwork. It is a group that regularly broadcasts their observing sessions.

https://www.nightskiesnetwork.ca/about

I cannot think of any better advice for someone like yourself than to stick your toe in the water for a few seconds before jumping in.


If an asteroid was to collide with Earth, how much warning would we have?

Asteroids come in all shapes and sizes, and generally the bigger they are the easier they are to detect.

Small asteroids, from a the size of sand grains (properly called meteoroids) to a few meters across hit the Earth everyday without causing harm and are undetectable until they hit (we call the trace as they hit the atmosphere a meteor) They cause no damage. For such objects we get 0 hours warning.

Somewhat larger objects can cause some local damage For example the Chelyabinsk asteroid caused damage from broken glass in Russia. Objects this large may be detected a few hours to a few days. The Chelyabinsk asteroid wasn't, and hit with no warning. We have observed some small asteroids just prior to them hitting the atmosphere.

We have found large numbers of objects whose orbit can bring them close to Earth. Every object we have found has an orbit that misses Earth.  With each new object that is located we can forecast its orbit for several centuries. However new objects are being found all the time. It is possible that a object will be discovered that has the potential to cause damage to a wider area. If the object happened to be on an orbit that had it collide with Earth then there could be less than a year's warning. That is very unlikely. It is slightly more likely that an object will be discovered to be on a collision course only after several orbits, in which case there would be years or decades warning.

Very large objects that could threaten the survival of humankind are rare. There are some such objects, (larger than Florence). We hope that we have now detected all very large near Earth asteroids, and discoverd them to be non-threatening (like Florence is not a current threat to Earth)

Finally a comet may be on a collision course with Earth. Comets that are spotted are usually detected a few months before they could threaten Earth. But comets are much rarer than asteroids.

In summary, between 0 hours and a century, with several decades being likely for an object that could do widespread damage.

The particular object referred to in the video is Apophis. It is roughly 400m asteroid that is not going to hit the Earth for as long as we can forecast. It will, however, pass close by in 2029, and again in 2036. In 2005 the uncertainty in orbit was such that we could not rule out a collision in 2036. Further observations in 2006 and after allowed and a more accurate orbit to be calculated, and showed that it would certainly miss.


Conventional matter to dark matter ratio, outside of galaxies

Let me see if I can answer at least some of this.

Yes, there is dark matter between galaxies. This is demonstrated by the fact that in galaxy groups and clusters, you need more dark matter than is found in the galaxies themselves to explain what's going on: in terms of why the groups or clusters are gravitationally bound in spite of the extreme velocities of the galaxies within the groups/clusters; why the X-ray-emitting gas in clusters is so hot and high-pressure and yet still confined to the cluster; and in order to explain the gravitational lensing of background galaxies by clusters. (Also supported by observations of things like the Sunyaev-Zeldovich effect in clusters.)

(In fact, the oldest evidence for dark matter was the extremely high velocities of galaxies within clusters as measured by Fritz Zwicky back in the 1930s. This is also one of the failure modes of Modified Newtonian Dynamics as an alternative to dark matter: even if you think you can get rid of dark matter within galaxies, you still end up needing dark matter within galaxy clusters, in addition to the galaxies themselves.)

And, of course, cosmological models indicate that dark matter accounts for about 25% of the critical density of the universe. In order to have that much dark matter, you have to have more dark matter than what's inside individual galaxies.
"If there is dark matter between galaxies. Do we know or believe: what is the ratio of dark matter out there?" I think the simplest way to answer this would be to take the dark-matter/baryonic-matter ratio for the universe as a whole, as determined by current cosmological observations and models. This gives a ratio of about 5:1.
"Galaxies exist in filaments and walls. Does dark matter do this? or is it evenly-spread?" Yes, dark matter does this. The reason we see galaxies in filaments, clusters, and walls is because the dark matter has gravitationally clumped into those structures; on this scale, the galaxies are basically along for the ride, following the gravitational pull of what the dark matter is doing. The reason for galaxies existing in filaments is because of the uneven distribution of the density of both regular and dark matter throughout the universe. An overdensity leads to the matter collapsing to form (for instance) a galaxy, but the overdensity can be asymmetric. This leads to the collapse happening faster in one particular direction, causing galaxies to form in filaments (see e.g. this summary paper for an overview).

This page has images from a large-scale cosmological simulation (the Millennium Simulation). Most of the images showing filamentary structure on different scales are showing the dark matter only.


Can we measure/detect GR light bending during solar eclipse using ordinary equipment?

The deflection you are looking for is $\sim 1.7$ arc seconds, which should be within the seeing limits at a good site. It should also be within the resolution limits of a telescope with $\ge200$mm diameter primary.

Having said that you will not be able to see it with the naked eye, you will need to be able to photography the star field during the eclipse and again when it is visible at night and then do some arduous plate measuring and data reduction to detect the shifts of stars in the field.

So amateur equipment is up to detecting the bending of light around the Sun in a total eclipse, but if by witness you mean see the stars move with the naked eye the answer is almost certainly no.


When was the atmosphere of Venus first observed?

The discovery is attributed to Mikhail Lomonosov, who inferred the presence of the Venusian atmosphere from observations of the transit of Venus in 1761, although this has been challenged by Pasachoff & Sheehan (2012) who argue that Lomonosov's observations do not match the phenomena observed during the 2004 transit, arguing that Lomonosov's telescope was not capable of detecting the Venusian atmosphere. They suggest that the strongest claim for detecting the atmosphere was by David Rittenhouse who observed the 1769 transit, although they assign the greatest credit to Johann Schröter (1796) who also took into account out-of-transit observations of the ring of light around the planet at inferior conjunction.

This challenge was itself challenged by Koukarine et al. (2013) who observed the 2012 transit of Venus using antique refractors, concluding that Lomonosov's experimental procedure was sufficient to detect the atmosphere and support the attribution of the discovery to Lomonosov.

An in-depth summary can be found in a review article by Vladimir Shiltsev in the Journal of Astronomical History and Heritage (2014).


