Are ocean water planets at Mars' atmospheric pressure possible? Can water planets exist whose atmospheric pressure at sea level is just above the triple point of water, say at about 690 Pa (0.1 psi)? If no, why not? Just curious. How thick or thin can atmospheres of ocean planets be?

There's a couple reasons why it would be difficult, but not necessarily impossible (and I'll toss in a hypothetical at the  end).   I don't know how to answer this questions with a lot of "in theoreys" and "this is unlikelys", but I'll give it a shot.  Maybe when the James Webb Space Telescope gets up and running we'll get a better look at planet's atmospheres and maybe even identify a planet somewhat like your scenario.  For now, exoplanet atmospheres are incredibly hard to measure, so there's mostly just models and educated guesses to work with.
One reason it would be tricky is because the vapor pressure alone approaches the triple point air pressure, so this means that if the planet got too warm too often, say, mid-day noon or during summer, assuming it has seasons, then the vapor pressure, just from the ocean would probably rise above the triple point and that's before you add any additional gases.   CO2, CH4, N2, O2 if you want to get creative and get rid of a lot of the CO2 by photosynthesis and lets not forget Argon which forms from decaying potassium 40 in the planet's crust.  Earth's atmospheric Argon alone is about twice the triple point pressure.
So you'd need a planet with oceans but virtually no other gases and that might be hard to do.   Planets can lose atmosphere to space by solar wind or jeans escape, but water is a light molecule and arguably gets removed from a planet more easily than gasses like CO2, N2, O2 or Argon.  That's not to say it's impossible that a planet might be sufficiently water rich that it loses atmosphere but retains water.  I just think it would be a rare occurrence.
A second potential problem is temperature variation, which means, if the planet is too warm, too much ocean becomes atmosphere and adds too much vapor pressure and if the planet is too cold, the oceans would be largely covered by ice.   Planets are dynamic and a thin atmosphere planet should have a wide temperature range, just as Mars does.   Latitude, day/night cycle or day side/night side if tidally locked and eccentricity could all play roles in significant temperature variation, and lets not forget, the planet's internal heat and potential volcanism.   There would be a narrow range where the planet could maintain both oceans and a very thin atmosphere.
A final consideration is what you mean by triple point.   Oceans are likely briny and astronomers as a rule, are more interested in the "triple point" of briny water not pure water, though a chemist would say that triple point applies only to pure substances not salt water or possibly ammonia water.
If you consider the triple point of salt water oceans, then it goes below freezing where ice sheets become possible, and with an expected natural variation in the planet's temperature, ice sheets become very likely.  That doesn't make your scenario impossible, but I think you'd need a cold planet mostly covered by ice with narrow equatorial regions of exposed ocean - just a thought.
The triple point of briny water is discussed here - a fun read.
And a chart with some thing like the triple point of salt water in this research gate question here.
If, in theory, Mars was given oceans, by some terraforming efforts, as a result, initially the water would evaporate but over time, it's combination of water vapor and CO2 would push it's atmospheric pressure further above the triple point, and as it's south pole releases more CO2 as it presumably gets a little warmer, Mars would move away from this scenario if it had enough liquid water.  It has too much CO2, though for now, the majority of it's CO2 is frozen.
And Ammonia/Water ocean is especially interesting as the freezing point of the correct ratio can dip below the freezing point of CO2, this drops the triple point significantly, below -80 C° but it may solve the CO2 problem.  Frozen CO2 at the poles with Ammonia/Water oceans and lots of water ice.  Most of the oceans would have a permanent water-ice cover.
Mars, for example, undergoes a variation in atmospheric pressure as it goes through it's seasons by as much as 25%.  (Scroll down to the Mars section).   That's mostly eccentricity driven and happens because Mars is further from the sun during it's Southern Hemisphere winter and this allows some atmospheric CO2 to freeze, which thaws again as the planet moves back towards it's perihelion and Southern Hemisphere summer.
Certainly a planet with oceans and triple-point like conditions could be designed if you could create your own planet, but in actuality, the likely mix of compounds in the atmosphere and temperature variation and general dynamics of a planet would, I think, make such planets rare, if they exist at all.   That said, observations have shown astronomers things that they didn't expect to see many times before, so there's a chance.
Hypothetical: a Mars like planet has ice at it's poles, but warms up enough to lose it's atmosphere.  Ice is stable even in a vacuum, provided it stays below freezing temperature.
After losing it's atmosphere, the planet warms up and it develops a magnetic field.  The ice melts and now it has oceans and a very thin atmosphere.   Perhaps not impossible.  It requires a few things to go just right, so, still unlikely, I think but who knows.


Communication for far away satellites Today in the tech age we have many communication technologies cell tower for cell  phones,  wifi, nuclear subs have some classified method of communication,  etc.  What communication technology is being used if a satellite probe is sent past Jupiter & transmits a signal back to scientists to monitor progress or review high definition detailed images?

That's the job of NASA's Deep Space Network, plus sites in other countries that perform similar functions. NASA currently has facilities in California, Spain, and Australia whose primary function is to communicate with devices sent beyond low Earth orbit.

In addition to these Deep Space Network sites, NASA and the US Department of Defense jointly operate the Tracking and Data Relay Satellite System network. This system comprises geosynchronous satellites outfitted with communications equipment, relay stations in White Sands, New Mexico and in Guam, plus a control station in Greenbelt, Maryland. The European Space Agency has developed a similar system that takes advantage of recent developments in long distance laser communications. These satellite-based systems are intended for use by vehicles near the Earth. More remote vehicles such as probes to Mars, Venus, and beyond generally use the Deep Space Network.


Is there any cosmological evidence that neutrinos travel at a speed other than c? I've been looking into neutrinos. One thing I'm puzzled about is how fast they travel.

See the Wikipedia SN1987A article where you can read this: “approximately two to three hours before the visible light from SN1987A reached Earth, a burst of neutrinos was observed at three separate neutrino observatories”. Some reports describe an initial burst of neutrinos which arrived 7.7 hours before the light. It's thought that the neutrinos were emitted before the light because of the way stars collapse, wherein photons take longer to get out. I don't have much of an issue with that. But what's interesting is that in 168,000 years travelling through space, the photons didn’t overtake the neutrinos. That suggests to me that neutrinos are massless, which ties in with what Fermi said in section VII of his 1934 paper. He plotted three curves and noted that the μ=0 curve matched the empirical curves. He concluded that “the rest mass of the neutrino is either zero, or, in any case, very small in comparison to the mass of the electron”.

The thing is that if neutrinos travel at exactly c they have no mass at all. And I can't find any evidence that they don't. I can find plenty of evidence for electrons travelling at speeds less than c. The neutrino is said to be a fermion like an electron. But I can't find any evidence of neutrinos travelling at speeds less than c.

Can anybody give me a reference to any cosmological evidence that neutrinos travel at less than c? Or any other hard scientific evidence? By evidence I don't mean an inference such as "neutrinos go missing so they must oscillate therefore they must have mass therefore they must travel at less than c". I mean evidence.

A paper by Wie et al. (2016) uses the coincidence between gamma rays from Gamma ray bursts and individual 3-30 TeV neutrinos to estimate the difference in propagation speed between neutrinos and photons.

They find
$$ \frac{|v-c|}{c} < 2.5 \times 10^{-18} $$

They say this is about 7 orders of magnitude more precise than any previously published limit. However, it does rely on the genuine coincidencce between the GRBs and these individual neutrino events, which does seem to be in question or even doubtful in some cases.

The mass limit this places on the neutrinos appears to be a rather uninteresting $<6.8$ keV, but it is a direct measurement of speed rather than an extrapolation of the speed from a mass limit and special relativity.


How does the length of the minor semi-axis of the moon's disc's inner terminator vary through the cycle? The curve that separates the bright and dark parts of the moon's disc as viewed from the earth is called the inner lunar terminator, which is semi-elliptical in shape. We can model the disc as a unit circle centred on the origin, in which the inner terminator runs from (0,1) to (0,-1) and its major semi-axis always has length 1.

How does its minor semi-axis vary during the lunar cycle? Please assume that orbits are circular.

In other words, if the point halfway along the inner terminator is X, what is the equation for the motion of X back and forth along the line-segment from (-1,0) to (1,0)?

Note that the shape of the visible disc is not a lune, except in the two trivial cases.

The moon is rotating wrt to the sun. As the moon rotates, the longitude of the terminator increases linearly. Treating the moon as being at "great distance, so the moon appears as an orthographic projection, the position of the middle of the terminator varies sinusoidally, it moves fastest when the moon is at first or third quarter.

In the orthographic projection, the longitude and latitude $(\lambda,\varphi)$ are mapped to (x,y) by
$$
\begin{align}
x &= R\,\cos\varphi \sin\left(\lambda - \lambda_0\right) \\
y &= R\big(\cos\varphi_0 \sin\varphi - \sin\varphi_0 \cos\varphi \cos\left(\lambda - \lambda_0\right)\big)
\end{align}$$

For points on the equator, and treating the central longitude $\lambda_0$ as the lunar prime meridian, and a scale of R=1, that simplifies to
$x = \sin(\lambda)$. As the terminator moves, its longitude increases linearly, so x varies sinusoidally.

Correcting for elliptical orbits, librations of the moon and a general perspective view would only marginally alter this.


Quasar mass and accretion rates This page on Wikipedia -- Quasars mentions that the "The largest known [quasar] is estimated to consume matter equivalent to 600 Earths per minute". However, there is no citation for this comment. How can I find out where this information came from? I've commented in the Talk section for the page.

Tricky to say for sure, but I would imagine it comes about from measurements of the luminosity and inference of the black hole mass in such systems.

The most extreme objects radiate at the Eddington luminosity, where gravitational forces on matter falling into the black hole are balanced by radiation pressure from the heated material closer in.

If infalling mass is converted to luminosity at a rate of
$$ L = \epsilon \dot{M} c^2,$$
where $\dot{M}$ is the mass accretion rate, $L$ is the luminosity and $\epsilon$ is an efficiency factor, which should be of order 0.1; then the mass accretion rate at the Eddington limit is given by
$$ \dot{M} = \frac{4\pi G M m_p}{\epsilon c \sigma_T} \simeq 1.4\times 10^{15}\frac{M}{M_{\odot}}\ {\rm kg/s},$$
where $M$ is the black hole mass, $m_p$ the mass of a proton and $\sigma_T$ is the Thomson scattering cross-section for free electrons (the major source of opacity in the infalling hot gas).

