How do astronomers know if a planet is orbiting a star, or it's simply another star in the background?

I think there are two parts to this question: How do we know that the transiting object is a planet, and how do we know that it's gravitationally bound to the parent star, rather than an interloper?

If the object is indeed orbiting a star, we should see periodic transits. If a candidate transit is observed, follow-up observations should also be able to observe it. A lack of further confirmation would be a strike against the idea that the object is indeed bound. An additional tool that's useful here is spectroscopy. If the body is orbiting, it should influence the motion of the parent star, because they'll orbit the shared center of mass. This means the star's motion with respect to an observer will change over time, causing a Doppler shift in its light. This shift can be measured throughout the body's orbit. If a periodic shift is observed - with the same period as the transits - then we have another bit of evidence for an exoplanet.

Now, the light curves of transiting exoplanets can look similar to those of eclipsing binaries - binary stars where the orbital plane of the components is aligned with the line of sight of the observer. A couple key elements can help astronomers differentiate between the two:

An eclipsing binary should display a secondary eclipse when the less luminous component passes behind the first. This is not likely to be prominent in the case of a transiting exoplanet.
The mass of the orbiting object can be determined by measuring the radial velocity of the star, as determined by looking at the Doppler shift. This is an easy way to distinguish an exoplanet from a star.

In short, an occultation of a background star by an interloper would not be periodic and would not be associated with Doppler shifts in the other star's spectrum. The mass of the orbiting body - as well as the presence or absence of a significant secondary eclipse - can help us determine its nature.

I'd like to note that we've directly imaged a number of exoplanets over the past decade. Beta Pictoris and HR 8799 are well-known examples. There are some difficulties with this, of course; planets are much less luminous than their parent stars, and so the light from the central star must be blocked during observations. Direct imaging is useful for planets in face-on orbits, where we're looking nearly perpendicular to the orbital plane. These exoplanets will, naturally, never transit their parent stars from our perspective, and so transiting planets aren't usually directly imaged planets. Additionally, direct imaging is also good for planets in orbits with large semi-major axes - which are also planets that are unlikely to yield strong transits from any position.


Some of the water evaporated, some of it froze and another good part of it went into the surface and is now subterranean. For more info read Mars' Vanished Water & Atmosphere --Where Did It Go? and MAVEN’s Quest – Where Did Mars Water Go?.

There's much still to learn and experiments in the future will help us to better understand the evolution of Mars' atmosphere.


Why is Starlink polluting the night sky a big concern if we have space telescopes?

It's a problem because there are still lots  and lots and lots of ground-based telescopes.
Ground-based telescopes are still (by far) the biggest optical telescopes, and the cost of space telescopes is prohibitive for many research projects. It will be a long time before a telescope anywhere close in size to the VLT can be launched.
Most space telescopes are specialist devices, observing in a particular part of the spectrum that is blocked by the atmosphere (so there are infra-red, ultraviolet and X-ray telescopes in space) Or doing a specific task (looking for exoplanets, or mapping the positions of stars)
Space is getting cheaper, but it will be a long time before it is as cost-effective as ground telescopes in the optical range. And so it will be a long time before all professional telescopes are in space.
So it is a problem for professional astronomers.


Do objects lose momentum as space expands

GR doesn't have global frames of reference, so we can't say whether a projectile launched from galaxy A slows down relative to galaxy A due to cosmological expansion when it's at a cosmological distance. However, suppose that galaxy A and galaxy B are both at rest relative to the Hubble flow. We can ask whether the velocity of the projectile relative to B, when it gets to B, is lower than its velocity had been relative to A, when launched from A.
There are a couple of easy ways to see that the answer is yes.
One is to consider the fact that ultrarelativistic massive particles have to have same behavior as massless particles. For example, people didn't even used to know that neutrinos had mass. So an ultrarelativistic neutrino, just like a photon, has to lose momentum and energy by the time it gets to B. If this holds for ultrarelativistic particles that have mass, then we expect it to hold as well for lower-energy particles that have mass, because we expect the behavior to vary smoothly with energy.
Another way to see this is that we know the universe cooled down as it expanded. This means that massive particles must have lost energy. We can't blame this on interactions, because actually the matter in standard cosmological models is an ideal gas. So the result must be the same, on the average, for a particle that just travels freely. If there was no such tendency for motion to settle down to the Hubble flow, then we wouldn't have a Hubble flow now.
It's not true, however, that redshift factors are the same for ultrarelativistic particles as for nonrelativistic ones. The effect is bigger for ultrarelativistic particles, which is why the universe is no longer dominated by radiation, even though it was at one time.


Go out in the daytime and practice lining up on the leaves of trees on a distant hill or some such. It's easier to find targets when they aren't against a nearly featureless black backdrop.
If you have a finder scope, likewise align it to the scope's view using a distant daytime target.

At night, look first for the moon to get a feel for how aligning works.


What did LIGO Actually See? (Gravitational waves discovery)

The actual image isn't much. I was able to find it from Science, and this is all it is:

It's just a ripple, seen at slightly different times from two different observatories. The shift fits perfectly by shifting it by the speed of light difference in their locations. Thus is the proof of gravity waves.

It should be noted that the reason there are two instruments is to provide a cross check against other vibration sources. Each observatory works by detecting vibrations on a 4 km scale, down to a very small order of magnitude (1/10,000 the width of a proton). When the two are compared, then one can assume the signal must have come from a non-local source, which only Gravity Waves fit that definition.


There is recycling of mass going on. As stars form they deplete the gas mass, but as they age they replenish the gas mass with a fraction through mass loss and supernovas.
Current thinking is that most halo formation has already occured, so the mass stays constant. There is perhaps a trickle of loss due to capture in massive objects (if there is some cross section between the DM particles and dense matter) including black holes, but this is vanishingly small over the stelliferous era (the next $\sim 10^{14}$ years or so). Over the degenerate era beyond this may matter more: if dark matter consists of particles and antiparticles they would slowly annihilate in the halo, and more rapidly inside the dense objects. Even if it doesn't annihilate, it will eventually very slowly fall into the central black hole in the black hole era due to gravitational radiation.
In the large, $M_{star}/M_{halo}$ begins low, as stars form it increases towards the current value and will keep on increasing across the stelliferous era (but still, it is $\ll 1$). Then around $10^{20}$ years in the future it will start declining as stars are randomly ejected from the galaxy or absorbed by the central black hole due to rare encounters, eventually becoming zero (or halving, if you still count the ejected stars). $M_{gas}$ will keep declining across the stelliferous era as it is used up, but infalling interstellar gas may slow the decline and make more stars for some time; in the degenerate era the small amount of remaining gas will keep on condensing on bodies or get ejected with them.
(See the excellent Adams & Laughlin paper for an overview of the long-term issues.)


No. Some days Venus is an evening star, visible for up to a few hours after the Sun sets. Some days it is a morning star, visible for a few hours before the Sun rises. Some days it is so close to the Sun in the sky that we can't see it at all (or even actually behind the Sun).

Venus can also be seen in daylight, if it is above the horizon, not too close to the Sun and the sky is clear. You need to know where to look for it, though, it's only a little brighter than the sky.


What were the 2 satellites I saw this morning?

The best way to figure it out would be to use the site "In the Sky" (https://in-the-sky.org/). You can enter time, date and location to identify satellites. This would at least be able to tell you if what you saw was a satellite.


How likely is it that the Sun will destroy our electric society?

I should give credit here to @honeste_vivere, who pointed out to me today that there have been recent studies excited by some extremely large coronal mass ejections that were classed as "near misses" in terms of causing major disruption.

Of particular interest to you would be the event of July 2012 discussed by Baker et al. (2013). I quote from the paper

" Had the season and time of day for this CME passage been right on striking the Earth, the world would have witnessed a storm larger (possibly much larger) than the 1859 Carrington event. This most likely would have had devastating consequences for many technological systems "

and then in the discussion

"
It is the opinion of the authors that our advanced technological society was very fortunate, indeed, that the 23 July solar storm did not occur just a week or so earlier.
Had the storm occurred in mid-July 2012, the Earth would have been directly targeted by the CME and an unprecedentedly large space weather event would have resulted. In fact, there is very legitimate question of whether our society would still be “picking up the pieces” from such as severe event [see NRC, 2008].
"

It looks likely that such  events are perhaps things that occur every few decades. Riley et al. (2012) suggest something as powerful as the Carrington event has a 12% chance of occurring in the next decade, with a 1% chance of something several times bigger. Even more powerful events are seen on other stars, where both flare and CME energies follow a power law relationship of the form $dN/dE \propto E^{-\alpha}$, where $\alpha \sim 2.5$ (Drake et al. 2013). Thus much larger, but rarer flares are possible and also seen on (usually) much younger and faster rotating stars. There may however be a very significant sporadic tail to the solar flare energy distribution too. Recent observations of candidate "superflares" on otherwise unremarkable solar-type stars have been reported, though the reality of these and the mechanisms are still being explored. Shibayama et al. (2013) There appear to be a group, consisting of 1% of stars that show (repeated) superflares more than 100 times the size of the Carrington event. These appear a little less likely around stars rotating as slow as the sun, but the bottom line is that, if there is nothing "special" about these stars, then these events occur every 800-5000 years (for events of energy 1e34 - 1e35 ergs) on a slow-rotating G-star. (Though you need to read carefully - it is probably a function of temperature and rotation). The authors note that superflares may be associated with the formation of very large starspots (or starspot groups). So that may be our clue to duck and cover.


Can Jupiter's bands be made out using a 15x70 pair of binoculars?

I use a 20x80 binocular on a tripod and I cannot make out Jupiter's bands. I believe that a 15x70 binocular will not be adequate.


Retrograde motion and Kuiper Belt Objects

No, this is not possible.
Kuiper belt objects are in orbit around the Sun and their average distance (semi-major axis) will be beyond the orbit of Neptune. This means that their orbital period will be much longer than Earth's (following Kepler's third law). In the most extreme case (further away) the motion of the object will be virtually non observable and the object will appear not to move with respect to the background stars apart from the apparent motion due to parallax.