Jupiter, Saturn, Uranus, Neptune system GM

In the JPL report on ephemerides DE430 and DE431,
on page 49, table 8 gives the mass parameters they used for the major planets.
They cite the
IAU 2009 system of astronomical constants.
There may be newer estimates, but to verify your own simulation against HORIZONS, you might as well use the same constants they did.


None. and that doesn't surprise me much.

Comets are thought to be Oort-cloud objects which by interaction either with other Oort-cloud objects or with objects passing the Solar system have been perturbed to venture into the gravitational reach of the outer planets, whence they are
flung into the inner Solar system via close encounters (gravitational slingshots) with one of the outer planets, in particular Jupiter and Saturn. For such a slingshot to be efficient, the encounter must be close when any binary (proto-)comet will most likly be disassociated by the gravitational tides of the sling-shooting planet.


What is the distribution of dark energy?

The actual model says that dark energy is distributed uniformly around all the universe and as you probably know it expands the universe with him so it stays distributing uniformly, there are no clouds or irregular forms because that wouldn't explain the actual universe model. Maybe in the future that change.


Could a rocky rogue planet get trapped in the orbit between Earth and Mars?

It is exceptionally unlikely.

Imagine that there was a planet already there. What could cause it to suddenly escape from the solar system?  It would have to be a massive event, such as a second passing rogue planet catapulting it out of orbit.

Running time backwards, for a rogue planet to be captured and neatly end up orbiting between Mars and Earth would require an unbelievably unlikely sequence of events, such as

  two rogue planets happen to pass between Earth and Mars at the same time and the interaction between the two results in a transfer of energy and momentum, which (by lucky chance) leaves one of the planets in a nearly circular orbit, (while the other escapes)".

Given how big space is, and in consequence, how rarely a rogue planet enters the inner solar system, this would make the above sequence of events practically impossible.

In this unbelievable scenario, the effects on Earth and Mars would depend on how closely the new planet passes. If we are close enough for significant effects on the planet, we are in a great deal of danger.


Do astronomers have an established, systematic way for saying what does or doesn't orbit what? (e.g. "Mars orbits Earth")

It is possible that there is nothing "official", there is just the technical use of language. For example Phil Plait notes that He incorrectly used the word "orbit" for the motion of the Hayabusa probe, which does not orbit the asteroid Ryugu. But hovers over the surface.

So what does Phil mean when he talks about orbiting?  I summarize the meaning thus: Object A is in orbit around object B if A moves around B (ie the True Anomaly of A increases from 0 to 360 degrees in fairly regular fashion) principally as a result of the gravitational field of B.

So although Mars does move around the Earth, its motion is not principally due to the gravitation field of the Earth, so it is not in orbit.  However the moon's motion is principally due to the gravitational field of the Earth. It is in orbit around the Earth.

The moon is also in orbit around the sun: It moves around the sun and the reason for this motion is principally the graviational field of the sun.

A body at the Earth Sun L2 point is not in orbit around the Earth (it is fixed relative to the Earth) It is in orbit around the sun. A body in the Earth Moon L2 point (which is probably very unstable) is orbiting the Earth and orbiting the sun. Its motion relative to the Earth is primarily due to the Earth (and secondarily due to the moon). Its motion relative to the sun is due to the gravity of the sun.

Cruithne is in a 1:1 resonance with the Earth, but its postion is not primarily due to the gravity of the Earth. It is in orbit around the sun, not the Earth.

Hayabusa is not in orbit around an asteroid, it is maintaining position by thrusters. Similarly, A Jumbo jet is flying not orbiting, it's motion is not mainly due to gravity.

This idea can be sharpened by the notion of a Hill sphere.
Inside the Hill sphere, a body will orbit. Outside it, the body will not orbit.


What is the relevance of the Tropics and the Polar Circles? Do they Exist on other planets?

The Tropics (Capricorn in the South and Cancer in the North) are the furthest point from the equator where the sun reaches a point directly overhead at any time during the year. The Arctic and Antarctic circles are the points at which the sun remains above the horizon for a full rotation of the Earth (i.e. 24 hours), commonly called "Midnight Sun". Any planet that has axial tilt while have similar lines.


Can I see Jupiter, Mars and Saturn through the National Geographic 76/350 Dobsonian Telescope?

I see that you have correct data for the magnification. Possible magnifications: 18x, 36x, 58x, 117x. The rule of a thumb is, that for optimal observing of the planets you set the magnification of 20x to 30x for every inch of the aperture. Your aperture in inches: 3. That means, that the optimal magnification is from 60x to 90x. But this rule applies to telescopes with better optics. I don't know how good the mirrors are, but I would recommend the setup with 6 mm eyepiece without a Barlow (= 58x). With this setup, you can observe the disc of the Jupiter, but also the rings of the Saturn. You will be able to see main craters and mare on the Moon.
If you place the Barlow on the scope, the magnification is too high for the small aperture. You should consider it only when observing near double stars. Magnification of 117x won't be useful with planets. Why? Look at the icon in the title bar of this Stack Exchange. When it is small like it is now, you can resolve the big letter A. But if you expand it to the whole screen, you would see only pixels and some form of A. Smaller picture is better, but not too small. Let this rule guide you.

So what is the answer? You will be able to see the Moon and its mare and some craters, but also Saturn with his rings. Jupiter's disc will be seen with ease, but harder will be Mars just on peaceful nights. What about Venus? Maybe better than Mars, but just small Moon like form. Mercury will be visible only as small dot. Neptun and Uranus are too dim to be observable with this telescope, but maybe you will be able to see them with some experience after 1 or 2 months (only like normal stars, maybe blueish). If Pluton is planet for you, it is far beyond the range of your scope. Even I couldn't see it never with my aperture of 250 mm.
But planets aren't everything. If you have budget for buying the solar filter (cheap ones for 10 dollars), you could point the telescope towards the sun, but beware! Don't mimic Galileo Galilei, who was observing the sunspots throught the telescope when the sun was low without solar filter! Check the page https://spaceweather.com/ for sunspots. Also, you could look at the globular clusters with the smallest magnification. Try Andromeda galaxy! It won't have some visible form, but it will be some cloud of brighter region. You could try also nebulae, for example M 57 in Lyra.
You see that you can do a lot with this telescope, but it won't be some spectacular sight and these are the only things you could observe. For more, you have to buy the better telescope. Hope you will have fun with it!


