How well would the Moon protect the Earth from an Asteroid?

It would be much better for Earth if the impactor hit the moon...

In this Worldbuilding answer, I used a paper on ejecta kinematics to do calculations for ejecta velocity upon impact. Without going into too much detail here, much of the ejecta from a large impactor would not exceed the moon's escape velocity of 2.38 km/s. You can examine Figure 7 from the linked paper which shows the logarithmic relationship between ejecta velocity and the edge of the crater. Only material within a few tens of meters of the edge of the formed crater can attain the velocity needed to escape the Moon's gravity.

The mass of the asteroid increases with the cube of its radius; while the mass of material within $n$ meters of the edge of the impact crater increases with radius; so it is evident that the the larger the meteor itself gets, the lower the danger of the any potential ejecta relative to the original impactor.

Furthermore, material ejected from the Moon has the potential to land back on the Moon, enter a stable orbit of Earth, or be ejected from the Earth-Moon system; so only a (probably small) fraction of Moon ejecta would threaten Earth.

Overall, fewer, smaller rocks are much better than large rocks when it comes to getting hit by things (except maybe for your car's windshield).

...unless the impactor knocked the Moon into Earth.

Of course, there is always the possibility that a very, very large impactor would knock the moon into a different orbit, potentially one that eventually impacts Earth. That would, obviously, be the worst case scenario. Significant changes to the orbit of the moon, even if these changes do not cause a collision with Earth, have the potential to do significant damage in terms of tides; not just ocean tides but also the poorly understood effect of tidal forces on Earth's mantle.

In any case; it is probably much better for a large impactor to hit the Moon instead of the Earth, but possibly much, much worse.


Signal-to-noise in inverse angstrom for spectroscopy?

A spectrum imaged onto a CCD or other detector consists of photons spread out along a wavelength axis. Signal-to-noise (S/N) depends on how many photons you have from the source, how many from the background, intrinsic noise from the electronics, etc. -- and what kind of bin you're adding up the photons in. If you summed up all the photons across the spectrum (over all wavelengths), you'd get one number with a very high S/N, but no spectral information. If you divided the spectrum into N different wavelength bins, you'd get a lower S/N per bin, but more spectral information.

What the S/N ratio you mention means is that if you divided that spectrum into bins that were 1 Angstrom wide, you'd have enough photons in each bin for a S/N of 10 in each bin (at least for some region of the spectrum, probably near the central wavelength). You can then have some idea how the S/N would improve for bins larger than 1 Angstrom in size, or how it would get worse for smaller bins. S/N per Angstrom is conventional for optical spectroscopy, both because Angstroms are traditional and because moderate resolutions spectrographs often have scales of a few tenths to several Angstroms per pixel along the wavelength direction.


How does large body gravity affect planet formation?

why didn't the sun's gravity keep the inner planets from forming?

The short answer is, the asteroid belt orbits the sun, it doesn't orbit Jupiter, and you shouldn't expect the same effect with 2 different relations.

The sun is the gravitational object that the Asteroid belt orbits.  Jupiter is not and it has a stabilizing effect on the material that orbits around it.  Jupiter has a (mostly) destabilizing effect on nearby material that orbits around the sun, though there is some stability with orbital resonance and trojan points.   The reason why that is gets a little bit mathy, but I'll cover the basics.

When a solar system forms, probably out of a combination of strong solar wind from a supernova and a dense enough gas cloud, most of the solar-system material forms the young sun and maybe 2%-4% remains in orbit and over time, spirals and flattens out into a disk.  (see cool video)

The sun's gravity stabilizes that disk.  There's no reason for it to disburse it, though the young and usually strong solar wind can blast it and clear out the closer regions of smaller particles and any ices.

Jupiter is entirely different.  Jupiter and other planets which form inteh disk, tends to eat everything in their path.  That's one of the definitions of a planet.  It clears out it's orbital area.  These planets on their own would remain in orbit around the sun, but they can interact over time with each other.   Pretty much anything in Jupiter's path, once Jupiter gets large enough, either gets eaten or gets thrown into a very different orbit, either further our or closer in.

The asteroid belt is (as Suhrid Mulay points out) a relatively low density region of space that doesn't have enough mass to coalesce into it's own planet, but it might not have always been that low density and that's probably only part of the reason.  The other part of the reason is the proximity of Jupiter which tends to disrupt anything that orbits too close to it and Jupiter may have at one point, moved in quite a bit closer to Mars before moving back out.  Tossing much of the material in that region far away.

Planets don't do well forming too close to other planets because they will gravitationally disturb each other.  A planet the size of Jupiter had pretty significant reach when it comes to disturbing other planets and planet formation.


Will LSST make a significant increase in the rate of astronomical event alerts?

Yes. The estimates are that LSST will produce about 10 million alerts per night (LSST Alert Distribution presentation) which will be at least a factor of 5x greater than the amount coming from ZTF currently. ZTF is an approximately 10% scale model of what the LSST alert stream will look as there is about 5x fewer alerts and the alert packets contain about 50-60% of the information that will be in the LSST alerts. The ZTF alert stream (Patterson et al. 2019) also uses the same software (Apache Avro, Spark and Kafka) as LSST will use, just running on less hardware.

These alerts (in Avro serialized binary data format) will not be sent directly to astronomers (and certainly not to cell phones), the plan is to send the alerts to a few "brokers" that will do additional filtering, classification and cross-matching to other catalogs to give additional information and context to the alert (which is basically a "this source on the sky changed in brightness by X amount or appeared"). There are several brokers in various stages of development of which the leading examples are ANTARES and Lasair.

Astronomers, or more likely their software systems, will subscribe to these brokers and add filters for the alerts to subset the stream down to the particular types of objects and science that interest them. These will most likely go into their own databases of interesting targets and astronomers or their software will decide on the small number of most interesting ones that will be sent out to other telescopes, some robotic for rapid response science case, some traditionally scheduled. The large number of alerts and triggers for follow-up has the potential to be very disruptive to the prior way of operating observatories, which is one of the reasons why we are working on system like the Astronomical Event Observatory Network (AEON) to handle and coordinate this.


How did the authors determine both the spatial size of gas cloud HCN-0.009-0.044 and its central mass at the same time?