The biggest supermassive black holes in the universe have $M \simeq 10^{10}M_{\odot}$ and thus the Eddington accretion rate for such objects is about $1.4\times 10^{25}$ kg/s or about 2.3 Earths/second or 140 Earths per minute. The difference between this estimate and the one on the wikipedia page could be what is assumed for the biggest $M$ or that $\epsilon$ is a bit smaller than 0.1 or indeed that the luminosity could exceed the Eddington luminosity (because the accretion isn't spherical).

Perhaps a simpler way to get the answer is to find the most luminous quasar and divide by $\epsilon c^2$. The most luminous quasar ever seen is probably something like 3C 454.3, which reaches $\sim 5\times 10^{40}$ Watts in its highest state. Using $\epsilon = 0.1$ yields about an Earth mass per second for the accretion rate.

So perhaps the number on the wikipedia page is a little exaggerated.


From where so much of energy would have arrived as was required for such a giant universe to have come into existence? If we convert the mass of all the stars and all the planets which must be weighing trillions and trillions of tonnes by the equation E = m.c^2 it gets converted into an unimaginable amount of energy. Would so much of energy not have gone into the evolution of the Universe? I am at a great suspense whether the Big Bang theory or any other theory at all answers from where so much of energy would have come that would have got converted into so many trillions and trillions of tonnes of the mass of all the galaxies put together. How we can believe in the big-bang-theory unless we know from where so much of energy would have arrived?

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

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

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

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


Practicing image processing and analysis I'm just getting into image processing and analysis using Python, and I'm looking for astronomical images (preferably raw, unprocessed images) to practice on. Can anyone suggest some places I can download public-access telescope images from?

I know you've already accepted an answer, but I feel that it doesn't provide a true representation of the type of image processing astronomers actually do. Actual image processing in astronomy is rarely done with JPG, PNG, or TIFF files. Instead, astronomers mostly work with FITS files and cameras/astronomical instruments actually produce FITS files when taking images. The advantage here is that FITS files can record much more information than just image information such as where/when the image was taken, multiple layers of the image, encoded WSC information, etc.

As part the mission objective of various organizations such as the National Science Foundation and NASA, all raw astronomical data taken in the United States by observatories funded by the United States are free and open to the public (after a limited time period in which the observing scientist gets exclusive rights to the data for their science objectives - usually several months to a year). This includes both ground (e.g. Keck, SDSS, Arecibo) and space observatories (e.g., Hubble or Spitzer). This is also true of many observatories funded by other countries as well.

A great place to start would be to head over to the Hubble Legacy Archive. Enter the site and try searching for an object or coordinates. For example, you can search for M101 (otherwise known as the Pinwheel Galaxy). You'll see a list of images come up and next to them, you can download the FITS file generated by Hubble when it took the image.

Another great resource would be the Sloan Digital Sky Survey. This is a systematic astronomical observation of as much of the sky as possible using a variety of instruments. You can practice looking at images and spectra. I should warn you that it will take a bit more effort to figure out where the data is and how to interpret it, but feel free to ask questions here when you run into trouble.

In general though, if you're looking for a particular type of image or data, find an observatory that records it and look for their publicly available data sets.


In practice, can very high magnetic fields significantly alter the dynamics of an astronomical-scale bodies? The direct effect of magnetic fields on the dynamics (as in the movements, trajectories etc of the bodies themselves, rather than lower-level phenomena like stellar winds and atmospheric degradation) of astronomical bodies in places like the solar system are so weak that they can be ignored without real consequence. But it seems intuitive that in the vicinity of bodies with extremely high magnetic fields - such as magnetars - that such fields would have a significant or even dominant effect dynamically, requiring electromagnetism to be considered alongside gravity in order to supply an accurate dynamical description of the system.

Considering the fact that the magnetic fields of such bodies range upward of 100 billion Telsa, the fact that the coupling constant for electromagnetism is some 40 orders of magnitude greater than that of gravity, and the fact that magnetic force drops off with the square of distance just as gravity does, it seems odd that such magnetic fields wouldn't have such an effect (at least on other bodies with magnetic fields, as well as bodies with a high content of ferromagnetic materials), but I've never heard any mention of such a thing. So what I'm wondering is if such behavior is expected in the vicinity of highly magnetized astronomical bodies, and if not, what am I overlooking here?

Where such an effect becomes important is during the accretion of matter from a regular (more-or-less) main sequence donor star onto a magnetized neutron star.
Here, the magnetic field at the NS can reach about 10$^{12}$ G and the rate of accretion about 10$^{13}$kg/s. On its way down the gravitation well towards the NS the material, by now mostly a plasma, will reach a point where the Lorentz-force will force it to spiral around and follow the magnetic field lines. So from that point onward its dynamics will be determined by the structure of the magnetic field, and the material will ultimately fall onto the magnetic poles, creating pulsars.
This forcing onto the magnetic field lines happens roughly when the magnetic pressure exceeds the ram pressure of the material. So,
$$
P_{\mathrm{ram}}=P_{\mathrm{B}} \\
\rho(r) \cdot(\dot{\vec{r}})^{2} = \frac{B(r)^{2}}{2 \mu_{0}}
$$
Here B(r) is a dipole field around the NS which drops off as r$^{-3}$ and so P$_\mathrm{B}$ drops off rapidly as r$^{-6}$, as Rob Jeffries  already mention in his comment.
The radius around at which the magnetic field becomes dominant is called the Alfven radius and is in the order of a few 1000 km.

This then looks something like this (Filippus et al., 2008):

I am not sure, however, if that counts as astronomical-scale, but this is the most dramatic effect of magnetic fields I could think of.


How is the Dating of Samples of Cosmic Objects Performed, Including Pre-Solar Dust? In the new Space.com article 7 billion-year-old stardust is oldest material found on Earth I had read about some pre-solar grains dating back to 7 billion years. What techniques are used in calculating age of these grains?
Are they collected first and then radiocarbon-dating is applied? If not, what other techniques might have been used?

Scientists recently identified the oldest material on Earth: stardust that's 7 billion years old, tucked away in a massive, rocky meteorite that struck our planet half a century ago.
This ancient interstellar dust, made of presolar grains (dust grains that predate our sun), was belched into the universe by dying stars during the final stages of their lives. Some of that dust eventually hitched a ride to Earth on an asteroid that produced the Murchison meteorite, a massive, 220-lb. (100 kilograms) rock that fell on Sept. 28, 1969, near Murchison, Victoria, in Australia.
New analysis of dozens of presolar grains from the Murchison meteorite revealed a range of ages, from about 4 million years older than our sun — which formed 4.6 billion years ago — up to 3 billion years older than our sun, researchers reported in a new study.

The BBC article pointed out by @uhoh in a comment answers the question nicely.

  To work out how old the grains were, the researchers measured how long
  they had been exposed to cosmic rays in space. These rays are
  high-energy particles that travel through our galaxy and penetrate
  solid matter.

  Some of these rays interact with the matter they encounter and
  form new elements. The longer they are exposed, the more of these
  elements form. The researchers used a particular form (isotope) of the
  element neon - Ne-21 - to date the grains.

  "I compare this with putting out a bucket in a rainstorm. Assuming the
  rainfall is constant, the amount of water that accumulates in the
  bucket tells you how long it was exposed," said Dr Heck.

  Measuring how many of the new elements are present tells scientists
  how long the grain was exposed to cosmic rays. This in turn informs
  them how old it is.

  Some of the pre-solar grains turned out to be the oldest ever
  discovered.

  Based on how many cosmic rays had interacted with the grains, most had
  to be 4.6-4.9 billion years old. For comparison, the Sun is 4.6
  billion years old and the Earth is 4.5 billion.

  However, the oldest yielded a date of around 7.5 billion years old.


Central temperature of a stellar model with combination of gas and degeneracy If we have a stellar density model $\rho(r) = \rho_c(1-r/R)$, where the star is composed of ions behaving as perfect gas and electron with non-relativistic degeneracy. The central pressure is due to both gas and degeneracy pressure. My final goal is to find central temperature of this model.

Now for the linear density model, I calculated the central pressure only for gas to be $P_c = \frac{5GM^2}{4 \pi R^4}$. And central temperature for only gas is $T_c = \frac{5G \mu M_{H} M}{12KR}$

I know the Pressure for non-relativistic degeneracy $P_{nrd} = C \rho^{5/3}$. I was wondering to find central pressure for degeneracy, may I use the Lane-Emden equations for polytrops and simply find central pressure and temperature from there? Then add the two temperatures to find the maximum central temperature for a gas+degenerate star.

For Lane-Emden model, they also consider density dependence on radial distance, but they express the density function as $\rho = \rho_c \theta^n$. Is it equivalent to $\rho(r) = \rho_c(1-r/R)$ for gas model?

If you have a gas of non-degenerate ions and degenerate electrons, then by definition you have an upper limit to the temperature, since if it were higher the electrons would not be degenerate.

Below this temperature, the electron degeneracy pressure will always greatly exceed any perfect gas pressure from the ions. Since degeneracy pressure does not depend on temperature, then the central temperature could be anything lower than the temperature below which the electrons can be assumed degenerate. For example, the mass, radius and central density of white dwarfs of a given mass will be very similar, although their central temperatures will change by more than an order of magnitude as they cool over billions of years.

If you wish to make the small corrections, either for partial degeneracy (the pressure is then slightly temperature dependent) or to include the ion pressure, then the only way to proceed is add the pressures and then solve the equations of stellar structure numerically.