Does the event horizon of a black hole increase or decrease by adding mass?

The radius of the event horizon ($r_\mathrm{s}$) is directly proportional to the mass of the black hole (M). More exactly:

$$r_\mathrm{s} = \frac{2 G M}{c^2}$$

The black holes whose merger was detected by LIGO would each have been about 90 km in radius, and after merger, a little less than 180km.

Read about the Schwarzchild radius


Can expansion of space blueshift galaxies with respect to one another?

Expansion will still redshift the light from these galaxies.

The universal expansion, at least by my understanding, isn't so much a force that pushes these galaxies further apart, but instead creates additional space everywhere, constantly. This includes the space between the galaxies that are approaching one another.

The light travelling from one galaxy to another will be affected by this expansion, since the space that light and matter occupy is also expanding. The result is that light "stretches" as it travels, lengthening its wavelength, because the space between one wave's crest and the next is expanding. Since that light continues to travel at a constant speed, these crests will never speed up or slow down to "catch up" with their original wavelength, so the lightwave gets stretched into a longer wavelength with lower energy - that is, it gets redshifted.

The light will still be blue-shifted by the galaxies' movement toward one another, but the Universe's expansion will not push them together faster. Rather, the Universe's expansion will universally (heh) increase the distance between all points in space, including the space between those galaxies. This expansion rate may not stop these galaxies from reaching each other, but it will always serve to increase the distance that they must travel to do so. Because of this, the Universe's expansion will always redshift light, just to a greater or lower degree based on initial distance.

For further clarification, the fact that there is a larger amount of expansion happening "behind" these galaxies has no effect on the expansion of the space between these galaxies. It's not so much a force that pushes all matter away from each other, but the literal creation of more space at all points in space all the time.

Creation of vs expansion of space

Now, there seems to be semantical ambiguity on this subject (or I may simply have an insufficient understanding of the subject - a very likely scenario) concerning the idea of dark energy as a force. Many interpret it (I think) as being a force that pushes matter away from other matter, but I don't think that's what it is. From my understanding, it seems more like a force that pushes space itself away from the surrounding space.

That is, if you were to somehow identify two "adjacent" points in space, their dark energy would push them away from each other. This of course means that those imaginary adjacent points are no longer adjacent, as there is now space between them. We can now identify new points in space between the two formerly adjacent points (note that in practicality, it is impossible for two points to be adjacent in this sense, as we can always divide the space between them further. This example is just a thought experiment).

Following this thought experiment, we find that, counterintuitively, this dark energy remains at a constant density. That is to say, these points are pushing the surrounding space away at the same rate as before, in spite of now being more spread out, and the "new" points between them are also now pushing the space surrounding them away at the same rate. Now, if the density of dark energy remains constant while the space it occupies is expanding, that can only mean that there is now more dark energy between the two original points than there was before.

This is why I like to say that space is "created" rather than simply expanded. It makes it easier for me to conceptualize. If space is thought of as the thing that creates dark energy (whatever it is), and that energy doesn't lose its density when space expands, then it makes sense to me to say that there is now more space there than there was before. To be honest though, I don't actually know what terminology is primarily used among those who regularly study this, just that this way of thinking makes it easier for me to conceptualize.

Disclosure: I am not an expert or even particularly knowledgeable on this subject. If I have made mistakes here, others should feel free to offer corrections. My answer is based on my own understanding gained from my own limited studies of this subject, and I won't pretend to even begin to understand the mathematics behind any of it.


How do we know Nemesis is not a black hole (or neutron star)?

If the Sun had been born in a relatively wide binary system with a star that was to become a black hole or neutron star via a supernova, then (a) it is quite likely that such a system would be disrupted by that supernova and we would not be in a binary system now; (b) there should be evidence of the supernova in the form of very high abundances of the daughters of certain short-lived radionuclides incorporated into solar system material. There is some evidence of the latter, but not I think enough for the Sun to have been in a binary system with such a star (though I might be checking this).

An alternative argument is that the Sun is captured in orbit by the stellar remnant at a later date. This avoids the supernova problems, but the capture process is inherently unlikely in our Galaxy once stars have left their birth environments, especially capture which is tuned rather precisely to yield just less than zero for the resultant system potential energy of a very wide binary. Capture by a "normal" star would in any case be much more likely than capture by a relatively rare compact object.


Brown Dwarf transcending past the sun with the naked eye

Possibly an aircraft or flock of birds, or maybe an insect really close to your eye, but more likely an illusion or misperception. It's not very "bright" to look directly at the Sun with your naked eye!


What is the orientation of planetary orbits?

No, the planet orbits do not have their major axis aligned.

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

You have Inclination and Longitude of the ascending node defining the plane, and Argument of periapsis defining the ellipse orientation in that plane.


Can there be a three-moon system where only two are visible most of the time?

Start with some basic calculations.  At any time, you can see a hemisphere, which is (gosh! :-) ) half of all space.  This means that no matter what you do the average presence (lit or not) at any time is  1.5 moons.  Since there's only one "sun," all moons present in the observable hemisphere will have an equal amount of the observable face illuminated.

Now, if we decide to have two of the moons orbiting each other in a binary fashion (and their net center is orbiting the planet), then 50%  of the time  these two moons will be in your visible hemisphere, and 50% of that time so will the third.  Conversely, 50% of the time these two moons will not be in your visible hemisphere, and 50% of that time neither will  the third.

All of the above assumes orbits similar to our Moon, i.e. roughly equatorial, around our own planet's rotational axis.  What if these 3 moons somehow manage to be circumpolar?  In this case they'll be permanently "quarter moons," visible whenever in our hemisphere of view.  Now make the third moon circumpolar with such huge ellipticity that it's not visible to the naked eye for most of the orbit.

So the only thing not possible in either  of the above scenarios is managing to make the first two moons visible for "most" but not "all" of the time.   Perhaps someone else can come up with a more esoteric solution.