The gas clouds are much larger than you think: From Fig. 1 of the paper (Takakawa 2019), the largest cloud (which they call the "Balloon") is roughly 10 arcsec across. Now they can't actually measure its exact distance, but its projected distance from the Galactic center is only 7 pc, and the authors argue that it's at least 5 kpc away due to absorption from molecular gas which is known at this distance (Sofue 2006), and since such hot and high density ($\sim 10^7\,\mathrm{cm}^{-3}$) molecular gas is abundant in the Galactic center, it is probably located there, i.e. at a distance of $R=8\,\mathrm{kpc}$.

The cloud's absolute size is thus
$$
D = 10'' \times 8\,\mathrm{kpc} = 80\,000\,\mathrm{AU} \simeq 0.4\,\mathrm{pc}.
$$

The mass is measured from dynamics, i.e. they measure the line-of-sight velocities of several clouds. These velocities depend both on the mass of the dynamical center (i.e. whatever object the clouds orbit), and on the total velocity of this object wrt. us. Since they only know the line-of-sight velocities and not the full, 3D motions, they cannot know with certainty the mass, but performing a chi-square fit of 3D models that fit the 2D observations, they determine the best-fit value of the mass to be $(3.2\pm0.6)\times10^4 M_\odot$.

Their Fig. 4 shows the model:

3D model of the motion of the clouds. The red cloud is the Balloon. Other observed clouds are shown in green, blue, and orange colors. The black dot indicates the position of the black hole. The actual observation is the 2D projection, seen in the bottom plane.

The model of the velocities also gives the actual trajectories of the clouds around the central mass. Because the radius of the Balloon's orbit is $\sim0.07\,\mathrm{pc}$, it means that this much mass must be concentrated within a region of maximum that size, and probably significantly smaller. This implies a density of $>2\times10^8 M_\odot\,\mathrm{pc}^{-3}$, at least an order of magnitude larger than even the most densely packed globular clusters, and since no luminous counterpart is observed at that location, the authors consider an intermediate-mass black hole as the most promising candidate for the gravitational source.


How to get the longitude of the perihelion to find the Heliocentric Equatorial coordinates of a planet

As opposed to what @user21 said, the longitude of periastron does depend on the longitude of ascending node (I was surprised by that myself).
In orbital dynamics, there exist the argument of periastron (symbol ω) as well as the longitude of periastron (symbol π [or ϖ but may lead to confusion) ), which is equal to the argument of periastron (ω) plus the longitude of ascending node (Ω). In other words:
π = ω + Ω
Now, the argument of periastron is equal to:
ω = arccos $\frac{\mathbf{n}\ ·\ \mathbf{e}} {|\mathbf{n}||\mathbf{e}|}$
If ez < 0 then ω → 2π − ω.
where:
$\mathbf{n}$ is a vector pointing towards the ascending node (i.e. the z-component of $\mathbf{n}$ is zero), and
$\mathbf{e}$ is the eccentricity vector (a vector pointing towards the periastron).
This formula can be found at https://en.wikipedia.org/wiki/Argument_of_periapsis (I call it instead the “argument of periastron,” as Meeus says: “The word ‘periapse’, used by some authors, is incorrect. The word perihelion means the point of the orbit that is closest to the Sun [from the Greek peri = near + helios = Sun]. Similarly, perigee is the point closest to the Earth [ge = Earth]. Therefore, ‘periapse’ would really mean the point closest to the apse; but this is ridiculous, because what is meant is the apse itself!”)
So, using the above formula, you should be able to figure out the longitude of the ascending node for each of your hypothetical bodies.
Have fun!


A little bit of caution is needed when approaching this subject as there are several surfaces in cosmological solutions that can be interpreted (or misinterpreted in some cases) as horizons. Not all of them appear in all comsological solutions and in some cosmological solutions some of these surfaces are only present for part of the evolution of the solution.

If you were to mean the particle horizon, which is the limit of the observable Universe. The cosmological red shift at the particle horizon is always infinite, when there is a particle horizon. No matter can ever leave the particle horizon (regardless of the cosmological model, as long as there is a particle horizon) as it expands radially outwards at a local speed of c relative to any matter on the horizon (of course to us it expands even faster than c as it is also receding due to expansion).

An ever-expanding Universe with a cosmological constant approaches a state similiar to de Sitter space in late times, however de Sitter space itself is not a realistic cosmological solution as it is devoid of matter. De Sitter space does not have a particle horizon and for that matter it doesn't have any singularities either.

In de Sitter space the cosmic event horizon, which is the surface beyond which events happening now cannot affect us in the future providing we stay at the same spot (this should not be confused with the particle horizon), sits at a constant distant to the observer that is inversely proportional to the square root of the cosmological constant.

In answer to your question then, de Sitter space is symmetric in time and space which actually gives rise to seemingly different physical interpretations of it. However the black hole information paradox is specifically the loss of unitariness in the time evolution of a black hole and you have not justified why you think an equivalent thing should happen in de Sitter space. The loss of unitariness in actually arises because  the event horizon disappears, but (naively at least) the information behind it also disappears. In de Sitter space the event horizon remains and even if it were to disappear by the cosmological constant miraculously going to zero we would still find the information behind it, so there is no analog to the information paradox in de Sitter space.

To take another example, consider Rindler coordinates in Minkowski spacetime, there is a horizon (the Rindler horizon) and there is radiation (Unruh radiation), however there is no information paradox as we know that unitariness is preserved in Minkowski spacetime and when the Rindler horizon is made to disappear by selecting a different coordinate system, all the information that was behind the horizon is still there.


How does moonrise/moonset azimuth vary with time?

The monthly variations are caused by the change in declination of the Moon as it travels (roughly) along the ecliptic.  This is caused by the tilt of the Earth's rotation axis, which results in an angle of about 23° between the equator (declination = 0°) and the ecliptic.  Around the constellations of Taurus and Gemini the ecliptic lies at much higher declination than near Saggitarius or Ophiuchus.  The same effect causes the change in rise and set times of the Sun during a year.

When you're in the northern hemisphere, a higher declination means that the transit altitude of the Moon or Sun is higher.  In fact, their whole daily (apparent - caused by the Earth's rotation) paths lie closer to the North star.  Because of that, the points where a daily path intersects with your local horizon lie further north, which you see as a change in rise/set azimuth of the Moon.

All this is true even without the 5° inclination of the Moon's orbit w.r.t. the ecliptic.  That inclination only mitigates the effect; the Moons declination does not exactly vary between -23 and +23° every month, but these extreme values themselves can lie between +/-18° or +/-28°, depending on whether the inclination effect increases (23+5°) or decreases (23-5°) the effect of the motion along the ecliptic.