How does one find the electric conductivy of a planet's atmosphere? I am working on an undergraduate project in which we study shaped structures in the atmosphere of the gas giants. We are investigating whether the magnetic field is the source of the instability generating the observed shaped structures. We therefore use spectral methods to make a stability study on the magnetohydrodynamics equations which in our case are

$$\frac{\partial{}\boldsymbol{B}}{\partial{}t}+\left(\boldsymbol{v}\cdot\boldsymbol\nabla\right)\boldsymbol{B}=\frac{1}{\mathrm{Re}_\eta}\boldsymbol\nabla^2\boldsymbol{B}+\left(\boldsymbol{B}\cdot\boldsymbol\nabla\right)\boldsymbol{v}$$

$$\frac{\partial{}\boldsymbol{v}}{\partial{}t}+\left(\boldsymbol{v}\cdot\boldsymbol\nabla\right)\boldsymbol{v}=\frac{1}{\mathrm{Re}}\boldsymbol\nabla^2\boldsymbol{v}+\boldsymbol{f-\nabla}p+\mathrm{N}\left(\boldsymbol{j\times{}B}\right)$$

where $\rm{}Re_\eta$ is the magnetic Reynolds number and $\rm{}N$ is the Stuart number expressed

$$ \mathrm{N} = \frac{\sigma B^2L_c}{\rho{}U}$$

where

$B$ is the magnetic field intensity;
$L_c$ is the characteristic length of the studied structure;
$\sigma$ is the flowing fluid electric conductivity;
$\rho$ is the fluid density;
and $U$ is the characteristic velocity scale of the flow.

Our model is now functional and the next step is to run the calculation with real values for physical parameters. We have all the data except the electric conductivity. Hence, I am  looking for this data (or a method to calculate an approximation). I thought that a first approximation could be made from the volume fraction of hydrogen and helium and their respective conductivity.

All ideas, avenues of research and scientific publication references are welcome. Thank you for help.

PS: I don't know much (yet) about astrophysics and the planet's atmosphere.

The conductivity of neutral gas is negligible, you need to estimate the free-electron conductivity in your gas.
For this, one could first consult a phase diagram, like referenced in this answer. There you'd see that for the atmospheric pressures and temperature for all gas giants, you will stay in the $H_2$-gas regime, so no exotic equation of state is necessary to describe the state of the visible atmosphere of the gas giants.

The updated models with and without Juno-data show that adjustments to structure models in the interior of the planet need to be made but the atmosphere, which you are interested in, is still an ideal gas.

To compute the amount of free electrons in the atmosphere, you need to compute the possible ions, and their ionization fraction. This gets really nasty for cold atmospheres, because molecules form. Mole fractions of those have to be computed by a chemical model, then ionized (possibly by a Saha-type ionization model) and translated into a conductivity (possibly by a Lorentz-Drude-type of model).

For your purposes that could get really involved, but I guess what you could do that is much simpler, is look at some published literature, e.g. here, just take and cite their conductivity profile.
However even in this source, the two models they use for the conductivity differ by 9 orders of magnitude in the neutral atmosphere, so you want to read carefully about their assumptions.


Do stars made by universe getting smaller or larger in mass? Do stars made by universe getting smaller or larger in mass compared to stars made 180 million years after the Big Bang? If they are getting smaller, is this due to the fact that the universe is expanding and the interstellar mass that made stars is getting less denser?

The size of stars depends on the amount of non-hydrogen elements ("metals") affecting the opacity of the gas that forms them. Metals make it easier to radiate away heat when a gas cloud is collapsing, making it easier for a large amount of gas to accrete (although it may mainly act by increasing the minimum mass of stars that form). There is a fair disagreement about whether the first stars ("population III") were very massive (hundreds of solar masses) or actually small, but I think the rough consensus is that very massive stars were likely. Once they emerged and died, they added metals to the interstellar medium and star formation eventually became more like the present.
The expansion of the universe does not have much effect once gas become trapped in dark matter halos and start to form galaxies.
Once the star has formed opacity and mass determines its structure and luminosity. It turns out that for increasing metallicity the maximum stable mass of a star decreases (section B.2). For the eventual metallicity at the end of the stelliferous era Adams and Laughlin estimate a maximum mass star as only 30 solar masses.


How do they know how far out Farout is? (2018 VG18) The Carnegie Science article Discovered: The Most-Distant Solar System Object Ever Observed begins:

Washington, DC— A team of astronomers has discovered the most-distant body ever observed in our Solar System.  It is the first known Solar System object that has been detected at a distance that is more than 100 times farther than Earth is from the Sun.
The new object was announced on Monday, December 17, 2018, by the International Astronomical Union’s Minor Planet Center and has been given the provisional designation 2018 VG18. The discovery was made by Carnegie’s Scott S. Sheppard, the University of Hawaii’s David Tholen, and Northern Arizona University’s Chad Trujillo.
2018 VG18, nicknamed “Farout” by the discovery team for its extremely distant location, is at about 120 astronomical units (AU), where 1 AU is defined as the distance between the Earth and the Sun.  The second-most-distant observed Solar System object is Eris, at about 96 AU.  Pluto is currently at about 34 AU, making 2018 VG18 more than three-and-a-half times more distant than the Solar System’s most-famous dwarf planet.
2018 VG18 was discovered as part of the team’s continuing search for extremely distant Solar System objects, including the suspected Planet X, which is sometimes also called Planet 9.  In October, the same group of researchers announced the discovery of another distant Solar System object, called 2015 TG387 and nicknamed “The Goblin,” because it was first seen near Halloween.  The Goblin was discovered at about 80 AU and has an orbit that is consistent with it being influenced by an unseen Super-Earth-sized Planet X on the Solar System’s very distant fringes.
The existence of a ninth major planet at the fringes of the Solar System was first proposed by this same research team in 2014 when they discovered 2012 VP113, nicknamed Biden, which is currently near 84 AU.

That's great! But the next bit confuses me.

The discovery images of 2018 VG18 were taken at the Japanese Subaru 8-meter telescope located atop Mauna Kea in Hawaii on November 10, 2018.
Once 2018 VG18 was found, it needed to be re-observed to confirm its very distant nature. (It takes multiple nights of observing to accurately determine an object’s distance.)  2018 VG18 was seen for the second time in early December at the Magellan telescope at Carnegie’s Las Campanas Observatory in Chile.  These recovery observations were performed by the team with the addition of graduate student Will Oldroyd of Northern Arizona University.  Over the next week, they monitored 2018 VG18 with the Magellan telescope to secure its path across the sky and obtain its basic physical properties such as brightness and color.
The Magellan observations confirmed that 2018 VG18 is around 120 AU, making it the first Solar System object observed beyond 100 AU.  Its brightness suggests that it is about 500 km in diameter, likely making it spherical in shape and a dwarf planet. It has a pinkish hue, a color generally associated with ice-rich objects.

Question(s):

How was the value of 120 AU actually determined?
or at least, how could it be determined if the explanation isn't readily available?
Is this measurement more of an orbit reconstruction, or a parallax measurement of current distance?

below: "Discovery images of 2018 VG18, nicknamed 'Farout,' from the Subaru Telescope on November 10, 2018. Farout moves between the two discovery images while the background stars and galaxies do not move over the one hour between images. Image is courtesy of Scott S. Sheppard and David Tholen." Source

It is a process of orbit determination.

The apparent position of any solar system body will change from night to night as a result of the combination of its actual motion around the sun and the motion of the Earth. A very distant body such as "Farout" will be moving very slowly.

To determine an orbit, a minimum of three observations is needed. From three observations it should be possible to fit a Keplerian elliptical orbit to the observations and from the orbit one can tell the object's position at any time. The method takes into account the effect of parallax as the position of the Earth is used in the determination of the orbit.

However there will be observational errors; the exact location of the body can't be seen, only a white blob on a photograph. This means that there can be errors in the orbit determined from the minimum of three observations. To counter this, further observations over a longer period of time can be taken. The process of orbit determination then becomes a process of finding the best fit orbit to the observations. The more spread out (in time) the observations are, the better is the orbit determined.


Does a magnet outside a black hole affect one inside it? If I had 2 really strong magnets, and I threw one into a black hole, while holding one outisde it at a safe distance, what would happen? Would the magnets affect each other in any way? Would one push/pull the other?

The tidal forces emerging from the black hole would at first break the magnet flung towards it into two pieces. Those two pieces would themselves undergo deformation and splitting, as the process of spaghettification will continue beyond the molecular, atomic and even sub-atomic levels, reducing everything that crosses the Event Horizon to a Singularity.

Note:

You won't actually see the magnet thrown towards the black hole crossing the event horizon, but that's a different matter altogether - please see this if it might interest you.

Now, magnetic attraction is essentially just an Electromagnetic Force, the range (not to be misunderstood as strength) of which is infinite - see reference.

In the context of the scenario stated in the question, the fate of attraction between a pair of magnets, one of which is thrown into a black hole, could be visualised by the following analogy: imagine holding a magnet in your hand and the other attached to a train which is accelerating downhill into a gorge. The two possible results are:

If the attraction is strong enough to pull the magnets together before the train falls into the gorge, then they both crash into the abyss together (assuming that the magnet attached to the train would easily be able to pull the one that's being held by hand). In case of a black hole, this would mean that both the objects (magnets) would be crushed to singularity. However, we might safely say that they both were "attached" before crossing the Event Horizon.
Or, the acceleration of the train reduces the attractive force infinitesimally, with the magnet in hand barely having any effect on the other magnet that crashed (and vice-versa). In a classical sense, this means that the force of attraction would still exist, even if it's reduced infinitesimally, to almost zero. However, with regards to a black hole, Event Horizon ensures that anything inside the horizon can no longer communicate with or influence anything outside the horizon (and vice-versa). This means, if the magnet somehow crosses the event horizon as a single piece, before losing its existence to singularity, its magnetic attraction would still never be able to cross the boundary defined by the horizon, even though the range of influence of the electromagnetic field is infinity. Ultimately, this would only last for a very brief moment and the magnet would be unable to prevent being pulverised to singularity, either by itself or by any attempts from any other entities outside of the Event Horizon.

Hope this helps.


