How many solar system objects that can really qualify as a KBO are likely to enter Pluto's Hill sphere each day?

The Hill sphere radius of Pluto is about $r$ = 6 million km.  Most of the Kuiper belt is in prograde motion around the Sun (like Pluto).  Pluto's average speed is under a lazy 5 km/s for an orbital period of about 248 years.  If the difference in orbital speed between Pluto and an average KBO is just 1 km/s, then Pluto will "sweep out" $\pi r^2 *86,400s *1km/s = 9.8 \times 10^{18} km^3= 2.9 \times 10^{-6} AU^3$ of the Kuiper belt per day.
The Kuiper belt is described as a donut extending from 30 to 55 AU.  The volume $V$ of a torus is $V=2\pi^2Rr^2$, where $R=42.5$ and $r=25$ are the major and minor radii, so the approximate volume of the Kuiper belt is $5.24 \times 10^{5} AU^3$.  Of course, KBO's are distributed more densely near the center of the torus, and exist increasingly rarely outside the torus.
So, Pluto might sweep out about $V_1 = 2.9 \times 10^{-6} AU^3$ of the $V_2=5.24 \times 10^{5} AU^3$ volume Kuiper belt per day.   That means there would have to be approximately $V_2/V_1 \approx 1.8 \times10^{11}$ (or 180 billion) total objects in the Kuiper belt for one to enter into Pluto's Hill sphere each day, on average.  It might even need to be much higher since Pluto's orbit is more inclined then the distribution of the Kuiper Belt, as shown here:

We only know of about 2000 KBOs according to this NASA site.  Pitjeva and Pitjev postulate only millions of objects in the Kuiper belt, so it may be unlikely that many KBOs enter Pluto's Hill Sphere radius each day.


Would it be possible to calculate the expected frequency of impact craters of all sizes on Earth

Yes. By looking at a nearby object that hasn't been subject to erosion and has been subject to the same position around the sun, you could estimate the distribution of impacts on the Earth in time and in size. The moon is excellent for this purpose.

Here's an article touching on the subject: http://adsabs.harvard.edu/full/1975LPSC....6.2597N


What will New Horizons be able to say about a magnetic field at Pluto?

I believe, I don't know if there is evidence of it however, that Pluto does not have an active core. An active core as I'm sure your aware produces the magnetic field, without a magnetic field neither Pluto nor Charon have a magnetosphere. If there was sufficient or any evidence at all to suggest otherwise, I believe it would have become a priority.

Theoretical models of Pluto are based typically on it's orbital speed, which suggest there is no field. However as you have stated we have witnessed tidal stresses producing activity, such as Io, which should be frozen like rock however is the most geological active body we know of it the solar system. One major difference, is we don't have images of neither Plutos nor Charons surfaces, if we had images of a smooth surface we could suffice the surface at least is active through tidal forces. Being all cratered and looking ~5 billion years old indicates the opposite. There just isn't any data as yet to say otherwise, and certainly wasn't when New Horizons was being designed.

As for what the craft could detect with it's instruments, the most indicative feature would be optical imagery and the dwarfs' surfaces. In essence, all of it's instruments could be used; ALICE for detecting atmospheric properties could certainly indicate a magnetic field, whether it's able to hold onto any material spewing from it's surface for example and any ionization of any atmosphere, which in turn is what PEPSSI does. The other great tool would be Ralph, mapping thermal imagery of the dwarfs. If there is indication of internal heating unexpected from gravitational compression of the core, or veins of thermal radiation could also help determine any activity in turn producing the magnetic field.

I hope this was what you were asking.


Why does Titan have an atmosphere while similar moons such as the Galilean moons don't?

As in real estate, so in astronomy:
Location, Location, Location
Where an object forms in a system will significantly impact its makeup.  For example, during the stage of planet(esimal) formation, due to the radiation of the (proto-)star, various chemicals and atoms will be expelled from the innermost areas, but may stay around long enough to be part of planet(esimal) formation further out.  This is the basic reason that the inner (rocky) planets, gas giants, and ice giants (and beyond that, Kuiper belt and Oort cloud objects) show significant compositional variations between each other and even within their own categories.
Galilean moons
Let's consider the impact of location on the Galilean moons of Jupiter.  The first one to consider is where in the solar system did the moon(s) in question form?  We're not 100% sure of the answer to this for any moon (even our own). But the current predominant theories hold that the Galilean moons formed from a disc of gas bound to Jupiter early during its formation.
Now when any astronomical body undergoes gravitational collapse, be it planet or star or whatever else, there must be a corresponding release of energy and heat.  This means that planets are pretty hot in their early lives, and the bigger they are the hotter they are and the longer it takes them to radiate it away (a volume versus surface area scaling consequence).  Jupiter, being the biggest and heaviest of the planets, also emitted the most radiation as a result of this process.  So much so that nearby moons would be scoured of various chemicals and gases, depending on how far away they were.  I bring this up in discussing the compositional variations among the Galilean moons in another answer.  Io, being the closest of the Galilean moons, was scoured of most of its ices, while the more distant Galilean moons have abundant and spectacular ices.  The magnetic field of Jupiter is even strong enough to strip Io of charged particles in its limited atmosphere, which are added via its vulcanism, and that ultimately results in sizeable radiation belts that substantially enhance Jupiter's magnetic field.
Titan
Titan is singularly unique, being the only (known) moon in the solar system to have a meaningful atmosphere.  Titan also has an unusually large orbital eccentricity, and is the only large moon around Saturn (in contrast with the Galilean moons of Jupiter, all large and of roughly similar sizes).  Moreover, the observations of the Cassini-Huygens probe suggest that Titan contains significant amounts of Oort cloud materials (its nitrogen in particular) that would not have been found in sufficient quantities in the Saturnian system at the time of Titan's expected formation.
The main theory for Titan's origins posits a similar disc accretion method as for the Galilean moons, and even that Saturn may have had a system of moons similar to the Galilean moons early on. Titan in this case is expected to have formed relatively further out. But then a series of large impacts disrupted the system in novel ways.  Impacts and viscous torques of the accretion disc caused any (large) inner moons to fall into Saturn, while Titan survives thanks to its more distant starting point and accretes some of the material of the impactors (some of which may have formed other moons, like Iapetus).  Some, perhaps all, of these impactors would be expected to have been Oort cloud objects, thus helping to explain the Cassini-Huygens data.  As long as these impacts occur after Saturn has lost its capacity to scour away atmospheres, Titan's accretion of their material then grants it its atmosphere and curious composition.


Is there an all sky meteor watch with spectroscopy?

You can do this kind of thing reasonably easily, even as amateur astronomer. See e.g. this link. There also have been programmes by the Americal Meteor society and there are a lot of published papers to that end, e.g. here
Two reasons it has not been done more widlely:
It's a difficult and very costly endevour to make high-resolution spectroscopy at very short time intervals - at the time you have for the meteor to show. There is no reasonable technique to justify a setup which points a large telescope within sub-second time to an arbitrary location as identified by a full-dome camera scanning the whole sky - to which you will have to attach the spectrograph.
At the same time there is little reason to believe that the outer material of meteorites which ablates in the upper atmosphere is different to what we find as meteorite on Earth. The surface is altered, and the processes are well-understood. Thermal conductivity and inertia are very small so that below this very shallow crust the material is unaltered. That said, missions like the Apollo missions as well as StarDust and similar did in-situ analysis of interplanetary material which has not been thermally altered in Earth's atmosphere.
General statistics on type of meteorites, including their chemical composition and origin can also be done with the finds from Earth's surface.
So in summary: the cost to gain ratio for such endevour might be much smaller than you seem to anticipate.


I believe the differences here are caused by the different timescales that are being used by SkyField vs NASA.

It seems that SkyField uses the proleptic Gregorian calendar for dates in the past. However, NASA used the Julian Calendar for dates before 1582, so for example, the eclipse on 0150-12-06 (Julian) falls on 0150-12-05 in Gregorian. Also, I would use ts.ut1 instead of ts.utc, as ut1 is the timescale used by NASA.

Using the geocenter can be a efficient way to check for eclipses worldwide. When observing from the geocenter, the penumbral shadow cone touches the Earth when the following condition is true:
$$\mu ≤ arcsin(\frac{r_m + r_e}{d_m}) + arcsin(\frac{R_s - r_e}{D_s})$$
Where $\mu$ is the angular separation between the Sun and the Moon, $D_s$ is the distance of the Sun, $d_m$ the distance of the Moon, and the three constant radii are:

Sun: $R_s = 695700 km$

Earth: $r_e = 6378 km$

Moon: $r_m = 1737.4 km$

This indicates that an eclipse, at least partial, is occuring somewhere on the Earth. The formula uses a spherical Earth as an approximation of its shape, so in some rare cases it can detect very near-misses by 20-25 km, but overall it is a good way to narrow down the search. Also, it is unnecessary to test this when the angular separation is outside the range 1.3° - 1.7°, since values below 1.3° always produce an eclipse, and values above 1.7° never produce one.

I would also like to point out that uncertainties, mostly in Earth's rotation speed, make it very difficult to know where eclipses occured on the surface of the Earth for dates in the far past. The uncertainty becomes very large outside the range BC2000 - AD3000. This is the reason why NASA hasn't computed eclipse paths outside that range. According to research, the uncertainty could be as large as over 4 hours (several thousands of kilometers for eclipse paths) for dates before BC4000. See this explanation on NASA's Eclipse Web Site.


What method is used to calculate the 'quality' of a solar/planetary image?

A little further research, and I found out that the basic principle, from which several variations derive, is to take the sum of the squares of the difference between adjacent pixels, to get a score.

The principle is that a higher quality picture has a higher probability that there will be significant differences in adjacent pixel values, i.e. that there can be a significant variation from pixel to pixel, whereas a lower quality picture will have similar valued adjacent pixels, as they have "blurred" together. Therefore, when taking the square of the difference of adjacent pixels and summing them together, higher quality pictures will show a significantly higher score, or summation.

I'm going to demonstrate this with a simplified example below, but I'm interpreting the meaning of "adjacent pixels" in a particular way, and I am hoping that if I am incorrect, someone will tell me and I can modify this answer appropriately.

Imagine we are imaging an object that exactly fills the frame of our small 16-pixel camera. In the middle of this object is a very dark square, that coincidentally happens to exactly fill the central 4 pixels of our camera. If the seeing for this object is perfect, then the camera will be exposed with a value of 255 for each of the light pixels, and 0 for each of the dark pixels.