Note: all this assumes the moons are small enough that their interactions do not lead to short-term chaotic behavior.


Heavens-Above shows two fairly bright objects passing over Nuremberg at that time on 2020-04-14:

a Resurs-O1 rocket moving south to north
an Iridium-4 rocket moving north to south

The next visible ISS passes at your location are between April 28 and May 3, all before dawn.


I'll venture a guess that the 7000 kilometers (your software's camera position) was with respect to the center of the Earth. Using 6371 km as a value for the Earth's radius, that would make the altitude only 629 km, which it turns out is quite a popular altitude for satellites in LEO.

For the simplifying case of midnight, one can draw a simple diagram. I get

$$h = R_{\textrm{Earth}} \left(\frac{1}{\cos(35°)} - 1\right)$$

which is about 1400 km at midnight, and the height will be lower before and after.

I'll wait to hear about the camera "height" and try to find another tool for you to use. Do you ever use Python programming language?

Red dots are Perth at local noon and midnight.


Is dark matter influenced by gravity of baryonic matter?

Dark matter attracts baryons not only by a "kind of gravity", but simply by gravity. Everything that has mass (even if it's only relativistic mass like light) attracts everything else by gravity, and thus baryons also attract dark matter.

The reason dark matter "organizes" galaxies is just that there is much more of it, roughly 7 times more, to be specific. So, on large scales, the dynamics of galaxies are dominated by dark matter. But because baryons can dissipate energy, it may collapse in clumps — such as molecular clouds — which are hence dominated by baryons.


Approximating density of hydrogen in [observable] universe

So to estimate the answer, you need to convert the mass density to number density by $n_H = \rho_H / m_H$. After getting the current number density you need to redshift it to finite $z$, here $z=6$. Keep in mind the density of hydrogen evolves $\sim (1+z)^3$ due to the expansion of the universe.

That said, the assumption 'let universe be completely made from hydrogen' renders this question highly unrealistic, hence I agree with @James K that this is not real world astronomy. In the real world, only about 5% of the total energy is made of baryons, among which Hydrogen abundance is $\sim 75\%$. Also, at redshift $z\sim 6$, star formation etc. lowers the neutral hydrogen gas density, c.f. recent observations here and here.


Does Uranus have a stronger gravitational pull on its moons than Earth does?

The "strength of gravity" (ie gravitational acceleration) is determined by the mass of the planet and the distance between the planet and the moon. The formula is
$$GM/r^2.$$

For the Earth (mass= 1 Earth, distance to moon=390000km) the acceleration is $0.003 ms^{-2}$
For Uranus and Titania  (mass =14.5 Earth, distance=435000km) the acceleration is $0.03 ms^{-2}$

In summary, the gravitational pull of Uranus on its moon Titania is 10 times greater than the gravitational pull of Earth on her moon.

The same formula applies to surface gravity:

For the Earth, (mass = 1, radius = 6370km) the acceleration due to gravity on the surface is $9.8 ms^{-2}$
For Uranus (mass =14.5 Earths, radius = 25400km) the acceleration due to gravity is $9.0ms^{-2}$.

The surface gravity on Uranus is less that that of Earth, due mostly to the relatively low density of Uranus meaning that you are much further from the centre of the planet when you are at the surface.

(These value vary due to the neither the Earth nor Uranus being perfectly spherical, and the effective gravity is also lower due to centrifugal effects)


Dew-heater bands (straps) do come in sizes even down to 1.25" and 2" eyepiece. Multiple vendors make these.  Dew-straps are something of a standard in that they are resistive heat-straps designed to run on roughly 12v power.  The typically use the (RCA) tuplip type plug.  The dew-strap and dew-controller typically do not need to be made by the same vendor to be compatible.
Check vendors such as AstroZap, Dew-Not, Kendrick Astro Products, and others.
Depending on your dew controller, you may need a controller with additional channels.  Multi-channel dew-controllers allow you to set different power levels for each channel (or at least for groups) since the amount of power needed to heat, say, the corrector plate on a large telescope is much more than the power needed to heat a tiny 1.25" eyepiece.
If you have adequate power, you can use a hair-dryer to defrost an eyepiece (although I prefer the dew-heater strap to prevent fogging.)


Why has Moving Cluster Method been successful only for Hyades?

Do you know this paper? Mamajek (2005). "A Moving Cluster Distance to the Exoplanet 2M1207b in the TW Hydrae Association".

I'll risk an opinion:
There is about a thousand clusters with kinematic measurements, but besides no more than half a dozen, all pm components are under 20 mas (milli-arcseconds) with errors more or less in the range 0.3-6 mas. Such errors are too high in relation to the pm values and as a result, the vectors will not converge.


The moon's rotation plane would precess (more) in the same direction as Earth's rotation, if the Earth spun faster.


Are the most distant known objects in the universe more than 14 billion light years away?

According to Wikipedia, GN-z11 has a redshift of 11.09, which, according to Ned Wright's cosmology calculator, corresponds to a light travel time of about 13 billion years and a comoving radial distance of about 32 billion light years.
The comoving radial distance is the distance measure with respect to which the diameter of the observable universe is about 46 billion light years, so that's the one you want in this context.
A lot of people don't understand the implications of cosmological expansion and think that if the light travel time is 13 billion years, then the distance to the galaxy must be (or must have been) 13 billion light years. They're sure enough of it that they'll report that distance as fact, without disclosing that they calculated it themselves. As a result, distances in popular articles generally can't be trusted. You can trust the redshift if it's reported, and you can probably trust times (either light travel time or time since the big bang).


Is the shape of observable universe and shape of space same?

When talking about curvature in (General) Relativity one usually refers to the metric interval and how this depends on the space-time coordinates.

For example, the metric interval in Minkowski space can be written
$$ ds^2 = c^2\ dt^2 - dr^2$$

This is an example of both flat space and flat space-time. It is flat space because if I want to find the distance between two events happening at the same time $t$, it is simply the integral of $dr$.

Similarly I can ask what the time interval is between two events and this is just given by the integral of $dt$.