Air molecules per cubic foot at a certain altitude above the surface The Earth's and other celestial bodies' atmospheres become thinner the higher you go. Imagine you were able to see atoms/molecules. At what altitude above the Earth's and Mars' surface would you see few enough molecules per cubic foot so you'd be able to count them? If you were, say, 500 miles (805 km) above the Earth's surface, how many air molecules per cubic foot (30 cm³) would you see?
Or, one can ask vice versa: at what altitude above the Earth would you see, say, a 100 air molecules per ft³? Which gas doesn't matter, I'm asking for all atmosphere molecules within the certain volume. And since even interplanetary space isn't perfect vacuum, what would it look like there, e.g. halfway between the Earth and Mars orbit?

The kinds of molecules / atoms and the density of those molecules / atoms as a function of altitude in the Earth's upper atmosphere depends very much on the time of day and the activity of the Sun.
Time of day: The Earth's upper atmosphere has a marked diurnal bulge. The density variation at a constant altitude can vary by an order of magnitude (a factor of ten) just due to the difference between night and day.
Solar activity: One poof from the Sun can make the Earth's upper atmosphere swell by not just one order of magnitude as is seen with the diurnal bulge but multiple orders of magnitude. Satellites in low Earth orbit need to use a good deal more propellant when the Sun's activity is high as compared to now when the Sun is rather quiescent.

To answer your question, there is no one answer to your question. This is part of the reason (perhaps the key reason) why predicting when a satellite in a degrading orbit will re-enter is imprecise. There are multiple models of the Earth's upper atmosphere that vary from fairly simple to ridiculously complex.
On the simple side, I prefer to use the US Naval Research Laboratory's mass spectrometer and incoherent scatter atmosphere model, or NRLMSISE-00 for short. One needs to provide latitude, longitude, date, time of day, and three parameters that indicate recent and current solar activity. And that's on the simple side.


Fate of Jupiter when our sun dies Five billion years from now, the sun will have grown into a red giant star, more than a hundred times larger than its current size.
  While this metamorphosis into the giant star will change the solar system, scientists are unsure what will happen to the third rock from the sun.We already know that our sun will be bigger and brighter, so that it will probably destroy any form of life on our planet.
  But whether the Earth's rocky core will survive is uncertain.
  At the end of its evolution, seven billion years from now, the sun will become a tiny white dwarf star.

(by Leen Decin, professor at the KU Leuven Institute of Astronomy (source))

When our sun runs through its last phase of its life it will expand and tremendous solar flares will flow out of it. Earth will end up getting baked and will lose its atmosphere.

Jupiter is a gas giant planet, a GAS GIANT. So,

What will happen in Jupiter? Will it evolve into a star since it contains hydrogen and helium? How will the atmosphere deal with the extreme heat?

Jupiter won't evolve into a star, it is not big enough. A body would have to have about 80 times the mass of Jupiter for there to be significant fusion occurring in the core. The end of life of the Sun won't change the mass of Jupiter.

Jupiter will continue to orbit the Sun as it evolves into a red giant. Although the solar wind will be much much more powerful, it won't have a significant effect on the overall mass of Jupiter.

When the Sun loses its outer layers it will, in the last million years or so of its life, lose about half its mass. This will have a significant effect on the orbit of Jupiter, causing Jupiter to migrate outwards.  While it is possible that some planet's orbits may become unstable and get ejected from the solar system, it is more likely that Jupiter will settle into a new, wider orbit around the white dwarf.


How is the mass of solar system objects calculated? When I go to the wikipedia page for any planet in our solar system, the exact mass and volume are reported. How is it possible to do this with them being so far away? I suppose after hundreds of years we can observe them going around the sun, but this doesn't really give any information since we need both the mass and distance from the sun for it to be of any use.

Planets are satellites of the Sun. Most planets have satellites.
Whenever you find a parent body with satellites, you get first good results about their masses and distances by applying Kepler's laws.
A second approach is using parallax measurements for distance estimates.
The distance to some of the planets can be measured very accurately by radar, e.g. for Venus.
The known planets of our solar system have been visited by probes. Using signal travel times, you can calculate the distance of the probe to Earth. One of the methods to determine the distance between probe and the planet is by photogrammetry.
Apply the measured data to the according system of equations. Usually more data than needed are available. Then you get an overdetermined system. A best solution can be found by minimizing appropriate squares.

Subtle deviations from Kepler's law can be used to infer masses of planets which are hard to measure in a more direct way.
Once you know the distance of a planet, use its apparent diameter(s) to calculate the actual diameter(s), and then the volume of a sphere or a spheroid.
With mass and volume you know the mean density of the planet.

Within our solar system, for very distant objects the apparent size of which cannot be determined you may estimate the surface albedo together with distance (by Kepler after knowing the mass of the Sun) and apparent magnitude to obtain a rough estimate of the size of the object.

Better size estimates for objects of small apparent size can be obtained by stellar occultations, meaning measuring the time a star is occulted by the object. Together with the angular velocity of the object you get data about its angular size.


Present distances between planet. How can I find them? It's "commonly known" how distant are our solar system planets from Sun. But we can't easily say that about planets, which distances can differ greatly, without some observations (or simulations, knowing their state in some moment in time).

How can I can check 'actual' relative distances or positions of planets?


How can I find the width of an orbit knowing some values? I need to find the width/x-radius of an orbit, which is ellipse shaped, knowing variables like the orbit's length(perihelion + aphelion in this case) and the eccentricity. I am not sure if this is even possible, I am pretty new to this. If there is already available data on this that you're aware of, it would solve my problem too.

To find the width (semi-major axis, see Wikipedia) of an orbit, you only need two of the following: perihelion, aphelion, and eccentricity, represented by the variables $P, A, $ and $e$, respectively, and $a$ is the semi-major axis.
If you are given perihelion and aphelion, you can just take the average of the two: $a=\dfrac{A+P}{2}$
If you are given perihelion and eccentricity, you can calculate $a=\dfrac{P}{1-e}$. Similarly, given aphelion, the width of your orbit is $a=\dfrac{A}{1+e}$.
If you are talking about the other kind of width (Semi-minor axis), represented by $b$, then the formulas are:

Given $P \text{ and } A: b = \sqrt{A \cdot P}$
Given $P \text{ and } e: b = \sqrt{-\dfrac{P(1+e)}{e-1} \cdot P}$
Given $A \text{ and } e: b = \sqrt{-\dfrac{A(1-e)}{e+1} \cdot A}$

I hope this helps.


How can a planet not rotate at all and yet have 1 year that is day and 1 year that is night? A mercury day if you consider its solar day is 2 mercury years.

Now if you consider just 1 half of mercury, in 1 of those years it is facing towards the sun and in another one of those years it is facing away from the sun.

Mercury is tidally locked. This tidal locking happens when the rotation rate slows down until it reaches 0.

How is it possible for a planet to not rotate and yet have 1 half of it facing towards the sun each year with that half changing every year? It would seem like it would have to rotate 180 degrees in a fraction of a second which I know is not possible.

Your premise is incorrect. We used to think Mercury was tidally locked, but since 1965 we now know it is in a 3:2 spin-orbit resonance, which gives it (long) days.

Anything you read that says it is tidally locked is old, or itself using old reference material.

Here are some quick sketches to show the orbit, relative orientation of the planet, and the sidereal and solar days. If you like it as an answer I'll redraw with a graphics tool instead of my iPad Pen app...


What are the lines in this Solar Cycle graph? What do the lines in this NASA graphic represent?

Why does Solar cycle 22 have a peak of nearly 300 (assuming that peak represents the maximum observations at the same time) and this link says the peak is 212)?

In the above plot by Hathaway,
the jagged line peaking at 284.5 in June 1989
is the monthly international sunspot number.
The smooth curve starting and ending a few years later is a
prediction
based on recent observations and statistics of past cycles.

The peak of 212.5 in November 1989 is for a
13-month smoothing
of the monthly number.
The Sunspot Index and Long-term Solar Observations (SILSO)
project at the Royal Observatory of Belgium
archives
these data and provides
this plot
of the monthly and smoothed sunspot numbers together.


Is there enough hydrogen left after a star dies so another star will have enough to light up? A star consumes quite a lot of hydrogen in its life, and is pretty much "vacuuming" everything in its vicinity. After it dies (eventually by supernova which will spread all its composition over light years), is there enough hydrogen left in that area to light up a new star? And will that star be more short-lived compare to its predecessor?

There are several misconceptions in your question.

First, a star does not vacuum everything in its vicinity. Rather it forms from a condensation in a gas cloud, which in turn collapses to a proto-star surrounded by a gas disc, which can contribute further material. Once formed in this way, a star typically does not acquire more gas (exceptions are symbiotic binary stars etc).

Second, a star with mass in excess of $\sim8$M$_\odot$ will (typically after a long time) suffer from a supernova, when most of its envelope is flung back into space. That gas is still mostly hydrogen, though enriched by 'metals' (non-primordial elements). However, the gas is hot and fast moving and hence not in a condition to form another star.

Third, the gas from the supernova will eventually mix with other gas and dissolve into the general pool of interstellar medium (ISM). Some of that may cool to form a molecular cloud (as gas cloud where molecular $H_2$ dominates), which in turn may become the site of new star formation.

We know that the Sun has been formed from enriched material, which is a mixture of primordial gas with the ejections of several supernovae.


Detecting anomalies from the sun from home I was wondering if there are any ways for some amateur to detect let's say solar flares or CMEs from home. What kind of hardware would you need in order to have chance of seeing/measuring those?

I have a background in programming software and I've gotten my feet wet in hardware if that's of any help.

I'm aware that the weather plays a "bit" of a factor and I am sure that I can't afford a satellite but anything is welcome!

The most extreme solar flares are visible as bright marks on the face of the sun and the most extreme CME are apparent as aurorae even at low latitudes.

Detecting CME as they leave the sun is difficult: you really need a spaceship to see that close to the sun.

Detecting CME as they interact with the Earth's magnetic field is possible.  You need to make a magnetometer.  There are several designs possible, from a "jamjar/laserpointer" design to fluxgate designs, and commercial detectors at various price points.

A magnetometer will respond to CME as they interact with the Earth's magnetic field, so they are not an early warning system, and you may find yourself detecting next-door's ironing, but at least the more advanced designs here demonstrate it is possible to get a signal that correlates with observations of the aurora.