Lastly, the rise/set azimuths do not correlate to lunar phase, as do the rise/set times.  Both phases and rise/set times are defined based on the Sun's position; we roughly define "noon" as the moment when the Sun is transiting in the south, and call the Moon "New" when it's in the same direction as the Sun, so there's no wonder that the New Moon transits around noon.  This is not true for the effects described above, and hence the corresponding periods are different.  The mean lunar period between twice the same equinox point (e.g. Aries) is called the Tropical month and lasts ~27.3d (this is roughly the same as the siderial month, but influenced by precession).  The effect of the inclination of the Moon's orbit has a mean period of one Draconic month, with a period of ~27.2d.  Instead, the Synodic month, the mean period between two cases of New Moon, lasts 29.5d, about 2.2d longer.  Hence, the declination and rise/set azimuths of the Moon do not correlate with the phases.


Amount of EM radiation (and particles) Earth "receives" from other planets in our solar system?

Of course all the planets have been observed by many telescopes at different wavelengths. Yes, they do emit energy at wavelengths other than in the visible part of the spectrum.

Broadly speaking, the peak of their output depends on their temperatures. Most of the light from the outer giant planets emerges in the infrared for example.

However, some planets also are strong radio emitters. For example Jupiter has a strong magnetic field that accelerates charged particles which then emit radio waves.

Neutrinos are hard to pinpoint - telescopes currently struggle to say exactly where they come from in the sky. The planets are not expected to be strong neutrino emitters.


What if a Black Hole passed through an Irregular type galaxy?

Black holes (even SMBs) are actually very small objects on an astronomical scale and in practical terms probably very little would happen.

There would be some perturbations of stellar orbits within the formation they were in, but unless there was a fairly close approach  very little would happen and on the scale of the entire formation the changes would affect only a small number of objects and not the overall "shape".

To put this in perspective, Sag. A* has a mass of about 4,000,000 solar masses.  Sounds big (is big :-) ).

But the Large Magellanic Cloud has a mass of about 10,000,000,000 solar masses (or about 2500 time larger than Sag. A*).  The dominant gravitational field is still that associated with the LMC itself, not that of Sag A*.

I think a good analogy would be a bullet fired into a lake.  Sure, it can cut a path through one isolated region and cause disturbance close to that path, but the bulk of the lake will be undisturbed apart from a slight ripple.

The Small Magellanic Cloud is still a pretty large mass at about 6,500,000,000 solar masses, so still roughly 1,400 times more massive than Sag. A*.  Again, we're talking bullet in lake territory.


The largest discrepancy in the history of science

There are roughly two possibilities: either there isn't a large vacuum energy, this would imply that there is something missing from quantum field theory. We don't know what it could be.

Or the zero point energy really is as large as QFT predicts, but there is something else that prevents it from having a large cosmological effect. We have no idea what that could be.

We might get more of an idea if we find out what Dark Matter consists of, or a good quantum theory of gravity we might know more.


Struggling to understand the phases of the moon

As noted in a comment, you should imagine the picture is looking down on the Earth, and remember that the actual moon is much further from the Earth, relative to their sizes.

Imagine a person standing on the left side of the Earth (in the diagram). The time would be midnight for them. They could see the moon when it is full, and they could see the moon when it is a gibbous moon (but the gibbous moon would be lower in the sky)  They could even see the moon at first quarter, but it would be on the horizon.  You can't see a crescent moon at midnight.

Now imagine a person standing in the top left on the Earth (in the diagram). The time for them would be about 9pm. (and night time) That person could see the half moon and even a crescent moon (but the crescent would be on the horizon)

It is only possible to see a crescent moon at dawn or dusk, you never see a crescent moon at midnight. At 6pm (at dusk) you can see the crescent moon. If you look to the moon, you see mostly the side in shadow, but a little of the light shining around the side, so it looks like a crescent.

Now you are also confused about the illumination side shown in the waning phases, This is because the picture shows the view of the moon that a person would see if they were standing on the Earth. Because that person is upside down (from our point of view) they see the moon reversed. In other words, from our position above the North Pole, we see the right side of the moon illuminated.  From their position, it is the left.

If the diagram still confuses, discard it and think in terms of angles.

From the perspective of the Earth, the sun and moon go around in circles at different rates, so the angle made between the sun and the moon changes. When the angle between the sun and the moon is about 90 degrees, then half of the near side of the moon is lit by then sun, and a half moon is visible.

When the angle is close to 180 degrees, then nearly all the moon is lit by the sun, and the moon is full.

When the angle is close to 0 degrees, then none of the near side will be lit, and the moon won't be visible.

When the angle is small, ie the moon is close to the sun, then a crescent moon will be visible. A crescent moon can only be seen if it is near to the sun, so the sun cannot be far below the horizon. This means that the a crescent moon cannot be seen late at night when the sun is far below the horizon, but at some times and locations the sun might not set until late in the evening, so the setting time of the moon might be quite late.

Nevertheless it is the angle between the sun and the moon that determines the phase of the moon, and nothing else.


How was Mars' very noticeable change in apparent magnitude explained by the Geocentric Model?

Epicycles: These result in the distance of the planet to Earth changing quite substantially over its orbit (in the geocentric model) and resulting in changes in brightness. With the right choice of epicycles, you can account quite well for the changes in brightness, certainly better than Ptolemy would have been able to measure it.
The less scientifically bent may have assumed it was just "in the nature of a planet" to be brighter at opposition (the god of war, putting out his power making his planet brighter just as it comes to rule the night sky)


Why moon sighted only in Kerala but not in rest of India?

Think of it like this:

Each hour after New Moon, the Moon is moving away from the Sun. Depending on the position of the Sun, Moon, and observer, it is unlikely to be visible until at least 15 hours after New Moon, and in some cases not until 24 hours after New Moon.
Someone will be the first to see the thin crescent after the New Moon. This occurs when the Sun is far enough below the horizon to make the sky "dark" and the Moon is far enough above the horizon (and away from the Sun) to be "clear" of the sky's glow. ("dark" and "clear" are relative. The sky will still be quite bright but not like the middle of the day.)
Observers that are farther east than the first person would not be able to see the Moon because the time would have been earlier. That is, the Moon would have been closer to the Sun and not visible.
Observers that are farther west should be able to see the Moon more easily because it is later, and therefore the Moon is farther from the Sun. In other words, the longitude affects how easy it is to see the new Moon.
Whether it is easy or difficult to see the new Moon also depends on the latitude because this can affect where the Moon is relative to the Sun and horizon. This variation changes throughout the year.

