Does Hawking Radiation really exist and Why "negative energy" in virtual particles?

Unless everything we know about general relativity and quantum field theory is wrong, Hawking radiation almost certainly exists.
It is also entirely undetectable. The thermal output of a 3 solar mass black hole by way of Hawking radiation amounts to $\sim 10^{-29}$ W. This, of course, is not only completely undetectable but the corresponding Hawking temperature, $\sim 2\times 10^{-8}$ K, is far, far below the temperature of the CMB; so any astrophysical black hole today would absorb many orders of magnitude more radiation from the CMB than it emits by way of Hawking radiation.
The explanation with virtual particle-antiparticle pairs and negative energy particles absorbed by the black hole originates, I believe, from Hawking's popular science book A Brief History of Time, but it really is at odds with Hawking's own groundbreaking paper on the subject (Nature 248, 1974). This is explained quite nicely in an informative article by Ethan Siegel, published last year online: https://www.forbes.com/sites/startswithabang/2020/07/09/yes-stephen-hawking-lied-to-us-all-about-how-black-holes-decay/?sh=52d60ad44e63 .
Adding to Ethan's explanation (which is well worth the read) I ought to mention two things. First, there is the closely related concept Unruh radiation: thermal radiation seen by an accelerating observer from the direction of his apparent Rindler-horizon, in spacetime that is seen as empty by an inertial observer. This alone stresses the important point that in quantum field theory, the concept of a particle becomes observer-dependent once we permit accelerating observers and/or gravitational fields. Second, the wavelength of Hawking radiation alone should tell us that it is not originating from the event horizon. For the aforementioned three solar mass black hole, the peak wavelength is almost 180 km. Comparing this to the less than 10 km Schwarzschild radius of the black hole, I think it is quite evident that this is not about particle-antiparticle pairs at the horizon itself.
Finally, regarding the images of the M87* supermassive black hole, first, these are not NASA images but images produced by an international collaboration called the Event Horizon Telescope project. Second, though they are often described as images of the black hole, that is not strictly correct. What you actually see is radio frequency emissions of infalling matter surrounding the black hole in the form of an accretion disk. The central, black shadow corresponds to the black hole's photon sphere (radius at which unstable closed photon orbits become possible). And it is important to remember that this is not exactly a photograph, but rather, a synthesized image that is fitted to a large number of radio frequency observations using an optimization method.


How to improve efficiency in periodicity detection?

AstroPy

Jake Vanderplas (Director of Open Software at the University of Washington eScience Institute) has written a Pythonic Preambulations post Fast Lomb-Scargle Periodograms in Python which by definition appears to be a good place to start to learn how to do Lomb-Scargle Periodograms... Fast... in Python.

I have not read it yet in full, but I've found other posts of his to be extremely useful.

He starts out by explaining the implementation in AstroPy which if you are going to be doing Astronomy, is probably a good thing to become familliar with.

SciPy

Alternatively, there is scipy.signal.lombscargle which you can try assuming you have a current version of SciPy installed but want to try something different than AstroPy or want to compare performance or results.

StackOverflow

I found over 40 posts containing 'Lomb-Scargle' in Stack Overflow which is a very good sign!

I notice that you've been active there and know how to ask well-received questions. Since your asking about speed and errors, StackOverflow is your go-to source! However, don't forget to be as clear as possible and quote all error messages and symptoms, and see if you can somehow add a short Minimum Complete and Verifiable Example or MCVE. That's hard when your problem results from a very large dataset I know.


Liquid water can in principle exist at many locations on present-day Mars, but there are a few interesting twists to the story.

At low elevation, the atmospheric pressure is high enough. The triple point pressure of H2O is 611 Pascal, which corresponds to about mid-elevation. At low elevations, such as the northern lowlands, the atmospheric pressure will exceed the triple point pressure, so the liquid phase of H2O becomes possible.
There will be substantial "sublimation cooling" or evaporative cooling. When the partial pressure of H2O is almost as high as the total pressure, vapor rapidly escapes into the atmosphere, which takes a lot of energy. If you put an ice cube at the equator of Mars, it will not melt, because the sublimation cooling is larger than the incoming solar energy, even at noon.
However, if that ice cube is covered with a thin layer of dust, or contains dust that will accumulate on top as the ice sublimates, vapor loss will be reduced. In this circumstance, ice can melt to form liquid water.
That leaves the question of how the ice cube would have gotten there initially. I call it the "source problem". Places on Mars that get hot have already lost all their ice, and places on Mars that have ice do not get warm enough. So, it is difficult to get liquid water on Mars, but it is possible in principle.
With salts, which lower the melting point, everything becomes much easier, although supplies of ice/frost and salt are needed. When water with salt evaporates, the salt concentration increases, so the solution is driven toward the eutectic point, where the melting point suppression is the strongest.


What if Earth and Moon revolved around each other like Pluto and Charon?

They do, but due to the ratio of masses being vastly different, they seem like they would not to do so as moon seems to rotate just around (the centre of) Earth.

The ratio of Earth and Moon's masses is $\frac{M_{Earth}}{M_{Moon}} = 81.3$ whereas for Pluto and Charon the same ratio is $\frac{M_{Pluto}}{M_{Charon}} = 8.09$.

Because the ratio for Pluto and Charon is relatively small, the center of the system – barycenter, around which the two bodies orbit – is somewhere on a line drawn between the two celestial bodies's mass centers. But for Earth and Moon, because Earth is proportionally much heavier, the system's barycenter does not reach outside of Earth, but instead is about 4,500 kilometers from the center of Earth (see the photo below, too):

  In cases where one of the two objects is considerably more massive than the other (and relatively close), the barycenter will typically be located within the more massive object. Rather than appearing to orbit a common center of mass with the smaller body, the larger will simply be seen to "wobble" slightly. This is the case for the Earth–Moon system, where the barycenter is located on average 4,671 km from the Earth's center, well within the planet's radius of 6,378 km. Source: Wikipedia - Barycenter

The major effect of this co-rotational system is that Earth seems to "wobble" on its orbit, as mentioned in the quote from Wikipedia above.

@JeppeStigNielsen makes a good point about differences in tidal locking in the comments below. In the Earth-Moon system, only Moon is tidally locked (which causes us to see only one face of it, so only roughly a half of it, from Earth), whereas in Pluto-Charon both of the bodies are tidally locked. Earth is not tidally locked because of the higher mass ratio between it and Moon, but the lower mass ratio Pluto-Charon system is, as the lower mass Charon has slowly changed Pluto's rotation to match its orbital movement.


Half lit means the Moon is at a quarter phase, either First Quarter or Third Quarter.

The original question asks "Does half-lit mean exactly 45 degrees?".  I assume this is asking if the angle $\alpha$ is 45 degrees. The answer is no. The angle $\alpha$ is found from trigonometry to be $cos(\alpha)=EM/ES$ where EM is the distance from the Earth to the Moon and ES is the distance from the Earth to the Sun. EM/ES is approximately 1/390 (=238000 miles/93 million miles), so $\alpha$ works out to be 89.85 degrees.

Aristarchus was trying to do the reverse calculation to find the distances. If the angle $\alpha$ could be measured accurately, the relative distances EM/ES could be calculated using the same formula. The problem is the difference between 89.85 degrees (the Sun is 390 times farther than the Moon) and 90 degrees (the Sun is infinitely farther than the Moon) is a very small angle and difficult to measure exactly. If the Sun were a lot closer, then $\alpha$ would be a smaller value, and the error in the measured angle of $\alpha$ would be restrict the value of EM/ES to a more meaningful range than "infinity".


An hypothetical timepiece on board of a spaceship. Please Help

On board of a hypothetical spaceship, I have a hypothetical clock consisting of 2 mirrors and a light beam is being reflected between the mirrors to measure the time.

The mirrors, clock and you are all in the same reference frame and your relative velocities are zero.

The speed the whole shebang is traveling at relative to something else is irrelevant.

So the clock on board the ship with you will calmly pass the seconds just as you'd expect an object at rest relative to you to do.

  Now if we hypothetically exceed the speed of light

We can't do that.

  what will I see is happening to my clock on board of my ship ?

Beyond the speed of light has no meaning I know of in this context.

  I just can't wrap my mind around this problem.

There's no problem : you can't travel faster than light : keep repeating that mantra and all will be well. :-)


What will happen when landing on Jupiter?

Jupiter does not have a "surface" and nor is there anything but an arbitrary division between interplanetary space and where its atmosphere begins.

The crushing pressure is its atmospheric pressure. The deeper into the atmosphere you go, the greater the column of gas that lies above you. It is the weight of this column of gas that is responsible for the rapid increase in pressure with depth.

The answer to your last question is most definitely addressed in the duplicate question about whether Jupiter is entirely made of gas. There is quite likely to be a liquid phase nearer the centre and there may be a solid core of order ten times the mass of the Earth. It is not a settled question.

The gas motions you talk about are essentially belts of weather systems in the upper layers of Jupiter's atmosphere. It is all most definitely gas that you can see.


How to interpret extreme ultraviolet/ x-ray data in the literature?

The EUVE catalog gives you a count-rate in a certain EUV band. To convert this into a flux, you need a flux conversion factor to go from counts per second to energy flux at the Earth per second. This conversion factor, also often known as an "ECF", depends on the intrinsic spectrum of the source and also how much absorption (known as the hydrogen column density) there is between Earth and the source. So there is no one answer that can be given.
I
Your best port of call is the PIMMS flux conversion tool, which has an EUVE module and can do these flux conversions for you if you know something about the spectrum and the absorption. I'd try it for you but I couldn't immediately see how to do it for EUVE, though it claims the information is there...


Do we know the fluctuations and constant temperature depth of the moon?

Yes, at some Apollo landing sites, it's been carefully measured! The astronauts make deep holes in the Moon's regolith and put in thermal probes that precisely measured temperature. These signals were radioed back to Earth by a solar-power transmitter, in some cases for years. These experimental packages were called ALSEP and included seismometers and passive retroreflectors and other "goodies".