Why is only a narrow chunk of sky darkened in this amazing Curiosity solar eclipse GIF from Mars? I found three NASA JPL images in the sequence PIA23133, PIA23134, and PIA23135. They were taken by the Curiosity Rover on Mars and show solar eclipses by Mars' two moons.

Details of these images are given in this answer to Puzzler: help understanding these amazing Curiosity eclipse GIFs

Question: Why is it that it's mostly a chunk of the right half of the sky that darkens during the eclipse in PIA23135?

Is this kind of effect visible in solar eclipses seen on Earth?

Enquiring minds want to know!

fyi in case you are wondering, per @MarkAdler's ultra-concise answer

  They use neutral-density filters to look at the Sun, which reduce the light by a factor of 100,000. The two Pancam cameras each have one neutral-density filter, with the left one filtering blue and the right one filtering red.

From https://photojournal.jpl.nasa.gov/archive/PIA23133.gif and

From https://photojournal.jpl.nasa.gov/archive/PIA23134.gif

From https://photojournal.jpl.nasa.gov/archive/PIA23135.gif

(a fairly slow) GIF:

Phobos' shadow on Mars is very narrow -- only about 15 miles across -- and it didn't quite pass across the Sun as viewed from the lander on that particular orbit.  This image (taken by Mars Global Surveyor) gives you an excellent picture of just how small it is:
 Credit: NASA/JPL/Malin Space Science Systems Source

The penumbral shadow of Phobos is visible on the landscape of Mars, as photographed by the Mars Global Surveyor. The center of the shadow was at approximately 10.9°N 49.2°W (Map of area Map zoom) at 04:00:33.3 UTC Earth time.
The image shows western Xanthe Terra on August 26 1999 at about 2:41 p.m. local solar time on Mars. The image covers an area about 250 kilometers across and is illuminated from the left. The dark spots on the crater floors are probably dark sand dunes. None of the craters pictured currently have names. Nanedi Valles is the meandering valley at the bottom right. The picture covers about 7°–15°N vertically and 52°–48°W horizontally. The vertical orientation of the image is 3.01° to the west of north; a north-pointing arrow superimposed on the image would point slightly to the right.

It only can darken a portion of the sky.


What is the difference between $E_\gamma$ and $E_{\gamma,iso}$ in gamma ray bursts? In the literature on gamma ray bursts I see several references to $E_\gamma$ and $E_{\gamma,iso}$ for example as in Piran (2005)

$E_{k,iso,52}$ is “isotropic equivalent” kinetic energy,
discussed below, in units of 1052 ergs, while $E_{k,θ,52}$ is the real kinetic energy in the jet i.e:
$E_{k,θ,52} = (θ^2/2)$$E_{k,iso,52}$. One has to be careful which of the two energies one discusses. In
the following I will usually consider, unless specifically mentioned differently, $E_{k,iso,52}$, which
is also related to the energy per unit solid angle as: $E_{k,iso,52}/4π$

My interpretation is that when an observation of a GRB is made, an energy $E$ is recorded. $E_{\gamma,iso}$ is the total energy that would have been emitted by the explosion if the energy was released equally in all directions and by considering our distance away from it. However, because we know that GRBs are tightly beamed, $E_\gamma$ is the corrected energy by taking $E_{\gamma,iso}$ and considering only a small fraction of it in a tight beam. However, this all seems a tad confusing and I can't see the point in talking about $E_{\gamma,iso}$ when we think they are very tightly beamed.
I would appreciate some clarification on what exactly $E_\gamma$ and $E_{\gamma, iso}$ are and where they come from.
Thanks

I will answer with what I now believe to be the correct answer. We see some gamma-ray burst emission hit our satellite which carries an amount of energy $E_{obs}$. If we say the burst was emitted equally in all directions (isotropic) then we derive a quantity $E_{iso}$. If we say that the burst was actually emitted in a tight beam then we derive $E_\theta$ where $\theta$ is the opening angle of the burst and we will find $E_{iso} > E_{\theta}$. $E_{iso}$ is used to compare burst energies in the literature without needing to find the opening angle.


Jacobi vs tidal radius for star cluster The tidal radius is defined in King (1962) as:

  the value of r (the radius) at which f (the density profile) reaches zero...

This also referred by King as the "limiting radius" written as (Eq 3):

$$r_t=R_{GC}\left(\frac{M}{2M_g}\right)^{1/3}$$

where $R_{GC}$ is the galactocentric distance, $M$ is the cluster's mass, and $M_{g}$ is the galaxy's mass.

This can also be written as Pinfield et al. (1998) Eq (12):

$$r_t = \left( \frac{GM}{2(A-B)^2} \right)^{1/3}$$

where $A$ and $B$ are the Oort constants and $M$ is the total mass.

Now, I've seen the "tidal radius" defined as:

$$r_t=R_{GC}\left(\frac{M}{3M_g}\right)^{1/3}$$

in Chernoff & Weinberg (1990) Eq 2. Notice the 3 in the denominator instead of the 2 as above. According to the authors:

  The tidal radius depends weakly upon the form of the mass distribution of the halo. The Eq. above is correct for the tidal stress of a Keplerian force field. On the other hand, for constant $v_g$, the factor $3M_g$ becomes $2M_g$.

This same quantity (using $3M_g$) is referred to as the "Jacobi radius" in Piatti (2015) Eq 4.

I've also seen the Jacobi radius written as:

$$r_J=\left(\frac{GM}{4\Omega^2-k^2}\right)^{1/3}$$

(where $G, M, \Omega$, and $k$ are the gravitational constant, the cluster
mass, the circular, and the epicyclic frequencies of a near circular
orbit, respectively) in Ernst et al. (2011) Eq (8).

According to Wikipedia, both the tidal and the Jacobi radius are the same thing.

My question: are these two radius the same quantity?

In the book Galactic Dynamics by Binney and Tremaine (second edition) there is a whole section explaining the difference between the Jacobi radius and the tidal radius (page 677-chapter 8).

Here, $r_J$ is defined as:

  $$r_J=  R_0\left(\frac{m}{3M}\right)^{1/3}$$
  The Jacobi radius $r_J$ (also, Roche or Hill radius) of an orbiting stellar system is expected to correspond to the observational tidal radius, the maximum extent of the satellite system. However this correspondence is only approximate.

Five reasons are subsequently given to explain why the correspondence is only approximate.


Why does moon's highest altitude change during the year I came across a question in which moon's highest altitude had to be calculated as seen from a certain latitude, also the month and the hour and the constellation in which moon would be in had to be calculated......

I could not understand why or how would moons path change during the year

The moon's orbit around the Earth is inclined at about 5° to the ecliptic
So as the Earth progresses in its own orbit around the sun, the moon's altitude also changes

During summers the part of the orbit on the other side of the Earth(night side) is below the ecliptic. While during winters it is the upper part which is on the night side

Source


Could we detect if Oumuamua asteroid wasn't an asteroid? If Oumuamua wasn't an asteroid, could we detect that it is not an asteroid?

Disclaimer: I do NOT suggest that Oumuamua is in fact an alien spacecraft, a giant coffin, or a statue of Weird Al.

Since it apparently has a very unusual shape, if this is something else, with a different material (neither rock nor ice), could we detect that something is unusual?

I think its a good question and has probably popped up in the back of many people's minds for milliseconds, both astronomers and lay people, though most of them would still be hesitant to admit it on an internet forum. And of course, our understanding of what is possible or even "normal" in the universe is vanishingly small.
edit 2:

Scientific American: Alien Probe or Galactic Driftwood? SETI Tunes In to 'Oumuamua
The Atlantic: Astronomers to Check Mysterious Interstellar Object for Signs of Technology
Breakthrough Initiative: Breakthrough Listen to Observe Interstellar Object 'Oumuamua'

From the NPR news item Astronomers Want To Know: Does This Interstellar Visitor Have A Message For Us?:

"Researchers working on long-distance space transportation have previously suggested that a cigar or needle shape is the most likely architecture for an interstellar spacecraft, since this would minimize friction and damage from interstellar gas and dust," the group said in its announcement.
"While a natural origin is more likely, there is currently no consensus on what that origin might have been, and Breakthrough Listen is well positioned to explore the possibility that 'Oumuamua could be an artifact."
"The possibility that this object is, in fact, an artificial object — that it is a spaceship, essentially — is a remote possibility," Andrew Siemion, a member of the initiative and director of Berkeley's Search for Extraterrestrial Intelligence Research Center, told The Washington Post on Monday.
"We don't want to be sensational in any way, and we are very realistic about the chances this is artificial," Yuri Milner, the Russian billionaire financing the project, told the Scientific American, "but because this is a unique situation we think mankind can afford 10 hours of observing time using the best equipment on the planet to check a low-probability hypothesis."
And they'll be checking on that hypothesis by scanning the object for possible artificial transmitters through a radio telescope at West Virginia's Green Bank Observatory.

edit 1: A few days after you've asked the question, a similar thought was uttered by the BBC correspondent Roland Pease in the Science in Action episode Interstellar Asteroid. Talking with astronomer Alan Fitzsimmons of Queens University, Belfast (who's image of X was featured in a NASA "Astronomy Picture of the Day"), Pease says:

You say ‘cigar-shaped”, that’s to me “rocket-shaped”. Is there any way you would know that this is an interstellar rocket?

So it looks like you're in excellent company!

The asteroid was discovered and has been observed mostly (if not completely) by it's reflection of radiation from the Sun. The Sun is roughly a blackbody emitter and most of the radiation is in the visible. Because of the dimness of the light even at maximum, there isn't much signal once dispersed spectroscopically.
There is a spectrum published in ArXiv Palomar Optical Spectrum of Hyperbolic Near-Earth Object A/2017 U1 on 2-Nov-2017 by Joseph R. Masiero though! Despite the low light level (V ~ +21) and expanded size due to poor seeing at low elevation resulting in a signal-to-noise ratio of order unity, the 5 meter Hale telescope (200 inch Mt. Palomar) was used to try to obtain some spectroscopic information. Ten 300-second runs (total of 50 minutes of collection) were used to produce the heavily binned spectrum below. I think those are 1-sigma error bars, so the spectrum doesn't say much more than "slightly red-ish".