A counterexample is seen when using the Schwarzschild metric, which is appropriate in the space outside a spherically symmetric mass. Considering only the radial and time components, the metric is
$$ds^2 = c^2 (1 - r_s/r)\ dt^2 - (1- r_s/r)^{-1}\ dr^2,$$
where $r_s = 2GM/c^2$ is the Schwarzschild radius associated with the mass $M$.

Now there is no simple relationship between $dr$ and the spatial separation between two events that occur at the same $t$, we must do an integral. Similarly, to work out the time interval between two events at the same $r$ we must do a more complicated integral. Both space and space-time are curved.

The evolution of the universe as a whole is governed by the Robertson-Walker metric.
$$ds^2 = c^2\ dt^2 - a(t)^2\frac{dr^2}{1 - kr^2},$$
where $k$ is a curvature parameter that may be -1, +1 or zero.

At the moment, the best measurements are consistent with a zero curvature - that the universe is spatially flat. That does not mean it is flat on much large scales than we can observe - that is still possible. A curved universe that was exponentially "inflated" shortly after the big bang may well mimic a flat universe even though the curvature is finite.

Note though that because the universe is inherently "lumpy", this spatial flatness is only true on scales much larger than planets, stars, galaxies and even clusters of galaxies.

Whether you could call space-time flat in a spatially flat Robertson-Walker universe I doubt. The reason for this is the time-dependent scale factor $a(t)$, which ensures there is a coupling between the $r$ and $t$ coordinates. It is this coupling that is responsible, for example, for the cosmological redshift of light.

A technical answer would need to look at the Einstein Tensor which characterises the curvature of space-time caused by mass and energy. In the Robertson-Walker metric, this tensor definitely has non-zero components, so you would have to say that even if space is flat, space-time is curved.


What was the definition of a planet before August 24, 2006?

I think this deserves a bit more of an answer because it's not quite so simple.  Yes, it's true that there was no formal definition of what a planet was prior to the 2006 vote, but essentially everyone knew what was and wasn't a planet, just as we know what is and isn't a star, though the boundary of when it stops being a very heavy gas giant planet and stars being a brown dwarf star, or likewise, the boundary between hottest brown dwarf and coolest red dwarf is basically arbitrary.   Definitions are useful but at times, inevitably arbitrary.
I'm reminded of Plato when he said

"Man is a featherless biped"

His students accepted the definition until Diogenes plucked a chicken and brought it into Plato's Academy, saying,

"Behold! I've brought you a man."

After this incident, Plato added "with broad flat nails".
Little did Plato realize that there were some bipedal apes that still qualified as human with his new definition.
A more recent definition that Man is the only animal that fashions his own tools, goes back to Thomas Carlyle, and it worked for a while, until it was pointed out by Louis Leakey, based on observations by his protégé, Jane Goodall, that some chimpanzees fashioned and used tools of their own making.   So, back to the definition drawing board.
The irony to both these stories is that no person of any intelligence would have confused a plucked chicken or a chimpanzee using a tool with a human.  We didn't need either definition, though, to be fair to Thomas Carlyle, I think his quote reads as much about praise for the benefits of making and using tools more than it's meant as an infallible definition of man.   My point?  Definitions can be problematic.
Lets consider the case of the comet, I mean, planet, I mean asteroid, I mean, dwarf planet Ceres.  It's been all 4 of those things depending on time period.  Ceres was a planet from 1801 until the 1850s when it was demoted to asteroid.  The largest asteroid, but still, one of a few large asteroids in the asteroid belt.  It wasn't just Ceres smaller size that did it in, but also, the discovery of other good sized bodies in the same general region of space.  These larger asteroids were closer to Ceres in size than Ceres is to Mercury.

And what about Mercury?  Surely Mercury has more in common with Ceres than it does with Jupiter and Jupiter and Saturn have moons larger than Mercury.
That said, neither Jupiter nor Saturn's large moons are even half the mass of Mercury.   Mercury also has a large Iron core.  It's core alone is more massive than Ganymede.    Mercury has a lot going for it that Pluto does not.  There are at least dozens, possibly hundreds of bodies in the solar system similar to Pluto and Ceres, but there are just 4 known bodies similar to Mercury. Mercury checks a lot of boxes.
Before 2006, there may have been no formal definition of a planet, but there was a general agreement of what a planet and there was debate.  Scientists argued for the removal of Pluto from the list of planets prior to 2006.
One of the problems was the excitement that was generated when Pluto was discovered.  The initial calculation estimated Pluto's mass to be about 6 times that of Earth. (I'm curious why they were so wrong).
Initial observations of Pluto's distance and brightness still gave it an estimated size similar to Earth and I remember when I was first learning about the planets in School, Mercury, not Pluto, was the smallest planet.  Pluto was - from memory, about the size of Mars.  Pluto's companion moon and high reflectivity may have given the illusion of being larger than it was.   It wasn't until 1978 when Charon was discovered that an accurate estimate of Pluto's size could be made and it was tiny.  Smaller than the Moon.
Pluto's size relative to Earth here.

Despite all the outrage over Pluto's demotion, some of it genuine, some of it comical, such as jokes made in The Big Bang Theory TV show blaming Neil deGrass Tyson, it was, to my thinking, a pretty clear cut decision that to this day remains surprisingly controversial.  And it makes for some good reading.
Two years of committees, debates since the 1990s and finally a vote of just 424 people in attendance.  That's almost congress-like in terms of argumentiveness and lack of efficiency, but in the end, I think the official definition of planets to be a good one, though it's not hard to imagine that it may run into problems.
A binary planet probably isn't impossible, but it raises the question of whether that qualifies.  And it's not hard to imagine if Mars was a bit smaller and the asteroid belt a bit larger - at what point would Mars no longer make the cut?  What if Mercury was half it's current size?
Ultimately, as the article points out.

"For astronomers this doesn't matter one bit. We'll go out and do
exactly what we did,"

Caltech's Mike Brown
I'll put it in my own words.  The exciting stuff is in the details, not in the categorizing.   Astronomers knew what a planet was prior to 2006 and Pluto was on the border even before it got demoted.  The new definition still isn't air-tight.  It doesn't have to be.