I'm going to quote verbatim this answer from Space Exploration SE since it fits so nicely here.

  tl;dr: The number is often given as 252 +/- 3 K (or about -21°C or -6°F), so @BowlOfRed pretty much nailed it using first principles!

    Without atmosphere, equilibrium temperature at 1AU is about -17C. I would expect that to be the temp "near" the surface around the equator. Colder as the latitude increases.

  Reading the references, it seems that the diurnal oscillation (what Earthies might call "monthly" or Lunies might call "daily") in temperature decays with a 1/e depth of about 30 centimeters, so by the time one reaches 1 meter, the temperature "stabilizes" temporally, but of course will increase with depth since the Moon's interior is hot. You can see this in Figure 2 of the last paper ("Subsurface temperature histories, covering 3.5 years...") where the oscillations in temperature at different depths along the probe are plotted.

  Further, some of this data can already be found in @KimHolder's question Does the temperature near the surface of the Moon rise rather quickly with depth? who quotes as NASA News article Down the Lunar Rabbit-hole which then quotes Mark Robinson of Arizona State University, principal investigator for the LRO camera.

    "The tunnels offer a perfect radiation shield and a very benign thermal environment," says Robinson. "Once you get down to 2 meters under the surface of the Moon, the temperature remains fairly constant, probably around -30 to -40 degrees C."

  Why the difference between the "ground truth" actual temperature measurements discussed below and the quote of a quote?

  (image source also here) My guess is that the -30 to -40 number is a mid-latitude number, while the following concentrates on nearer to equator values. There's no physical way the temperature can drop by 10 or 20C in 1 meter of increasing depth, so the difference is more context and location.

  I'm not an expert on planetary science, but the oft-cited number for an average sub-surface temperature about 1 meter below the surface seems to come from the following papers:

  The Apollo 15 lunar heat-flow measurement, Langseth, M.G., Clark, S.P., Chute, J.L. et al. The Moon (1972) 4: 390. https://doi.org/10.1007/BF00562006

    Abstract:

    The heat-flow experiment is one of the Apollo Lunar Surface Experiment Package (ALSEP) instruments that was emplaced on the lunar surface on Apollo 15. This experiment is designed to make temperature and thermal property measurements in the lunar subsurface so as to determine the rate of heat loss from the lunar interior through the surface. About 45 days (1 1/2 lunations) of data has been analyzed in a preliminary way. This analysis indicates that the vertical heat flow through the regolith at one probe site is 3.3 × 10−6 W/cm2 (±15%). This value is approximately one-half the Earth's average heat flow. Further analysis of data over several lunations is required to demonstrate that this value is representative of the heat flow at the Hadley Rille site. The mean subsurface temperature at a depth of 1 m is approximately 252.4K at one probe site and 250.7K at the other. These temperatures are approximately 35K above the mean surface temperature and indicate that conductivity in the surficial layer of the Moon is highly temperature dependent. Between 1 and 1.5m, the rate of temperature increase as a function of depth is 1.75K/m (±2%) at the probe 1 site. In situ measurements indicate that the thermal conductivity of the regolith increases with depth. Thermal-conductivity values between 1.4 × 10−4 and 2.5 × 10−4 W/cm K were determined; these values are a factor of 7 to 10 greater than the values of the surface conductivity. If the observed heat flow at Hadley Base is representative of the moonwide rate of heat loss (an assumption which is not fully justified at this time), it would imply that overall radioactive heat production in the Moon is greater than in classes of meteorites that have formed the basis of Earth and Moon bulk composition models in the past.

  In-situ measurements of lunar heat flow (also here) in Soviet-American Conference on Geochemistry of the Moon and Planets, Langseth, M. G., and S. J. Keihm, NASA Spec. Publ., NASA SP-30, 283 – 293, 1977.

  The number is often given as 252 +/- 3 K (or about -21°C or -6°F) as for example in Latitude variation of the subsurface lunar temperature: Lunar Prospector thermal neutrons R. C. Little W. C. Feldman S. Maurice I. Genetay D. J. Lawrence S. L. Lawson O. Gasnault B. L. Barraclough R. C. Elphic T. H. Prettyman A. B. Binder
  First published: 29 May 2003 https://doi.org/10.1029/2001JE001497

  Some of the experimental and theoretical details behind the data and its analysis can be read in Revised lunar heat-flow values, Langseth, M. G., Keihm, S. J., & Peters, K., Lunar Science Conference, 7th, Houston, Tex., March 15-19, 1976, Proceedings. Volume 3. (A77-34651 15-91) New York, Pergamon Press, Inc., 1976, p. 3143-3171.


Standard practices and models for generating the ephemerides of the solar system

There are a number of different organizations that produce ephemeris models of the solar system. Even within one organization, there a number of different ephemeris models.

The three leading organizations that produce solar system ephemerides are

The Jet Propulsion Laboratory (JPL) in the USA, which produces the Development Ephemeris (DE) series of solar system ephemerides (and also ephemerides for many of the smaller bodies in the solar system);
The Institute of Applied Astronomy at the Russian Academy of Sciences, which produces the Ephemerides of the Planets and Moon (EPM) series of solar system ephemerides; and
The Institut de Mecanique Celeste et de Calcul de Ephemerides (IMCCE) in France, which produces the INPOP series of solar system ephemerides.

There appears to exist a friendly spirit of competition and cooperation amongst these three groups. They apparently share measurements and ideas amongst one another. They do not exactly agree on results (which is a good thing). Each group regularly updates their respective ephemeris models as new measurements are gathered, and as numerical integration techniques and computing capabilities are improved.


Determining North-South Line Via Watch Method: Theory & Reason

When the Sun is on the meridian and is due South (local noon in Northern Hemisphere) your hour hand will point to (or close to 12) - assuming your local noon and civil noon are similar! (Daylight saving means 1pm fills this role).

The Sun crosses the sky in 24 hours, the hour hand travels round the watch in 12 hours. Hence the watch hour hand is twice as fast, so when the hour hand has reached 2pm, the Sun has only gone half as far (ie "to 1pm"), so if you point the hour hand towards the Sun in this case, the meridian is an hour further behind rather than 2 hours.

By the time we get to 4pm the Sun will have travelled "two watch hours" away from the meridian and so on.


Question: calculating orbit using another orbit

As a hint: a force directed inward does not change the angular momentum, and therefore does not change the semi-major axis nor the period. It's not clear which angle you're talking about, but knowing that the semi-major axis did not change should let you set the problem up as a simple geometry problem.


Why are solar telescopes built on lakes?

To amplify andy256's comment, the problem that solar telescopes face is that heating of the surrounding ground during the day gives rise to turbulence in the air near the ground, making the observing conditions worse (think of the heat shimmer just above the surface of hot pavement or a hot road -- that's turbulence bad enough for your naked eyes to notice). Since water has a high thermal inertia, it doesn't heat up as much as land does, so you get less turbulence. (Another approach is to put the telescope on top of a tower, above the worst of the turbulence.)

Note that Big Bear is at an altitude of 2,000 meters, so it is at a moderately high altitude.


Estimating Sun luminosity from first principles

The thermal energy of the Sun is something like
$$E \sim  \left(\frac{3k_BT}{2}\right)(N_i + N_e)\ , $$
where $N_i$ is the number of ions and $N_e$ is the number of electrons and the interior temperature $T \sim 10^7$ K.
To first order we can consider the Sun to be made of hydrogen, so $N_e = N_i \sim M/m_p$, where $m_p$ is the mass of a proton.
Putting in the numbers gives $E \sim 5\times 10^{41}$ J.
The typical mean free path of a photon in the solar interior is $l\sim 10^{-3}$ m. The solar radius is $R\sim 7\times 10^{8}$ m. Since the photons get out by a "random-walk" diffusion process, then it takes about $(R/l)^2$ steps, each of which takes a time $l/c$.
Thus the diffusion timescale
$$ \tau \sim \left(\frac{R}{l}\right)^2 \left(\frac{l}{c}\right) = \frac{R^2}{lc} = 1.6\times 10^{12}\ {\rm s}$$.
The ratio of these two numbers gives $3\times 10^{29}\ {\rm W}$, so about 3 orders of magnitude larger than the solar luminosity.
This ratio does not give the solar luminosity.


Why do the planets in the Solar system stay in the same orbital plane?

Answer to the NEW question:

the Angular Momentum Conservation Law states that, for any moving body, its angular momentum does not change unless you exercise an external force different from the central force.

For an orbiting body like a planet, this means that Sun's gravity, being the central force, does not modify Angular Momentum, but any other external force will do.

Examples of external forces are collisions or the forces made by Jupiter on another planet, or by Neptune on Pluto.

After the Solar System was formed, these external forces are quite small, and thus does not change greatly the Angular Momentum of any major body. But you can see how passing near a body can alter a comet's orbit.

Moreover, the external forces made by bodies that are in the same plane as an orbiting body does modify the value of its Angular Momentum, but not the direction. This causes that  the orbiting body changes its orbit but can not not change planes.

So if you add small forces from objects in the same plane, you end up with no changes to planes.


Should this photo of the sun's surface actually be white?

The term "color" is a label that humans have assigned to denote the ratio between the intensity at various wavelengths in the three different wavelength bands, or regions, that the human eye is able to perceive. These bands are centered roughly at 430, 545, and 570 nm, but are quite broad and even overlap:

Human cone response, normalized to the same height. In reality, the response of the blue cones is significantly smaller, and the green is somewhat larger (from Wikipedia).

If an object emits light only at, say, 450 nm, the ratio is roughly 0.1:0.2:1 (in the order R:G:B); it then looks a special way to us, and we call it "blue", or maybe "violet". If it emits at 550 nm, or 650 nm, we call it "green" or "red". An object that emits light in a more continuous spectrum that covers the region 500–600 nm, we'd name something like orange-/brown-/olive-ish, depending on the exact spectrum.