These objects will still show apparent retrograde and prograde motion because of the movement of the Earth around the Sun. This prograde and retrograde motion can also be observed in nearby stars where it is used to measure the parallax and thereby the distance to the star. See the figure below for the apparent motion of a star as measured by Hipparcos.

  I imagine that if you were to observe a KBO that was in the direction of Earth's motion then there would be no contribution from the Earth's velocity on the observation and you would be able to see the object in prograde motion, no matter how slow it was travelling.

The direction of the Earth's motion is constantly changing (changes 360° in a year) so the time at which a KBO will be in the direction of the motion of the Earth will be very short.


Why don't we orbit the center of our galaxy?

There are a number of misconceptions with this question.

First off, the "gravitational pull" (which I'm interpreting as gravitational force) by the Milky Way on the Earth is seven orders of magnitude smaller than that exerted by the Sun.

Secondly, gravitational force is the wrong metric. The Newtonian gravitational force exerted by the Sun on the Moon is more than twice that exerted by the Earth. Yet all but a tiny, tiny minority of professionals who deal with the solar system will say that the Moon does indeed orbit the Earth.

Thirdly, it's not correct to look at "orbit" as being a mutually exclusive concept. That the Moon does indeed orbit the Earth doesn't mean that it doesn't orbit the Sun. It does. The Moon not only orbits the Earth and Sun, but it also orbits the Milky Way, the Local Group, and the Local Supercluster.

So what is the right way of looking at orbits amongst a hierarchy of masses? One issue that needs to be addressed is what the word "orbit" means. Note that people do write and talk about parabolic and hyperbolic "orbits". I'm assuming that "orbit" in the context of this question means a bounded trajectory, which would rule out calling a hyperbolic or parabolic trajectory an "orbit".

That leads to the first of three characteristics of whether object A can be said to be orbiting object B: Object A can be said to be orbiting object B only if object A is gravitationally bound to object B. The appropriate metric here is energy, at least for cosmologically small distances. (Note: The distance between the Milky Way and the Andromeda Galaxy is very small in a cosmological sense.) From the perspective of a frame centered at object B, the total mechanical energy of object A needs to be negative he mechanical energy to be able to say that object A is (at least temporarily) orbiting object B.

Bounded orbits remain bounded forever in the two body point mass problem. That may not be the case when other larger objects are involved. This leads to a second criterion: Object A can be said to be orbiting object B only if the gravitational influences of larger objects on object A are but small perturbative effects that don't result in instabilities. The appropriate metric here is an energy-based sphere of influence. The two most widely used spheres are the Hill sphere and the Laplace sphere. For example, because the Moon is gravitationally bound to the Earth and because the Moon is well inside the Earth's Hill and Laplace spheres with respect to the Sun, it is appropriate to say that the Moon does indeed orbit the Earth.

What about the Sun and the Milky Way (and even larger objects such as the Local Group)? Can the Moon also be said to orbit those? The answer is yes. One way to look at it is to ask what would happen if the Earth suddenly disappeared. Would the Moon's path around the Sun change much? The answer is no. The Moon would still be orbiting the Sun at about one astronomical unit. The same applies to the Earth with respect to the Milky Way. The Earth's trajectory about the galaxy would not change by much at all if the rest of the solar system magically disappeared. This point of view however leads to some troubling issues. Planets (and stars and galaxies) don't magically disappear. Moreover, if Saturn somehow did magically disappear, its rings and all but its very outermost moons would be on hyperbolic trajectories with respect to the Sun. A way around this magic is to say that if object A orbits object B, and object B orbits object C, then object A also orbits object C. "Orbit" is not a mutually exclusive concept.


How many Supernovae have been Imaged in the Hubble Ultra-Deep Field?

Supernovae probably occur about once per century in our own (quite large) galaxy.

If we were to assume similar star formation rates and a similar stellar birth-mass distribution in other galaxies and that the galaxies in the HST-UDF were similar to our own, then we could also assume that the supernova rate was similar.

The UDF was taken over about 4 months, or 1/300 of a century. So one might expect a supernova to occur in about 1/300 galaxies - so about 300 of the 10,000 galaxies observed will have hosted a supernova during the period that the image was built up.

Of course, even if this figure is correct (see below), many of these may have been missed because the galaxies were too far away or the supernovae were obscured by dust, or because they brightened and faded between observations (there was a 6 week gap in the observations). On the other hand, the UDF was imaged later by HST with other instruments, leading to further opportunities to identify supernova candidates.

The calculated number could be awry in many ways. Most galaxies are smaller than the Milky Way, so one might expect fewer supernovae. On the other hand, most of these galaxies are far away and the star formation rate was likely higher in the past, leading to more supernovae. On the other hand, the supernova rate will be higher in some types of galaxies (Starbursts), but lower in others (ellipticals). Without a very detailed demography of the types, metallicity, masses and ages of the galaxies, and a detailed understanding of how the supernova rate varies with these parameters, your question is difficult to answer.


How does water exist within the frost line of a star?

When you talk about the frost line you are talking about a protostar, and its protoplanetary disk, not an actual star. So, there is no empty space nor strong solar winds.

Inside the frost line, water does not condense into ice grains, but the individual molecules could as well become attracted by the planets-to-be.


How do accretion disks around black holes evaporate via winds?

Not an expert, but here are my 2 cents until there will be a better answer:
Winds are not directly responsibly for the evaporation. Rather can they consist of evaporated material.
The cause for the evaporation of gas out of the accretion disk is primarily heating from the much hotter corona. Or as Liu & Taam put it:

The ions in the corona are directly heated by viscous
dissipation, partially transferring their energy to the electrons by
means of Coulomb collisions. The energy gained by electrons
cannot be effectively radiated away and is conducted to the
lower, cooler, and denser coronal layers by electron–electron
collisions. In the transition layer, the conductive heat flux
is radiated away through bremsstrahlung only if the number
density in this layer reaches a critical value. If the density
is too low to efficiently radiate the energy (lower than the
critical value), a certain amount of lower, cooler gas is heated
up, whereby the bremsstrahlung cooling rate is raised, until
an energy equilibrium is established between the conduction,
radiation, and heating of the cool gas. The transfer of gas
from the disk to the corona, to establish the equilibrium, is
called evaporation.

So, all you need for evaporation is a hot corona above the accretion disk and more heat transfer as you can radiate away. Then gas will "boil off" and the corona will "dig into"   the disk until a critical density and equilibrium is reached.
For all this you do not need any winds. A wind will originate from the disk or the corona when gas is further accelerated outwards. This acceleration can be primarily radiation or line-driven when the intense radiation from the central regions illuminates the outer parts of the disk (see here). This also works for AGN. See e.g. these simulations.
Alternatively, these winds can also be (partly) magnetically or thermally driven. Which of these mechanisms dominates depends on the type of source (e.g., AGN/XRB), state and position on the disk. (Someone else might be able to give details here.)
And because images are always nice, here is a graphic how one can imagine the geometry for an AGN (from this paper):


Does the galaxy look like a Web or a Tree?

The process where you link random points within distance $r_0$ into a graph is known as continuum percolation. As $r_0$ increases from 0 at first there are just isolated clusters (binaries, randomly very close stars) that gradually merge. At a critical distance of a few parsecs these clusters mostly merge into one major galaxy-spanning cluster, and above that distance this cluster just gets more densely connected. The big network at the critical range has a scale-free structure of local clusters, bridges, and trees: neither a proper web nor a tree.
Some illustrations from my book manuscript:

It should be noted that these pictures assume a constant star density. In reality the stellar density varies, so for a given $r_0$ there is a part of the disk and bulge that are connected and parts of the outer galaxy and halo that are not connected.


where to obtain ephemeris data for geodesy satellites?

The JPL Horizons website gives ephemerides for a large number of Earth satellites as well as deep space spacecraft, planets, moons, asteroids, etc.

If you configure it as shown below, selecting the units you'd like, the full state vectors (x, y, z, vx, vy, vz), use geocentric coordinates (@500) you will get the following output. You can choose various time step sizes (down to 1 minute) and time frames. You can choose direct download as csv text rather than HTML if you like - recommended for long files.

It looks to me that because of the special status and use of the LAGEOS satellites, the ephemeris only predicts a relatively short period into the future, unlike other spacecraft which have predictions for years sometimes.

This should get you started! If you have more questions about this post as comments and I or someone else will try to help.

Here are the Table settings. I believe that the ICRF/J2000.0 with Ecliptic and Mean Equinox settings will give you something close to GCRS but right now I am not sure how close. I'll keep reading and ask for help in comments.

Here is a sample output from the above configuration. It scrolls left-right and up-down.

Revised: Sep 11, 2017                 LAGEOS-2                         -122195

LAGEOS-2 (1992-070B, "Laser Geodynamics Satellite", NASA/ASI)
60-cm diameter sphere, 426 corner cube reflectors (VIS), 4 germanium (IR)
Deployed: 1993-Oct-22 from Space Shuttle Columbia (STS 52)