This is how the differences between adjacent pixels would be calculated (I'm assuming only horizontal and vertical differences, not diagonal):

However, movement of the camera or poor seeing resulted in the light being spread more between the pixels, the calculations of the differences between adjacent pixels would be different:

The score, that is the sum of the squares of the differences, for the first (perfect) image is 520200.

The score for the blurred image is 304200. Therefore the sum of the squares of the differences between adjacent pixels suggests that the first picture is the better one.

This strategy isn't perfect, obviously. The presumption is that better images will exhibit a better score due to larger differences between adjacent pixels, but this is an assumption.

The squaring of the difference is intended to provide greater weight to larger differences, but I came across a comment on the PIPP support website that says that PIPP has modified the basic algorithm to give still greater weight to sharper contrast changes over lower contrast changes - so maybe squaring the differences isn't enough sometimes.

Apparently this quality measurement strategy doesn't work when the object is small in size on the sensor, and may be bright or even over-exposed. The above methods may select blurred and smeared frames over frames that are actually better. In this case, a strategy that looks at the histogram of values for the frame and selects the one with the higher peak, may produce a better selection of appropriate 'best quality' frames. It assumes that the histogram will have a higher count of frames sharing a pixel value across a smaller range, rather than spreading the light out across a wider range of pixel values.

An alternative quality measurement strategy simply adds up the values of all the pixels in an image. A higher score will indicate more light got through to the sensor, which implies that less cloud (for instance) got in the way.


Which planets, nebula, and other objects can be seen using Celestron Astromaster 130EQ

Okay, so that's a 130 mm newtonian with an f/5 focal ratio. Per the manufacturer's website, it comes with two eyepieces, 10 mm and 20 mm, giving 66x and 33x magnifications respectively.

First off, make sure your scope is collimated. The procedure is described in their generic Telescope Maintenance document, and also more specifically in the Astromaster Manual. See links on the site. The performance of your scope when observing planets depends a lot on collimation. I made it a habit to check collimation every time before I use the scope; I'm so used to it that it only takes 5 minutes now.

The topic of collimation is huge, just google "telescope collimation" and learn. I'm still learning new things after years in this hobby.

Once the scope is collimated, you could try to push magnification even higher, by using a 2x ... 3x barlow with either of the eyepieces - that would double ... triple magnification. Please be aware that more magnification is not always better; for any observation there's an optimal magnification, not too big, not too small. But I doubt your scope would do well above 200x, even though the theoretical limit is 260x. Probably not worth trying anything beyond the high 1XX values - that is, for your eyepieces, a 3x barlow.

For the Moon, any magnification works. Low magnification gives you broad images with lots of stuff to see. High magnification zooms into particular details. Good collimation is critical.

For planets, it depends.

Saturn looks good at any magnification. If you can see the rings, the shadow of the rings on the surface of the planet, and the main split inside the rings (the Cassini division) then you're doing fine. It doesn't have a lot of detail to justify very high magnification, unless you're looking for the finer ring divisions, but your scope can only see Cassini.

Jupiter is a bit tricky. It has enough fine detail to justify high magnification, but it has low contrast - and then as you increase magnification it gets more washed out, more dark, and hard to see. "Medium-high" magnification works best. Just try different eyepieces, with and without barlow, and see what works best for you. It also helps to keep your eye at the eyepiece, keep tracking the planet, and let your eye adapt - in time your brain/eye system will start pulling more detail out of the image. It takes me at least 2 ... 3 minutes to start seeing the really fine stuff there.

Mars - don't even bother unless it's at opposition. It only looks good when it's closest to Earth. The next Mars opposition is in July 2018, so there's a bit of wait until then (will start to look good a few months before that). Even then, you need several things to see Mars well:

very high magnification
excellent collimation
a good instrument
excellent seeing (the inverse of turbulence)

Otherwise it just looks like a red dot.

Seeing will affect all solar system objects (Moon and planets). When seeing is bad (when atmosphere is turbulent), any object will look bad. It depends on the weather - not just weather at ground level (what you think of as "weather") but weather in the whole thickness of the atmosphere, all the way up (the stuff only airplane pilots usually care about). If you live in North America, the Clear Dark Sky site can give you a prediction of a few parameters including seeing. When seeing is bad, high magnification is worthless (the image is fuzzy and keeps shaking like jello).

Another thing that helps: if there's a large temperature difference between inside and outside, leave the scope outside for a while to catch up with the air temperature. This will reduce internal turbulence and improve image quality on planets. I always leave my scope at "outside" temperature for about 1 hour before observing. When there is no temperature difference between inside and outside, then don't worry about this. This is like seeing (turbulence), but inside your scope.

Please note that light pollution has absolutely no effect on solar system objects (see below re: light pollution). Even when light pollution is very high, the Moon and the planets still look the same.

Dark adaptation does not matter for solar system objects (see below re: dark adaptation). You don't have to be in a dark place when observing them. I've looked at Saturn from a sidewalk under the bright lights downtown (city center), with cars and people passing by - no problem.

Deep sky objects (nebulae, star clusters, and galaxies), a.k.a. DSOs, are different.

These are objects with low brightness. Because of that, your eye cannot see details or colors. Collimation is less important. Seeing also doesn't matter much. Collimation and seeing affect fine details, but you can't see fine detail in DSOs.

What does affect them is light pollution - the light from all sources around you, glowing in the dust, water drops, etc in the atmosphere. This is why it's hard to see DSOs from the city. If you can drive 1 hour away from the city, to a place far from sources of light (cities, factories), then DSOs are easier to see.

Google "light pollution map" and you will find several sites that give you light pollution intensities in your area.

But even from the city, there are many DSOs that you can see. M31 (the Andromeda galaxy), the Orion nebula, the M13 cluster, the Dumbbell nebula - all these are clearly visible in an instrument like yours even from very light-polluted places. I've watched M13 many times from the middle of the Silicon Valley.

The trick is how to find these objects. Get a book called "Turn left at Orion" - it's a guide for beginners showing how to find many DSOs. Get a good planisphere (star chart) like this one and keep it next to the scope when you're observing DSOs. Start with something easy like the Orion nebula.

Also you could google "how to find the Orion nebula" (or M13, etc) and you'll find online guides which are sometimes helpful.

Usually you don't need high magnification for DSOs. Low and medium magnification work well - but it depends; e.g. I've watched M13 in high-ish magnification and it looks good. What really does help here is eyepieces with a wide field of view, because it makes it easier to find the objects. I'm guessing your eyepieces are of the Plossl design, with a 52 degree FoV, which is fine, but more is better for DSOs. Now don't go ahead and buy some fancy expensive eyepieces - keep them proportional in price with the scope, and the original ones should be fine for now.

Another thing that really helps with DSOs is telescope size - bigger is always better. That's also true for solar system objects, but it's even more important here.

Finally, make sure your eyes are dark adapted before looking for DSOs. Observe from a dark place, and stay in the dark for at least 15 ... 20 minutes before observing. Do not use anything with a screen (smartphone, computer) because that destroys your dark adaptation. Do not use flashlights. There are flashlights for astronomers, with a red LED and a dial button, and you need to dial down the light to the least amount that allows you to see and read the star chart. Very weak red light will not destroy your dark adaptation. You could also improvise a red flashlight, but make it as weak as possible (or put a dial button on it).

Clear skies to you and good luck!


How do we know that comets definitely mase and not just fluoresce?

I'm not sure if this is what you're asking but:
fluorescence:   excited electrons randomly emit photons.
Superfluorescence:  excited electrons emit photons, and there is some self-stimulation, leading to a strong short pulse but little coherence.
Superradiance:    strong, shorter pulse,  much self-stimulation , high coherence in output.
Lasing/Masing:  almost always requires a cavity of some sort, leading to directional coherent output.
I cannot tell exactly what the authors of the article consider qualifying as Maser vs. Superradiance.  possibly the  material volume in question (on the comet) is elliptical enough to lead to a more-or-less directed beam.


Why the earth orbits the sun from X to -X and not the opposite

The thing that keeps the Earth from falling into the Sun is that it orbits the Sun, once a year. We usually don't say that the Sun orbits the Earth.
Orbits have a direction or sense in space. What that direction is called is an issue of history and linguistics,
The Earth rotates around its own axis, this rotation also has a sense, and the same caveats about what this is called apply.
It is the daily rotation of the Earth around its axis and the tilt between this axis and the axis of the Earth's orbit around the Sun that determines for a given spot on the Earth where along the horizon the Sun will rise and set.
If nothing else changed except the Sun suddenly started rising where it normally sets, and setting where it normally rose, that would mean that the Earth started to rotate in the opposite direction around its axis.
We would call that retrograde rotation and it would be a surprise.
It's generally believed that the sense of rotation of most (but not all) planets around their axes is similar to the rotations of the planets around the Sun is evidence of the rotation of all the matter that formed the Sun and the solar system. The Moon also rotates with the same sense around the Earth and the Sun rotates with the same sense around its own axis.

What would happen if that were the case?

This is a great question and is hard to answer. It depends a great deal upon if the Moon was also now in the opposite sense or not.
But I can say one thing that would happen is that people would wonder why the Earth was rotating in the opposite sense of most other things in the solar system!


Could dark matter assist with star formation?

Strange you should ask - I am currently working on a paper on NGC 2516 - "the southern Pleiades" and looking at the dynamical status of the stars compared with the distribution of visible mass.

Our conclusion is that the radial velocities of the stars are in virial equilibrium, with velocity dispersions that are entirely consistent with the mass that is present in the stars that are visible. So, no need for dark matter there - you could probably rule out any contribution of more than about 50% of the visible stars. (NB the point of the work is not to look for dark matter, is is not expected - see below).

This is one of the first very detailed measurement for an open cluster (like the Pleiades), because the velocity precisions required have to be considerably better than the velocity dispersion. And for a sparse cluster like the Pleiades, this is $<1$ km/s, which is challenging.

There is one old paper by Jones (1970) that uses proper motions in the Pleiades to estimate a dynamical mass based on virial equilibrium of $690\ M_{\odot}$, which compares with $470\ M_{\odot}$ from directly counting stars. But Jones points out that the latter number is a lower limit (they certainly couldn't see very low mass stars and brown dwarfs in their study) and the former number has large uncertainties. So there is no real evidence for any "dark matter".