It looks like Kerala is on the west coast, so they may have been in the position to see the Moon in that 15 to 24 hour time range whereas it would have been too early for people further to the east to see it.

Kerala is also farther south than the rest of India. I have not done the calculations for this New Moon, but it is possible that being south and west increased the possibility of seeing the Moon before people further north or east would see it. If this is the case, then Kerala would have been in the best place to be the first to see the new Moon -- assuming the sky was clear :-).

The image shows that, according to the calculations by https://www.moonsighting.com/visibility.html, the new moon might be visible from only the far South-Western coast of India.


How does Io's atmosphere behave locally near volcanic plumes?

My impression is that for answering your question, one would actually have to run simulations, ideally so-called global circulation models (GCM). If this is for a research project, the MIT GCM would be a good candidate to set up for the atmosphere of Io.
Of course, there are research groups working on that issue, e.g. the team of Prof. Goldstein at the University of Texas in Austin, and probably some more.
If you think this is like using a sledgehammer to crack a nut, you could also think about local plume modeling, I found a paper entitled A new strategy for the estimation of plume height from clast dispersal in various atmospheric and eruptive conditions which seems to come along with Matlab code:

Highlights [of the quoted paper]

A new revised numerical strategy for the determination of plume height is presented.
Accounting for wind, particle shape, gravitational spreading and climate zone.
Wind significantly affects the clast support envelope of weak and transitional plumes.
The new method is available through a Matlab package and a series of nomograms.

Edit/ Warning. Jean Marie Prival says

Being a bit familiar with the Rossi et al. (2019) plume model, I don't think it can apply here: it needs isopleth data as input, i.e., it needs maximum grain-size data from multiple locations in the field... (It is generally used to infer maximum plume height of past eruptions for which we don't have direct plume observations/measurements, but only tephra deposits.) Also, the model is parametrized for the Earth: things like particle density, tropopause height, etc., which are all different on Io

Last but not least, I would aim for some back-of-the-envelope estimates. For such a Fermi estimate we have to make some assumptions about the plume dynamics, and we also need some numbers. For that purpose, I found an Wired article by Erik Klemetti: Just How Hot is Io's Volcanic Surface? which might help us to estimate the state of the atmosphere, given by $T$, $p$, and humidity on Earth and probably ${\rm SO_2}$-content on Io.
Up to now I sketched 3 approaches to solve the problem in prinicple, without giving an answer. That's why I want to close my post by quoting the above mentioned Computational Fluid Physics Laboratory from U Texas which shows the result of their simulations.

For me it looks like the plume material shoots out of the atmosphere (and probably carying along parts of the atmosphere with it), before slowly setteling down on Io's surface. The plume height in the above picture is about ${145 \rm km}$, other websites report plumes of ${300 \rm km}$ height. It is a global phenomena which influences the overall atmosphere, probably causing wave-pattern all over the moon's atmosphere.
Related

I asked Characteristics of Venus' jet streams? which about the atmosphere on another object, but touches the same issue of running GCMs for other celestrial objects with atmosphere.
I also found The collapse of Io's primary atmosphere in Jupiter eclipse pretty interesting. The following paragraph supports my hypothesis that the volcanic plumes shoot out into the exosphere, as well as your statement, Ioannes, that the atmosphere is highly variable.

If the predominately ${\rm SO_2}$ atmosphere is in vapor pressure equilibrium with the large reservoir ${\rm SO_2}$ ice that covers much of Io's surface, as Io enters eclipse and is no longer heated by the Sun, the decreasing surface temperatures will cause the primary atmosphere to collapse out onto the surface [Saur and Strobel, 2004]. The atmosphere, with a scale height of 10 km, would transition from a collisionally thick atmosphere, with a mean free path (l) = 1 m, to an exospheric atmosphere (l ~100 km). Conversely, if the atmosphere remains constant during eclipse, it would point to a volcanically supported atmosphere.


Is there a complete edition of the Cambridge Catalogue of Radio Sources (esp. 3C) available digitally?

You can use VizieR to retrieve the whole catalogue (3C and revised 3C).

VizieR is an online astronomical catalogue that allows you to access a quite comprehensive list of catalogues (16,929, as of today); you can easily query and extract all kind of data from this online tool.


Accelerating universe expansion and standard candle

Most type Ia supernovae are thought to arise from the thermonuclear detonation of white dwarfs that are composed almost entirely out of carbon and oxygen.

These white dwarfs are the cores of relatively low-mass stars that have lived their lives, gone through stages of core hydrogen and helium burning, leaving behind degenerate carbon/oxygen cores that become cooling white dwarfs after the outer envelope has been shed during the asymptotic giant branch and planetary nebula phases. As such, their composition, at least to first order, is almost independent of the initial composition of the star from which they were formed. That is, even if the progenitor star had a very low initial metal content, the white dwarf produced would still be almost exclusively a carbon/oxygen mixture, which had a similar Chandrasekhar mass and a similar explosive potential.

It is well known however that not all type Ia supernovae are the same. It has long been known that their light curves are subtley different and there is a so-called stretch factor that can be applied to get a "corrected" peak magnitude. a.k.a The width-luminosity relation.

More recently there has been a realisation that type Ia supernovae could arise from both accretion or mergers and there is clear evidence that the amount of radioactive Ni varies from explosion to explosion. A very recent paper by Milne et al. (2015) has however challenged the view of metallicity independence. They claim there are two populations of type Ia SNe, connected with progenitor metallicity, and that these populations become more apparent at high redshift when looking at rest-frame ultraviolet emission. The gist of their conclusions is indeed, as your question supposes, that this may go some way to ameliorating (but not eliminating) the need for dark energy.


In this image, where would the antumbra be?

There is no antumbra in this image.

An antumbra only exists when the moon is further from the Earth than in that image, so that the moon appears smaller in the sky than the sun.

In that case, when the moon is perfectly aligned with sun, a ring of sun is visible behind the moon. The region in which the ring is visible is called the antumbra.

In the image, the moon is closer to the Earth, so when the moon and sun are aligned, the moon entirely blocks out the sun, and so there is a region of umbra.