*******************************************************************************
Ephemeris / WWW_USER Mon Sep 11 21:51:05 2017 Pasadena, USA      / Horizons
*******************************************************************************
Target body name: LAGEOS-2 (spacecraft) (-122195) {source: LAGEOS-2}
Center body name: Earth (399)                     {source: DE431mx}
Center-site name: BODY CENTER
*******************************************************************************
Start time      : A.D. 2017-Jan-01 00:00:00.0000 TDB
Stop  time      : A.D. 2017-Jan-05 00:00:00.0000 TDB
Step-size       : 1440 minutes
*******************************************************************************
Center geodetic : 0.00000000,0.00000000,0.0000000 {E-lon(deg),Lat(deg),Alt(km)}
Center cylindric: 0.00000000,0.00000000,0.0000000 {E-lon(deg),Dxy(km),Dz(km)}
Center radii    : 6378.1 x 6378.1 x 6356.8 km     {Equator, meridian, pole}
Output units    : KM-S
Output type     : GEOMETRIC cartesian states
Output format   : 2 (position and velocity)
Reference frame : ICRF/J2000.0
Coordinate systm: Ecliptic and Mean Equinox of Reference Epoch
*******************************************************************************
            JDTDB,            Calendar Date (TDB),                      X,                      Y,                      Z,                     VX,                     VY,                     VZ,
**************************************************************************************************************************************************************************************************
$$SOE
2457754.500000000, A.D. 2017-Jan-01 00:00:00.0000,  5.049112760756619E+03,  9.246781656953866E+03,  5.722633641721932E+03, -2.978199703429669E+00,  3.689975896939779E+00, -3.350502047340569E+00,
2457755.500000000, A.D. 2017-Jan-02 00:00:00.0000, -6.140907600397200E+03, -8.148707798015072E+03, -6.918983669661846E+03,  2.434461456961566E+00, -4.250750074318191E+00,  2.808377534161328E+00,
2457756.500000000, A.D. 2017-Jan-03 00:00:00.0000,  6.849580412560958E+03,  6.106324592083777E+03,  7.740446107133596E+03, -1.906840987637830E+00,  4.973834339923835E+00, -2.290764139470047E+00,
2457757.500000000, A.D. 2017-Jan-04 00:00:00.0000, -7.504132672898423E+03, -4.707277560275819E+03, -8.539479988453819E+03,  1.304309397636298E+00, -5.242849004192330E+00,  1.679103108590806E+00,
2457758.500000000, A.D. 2017-Jan-05 00:00:00.0000,  7.740851830184564E+03,  2.245705288312970E+03,  8.942780571589567E+03, -6.155844902136471E-01,  5.667380276723395E+00, -9.687665040214430E-01,
$$EOE
**************************************************************************************************************************************************************************************************
Coordinate system description:

  Ecliptic and Mean Equinox of Reference Epoch

    Reference epoch: J2000.0
    XY-plane: plane of the Earth's orbit at the reference epoch
              Note: obliquity of 84381.448 arcseconds wrt ICRF equator (IAU76)
    X-axis  : out along ascending node of instantaneous plane of the Earth's
              orbit and the Earth's mean equator at the reference epoch
    Z-axis  : perpendicular to the xy-plane in the directional (+ or -) sense
              of Earth's north pole at the reference epoch.

  Symbol meaning:

    JDTDB    Julian Day Number, Barycentric Dynamical Time
      X      X-component of position vector (km)
      Y      Y-component of position vector (km)
      Z      Z-component of position vector (km)
      VX     X-component of velocity vector (km/sec)
      VY     Y-component of velocity vector (km/sec)
      VZ     Z-component of velocity vector (km/sec)

Geometric states/elements have no aberrations applied.


Is the universe expanding faster than speed of light?

According to Hubble's law

  For [proper] distances $D$ larger than the radius of the Hubble sphere $r_{\mbox{HS}}$ , objects recede at a rate faster than the speed of light ... $$r_{\mbox{HS}}=\frac{c}{H_0}$$

With $H_0=67.8\mbox{ }(\mbox{km}/{\mbox{s}})/\mbox{Mpc}$ the Hubble constant according to the Planck mission, $c=299,792,458\mbox{ m}/\mbox{s}$ the speed of light, and one Megaparsec $1\mbox{ Mpc}=3.0857\cdot 10^{22}\mbox{ m}$, we get
$$r_{\mbox{HS}}=\frac{c}{H_0}=\frac{299,792,458\mbox{ m}/\mbox{s}}{67.8\cdot 10^3\mbox{ }(\mbox{m}/{\mbox{s}})/3.0857\cdot 10^{22}\mbox{ m}}=1.3644\cdot 10^{26}\mbox{ m}.$$
That's $$1.3644\cdot 10^{26}\mbox{ m}/1.3644\cdot 10^{26}\mbox{ m}= 4,421 \mbox{ Mpc}$$
or $r_{\mbox{HS}}=14.422$ billion lightyears (1 pc = 3.26156 ly).

Hence objects further away than this proper distance (what you would measure with a chain of rulers) recede faster than the speed of light. The distance is also called Hubble length.

You may find 13.8 billion lightyears elsewhere. That's calculated with $H_0=70.4\mbox{ }(\mbox{km}/{\mbox{s}})/\mbox{Mpc},$ according to the estimates of 2010, based on WMAP data, or with even older data. The exact value is not known.

Since the Hubble constant isn't really constant over time, the universe is thought to be a little younger (13.8 billion years) than the 14.422 billion years you get by just dividing the Hubble distance by the speed of light.


Which constellation would appear to revolve around Denab,Vega or Thuban when axis of the earth would point toward them?

When we look at the Polaris from the earth, the Big Bear appears to be revolving around it.

Ursa Major is relatively close to Polaris in angular distance, so this motion is easier to see.  But in fact all the constellations appear to revolve around the axis.  The closer they are, the tighter the arc and the easier this motion is to see.  If they are far away and you are not at very high latitudes, they will be obscured at some point by the horizon and the motion is hidden.

So for your question, look at a starmap and the closer a constellation is to the future axis, the more it will behave like Ursa Major today.


What happens if a white dwarf collides with a star like our sun or a red dwarf?

Fun question.   Without doubt it would be very violent and spectacular, but not much is known about stellar collisions and only a few have ever been observed.

Most stellar collisions happen due to tight orbits where the stars spiral in towards each other or perhaps, 3 or more body chaotic star systems with unstable orbits that lead to an impact.   Space is so large that actual star on star collisions are exceptionally rare.

A spiral in towards each other is almost certainly more common, it also quite different in terms of outcome.  Much of the gravitational kinetic energy of the collision goes into angular momentum, so, while still spectacular, it's quite a bit less explosive than a direct head-on collision.

Black hole/black hole collisions have been observed by detection of gravity waves when two black holes orbit around each other and the orbit decays.

Neutron Star/Neutron star collisions may be responsible for the very heavy elements in the Universe and White dwarf-White dwarf collisions may lead to a supernova similar to a type 1a Supernova (per Wikipedia source).   All these collisions.

In a tight orbit, a white dwarf will steal material from a near-by star and if the orbit is close enough, lead to a decaying orbit and collision, but that would happen over time.  If the white dwarf reaches the Chandrasekhar limit in that scenario, it will undergo type 1a supernova.  If there's not enough mass for that, it should eventually absorb the nearby star, leading to a new star, admittedly a strange one with very high metallicity and perhaps abnormally hot and short lived for it's mass, but it should basically be a star once the collision settles down, but probably a star that burns both hydrogen and helium, but wouldn't collect enough helium for a helium flash.   I don't see any other alternative with that kind of below Chandrasekhar mass spirally in collision than it being a very metallic star, hot for it's mass and with a shorter lifespan.

Now a head on collision - that's very different and, I'm basically guessing, but you have a white dwarf star - about the size of Earth with a mass probably at least 30%-40% of the sun.   White dwarfs don't have a specific minimum size, but for a star to go white dwarf within the 13.6 billion years or so that stars have been around, that star would have needed enough mass to have completed lifecycle and in the process, it loses some mass, so perhaps, 30% of the mass of the sun might be the current lowest possible mass of a white dwarf, but as billions of years pass, that low end will grow smaller.

But if we take a low-mass white dwarf at 30%-40% the mass of the sun, it's still a super-dense object and crashing that into a star at their combined escape velocity, not the more common decaying orbital spiral, that's a messy calculation.   I'd only be guessing.

Escape velocities aren't additive, but still, white dwarf and small star would collide in this theoretical at an impressive velocity.    If it was a glancing blow, the white dwarf would almost certainly punch straight through, blowing a lot of matter off the star and perhaps affecting it's rotation rate.   It would be an impressive site.

If it was a direct hit, I don't know if the white dwarf would punch through the star or not.   It might still punch through.   The White dwarf would pull the surface of the star towards it as it got close and that would be the first contact between the two.   The Sun's escape velocity is about 618 km/s and the mutual gravitation the White dwarf would be traveling to the sun a bit above that, but the White dwarf's surface gravity is much greater, so the mater pulled off the sun into the white dwarf would be hitting the white dwarf at a few thousands of km/s.  (Small white dwarf, so lets say 3,000 km/s).

That's still far too slow for fusion to happen, so this is basically still just collision mathematics.   As the white dwarf plowed deeper into the star, towards the much hotter core, the heat and velocity of impact probably would increase fusion rates, but if it just passed right through the star that would be a pretty temporary and minimal effect.   The larger effect would be the disruption to the star after a very dense planet sized object punched it's way through it.

If the star was viscus enough to stop the white dwarf from making it all the way through, then the combined mass and dense center and added heat from impact should significantly increase the rate of fusion of the original star, even though some material would likely be blown off, the net result should be a heavier star that burns hotter and faster and if the new mass is comfortably above 1.4 solar masses, you likely have a super-nova in the making.   But as to whether or not a white dwarf would be to punch all the way through a star upon direct collision - I have no idea.  I guess that it would, but that's just a guess.

A fun related question is, how long would it take the white dwarf to emerge from the other side, assuming it's a relatively center hit and assuming it passes through (which it might, but I'm not sure if it does).   I'll try to give that a look too.

Anyway, just my 2 cents on this question.   Corrections welcome.


What is the bluest narrow band filter mounted on ground based telescopes?

You probably will have difficulty imaging Ly$\alpha$ from a z=1.7 quasar at 330nm.

Says P. Veron: The lower limit [z=1.8] is set by the ultraviolet transmission of the atmosphere... between 3300 and 3600Å.


Can earth escape sun's gravity with the help of a black hole heading towards our solar system?

There is no need to hypothesise a black hole. A black hole's gravity is indistinguishable from that of a similarly sized star, except in the region closely surrounding the event horizon.

Black holes can't be only 1 solar mass, there is no known mechanism that could cause a single solar mass object to undergo complete collapse. Black holes have at least three times the mass of the sun.

So could a massive star disrupt the orbit of Earth? Yes it could. If the star were to pass by the Earth, it could cause a slingshot effect, transferring some of it's momentum to the Earth, which could cause Earth to leave solar orbit.

In the precise situation that you describe, Earth would initially be travelling in a straight line, but then would follow a complex and possibly chaotic path, as the sun and the black hole pass each other. It is quite possible that the Earth could be ejected, or transferred to an orbit of the black hole in that case.