The dynamics of globular clusters are easier to measure - they are more massive and have much higher velocity dispersions. The general picture is reasonable agreement between velocity dispersions and inferred mass distributions from counting the visible stars. There are of course uncertainties, but there can be no large dark matter component. Limits as small as $<6$% dark matter have been set in some clusters (e.g. Ibata et al. 2012).

The first Gaia satellite results, due in about 18 months time, will blow this field apart. We will have exquisitely precise tangential velocities for $\sim 1000$ stars in the Pleiades.

Dark matter is not expected to play a role in small scale (star cluster) formation. When clusters form, the dissipation of energy in gas interactions is what allows them to end up as bound systems. Dark matter is dissipationless. As the escape velocity of these clusters is 1-10 km/s, compared to velocities of hundreds of km/s for datk matter, there is not even expected to be any gravitational concentration of dark matter. This is quite different from galaxy formation, which is associated with pre-existing dark matter structures.


How are non-visible electromagnetic waves detected that originate from space?

All electromagnetic radiation is fundamentally indistinguishable - i.e., all rays of light (electromagnetic radiation) are composed of photons - and they only vary in their properties, such as the energy and phase of the light. Thus, detecting light works fundamentally the same way for any electromagnetic radiation, e.g., the light is typically concentrated by some kind of system (for example, in telescopes and in our eyes this system is composed of lenses), though this is not completely necessary always, and then the light is absorbed by a detecting device; however, in practice light of certain energy/frequency requires different detecting devices. In your eye/brain system, the detection device is an array of the photoreceptor cells (rods and cones) in the retina. For observing light from objects out in space, we use telescopes, where there is a range of such detection devices that have been invented over the years.

(https://www.nasa.gov/directorates/heo/scan/communications/outreach/funfacts/txt_radiowaves.html)
Modern telescopes utilize photodetectors (e.g., a CCD chip) that convert light into electrical current which can be read into a computing device and interpreted, and many more complicated optical systems (e.g., adaptive optics) are used to study the light received from distant objects in space. These detectors/receptors are based on solid-state physics (another example of how quantum theory helped revolutionize astronomy research), since they usually are composed of a P-N junction (which is just one application of P-N junctions for harvesting light and converting it to energy). As explained above, all electromagnetic telescopes fundamentally use photodetectors, but how the light is harvested at the photodetector depends on the energy of the light.
The wiki article for solid-state photodetectors explains the various types and the mechanisms of their operation, if you have more questions about this please start a new question. Note, that biological photoreceptors operate on different mechanisms than solid-state ones.
For visible light, obviously, the telescope focuses the light onto the viewer's eye! But you asked about non-visible light specifically. Visible light is only a very small sliver of the electromagnetic spectrum. On the low-frequency end of the spectrum are radio waves, which require very large antennas to collect enough radio energy to study them and extremely sensitive receiving equipment.
As the energy (frequency) of the light increases we next have infrared which is often detected with the same telescopes used to detect visible light, just with a different photoreceptor. The light that has higher energy than visible light - UV, X-ray, an gamma-ray - where most of the light at these wavelengths is absorbed by the Earth's atmosphere, so observations at these wavelengths must be performed from the upper atmosphere or from space.


It looks like @planetmaker's suggestion of using WSL will work.
On my Windows machine I've had Ubuntu 18.04 installed for a while, so have tested using that; but this should work with the latest version of Ubuntu as well.
Here goes:

Start Ubuntu from the Windows Start Menu
In the Ubuntu terminal window which appears, type sudo apt install iraf
The sudo command will ask for your password, so type it in and press enter.
Now you'll see a whole load of text, like this:

Reading package lists... Done
Building dependency tree
Reading state information... Done
The following additional packages will be installed:
  blt cpp-7 fontconfig-config fonts-dejavu-core g++-7 gcc-7 gcc-7-base iraf-noao javascript-common libasan4 libblas3
  libcfitsio5 libcilkrts5 libcminpack1 libdrm-amdgpu1 libdrm-common libdrm-intel1 libdrm-nouveau2 libdrm-radeon1
  libdrm2 liberfa1 libfontconfig1 libfontenc1 libfuntools1 libgcc-7-dev libgfortran4 libgl1 libgl1-mesa-dri
  libgl1-mesa-glx libglapi-mesa libglvnd0 libglx-mesa0 libglx0 libice6 libjbig0 libjpeg-turbo8 libjpeg8 libjs-jquery
  libjs-jquery-datatables liblapack3 libllvm10 libpciaccess0 libsensors4 libsm6 libstarlink-ast-err0 libstarlink-ast0
  libstarlink-pal0 libstdc++-7-dev libtcl8.6 libtiff5 libtk-img libtk8.6 libubsan0 libwcs5 libwcstools1 libx11-xcb1
  libxaw7 libxcb-dri2-0 libxcb-dri3-0 libxcb-glx0 libxcb-present0 libxcb-shape0 libxcb-sync1 libxcomposite1
  libxdamage1 libxfixes3 libxft2 libxi6 libxinerama1 libxmu6 libxpa1 libxpm4 libxrandr2 libxrender1 libxshmfence1
  libxss1 libxt6 libxtst6 libxv1 libxxf86dga1 libxxf86vm1 python3-astropy python3-numpy python3-pluggy python3-ply
  python3-psutil python3-py python3-pyraf python3-pytest python3-pytest-arraydiff python3-pytest-astropy
  python3-pytest-doctestplus python3-pytest-openfiles python3-pytest-remotedata python3-stsci.tools python3-tk saods9
  saods9-doc saods9-tclpackages tcl tcl-signal tcl-xpa tcl8.6 tcllib tclxml tk tk-html1 tk-mpeg tk-table tk8.6
  tk8.6-blt2.5 tkblt tkcon tksao x11-common x11-utils xbitmaps xterm
Suggested packages:
  blt-demo gcc-7-locales g++-7-multilib gcc-7-doc libstdc++6-7-dbg gcc-7-multilib libgcc1-dbg libgomp1-dbg libitm1-dbg
  libatomic1-dbg libasan4-dbg liblsan0-dbg libtsan0-dbg libubsan0-dbg libcilkrts5-dbg libmpx2-dbg libquadmath0-dbg
  iraf-dev apache2 | lighttpd | httpd lm-sensors libstdc++-7-doc libtk-img-doc libxml2-utils python-astropy-doc
  python3-astropy-affiliated python3-beautifulsoup python3-h5py python3-matplotlib python3-pandas python3-scipy
  python3-tz gfortran python-numpy-doc python3-nose python3-numpy-dbg python-ply-doc python-psutil-doc subversion
  ipython3 python3-urwid tix python3-tk-dbg python3-pyds9 xpa-tools tcl-tclreadline tcllib-critcl mesa-utils
  xfonts-cyrillic
The following NEW packages will be installed:
  blt fontconfig-config fonts-dejavu-core iraf iraf-noao javascript-common libblas3 libcfitsio5 libcminpack1
  libdrm-amdgpu1 libdrm-intel1 libdrm-nouveau2 libdrm-radeon1 liberfa1 libfontconfig1 libfontenc1 libfuntools1
  libgfortran4 libgl1 libgl1-mesa-dri libgl1-mesa-glx libglapi-mesa libglvnd0 libglx-mesa0 libglx0 libice6 libjbig0
  libjpeg-turbo8 libjpeg8 libjs-jquery libjs-jquery-datatables liblapack3 libllvm10 libpciaccess0 libsensors4 libsm6
  libstarlink-ast-err0 libstarlink-ast0 libstarlink-pal0 libtcl8.6 libtiff5 libtk-img libtk8.6 libwcs5 libwcstools1
  libx11-xcb1 libxaw7 libxcb-dri2-0 libxcb-dri3-0 libxcb-glx0 libxcb-present0 libxcb-shape0 libxcb-sync1
  libxcomposite1 libxdamage1 libxfixes3 libxft2 libxi6 libxinerama1 libxmu6 libxpa1 libxpm4 libxrandr2 libxrender1
  libxshmfence1 libxss1 libxt6 libxtst6 libxv1 libxxf86dga1 libxxf86vm1 python3-astropy python3-numpy python3-pluggy
  python3-ply python3-psutil python3-py python3-pyraf python3-pytest python3-pytest-arraydiff python3-pytest-astropy
  python3-pytest-doctestplus python3-pytest-openfiles python3-pytest-remotedata python3-stsci.tools python3-tk saods9
  saods9-doc saods9-tclpackages tcl tcl-signal tcl-xpa tcl8.6 tcllib tclxml tk tk-html1 tk-mpeg tk-table tk8.6
  tk8.6-blt2.5 tkblt tkcon tksao x11-common x11-utils xbitmaps xterm
The following packages will be upgraded:
  cpp-7 g++-7 gcc-7 gcc-7-base libasan4 libcilkrts5 libdrm-common libdrm2 libgcc-7-dev libstdc++-7-dev libubsan0
11 upgraded, 108 newly installed, 0 to remove and 277 not upgraded.
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After this operation, 526 MB of additional disk space will be used.
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Watch as it downloads and installs over one hundred packages...
When it's done then you can run the new program by typing irafcl and pressing enter:

This package contains tasks for viewing and manipulating 3d images.
It is a pre-release version, and does not reflect the ultimate
partitioning of n-dimensional image tasks within IRAF

  NOAO/IRAF Debian 2.16.1+2018.03.10-2 Wed, 14 Mar 2018 09:02:35 +0000

  This product includes results achieved by the IRAF64 project in 2006-
  2009 directed by Chisato Yamauchi (C-SODA/ISAS/JAXA).

  Welcome to IRAF.  To list the available commands, type ? or ??.  To get
  detailed information about a command, type `help <command>'.  To run  a
  command  or  load  a  package,  type  its name.   Type  `bye' to exit a
  package, or `logout' to get out  of the CL.

  Visit http://iraf.net if you have questions.

  The following commands or packages are currently defined:

      dataio.     images.     lists.      obsolete.   proto.      utilities.
      dbms.       language.   noao.       plot.       system.

ecl>

Now, although it's installed and runs its command-line interface, I think that you'll encounter various issues when trying to run it: among the packages installed were some which require X Windows, and those won't work when running under Windows (you'll see error messages talking about being unable to access devices, or "Can't open display. DISPLAY is not set").
I advise you to test anything 'graphical' - that is, anything that tries to draw graphics to a display - before you invest too much time.
I think you might end up having to install a virtual machine instead.