The Sun emits photons at all wavelengths, but not in an equal amount at all wavelengths. The particular ratios between the three bands that we can see, we have labeled "white". However, when the Sun's light enters our atmosphere, some of the light is absorbed, especially at the blue wavelength. Filtering out the blue results in a spectrum that looks more orange to us. The figure below shows the Sun's "true" spectrum (in yellow), and the spectrum seen from the surface of Earth (in red):

The Sun's spectrum measured outside our atmosphere (yellow) and at sea level (red) (modified image from Wikipedia, with data from Global Warming Art).

Sometimes we want to observe the Sun in a wavelength region that is invisible to humans, for instance in UV or X-rays. This can be done with a telescope and a detector that is sensitive to light in that particular region, but in order for us to see it, we represent the image with a color that we can see. The image in the top of the link you provide is taken with the European spacecraft SOHO's instrument EIT at 19.5 nm, which we call "extreme UV", bordering on soft X-rays. Since this is invisible to humans, they arbitrarily chose to represent it using green. They might as well have chosen pink or brown.

The Sun in Extreme UV, during a particularly violent solar flare (from the SOHO gallery).

Several of the photos in your second link are images taken by the Japanese space telescope Hinode, which observes both in the optical (i.e. visible by humans), X-rays, and far UV. If these are shown in orange, again it's just to make them visible to us, and you may say that they have been "doctored to meet our expectation". In this way, I like better when they choose a color such as all green, so we know it's "false color".


Space telescope located in outer solar system

The disadvantages would likely outweight the advantages.

It's cold out there. This makes it easier to keep an infra-red telescope cool
The sun's just a super-bright star.  This means more of the sky is visible and not in the glare of the sun. However you orbit so slowly that there will be a few objects that you won't be able to image because they are behind the sun for years.
There's no Earth to get in the way. The Earth blocks half the sky for Hubble. Putting a telescope on Pluto would have the same problem.
There's no space debris there (yet)

There may be other advantages, but compared to the immense difficulty of putting a probe into an 250 year orbit (you need to get it out there, with enough fuel left to put it into orbit) And the time it takes (you can't take much advantage of slingshots because you need to be in an orbit not just on an escape trajectory). And all the other disadvantages

Saw something interesting, can't just get hubble to take a look, because there is a five hour light time.
No power, sun is too weak so depend on a box of plutonium or similar.
No nice high resolution images, sorry those take too long to download. Low resolution only.
We've never landed anything as complex and fragile as Hubble on any other body, And the landing has to be fully automatic.

And you don't get to see anything new because space is just as clear from Hubble's orbit as it is from Pluto.


Could the Sun and our Solar System one day leave our galaxy?

It's certainly possible for a star system to be stripped away if its galaxy interacts gravitationally with a second galaxy. Tidal forces between the two can disrupt stellar orbits and form tidal tails and other possibly transient structures. This won't happen on human timescales - no satellite galaxies are anywhere close enough to us to pose a threat - but in several billion years, the Milky Way will collide with Andromeda in a process that will completely disrupt the structures of both galaxies.
It's been suggested that the Solar System may be stripped from its present galactic orbit during the collision. Cox & Loeb 2008 suggested that there is a  12% chance that we will be swept into a tidal tail >20 kpc from the center of the merger remnant, and a >50% chance that we will end up >30 kpc from the center, will in the outer halo - compare both of these with our current distance of ~8 kpc from the galactic center. Andromeda could even strip us away on its first pass at the Milky Way, though the authors regard this as unlikely.
All of these outcomes, of course, do still end with the Solar System being bound to the remnant. It's certainly true that we could be ejected from the system entirely, but that probability appears low compared the odds that we will simply move into a comparatively wide halo orbit.
If you regard the scenario where we're temporary bound to Andromeda as leaving the galaxy, then yes, that's indeed a possibility. Unfortunately (or fortunately?), we will soon coalesce once again with the Milky Way on timescales of ~1-2 billion years.
As an aside: There are other ways for a star to be ejected from a galaxy. One of the more common involves an encounter with a supermassive black hole. If a binary star approaches the black hole, the resulting three-body encounter will disrupt the system; one star may be captured by the black hole, while the other may be ejected at speeds greater than the galactic escape velocity. We know of several dozen of these so-called hypervelocity stars, including US 708 and S5-HVS1.


Where is the sun relative to my lunar rover?

If you like Python then see Skyfield; Observing from a Moon location which I believe will do exactly what you need! Also see for example this answer to Ephemeris for lunar body-centered body-fixed coordinates? written by Skyfield's author.

For example, you posted your question at 2020-04-20 15:21:36 UTC. Let's say your rover has just driven up to Surveyor III at 3° 0′ 58.03″ S, 23° 25′ 4.48″ W  (-3.01612°, -23.41791°) (read more about Surveyor III in How did Apollo-12 manage to land next to Surveyor-3? First “Space-Tourists”?)

At that moment the Sun was below the horizon. The output of the script below is

RA:  01h 54m 39.28s
Dec:  +11deg 45' 38.6"
alt:  -38deg 30' 34.6" degrees (below the horizon)
az:  265deg 50' 21.5" degrees around the horizon from north

  I can get the coordinates of the Earth-Moon barycenter using VSOP2013

factoids: VSOP is what PyEphem uses under the hood. It is currently in maintenance mode maintained by the same person who wrote Skyfield.

#!/usr/bin/env python
# -*- coding: utf-8 -*-
# adapted from https://rhodesmill.org/skyfield/planetary.html

from skyfield.api import PlanetaryConstants, Loader

load = Loader('~/Documents/fishing/SkyData')  # avoids multiple copies of large files
ts = load.timescale(builtin=True)
t = ts.utc(2020, 4, 20, 15, 21, 36)

eph = load('de421.bsp')
earth, moon, sun = [eph[thing] for thing in ('earth', 'moon', 'sun')]

pc = PlanetaryConstants()
pc.read_text(load('moon_080317.tf'))
pc.read_text(load('pck00008.tpc'))
pc.read_binary(load('moon_pa_de421_1900-2050.bpc'))

frame = pc.build_frame_named('MOON_ME_DE421')
Surveyor_III = moon + pc.build_latlon_degrees(frame, latitude_degrees = -3.01612,
                                              longitude_degrees = -23.41791)

apparent = Surveyor_III.at(t).observe(sun).apparent()
ra, dec, distance = apparent.radec()
print('RA: ', ra)
print('Dec: ', dec)

alt, az, distance = apparent.altaz()
if alt.degrees >= 0:
    print('alt: ', alt, 'degrees (above the horizon)')
else:
    print('alt: ', alt, 'degrees (below the horizon)')
print('az: ', az, 'degrees around the horizon from north')


Up to 384 minor planets (including Pluto) qualify as planets?

That actual IAU Resolution B5 adopted at the IAU General Assembly in 2006 states:

(1) A planet is a celestial body that:
(a) is in orbit around the Sun,
(b) has sufficient mass for its self-gravity to overcome rigid body forces so that it   assumes a hydrostatic equilibrium (nearly round) shape, and
(c) has cleared the neighbourhood around its orbit.

The last part being the neighbourhood around its (whole) orbit is the important part; it's not enough for Pluto to dominate its family of moons (and this last part is debatable as Pluto and Charon orbit a common barycenter (center of mass) which is between Pluto and Charon since Charon is so massive, compared to Pluto).
Since Pluto orbits in the Trans-Neptunian region with a similar orbit to other bodies, some of which are similar or larger in size and mass (Eris), and cross the orbit of Neptune, it can't really be said have to "cleared the neighbourhood around its orbit". Indeed Pluto is the prototype of the Plutinos; object which have a 3:2 mean motion resonance with Neptune (orbiting twice for every three orbits Neptune makes).


Safety while using a telescope to view a total solar eclipse

Yes, you can remove the solar filter during totality - otherwise you will have no way to see the corona (or anything at all) through the telescope during totality.
The dangerous time is (as you also say yourself) towards the end of totality - and it's easy to get lost in time and the joy of the moment. Set a timer to wake you seconds before end, have others observing with you do the same. You absolutely must put back the filter in time prior to end of eclipse or risk your eye sight looking through. If you (only) use it to take images, it's not as critical as it's not your eye being fried but your camera, but don't go beyond diamond ring without filter and camera attached either.
Thus either case: Put back the filter preferentially before the end of eclipse.


What is Earth believed to have looked like before the collision with Theia?

Greenwood et al suggest that Earth had a lot of its water (maybe upto 70%) before Theia, but I can't find anything definite saying that it was liquid, although several sites reporting on the Greenwood paper assume that Earth was cool enough to have an ocean. It almost certainly had a crust, though. And immediately after the impact it had a magma ocean for quite a while. ;)


Why does the Fourier transform of this CMB image have a hole in it?

Having now looked at the paper by Aiola et al. (2020), it emerges that for that map, they filtered the data to exclude low frequency multipoles with $|l|<150$, corresponding to about 1 degree. This filtering was done to all the maps in the paper and will be responsible for the dramatic "hole" in your Fourier transform.
As for the high frequency rings, well they might correspond to high $l$ peaks in the angular power spectrum of the CMB, but other things to consider:
I don't know if the image has already gone through some deconvolution process. If not, then your CMB signature will also be multiplied by the FT of the angular response function of the telescopes (probably a Bessel-like function with some high frequency features).
The finite size of the map will introduce apodisation effects, whereby your Fourier transform is the convolution of the CMB transform with the FT of the apodisation function (a narrow 2D ${\rm sinc}$ function.
You also have the pixel size to worry about. Pixelation will be convolution of the true signal with a square top hat kernel. This will also introduce a multiplicative function in to the net Fourier transform of the FT of the pixel kernel (i.e. multiplying by a broad 2D ${\rm sinc}$ function). I think I see this as a cross-shaped pattern on the image.


How many quakes have been detected by InSight?