Fortunately the gaps between stars are so vast (and black holes even more widely spaced) that in the nearly 5 billion years of the Earth, not one has come close enough to cause major disruption to Earth (we know because we are still here).


Which star / galaxy is moving away from us the fastest?

When a galaxy recedes from us, the light we see from it is redshifted. For galaxies at cosmological distances, this redshift is fundamentally different from a Doppler shift; whereas the latter is due to a velocity difference between the emitter and the receiver, a cosmological redshift is due to photons traveling through an expanding space$^\dagger$.
Hence, as @uhoh comments, your question is equivalent to asking "Which galaxy has the highest measured redshift?". Redshift is arguably the single most important concept in astronomy, and indeed we do catalogue the redshift of all galaxies, if possible. For our adopted cosmological model, the cosmological redshift can be translated to both a recession velocity, a distance, and an age of the Universe when the light was emitted.
The answer is GN-z11 (Oesch et al. 2016) which has a redshift of $z=11.09$. This corresponds to a distance of $d = 32.2\,\mathrm{Glyr}$ (i.e. billion light-years), and hence, by Hubble's law, to a recession velocity of
$$
v = H_0\,d = 670\,000\,\mathrm{km}\,\mathrm{s}^{-1},
$$
or more than twice the speed of light. Furthermore, the light we see today was emitted when the Universe was only 410 Myr (i.e. million years), or 3% of its current age.
You may think that "twice the speed of light" violates the theory of relativity, but this velocity is not a velocity through space. Both our galaxy (the Milky Way) and GN-z11 move through space at modest velocities of a few 100 km/s. The recession is merely due to space expanding, and space is allowed to expand at whatever rate is wishes.

$^\dagger$"Fundamentally different" might be a too strong statement, since the cosmological redshift can be interpreted as infinitely many infinitesimally small Doppler shifts. However, a hypothetical scenario that emphasizes the difference between the two types of redshifts is the following:
If an emitter and an observer are stationary wrt. each other when the emitter emits a photon, then start moving away from each other while the photon is traveling, then stop again before the observer receives the photon, the observer would measure zero redshift.
On the other hand, if space is static when the emitter emits the photon, then while the photon is traveling suddenly expands by a factor of, say, four, and then is static again when the observer receives the photon, the observer would measure a redshift of $z+1 = 4$.


How can you determine the date using nothing but a star chart from the day and the time?

First determine the North (South) direction by drawing a line from the center through Polaris (Sigma Octantis). The other principal directions follow from that.
You need to obtain the coordinates of a well-recognized star in equatorial coordinates (right ascension and declination), not too near the celestial pole. Given your chart, it's no problem as it shows the sky for a location near or even exactly at the equator.
As it's a question from a competition I will only sketch the path briefly:  The stellar coordinates allows you to calculate the local hour angle of it and from that the sidereal time. You then have to find the date where the sidereal time matches your given mean solar time. Depending on accuracy and possibly further details given in the task, you will also have to consider the equation of time.
OLD ANSWER where the time was not yet given:
A star chart is not unique, it's ambiguous for date vs. time. Just as you state: it could be from 22h today. Or it could be the sky at 21h in 14 days or 20h in a month. Generally every half month you have to remove one hour from the clock and it is identically valid.
Obviously a star chart is of limited use for times where the Sun is over the horizon or for the early twilight time where you still don't see any stars.
So what you can do is figure out for what date / time combinations the star chart is valid. In year books usually several of those are given for the star charts printed. Assuming that people usually observe in the early evening hours, one can get an estimate for the intended month though (stating the observation time).
Thus practically I'd use a programme like Stellarium and try find a month where for the chosen observation hour the night sky looks like shown. Mind also it is WAY more accurate to find North (South) direction in a star chart by drawing a line from the centre through Polaris (Sigma Octantis). All other directions follow.
You can break ambiguity of time and date, if your chart shows solar system objects like planets which change their position with respect to the stars.


The short answer is no.

We don't even think the census of stars within 30pc is complete, let along have all the other information you want.

For all bright stars (say V<10) and close, there are decent parallaxes and proper motions in the Hipparcos and Tycho catalogues, so you can compile tangential velocities for these. A lot of these will also have radial velocities, but there is no recent uniform catalogue of these. Things like the General catalogues of radial velocities are getting a bit old. The Hipparcos input catalogue might be a reasonable source.

Magnetic fields have been measured on relatively few stars and usually those which are much younger and more active than the Sun.

There is no way of directly measuring stellar ages. This is one of the big problems in astronomy. Estimates can be made for stars in the field but they are rarely more precise than 20-30% and their accuracy, especially for older stars is open to question. I know of no uniform compilation that would help.

If you were to restrict your sample to even closer to the Earth, then you might find that the RECONS database for stars closer than 10pc might be of use.


how to plot sky map like www.nakedeyeplanet.com with dates tags with planetary motion path

The Python program Skyfield can generate the data for most of this. It can for example generate curves for planet trajectories, provide star positions and magnitudes from a catalog, and even generate boundaries for the constellations.

https://rhodesmill.org/skyfield/api.html

Then Matplotlib or Plotly can be used in the same python script to make plots as nice-looking as you like.

There may be other canned programs that can do this as well, just for example Stellarium and Celestia.

For further resources, see answers to Where can I find/visualize planets/stars/moons/etc positions?

and the animation in this answer


Is it possible for two stars to be in a horseshoe orbit around a much larger star?

Yes, it is possible, however such orbits are likely to be unstable.

For example in the paper

  Ćuk, Matija, Douglas P. Hamilton, and Matthew J. Holman. Long-term stability of horseshoe orbits. Monthly Notices of the Royal Astronomical Society 426.4 (2012): 3051-3056. doi:10.1111/j.1365-2966.2012.21964.x

it was found that if the mass of second body becomes larger than 1/1200 of the first body, horseshoe orbits become unstable. (Third body was assumed to have negligible mass).

Nonzero mass of the third body was investigated recently in a thesis:

  Balaji, Bhaskaran. Mass ratios in stellar triple systems that admit horseshoe orbits. Diss. Massachusetts Institute of Technology, 2016. abstract, pdf

and it was found that in addition to constraint of Ćuk et al. nonzero mass of the third body  further decreases stability of such systems, so that the lifetime of horseshoe orbit $\tau \propto m_3^{-1}$.

For stars only coupling of more massive giants with lighter  red dwarfs would fulfill  first constraint and then the third star would never have 'negligible' mass, one could conclude that horseshoe orbits when all three bodies are stars are always unstable. And for larger masses of smaller stars the lifetime of such orbits would be too short to expect such orbits in our vicinity.

However it would probably be possible to have stable horseshoe orbits of two stars (say a blue giant and red dwarf) and a planet.


Is seeing the apollo moon Landers via earth telescope that hard?

Yes, it still is.

If the resolution of an interferometric telescope is really 0.001 arcseconds, that translates to 4.85E-09 radians. At a perigee of 362,600 km that's 1.8 meters.

The top part of the Lunar Modules is about 4.2 meters wide which is a little more than double the resolution, so if there is some good contrast then it might appear in a reconstructed interferometric image.

However, those optical interferometers are not designed for imaging of extended objects. They only have a few telescopes and so are mostly limited to resolving more than one point-like object from each other, such as a double star or a star and an exoplanet.

With so few telescopes in an optical interferometer, a complex image like lunar terrain would be difficult or impossible to reconstruct without a model. For example, the images of the accretion disk of a black hole by the event horizon telescope required some input modeling to interpret the interferometric data.

Near sunrise or sunset however, the shadow of a lunar lander will be much larger. The lander is about 7 meters tall, so the shadow could be 4 or more meters wide by several tens of meters long. Since the Moon's orientation, as well as that of the lander and the local terrain is extremely well documented, then at each moment you could estimate what the shadow would look like in size, shape, and orientation. With that model, an interferometric image based on a few telescopes could be used to confirm the shadow.

However, using a model like that won't appease conspiracy theorists.

 Source


Free source of printable star charts in the format of the Millennium Star Atlas?

OK, I think that I've found the answer - it looks like Cartes du Ciel will do what I'm looking for (although I haven't yet figured out the simplest workflow).

Click on the icon for Equatorial Coordinates

Then click on the icon for 'configure the program'

Under the Colour tab, click on 'black on white'

Under the Lines tab, make sure that 'show grid label' and 'show coordinates grid - line' are checked.

Zoom in to the desired magnification

Click on printer icon


If the Sun were bigger but colder, Earth would be hotter or colder?

The equilibrium temperature of the Earth, $T_E$, scales roughly as $L^{1/4}$, which is proportional to $R^{1/2} T$, where $L$, $R$ and $T$ are the solar luminosity, radius and temperature.

The actual approximate relationship is derived by equating the power received by the Earth, which is proportional to the solar luminosity $L$, with the power radiated by the Earth, which is proportional to $T_E^4$ for a blackbody. Hence $T_E \propto L^{1/4}$.

So the answer to your question depends on by how much you increase the radius compared with the decrease in temperature.

There will be second order effects that do depend on the spectrum of radiation from the Sun (and therefore its temperature) compared with the wavelength dependence of the albedo and emissivity of the Earth. So I will post a better question...


any unique phenomenon when the cores of two planets are aligned with their star's core?

Referring to your #gravity tag this would only create a peak in the planets tidal forces. But this would be a visible effect in case the planets would be big and close enough... Last night happened that with Sun-Earth-Moon although they were not aligned as you say.

To sum up: a proper alignment would increase the effect of tidal forces.


What is the mass limit in a stellar accretion disc?

The accretion disc is formed by material in orbital motion against a central body, which can be a star.

The size, mass and other characteristics are usually determined by the central object, in this case the star.

In general, the protoplanetary accretion discs are the largest ones (with the largest mass) and as the age on the central star increases, the average size decreases.

This is because as the age of the system increases, more mass is drawn into the system and less is left out in the disc itself.

Once the central star crosses the Chandrasekhar limit, it will become unstable and become a supernova and once it passes the Tolman–Oppenheimer–Volkoff limit it will become a black hole.

The physics of black hole accretion disc is entirely different and is possible to derive a relation between the black hole mass and size of the accretion disc. The relation appears logarithmic.