Which software is appropriate to integrate radial velocity with respect to time

I'm not sure what you are actually trying to do. Radial velocity isn't generally measured, it is computed from the projected velocity. But if I take you at your word, that your raw data is radial velocity, then that simplifies things.

If you know the period, $P$, and you know the $v$ AND if you know the theoretical relationship between them $v = f(p)$ then you can use linear least squares to compute the adjustable parameters of $f(p)$ by plotting $v$ (measured) vs $v$(calculated).

Any 'best fit' equation is able to interpolate values of $v$ so all you need to divide up the period into small intervals and approximate the area under the curve using its two end-points (it is rarely necessary to use parabolic fits that also use the mid-points of those small intervals).

Thus the problem of integration becomes one of simple geometry (you have a tall thin rectangle with a triangle on top of it for each interval). Let's think about a star with a 30 day period. That's ~40,000 minutes, so I'd probably use 5 or 8 minute intervals ...starting at minute 0 what's the area under the curve from t=0 to t=8 minutes? then from 8 to 16, all the way up to 43,200th minute. Sum of those areas is total area. An Excel spreadsheet can easily do that.

Actually, if you also use 5 minute intervals and compare the results you'll see how much better (hopefully NOT much) a smaller interval is...analytical integration is this taken to the limit of the interval approaching zero, but in real life, numerical integration is necessary for almost all 'interesting' problems.  Here are 2 links, fwiw:

https://www.physicsforums.com/threads/finding-periodic-best-fit-equation-for-data-set.715638/
http://www.aanda.org/articles/aa/pdf/2004/08/aah4718.pdf ← see page 537 and references noted there.
The Institute for Astronomy (IFA) at the University of Hawaii has several pdfs on BW that may be helpful


What is the current accepted theory for Neptune's immense wind speed?

Disclaimer: Most of what is written here is based highly on theory and speculation. Voyager 2 is the one and only trip we have taken to Neptune and most of these theories are built upon the data it has sent back. Leave large room for error

Is Neptune windy?

Before we start answering why Neptune is windy we should first specify if Neptune is even windy in the first place. What proof do we have?

The Great Dark Spot

One observation was of a large abhorrence on the surface of Neptune aptly titled "The Great Dark Spot". Repeated viewings placed the movement of the GDS to be about 700mph.

Sourced from: http://nineplanets.org/neptune.html

Keck Telescope

A lot of observations and images of Neptune have come from the Keck II Telescope, which is able to observe other markings moving at similar to faster speeds on Neptune.

Sourced From: http://www.keckobservatory.org/recent/entry/keck_images_of_neptune_best_ever_captured1

What causes the winds on Neptune?

It's theorised that there are two potential causes to Neptune's winds. Shallow processes in the outer atmosphere or much deeper atmospheric changes that extend into the interior.

http://www.space.com/21157-uranus-neptune-winds-revealed.html (paraphrased for simplicity) Claims that

  The researchers found the windy layers of Neptune occupy the outermost 0.2 percent of It's mass, suggesting that shallow processes drive those winds, such as swirling caused by moisture condensing and evaporating in the atmosphere.

What is the main source of energy for these winds?

So now you are probably wondering where Neptune get's this energy from as it is so much further from the Sun than we are.

http://nineplanets.org/neptune.html Claims that this is due to Neptune's own internal heat source radiating twice as much energy as it receives from the Sun.

Why does Neptune have stronger winds than Earth then?

So theoretically, if it was down to energy, it would make sense that Earth, being so much closer to the Sun, would have much more energy required to reach these speeds.

Yes and No.

The Sun is the main provider for our energy but it also creates turbulence in the atmosphere. It's cooling and heating of the layers keeps air constantly rising and falling, disturbing natural air flows and stemming the size of our storms.

On top of this http://www.space.com/18922-neptune-atmosphere.html claims that 80% of Neptune's atmosphere is molecular Hydrogen.

Hydrogen being the smallest and lightest element requires less energy to reach high speeds than our larger, nitrogen based atmosphere. This may not be a main factor to consider but it is certainly contributing in no small amount.


Could the earth become a star if more mass was added?

Regarding the title: Yes.

  Does this mean that the star started off as a planet?

Yes, a star could technically start out as a planet, if it accreted enough mass. However, this is extremely unlikely, since the planet would need to be 80x the mass of Jupiter for it to undergo nucleosynthesis.

  Stars require hydrogen fusion and earth has little H. Could the earth become a star if more mass was added, but its relatively low abundance of hydrogen remained the same?

The most important components for a star is the composition and mass. As it stands, Earth doesn't have nearly the concentration of hydrogen for a star to form, even if we kept adding mass.

Most stars work by fusing the lightest element, hydrogen. To fuse heavier elements requires a much, much higher temperature, since the more protons the element has, the more it will repel other ones instead of fusing. Yes, massive enough stars can fuse heavier elements.

However, Earth would inevitably grow in hydrogen, so it would eventually have a significant amount. Above a certain size, given ambient conditions (like radiation pressure, light intensity from the star, etc), Earth would become massive enough that it starts capturing hydrogen gas from the interplanetary medium, and becomes a gas giant (as called2voyage pointed out). Where this happens isn't precisely known, but it's roughly 10x Earth's mass. This means that, theoretically, a terrestrial planet could form a star if you increased its mass ever further.

Now, I can hear you thinking, "What if Earth were surrounded by a vacuum instead? If we kept adding mass, would it become a star?" Well theoretically, the planet would eventually reach temperatures where most of its elements could fuse. For example, the "star" could sustain itself on oxygen and silicon fusion, but not on iron fusion.


Why is the Solar Helical (Vortex) model wrong?

It isn't correct, because a vortex is not a helix, and so while the planets do trace a helical path as they move through the galaxy, this is not evidence of a vortex.

Yes, the sun actually is moving through space, as it traces a path around the centre of the galaxy. The whole mass of the solar system moves with it, so the planets are not left behind as the sun moves.

Rhys Taylor and Phil Plait have comprehensive smackdowns debunking this vortex idea and other misunderstandings/delusions by the author.


If Pluto is tidally locked with Charon, is its rotation still slowing down?

Short answer.  Pluto's rotation is 100% governed by Charon, or, more accurately, the Pluto-Charon system.  The sun is irrelevant to it's rotation.

Small point to add.  Pluto's rotation is pretty much governed by the Pluto-Charon orbit.  If Charon moves closer to Pluto, Pluto's rotation will correspondingly increase.   If Charon moves further away, Pluto's rotation will decrease.  That's how tidal locking works.   I answered the question in regards to the orbit changing, not the rotation changing, but in a tidally locked system the rotation period follows the orbital period.

Long answer:

  I know that a a planet or moon's rotation will slow down or speed up
  until it is tidally locked with a body,

Tidal locking doesn't always happen.  For distant orbits the tidal force that makes tidal locking happen is often very weak and very slow, often requiring more time than the lifespan of the star in that solar system.

Also, for high eccentricity orbits, classic tidal locking is impossible though resonant orbits like Mercury's 3:2 around the sun are probably fairly common.   And many moons will escape their planet or crash into their planet long before mutual tidal locking occurs, though it's not uncommon for moons themselves to be tidally locked.   Also small moons don't have sufficient gravity to pull the object they orbit into mutual tidal locking.

For mutual tidal locking, like Pluto-Charon, there needs to be respectible mass ratio or, perhaps an initial formation where orbital speed and rotation are very close, so it doesn't take much force to lead to mutual tidal locking.  Our moon, at 1/81st the mass of the Earth, would eventually cause the Earth to tidal lock to it, but estimates say it will take longer than it will take the sun to go red-giant for that to happen, and the expected outcome of our moon is that it will crash into the Earth before there is mutual tidal locking.

So, mutual tidal locking like pluto-charon is actually pretty rare because you need the smaller object to have a significant percentage of mass to the larger one.   One-way tidal locking is quite common, but your question addresses mutual tidal locking.

  is Pluto's rotation being affected by the sun or is it being forced by
  Charon just to to be tidally locked with Charon?

For a quick approximation, the tidal force can roughly/bad approximation be estimated by the size of the object in the sky times it's density.   There's other factors, ofcourse, like the size of the object receiving the tidal force, but for a rough approximation, the size in the sky gives an estimate.   For example, the sun and moon each take up about 1/2 a degree of arc when viewed from the surface of the Earth, so they have similar tidal forces due to having similar size.   The major difference being that the Moon is about 2.4 times more dense than the sun so it's tides are correspondingly about 2.4 times as strong.

This visual size approximation doesn't work for very-near-earth orbits, as it's the size that would appear from the center of the Earth, not the surface that should be applied.

Charon from the surface of Pluto is huge.   It spans several degrees of arc, over 10 times the size the Moon appears from the Earth.   The sun, from Pluto is tiny about 1/30th the size it appears from Earth, which is only about 50% larger than Venus appears in the sky at it's largest, or, about 1 minute of arc, but even at that distance the sun is still bright enough to cast visible shadows and would probably damage your retina if you stared at it.   It would be the most impressive light in the sky from Pluto, but still tiny, so the tidal effect on Pluto-Charon would be quite small.  (I'd prefer not to do the math, my math tends to run into errors and I'm not good with mathjax).

But the sun's tidal influence on the Pluto-Charon system is close to negligible, It can be calculated and it's not zero, but it's very small.   Over a very very very long time, the Sun's tidal force will sap energy from the Pluto-Charon system and draw them closer towards each other, but at that distance, billions maybe trillions of years.  Pluto is probably more likely to meet a different fate well before then, such as crashing into Neptune or being thrown out of the solar-system before the Pluto-Charon system becomes unstable.

It would be different if Pluto-Charon orbited at Mercury or Venus distance (ignoring the obvious melting of ices and outgassing in a warmer environment).  That close, the sun becomes a major player in the future of the mutually tidally-locked systems, likely causing them to crash into each other, but at Pluto distance, there's virtually no effect.

When a system is mutually tidally locked with a nearly circular orbit, then it's very stable.  If there's no tidal locking or one way tidal locking like Earth-Moon, then there's a tidal bulge that leads or follows the moon and that creates a force that can slow down or speed up the planet and push the moon away or draw it in closer.   With tidal locking, there's no leading or following with the tidal bulge, so the orbit is less prone to change.   There's always some orbital variation and some orbital energy converted to heat due to no orbits being perfectly circular, but mutual tidal locking is still a very stable orbital system with less change than most, as long as there isn't a 3rd party forcing change, such as a debris cloud creating drag or outgassing or orbiting close to a massive body that saps energy from the tidally locked orbit.