Do you know about any Image Processing Challenges known in Space Telescope like Hubble?

First, I'll talk a little bit about the state of what is being done: (according to my knowledge)

In Astronomy on-board processing is usually avoided if possible.

The reason for this is to have the pure data as unspoiled as possible, because instrument effects, cosmic rays, the light from Earth's corona, etc. can introduce many systematic errors into the data. If that would be removed on-board and then sent to Earth, then there would be no control later on what the actual, original data was.
This became relevant e.e. for the exoplanet-finding Kepler(see also the technical data there) space telescope where the photometric data was showing very unexpected behaviour at first, before the reasons for it were realized, obviously long after the spacecraft was designed and built.

Thus, the image processing challenges are usually resolved on the ground stations and with the astronomers on their big computers. Also the interpretation of the data, after being cleaned, debiased, etc. is non-trivial.

As however the downlink speed depends on the distance $d$ to Earth like $\sim \frac{1}{d^3}$ so satellites that orbit far away do have to apply some clever ways to reduce data sometimes. The Hubble is not a candidate for this, as it orbits Earth at only some 500 km a.s.l, its data can be transferred to Earth without problems.

The Gaia satellite however, orbits the second Earth-Sun Lagrange point and is thus farther away from Earth, 1.5 million km. This mission is an astrometric mission, meaning it's there to measure positions and velocities of stars to extreme precision. If all the pixels of all onboard systems would continuously be read out and produce data, that would be too much to handle and some would be lost. So Gaia has a pre-detector that tells the main detector "Hey a star is coming". The main CCD then creates a moving window of 12x12 pixels from which all electrons are read-out and accepted as data. So that's some limited processing there.

In terms of data transfer, missions like New Horizons (Pluto encounter last year) or the Voyagers, are too far out to simply send their data over. Therefore the Deep Space Network exists, to look out for radio signals from those probes and actively amplify them.

What you could do:
So now that I've spent time explaining that on real spacecraft it's not being done what you want to do, we can bend the rules and just try to think about what's being done on the ground (if you want to):

Image calibration: A detector taking an image wants to be very well understood, before claiming any interesting discoveries or measurments. Part of this process is taking a dark image to characterize what your CCD is doing when nothing is measured. Then a flatfield image is done, an image of an ideally uniformly white surface, in order to characterize pixel-to-pixel variations. More. For the measured image then, the dark image is substracted and divided through the flatfield image.
Cosmic ray removal: Cosmic rays fly through your camera in a random direction at random times (in space as well as on the ground). They produce streaks in CCD cameras, and can be automatically removed by taking the same image three times, and discarding one of them or only a pixel-region if a dramatic intensity streak somewhere is measured.
Measuring a point-spread function:  Stars are so far away, that they appear as point sources. However through instrumental effects this ideal point will always be spread out into a finite shape, called the PSF. Because this function can be easily a few pixels wide, it is very important to know the shape of the PSF to be able to assign photons to a star that have been measured in off-center pixels. Knowing then a star's photon output and it's distance one can calculate the luminosity. But even that can't be done on an on-board chip. What you could to is to try to automatically fit an Airy-function to the PSF.

Ok so that was just a few ideas, but I'm not an observer, there may be better ones out there. Also you got keywords not to google further.


Why are the natural satellites (moons) of all planets solid?

To answer this, we have to consider the definition of an atmosphere. A popular way of looking at it is to think of an atmosphere as a layer of gases surrounding a body. by that definition, we can say that gas giants are really just planets that are massive enough to accrete substantially large atmospheres (because deep down, they have a rocky core - although not a core similar to the "rocky" inner planets in our solar system). So now we can reduce the question to "Why do most natural satellites not have extremely large atmospheres?"

This comes down to mass. The more massive an object is, the more gravitational pull it has on objects around it. This means that a very massive object traveling through a region of gas would attract the gas more than a less massive object would. In the early Solar System, the more massive bodies far enough out in the protoplanetary disk managed to gain a substantial amount of matter via core accretion, as gas and dust were collected by a rocky core. Moons are unable to do this because they are very low-mass; therefore, they cannot become gas giants (in fact, many moons around giant planets may be the remains of the material the planets never accreted).

Gas dwarfs, low-mass gas planets with small envelopes, are certainly possible. However, they would have to have radii at least 1.5-2 times the radius of the Earth, and masses scaled appropriately. Moons are nowhere near that size.


What is supposedly in the center of the Milky Way?

Photos of the galactic center aren't too bright because of all the gas and dust between us and it. For example (in infrared):

I'm guessing, though, that you're talking about other galaxies, because there are no views of the galactic center of the Milky Way face-on. Although the galactic center is pretty luminous, just not in the wavelengths we're used to.

Anyway, the galactic center has lots of stars - many massive, but a handful like our Sun. They're young, though, which is odd, as there isn't much star formation happening.

The main attraction, though, is Sagittarius A*, a radio source that is apparently a supermassive black hole. It's actually a subset of the more complex radio source Sagittarius A. So far, astronomers think that most galaxies have supermassive black holes in their centers.

We know that it has to be really massive because of how it perturbs the orbits of stars nearby:

We know it's massive and we know it's a very strong radio source. A supermassive black hole is just about the only thing it could be.

Courtesy of WayfaringStranger: An awesome picture from NASA:

In this paper by Ghez et al., the orbits of stars were monitored and a mass of the central object was determined to be $4.1 \pm 0.6 \times 10^6 M_{\odot}$. This paper discusses the properties of the star cluster itself.


How does a gravity slingshot actually work?

Here's an intuitive understanding without math or physics explanations (others will provide that stuff here):

You are right that approaching and leaving the vicinity of a planet in itself adds up to zero effect. Gravity assist is the effect of being "dragged along" with the movement of the planet. If a spaceship approaches the planet from behind in its orbit, it will be dragged along and accelerated. If a spaceship approaches from infront of the planet in its orbit, the spaceship will slow down as the moving gravity field of the meeting planet pulls it backwards.