Any natural or artificial structure around a star would feel the gravity of the star and consequently, it is the mass of the star that determines amount of the counteracting force (rotation, radiation pressure etc.) required to maintain stability.

For example, as the stellar mass increases, the rotation speed must increase if the orbit is maintained at the present position or the orbit must become larger. In case of stellar radiation pressure is used to counteract gravity, as the mass (and gravity) increases, the system becomes unstable (as the ratio of radiation pressure to gravity is constant).


There are far fewer than 0.2% of stars that are as massive as $8M_\odot$. It is possible you've confused it with the percentage of stars born with $M>8M_\odot$.
The thing is, those massive stars die very quickly, but most stars, those with masses $<1M_\odot$, are still alive. So the fraction of massive stars in existence now is much lower. i.e. Almost all the massive stars ever born have already died, but almost all the low mass stars ever born are still alive.
What matters for the present supernova rate is the current birth-rate of massive stars.
The current star formation rate in the Milky Way is something like 1 solar mass per year. The way the birth distribution of stellar masses works out, most stars born are of order $0.3 M_\odot$, but about 0.2% are $>8 M_\odot$, which means, if you work through the numbers, that you need round about $150 M_\odot$ of star birth to produce one massive star - which means you get a type II (core collapse) supernova every 150 years. These numbers are uncertain by factors of two.
Another factor to bear in mind is that we don't "see" a supernova every 50 years in the Milky Way. In recorded history, it is more like one every 300 years. However, the majority of supernovae would have been totally hidden from an observer on Earth by dust in the Galactic plane (where star formation occurs). Thus the true rate in the Milky Way might be closer to 1 in 50 years.
Finally, note that some of these supernovae are caused by exploding white dwarfs, not the core collapse of massive stars. Thus the rough number of one every 150 years is probably consistent with the historical observational record.


Have we ever observed a body, such as a large asteroid, "hitting" the Sun?

Yes

and here's a video of "a Giant Comet Hitting the Sun":

https://www.youtube.com/watch?v=Mat4dWpszoQ

The impact occurred sometime during May 10-11, 2011.  The comet was not named but believed to be a member of the Kreutz family of comets

Many close calls

Before this spectacular plunge we had witnessed several other comets graze (come close without hitting) the Sun.

I've included a nice picture of comet Lovejoy after its close approach of the Sun brought it through the Sun's corona around December 15, 2011:

What happens when a comet hits the sun?

It acts like a supersonic snowball in Hell

  If a comet is big enough and passes close enough, the steep fall into
  the sun’s gravity would accelerate it to more than 600 kilometres per
  second. At that speed, drag from the sun’s lower atmosphere would
  flatten the comet into a pancake right before it exploded in an
  airburst, releasing ultraviolet radiation and X-rays that we could see
  with modern instruments.

  The crash would unleash as much energy as a magnetic flare or coronal
  mass ejection, but over a much smaller area. “It’s like a bomb being
  released in the sun’s atmosphere,” Brown says. The momentum propelled
  by the comet could even make the sun ring like a bell with subsequent
  sun-quakes echoing through the solar atmosphere.


What are the craters on the moon with peaks in the middle of them?

Nice picture! Those peaks are actually a very common feature in craters formed by large impacts. Craters with these features are known as "complex craters". Basically what happens is that when the meteor hits, it pushes in the surface. Then the underlying mantle produces an "elastic rebound" and pushes outwards. This is a similar effect to what happens when you hit a drum. Your drumstick rebounds off the drum, and the drum begins to vibrate. Since rock (even molten rock) is much more dense than a drumskin, it does not vibrate very much. The vibrations solidify, resulting in a crater with one central peak.

How exactly this happens is still being investigated. This paper provides a pretty good explanation of what happens, but it's quite technical.

Even larger impacts will produce ring structures. These are relatively rare on the Earth, since such structures are quickly weathered away. A good example of this is the Chicxulub crater off the Yucatan peninsula (the impact that killed the dinosaurs).


What is faster than a supernova explosion?

I am assuming you are talking about how long the event lasts. See question on this site "How long does a supernova explosion last?". The answer is that if you are talking about how long it brightens then dims (its light curve), then it typically takes a couple of days to get to full brightness then weeks to months to fade. Core collapse can be a fraction of a second (keep in mind there are different kinds of supernovae).

The "fastest" pulsar found so far spins 716 times a second. If you really want fast, how about the standard for atomic clocks? The used transition is between two ground states of the element caesium that are very close together and lasts about $1\times 10^{-10}$ seconds. There are even better atomic clocks now.


Can tidally locked planets have liquid oceans on their night side, and would such a planet be habitable?

Short Answer:

With current knowledge nobody knows.  Some calculations suggest that it might be possible for a tidally locked planet to be habitable and to have life.

Long Answer:

The first stop searching for information about the habitability of tidally locked planets in red dwarf star systems might be the Wikipedia article Planetary Habitability.

  Astronomers for many years ruled out red dwarfs as potential abodes for life. Their small size (from 0.08 to 0.45 solar masses) means that their nuclear reactions proceed exceptionally slowly, and they emit very little light (from 3% of that produced by the Sun to as little as 0.01%). Any planet in orbit around a red dwarf would have to huddle very close to its parent star to attain Earth-like surface temperatures; from 0.3 AU (just inside the orbit of Mercury) for a star like Lacaille 8760, to as little as 0.032 AU for a star like Proxima Centauri[76] (such a world would have a year lasting just 6.3 days). At those distances, the star's gravity would cause tidal locking. One side of the planet would eternally face the star, while the other would always face away from it. The only ways in which potential life could avoid either an inferno or a deep freeze would be if the planet had an atmosphere thick enough to transfer the star's heat from the day side to the night side, or if there was a gas giant in the habitable zone, with a habitable moon, which would be locked to the planet instead of the star, allowing a more even distribution of radiation over the planet. It was long assumed that such a thick atmosphere would prevent sunlight from reaching the surface in the first place, preventing photosynthesis.

  This pessimism has been tempered by research. Studies by Robert Haberle and Manoj Joshi of NASA's Ames Research Center in California have shown that a planet's atmosphere (assuming it included greenhouse gases CO2 and H2O) need only be 100 millibars (0.10 atm), for the star's heat to be effectively carried to the night side.[77] This is well within the levels required for photosynthesis, though water would still remain frozen on the dark side in some of their models. Martin Heath of Greenwich Community College, has shown that seawater, too, could be effectively circulated without freezing solid if the ocean basins were deep enough to allow free flow beneath the night side's ice cap. Further research—including a consideration of the amount of photosynthetically active radiation—suggested that tidally locked planets in red dwarf systems might at least be habitable for higher plants.[78]

https://en.wikipedia.org/wiki/Planetary_habitability#Red_dwarf_systems1

This article links to a longer discussion of the habitability of red dwarf star systems.

https://en.wikipedia.org/wiki/Habitability_of_red_dwarf_systems2

And of course the various sources listed in those articles are worth studying.


How to calculate B-V colour index value percentage difference

$B-V$ corresponds to the base 10 logarithm of a flux ratio.

$$B-V = -2.5 \log \left(\frac{f_B}{f_V}\right)$$

So trying to guess what you are trying to calculate, it is the percentage change in the blue to visible flux ratio?

In which case the percentage change is
$$ p = \frac{ 10^{-(B-V)_2/2.5} - 10^{-(B-V)_1/2.5}}{10^{-(B-V)_1/2.5}}\times 100$$

A percentage change can of course be negative.


Why isn't the star that created the black hole a black hole?

You are correct in saying that a star loses a lot of its mass in a supernova. However, there is a reason why the star still becomes a black hole. Actually, I suppose the question here is "Why doesn't a star become a black hole before it even undergoes a supernova?"

There is a reason for a supernova (I'll assume you're talking about type II supernovae, which result from incredibly massive stars). Stars undergo nuclear fusion, and this leads to "thermal pressure", which counteracts the force of gravity. Without this pressure, gravity would indeed make a sufficiently large star collapse upon itself. Gravitational collapse occurs when there is not enough pressure to counteract gravity; the result is a spectacular supernova. So stars only become black holes (or other compact objects, such as neutron stars) when they cannot produce enough energy to counteract the force of gravity due to their own mass.

As for the first part of your question (sorry for answering in reverse), light in the area of a black hole cannot escape if it is inside its event horizon or on a trajectory towards it. The radius of the event horizon for a non-rotating black hole is its Schwarzschild radius, which is proportional to the mass of the black hole. The reason this is non-applicable in stars is because the Schwarzschild radius in stars is deep inside its interior, and there is not a strong enough gravitational field to produce an event horizon to trap any light near it.

Thermal pressure reference: https://en.wikipedia.org/wiki/Gravitational_collapse

I hope this helps.


nonagesimal in Kepler's parallax computation

It looks like the assumption regarding the 90° point is that the amount of the ecliptic visible is 180°. So the 90° direction is the halfway point.

The nonagesimal is on the ecliptic, so the altitude approximates the arc. The wording in the initial part has that correct, but then it goes on to say "the same as the angle" - this seems to be a mistake. The arc is directly related to the angle, yes, but it is not the angle.


What is the temperature of the solar atmosphere (the corona) and how is it measured?

This is a rather broad question and this will not be a fully comprehensive answer.

There is no single temperature to the solar corona. The coronal temperature varies by an order of magnitude from place-to-place. It is hottest ($\sim 10^{7}$ K) in magneticloops undergoing flares, which tend to be anchored in low latitude regions. It is coolest (a bit less than $10^{6}$ k in coronal holes which are usually at high latitudes. It is hard even to come up with some average number, because it is not altogether clear how the various pieces should be weighted in any average. It is also cyclic - in that magnetic activity, and average coronal temperatures vary with all the other magnetic phenomena on the Sun with a roughly 11-year period. If you weight it by emission measure, which is density squared times volume, then the solar corona is between $10^{6}$K at solar minimum to about $3\times 10^{6}$K at solar maximum.