This is the latest catalogue from seismic events recorded by InSight: Mars Seismic Catalogue, InSight Mission; V5 2021-01-04. This report includes seismic events up to October 12, 2020. These documents get periodicaly published, so keep an eye on the page pointed out by @planetmaker on the comments. The robot has measured a total of 1759 Marsquake events as of the report date. Keep in mind that 1034 of those events are of suspicious quality, meaning that the signal may not be attributable to a seism. Setting dubious events aside, in the period of 686 days, the mission  has detected about one quake per day. I could not find any website with live data.


Undisturbed planetary alignment after galactic collision

The short answer is that they don't. If an even larger SMBH passed through the centre of the Milky Way and "swept up" the existing SMBH, all the stars and planets that are close to it (within a few tens of light-years, maybe more) would experience huge disruption and might end up orbiting the new black hole and being carried away, or flying free, or whatever.

The stars and planets much further away, which make up the bulk of the Milky Way wouldn't be much disturbed, because they are held in their orbits by their gravitational attraction for each other, and for the dark matter halo which suffuses the galaxy and only a small contribution comes from the gravity of the SMBH.


Does not sound astronomical. But in order to test for astronomical phenomenon you need to state date time (time zone) and location, and direction (roughly). Of course if it was cloudy. The free software Stellarium (stars planets, satellites + ...), in conjunction with Heavens Above (up to date satellite info) can answer most of these types of questions.


It looks like you are asking about rubble piles, asteroids that are made up of a large number of different sized objects that are weakly held together by gravity. A few of the component objects are large, but most are very small (down to grains of sand).

By way of analogy, think of playing pocket pool. Rack the 15 target balls but leave the rack on. The cue ball bounces off when you strike the rack with the cue ball, but that's about it.

Now let's try again, but this time with the rack removed. To make the target balls a even more rubble pile-like, we'll use a piece of cardboard to add a bit of space between the target balls. Now something very different happens when the cue ball strikes the target balls. The balls are only loosely connected to one another. This makes them very good at absorbing the momentum of the cue ball and at distributing the energy and momentum throughout the rack. Given a low to moderate cue ball velocity, the collision will be close to purely inelastic.

And that's how you get a rubble pile. It's one inelastic collision after another after another. Over the course of 4.6 billion years, you have a pile of dust and sand with some rocks and a few large boulders mixed in.


I will edit this with the appropriate equations and timeframes in a few hours when I get back to my office. For now I will address the foundations of your questions with some basics.

1: The radius of the frost line is determined by the star itself (edit: More precisely the nebula that the star forms from.), more in-depth answer will come with edit.

2: The fact that most of the data for these systems is from methods that favour large planets close to the parent star, means that our selection isn't biased, but the method instrument combination is. Another possibility is that migration and loss of angular momentum could be the culprits in this scenario. It is also worth noting, that in the next decade there should be significant data to state whether our solar system is actually common, an anomaly or just younger than many of the other solar systems observed thus far.

3: Larger gas giants "Jovian" planets like all orbiting objects migrate inwards as they loose angular momentum. During the oligarchic growth phase of a soon to be large planet the forming core accrets a lot of material while clearing "all" of the material within it's orbit and associated jeans radius, as the core migrates inwards the supply of material is once again increased.

In short that is why there are "Gas Giants" within the frost line, things migrate inward as they loose angular momentum, and "Ice Giants" form beyond the line because they took too long to migrate inward and gather enough material to move to the next level. Also, if the protoplanet takes too long to accret a ~10-15 earth-mass core, all of the accretable material will be ejected from the system by the star before any significant atmosphere can be obtained.

The dwarf planet part, has to do with the amount of material and how long the protoplanet took to accret it. In that respect, the ice giants and the dwarf planets share a common shortcoming.

Update:

I find myself overwhelmed at work and unable to fully explain the entireties of the process at this time, please forgive me.

However, I do offer a link to an upper division astronomy class' notes that gives a very good explanation of what I believe will help you with your question.


How long do other planets in our solar system have total solar eclipses?

TL;DR: Last for couple of seconds except for Jupiter where it lasts for minutes.

Mars

Phobos and Deimos are too small to totally eclipse sun but there are annular eclipses. See this video which Curiosity Rover took on Mars September 13, 2012 where Phobos is eclipsing Mars.

Jupiter

Only four moons are large enough to create totality: Io, Ganymede, Europa, and Callisto. But three can pass between Jupiter and the Sun at the same time. A triple eclipse happens on the gas giant once or twice every 10 years. Last for some minutes. For more information, see OP's answer.

Saturn

Seven of Saturn’s 62 moons (Janus, Mimas, Enceladus, Tethys, Rhea, Dione and Titan ) are big enough to create a total solar eclipse. But because of Saturn axial tilt of 26.7 degrees, solar eclipses on Saturn are much more rare than solar eclipses on Jupiter (only occurs once every 15 years). This article discussed about eclipses by Pandora and Epimetheus (the former one lasted for split second and the latter one lasted for 15 seconds)

Uranus

Twenty-seven moons circle Uranus, but more than half of them are too small or too far away to completely cover the Sun.  Also, eclipses can occur only when the ring plane-crossing of Uranus (equinox) occurs, approximately every 42 years, with the last crossing being in 2007/2008.

Neptune

Neptune’s six inner moons and Triton can create a total solar eclipse. Because of the planet’s distance from the Sun, Neptune only receives 1/900th of the light Earth does. At 0.018 degrees in apparent diameter, the Sun looks like a dot. That’s why Neptune’s solar eclipses start and end in a matter of seconds. Also, eclipses of the Sun from Neptune are even rare due to the planet's long orbital period and large axial tilt of 28 degrees. In addition, the largest moon, Triton, has an orbital inclination of about 25 degrees to Neptune's equator and has a retrograde orbit. So, eclipse occurs for only a couple of seconds.

Pluto

Pluto also experiences total solar eclipses but every 120 years, Pluto and its largest moon, Charon, take turns eclipsing each other once a plutonian day: that’s approximately 6.4 Earth days. But because Pluto only shows one side of its face to Charon, solar eclipses only happen on that half of the planet.
Source: https://astronomy.com/news/2017/06/total-eclipses-planets


Smallest mass of star to be a black hole?

Stars with 6 Jupiter Masses to 90 Jupiter Masses but are not able to heat up sufficiently to enable fusion become brown dwarfs, and die brown dwarfs.

White dwarfs form when stars have a mass range of about 85 ± 4 Jupiter Masses to 8~9 Solar Masses, or when a star sufficiently heats up to enable fusion.

Neutron stars generally form when the star is >9 Solar Masses to about 24~25 Solar Masses.

Also, as mentioned in the comments, neutron degeneracy pressure doesn't support the neutron star as a whole. Although it has some effects, a neutron star is supported by not only neutron degeneracy pressure but also by repulsive nuclear forces as a neutron star is comparable to that of an atomic nucleus.

As for your blitzar definition, I doubt 4 Solar Masses is the limit. Let's say that the rotation of a given neutron star is a little less than the speed of light. Supposing that if the neutron star doesn't fling itself apart, the mass limit would be much higher, because the centrifugal force forcing the neutron star is much, much more stronger. A blitzar's maximum mass is 18% more than the maximum mass for cold, nonrotating neutron stars.
For your question on blitzars, the star would need to spin rather slowly to avoid forming such, because a blitzar's spin is from the amplification of the original star's spin as is shrinks.

Black holes form when a star is greater than 25 Solar Masses to the point where stars are so massive that they form a black hole itself before even forming a stable star. Such is an example of a quasi-star and is theoretically real in the early universe.


Stars really are immensely far, however it's a common misconception that the stars that you can see are millions of light years away. Most of the visible stars are a few tens to a few hundreds of light years away.

However it is possible that there are stars that have exploded in a supernova: Eta Carinae looks to be approaching the end, and is 7500 light years away (so we see it as it was 7500 years ago). It's one of the most powerful stars in the milky way, but it is so distant that it looks like a faint star.

It is impossible for matter from an Eta Carinae supernova to reach the solar system before light from the supernova, because matter must be slower than light. No matter how far the star is, whether near or far, light beats matter in any race. In fact, Eta Carinae is too remote for any matter from its eventual supernova to ever reach the solar system.

The "starstuff" of which we are made was present in the molecular cloud that collapsed to form the solar system about 4.7 billion years ago, that molecular cloud was enriched with material from ancient stars that died during the 8 billion years that elapsed before the formation of the sun.


How and when did we learn about the universe?

Throughout man's history, we have been learning things about the universe we live in.

Probably one of the most interesting discoveries that let us know how amazingly large the universe really is was not really very long ago, only in the 1920s when Edwin Hubble demonstrated that the Milky Way was not in fact the total sum of the universe, but just a galaxy, and there were other galaxies beyond our own, much further away than any of the stars that we see in the night sky with our naked eyes. This fundamentally changed our perspective of our place in the universe.

We have subsequently made many more significant discoveries about the nature of the universe, and many of those are covered in other questions already. If there are specific details you would like to know about, raise a specific question. The whole topic cannot be broached in one answer.


The axis points in pretty much the same direction all the time, though it does wobble slightly, ranging between 22.1 degrees and 24.5 degrees every 41,000 years.   (Source)

Perpendicular between two lines is a 2 dimensional property, so, while the axis is never "straight up" so to speak, there is a 90 degree angle between the axis and the line between the center of the sun and the center of the earth during  the spring and fall equinox.  Those are the only times.


What are the RAs of the boundaries between traditional zodiacal signs?

TL;DR: Calculating the right ascension (i.e. converting from ecliptic to equatorial coordinates) requires some spherical trigonometry, but fortunately there are internet tools that will do this, such as here. The resulting RA coordinates are:
Aries 0h 0m 0s; Taurus 1h 51m 39s; Gemini 3h 51m 16s; Cancer 6h 0m 0s; Leo 8h 8m 44s; Virgo 10h 8m 21s; Libra 12h 0m 0s; Scorpio 13h 51m 39s; Sagittarius 15h 51m 16s; Capricorn 18h 0m 0s; Aquarius 20h 8m 44s; Pisces 22h 8m 21s.