See also Ye et al. 1I/2017 U1 (`Oumuamua) is Hot: Imaging, Spectroscopy and Search of Meteor Activity for further background:

The moderate spectral gradient of ‘Oumuamua indicates that its surface is devoid of ultrared material that is common on outer solar system objects like KBOs. Various classes of KBOs are typically very red in color, with a
spectral gradient $\gtrsim (\text{100 nm})^{-1}$, likely due to the irradiation of organic material resulting from the bombardment of energetic particles (Brunetto et al. 2006). The less reddish color of ‘Oumuamua suggests that the object was either formed close to its original central star, or has lost its ultrared material due to close approach(es) to its original or other stars. It is difficult to say which scenario is more likely due to the chaotic nature of small body dynamics.

According to Sky and Telescope's Update on `Oumuamua, Our First Interstellar Object:

The object had already dimmed to 20th magnitude by the time the Pan-STARRS 1 telescope swept it up, so observers worldwide had to scramble to get major facilities to gather more observations before the one-of-a-kind interloper becomes impossibly faint.
One lucky break came immediately after the MPC's initial announcement. Observer Joe Masiero (Jet Propulsion Laboratory) just happened be in the middle of a run with the 5-m Hale Telescope at Palomar Observatory, and he quickly obtained a spectrum that's reproduced below and explained in more detail here. There's no obvious absorption that would signify the presence of particular compounds or minerals.
But the gentle slope, increasing toward the red (right) end, suggests that `Oumuamua is not a particularly good match to objects in the Kuiper Belt (which tend to be much redder). Instead, it's the kind of spectrum that a rocky surface would exhibit after being "weathered" by long-term exposure to space radiation.

So I'd say that the answer to your question is "No", there's very little besides its orbit and light curve and approximate color that has been or probably could have been detected. With more warning, and serious interest, it's possible an observation campaign could have been organized to get more aperture area and time to make a better spectrum, but since S/N only scales as the square root of those, a lot of other observations would have had to have been rescheduled to make it happen.


What type of telescope can show cliffs on the moon surface from a city location? I am a complete novice in this world and I would like a telescope where I could have enough power to view the moon's surface, I don't expect to see the flag or the hover sending signs to me, but perhaps enough to view cliffs, and other details from the surface of it and perhaps from other places.

I honestly don't even know what kind of telescope to purchase. I live in a capital city, so there is a lot of light around here. My apartment is high with a balcony, which is where I would place it and there is nothing in front of me from there.

What kind of telescope would fit my needs?
Are there reputable online stores that I could consult?

I have a $500 budget for this and honestly, I have no clue about specs or anything, I am a completely novice in this field.

The main things to look for are:

Decent optics (nearly anything except those with plastic lenses).
A steady mount that points where you want to, and moves smoothly. An altitude-azimuth mount is fine because with practice you can guide at high powers.
The eyepieces should be 1 1/4 inch size or more expensive 2 inch. The older 0.96 inch are very hard to find. This lets you upgrade eyepieces later.

The larger the aperture the better. Don't worry about magnification.

Hint: find your local astronomical society, ask there, go to one of their "star nights", they may even have telescopes on loan.


How did they measure the distance to the center of our galaxy to 0.3% accuracy? The abstract of A geometric distance measurement to the Galactic center black hole with 0.3% uncertainty (Open Access) says:

  We present a 0.16% precise and 0.27% accurate determination of R₀, the distance to the Galactic center. Our measurement uses the star S2 on its 16-year orbit around the massive black hole Sgr A* that we followed astrometrically and spectroscopically for 27 years. Since 2017, we added near-infrared interferometry with the VLTI beam combiner GRAVITY, yielding a direct measurement of the separation vector between S2 and Sgr A* with an accuracy as good as 20 μas in the best cases. S2 passed the pericenter of its highly eccentric orbit in May 2018, and we followed the passage with dense sampling throughout the year. Together with our spectroscopy, in the best cases with an error of 7 km s−1, this yields a geometric distance estimate of R₀ = 8178 ± 13stat. ± 22sys. pc. This work updates our previous publication, in which we reported the first detection of the gravitational redshift in the S2 data. The redshift term is now detected with a significance level of 20σ with fredshift = 1.04 ± 0.05.

and figure 2 shown below provides an example of the error analysis post fitting.

If you have the angular size of an orbit and a set of radial velocity measurements, even if it was viewed edge-on and the inclination was known, wouldn't this give only a parametric relationship between R₀ and the mass of the central object?

I don't understand how they were able to state a specific value for R₀, I'm probably missing something obvious but I can't figure it out. According to this answer:

  We have reasonably good measurements of the mass of Sagittarius A*, thanks to measurements of the movements of stars like S0-2 over several decades. It's been well-established that the mass of the central object is $M\approx4\times10^6M_{\odot}$;

so to me the whole situation appears on the surface to be circular referencing. I'm sure it isn't of course, I'm just missing something.
Is the measurement really to the "Galactic center" as the first sentence of the abstract states, or really the distance to Sgr A* around which S2 orbits?
"bonus points" Why is every instance of "Galactic" in this published paper (not a preprint) capitalized?

  Selected posterior densities as obtained from the MCMC sampler with N = 200 000, here for the noise model data set. The contour lines mark the 1, 2, and 3σ levels. We only show the diagrams with the strongest correlations. All parameters are well determined (see Appendix D). Source

The orbit of star S2 is completely determined by the astrometric observations. i.e. One has the orbital period, the angular scale of the orbit and the inclination of the orbital plane. With this information alone, the distance and central mass are degenerate. If you know the distance you get the mass and vice versa.

However, here the mass is not needed. Full knowledge of the orbit means that you can predict the radial (line of sight) velocity of the star at any point in its orbit. The conversion from the angular scale of the observations to an absolute velocity, depends (linearly) on the distance to the source.

Hence they fold a set of radial velocity measurements into their analysis and get an estimate of the distance.

The distance is that to Sgr A*. The paper discusses that this is compatible with other measurements of where the dynamical centre of the Galaxy is.

The Galaxy is a proper noun (like the Sun), referring to our Milky Way, and hence is capitalised. When referring to the Galaxy it is conventional (though not strictly necessary I think) to use the word Galactic to indicate a property of the Galaxy (as opposed to some other galaxy).


Why do neutrinos take time to scatter? After a supernova neutrinos get released by electron capture,and they escape in a flood.

My guess is because they have tiny masses and get momentum from momentum conservation.

  However, the huge number of neutrinos it emits carry away so much energy that the temperature of an isolated neutron star falls within a few years to around $10^6$ kelvin.

Reference : Wikipedia,Original(wikipedia reference)
Wouldn't most of the neutrinos scatter after supernova almost instantaneously before a formation of a neutron star?

You are barking up the wrong tree. The passage you quote is talking about how neutron stars cool after they have formed. Neutrinos are emitted in a burst lasting a few seconds after core collapse but then continue to be produced whilst the neutron star is hot.

Neutrino production is indeed fearsomely high in the first fraction of a second after core collapse. However, the hot, dense material is opaque to neutrinos (mean free paths of 10-100 m) and they diffuse outward in much the same way as photons from the Sun. But the timescale for this is only about 10 seconds. Once the neutron star has cooled to about $10^{10}$ K, the fermions (neutrons, protons, electrons) in the neutron star become degenerate and the neutron star becomes transparent to neutrinos because only the small fraction of fermions within $kT$ of their Fermi surfaces can interact with thermal neutrinos and the neutrino mean free paths increase as $\sim T^{-2}$ and the production rate of neutrinos falls as $T^6$.

Still, the neutron star would then radiate away all of its thermal energy within a further 10 seconds, but the main neutrino-generating reactions - cycles of beta and inverse beta decay (aka the Urca process) become blocked (except perhaps right in the core at the highest densities) by an inability to simultaneously conserve energy and momentum in the degenerate gas. Instead, neutrinos continue to be generated by the modified Urca process at a much slower rate, using "bystander" baryons to conserve momentum.
$$ n + n \rightarrow n + p + e + \bar{\nu}_e$$
$$ n + p + e \rightarrow n + n + \nu_e$$

Though the efficiency of this process is low, so is the thermal energy contained within degenerate gases. Thus rapid cooling still occurs. I am not clear where Wikipedia gets its information from, but cooling to even a surface temperature of $10^{6}$K (the interiors are 1-2 orders of magnitude hotter) is likely to take longer than a few years - somewhere between 100 and $10^4$ years, unless direct URCA processes are somehow permitted (e.g. in quark matter, see Yakovlev & Pethick 2004, for a thorough review of neutron star cooling).


What's the rationale behind the false colours in solar observation photographs? Images from SOHO, SDO and other sun observatories are often coloured differently for different wavelengths or temperatures:

http://www.nasa.gov/mission_pages/sunearth/science/Sun-Wavelength-Chart.html has statements like "This light is emitted from the upper transition region and the chromosphere and are typically colorized in red."

How is the convention of which temperature or wavelength gets which colour established?

Is it simply so that scientists can immediately tell what temperature or wavelength a picture shows, or are the colours optimized to show particular features?

And why is there a white ring or circle in the sunshade in some SOHO imagery, and why different sizes compared to the disc in the two photos above?

Some of the images shown in your question are not colored or colorized, it is not a false color; it's the real color. The calculations that are made to produce the false colors that are used on colorized image datasets are shown in the aia_lct.pro file, where the exact RGB values are provided.

Where a false color is chosen it is done so (for SDO and some other images) using the AIA Standard Color Tables (which are generated in the aia_lct.pro file).

Information on the "VSO's Guide to SDO Data Analysis" (.PDF), on page 46, describes use of the subroutines in the sdo/aia/idl/calibration/aia_colors.pro source code file from http://www.heliodocs.com/ or view "Installing Solarsoft" for ready to run versions and more info.

See Section 6.6.1 "Using AIA Standard Scaling and Colors" for additional information about the aia_colors.pro function. The Introduction of that Guide contains additional information and links about the SDO mission.