No orbits are stable forever, and considering all the factors, Pluto and Charon are close enough that they exchange some atmosphere and as the sun grows more luminous over time, that exchange will probably increase, and likely both are losing mass slowly over time due to the solar wind.    This very gradual loss of mass, I would think, would cause the orbit to expand, but very slowly.   With no loss of gas by the solar wind, the Pluto-Charon orbit should very slowly contract due to tidal forces from the sun and some orbital energy being converted into heat, but both of those effects are very slow.   Mutual tidal locking is for the most part a very stable system.

Hope that wasn't too long, but that's the gist of it.   My answer could probably be improved with some math and some linked references but that would make it even longer.


Just to clarify, inner planets usually refers to Mercury, Venus, Earth, Mars.    But in relation to your question, the "inner" planets are just Mercury and Venus and the outer planets are Mars and out, because the +1 / -1 formula is in relation to whether the Earth is passing the planet as it orbits the sun or being passed by the planet.

Someone else can do the math if they like, the gist is explained here.  Because it's the motion relative to the stars in the background that you're tracking, anything that orbits closer to the sun and faster than Earth would move in a different direction and require a different formula than orbits further out and slower.

If your asteroid remains outside the Earth's orbit, then it would obey the outer planet formula.

If your asteroid is an Earth crosser, which isn't uncommon, then it would have a different type of orbital period that would follow neither formula, and I'm not sure what formula would apply for an earth-crossing asteroid.


I'm not entirely sure what your source of confusion is, so I'll just describe how temperature and degeneracy relate and hope that clears it up.

Fermions

All particles which are fermions are subject to the Pauli Exclusion Principle. This means no two fermions can occupy the same quantum state. They will resist doing so. What that really means, doesn't matter too much so for now, just accept it as an axiom.

Electrons are fermions. A star, before it becomes a white dwarf, is filled with electrons just minding their own business. Then, as the star begins to form itself into a white dwarf, the crushing pressure of gravity squeezes all the electrons into a tight space. Gravity tries to push these electrons into the same quantum state, but the Pauli exclusion principle forbids this.

Fermi Energy

Note: The following is a gross over-simplication, but it gets the idea across without being too terribly wrong. Don't take it as gospel truth.

You can define a set of all the possible quantum states. For simplicity, I'm just going to call them $\mathrm{state}\ 1$, $\mathrm{state}\ 2$, ..., $\mathrm{state}\ 67$, ..., $\mathrm{state}\ 10^{10483}$, etc. (ordered by energy). From above, we know each state can be occupied by only one electron. In general, these electrons will occupy a variety of non-sequential states. Pre-white dwarf, we might order the electrons by state and say

$\mathrm{electron}\ 1$ is in $\mathrm{state}\ 43$,
$\mathrm{electron}\ 2$ is in $\mathrm{state}\ 12084$,
$\mathrm{electron}\ 3$ is in $\mathrm{state}\ 4.187\times10^{78}$
etc.

As the electrons are squeezed in the white dwarf, they'll try to get as close to the same quantum state as possible without violating the Pauli exclusion principle. A fully degenerate star would have electrons in the following states:

$\mathrm{electron}\ 1$ is in $\mathrm{state}\ 1$,
$\mathrm{electron}\ 2$ is in $\mathrm{state}\ 2$,
$\mathrm{electron}\ 3$ is in $\mathrm{state}\ 3$
etc.
$\mathrm{electron}\ n$ is in $\mathrm{state}\ n$

In this completely degenerate state, all electrons have been forced to occupy all the lowest energy states. The Fermi energy is the energy of the last electron in the highest state. If all electrons happen to be squeezed such that they're all below the Fermi energy, then we consider the object degenerate.

Temperature and Degeneracy

Gravity might try to squash all the electrons down into the lowest quantum states, but the ability of gravity to achieve this depends on the temperature. Electrons at a higher temperature have higher energies and it becomes harder for gravity to suppress them. With a certain set of assumptions1, one can say that for a star of a given temperature, $T$, and a given density, $\rho$, it will be degenerate if2:

$$\frac{T}{\rho^{2/3}} < 1261\ \mathrm{K\ m^2\ kg^{-2/3}}$$

From this equation, you can hopefully see that being colder actually helps a white dwarf be degenerate. The two ways to make a star (or anything really) become degenerate are to cool it down and increase the density. For white dwarf formation, the fact that it's degenerate is more a function of the high density than it is the cool temperature.

1Assumptions and preceding equations are in An Introduction to Modern Astrophysics, 2nd ed., pg. 566.

2As a preemptive measure, this equation does not apply universally. You cannot apply it to yourself and declare yourself degenerate!


What fraction of galaxies host active galactic nuclei?

As Pela said in the comment, it is highly redshift dependent so there is no straight answer. "Some" is the safest option.


A detached black hole is simply a black hole that is not part of a close binary pair.
There are two ways that we can detect stellar mass black holes. Firstly when they are close to a star, and matter is flowing from the star into a disc around the black hole.  Matter in the disc will orbit at substantial fraction of the speed of light, and friction will cause it to heat up to extreme temperatures, hot enough to emit X-rays.  If we can detect these X-rays and we can work out the mass of the object, we can deduce the existance of a black hole.
The second way is through gravitational wave astronomy, detecting the gravitational waves emitted when two black holes merge.
Detached black holes, that don't have a partner, are much harder to detect.  As there would be no disc of hot matter they would not emit x-rays.  In fact they would not emit anything. A detached black hole could be part of a binary system, but far enough from the star that no matter transfers from the star to the black-hole.  Such a black hole could be found by observing the radial velocity of the star. The gravity of the black hole will cause the star to orbit around it, the forwards and back motion of the star can be detected and the mass of the object can be deduced.  A large but invisible mass is a likely black hole.  See A detached stellar-mass black hole candidate in the globular cluster NGC 3201


The spectrum of a star is almost certainly a unique fingerprint. Even though stars are born in clusters, formed from the reasonably chemically homogeneous environment of a giant molecular cloud, there are likely to be small differences in their local environment. Furthermore their formation environments, and the stars themselves at later times, can be polluted by external events (e.g. nearby supernovae). Even if their chemical abundances were identical, their spectra would not be because of differences in their mass, rotation and hence their temperatures.

But, the differences between stars may be smaller than the measurement precision, rendering their spectra effectively indistinguishable.

To give any sort of definitive answer requires some definition of what types of stars are being observed, their temperature, gravity and chemical abundance dispersion, at what spectral resolution and with what signal to noise ratio in what waveband.

At the moment I would say this is impossible for any star in our Galaxy. The differences in the spectra between one star and its most similar "twin" (the word might be appropriate here, as they were likely born in the same stellar nursery) are far smaller than we can distinguish with current data and techniques.


Calculate angle of inclination for a certain angle in a planet's orbit?

In short, no, you don't have to solve Kepler's equation. If you define a viewing longitude and you know the full set of Keplerian orbital elements, you can calculate if a planet will show a transit.

Let me first clear up some of the jargon, though. The term inclination is traditionally reserved for the angle between the orbital plane and the reference plane. So unless the entire orbit is wobbling (for instance due to precession) the inclination will be constant for all positions of the planet.

The angle that varies as the planet moves through the reference plane is the latitude. However, in this particular problem calculating the latitude isn't particularly useful. The more practical coordinate to look at is the height of the planet away from the plane, I'll call this $r_z$. If along your observers line of sight you find $r_z$ to be smaller than the solar radius, the planet will show a transit.

Working out a fully general elliptical orbit in 3 dimensions is somewhat tricky, so I'll start out with a simple Cartesian coordinate system aligned with the orbital plane. In subsequent steps I work my way back using a number of coordinate transformations.

First a list of the Kepler parameters and their meaning:

$\nu$ - true anomaly - The anglular position with respect to perigee in the orbital plane.
$\omega$ - argument of periapsis - The angular position of perigee with respect to the ascending node in the orbital plane.
$\lambda_{asc}$ - the longitude of ascending node in the invariable plane ($\Omega$ in the picture).
$i$ - inclination - the angle between the orbital plane and the invariable plane, such that $\lambda_{asc}$ gives the rotation axis.

Now, in the orbital plane, the radius of the planet orbit is given as
$$
r(\nu) = \frac{a(1-e^2)}{1+e\cos(\nu)}
$$
If I define the x-axis to align with the ascending node, I can write the position vector as
$$
\vec{r}_{orb} = r(\nu)
\begin{bmatrix}
\cos(\omega+\nu) \\
\sin(\omega+\nu) \\
0
\end{bmatrix}
$$

To move to a Cartesian coordinate system aligned with the invariable plane I have to apply a rotation of angle $i$ about the x-axis, so
$$
\vec{r} = R_x(i)\vec{r}_{orb}= r(\nu)
\begin{bmatrix}
\cos(\omega+\nu) \\
\sin(\omega+\nu)\cos(i) \\
\sin(\omega+\nu)\sin(i)
\end{bmatrix}
$$
So now I have an expression for $r_z$, I just need to know how $\nu$ relates to the planet longitude in the invariable plane, $\lambda_{planet}$.

I can find the longitude of the planet relative to $\lambda_{asc}$ as
$$
\lambda'_{planet} = \arctan( \frac{r_y}{r_x} )
$$
such that the actual longitude is
$$
\lambda_{planet} = \lambda_{asc} + \lambda'_{planet}
$$
Inverting this last expression gives
$$
\nu(\lambda) = \arctan( \frac{\tan(\lambda - \lambda_{asc})}{\cos(i)} ) - \omega
$$

Now I have all the equations to answer the question:
If the planet is along the observer's line of sight, then is has a true anomaly
of $\nu(\lambda_{obs})$, which gives a height above the plane
$$
r_z(\nu) = r(\nu)\sin(\omega+\nu)\sin(i)
$$
If $|r_{z}| < R_{sun}$ the planet can be seen in transit.


Why different specifications for telescopes and binoculars?

With a binocular, all its optical components are fixed -  the user can't change them.  What's important for the user to know is the size of the front lens, which determines the brightness (and in theory sharpness) of the image, the magnification, and the field of view.  These are all useful things to know.

A telescope has an interchangable component, namely the eyepiece.  The choice of eyepiece determines:

The magnification (= focal length of telescope divided by focal length of eyepiece)
The field of view, which depends on magification and design of the eyepiece
The exit pupil size (diameter of objective divided by magnification)
and a few other things

Hence, based on a particular eyepiece type you can work out the same information as you get with binoculars.