The temperature of coronal plasma can be measured using X-ray spectroscopy. If you assume that the plasma can be represented by some equilibrium temperature, then you can make a prediction of what the X-ray/EUV spectrum will look like. To some extent the spectrum will also depend on the composition of the plasma and also on its density, but these produce comparatively minor differences compared to changes in temperature and can be simultaneously estimated in most cases.

For stars you essentially do the same thing but with worse data. Here you are getting some sort of average temperature for the whole corona - there is no spatial resolution. X-ray satellite observatories like Chandra and XMM-Newton have grating spectrographs and detectors with energy sensitivity that can record X-ray spectra from stars. You then use software like XSPEC along with an optically thin plasma model, such as that of Mewe-Kaastra-Liedahl, (and not forgetting to include absorption by the interstellar medium) to get the best fitting temperature for the measured spectrum.

As far as how the Sun compares with other stars, an excellent starting point is the review by Guedel (2007). A short summary would be that the temperature of a stellar corona depends on it's rotation rate - the faster it rotates, the hotter it is. Because (single) stars lose angular momentum and hence rotate more slowly as they age, then both X-ray luminosity and (average) coronal temperature decrease with age.

Telleschi et al. (2005) studied solar analogues at a variety of ages. They found a relationship that the average temperature
$$\langle T \rangle = 1.16\times10^{7} P_{\rm rot}^{-0.48}\ {\mathrm K},$$
where the rotation period $P_{\mathrm rot}$ is in days.

Given that rotation period and age are linked by a "gyrochronology" relation roughly of the form $P_{\rm rot} \simeq 27 (t/4.5\ {\rm Gyr})^{1/2}$, for $t \geq 100$ Myr, this means that
$$\langle T \rangle = 2.4\times10^{6} (t/4.5\ {\rm Gyr})^{-0.24}\ {\mathrm K}.$$
At ages below 100 Myr, the coronae "saturate" at the luminosity and temperature of a 100 Myr star.

In general, stars with photospheric temperatures cooler than the Sun tend to have hotter coronae, whilst those with warmer photospheres have cooler coronae. Stars with spectral types A basically don't have coronae, and hotter O and B stars have coronal winds that are heated by completely different mechanisms to the solar corona.


How do scientists differentiate between a total and a partial transit of a planet passing in front of its star?

The shape of the transit is different if there is a partial transit, not just the depth.
As a planet passes fully in front of the star, the brightness is almost constant while it is transiting (bar the limb darkening effect). Hence you get a "bucket-shaped" eclipse with a wide, flat bottom.
If the transit is partial then you get a more "U-shaped" and finally a "V-shaped" eclipse because there is a singular minimum point when the maximum area of star is obscured.
In this way you have both the shape and depth of the transit to work with and this can give you both the inclination (or rather the "impact parameter") of the exoplanet orbit and the exoplanet size.


Why is the Earth called terrestrial? Can we call the Earth an ocean planet?

There really isn't a lot of water on the Earth. The image below, from the USGS science school, graphically illustrates the volume of the water on (and in) the crust of the Earth, relative to the volume of everything else.

The bigger blue ball is all the salt water, from all the oceans & seas, the small one is all the fresh water, from the rivers, lakes and aquifers. You may need to zoom in to see the fresh water ball...

There is more water dissolved in the mantle, but we can't get at it (with current technology), apart from what's released in volcanic eruptions.


During solar eclipses on Jupiter, can the moon(s)' shadow(s) on the surface be seen from Earth with a telescope?

Yes, you can see it (I have seen Io's shadow on Jupiter and we were happy it was visitor night so that we could share the view with guests)

A 50cm mirror and 125x magnification allows you to see it when the air is not too disturbed and when you know where and when to look. Likely a  somewhat smaller telescope will do, too, as light sensitivity is not too crucial - more the resolution and magnification so that you can still see the tiny black dot of the shadow. The shadow has about the same diameter as the width of some major clouds bands on Jupiter.

Stellarium is a great tool to find the right time to look at Jupiter to find these transits.

Find a local amateur observatory and they sure will be happy to share this view with you (possibly not before late northern summer due to the rise time of jupiter in 2020)


If the hammer and feather move at the same speed why do comet and the tail particles move at different speeds?

The tail of a comet is not actually "slowing down and falling away" from the comet, like you might expect to see when smoke streams out from behind a moving object on earth. The tail of a comet is actually being pushed away from the sun by the solar winds and radiation. That's why the tail of a comment always points away from the sun, and doesn't stream out behind it.

If there were no other forces at work, the coma would actually travel at about the same speed at the comet. As the material surrounding the core of the comet vaporized, it would form a coma ("atmosphere") made up of sublimated gas mixed with dust, moving along  at the same speed (more or less(1)) as the comet… like you observed in the video.

(1) In reality, without the solar winds, the coma of the comet would continue to move outward, as its expulsion velocity would easily overcome the comet's weak gravitational pull.


What to look for in a tripod for binoculars?

Get the sturdiest tripod that you can!

My 20x80 binoculars are sitting on a ball and socket mount on a very heavy tripod. I use the Benro BH1 ball mount, which is an absolute joy to aim. This mount is rated for 6 KG, almost three times the actual weight of the binoculars. With high-powered binoculars the aim must be adjusted every minute or so, therefore you will want an easy to adjust mount that is far from the limits of its capacity. You really can't beat the ball and socket design for this purpose in my opinion. I need to photograph the setup for another question and when I do I'll check what model the tripod is.

Look also at the tripod of the user who asked the above-linked question. These large binoculars need the absolute heaviest duty tripod that you can reasonably afford. You will need all-metal construction and I would suggest a hook as well to hang a weight from, this will make the tripod steadier. You will have a hard time going overboard, so if you find two tripods with a similar price but one is heavier-duty, get that one.


My local night sky is so bright that I can only see the Moon, so would a low-cost telescope be any use?

There is hope!
You may be able to see some stars through a telescope even though you can't see them with your eye.
For extended objects like sky brightness, nebulae, and the Moon and resolved planets, magnification reduces sky brightness, while aperture increases it.
For unresolved (point-like) objects like stars and unresolved planets, as long as focus is good magnification does not reduce its brightness but aperture still increases it.
However
If you can't see stars or planets by eye, it is going to be hard to point your low-end telescope at them, and searching randomly has a low probability of success.
In this case
I recommend first getting your hands on a pair of binoculars with low magnification (see if you can borrow some somehow, don't buy them yet!), something you can hold in your hands comfortably. Then get a star chart or use a website like in-the-sky.org and going to the planetarium page to view a simulated night sky, and find a bright star or planet and note where it is relative to the horizon. You might use a map or compass to find out where your North, South, East and West are so you can compare the sky map to your local surroundings.
Once you are able to find stars or planets this way, you can ask yourself if it's worth it to get a telescope and go through the same exercise.
You can also ask yourself if a pair of binoculars might be a better investment.
Also, please read answers to I live in an area with a lot of light pollution, how can I view the stars without building an observatory? for some more advanced options.


Where should I be looking to see Neowise in southern Sweden?

The comet's position changes from night to night.
It was in Auriga earlier this month but is now in Ursa Major.
Here is a finder chart generated by Stellarium, with cyan dots indicating the position of C/2020 F3 at 01:10 GMT+2 from July 22 to 31.


Are trojans in L5 more likely than in L4?

There is a symmetry in the rotating gravitational field, which means that capture of an asteroid to the L4 is just as likely as to the L5 Lagrange point.

In the case of Mars the split is 1:6, and a simple binomial model suggests a probability of 0.0625 (the probability of a 1:6 or 0:7 split given the hypothesis that they are distributed randomly is 0.0625) This doesn't give a reason to suppose there is anything unusual happening, and as noted in the comments, there is no favouring of L5 when other planets are considered.

As noted by UserLTK: "Earth has an L4 asteroid, none in L5. Uranus also has an L4, no L5 and Neptune has more L4 than L5 objects."

The conclusion is that No, the greater number of L5 martian trojans is just a random effect.


what determines the density of atmospheres

Most simply, the density is determined by the number of molecules, the surface area of the planet, the temperature, and the gravity.  But it sounds like what you really care about would be called the "column density", which depends only on the number of molecules and the surface area, and you are wondering why there is not some direct correlation between the two.

I would say that the reason you don't see a clear correlation is that atmospheric gas generally comes through the surface of the planet, so all else being equal, you'd expect all planets to have the same column density.  But all else is not equal, because you have issues of how volcanic is the interior (so how much gas gets liberated), and how much escape into space happens.  The amount of volcanism is higher for bigger planets (they take longer to cool), and bigger planets also have stronger gravity and cling better to the gas, so one might generally expect larger planets to have higher column density (and indeed the Earth and Venus have higher column densities).  But a higher temperature will lead to escape, and a lower temperature will lead to freezing out many of the species, so temperature is a key factor as well.  Magnetic fields can also play a role, by protecting the atmosphere from being stripped by the stellar wind, and it's still not real clear what planets will have strong fields (Earth) and what planets won't (Venus, whose interior is almost the same as the Earth but doesn't spin much).  Finally, you can have even more complicated effects, such as the way Earth's oceans have dissolved and locked up carbon dioxide.  So when there are a lot of potential factors, you can expect a lot of variation.


Do Pluto and Charon have unusual Lagrange points?

The L1, L2, and L3 points are unstable in any orbital system. (source)

The L4 and L5 points of a pair of bodies are only stable if the larger of the bodies is at least 25 times as massive than the smaller (source).  The ratio of the Pluto/Charon system is only 8.7.  Because of this, none of the Lagrange points are stable, and an object orbiting at any of them will require active station-keeping to compensate for perturbations in the orbit.


Determining the shape of ʻOumuamua via stellar occultation

Observing star occultations is dependent on the object passing in front of a star.

During its short time in the inner solar system, ʻOumuamua didn't pass in front of a star (from the point of view of Earth) It is now beyond the orbit of Saturn. It became too dim for Hubble in about January 2018. There are no forecast occultations of any star (and its orbit is probably not well enough determined to predict any, given the observation of non-gravitational and not well-understood accelleration). Any occultation that did occur would only happen for a very short time and only be visible in a very narrow strip of the Earth. We can't set up large telescopes every 100m on Earth, and constantly observe all the (very dim) stars that might possibly dim further for half a second.