Here is the Description of the colors NASA uses for different wavelengths (using the AIA Standard Color Tables):

  view-source:https://sdo.gsfc.nasa.gov/data/aiahmi/

Telescopes

<select name="telescope" id="telescope" class="form-control">
    <option value="aia_0171">AIA 171 (gold)</option>
    <option value="aia_0193">AIA 193 (bronze)</option>
    <option value="aia_0304">AIA 304 (red)</option>
    <option value="aia_0211">AIA 211 (purple)</option>
    <option value="aia_0131">AIA 131 (teal)</option>
    <option value="aia_0335">AIA 335 (blue)</option>
    <option value="aia_0094">AIA 094 (green)</option>
    <option value="aia_1600">AIA 1600 (yellow/green)</option>
    <option value="aia_1700">AIA 1700 (pink)</option>
    <option value="hmib">HMI Magnetogram (gray)</option>
    <option value="hmibc">HMI Colorized Magnetogram (colorized)</option>
    <option value="hmii">HMI Intensitygram (gray)</option>
    <option value="hmiic">HMI Intensitygram (orange)</option>
    <option value="hmiihi">HMI Intensitygram High Cadence (orange)</option>
    <option value="hmiif">HMI Intensitygram Flat (orange)</option>
    <option value="hmid">HMI Dopplergram (gray)</option>
    <option value="COMP211193171">AIA Composite 211, 193, 171</option>
    <option value="COMP304211171">AIA Composite 304, 211, 171</option>
    <option value="COMP094335193">AIA Composite 094, 335, 193</option>
    <option value="COMPHMI171">Composite AIA 171, HMI Magnetogram</option>
</select>

Here is a webpage showing the images in your question in various sizes, including a 48 hour movie for each.

The full Electromagnetic Spectrum:

The Visible Spectrum, including Fraunhofer Line Letters and Wavelength Numbers:

The Fraunhofer lines are typical spectral absorption lines. Absorption lines are dark lines, narrow regions of decreased intensity, that are the result of photons being absorbed as light passes from the source to the detector. In the Sun, Fraunhofer lines are a result of gas in the photosphere, the outer region of the sun. The photosphere gas is colder than the inner regions and absorbs light emitted from those regions.

Here is the wavelength's frequency for each Element of interest:

Each Filter accepts a narrow band which has a specific use and specific properties. Filters are more expensive when the bandwidth is very narrow, single frequency Filters are unnecessary and unaffordable.

Different frequency ranges of the spectrum are used to resolve different aspects of the object being observed, similarly different bandwidths are used to further enhance those features.

Listing the bandwidths for every filter and the expected results would be exhaustive, one website that gives such info for each of their filters is Daystar where their filter comparison page indeed confirms that the Sun can be viewed in black and white using a monochrome sensor. It's when you want to combine multiple channels that use of color becomes necessary.

Here is an example of the differences between bandwidths for the Hydrogen Alpha Filter (a deep-red visible spectral line in the Balmer series with a wavelength of 656.28 nm):

Filters with 0.8 ångström bandwidths will reveal prominences in high contrast
Filters with 0.7 ångström bandwidths will reveal prominences in high contrast and occasionally, surface texture. Prominences are generally larger with .7Å filters than in narrower bandwidths
Filters with 0.6 ångström bandwidths will reveal some surface contrast as well as prominences.  A .6Å filter can be a good compromise for those having a difficult decision
Filters with 0.5 ångström bandwidths will reveal better surface contrast as well as great prominences
Filters with 0.4 ångström bandwidths will reveal excellent surface contrast as well as fine chromosphere detail
Filters with 0.3 ångström bandwidths will reveal superior surface contrast above any other filter. Views are generally somewhat dimmer.  Expect "Pencil-Thin" details on surface and prominence features

It is bandwidth that is used to derive contrast, not coloring.

The SDO (Solar Dynamics Observatory) is a sun-pointing semi-autonomous spacecraft that allows nearly continuous observations of the Sun. Unlike the human eye which uses rods and cones, the SDO uses 3 instruments:

The AIA (Atmospheric Imaging Assembly) images the solar atmosphere in multiple wavelengths to link changes in the surface to interior changes. Data includes images of the Sun in 10 wavelengths every 10 seconds.
The EVE (Extreme Ultraviolet Variability Experiment) measures the Sun's extreme ultraviolet irradiance with improved spectral resolution, "temporal cadence", accuracy, and precision over preceding measurements made by TIMED SEE, SOHO, and SORCE XPS.
The HMI (Helioseismic and Magnetic Imager) studies solar variability and characterizes the Sun's interior and the various components of magnetic activity. HMI produces data to determine the interior sources and mechanisms of solar variability and how the physical processes inside the Sun are related to surface magnetic field and activity.
The Magellan craft used Synthetic Aperture Radar uses radio waves and not visible light to image map objects. Magellan synthetic aperture radar mosaics from the first cycle of Magellan mapping are mapped onto a rectangular latitude-longitude grid to create this image. Data gaps are filled with Pioneer Venus Orbiter altimetric data, or a constant mid-range value. Simulated color is used to enhance small-scale structure. The simulated hues are based on color images recorded by the Soviet Venera 13 and 14 spacecraft. The image was produced by the Solar System Visualization project and the Magellan science team at the JPL Multimission Image Processing Laboratory.

Similarly SDO data outside the visible spectrum is colorized to allow invisible portions of the electromagnetic spectrum (and Magnetometer recordings) to be visualized by the human eye in a readily digestible image rather than serving the general public a large dataset of numbers.
The visible spectrum is the portion of the electromagnetic spectrum that is visible to the human eye. Electromagnetic radiation in this range of wavelengths is called visible light or simply light. A typical human eye will respond to wavelengths from about 390 to 700 nm. In terms of frequency, this corresponds to a band in the vicinity of 430–770 THz.

The spectrum does not, however, contain all the colors that the human eyes and brain can distinguish. Unsaturated colors such as pink, or purple variations such as magenta, are absent, for example, because they can be made only by a mix of multiple wavelengths. Colors containing only one wavelength are also called pure colors or spectral colors.

  Q: How is the convention of which temperature or wavelength gets which colour established?

  Is it simply so that scientists can immediately tell what temperature or wavelength a picture shows, or are the colours optimized to show particular features?

The colors are not optimized to show particular features or provide contrast. The aia_lct.pro file is where the exact RGB values are calculated, see top of this answer.

  Q: And why is there a white ring or circle in the sunshade in some SOHO imagery, and why different sizes compared to the disc in the two photos above?

The coronagraph blocks out the Sun with an occulting disk to reveal the much fainter structures in the Sun's atmosphere. Sometimes multiple images are combined to remove the disk and provide a unoccluded image.

The Large Angle and Spectrometric Coronagraph (LASCO) in the Solar and Heliospheric Observatory (SOHO) doesn't use the latest technology, sometimes there are disturbances in the images.

There are two kinds of disturbances that repeatedly occur:

Blackouts and Whiteouts, in broken lines, circle-like shapes, or over the whole picture. They are caused by the electronics box. There has never been a firmware update, since it was judged as too sensitive changing the flight-software.
Black and white pixels, occurring in patterns, without pattern or alone. Those "missing blocks" are telemetry dropouts, caused by radio interference or a disturbance in the data transfer to Goddard Space Flight Center.

[The section below might benefit from some more editing if the OP requests it.]

  AIA 193

  This channel highlights the outer atmosphere of the Sun - called the corona - as well as hot flare plasma. Hot active regions, solar flares, and coronal mass ejections will appear bright here. The dark areas - called coronal holes - are places where very little radiation is emitted, yet are the main source of solar wind particles.

  Where: Corona and hot flare plasma

  Wavelength: 193 angstroms (0.0000000193 m) = Extreme Ultraviolet

  Primary ions seen: 11 times ionized iron (Fe XII)

  Characteristic temperature: 1.25 million K (2.25 million F)

  AIA 304

  This channel is especially good at showing areas where cooler dense plumes of plasma (filaments and prominences) are located above the visible surface of the Sun. Many of these features either can't be seen or appear as dark lines in the other channels. The bright areas show places where the plasma has a high density.

  Where: Upper chromosphere and lower transition region

  Wavelength: 304 angstroms (0.0000000304 m) = Extreme Ultraviolet

  Primary ions seen: singly ionized helium (He II)

  Characteristic temperature: 50,000 K (90,000 F)

  AIA 171

  This channel is especially good at showing coronal loops - the arcs extending off of the Sun where plasma moves along magnetic field lines. The brightest spots seen here are locations where the magnetic field near the surface is exceptionally strong.

  Where: Quiet corona and upper transition region

  Wavelength: 171 angstroms (0.0000000171 m) = Extreme Ultraviolet

  Primary ions seen: 8 times ionized iron (Fe IX)

  Characteristic temperature: 1 million K (1.8 million F)

  AIA 211

  This channel (as well as AIA 335) highlights the active region of the outer atmosphere of the Sun - the corona. Active regions, solar flares, and coronal mass ejections will appear bright here. The dark areas - called coronal holes - are places where very little radiation is emitted, yet are the main source of solar wind particles.

  Where: Active regions of the corona

  Wavelength: 211 angstroms (0.0000000211 m) = Extreme Ultraviolet

  Primary ions seen: 13 times ionized iron (Fe XIV)

  Characteristic temperature: 2 million K (3.6 million F)

  AIA 131

  This channel (as well as AIA 094) is designed to study solar flares. It measures extremely hot temperatures around 10 million K (18 million F), as well as cool plasmas around 400,000 K (720,000 F). It can take images every 2 seconds (instead of 10) in a reduced field of view in order to look at flares in more detail.

  Where: Flaring regions of the corona

  Wavelength: 131 angstroms (0.0000000131 m) = Extreme Ultraviolet

  Primary ions seen: 20 and 7 times ionized iron (Fe VIII, Fe XXI)

  Characteristic temperatures: 10 million K (18 million F)

  AIA 335

  This channel (as well as AIA 211) highlights the active region of the outer atmosphere of the Sun - the corona. Active regions, solar flares, and coronal mass ejections will appear bright here. The dark areas - or coronal holes - are places where very little radiation is emitted, yet are the main source of solar wind particles.