What will happen to the shape of a galaxy when a super massive black hole lying in its center dies(evaporates out)?

Presumbably we rotate beacuse of the BH.

No. The galaxy is being held in one piece due to its own total gravity. The black hole is only a small fraction of that. Basically, the BH doesn't matter.

  When the black hole dies in our galaxy

The BH will probably be the last thing left of our galaxy at the end. And even then it will take some incredibly long time for it to evaporate. BH evaporation for very large BHs is basically the slowest process you could imagine.

  It will be some irregular shape not spherical?

The galaxy is not spherical. Its shape is rather more like a round disk (with some irregularities and some features like arms, etc).


Why in my area does NEOWISE appear so dim and small?

There's one main reason - and maybe some other reasons as well:
The primary reason is the date the photos were taken you compare your photo with. Two weeks ago (10th July 2020), the brightness of the comet was around 1 - 2 magnitudes - that is comparable to the brighter stars and it was readily visible with the unaided eye on onset of nautical twilight. A week later already (17th July) it moved further from the sun and it was only around 3.5 magnitudes (personal guestimate from comparison with surrounding stars like some from Canis Minor) - thus it needed already a dark sky to be nicely visible. Now is already another week past and its heliocentric distance increased further, its brightness dropped to something comparable to the stars just barely visible with the naked eye. Conversely its activity also reduced dramatically with increasing heliocentric distance. As it meanwhile has past its perihelion as well as closest approach to Earth, it's meanwhile both effects adding up: geometry and much less material to scatter the sunlight due to reduced activity.
See this animation for a trajectory.
While the pure physics above are the main reason for the difference, other reasons might be the amount and skill of post-processing done on images. People might have stacked several photos in order to gain a higher noise-to-contrast ratio and applied a logarithmic brightness curve or similar.


Why does the Moon appear gray when passing between the Sun and the Earth?

That's what it really would look like if you were there with DSCOVR. The albedo of the Moon is only about 0.136, about half of the Earth's average albedo. Of course the part with clouds is higher.

I was shocked too, but it was explained in written copy that accompanied  the release of the original image.

  Shouldn't the Moon appear as bright as a full Moon seen at midnight from Earth?

It does. If the moon were a diffuse, white ball, a full moon would be about seven times brighter!

If you watch the image or GIF, the Moon is roughly the same brightness as central Australia or the Sahara region.

Phil Plait explains well in Bad Astronomy.

There's a lot to read here.

EDIT: I just ran across these images of astronauts on the surface while reading this answer. Their suits are not 100% white to begin with, but the Lunar soil - at least in these locations - is significantly darker. It is close to the same color as the (presumably) nearly-black radiator fins for the heat sink of the RTG unit (2nd photo) at the astronaut's foot.

above: "Buzz Aldrin carries the EASEP." from here

above: "Astronaut Alan L. Bean from Apollo 12, put the Plutonium 238Pu Fuel from the Lunar Module into the SNAP 27 RTG" from here.

above: Image from NASA/NOAA from Bad Astronomy


Researching mechanics without “fixed stars”

Say I have a single observatory, and the only things I measure are the times and azimuth angles for the high magnitude solar system bodies (the Sun, Moon, Mercury, Venus, Mars, Jupiter, and Saturn) as they ascend into the sky and descend out of it (for a fixed elevation angle).
If I gather these for a few years, I will have enough data to have an overdetermined system to calculate the latitude of my observatory, the Earth's axial tilt, the Earth's rotation rate, and the orbital elements of all the bodies.
The trick would be to realize that all these bodies (including the Earth) can be approximated by rotating spheres that are revolving around the Sun (with the Moon also revolving around the Earth).  Solar and Lunar eclipses certainly would have helped with this notion.
Though our own ancient astronomers used the relatively fixed locations of the stars as a reference to better track the motions of the planets, they certainly aren't necessary. In fact, perhaps we would have understood the motion in our Solar System sooner in our development if we hadn't been distracted by the rest of the Universe.


Why is the track of the subgiant stage almost horizontal on the HR diagram?

I will replace the previous answer to focus on the "subgiant" branch prior to red giant, rather than the pre-main-sequence or the "horizontal branch" of core helium fusion.  Those are other times that the luminosity is constant, but this question is about the subgiant branch, which I missed before.

The reason the luminosity is nearly constant on the subgiant branch is related to the "mass-luminosity relation" of pre- and main-sequence stars.  It is due to radiative diffusion and how it leads to a luminosity that depends only on mass, for a given composition.  If you compare to pre-main-sequence tracks, you should find that the subgiants more or less retrace that prior evolution, just with a somewhat higher luminosity because many of the electrons have been swallowed into neutrons, reducing the opacity and increasing the all-important rate of radiative diffusion.  It's essentially just a helium-dominated mass-luminosity relation, instead of hydrogen-dominated, as the radius rises due to the details of how the interior is evolving.

The reason the luminosity eventually rises steeply on the red giant branch is that as the degenerate core starts to build in mass, it starts to control the temperature of the fusing region, and this significantly changes the internal structure in ways that get into explaining red giants.


Are there predictions of the distribution of matter at the end of time

The basic answer to your question is "yes to all of the above, in different eras".

If we use the concordance $\Lambda$CDM model, with the cosmological constant remaining constant, then the universe will keep on expanding at an accelerating rate forever. The first big effect of this is that gravitationally bound clusters of matter such as galaxy clusters will become separated by ever larger distances (but they themselves are not changed noticeably by this). The current cosmic web of galaxies and voids disintegrates into little separated droplets over the next few hundred billion years.

Inside these "island universes" local gravitational interactions tend to make galaxies merge with each other, eventually ending up as a single big elliptic galaxy.

Over long time (around $10^{20}$ years) close-encounters between stars lead to some stars being flung out of the galaxy into the outer darkness, and the others bunching up closer to the central black hole. Eventually this reduces the galaxies to big black holes and scattered stars. The scattered stars cool down to low temperatures.

If proton decay happens, then somewhere beyond $10^{33}$ years or so the scattered stars quietly and slowly evaporate into elementary particles. The black holes are also expected to evaporate, and after $\sim 10^{99}$ years or so the only thing left will be separate stable elementary particles - electrons, positrons, neutrinos, photons, maybe some stable dark matter. Due to the expansion each particle will be alone in its visible universe.

A good overview is Adams and Laughlin's "The Five Ages of the Universe" (or their paper, for more detail).


What is the final destiny of a neutron star?

Neutron stars have extremely small heat capacities. That is because they consist largely of degenerate fermions and the heat capacity is further suppressed if, as expected, those fermions are in a superfluid state.

This has (at least) two consequences:

(a) they cool down extremely rapidly - neutrino emission processes are highly effective, in the first $10^5$ years or so of a neutron star's life, at reducing its interior temperature to a few $10^7$ K and the surface temperature to $<10^6$ K. After that, the dominant cooling process is photons emitted from the surface ($\propto T^4$) and neutron stars rapidly fade from view thereafter.

(b) However, the low heat capacity also means that it is easy to keep a neutron star hot if you have any way of adding energy to it - such as viscous dissipation of rotation by friction, accretion from the interstellar medium or ohmic heating by magnetic fields.

No isolated neutron star surfaces have been measured with temperatures much below $10^6$ K - i.e. all observed isolated neutron stars are at young ages. The situation is summarised in section 5.7 of Yakovlev & Pethick (2004). Without any reheating, a neutron star would reach 100K in only a billion years - this is already utterly invisible. The reheating mechanisms must play some role for older neutron stars, but as Yakovlev & Pethick state:  "Unfortunately, no reliable observational data on the thermal states of such stars are available". In conclusion, nobody really knows at the moment what the long-term ($>10^6$ years) fate of neutron stars is in terms of their temperature.

The situation with regard to spin and magnetic field is more secure. There are not the same mechanisms available to spin-up an isolated neutron star or regenerate their magnetic fields. Both are expected to decay with time and indeed the spin-down rate and magnetic field strength are intimately connected, because the spin-down mechanism is the emission of magnetic dipole radiation. The magnetic field decays through the generation of currents that then ohmically dissipate (providing a source of heat) or perhaps more rapidly via currents generated by the Hall effect or through ambipolar diffusion.

For pure magnetic dipole radiation, one predicts $\dot{\Omega} \propto \Omega^3$. For typical surface magnetic field strengths of $10^8$ T, pulsars spin down to periods of around a few seconds in less than a million years, at which point the "pulsar activity" switches off and we can't see them any more, unless they are in binary systems and accreting matter in order to spin them up again. Unfortunately, there is very little observational evidence to pin down how fast magnetic fields decay (because we don't see old, isolated neutron stars!). The decay of B-field cannot be very fast, certainly timescales are longer than $10^5$ years. Theoretical estimates of B-field decay timescales are more like billions of years. If this theory is right then neutron stars would continue to spin down very rapidly even after the pulsar mechanism has ceased.


What are the small things, look like stars, moving in the sky?

A starlike point visibly moving across the sky is a satellite. The biggest and brightest is the international space station, but there are many others.

They can be seen any night that is clear, usually shortly after dusk, or before dawn. http://www.heavens-above.com/ allows you to see predictions of when satellites will be overhead.


Why doesn't the equipartition theorem disallow spinning dust?

The first relevant point is that the equipartition theorem only formally applies in thermodynamic equilibrium, which requires that all temperatures be the same.  Since the spinning dust only happens when there are temperature differences, holding to the analogy with a heat engine mentioned in the source, we should not expect that theorem to be the whole story.  Granted, we often apply that theorem even when there are temperature differences, so the issue boils down to, when is the behavior more or less the same when all temperatures are the same, and when is the behavior fundamentally a heat engine?  I will assume the source is correct that this is heat engine-like behavior for spinning grains.

Once you have a heat engine in operation, the sky is the limit on how much energy you can put in any given mode (as long as you can maintain the temperature difference), because a heat engine does work, and work energy never needs to be equipartitioned, you can partition it any way you like by setting up an appropriate apparatus.  The dust particle is therefore a kind of apparatus for doing work, in the presence of temperature differences, and channeling the work energy into rotation.