By contrast, Ultima Thule is a much larger object, with a well-known orbit, and the possibility of forecasting the exactly when and where an occultation of a reasonably bright star will occur, and set up observation stations on the narrow strip that is predicted.

So while an occultation could have told us about its shape, none occurred and none are expected.


Enhanced Star-Gazing with Special Glasses

Stereo Vision?
Optically, you'll need something like a couple big Fisheye lenses mounted on a helmet.
Looking straight through such optics yields horrible distortion, so you'd have to feed the lens' output into CCD/CMOS sensors, and massage that into flattened images your eyes can handle, on a video screen. Perhaps Google Cardboard?
Probably doable and light enough to wear.
A straight optical approach to correcting the fisheyes would probably end up too heavy to lift.


What is meant by "Radial Direction" of a galaxy?

'Radial direction' typically means from the center, moving outwards. Typically, these kinds of studies will break a galaxy up into several annuli, or rings, and examine the star formation activity within each ring. (Imagine drawing a series of concentric circles, each one a little bigger than the last.) Then, if the star formation rate or star formation history changes as a function of radius, those changes can be more easily quantified.

It is often difficult to do this well because galaxies are not perfectly symmetric. The LMC, for instance, is a dwarf irregular galaxy, meaning its structure is not nearly as orderly as some other galaxies (e.g. The Milky Way, which is a spiral). The LMC does contain more stars near its center, and it even has a bar, but it is difficult to discern much more structure than that. To define the center of such a galaxy, the total light profile is measured and the centroid calculated.

Since we are rarely fortunate enough to be looking at another galaxy perfectly face-on, we also have to consider the effects of inclination on our observations. (If you were to hold up a plate and angle it a little bit, it would begin to look like an oval, but that doesn't mean it isn't still actually round.) If they have taken inclination into account, they should reference the "deprojected radius." That means that they know that the inclination to our line of sight makes some parts of the galaxy look closer to the center than they actually are, and they corrected for that effect.


Is axial tilt direction correlated with semi-major axis direction?

It's mere coincidence. In about 10,000 years, perihelion will coincide with the northern hemisphere summer solstice.

There are two precessions of interest here. One is the 26,000 year period axial precession of the Earth's rotation axis, caused mostly by the Moon and the Sun; the other is the 112,000 year period apsidal precession of the Earth's orbit, caused mostly by Jupiter and Venus (but also with a pinch of help from general relativity). This result in three different concepts of a year:

The sidereal year is the mean time it takes the Earth to make a complete orbit with respect to inertial space ("the fixed stars").
The tropical year (aka solar year) is the mean time it takes for the Sun's ecliptic longitude to increase by 360 degrees, and also the time between one vernal equinox to the next. This is the year used as the basis of our calendar. This year is shorter than the sidereal year by about 20 minutes and 25 seconds.
The anomalistic year is the mean time between one perihelion passage of the Earth to the next. This year is longer than the sidereal year by about 4 minutes and 43 seconds, and hence longer than the tropical year by about 25 minutes and 8 seconds.

That 25+ minute difference between the anomalistic and tropical years means that time of perihelion passage advances by about a day every 57 years.


Would a naked eye supernova become a disc in a matter of days?

The human eye has a resolution of roughly 2 arc minutes, or about 0.5 milliradians (15 pixels across a full moon). The distance to Betelgeuse is a bit uncertain, but lets say 600 light years, which is near enough. So to show a disk simply because of the size of the source, it would need to be about 0.3 light years across.
The source can't grow faster than the speed of light, and it reasonably might grow that fast as the radiation from the supernova heats the dust and gas already shed by Betelgeuse. If so it would take about 0.15 years or about 8 weeks to grow large enough to (just) show a visible disk. The actual material ejected in the supernova explosion would move at a fraction of that velocity.
One thing that might happen is that scattering in the Earth's atmosphere might make a very intense point source look larger. Light would arrive at your eye from a number of directions close to the point source, having scattered on its way through the atmosphere. Effects in the eye might also make a very bright point source look larger.


Could images of (very) nearby targets be recovered from Event Horizon Telescope data?

The question asks about "targets" but doesn't specify black holes. However the other "targets" are related to infall of black holes.

Nonetheless, I think the answer is a solid "yes" either way.

Data is recorded separately at each site. Hours and hours of digitization of one or more 2 GHz wide data channels around 230 GHz, down converted to baseband and synched using a local atomic clock. See this answer and this answer

Any signal reaching the feed horn of each radio telescope is going to be recorded in its data stream. So if there is another source nearby that is covered by the radiation pattern of all of the dishes in the EHT collection, then it certainly could be imaged to some (indeterminate) level of resolution, with (at least) the following caveats:

You have to run the numerical correlations again, with a new range of time-offsets for each telescope. If the source is much brighter, you wouldn't have to use the whole dataset. But if it's just as dim or dimmer, you will have to correlate lots of data.
If the source is far away from the imaged event horizon, there may be several different types of calibrations for each dish and site that have been optimized, and may not work as well for a substantially removed object. The existing data may result in slightly lower resolution far from the target, depending on how calibrations were done. In addition to signals of the black hole, signals from other sources were recorded to assist in calibration of the whole array.

I need to read further on that, I'll update the part about calibration tomorrow.


There is absolutely no possibility that our Sun is part of a "binary star system". Even the least luminous star (with a mass of about 8% that of the Sun) would have an apparent brightness and proper motion that would easily have led to its discovery. See for example Is Sun a part of a binary system? and  Is it possible that the Sun has a binary partner (the Nemesis Theory) that has eluded detection?


What do the words "p-type" and "s-type" mean?

"Satellite type" and "Planet type".
The terms seem to have been coined by Rudolf Dvorak in 1982 paper "Planetenbahnen in Doppelsternsystemen"

Due to the fact that quasiperiodic orbits exist around stable orbits 3 different types of possible planetary orbits are found: S-types (satellite type orbit around one primary), P-types (orbits around both primaries), and L-types (librating orbits in the vicinity of L4 or L5).

It's not immediatly clear from this whether "P-type" means "planetary type" or "primary type" (and the full paper would be in German and doesn't seem have been digitised).  However in an English paper from 1986 Critical orbits in the elliptic restricted three-body problem.  Dvorak is explict:

There are three types of stable motion for a third massless body that are interesting as regards this problem: the P-type (planet type) surrounding both primary bodies, the S type (satellite type) orbiting one of the primaries and L-type (Librator type) librating around the Langrangian equilibrium points ...


There's a few parts to this question, I'll give it a shot.  Also, since you're asking about the geothermal energy of the earth, I'm going to ignore radiation from space and radioactive van allen belts (Jupiter has a very radioactive van Allen belt).   Also, as voyager discovered when it left the solar system, there's more radiation in the form of cosmic rays out there, that the sun effectively shields most of the planets from.

But, because you mention the Earth's geothermal heat from radiation, so I'll focus on that.

Here's a very well answered similar question about radioactive elements inside the earth.  what percentage of the Earth's core is Uranium

  So as I understand it the source of most geothermal energy on Earth is
  due to radioactive decay.

A healthy share of Earth's core heat is from the planets formation.  This is a pretty good article on that subject - http://physicsworld.com/cws/article/news/2011/jul/19/radioactive-decay-accounts-for-half-of-earths-heat, it says that of the current heat, about 50% is from the energy left from the formation of the planet and about 50% is from radiation - so the radiation is very important, but it's probably not the cause of "most" of the heat in the earth.

As far as I know, no similar studies have been done on any of the other planets.

  Jupiter is the most massive planet but if the bulk of it's mass comes
  from gaseous elements instead of heavier ones like uranium or whatever
  than it being the most massive might not necessarily mean it has the
  most radioactive material.

That I pretty much agree with.  The two primary differences between the 4 gas giant planets (or, two gas giant, two ice giants if you like) and the 4 inner rocky planets is that the gas giants formed outside the frost line and key difference #2, the 4 outer planets are much larger, so they were more able to hold onto more hydrogen and helium, which are the 2 most abundant elements in the solar system.

Obviously, the 4 outer planets have a lower percentage of heavy elements overall.    As to their percentage of Uranium - well, your guess is as good as mine, but, probably a lower percentage of radioactive elements than the inner planets seems reasonable to assume.

Also, any radioactive elements in the 4 outer planets would likely be well below their thick gaseous surfaces, so it would be difficult to detect, outside of Neutrino detection I suppose, and more trouble than it's worth to try to collect.

The 4 inner planets (and lets throw the moon in there as well, and asteroids if you like) are more interesting I'm assuming David Hammen's answer is correct in the linked question above, and that most of the Uranium on earth is in the crust or mantle, the highest percentage in the crust, and so that's probably true for any other rocky planet that formed and was, at one time, molten enough to circulate, so all the inner planets probably have similarly small amounts of uranium and other radioactive elements as Earth.

Mercury may have had much of it's crust and some of it's mantle blown off in a giant collision, and if that's the case, Mercury might have lost a portion of it's radioactive elements.  Brief summary of that here, point #6:  http://messenger.jhuapl.edu/faq/faq_science.html

unrelated footnote, but if Mercury's crust was blown off, it might be a good planet to mine for rare earth elements, or maybe asteroids will be easier, who knows, but I've always been curious of the mining possibilities on Mercury.  Abundant solar energy too, but it's a hard planet to fly to because of the sun's gravity, so, maybe asteroids are the way to go for space mining.

The moon was also formed by a giant impact, so, the moon was formed by material from Earth's but it has a much higher percentage of mantle than the Earth, so it could have a higher percentage of Uranium and other radioactive elements than the earth.  How well the Moon's material was circulated in it's cooling process is another question.   Being smaller it probably cooled more quickly.

Venus and Mars, I see no reason why they shouldn't have similar percentages of radioactive elements than Earth.

 Source:  https://en.wikipedia.org/wiki/Internal_structure_of_the_Moon

One way to look at this question is to investigate how quickly the other inner planets cooled.  Far from exact, but that probably tells us something, since heat generated by radioactive decay inside the planet would probably lead to volcanism at some point.  Even the Moon appears to have had volcanic eruptions as recently as 33 million years ago, so some ongoing internal heating seems likely.