Explanation and background

Since traditional (Western) astrology uses the same tropical year as astronomical celestial coordinates, the tropical zodiac commences with Aries at 0˚ longitude on the ecliptic, and each subsequent zodiac sign (NB not the constellation, which is different) commences exactly 30˚ further along the ecliptic. So, the zodiac division of Taurus is from 30.00˚ to 59.99˚, Gemini from 60.00˚ to 89.99˚, and so on. It's then just a matter of converting ecliptic longitude to celestial coordinates (right ascension).

Further useful information on the relationship between the ecliptic and right ascension can be found in Mike G's answer to "Effect of the obliquity of the ecliptic / tilt of the Earth on the equation of time".

Both the traditional tropical zodiac and modern astronomical coordinates are based on the northern hemisphere vernal equinox. This starting point, also known as the First Point of Aries, is one of the two points at which the celestial equator crosses the ecliptic.

When Hipparchus defined the First Point of Aries in 130 BCE, it actually aligned with the border between the constellations of Pisces and Aries. However, due to the precession of the equinoxes, the tropical zodiac's starting point has a retrograde movement along the ecliptic at the rate of one degree every 71.5 years. As a result, the First Point of Aries is currently deep in Pisces! Using the current official IAU constellation boundaries, the point of 0h RA will cross into Aquarius in the year 2597, so we have quite some time to wait for the "Age of Aquarius".

Note that not all astrological traditions are based on the tropical year. Hindu astrology, for instance, is based on the sidereal zodiac - i.e. it aligns permanently with the visible stars, regardless of the Earth's axial precession. Some modern Western astrologers also use a sidereal zodiac, in which sidereal Aries currently begins on 15 April.


Are the darker blue/black areas in this picture what the atmosphere of Jupiter looks like without clouds?

userLTK's comment is correct, you cannot see very deep at all.

Information about the image

First off, there are three imagers on the Juno spacecraft: JunoCam, UVS, and JIRCAM. You can see a full list of instruments here. JunoCam is described as a:

  visible-light camera/telescope of the Juno Jupiter orbiter, a NASA space probe launched to the planet Jupiter on 5 August 2011. It was built by Malin Space Science Systems. The telescope/camera has a field of view of 58 degrees with four filters (3 for visible light).

The image you're seeing was taken by JunoCam and is a visible image. In fact, this image is combined by images taken over multiple days and from multiple orbits to get different vantage points and show all areas lit up. It isn't however a mosaic $-$ the 58 degree field of view allows JunoCam to image the entirety of Jupiter in one image. Instead, this image is simply the carefully crafted combination of many different images of the south pole.

FWIW, UVS is an imager in the ultraviolet and JIRAM is an imager in the infrared.

How deep are you seeing into Jupiter?

Not very deep at all. Because the image is in the visible spectrum, the depth into Jupiter before it becomes too opaque is going to be very, very short. You may not even be seeing down to Jupiter's "surface" where the atmospheric pressure equals 1 bar.

Incidentally, JIRCAM can see down to about 5-7 bars if its lucky, enough to penetrate the very upper layers of the atmosphere.

Infrared images of the poles

Obviously I can't be certain what you saw before concerning infrared images of Jupiter's poles, but my guess is they may have been something like the image below. This image was taken by JIRAM


Did Einstein supported Big Bang Theory cosmological model?

Einstein's opinions were not static, and he lived at a time when there were several competing theories and not much observational evidence.

Einstein introduced a cosmological constant $\Lambda$ into his equations, the purpose of which was to allow for a steady cosmos with no expansion or contraction. He is said to have later regretted this addition. He was aware of Hubble's observations of galatic redshift, and Lemaître's "hypothesis of the primaeval atom". But along with most physicists in the 1920s he did not support it. He suggested an oscillating model in the 1930s, but rejected it later (apparently because it didn't solve the problem of entropy)

His final opinion then seems to have been "we don't know".  When he died in 1955 there was no consensus, and Einstein was by then an old man and not actively involved in research physics.


Can I take logarithms of magnitudes (which are already log)

You can, the question is whether you should.

Is there a theoretical reason why you think that magnitude should vary exponentially with redshift? If not then you may be falling into a fallacy.

You see with a polynomial with sufficiently many terms, I can get a curve that fits your data perfectly, r=1, Does this my polynomial is a better model? Probably not.  Just throwing functions at your data and choosing the one that gives the biggest r isn't mathematical modelling of data.

I would question why you are using magnitude at all? You seem to be finding that "close objects appear brighter" which isn't a new discovery.


Can we find out whether early Venus was Earth-like or not?

My ad-hoc opinion: This wouldn't be the first step of Venus exploration. Geologic in-situ investigation of the resurfacing hypotheses would already be a very challenging mission.
Might be, one could find some metamorphic remnants which have survived the last resurfacing, and one could determine the age of rocks. Might be there exist some old layers below the surface, which haven't been molten up completely.

Some kinds of crystals are more resistant to heat than others. They may also contain some records of more ancient epochs. Comparing their isotopic ratios might tells something about the ancient atmosphere, or at least about ancient geology. But conditions on the surface of Venus are very hard for probes.

More feasible would be a detailed isotopic analysis of Venus' atmosphere, as e.g. proposed as VDAP mission concept, or by follow-up balloon missions. A submarine-like ballon diving into the hellish conditions of the lower atmosphere or even to the surface for a short period might be an approach.

A TLS (tunable laser spectrometer) has been suggested to analyse the atmosphere.

May be an additional mass spectrograph would be useful. If short dives close to the surface become feasible, cameras would make sense. A LIBS (laser induced breakdown spectroscope) would make sense only on direct contact with the surface due to the atmospheric absorption. APXS wouldn't probably work due to the heat, unless samples could be taken for analysis higher into the (cooler) atmosphere. IR spectroscopy could possibly work, if cooling can be achieved somehow (e.g. by insulation and adiabatic decompression). If surface samples can be taken during short surface contact, more ways to analyse rocks would become possible. The more instruments, the more weight, the larger the ballon, the larger the probe, the more expensive. But I'm far from elaborating technical or scientific proposals here for space agencies.

Hence VDAP with a small TLS and a QMS (quadrupole mass spectrometer) - as already proposed by hugely experienced scientists - would be a good start.


Could a cryo-volcano be the reason behind this colour difference in Iapetus's hemispheres?

Reposting my own answer from Space Exploration on nearly identical question:

The color dichotomy of Iapetus is due to the darker half, the Cassini Regio, being a result of the moon's accumulation of the dust in the Saturn's largest, yet extremely tenuous, diffuse dust ring called the Phoebe ring depositing onto the moon's leading hemisphere, while the trailing hemisphere's surface remains uninfluenced by this source of dust.

Phoebe ring was only observed first in late 2009 by the Spitzer telescope in the infrared range, and is believed to have been caused by bombardment of micrometeorites of Phoebe, another Saturn's moon that the prevailing theories suggest is a captured Kuiper belt originating planetesimal orbiting Saturn in a retrograde, 173° to the ecliptic inclined orbit. It is the largest discovered ring of the Saturn, stretching from calculated 59 to 300 Saturn's radii (observed between 128 and 207 Saturn's radii by the Spitzer telescope), is tilted 27° to Saturn's inner rings and with an orbital inclination of 175°, suggesting its origins and again retrograde orbit:

    The Phoebe ring was discovered by NASA's Spitzer Space Telescope, detecting it in the infrared spectrum. (Source: Wikimedia)

Iapetus, being in a prograde orbit with a 17.28° inclination to the ecliptic, is directly in the path of this retrograde dust ring at an average distance from Saturn of 215 Saturn radii. Since it's tidally locked to Saturn, it would always face same side towards any debris and dust in its path. Joseph A. Burns, Cornell's Irving Porter Church Professor of Engineering and professor of astronomy said in a press release for Cornell Chronicle:

  "The ring of collisional debris that has come off Phoebe and its
  companion moons is out there, and now we understand the process
  whereby the stuff is coming in. When you see the coating pattern on
  Iapetus, you know you've got the right mechanism for producing it."

So this is the theory. Iapetus is in a prograde orbit intersecting at inclination Phoebe dust ring in a retrograde orbit around Saturn, and since Iapetus is tidally locked and always hitting this dusty path head on, only one hemisphere (the darker one, called Cassini Regio) is covered in Phoebe's dust, while the other hemisphere (the lighter one, called Roncevaux Terra) remains free from it, unveiling this Saturn moon's true surface.

    Artist's conception showing nearly invisible and largest Saturn's Phoebe ring (Source: Wikimedia)

Same mentioned Cornell Chronicle press release explains why the transition from the dark hemisphere to the whiter parts of the moon isn't seamless, but rather a mottled, patchy array of bright and dark spots:

  Small, white craters that dot Iapetus' darker half indicate a veneer
  of dark dust, only meters deep, covering a white, icy surface that
  matches the rest of the satellite. The imaging data also revealed that
  all the materials on the leading side are much redder than the
  shielded and brighter trailing side -- another indication that the
  leading side's dust came from elsewhere.

  The pattern, the scientists say, supports a theory described in a
  companion paper in Science that the darker parts of the moon tend to
  heat up when struck by sunlight, encouraging the ice to evaporate
  underneath.

  This causes any dark spots to get even darker, creating the mottled
  look.

           Mosaic of Iapetus taken by Cassini probe during NASA's Cassini Solstice mission, showing the bright trailing Roncevaux           Terra hemisphere with part of the dark area Cassini Regio appearing on the right. (Source: Wikimedia)


Could there be a closer star to Earth than the Alpha Centauri triple star system, excluding the Sun?