In the calculation of GMST, what are these constants?

As the comments have said, the second and third numbers are the ratio of sidereal to solar days but it is split into two parts from the original formula which gives GMST at 0h UT1 by including the hours part directly.

The first number comes from the definition of sidereal time as "in general terms, sidereal time is the hour angle of the vernal equinox" (Explanatory Supplement to the Astronomical Almanac, 1992, p48 (referred to on p77 of the 3rd Edition of the Explanatory Supplement)). This is bound up with the original definition of the right ascension of the fictitious mean sun which moves around the celestial equator at a fixed rate which was due to Simon Newcomb in 1895 (Astronomical Papers Prepared For The Use Of The American Ephemeris And Nautical Almanac Volume 6 (1898). This was based on noon and shifting it by 12 hours to the definition used now at converts the 18.64606 to 6.646065. Differences in the measured values of the precession constants and changes in the star catalog reference systems from FK4 to FK5 have resulted in changes in the values from the original.

Aoki et al. 1982 discuss this in more detail and added a $T^3$ term (their equation 13, although note the values there are in seconds not hours as in the original USNO version above) for what became the IAU 1984 system. Capitaine et al 2000 added a $T^4$ term along with another 32 more terms to the so-called equation of the equinoxes which is added onto the formula for GMST to get GAST (Apparent Sidereal Time). The equation of the equinoxes takes account of the accumulated precession and nutation as the fictitious mean sun is not a very good reflection of reality for precise work. Because of these increased complexities and coupling between the Earth's rotation and the background reference frame of the stars needed for precise work, the IAU did away with sidereal time in 2006 and replaced it with Earth Rotation Angle (ERA) which is a simple linear function of UT1 time.


The key is to compare the range of pulsar periods and ther behaviour with the typical dynamical timescales of stars. Pulsar periods range from just less than $10^{-3}$ s to $\sim 10$ s. And $\dot{P}$ is positive - the periods are getting longer for most pulsars (and all of them that aren't in binary systems).

The dynamical timescale is $\tau \simeq (G\bar{\rho})^{-1/2}$, where $\bar{\rho}$ is an average density for the object.

For a star like the Sun $\bar{\rho} \sim 10^{3}$ kg/m$^3$ and $\tau \sim 1$ hour.

For a white dwarf star $\bar{\rho} \sim 10^{10}$ kg/m$^3$ and $\tau \sim 1$ second.

For a neutron star $\bar{\rho} \sim 10^{18}$ kg/m$^3$ and $\tau \sim 10^{-4}$ seconds.

Whatever mechanism you choose to cause the pulsar phenomenon, it cannot happen any faster than a few times this dynamical timescale if it is something involving the whole star.
This basically rules out pulsations from normal stars and white dwarfs because they could not have pulsation periods that are short enough to explain most pulsars.

Next one might consider binarity, but again, the minimum orbital period, where the components are touching, is a few times the dynamical timescale and some sort of phenomenon involving binarity is ruled out by the long dynamical timescales of normal stars and white dwarfs.

Next one considers rotation; again, simple Newtonian physics can be used to show that an object will rip itself apart if it spins any faster than the dynamical timescale, so again this rules out normal stars and white dwarfs.

So we are left demanding that whatever object causes the pulsar phenomenon is highly dense. In addition there is microstructure in the pulses which indicates the entire source is varying on timescales shorter than a ms, which means the entire emitting region must be much more compact than about 300 km (i.e. $c\times 10^{-3}$s).

So why a rotating neutron star? I think the argument here is that if this were some kind of oscillatory motion you might expect the period to be stable. For some kind of binary phenomenon you would expect the period to get shorter, due to the emission of gravitational waves. The rotation of a magnetised neutron star and the loss of power by magnetic dipole radiation can explain why pulsar periods get longer.


Can you create a black hole using kinetic energy?

You can't make a black hole simply by accelerating a particle, but it has been theorized that smashing some together with ridiculously high energies could produce micro black holes. This, as far as I know, has never been known to happen at particle accelerators, simply because they are not powerful enough.


I saw TWO moving stars going at the same speed! What could this be?

The International Space Station (ISS) and several other satellites can be seen early in the morning or late in the evening, even from urban locations.

Go to this site.

https://www.heavens-above.com/main.aspx

Enter your location then click on ISS. The site will give you a list of the times when the ISS will be visible and the area of the sky that it will move through.

The periods when the ISS or other satellites are visible come in cycles. If nothing is listed as visible in the ten day period wait a week then try again. It might take several weeks. If you live somewhere between 50 deg North latitude and 50 deg South latitude the ISS will be visible sooner or later.

Once you have seen the ISS passing overhead you can compare it to what you saw on your camping trip.

Some people are using electronic eyepieces to view the sky from light polluted areas. They often broadcast their observing sessions on the NightSkysNetwork

https://www.nightskiesnetwork.ca/

If you live in a light polluted area and are interested in astronomy you might find the site interesting.


How long would it take to reach the edge of the reachable universe?

Jonathan's answer is essentially correct, but as Rob Jeffries comments, he doesn't take into account that the Universe is expanding during the journey.

The edge of the observable Universe is 47 billion lightyears (Gly) away. Even if you are a lightbeam, you cannot reach that point. The farthest you can go if departing today is roughly 5 Gpc, or 17 Gly, but this journey would of course take infinitly long (or else it wouldn't be "the farthest you can go"). This distance is probably what the linked article is referring to (I didn't read the article; it's very, very long).

So, in order for the answer to be any fun, you have to freeze the Universe, using magic, which is what Jonathan's calculator is doing. Here I'll just provide the analytical solution: In that case, the proper time $\tau$ (i.e. the time experienced by the traveler) to reach a distance $x$ when traveling at a constant acceleration $a$ is
$$
\tau = \frac{c}{a} \cosh^{-1} \left( \frac{ax}{c^2} +1 \right),
$$
where $c$ is the speed of light. If you wish to decelerate after having reached halfway, you just divide $x$ by $2$ and multiply the result by $2$.