How and why that heat engine operates and picks rotation to channel the work is a much more subtle issue.  For example, do we need temperature differences across the dust grain itself, or just between the grain and the surroundings?  If the "hypothetical" Crooke's radiometer (meaning, one that works quite differently from actual Crooke's radiometers and actually spins the opposite direction) is a good analogy, then the spin is from temperature differences across the dust grain itself, which warm the gas in the surroundings of the dust grain, producing gas flows that maintain constant pressure.  Those flows require the gas to receive torques, and conservation of angular momentum requires an opposite torque on the dust grain.

To get torque like that, it seems that you not only need a temperature difference across the dust grain, but it also has to have a kind of "quadrupole" character-- hot-cold on one side of the grain, cold-hot on the other, producing a "handedness" to the temperature structure in the grain.  So we immediately see that one cannot get spin until one has a large enough system to be able to support the concept of "temperature difference" across the object, and molecules are generally not thought of that way-- though maybe in the case of very large molecules, they could be.  So the question then comes down to, how large does a molecule need to get before it can support a concept of spatially nonuniform temperature? I don't know how big that needs to be, but my guess is, very big indeed-- there are certainly polymers that can get very large, but the molecules talked about in astronomy are generally not polymers.

As for the sign of the rotation, that can be very difficult to figure out!  It seems to depend on how the temperature differences get created in response to absorption of a radiation field (or other interactions with the environment, perhaps thermal contact with warmer gas).  Generally speaking, concave sides warm up more quickly than convex, so cause the gas in the concave part to be warmer than in the convex, so a flow sets up from concave to convex.  That acts like a jet engine that pushes the concave side forward.  That this is subtle can be seen from the fact that the higher pressure gas would seem to push the other way on the dust, but it's like the way a sail on a sailboat works-- follow the deflection of the air and conclude that the boat deflects the opposite way.

If this is all correct (see the Wiki on Crooke's radiometers for background information), then I would say the shape of the dust grain you drew produces spin in the direction that you indicated, but that no astrophysical molecules are large enough to produce the same effect.


Could the word "multiple stars" include binary stars?

I'm not aware of any official (i.e., IAU) definition of a multiple star system.  However, as someone who used to do research in that field, I would interpret multiple star system as encompassing systems with two or more stars.  So a binary system, a triple system, and a quadruple system are all multiple star systems.


How can a supernova cause a gravitational wave?

The key here is that to produce gravitational waves, the supernova explosion must not have perfect spherical symmetry. Technically, what you need is an "accelerating" gravitational quadrupole moment.

It is quite unlikely that supernovae explosions will be symmetric. Asymmetries can be produced by binary companions, magnetic fields, rotation...

The characteristics of a "burst" source of gravitational waves will be quite unlike the signature of coalescing binary black holes. The magnitude of the signal is expected to follow the collapsing core of the star and reach a peak at something called the "core bounce", where the equation of state of the gas "stiffens" and a shock wave propagates outwards. The signal is thus likely to gradually increase in amplitude for some tens of milli-seconds, followed by an abrupt spike at the bounce, followed by a "ring-down" as the core (a proto-neutron star) settles down towards a spherically symmetric configuration over the course of a further 10 milli-seconds or so.

The frequencies of the waves will have a broad range governed by the characteristic frequency implied by $(G\rho)^{1/2}$, which is the inverse of the freefall timescale, where $\rho$ is the density at the time when the GWs are emitted. For a typical core collapse, the densities could range from $10^{14}$ to a few $10^{17}$ kg/m$^3$ over the course of the event, producing GW frequencies of tens to thousands of Hz - exactly in the range at which aLIGO is sensitive.

Because GWs pass through the envelope of the star unimpeded then the GW signature of a supernova probes the very heart of a supernova explosion and should be seen some hours before the visible signature is apparent. On the other hand any neutrino emission from a core collapse supernova should be detected at a very similar time to the GW waves. Any delay here could fix the currently poorly known neutrino mass.


Which constellations are straight up and straight down?

Crap: I've just realized the answer is given on that Wikipedia page!

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

Looking up from here: Coma Berenices

Looking down from here: Sculptor

Hope this helps someone googling.


Does the Moon's magnetic field affect Earth's magnetic field?

So would the Moon's magnetic field affect the Earth's magnetic field, just as its gravitational pull affects Earth's gravitational pull for oceans?

Yes, but only slightly. Firstly, magnetic fields can superimpose, so the field at any point is the sum of the field due to the Earth and the field due to the moon.

However, the moon is rather far away (and has a weak magnetic pole strength), so the magnetic field due to the moon on Earth's surface is nearly negligible (magnetic field also decreases as an inverse-square law)

In addition, the magnetic field of the Moon may bolster or erode the Earth's field as magnets moving relative to each other tend to either lose magnetization or become stronger. But this process has a negligible effect when we take the Moon and Earth.


Does The ISW Effect Predict Net Redshift Over Distance?

The (late time) ISW is caused by the evolution of cosmic structures as photons of the cosmic microwave background traverse them on their way to our detectors. It may cause a redshift or blueshift with respect to the redshift predicted for a homogeneous expanding universe.

A bit more detail: If a photon "falls" into a potential well, its frequency and energy increase. But as it exits the potential well it loses the same amount of energy. In a similar way, if a photon crosses a "void", it loses and then gains energy.

However, if the universe contains dark energy (the ISW does not work in a flat universe containing just matter) and is expanding, then the structures the photons traverse are evolving with cosmic time such that the process is asymmetric. A potential well is "smeared out" a little by the expansion, and the photon does not lose as much energy exiting, as it gained on entering. Ditto for a void, but in reverse.

The net effect is small blue- and redshifts (of order 1 part in 100,000 at low spatial frequencies) with respect to the overall redshift you would expect anyway from the cosmic expansion of a homogeneous universe. These shifts are expected to be correlated with the cosmic structures in the same line of sight. The ISW has been detected (albeit with low significance) for the past decade, by cross-correlating CMB spatial variations with catalogues that map out the large scale cosmic structure (e.g. high redshift radio galaxy catalogues). See Nishizawa (2014) for a review.


The Earth's equator is not in the same plane as the planets. The planets, including the Earth orbit the sun in a plane called the ecliptic. The equator of the Earth is tilted by 23.5 degrees to the ecliptic.

This means that the planets will be seen on the same great circle of the sky that the sun is on.  The planets move relative to the sun and the stars, and so when the ecliptic is high in the sky, the planets will appear high above the horizon. It will tend to be during winter that the planets will appear highest.


What kind of radiation do supernova remnants emit?

Young supernova remnant release a substantial amount of radiation as they expand into the interstellar medium. In the free expansion phase, the outgoing shock wave heats up matter to $\sim10^6$ Kelvin, producing thermal x-ray emission. Throughout their lives, the remnants are responsible for quite a lot of non-thermal emission, including synchrotron radiation from relativistic electrons. This synchrotron emission stretches throughout the electromagnetic spectrum, including at radio frequencies. There is, additionally, non-thermal gamma ray emission through several mechanisms, including inverse Compton scattering and hadronic interactions.

Supernova remnants are also thought to be one of the main sources of galactic cosmic rays. Protons are accelerated through the shock front to energies of up to a few petaelectronvolts. Cosmic ray production may be associated with gamma ray emission if there is additional dense gas for the cosmic rays to interact with.

This brings us to the final point of your question: yes, supernova remnants are responsible for some neutrino production. Those cosmic ray/gas interactions produce (in addition to those gamma rays) neutrinos through hadronic channels and the decays of short-lived intermediate particles, including pions and kaons. This also means that probing the gamma ray spectrum of a supernova remnant can tell us something about its neutrino spectrum.


Birthdate differences calculation relative to earths orbit duration

The sidereal year is 6 hours and 9 minutes longer than a 365 day calendar year. To simplify your question, assuming you an your friend were born on the same date and time in 1983 and 1984 you would be 365 days, 6 hours, and 9 minutes older than your friend.

Since there is no leap day between the two times you are considering, it doesn't affect the calculation.


Star formation in a galaxy can end for a number of reasons:

Ram pressure affecting a galaxy falling into a cluster can strip away gas, which can end most star formation in the galaxy itself.
The winds of massive stars (as well as supernovae shocks) can eject cold gas via radiation pressure.
Mergers with other galaxies can also eject gas, although these interactions often lead to an increase in star formation, briefly.

As far as I'm aware, how strongly each of these phenomena contribute is still an open question, but we do know that they either eject gas or heat it up (and therefore make it unsuitable for star formation, which requires cold gas), thereby slowing or even effectively ending star formation in a given galaxy. A gas cloud is supported against gravity by thermal pressure (and occasionally magnetic fields). Clouds that are hot have higher thermal pressures and therefore can resist gravity; cold clouds cannot. Cold, dense clouds are therefore the best places for star formation (more gravity and less pressure).

As for the amount that can be ejected, the exact amount depends on a) the initial gas quantity, b) the mechanism, and c) likely a large number of other free parameters, so I don't think I can answer that for the general case.

Regarding if stars can form with little hydrogen: If hydrogen was ejected, so would other elements, so it's not as if all the hydrogen would be blown away and helium, lithium and others would remain. Hydrogen-poor stars exist, but they're older, evolved stars, not young ones.


What is the minimum size of a ball of gas to become a star?

There is no equation, you need a detailed model of the interior physics of very low mass stars. Very roughly, you can say that
hydrogen fusion occurs when the central temperature exceeds about $10^{7}\ K$ (the density dependence is secondary) and that from the virial theorem, the central temperature is given approximately by $T \simeq 1.6\times 10^{7} M/R$, where the mass and radius are in solar units. $M/R$ decreases slowly down the main sequence towards lower mass stars, but then what happens is that electron degeneracy pressure becomes important and less massive objects do not become much smaller than Jupiter-sized and thus $M/R$ decreases and fusion stops - or rather it never starts.

In detail the minimum mass for hydrogen fusion in a manner that is capable of sustaining a star in equilibrium against gravitational contraction is about $0.075M_{\odot}$. With an uncertainty of about $0.002M_{\odot}$.

It is slightly more complicated than this, since at lower internal temperatures the deuterium in a star can fuse. This will happen during the early life of any ball of gas more massive than about 13 Jupiter masses.

Your assumptions about the relative H and He abundances are completely wrong. Even in the early universe the gas is $\sim 25$% He by mass.


How to know when the moon is visible on a particular night?

Google-fu is correct... It really depends upon where you are, for an exact time. However, if you understand that there are 8 named phases and they typically have a "standard" rise time, you can know approximately what time and what phase will be visible. The one caveat is that you need to look at the moon to know where to start.