That image shows a total eclipse. An antumbra is present only in an annular eclipse.


What is equation of straight line for a ideal blackbody on the color color diagram?

Use Planck's law: $$F(\lambda,T) = c  \lambda^{-5}  (e^{d/(\lambda  T)} - 1)^{-1}$$.

Since magnitude is defined as, $M = -2.5 \log_{10}(F) + \textrm{constant}$, the color index $A-B$ is
$$(A - B) = -2.5 \log_{10}(F(\lambda_A,T)/F(\lambda_B,T))$$.

To answer your question, use an appropriate effective wavelength of $U,B,V$. Link $U-B$ and $B-V$ via temperature, you should get $U-B = f(B-V)$.


What is projected separation and how can I make sense of its unit (h^-1 kpc)?

It means the separation between the two objects if they were both at the same distance away. This separation is found by multiplying the angular separation (in radians) by the distance to the objects.

Often, the distance to extragalactic objects depends on the assumed value of the Hubble parameter. In this case, $h$ is a dimensionless number corresponding to $h=H_0/100$, where $H_0$ is the Hubble parameter in km/s per Mpc.

The current best value of $H_0$ is about 70 km/s per Mpc, so $h\simeq 0.7$.


If a black hole has a mass similar to a star, why does it have a different gravitational pull?

Actually, it doesn't. If a black hole and a star have the same mass, their gravity is the same. As the formula for the attraction (gravity) between two objects is given as $F=\dfrac{GMm}{r^2}$, where $G$ is the gravitational constant, $M$ is the mass of the black hole/star, $m$ is the mass of the smaller object, and $r$ is the distance between the two. Even though a black hole is tiny and a star is much, much bigger than a black hole, the radius of each object does not matter. It only depends on mass and distance.
The reason why it seems like the gravitational pull is stronger is because the black hole indeed is much smaller than the star with the same mass. If you had a $1 M_\odot$ star and a black hole of the same mass, at distances less than $1 R_\odot$ would mean that you would be inside the star,  and still outside the black hole. Therefore, you would feel gravity from the black hole and accelerate toward it, while (assuming you weren't vaporized by the Sun's internal temperatures, you would experience drag from the hydrogen gas, as well as (even if the gas was essentially dragless) more mass being concentrated in the other direction (away from your vector of falling) as a star is not a point-like particle, its mass is distributed throughout the star.


See Where can I find/visualize planets/stars/moons/etc positions? for an extremely general answer on how to compute positions, but the files you're looking for specifically are the "leap second kernels" at http://naif.jpl.nasa.gov/pub/naif/generic_kernels/lsk/

There was a mild kerfuffle on spice-discussion lists when NASA failed to update the kernels after the end-of-2016 leap second was announced, but they have updated it now.


How do you calculate the lookback time distance to a given galaxy?

No you can't. Other information is required.

For low redshifts - let's say smaller than 0.1 - and by that I mean the wavelength increases by 10 percent, you might get away with using Hubble's law to estimate the distance and then get the look back time by dividing by the speed of light.
$$ t \simeq \frac{\lambda - \lambda_0}{H_0 \lambda_0},$$
where $H_0$ is the present day Hubble parameter of about 70 km/s per Mpc, $\lambda$ is the measured wavelength and $\lambda_0$ is the rest wavelength.

So far so good, you just need to know $H_0$. However for larger redshifts it gets horribly complicated because the Hubble parameter changes with time in a way that depends on the curvature of the Universe and hence on the cosmological parameters defining the matter density and dark energy density.

There is indeed a highly complicated, non- linear formula involving integrals that I will look up a reference for. But possibly the better way to proceed is to use a simple look-up table produced from such calculations with certain assumed values for ratios of matter and dark energy densities with respect to the critical density; a.k.a. $\Omega_M$ and $\Omega_{\Lambda}$.

The plot below is an example taken from http://www.astro.caltech.edu/~eran/MATLAB/Cosmology.html which shows look back time versus redshift for two different cosmologial models (but with the same value of $H_0$). The curves are very different at high redshifts, but converge at small redshifts.

Here is a cosmological calculator that can do the job for you. Enter the redshift and your assumptions about the Hubble parameter and other cosmological parameter and it will tell you the age of the universe at that redshift as well as the lookback time.


How much would you weigh if Earth stopped orbiting the Sun?

In a sense, this is one of those "If I break the laws of physics, what do the laws of physics say will happen" kinds of questions.

Ignoring that, a 150 pound person would still weigh about 150 pounds if the Earth miraculously stopped orbiting the Sun. Almost all of a person's weight results from resistance to the Earth's gravity field. The tidal gravitational forces from the Moon and Sun are on the order of 10-4 ounces of force.

Note: I'm answering the question from the perspective of "what would a bathroom scale say". A doctor's scale would say a 150 pound person weighs 150 pounds regardless of whether the person is standing on the Moon, the Earth, or a very massive planet. A doctor's scale measures mass. A bathroom scale measures apparent weight.

So what would a bathroom scale say? If the Earth is at one astronomical unit from the Sun, whether the Earth is orbiting at 30 km/s or falling straight toward the Sun doesn't matter one iota. All that matters is that the Earth is one AU from the Sun. In both cases, that the Sun is 1 AU distant makes a 150 pound person weigh about $1.8 \times 10^-4$ ounces less at noon or at midnight than at sunrise or sunset.

The difference between the orbiting Earth and the in-falling Earth is that these tidal forces are more or less constant in the case of the orbiting Earth. They grow ever larger in the case of the in-falling Earth as it gets closer and closer to the Sun.

These tidal forces become quite large by the time the surface of the Earth just touches the surface of the Sun. Now our 150 pound person would weigh only 73.5 pounds per a bathroom scale if he was standing at that point where the surface of the Earth just touches the Sun. He would also weigh about the same at the point on the Earth furthest from the Sun. He would however weigh considerably more, about 188 pounds, if he happens to be at a point where the Sun is on the horizon.

If, on the other hand, some mysterious, magical force makes the Earth stop orbiting the Sun and makes this change permanent (i.e., the Earth stops moving with respect to the Sun), a 150 pound person closest to the Sun would weigh about 1.4 ounces less, while a 150 pound person furthest from the Sun would weigh about 1.4 ounces more. A 150 pound person who sees the Sun at the horizon would still weigh about 150 pounds in this scenario.