It's begining to look like there aren't as many brown dwarfs as was once thought:

  NASA's Wide-field Infrared Survey Explorer, or WISE, has been turning up a new crowd of stars close to home:

  ...

  Previous estimates had predicted as many brown dwarfs as typical
  stars, but the new initial tally from WISE shows just one brown dwarf
  for every six stars. It's the cosmic equivalent to finally being able
  to see down a mysterious, gated block and finding only a few homes.

Reasonably current data puts the current closest brown dwarf at 6.5LY (2 parsec) and less than 0.08 solar masses.

A nearer one could show up, but it's not looking likely.


Short answer is "The Sun".

As Conrad notes, since when you include the effects of Jupiter, the Earth orbit is non-keplerian, notions of centre are not really defined. But you can ask, in Newtonian gravity, where the Earth's acceleration vector points. Now the acceleration of the Earth is due mostly to the Sun, partly to the moon and slightly to the other planets.

Let's ignore the moon (ie consider the motion of the Earth Moon barycentre)

The acceleration due to the sun is 4 orders of magnitude greater than that due to Jupiter. So if Jupiter is at right angles to the Earth, the acceleration vector of the Earth is slightly pulled away from the centre of the sun. But not by much, in fact it points to a point about 4000km from the centre. The sun has a radius of 700000km, so the point the Earth is orbiting is well inside the sun, and isn't the Sun Jupiter barycentre.

To see why the Earth doesn't orbit the barycentre, consider the motion of the Sun in a three body system (Sun-Jupiter-Earth) The Earth doesn't orbit the Sun-Jupiter barycentre just as the Sun doesn't orbit the Earth-Jupiter barycenter.


Can exoplanets be found using neutrino detectors?

The answer is presumably no.
Let's assume that the nearest possible exoplanet is in the Proxima Centauri system, and it presents an electron antineutrino luminosity identical to that of Earth. The flux from the exoplanet at the KamLAND detector would then be weaker than the flux from Earth by a factor of roughly
$$\left(\frac{\text{Radius of Earth}}{4\;\text{light-years}}\right)^2\approx2\times10^{-20}$$
thanks to the inverse-square law. So the intrinsic signal strength is likely to be essentially zero.
More concretely: the KamLAND collaboration's 2013 measurement of the Earth's geoneutrino flux was approximately $\sim3.4\times10^6\;\text{cm}^{-2}\text{s}^{-1}$, which isn't a lot considering the low probability that a given neutrino will interact with the detector. Indeed, the group reported only $116^{+28}_{-27}$ events in their period of monitoring - and some of that was during a time where nearby nuclear reactors were off and therefore not producing neutrino noise!
Any signal from an exoplanet would be tiny and easily lost in the background of all of the other possible neutrino sources.


Currently proposed ways to distinguish between a really big planet and a really small star

The dividing line between star and brown dwarf is the mass at which hydrogen fusion via the pp-chain occurs at its core. A brown dwarf below this mass limit will never reach a sustainable equilibrium due to this reaction and will continue to cool as it gets older. A star will reach the long-lived main sequence.

The dividing line between planet and brown dwarf is much less well defined. Some want the definition to be mass-based, and there is a mass-based dividing line at about 13 Jupiter masses between brown dwarfs that fuse deuterium in their interiors (which provides only a brief hiatus in their cooling) and "planets" that do not.

Others argue that any definition of a planets should be based on whether they formed around a star, but this is problematic, because such objects may then be disrupted or escape and also it is not clear where a binary brown dwarf becomes a star plus planet.

The reason that some campaign for a different definition is that the composition and structure of a planet formed around a star could be different to that of a planet that collapses from a small gas cloud. According to the "core accretion model" for gas giant formation, the core of such a planet would consist of solid ice or rocky material, whereas it would be gas all the way down in a less massive version of a brown dwarf. However, the waters are further muddied by ideas that large planets can also form by direct collapse  in the disc of a forming planetary system.


Has a “white hole” theory been advanced to explain anomalous star formation in Sagittarius A?

I'm going to say that the answer to the main question is "no", for two reasons.

The Introduction to this recent paper by Naoz et al. mentions a variety of different suggested models and explanations, none of which is based on anything like alternate cosmological models. (The paper itself argues that the apparently problematic distribution of stellar orbits may be an artifact of failing to account for the presence of binary stars, so that the system isn't really anomalous at all.)
It's not at all clear to me how an alternate cosmological model like the one you link to would be even remotely relevant to the details of star formation in the Galactic Center. In other words, I wouldn't expect anyone to be trying to explain star formation near Sgr A* using alternate cosmologies or extensions to GR (the star formation took place mostly well outside the region where GR becomes noticeably different from Newtonian gravity, so modifications to GR would be even less relevant).

As for the second question: we simply don't know the direction or amount of the central supermassive black hole's spin (yet).


From which exoplanets is our sun the brightest star on the night sky?

I wrote https://github.com/barrycarter/bcapps/blob/master/ASTRO/bc-solve-astro-13115.m to solve this with the full results at: https://github.com/barrycarter/bcapps/blob/master/ASTRO/brightest-stars-from-other-stars.txt.bz2

As viewed from RigelKentaurusB, the Sun has magnitude 0.48,
and is the 10th brightest star in the sky. This is the only
star system with known exoplanets where the Sun even makes
the top 10.

From EpsilonEridani, the sun shines at magnitude 2.37,
making it the 79th brightest star in their sky. Things go
rapidly downhill from there:

79 EpsilonEridani Sun 2.36646
207 HIP113020 Sun 3.18865
208 HIP85523 Sun 3.11216
247 HIP106440 Sun 3.29606
396 HIP15510 Sun 3.74083
423 HIP74995 Sun 3.81444
685 Fomalhaut Sun 4.25746
878 HIP64924 Sun 4.48202
949 HIP109388 Sun 4.54456
972 HIP99825 Sun 4.55679

This is based on Mathematica's list of 552 exoplanets
orbiting 464 stars, and a total list of 88,637 stars.

If the naked eye visibility limit is magnitude 5.5, the Sun
would only be visible from 28 of these stars:

10 RigelKentaurusB Sun 0.476137
79 EpsilonEridani Sun 2.36646
208 HIP85523 Sun 3.11216
207 HIP113020 Sun 3.18865
247 HIP106440 Sun 3.29606
396 HIP15510 Sun 3.74083
423 HIP74995 Sun 3.81444
685 Fomalhaut Sun 4.25746
878 HIP64924 Sun 4.48202
949 HIP109388 Sun 4.54456
972 HIP99825 Sun 4.55679
1086 HIP57443 Sun 4.65677
1126 HIP21932 Sun 4.69871
1381 HIP57087 Sun 4.87838
1426 HIP83043 Sun 4.89689
1409 Pollux Sun 4.90049
1617 HIP10138 Sun 5.01833
1649 HIP57050 Sun 5.04143
1676 HIP3093 Sun 5.05658
1716 HIP48331 Sun 5.06528
1831 Gl317 Sun 5.13093
2104 HIP22627 Sun 5.24572
2264 Rho1Cancri Sun 5.31849
2263 HIP40693 Sun 5.32723
2397 HIP27887 Sun 5.36943
2355 HIP80337 Sun 5.37669
2382 GJ1214 Sun 5.3879
2689 UpsilonAndromedae Sun 5.47503

If the limit is magnitude 6.5, 33 more star systems can see the Sun:

2838 Alrai Sun 5.52682
2926 HIP53721 Sun 5.571
3399 HIP79431 Sun 5.69393
3637 MuArae Sun 5.74863
3679 HIP113357 Sun 5.76061
3783 TauBootis Sun 5.7949
3941 HIP98767 Sun 5.83457
4071 HIP85647 Sun 5.86725
4318 HIP71395 Sun 5.92513
4580 HIP6379 Sun 5.95739
4783 IotaHorologii Sun 6.01138
4835 HIP7978 Sun 6.02527
4867 RhoCoronaeBorealis Sun 6.03887
5016 HIP1292 Sun 6.0583
4949 NN3634 Sun 6.08757
4993 HIP55848 Sun 6.10352
5252 HIP49699 Sun 6.11214
5300 HIP83389 Sun 6.11214
5031 HIP65721 Sun 6.11803
5343 HIP79248 Sun 6.12236
5385 PiMensae Sun 6.12986
5453 EpsilonReticulii Sun 6.13303
6158 HIP98505 Sun 6.251
6351 HIP113421 Sun 6.30305
6436 Nu2CanisMajoris Sun 6.31593
6343 HIP99711 Sun 6.32197
6737 Hamal Sun 6.35637
6516 HIP64457 Sun 6.37975
6867 HIP58451 Sun 6.4205
7614 HIP25110 Sun 6.43775
7525 HIP109378 Sun 6.46941
7538 HIP54906 Sun 6.47034
7957 HIP96901 Sun 6.48193

OLD ANSWER FOR REFERENCE (note that Sirius does not have known exoplanets, so the Sirius solution I propose below wouldn't work):

As others have noted, since Sirius is 25 times more luminous than our Sun (though still not luminous as Canopus or Rigel), so it would be the brightest star for most nearby star systems, and all the exoplanets we've found are fairly closeby.

The only possible exception would be an exoplanet orbiting Sirius itself, since Sirius would be considered their sun, and not a star in the night sky.

Unfortunately, Sirius' smaller companion, Sirius B, would be much brighter than our Sun there. I'm also pretty sure Sirius B would be in the night sky (not just in the daytime sky), so it would count.

I'd like to poke around a bit more before declaring this answer complete. I think I can show that any point where the Sun is brighter than Sirius must be less than 4.3 light years (and probably much less) away from our solar system, so, unless we discover an star system MUCH closer than Proxima Centauri, there are no exoplanets where our Sun is the brightest star.


Angular speed of a planet across the sky

There is no simple formula.

First, it's not true that the angular speed of a planet relative to the sun can be found from its orbital period.  This is because planets don't move in circles, but in eclipses and the speed of the planet changes, as described by Kepler's second Law.