If you plug in the $x=15\,\mathrm{Gly}$ you request, you get roughly 45 years. To get to the edge of the Universe at 47 Gly actually only takes a few years more. The reason for this is simply that traveling at 1G gets you to (almost) the speed of light in only a couple of years, and hence you experience (almost) no time, no matter how far you go.

The time experienced for the Earthlings for the traveler at constant acceleration is given by
$$
t(\tau) = \frac{c}{a} \sinh \left( \frac{a \tau}{c} \right),
$$
which works out to 15 Gyr for the 15 Gly, and 47 Gyr for the observable Universe. The reason is simply that the traveler, from the point of view of the Earthlings, extremely fast reaches a speed which is almost the speed of light.


How many space telescopes are currently active?

More than 20 if the Wikipedia's List of Space Telescopes is accurate. I extracted the active ones, and removed duplicates (to the best of my knowledge):

Swift Gamma Ray Burst Explorer
AGILE
FGST
IKAROS
NuSTAR
Astrosat
Insight (Chinese: 慧眼)
Спектр-РГ (Spektr-RG)
The famous Hubble Space Telescope, HST, see hst
STSat-1
IRIS
Hisaki
Lunar-based ultraviolet telescope
BRITE
NEOSSat
TESS, see tess
CHEOPS, see cheops
Odin
IBEX
AMS-02
DMPE
Chandra, see chandra
XMM Newton, see xmm-newton
NICER (Mounted on the ISS).

Interestingly, the majority is satellite based, but there is e.g. also a moon-based one.
In addition to that, there are space crafts whose Instruments have been temporarily or permanently been repurposed to astronomy. For instance, you could look at the Moon with METEOSAT.


Frequency of transit in planar and non-planar planetary systems

This is unfortunately not true.
If you already look at our solar system, even this is not co-planar.
The few degrees inclination that the planets have to each other would already be enough to make some of them undetectable, when others are detectable by transits.

The answer to your question therefore is not so much about the transit probability, but more about what the mutual inclinations are of the planets. Those can be stirred up by close-planet encounters or resonances after the gas disc dissipates in which the planets form.


There is nothing wrong with your maths, but it isn't relevant to the question of the frequency of impact.

Now you ask about comets, but do you also want to include asteroids, since these are rather more common.

You also ask about catastrophic, and this is a poorly defined word. A Tunguska scale event is a local disaster, but has no significant global effects.

The Torino scale classifies objects according to their threat. A "10" on the Torino scale signifies a object carrying an energy of more than 10000 Megatons of TNT and capable of causing a global catastrophe. Such events occur, on average of less than once in 100,000 years.

Within the next 100 or so years the probability is considerably lower than $10^{-5}$ in each year, as many of the potentially hazardous objects have been observed, and their orbits shown not to intersect with Earth. So while a naive estimation would suggest that the chance of an impact in the next 100 years is about 1 in 1000, the actual probability is rather less than that.

If you restrict to only comets, the probability is reduced further.


Any object's mass warps the space around it towards the center of the object which causes other objects to 'fall' in towards that object or towards each other. Bigger objects have more mass therefore they warp space to higher degree causing more gravity. It doesn't matter what a substance is made of as long as it has mass. If it has mass it causes gravity.


You will always get a "stable" orbit if the stars have less than escape velocity relative to each other. (unless you are modelling the stars as having non-zero radii so they can collide)  The stars will enter into elliptical orbits around a barycentre.
But I guess you want a circular orbit. For a circular orbit the speed $v$ is given by
$$v^2=GM/r$$
where r is the orbit radius (from the centre of mass), G is Newtons gravitational constant, and M is the reduced mass
$$M= \frac{m_1m_2}{m_1+m_2}$$
for stars of mass $m_1$ and $m_2$.
You will find it convenient to use astronomical units (not SI units), so distance in AU, time in years, mass in "solar masses"  In these units $G=(2\pi)^{2}$ and you avoid difficulties with very large and very small values. (in these units the speed of the Earth is $2\pi$ AU/year which makes sense)


It is exactly the point of no return for light. In other words, it is the point (actually more like a sphere) in which the escape velocity from the Black Hole gravity reaches $c$.


What would the pressure and temperature of gas be, right above Jupiter's gas/liquid boundary?

Technically there isn't really a gas-liquid boundary because temperatures are well above the critical point of hydrogen (33K and about 18bar). It's a supercritical fluid. There are important changes at various depths though including the bottom of the circulating winds, the transition to metallic hydrogen and (it now appears) an increasing density of dissolved/suspended heavier elements making up a somewhat diffuse core. There is lots of "pre-Juno" information easily accessible, including temperature and pressure at the transition to metallic hydrogen (10000K and 20GPa). This source gives a pressure of 100kBar at the bottom of the circulating winds layer (about 10 GPa), but I can't find a temperature.

I just found this very interesting diagram: from these lecture notes which shows how the expected temperatures and pressures inside all the giant planets relate the various phases of hydrogen. Notice how all of them except Jupiter do cross the short curved black line at the left, so that they do have a gas-liquid boundary at pressures of a few atmospheres and temperatures of a few tens of Kelvin


GW190521's Luminosity Distance is greater than the age of the universe - how?

Luminosity distance is simply the distance that the object would have to be at in a static special-relativistic (or Newtonian) universe to have that luminosity. It's not the distance in the expanding general-relativistic universe that we actually inhabit.
Out to distances of roughly 1 billion light years, spacetime is roughly flat, the Hubble speed is not too large, and the luminosity distance is not too inaccurate. At substantially larger distances, it's way off.


The key to answering your question is to know the albedo of the asteroid compared to the Moon. The Moon is rather dark (reflecting 7% to 12% of the light if I remember correctly, about as black as worn asphalt). I think, but do not know with certainty, that the asteroid would not be much darker.

For the dimensions that you give, the asteroid would be approximately the same apparent size as the moon: 0.45 x 0.65° compared to the Moon's 0.5°.

So both have approximately the same albedo and approximately the same size. Therefore I say that the asteroid would be visible in the daylight.

Also this is a sci-fi story, correct? If you can get the asteroid into that orbit, then anything is possible. :-)