Could the Sun's bending of light be measured on photographic plates before Einstein's prediction?

I suggest reading the paper on the 1919 expedition to get a clearer picture of why they did it at that time and why it hadn't been done before. From reading chapter 2 I think the main reason was the astrophotography equipment required for the experiment and the alignment of bright enough stars close to the sun to observe the effect.

Of course before Einsteins prediction in 1911 no one had any specific reason to observe the stars close to a solar eclipse before and during totality, it might seem obvious now but that's hindsight for you.


If objects don't move when the universe expands, how can the expansion result in redshift?

The redshift is indeed not caused by the movement of the objects, but by the expansion itself. This is a theoretical result from considering the FLRW metric for a light ray traveling on a null geodesic.

A derivation can be found here.

The cosmological redshift $z$ is given by $z+1 = 1/a$, where $a$ is the scale factor (the size of the Universe) at the time the light was emitted. Notice that this result doesn't care how the Universe expanded. In principle, you could imagine a universe which was static when the light was emitted, and static when the light is observed, but expanded by the factor $a$ at some point in between, and you'd observe the same redshift as if that universe had expanded gradually in the meantime. This is in sharp contrast to a normal Doppler shift, where in such a hypothetical case, the would be no observed redshift.

Note also that galaxies receding faster than the speed of light $c$ is no hindrance for us to see them; all galaxies with a observed redshift larger than $z\simeq1.4$ recede faster than $c$. The most distant observed galaxy GN-z11 has a redshift of $z=11.09$, and recedes at $v\simeq2.2c$. When it emitted the light we see, it receded even faster, namely at $v \simeq 4.3 c$. See the ant on a rubber rope "paradox" for an intuitive explanation.


Why were the Magellanic Clouds named that way?

Michel Dennefeld has written an arXiv paper, "A history of the Magellanic Clouds and the European exploration of the Southern Hemisphere" which goes into the history of the European discovery of the Magellanic Clouds and the history of the name. The term Magellanic clouds appears to have originated with sailors, as noted by Edmond Halley in his 1679 "Catalogus Stellarum Australium":

His work is known because of
the inclusion of a new (fleeting) constellation, Robur Carolinum, to please King Charles II. He
mentions in an appendix that:

Proxima, ex iis quas observavi, est in Cauda Apodis, in distancia paulo ultra 8 graduum, Duae
Nubeculae, quae a Nautis Nebulae Magellanicae appellantur, exacte referent Galaxiae
albedinem & Telescopio inspectae, hinc inde Nebulas Parvas & exiguas Stellas ostendunt…

It appears it was clear to him that the name Magellanic Clouds was used only by sailors, and
that those Clouds resemble the Milky Way.

Dennefeld goes on to describe the gradual spread of the term "Magellanic Clouds" beyond the nautical community, e.g. in Nicolas de La Caille's 1755 catalogue:

But his words,
“que nous appelons nuées de Magellan”, seem to indicate that around that time the
association of the Clouds with Magellan was already spreading beyond the nautical
community, even if the scientific term was still simply Nubeculae or les Nuages.

The astronomical usage switched from "Nubeculae" around the time that Latin was abandoned as the language of science, and the first usage in the scientific literature seems to have been by John Herschel:

John Herschel (1847), in a section however entitled "On the two nubeculae or
Magellanic Clouds", talks only about Nubeculae. Only in the accompanying figure (Fig. 8),
where he gives his visual observations (figure where it is however hard to recognise the
Clouds), does he mention again, "The two Magellanic Clouds as seen with the naked Eye".
So he uses both the scientific denomination and the more public name, and Herschel seems
to have been the first one to use the name Magellanic Clouds in a scientific publication.

So it appears to be that the name became Magellanic Clouds because this is the name that was originally used by European sailors, that then gradually became the common name as Latin was abandoned.


How large can a ball of water be without fusion starting?

You really need a full-blown stellar evolution model to answer this precisely and I'm not sure anyone would ever have done this with an oxygen-dominated star.
To zeroth order the answer will be the similar to a metal-rich star - i.e. about 0.075 times the mass of the Sun. Any less than this and the brown dwarf (for that is what we call a star that never gets hot enough at its centre to initiate significant fusion) can be supported by electron degeneracy pressure.
A star/brown dwarf with the composition you suggest would be a bit different.  The composition would be thoroughly and homogeneously mixed by convection. Note that other than a thin layer near the surface, the water would be completely dissociated and the hydrogen and oxygen atoms completely ionised. Hence the density of protons in the core would be lower for the same mass density than in a "normal star".  However, the temperature dependence is so steep I think this would be a minor factor and nuclear fusion would be significant at a similar temperature.
Of much greater importance is that there would be fewer electrons and fewer particles at the same density. This decreases both the electron degeneracy pressure and normal gas pressure at a given mass density.
The star is therefore able to contract to much smaller radii before degeneracy pressure becomes important and can thus reach higher temperatures for the same mass as a result.
For that reason I think that the minimum mass for hydrogen fusion of a "water star" would be smaller than for a star made mainly of hydrogen.
But how much smaller? Back-of-the-envelope time!
Use the virial theorem to get a relationship between perfect gas pressure and the temperature, mass and radius of a star. Let gravitational potential energy be $\Omega$, then the virial theorem says
$$ \Omega = -3 \int P \ dV$$
If we only have a perfect gas then $P = \rho kT/\mu m_u$, where $T$ is the temperature, $\rho$ the mass density, $m_u$ an atomic mass unit and $\mu$ the average number of mass units per particle in the gas.
Assuming a constant density star (back of the envelope) then $dV = dM/\rho$, where $dM$ is a mass shell and $\Omega = -3GM^2/5R$, where $R$ is the "stellar" radius. Thus
$$\frac{GM^2}{5R} = \frac{kT}{\mu m_u} \int dM$$
$$ T = \frac{GM \mu m_u}{5k R}$$
and so the central temperature $T \propto \mu MR^{-1}$.
Now what we do is say that the star contracts until at this temperature, the phase space occupied by its electrons is $\sim h^3$ and electron degeneracy becomes important.
A standard treatment of this is to say that the physical volume occupied by an electron is $1/n_e$, where $n_e$ is the electron number density and that the momentum volume occupied is $\sim (6m_e kT)^{3/2}$. The electron number density is related to the mass density by $n_e = \rho /\mu_e m_u$, where $\mu_e$ is the number of mass units per electron. For ionised hydrogen $\mu_e=1$, but for oxygen $\mu_e=2$ (all the gas would be ionised near the temperatures for nuclear fusion). The average density $\rho = 3M/4\pi R^3$.
Putting these things together we get
$$h^3 = \frac{ (6m_e kT)^{3/2}}{n_e} = \frac{4\pi \mu_e}{3}\left(\frac{6 \mu}{5}\right)^{3/2} (Gm_e R)^{3/2} m_u^{5/2} M^{1/2}$$
Thus the radius to which the star contracts in order for degeneracy pressure to be important is
$$ R \propto \mu_e^{-2/3} \mu^{-1} M^{-1/3}$$
If we now substitute this into the expression for central temperature, we find
$$ T \propto \mu M \mu_e^{2/3} \mu M^{1/3} \propto \mu^2 \mu_e^{2/3} M^{4/3}$$
Finally, if we argue that the temperature for fusion is the same in a "normal" star and our "water star", then the mass at which fusion will occur is given by the proportionality
$$ M \propto \mu^{-3/2} \mu_e^{-1/2}$$ .
For a normal star with a hydrogen/helium mass ratio of 75:25, then $\mu \simeq 16/27$ and $\mu_e \simeq 8/7$.  For a "water star",  $\mu = 18/11$ and $\mu_e= 9/5$. Thus if the former set of parameters leads to a minimum mass for fusion of $0.075 M_{\odot}$, then by increasing $\mu$ and $\mu_e$ this becomes smaller by the appropriate factor $(18\times 27/11\times 16)^{-3/2} (9\times 7/5\times 8)^{-1/2} = 0.173$.
Thus a water star would undergo H fusion at $0.013 M_{\odot}$ or about 13 times the mass of Jupiter!
NB This only deals with hydrogen fusion. The small amount of deuterium would fuse at lower temperatures. A similar analysis would give a minimum mass for this to occur of about 3 Jupiter masses.