  Where: Active regions of the corona

  Wavelength: 335 angstroms (0.0000000335 m) = Extreme Ultraviolet

  Primary ions seen: 15 times ionized iron (Fe XVI)

  Characteristic temperature: 2.8 million K (5 million F)

  AIA 094

  This channel (as well as AIA 131) is designed to study solar flares. It measures extremely hot temperatures around 6 million Kelvin (10.8 million F). It can take images every 2 seconds (instead of 10) in a reduced field of view in order to look at flares in more detail.

  Where: Flaring regions of the corona

  Wavelength: 94 angstroms (0.0000000094 m) = Extreme Ultraviolet/soft X-rays

  Primary ions seen: 17 times ionized iron (Fe XVIII)

  Characteristic temperature: 6 million K (10.8 million F)

  AIA 1600

  This channel (as well as AIA 1700) often shows a web-like pattern of bright areas that highlight places where bundles of magnetic fields lines are concentrated. However, small areas with a lot of field lines will appear black, usually near sunspots and active regions.

  Where: Transition region and upper photosphere

  Wavelength: 1600 angstroms (0.00000016 m) = Far Ultraviolet

  Primary ions seen: thrice ionized carbon (C IV) and Continuum

  Characteristic temperatures: 6,000 K (11,000 F), and 100,000 K (180,000 F)

  AIA 1700

  This channel (as well as AIA 1600) often shows a web-like pattern of bright areas that highlight places where bundles of magnetic fields lines are concentrated. However, small areas with a lot of field lines will appear black, usually near sunspots and active regions.

  Where: Temperature minimum and photosphere

  Wavelength: 1700 angstroms (0.00000017 m) = Far Ultraviolet

  Primary ions seen: Continuum

  Characteristic temperature: 6,000 K (11,000 F)

  AIA Composite 211, 193, 171

  This image combines three images with different, but very similar, temperatures. The colors are assigned differently than in the single images. Here AIA 211 is red, AIA 193 is green, and AIA 171 is blue. Each highlights a different part of the corona.

  AIA Composite 304, 211, 171

  This image combines three images with quite different temperatures. The colors are assigned differently than in the single images. Here AIA 304 is red (showing the chromosphere), AIA 211 is green (corona), and AIA 171 is dark blue (corona).

  AIA Composite 094, 335, 193

  This image combines three images with different temperatures. Each image is assigned a color, and they are not the same used in the single images. Here AIA 094 is red, AIA 335 is green, and AIA 193 is blue. Each highlights a different part of the corona.

  [Webpage had no Info Popup for: AIA 171 & HMIB]

  HMI Magnetogram

  This image comes from HMI, another instrument on SDO. It shows the magnetic field directions near the surface of the Sun. White and black areas indicate opposite magnetic polarities, with white showing north (outward) polarity and black showing south (inward) polarity.

  Where: Photosphere

  Wavelength: 6173 angstroms (0.0000006173 m) = Visible (orange)

  Primary ions seen: Neutral iron (Fe I)

  Characteristic temperature: 6,000 K (11,000 F)

  HMI Colorized Magnetogram

  This image comes from HMI, another instrument on SDO. It shows the magnetic field directions near the surface of the Sun. White and black areas indicate opposite magnetic polarities, with white showing north (outward) polarity and black showing south (inward) polarity.

  Where: Photosphere

  Wavelength: 6173 angstroms (0.0000006173 m) = Visible (orange)

  Primary ions seen: Neutral iron (Fe I)

  Characteristic temperature: 6,000 K (11,000 F)

  [Webpage had no Info Popup for: HMI Intensitygram - colored]

  [Webpage had no Info Popup for: HMI Intensitygram - Flattened]

  [Webpage had no Info Popup for: HMI Intensitygram]

  [Webpage had no Info Popup for: HMI Dopplergram]

  [Webpage had no Info Popup for: HMI Intensitygram - High Cadence]

  EVE/ESP and EVE/MEGS-P

  This plot shows time series of 5 strong EUV emission lines, showing how bright the flare is emitting at these various emissions. Also shown is a Dark value, which is a detector that is blocked from seeing the Sun, which shows energetic particles from the Sun that can penetrate the EVE instrument and cause false counts. This Dark diode will increase during solar storms.

  Where: Solar Irradiance (full Sun)


Does natural satellite(s) of a planet affects its orbital velocity around a star? Please excuse me for a lay question.

As we know, to calculate the orbital velocity, we take the mass of the orbiting body, the mass of the body being orbited and the distance between the two bodies into account.

However, we don't really care about the natural satellite(s) (of a planet) in the calculations. Why are they insignificant? Why isn't the whole system involved in the calculations?

Yes, it does affect the planet's orbital velocity. Both planet and satellite move around the planet-satellite center of mass. That center of mass is orbiting the star with some constant velocity (let's assume circular orbits to keep things simple) but if you're only looking at the planet, then the orbital speed around the star will wobble up and down around the because of its additional motion around the planet-satellite center of mass.

For most planet-satellite systems this effect is very small since usually the satellite is very, very light compared to the planet. Only when you have a genuine double system like Earth-Moon or Pluto-Charon does this effect start to matter. For transiting exoplanet systems they're trying to infer the existence of exomoons by exploiting the effect this will have on the transit timing of the planet. So far no exomoons have turned up though.


Lunisolar calendar: what phase was the moon Y years ago on the spring equinox? How would I go around answering these questions (e.g., for $Y=10,000$ -- IOW, historical observations are not available and the time is too long ago to rely on a fixed calendar):

How many days ago ($D$) was the spring equinox $Y$ years ago?
What phase (how many days from conjunction) was the moon on that day $D$?

Here is my "preliminary research":
Methodology
I suppose I could solve the 3 body problem (Sun, Earth, Moon) numerically backwards in time, but I suspect that the numerical errors and gravitation perturbations from other planets will accumulate rather quickly (how quickly?)
I could do arithmetic with the durations of lunation and year, but it appears that they are not as stable as I wish they were (are they?)
Software
Online calculators (NOAA, Stellafane) do not go that far back.
Presumably they rely on a version of Almagest but, again, it is not clear if these formulas are reliable for 3-5-10k year calculations.
Question:
How would I answer 1 & 2 above?
I am interested in both methodology (how it is done in general) and implementation (e.g., a program that would compute those numbers for me, given the value of $X$). It's okay to answer only a part of this, I will ask another, follow-up question.
PS. I asked this question on Math.SE and was told to "ask an astronomer".

This is a challenging problem and I can't give you a definitive answer, but here are some thoughts, mostly on the methodology rather than implementation.
First, you are right that fundamentally the way to do this is to integrate the equations of motion backward in time, taking into account the relevant forces.  You are also right that only considering the Sun, Moon, and Earth is inadequate.  The farther back in time you want to go, the more precise your model needs to be, so you need to include the influence of the other planets.
However, just considering the gravitational forces of the various bodies, but treating them as spherical masses, is also inadequate for your question.  For different reasons for the two parts, you have to take into account the fact that the Earth isn't a perfect sphere.
For the equinox, this is because of precession.  For determining the date of the equinox, it's not enough to know where the Earth is in its orbit; you also have to know where the pole is pointing.   There are detailed models of the Earth's precession available, but I don't know how far back they go before uncertainties become large.
For the moon phase, you have to take into account the fact that interactions between the Moon and the Earth's equatorial bulge are gradually transferring angular momentum from the Earth's rotation into the Moon's orbital motion, causing the Moon to slowly move away from the Earth (and thus causing the length of the lunar month to increase slowly). Again, this has been measured and could be incorporated into a model.  The rate of orbital period change is such that 10,000 years changes the Moon's period by several seconds.  That doesn't sound like much, but the cumulative effect is several full orbits' difference over that time period.  In other words, you'd be completely wrong about the Moon's position in its orbit on a given date if you didn't take it into account.
I suspect that there are other effects that would need to be taken into account as well to get an accurate prediction in the distant past, but those are the two that come to mind.
Regarding implementations, currently JPL Horizons is what I would consider the state of the art (though there may be others of comparable accuracy).  Their time horizon is 10,000 years (i.e. you can't request positions farther back than that).  I don't know if that limit is set by accuracy (which will certainly get worse as you go farther back) or by computational resources.


The role of 'the interference of waves' in VLBIs' I've been trying to find out how VLBIs work. It says in this book I've read that many radio telescopes around the world work in tandem to use the interference of waves to get a clearer picture of celestial bodies, which is how they got the black hole picture.
But what I've found only talks about the difference in the time that radio waves takes to arrive to different points on the globe and how that is calibrated. Not a single word about the interference of waves.
Wikipedia's Very Long Baseline Interferometry; Method has a short discussion, but does not really explain the basic principle of how delays and interference are used to build up an image.

You can think of the object being photographed as a collection of light-emitting points. The light emitted from each point is spherically symmetric, at least over the small angle that we see.
If you sample one of those point sources at two points, and the source lies exactly on the perpendicular bisector of the line segment between the sampling points, then light of the same amplitude and phase will reach both sampling points at the same time. If the source is slightly off that plane, then the matching light will reach the sampling points at slightly different times. This allows you to disentangle the light coming from different directions.
I don't know anything about real-world VLBI, but in principle the way you extract an image from the raw data is by simulating a camera. You introduce electromagnetic waves with the strengths you measured at the spacetime locations where you measured them, and simulate them reflecting off a planet-sized mirror (or refracting in a planet-sized lens) and hitting a detector. You can show that light from any point in your source object will be focused to a point on the detector by your mirror/lens, i.e. it constructively interferes there and destructively interferes elsewhere on the detector. By linearity, the sum of the light from all the source points forms an image of the source.


How tall are the "fractured" linear structures on Europa's surface? What are the altitude differences in the high resolution part of this image of Europa?

How high do those ridges rise over their neighboring valleys? And how large is the area shown in this image?

And are the craters believed to have been formed by impacts or some kind of "eruptions" from below? It looks to me as if the areas which have the most (tiny) craters is on top of and thus newer than the less cratered areas. Well, here and there anyway.

Apparently they have been inferred to be "220 meters". See the references in:

S. D. Kadel, S. A. Fagents, "TROUGH-BOUNDING RIDGE PAIRS ON EUROPA", , Lunar and Planetary Science XXIX. (PDF)