When describing the motion of a planet in the sky you need to consider the elliptical motion of both the Earth and the planet, the superposition of these motions causes the planets to make looping motions in the sky, sometimes slowing down and moving backwards

The image shows how Mars would have appeared to loop and move backwards in 2003 and 2005.

The motion in the sky can be calculated, but not by a simple formula. You start with the orbital elements of the planets, solve Kepler's equation to determine the position on their elliptical orbit. Solve some trigonometry to determine the position of each planet relative to the sun, then some vector calculations and spherical geometry to find the position of the planet relative to Earth.

There is a guide online for writing a computer program to do this.


What Causes the Large Radiation Fields Around Jupiter?

As you stated, we have not been able to simulate the pressure and temperatures required to generate those that are believed to exist in Jupiter's interior other than in short-lived shockwave experiments, according to the NASA webpage A Freaky Fluid inside Jupiter?, observing that

  "Liquid metallic hydrogen has low viscosity, like water, and it's a good electrical and thermal conductor," says Caltech's David Stevenson, an expert in planet formation, evolution, and structure. "Like a mirror, it reflects light, so if you were immersed in it [here's hoping you never are], you wouldn't be able to see anything."

Going further, according to the article Jumpin' Jupiter! Metallic Hydrogen (Lawrence Livermore National Laboratory), discuss the shock wave results, finding the level at which hydrogen metallises as being

  from 0.9 to 1.4 Mbar, resistivity in the shocked fluid decreases almost four orders of magnitude (i.e., conductivity increases); from 1.4 to 1.8 Mbar, resistivity is essentially constant at a value typical of that of liquid metals. Our data indicate a continuous transition from a semiconducting to metallic diatomic fluid at 1.4 Mbar, nine-fold compression of initial liquid density, and 3,000 K.

The findings from the researchers above are summarised in the diagram below

The source is the Jumping Jupiter link above.


The answer depends on your definition of a day.

If you define a day as we usually define it in the Earth (time between the Sun is at noon or average time between sunrises, a 24 hours day in Earth), the length of a day in the Moon is the synodic period of the Moon and it takes 29.530589 days (29 d 12 h 44 min 2.9 s).

Anyway, if you define a day as rotation on own axis, that is, time between a far away star being at noon, you have what in Earth is called sidereal day (23 hours, 56 minutes in Earth). In the moon the day defined this way would be 27.321661 days (27 d 7 h 43.19 min 11.5 s).

Please notice that since the Moon is tidally locked to the Earth, rotation periods equal orbital periods.


Observing the motion of the solar system and galaxy through space?

All motion is relative. There is no absolute space which can be said to be the correct reference frame in which we must measure all motion. Hence, for any object you're free to construct any reference frame that does, or does not, give that object a helical motion.
Reference frames
Of course, some reference frames make more sense than others. When you cycle to work, it makes sense to measure your speed relative to the bike lane. The reference frame of the Sun makes sense if you're concerned with planets' orbit. In this frame, Earth has no helical motion, only circular motion. Or you could call it helical motion with $v_z = 0$, if you wish.
On the other hand, the Sun and its neighbors follow a path through the Milky Way; the so-called Local Standard of Rest, measured by observing nearby stars and star-forming regions. In this frame, Earth has a (non-$v_z\!\!=\!0$) helical motion.
Similarly, in the reference frame of Milky Way's path toward, say, Andromeda, the Sun has a helical motion. Another star will have a similar, but different, motion.
Comoving coordinates
That said, there is one reference frame which has a special status, namely the Universe's comoving frame. These are the coordinates that expand along with the Universe, and they are the coordinates in which everything on average is at rest. If you hear the phrase "Milky Way's motion through space", this is probably what is referred to. Since time flows differently in different frames, it's also the frame of choice when you talk about the age of the Universe. Likewise, since lengths are contracted in moving frames, the comoving frame is the frame in which we measure cosmological distances.
It's important to realize, however, that motions, positions, etc. measured in the comoving frame are no more "correct" than in any other frame.
The cosmic microwave background
But how can we measure a velocity in this reference frame, if all motion is relative?
One way would be to measure the motion of a million galaxies with respect to us, and take the average. But a much easier method, which is in fact how we do it, is to measure the dipole of the cosmic microwave background (CMB). This radiation was emitted when the Universe was very young, and has the special property that it was emitted isotropically (i.e. the same in all directions) with a near-perfect Planck spectrum.
Hence, if you standing still in the comoving frame, you will see exactly the same CMB spectrum in all directions, whereas if you're moving you will observe the CMB to be blueshifted in front of you, and redshifted behind you. This difference between the two opposite directions is called the dipole, and its value was measured by the Planck satellite (Planck Collaboration 2020) to imply a total velocity of the Sun in the comoving reference frame of
$$
v_\odot = 369.82\pm0.11\,\mathrm{km}\,\mathrm{s}^{-1},
$$
toward the Galactic coordinates
$$
\{\ell,b\} = \{264.021^\circ\pm0.011^\circ, 48.253^\circ\pm0.005^\circ\}.
$$
Note the amazing precision, with an uncertainty in speed and direction of only 400 km/h and less than an arcminute, respectively!
This velocity is important to know, since it is used to correct all measured cosmological velocities.


Can a gas giant have an other gas planet as satellite?

It would seem so.

Think about brown dwarfs. At the lower end of the mass spectrum, they're only a couple dozen times that mass of Jupiter, and only a couple times the mass of Hot Jupiters. Some brown dwarfs have been found to have planets. Taking a look at some examples:

2M1207: Roughly 25 Jupiter masses, this brown dwarf has a planetary-mass object orbiting it, 2M1207b. This object has between 3 and 10 times that mass of Jupiter, and is fairly hot, although it orbits far from the star.
2MASS J04414489+2301513: Roughly 20 Jupiter masses, with a companion on the scale of 5 to 10 Jupiter masses that could be a planet.

Note that these brown dwarfs are very low-mass, only a bit more massive than hot Jupiters. They can have orbiting gas giants - why couldn't high-mass hot Jupiters have the same?


What is the definition of a "pole" of a celestial body?

You are mixing the rotation of the body around a barycenter with its moons, and the rotation of the body around its own center.
For a bound rotation like Pluto and Charon both have to have the same rotation period - yet both have to be present. Thus Pluto's axis of rotation of course intersects its surface - it rotates around its own axis at exactly the same speed as it orbits around the common barycentre with its moon Charon. Were it different, it would not be a bound rotation.
The North pole usually is defined as the place where the body's rotation axis (about it's own center of mass) with the right-hand-rule intersects the surface (thus the vector of the rotational momentum in positive direction intersects the surface when placed in the centre of rotation).
There is virtually no non-rotating body.


You should go to the site https://fits.gsfc.nasa.gov/ where you can read about the FITS data format. This site also has utilities for examining and viewing the data. Most of your concerns about this data should be answered by the documentation on this web site.


How does solar activity (e.g. flares, coronal mass ejections) change over the life cycle of a Sun-like star?

As in the Earth and more than in the the giant gaseous planets the Sun's magnetic field is generated like in a dynamo. A few facts about it:

The rotating plasma inside the Sun acts like a dynamo creating it's
magnetic field.
The plasma is ionized so all interact with the magnetic field and
vice-versa.
The sun as a fluid body also has Differential rotation, that
means the equatorial part of it rotates faster than the poles. This provokes what is known as the Solar cycle every 11 years.

At the begining of the Solar cycle the Sun's magnetic field is pointed North-South as in the Earth. During the cycle the differential rotation makes the plasma interact with the magnetic field causing it to be displaced. This makes the first the Sun's spots appear near the poles. As the cycle advances the magnetic field get more and more scrambled, it's orientation changes to west-east and the spots appear closet to the equator. Finally the magnetic field can't hold the "pressure" and "breaks" like a rubber band creating the biggest spots and flares. The magnetic field stabilizes again but with a south-north polarity and the cycles begins again.

Take a look at this gif image that shows how the cycle go.

And in this image you can see what is know as the butterfly diagram that shows how the Sun's spots are located closer to the equator as the cycle advances. And it lasts 11 years.


The factor k, the fraction of sky you cannot see, is not adjusted because of the ground or your peripheral vision. When on a flat plain (a "sea-level" horizon) and none of the sky is covered by clouds, buildings, or trees, then k=0. (Therefore, F=1, and the Zenithal Hourly Rate is not diminished.)

I do not know how the light pollution map was created and whether it accounts for the humidity. The humidity can have a significant effect on the limiting magnitude. In other words, if it is a dry day and the air is clear, you can see fainter stars than if it is a humid day and the sky is murky. If your location is clear (so you have blue skies during the day and black skies at night), then assuming m=6.5 is realistic. (Note that when you click on a location on the light pollution map, it will give the Bortle classification. Clicking the link for the Bortle class will give the expected limiting magnitude.) Of course, the limiting magnitude also depends on your eyesight. Younger people can see fainter stars than older people. (I just tried the map for my location, and it predicts 5.5 to 6.0. I can usually only see stars in the range of 5.0 to 5.5 because of the light pollution and humidity.)


How to calculate a planet's apparent size when the planet is viewed from a moon in orbit around it?

By a straightforward bit of trigonometry, if the distance from the observer to the (centre of) the planet is x km, then the radius, r km, of the planet subtends $\arcsin(r/x)$, and so the angular size of the planet is twice this: $2\arcsin(r/x)$

For Io x = 420,000 km, and the radius of Jupiter is r = 70,000 km, so the angular size is
$2\arcsin(1/6)=19^\circ$. For comparison, the moon from the Earth has an angular size of about half a degree.

Stellarium can be set up to view from other worlds of the solar system


How do they know the TNT equivalent of the meteor explosion over Vermont?

According to NASA Meteor Watch Facebook page, they:

obtained infrasound measurements from 3 nearby stations - the
amplitudes and durations of the signals put the energy of the fireball
fragmentation at 440 pounds (200 kilograms) of TNT.