No, the Main Sequence is more like a starting line. Most stars spend a long time on one point of it (10 billion years for the sun) while they fuse hydrogen into helium. Then they wander off it.

In this diagram the black line is the Main Sequence. The colored lines show temporal sequences. This progress through time of an individual star is called its evolutionary track.

The numbers along the black line are solar masses (1=the sun). This diagram evolved through a sequence of images by Wikimedia users Rursus, G.A.S, and Jesusmaiz.


Do extraterrestrial landscapes or planets count as nature?

ARTFL Nature:

  The existing system of things; the world of matter, or of matter and mind; the creation; the universe.

That definition includes the surface of Mars, and the core of Betelgeuse.


A quasar is simply an active galactic nucleus (AGN) that is viewed from a particular angle; see the picture below, in which quasars are labeled "QSO". This is really a remarkable figure because historically all of the names in the figure were thought to correspond to different types of objects, when really they all refer to the same thing!

Your question really shouldn't be "Was there ever a quasar in the Milky Way?", since the dotted line in the figure would correspond to the Galactic plane and we would not see Sagittarius A* (the Milky Way's super-massive black hole) from the correct angle. A better question might be, "Has Sagittarius (Sgr) A* ever been active?" The answer to that question is yes; according to this page it was probably active (very bright with a jet) about 10,000 years ago. However, at the moment, it isn't really doing anything, since it isn't currently accreting anything (to put it plainly, it isn't eating anything, so it doesn't have enough energy to be active). However, many astronomers (myself included!) are anxiously waiting for a cloud of gas called G2 to fall into Sgr A*. We are hoping that Sgr A* will burp or do something interesting.


Why does the Hertzsprung–Russell diagram's x-axis go from large temperatures to lower?

Originally, what was plotted was luminosity against colour, and by colour I mean the wavelength of the peak intensity. Short wavelengths on the left, and long on the right, as you would expect.

Now since stars emit (nearly) black body radiation, there is a close relationship between colour and temperature. I suspect that the reason that the x axis isn't inverted when temperature is used is just inertia:- We have always plotted out H-Z diagrams like this and we will keep doing so.


Is the gravity on Kepler 39b higher than on the Sun?

You are probably overlooking why the radius is what it is, or equivalently why the planet is denser than the Sun.

The Sun is undergoing fusion in its core, which releases energy, which causes it to expand (though it achieves an equilibrium state, as expansion decreases rate of fusion which leads to contraction, which leads to accelerated fusion which leads to expansion...).  Jupiter-like bodies, however, don't produce heat from fusion (except maybe some deuterium fusion very early in their lives); they slowly cool and contract over time.  So such a body can compress into a denser state, allowing it achieve higher surface gravity (give or take difficulties in defining "surface" for a gaseous body).

Eyeballing the calculation, you can see that the sun is 10 times the radius of Jupiter and 1000 times the mass, so it's surface gravity should be about $1000/10^2=10$ times greater than Jupiter's, which is what you find.  But Kepler 39b is 18 times more massive than Jupiter and only marginally larger. That's a big uptick in density. So that it's surface gravity would be about $18/(1.25)^2=11.52$ times stronger than Jupiter's (also what you find).  And $11.52>10$, so Kepler 39b has higher surface gravity.


Pinhole Projector: Can I use a specific solar optical eye piece to focus the Transit of Mercury?

The whole point of a pin hole projector is that it doesn't need any optics just a pin hole.

If you are after magnification then that is related to the distance you project the solar image behind the pin hole, with a projection distance of 1000mm the solar disc will have a diameter of about 9mm. Increasing the distance will increase the size of the image but also reduce its brightness. Increasing the size of the pin hole will increase the brightness but reduce the sharpness of the image.

I suppose you could use a handheld magnifying glass to look at a small image but would not personally recommend it as unless you are extremely careful there is always a risk of fire or even blinding yourself or someone else when using a lens near an unfiltered solar observation. There is a famous story of an experienced astronomer setting his beard on fire because he left the cap off his telescopes finder.


Birth and Evolution of Millisecond Pulsars

The standard model for the formation of millisecond pulsars is the recycling scenario.

The starting point of this model is a stellar binary system consisting of a neutron star and a secondary 'regular' star, where the neutron star is a normal pulsar with a rotational period of roughly $\sim7$ seconds.