There are tons a amateur radio astronomy clubs and groups.  Radio Astronomy was basically started by an amateur named Grote Reber, where he basically did the experiment that you are pondering to do in 1937 in his backyard in Wheaton, Illinois.  You can visit that exact instrument at the NRAO facility in Green Bank, West Virginia.  It was named a National Historic Landmark in 1989.

Just Google Amateur Radio Astronomy

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

http://www.nrao.edu/whatisra/hist_reber.shtml


How much star can star fusion fuse if star fusion can fuse hydrogen?

There's a rough equation based on a star's mass but the age and metalicity of the star are factors too.   As large stars grow older they grow hotter on the inside and as they grow hotter, they fuse hydrogen faster, but they also become less massive over their lifetimes and have less and less hydrogen over time, so there's two opposing factors.

A rough equation is found here:

http://www.astronomynotes.com/evolutn/s2.htm

  a star's luminosity increases with mass as $(\text{the star's mass})^p$. The
  value of the exponent p varies between 3 and 4. For the rare massive
  stars (M* > 30 Msun), p = 3 and for the more common low-mass stars (M*
  < 10 Msun), p = 4

Loosely speaking, the luminosity is roughly equivalent to the rate of hydrogen fusion though the hydrogen fusion takes place inside the star and it takes some time for the energy to make it's way to the outside of the star and radiate away, the age of the star is far greater than the time it takes the newly created energy to escape, that it's still a pretty good ratio direct, luminosity to hydrogen fusion rate.

A star with 2 solar masses is about (4th power) 16 times as luminous, which means it fuses hydrogen about 16 times as fast.   When you get into significantly larger stars, the 4th power rule gradually becomes the 3rd power.

But as far as age being a factor, our Sun during it's main sequence will be about twice (or a bit more than twice) as luminous when it's 8 or 9 billion years old as it was when it was 1 billion years old, which corresponds to a gradual doubling of hydrogen fusion rate.  This happens (basically) because the core retains heat and more heat makes fusion more likely to happen, even as the amount of hydrogen is significantly lower in the later stage of it's main sequence than the early stage.


Quoting http://eclipse.gsfc.nasa.gov/transit/catalog/VenusCatalog.html

"When a transit of Venus occurs, a second one often follows eight years later. This is because the orbital periods of Venus (224.701 days) and Earth (365.256 days) are in an 8 year (2922 days) resonance with each other. In other words, in the time it takes Earth to orbit the Sun eight times, Venus completes almost exactly thirteen revolutions about the Sun. As a result, Venus and Earth line up in the same positions with respect to the Sun. Actually, the two orbital periods are not quite commensurate with each other since Venus arrives at the eight year rendezvous about 2.45 days earlier that Earth. After the third eight-year cycle, Venus arrives too early for a transit to occur."


Light from the solar photosphere is observed to have a gravitational redshift. The expected value, based on General Relativity is equivalent to a doppler redshift of 633 m/s. It is difficult to measure the exact value because it is obscured and blurred by the turbulent motions in the Sun's atmosphere.

As far as I am aware, the estimated values are entirely consistent with General Relativistic predictions, though the precision is not high enough to make this an especially sensitive test (e.g. $698\pm 113$ m/s Takeda & Ueno 2012).


For how many bodies can there be a stable orbit with no very heavy central body?

It is not unusual to have pairs of pairs or even more complex stellar systems.  Two stars make a stable pair.  A pair of these make a stable system and a pair of these would also make a stable system.  It is, however, necessary for stability for each pair to be separated by roughly 10 or more times the separation of the previous level.  This could continue ad infinitum were it not for the tidal field of neighboring systems or the Galactic potential limiting the overall size.


Check this wiki link out. "The name Sgr A* was coined by Brown in a 1982 paper because the radio source was "exciting", and excited states of atoms are denoted with asterisks."

There is no unified naming system for black holes. They are usually named after their host galaxy. Others are identified by the name of the survey in which they were observed. A few black holes are catalogued by their constellation and the order in which they were discovered

EDIT: The 1982 paper by Brown and Lo can be found here.


There are many other variations on telescope designs others than the ones you have mentioned. They differ in the light path, the shape of the optics utilised, the use of mirrors and/or lenses, and the physical structure.

A Schmidt-Cassegrain is a particular type of Cassegrain reflector. A Cassegrain reflector uses two mirrors, a primary and a secondary. Both are curved, and the secondary is placed in front of the primary, which blocks some of the light to the primary. The primary directs light onto the secondary, and the secondary directs light through a hole in the centre of the primary. This 'folds' the lightpath, resulting in a short telescope, even at quite wide apertures.

A Schmidt-Cassegrain is a Cassegrain with particular design details. The primary mirror has a spherical curve, and the secondary usually has a spherical curve, though some manufacturers modify the design with a parabolic secondary to improve the performance. A key characteristic is the addition of a lens to correct for the optical aberrations produced by the spherical mirror(s). This lens is also used to mount the secondary mirror, so no vanes are required to hold it in place.

A Galilean telescope is a refractor, and so it uses lenses exclusively instead of mirrors. Lenses may be able to out-perform mirrors, but it is easier to manufacture high-quality large mirrors than large lenses.

Newtonian telescopes are also reflectors with two mirrors, but one is flat. It also blocks some of the light in the centre of the objective, but it is held in place by vanes instead of a lens. The flat mirror angles the light out the side of the telescope near the objective end of the telescope. While this may frequently place the eyepiece in a more convenient position to access it, especially when aiming near the zenith, it is also necessary to rotate the telescope in the mount to get the eyepiece in the right place.

There are pros and cons to all telescopes, and selecting a telescope requires knowing what you will be doing with it, and choosing something that complements that purpose well with its pros, and that has cons that are not relevant.


Can two detectors for gravitational waves be combined into a very long baseline interferometer?

They do use two or more detectors to triangulate the origin of a source. Most recently the Advanced Virgo detector in Italy was used in conjunction with the two LIGO detectors in the USA. This has been published in Physical Review Letters (Phys. Rev. Lett. 119, 141101 – Published 6 October 2017).

With two detectors the sky area for "pinpointing" the source is quite a large area, but with three detectors the area is reduced by a factor of about 20.


How to calculate geocentric transformation to heliocentric coordinates?

Earth's position as seen from the Sun is directly opposite the Sun's position as seen from Earth, at the same distance.
In ecliptic coordinates,
$$ \begin{align}
l_\oplus &= \lambda_\odot \pm 180^\circ \\
b_\oplus &= -\beta_\odot \\
r_\oplus &= \mathit{\Delta}_\odot
\end{align} $$
The heliocentric position of the Sun is always at the origin ($r=0$) by definition.


Over the year, the sun follows a path known as an annalemma. This is a figure eight shape and is caused by the combination of the Earth's rotation around the Sun and its tilt relative to the plane of the equator.

The analemma can be observed by pointing a camera at the sun, and at the same time each month, take a picture and over lay the images:

Taking a picture every day for a month (in September) would mean that you'd see it lower in the sky, curving slightly as it followed the figure of eight shape of the analemma.

Note:

An analemma can be different on different planets. For example, a Martian analemma is different because its tilt and shape of orbit is different to that of Earth's:

Mercury: Because orbital resonance makes the day exactly two years long, the method of plotting the Sun's position at the same time each day would yield only a single point. However, the equation of time can still be calculated for any time of the year, so an analemma can be graphed with this information. The resulting curve is a nearly straight east-west line.
Venus: There are slightly less than two days per year, so it would take several years to accumulate a complete analemma by the usual method. The resulting curve is an ellipse.
Mars: teardrop
Jupiter: ellipse
Saturn: technically a figure 8, but the northern loop is so small that it more closely resembles a teardrop
Uranus: figure 8 (Uranus is tilted past "sideways" to an angle of 98 degrees. Its orbit is about as elliptical as Jupiter's and more elliptical than Earth's. )
Neptune: figure 8

For more, read this.


Closest stable orbit about the sun of an asteroid?

There's no limit defined by orbital mechanics.  Newton's law of gravitation allows for any perihelion that isn't actually inside the sun.
The Parker Solar probe is in a stable orbit (discounting planned interactions with Venus) with a perhelion of 6.2 million km (at a temperature of 1400⁰C)  But there is no reason that an object couldn't go closer. There is an asteroid 2006HC4 with a perihelion of just 10.2 million km. Its aphelion is beyond the orbit of Mars
Gravitational effects are not the only consideration for stability. E.g. the Yarkovsky effect will tend to deplete the vulcanoid region (Vokrouhlický et al. 2000), YORP effect can spin up asteroids to breakup velocity, radiation pressure removes dust.
However getting into such an orbit is difficult.  If you are a main-belt asteroid, you would need a huge push to get you into a sungrazing orbit.  Comets from the Oort cloud don't need so much of change in velocity (as they are already moving slowly) and so most bodies that we see falling close to the sun are made largely of ice, and are disrupted by the sun's heat.


Accurate catalogue of the Stars according to Hemisphere

Note that astronomers generally are not concerned with hemispheres, so it is unlikely that you will find a catalog that is only for a specific hemisphere. Catalog generally cover the entire sky. Also, most catalogs are for either stars or deep sky objects (galaxies, nebulae, and so on). I have not seen a catalog that includes both stellar and non-stellar.
For stars, I suggest the Yale Bright Star Catalog. It includes around 9000 stars, some of which are slightly too faint to see with the naked eye.
For nonstellar objects, I suggest the NGC Catalog.


What if we throw two solid objects parallel in space? Do those two objects have any chance to collide with each other?

It all depends on the direction you "throw them" and the space-time they find themselves in.

Case 1 - in a perfectly empty universe (other than your footballs) the two would eventually collide. It doesn't matter what initial velocity you give them. The only thing that matters is there separation and their mass. You can just use Newtonian gravity to compute the two footballs' acceleration toward one another -- and hence the time for them to collide.

Such an 'empty' universe as above is called a Minkowsky space-time.

If you put these footballs on a trajectory into our real universe, well we know that space-time is curved by the matter/energy (just energy density in general) that occupies it. Hence a trajectory that starts out parallel will invariably end up "not parallel". The balls will either collide or diverge depending on the geometry of the space-time they find themselves in. The "geometry of the space-time" is (again) completely dependent on the distribution of "stuff" (matter/energy) in the universe as related to the location of the footballs.

In short, the footballs will follow the geometry of the space-time they find themselves in. That geometry (in our real universe) means they will not remain on parallel paths.