Here is an image of the infrasound:

From this news story,

A man in Canton, Maine said there was a delay of several minutes from
the time he saw the fireball until heard a loud boom. He said it shook
his home unlike anything he experienced before.

It all makes me wonder How loud was the asteroid that killed the dinosaurs?
Further reading on infrasound detectors monitoring space:

What is used to monitor objects burning up in the atmosphere on descent?
this answer to What local pressure changes would a starship launch on Mars cause?

and others:

What efforts have been made to separate the microseismic events related to wind-related ocean waves from the components of volcanic origin?
Project Mogul: Does upper atmosphere indeed have a sound waveguide channel?


I'm going to try to take a stab at answering this. With our current technologies, detecting exomoons can prove hard however there are various techniques being used today such as:

Analyzing data from the Kepler Spacecraft
Dynamic effects – the exomoon tugs the planet, which causes deviations in the times and durations of the host planet’s transits. This is similar to the radial velocity technique for detecting exoplanets. Source: UniverseToday
Transit effects – the exomoon may transit the star immediately before or just after the planet does. This will cause an added dip in the observed light. See this video for a great demonstration. This is similar to the light curve technique for detecting exoplanets. Source: UniverseToday
Gravitational Microlensing - which is a technique used to detect exoplanets like you stated above however it may also reveal signs of an exomoon. Read this source

I found this information doing some simple googling, feel free to edit or add to.


Why we can't see our Sun as it was a few years ago?

Our Sun moves through the galaxy much slower than light does. So the light that was emitted last year by the Sun is already very far away from the Earth. Therefore, we don't receive any light from (i.e. we don't see) the Sun as is was last year.

As an analogy, image walking along and shooting bullets in all directions. The bullets are going much faster than you can go by walking, so you won't ever get hit by your own bullets. But someone walking next to you can get hit by the bullets you are shooting out.


What's the relationship between Moon phase and the Milky Way visibility?

Moon Phase
With the moon phase, it's going to be in Waning Gibbous on March 9th, so it will be reflecting a lot of light back to earth. Any light that interferes with the sky will make dimmer objects harder to see.

The Moon's reflectivity figures taken from here:

Fraction of the moon: 0.61371425766053
Illumination of the moon disk: 88 %

Location
It will also depend a lot on your home location (hemisphere) as to the position of the Milky Way at different times. Your proximity to any light pollution will of course have an impact, as living on the outskirts of a city myself, it is impossible to view the Milky Way. Only when in very rural areas on clear nights have I had a good view of it.
New Moon
On the 20th March there is a New Moon, so this may be a more opportune time to try to capture it.


What would an asteroid inside of Mercury's orbit be called?

The name used for such objects is vulcanoids. This term refers to the hypothetical intra-Mercurian planet Vulcan, which was proposed by Urbain Le Verrier (best known for his successful prediction of the existence of Neptune) to account for the anomalous precession of Mercury's orbit that is now explained by general relativity.
Note that the expected number of vulcanoids is low: the Yarkovsky effect should  clear out objects smaller than 1 km on ~gigayear timescales (Vokrouhlický et al., 2000), while the YORP effect would be effective at destroying larger objects by spinning them up until they break apart (Collins, 2020). Observational constraints from NASA's STEREO mission suggest the number of objects larger than 1 km may be no more than 76 (Steffl et al., 2013).


Can we find rocks from Venus or Mercury on Earth?

You can think of it in terms of Hohmann transfer orbits, which define the minimum $\Delta v$ that needs to be applied to bring something from one orbital radius to another orbital radius when orbiting a massive body. This calculation takes into account that the two objects have Keplerian orbits where the objects begins with at least the orbital speed of the initial orbital radius.

The Hohmann $\Delta v$ is given by
$$\Delta v = \sqrt{\frac{GM_{\odot}}{r_1}} \left(\sqrt{\frac{2r_2}{r_1 +r_2}} -1 \right),$$
where $r_1$ and $r_2$ are the initial and final radius from the Sun.

Conceptually what you have to do is give a rock enough energy to escape from the planet and then give it an additional $(m/2)(\Delta v)^2$ of kinetic energy to get it to transfer into the other orbit. If the ejection speed is $v_{ej}$ then
$ v_{eg} > \sqrt{(\Delta v)^2 + v_{esc}^2},$
where $v_{esc}$ is the escape velocity.

The numbers for Mercury $\rightarrow$ Earth are $\Delta v = 9.2$ km/s and for Mars $\rightarrow$ Earth $\Delta v = -2.6$ km/s (you have to slow it down to allow it to fall inwards).

The escape velocities for these planets are 4.35 km/s and 5 km/s respectively (so almost the same).

This means you need to give a rock more kinetic energy to get it to Earth from Mercury as from Mars. In the case of Mars, the transfer kinetic energy is almost negligible once the rock can escape Mars' gravity. In the case of Mercury, the rock needs to be given an initial ejection velocity of $> \sqrt{9.2^2 + 4.3^2}= 10.1 $ km/s. This compares with $> \sqrt{2.4^2 + 5^2}= 6.5$ km/s for Mars. At lower ejection speeds most of the ejected objects will be reaccreted by the planet.

Against this, the leading theory to explain migration of rocks between planets is high velocity impacts. Objects falling from much further out will hit Mercury with greater speeds than Mars and impart greater energies to the ejecta.

A paper addressing the possibility of Mecurean meteorites was presented by Gladman & Coffey (2008). They concluded that once ejection speeds are large enough ($\sim 10$ km/s) to produce Earth-crossing ejecta, that significant accretion of meteorites should take place. Several per cent of high speed ejecta should impact the Earth (or its atmosphere at least) within 30 million years. This compares with an efficiency a factor of 2-3 higher for Mars.

There are various reports and speculations that at least one meteorite in existing collections (NWA7325, pictured) may have come from Mercury. See here for example. It appears that the main problem is getting agreement on what the chemical signatures of such meteorites are.

Accretion of material from Venus is a different matter. The required ejection velocities are higher because the escape velocity for Venus 10.4 km/s. But more importantly, drag in the dense Venusian atmosphere would prevent anything emerging from the planet with anything like these speeds.


Are the stars seen at the equator different throughout the year?

Yes, the constellations that are visible at midnight on Jan 1 are different than those visible at midnight on July 1 (6 months later). The website Stelvision has a star chart that you can customize for date, time, and location. This is due to the Earth's revolution around the Sun and the direction that an observer is facing at midnight.

From a given location, a given star or constellation rises, transits (when an object is due south), and sets 2 hours earlier each month.

In generally, the above is true for all latitudes. Of course, stars/constellations that are circumpolar never rise and set, and from the Earth's pole all objects are circumpolar. But even in this case, if you would pick a direction to look at midnight on Jan 1, those constellations would be behind you (180 degrees away) at midnight on July 1 (assuming you could see them in the daylight on one of those dates ;-).


From that location, the easiest time in the morning to see the comet is between 4:50 and 5:10 AM EDT.
I've seen it at similar times relative to nautical dawn at my location.
Much earlier than that, the comet may be too low, especially if nearby trees obscure the horizon.
Some software mishandles daylight saving time.
As the comet shifts eastward, nautical dusk around 10:00 PM will be a good time in the evening to look for it.
This Sky & Telescope article provides both morning and evening finder charts.


How are galactic rotation curves determined from radial velocity measurements?

Imagine a spiral galaxy where the rotation of gas and stars within the disk is close to circular (generally true for gas clouds, though less true for stars) in a single plane (the disk), and imagine that it is tilted at some intermediate angle with respect to your line of sight. A circular disk galaxy will then project to form an apparent ellipse, with a major axis and a minor axis (see figure).

If you look at points which lie along the minor axis (say, points b1 or b2), you will see stars and clouds which are moving perpendicular to your line of sight, so they have zero radial velocity relative to you, and there will be no Doppler shift.

If, instead, you look at points which lie along the major axis (say, points a1 or a2), your line of sight will intersect gas clouds and stars which are moving (mostly) along your line of sight (either towards you or away from you), so you will observe the maximum possible radial velocity and thus the maximum possible Doppler shift (either positive = velocities away from you, or negative = velocities toward you).

There's one further factor, which is the effect of inclination. If the galaxy is genuinely face-on (inclination $i = 0$), then you won't seem any Doppler shifts, because all the motion is in the plane of the sky, perpendicular to your line of sight. (In practice there will really be some motions up and down with respect to the galaxy plane, but this won't tell you about the rotation curve.) If the galaxy is edge-on, then you get the maximum possible Doppler shift, which gives you the rotation velocity $V$ directly. If the galaxy has an in-between inclination, then the radial velocity is the part of the cloud/star velocity in the direction of your line of sight, which is $V \sin i$. So if you can determine the galaxy's inclination, you can divide the observed velocity along the major axis by $\sin i$ to get $V$.

Up until relatively recently, the practice (in optical astronomy) was to use a long-slit spectrograph and orient it so that the slit lies along the galaxy's major axis. Then at each point along the major axis, from the center outward, you measure the Doppler shift, apply the $\sin i$ inclination correction, and then you have the rotation curve: $V$ as a function of radius $r$ along the major axis.

If you have fully two-dimensional spectroscopy (e.g., from a radio telescope interferometric array observing emission from atomic hydrogen or molecular gas, or an optical or infrared integral field spectrograph), then you can use the data at intermediate locations within the disk (not just along the major axis), where the observed velocity is $= V \sin i \cos \alpha$, where $\alpha$ is the angle between the major axis and the location you're looking at (so for locations on the major axis, the observed velocity is just $V \sin i$). Then you can fit a model to the set of observed velocities (which you get from the Doppler shifts) and work out $V$ as a function of radius. (The standard approach is called a "tilted ring" model, where you assume the velocity is a simple function of radius, but allow for the possibility that the disk might be warped, so that the orientation -- including the inclination -- might change with radius.)