Volcanoes on the moon

Size is also a key factor in what keeps the inside of a planet hot, and in some cases, tidal heating (Io, Enceladus), but that's not really relevant to this question.

Neither Mars nor Venus are volcanically active like the earth, but Olympus Mons may have erupted as recently as 25 million years ago, and there is evidence of large lava flows on Venus.   As to how big a role radioactivity plays in heating up the cores of those 2 planets, I don't know precisely how much.

Overall, I think this is a good question on planet cores and radioactivity, as the inside of the planet is just as interesting as the surface, well, to me, at least.   I know this doesn't answer your question specifically as to which is the most radioactive, cause I don't know, but I like the question.


You need to define an agreed standard for orientation of axis and origin.

What co-ordinate system you choose as a standard depends on your purpose.

Here are some examples and links : Celestial coordinate systems.

Typical reference points are defined using very distant objects whose position can be measured accurately ( typically quasars ).


Why is detecting brown dwarfs difficult?

Brown dwarfs have historically been difficult to detect (directly) simply because of how faint they are. Typical luminosities may range from $10^{-3}L_{\odot}$ to $10^{-5}L_{\odot}$ depending on spectral type. Any object that dim will be tough to find, regardless of spectral type or what sort of telescope you're using - you can have the largest, highest-resolution telescope in the world, but it can't make the source intrinsically brighter.
This can be mitigated in part by looking for brown dwarfs in young stellar clusters, when they're at the most luminous point in their lives - old brown dwarfs are quite difficult to detect directly. The downside to this is that young, high-mass brown dwarfs are not easy to differentiate from young, low-mass stars; there's much less of a distinction early in life. The development of the lithium test in the 1990s helped make strides in this area, but in many cases it's not conclusive.
Another reason that brown dwarfs weren't confirmed observationally until a couple decades ago was simply inadequate equipment; adaptive optics wouldn't come into use until the 1990s, even though we may take it for granted today. Additionally, we've since made advances in indirect detection techniques, such as the radial velocity method and (to, I think, a lesser extent) changes in astrometric measurements. The radial velocity method has been used widely to detect massive exoplanets, and as brown dwarfs are more massive than exoplanets, they should produce stronger signals (assuming the same semi-major axis).
The bottom line, though, is simply that brown dwarfs are extremely faint. At the same time, WISE, 2MASS and other surveys have discovered hundreds (thousands?) of candidate and confirmed brown dwarfs. Granted, given the expected brown dwarf population of the galaxy, that's not a lot, but it's not nothing.


I'm having trouble understanding the basic model for how matter bends space.

No wonder, because it doesn't. It curves spacetime. But spacetime isn't the same thing as space. See this Baez article: "Similarly, in general relativity gravity is not really a 'force', but just a manifestation of the curvature of spacetime. Note: not the curvature of space, but of spacetime. The distinction is crucial". Space isn't curved where a gravitational field is. Instead spacetime is.

  The classic illustration is the canvas and the ball but that model needs a reference frame where one can define a downward direction to work.

That's right, it uses gravity to "explain" gravity, so it isn't ideal. However there is a way to understand it, and IMHO it's simpler than you think.

  For the life of me I can't figure out why matter just doesn't "sit" in space.

It does just sit in space. And it causes gravity because it's a concentration of energy that "conditions" the surrounding space. It alters it, and this alteration diminishes with distance. Remember the stress-energy tensor, remember that stress is directional pressure, and think of matter as exerting an outward pressure in all directions in the surrounding space. Something like this:

Earth image courtesy of NASA, lines drawn by me

Then see the Einstein digital papers, where Einstein said a gravitational field is a place where "space is neither homogeneous nor isotropic". This inhomogeneity isn't linear. You can plot it, using say optical clocks dotted throughout an equatorial slice through the Earth and the surrounding space. You plot lower slower clocks as lower down in a 3D image, and higher faster clock rates higher up. When you've finished your plot looks like this:

CCASA image by Johnstone, see Wikipedia

The upper image is a bit like looking at the lower image from underneath. But anyway, your plot is curved, and you made it using clocks in space, so we can say spacetime is curved. Note that whilst we started with a two-dimensional equatorial slice and ended up with a three-dimensional curved plot, there is no extra dimension in any real sense. All we're depicting is the "coordinate" speed of light. This might sound unfamiliar but again see the Einstein digital papers: "the curvature of light rays occurs only in spaces where the speed of light is spatially variable". Light doesn't curve because your plot of optical clock rates is curved. It doesn't curve because spacetime is curved. Einstein never actually said that. He said it curves because the speed of light varies. Hence the tilted light-cones on John Norton's course. Light curves wherever your plot isn't flat. Where the speed of light varies. It's actually rather like sonar:

Image courtesy of FAS and the US Navy, see course ES310 chapter 20


Are a black hole's jets caused by the black hole's magnetic field?

You appear to have a thing about linking the magnetic field generated by Jupiter to that around a black hole (note that a field does not "belong" to an object) - the two things are entirely different, the only connection being that magnetic fields are generated in both cases by currents caused by the motion of charged particles.

In Jupiter's case, the magnetic field is (likely) generated via a dynamo process, involving currents that flow in its metallic hydrogen interior.

In the case of a black hole, as material travels inwards, it forms an accretion disk around the black hole (because angular momentum must be conserved), in which magnetic field can be generated and amplified. The twisting of the magnetic field lines around the rotation axis of the black hole could create a kind of magnetic funnel lined up with the rotation axis, along which charged particles can be accelerated by a strong magnetic field gradient. The field lines would not look like you have shown for Jupiter, they would be twisted into a helical topology along the rotation axis. This is part of the Blandford-Znajek effect, which I do not claim to properly understand.

So the answer to your question is maybe, but its very much a topic of contemporary research.


Has any of the orbiters or rover on Mars captured image of comet NEOWISE?

Not yet, as far as I can tell.  The comet looks very different from Mars and while I haven't done the math, I don't think it has an apparent size from the rovers' perspectives to make it look like anything more than a dot.  Since Neowise is moving away from the Sun, the length and brightness of its tail is diminishing rapidly.


How many exoplanets have been found around binary stars?

Go to the exoplanets.org database.

Top left hand dropdown menu - select "Stars"

You get a table listing all the known stars with exoplanets.

Click on the big plus sign in the top right hand of the table and select "Binary Flag"

This will now show a boolean flag that indicates if the star is known to be part of a binary (or multiple) system.

In the FILTER window at the top of the page, just type "BINARY" and the table will just show binary stars.

It is possible to export the table as a csv file using the "export" button at the top right. Counting the number of lines with the BINARY flag indicates that as of December 2019, there are 511 known exoplanets around binary stars.


"Galaxies and nebula" are faint and spread out, and for that very little matters as much as aperture!  Back before light pollution was a a serious problem, 6 inches (150 mm) was generally acceptable, these days I tend to recommend 8" or 10" (200 or 250 mm) to just to get through the sky glow.

The 5" scope you are thinking about will be very nice on the Moon, and Saturn/Jupiter and some larger smudges should be detectable but... the view you'll see of "Galaxies and nebula"  is not likely to knock your socks off.


The Solar Motion and the peculiar velocities of stars

When you talk about the motion of anything, you need to first define what "not moving" means.  For example I may be sitting in a chair (and not moving relative to the chair), but I know that the Earth is rotating, and orbiting the sun, and the whole solar system is orbiting in the Milky Way.

When I look at the motion of stars, I know that (nearly?) all the nearby stars are also orbiting the Milky Way. But I want to ignore that overall motion, and only consider the motion of the stars relative to each other.

It is possible to measure the motion of a star relative to the observer. We also know the motion of the observer relative to the sun (since we know how fast the Earth spins and how fast it orbits the sun) So by simply subtracting one motion from the other we can find the motion of the star relative to the sun.

But instead of using the sun as a fixed frame of reference, I can look at the local stars, find their average motion relative to the sun, and subtract that motion from each individual velocity. This will give a motion relative to the motion of the stars around the Milky Way. I know the average will be zero because I have chosen to subtract a certain amount from each star's velocity to make the average zero. It is zero "by construction"

(simple example Consider the numbers "3,7,2,3,5". The average is 4. If I subtract 4 from each number I get (-1, 3, -2 -1, 1) with average = 0)

The motion of a star, when you have subtracted the average motion of all the local stars, is called the peculiar motion


Mass of a potential black hole in a binary system

Your first equation is incorrect. The left side has dimensions of mass, the right side has dimensions of mass × time × length-1. The velocity semi-amplitude (usually denoted $K$ rather than $v$) should be raised to the third power.

As noted by @PM-2Ring in the comments, your second equation is also incorrect as you didn't expand the $(m+M)^2$ term correctly.


Calculating the time a star is at my local meridian based on its right ascension

The sun's position is irrelevant to when a star will be in its meridian; the Sun's coordinates in the RA/dec grid does change over the year. A star's position is fixed in that system and it will be always at the exact same sidereal time that it culminates (as that's the definition of sidereal time).
The difference of the sidereal day to the mean solar day is ~3:56 minutes (366 earth revolutions in a sidereal year, 365 in a solar year), thus a star culminates (and rises and sets) that much earlier each day.
If you compare it to the true solar time (which determines when the Sun is in the meridian) as you try, you also have to account for the difference of the mean solar time and the true solar time due to the eccentricity of Earth's orbit; this phenomenon visible as the analemma (which basically visualizes the equation of time) accounts for a difference of the true noon which varies by about +-20 minutes throughout the year - and can explain the differences you see.
This means that the Sun culminates at different times throughout the year. On March 21st it may be 12:00h. But that is not exactly true 2 month earlier or later, it might be 11:40h or 12:20h.
I think the explanation in the video as linked in your comment as "hours behind the sun" can be a bit mis-leading, and is only exactly true for that one exact date (and defines how the two coordinate systems align with respect to eachother) - but he ignores the equation of time and assumes the mean solar time to be used while his words "behind the sun" imply true solar time. For the task at hand, where to visually find a certain constallation, that is good and accurate enough, though - so not wrong, just mathematically you will have to consider effects he graciously left out.