Because a neutron star is such a compact object, its gravitational influence is very large ($F_g \propto M/R^2$). If the two stars are close enough, the outer layers of gas may be pulled of the companion star and flow towards the neutron star. All this matter will settle into an disk structure around the neutron star and start orbiting faster and faster as it draws closer to the surface ($v_{orbit} = \sqrt{GM/R}$).

By the time this matter touches down on the neutron star surface its orbital period can be as fast as a millisecond, certainly significantly faster than the neutron star. This bit of motion (or more precisely, the angular momentum) is then transferred to the neutron star (this process is called accretion), causing it to rotate slightly faster (spin-up).

After some $0.1\sim0.2 M_{\odot}$ (where $M_{\odot}= 2 \times 10^{30}$ kg gives the mass of the Sun) of material is transferred to the neutron star, its rotational period will have decreased from a few second down to milliseconds.

The transfer of this much matter takes a very long time, roughly $10^8$ to $10^9$ years. After this time, and with all the redistribution of mass, the binary system is expected to widen which causes the mass transfer to stop. As the region around the neutron star clear up, the radio emission mechanism turns back on (this mechanism does not work during the accretion stage) and we are left with a millisecond pulsar.

From this point on the neutron star will behave like a regular pulsar: over the course of a very long time it will convert its rotational energy in radio emission and slow down until it becomes to slow to act as a pulsar.

As for the evidence for this model; such binary systems containing an accreting neutron star emit massive amounts of X-ray emission (hence called X-ray binaries) and have been known to exist since the 60's, long before the first millisecond radio pulsar was discovered (in 1982). This entire scenario was proposed quickly after, but could not be confirmed for a very long time.

It was only in the late 90's that a neutron star was found whose X-ray emission was pulsed at millisecond periods (see this, and this), confirming that these neutron stars are indeed spinning up. The final piece of evidence; a millisecond pulsar which switches back and forth between X-ray and radio emission was only found very recently.


1) Nearly all black holes are Kerr black holes.

2) There are roughly 100 million black holes estimated in our galaxy.

3) There is a confirmed black hole about 7800 light years away.

4) There is a potential black hole about 1500 light years away.

5) It is very likely that there are black holes even closer than that, but they are difficult to detect.

Some of your assumptions are wrong, but for now the closest known potential black hole is still a pretty good distance away. Traveling that far with our current technology is impossible, and overcoming that hurdle is not trivial. Regardless, using a Kerr black hole for time travel may not be feasible anyway.


Anything which casts a shadow can become a sundial, as JH's comment points out.

The other thing you need to know is the latitude of the building, as that affects the change in shadow direction per hour of elapsed time.  Worse, due to the eccentricity of Earth's orbit (and rotation axis), there's an offset between clock time and sundial time which varies over the year.  You can get most of the formulas at the Wikipedia page .

There's a handy calculator at blocklayer.com


Could evaporating hot Jupiters have metallic hydrogen on their surfaces?

Metallic hydrogen is an odd substance. When you push hydrogen atoms very close together, their electrons can come free, and move around, instead of being tightly bound to the atomic nuclei. As this form of hydrogen would conduct electricity, it behaves like a metal. At least this is the theory. Nobody has been able to produce enough pressure to actually make any metallic hydrogen in the lab.

However, the pressure inside Jupiter should be high enough to form metallic hydrogen. In extrasolar planets there could be large amounts of metallic hydrogen too.

However as soon as you release the pressure from metallic hydrogen, it turns back to normal molecular hydrogen. So it could not exist on the surface of a "hot jupiter", even one from which the outer layers had been stripped away by the solar wind. The metallic hydrogen that had been in the interior would change back into molecular hydrogen as it approached the surface.


If a star orbited another star, how would that affect parallax measurements of the other star from the vantage point of the first?

The other star would have a significant proper motion.

Over the course of one orbit, it would appear to travel right around the sky once. The variation in position due to the Earth's orbit of the sun would be superimposed on top of this motion.

Except that the proper motion of the star, and its parallax would be much greater than for any real star, this is not really different to the way in which the parallax motion of a star is superimposed on its proper motion.


Refraction by Saturn's atmosphere - how dense is it here?

This is a hard question to answer without hard numbers. When you look at this image, you can clearly see that the refractive power of the atmosphere changes as you look closer to the limb of Saturn. The rings start off being refracted slightly, then get more refracted as you get closer to the limb. That essentially implies the portion of the ring closer to the limb is traveling through lower and denser regions of Saturn's atmosphere and thus being refracted more. In that sense, there isn't a single answer here since there is a range of refraction going on.

My first approach in considering this problem is to consider optical depth. In case you don't know what this is, optical depth, usually denoted by $\tau$, is a dimensionless number which is a measure of how much material light must pass through and interact with when traveling through a medium. The way optical depth is defined is such that a medium with $\tau\lt1$ means light can get through, while $\tau\gt1$ means light cannot get through. A more conceptual way of thinking about optical depth is to say that it represents, on average, how many atoms a particle of light will hit when trying to travel through a medium. That is, if $\tau=0.5$, then about half the light particles will hit something and be lost, while the other half will get through unscathed. If $\tau = 2$, then on average the light particles will hit two atoms and most light will be scattered/absorbed out of your line of sight, making you unable to see through that medium.

A simplified definition of optical depth is given by

$$\tau=\sigma N \ell$$

where $\sigma$ is the collision cross section of the atoms in the medium, in $m^2$, $N$ is the number density of atoms in the medium, in $m^{-3}$, and $\ell$ is the path length of light through that medium, in $m$. So let's finally take a look at your questions.

  Roughly speaking, what is the density where this is taking place? Is it rarefied compared to earth but simply a very very long path length, or is this happening at a density far larger than Earth's surface atmosphere?

Ultimately, you're asking what $N$ and $\ell$ are such that $\tau=1$, since this would be the point where the refraction stops and you don't see any more of the rings. Unfortunately, this is difficult to answer based on that image alone. At best you can look up a radial density profile for Saturn such as in Nagy et al. 2009 (ref. fig. 8.1). Note in that paper, they refer to height as the height above the 1 bar pressure level. This paper also gives you an idea for common atoms within the upper atmosphere and you can look up $\sigma$ for these atoms. Predominantly though, I imagine hydrogen will dominate the interactions so using the cross section of hydrogen will be a good approximation. Then assume some sort of approximate path length $\ell$ based on the geometry (maybe use the formula for chord length). Putting actual numbers into this problem could be a research project all to its own, but you can play around with what you consider to be good values for $N$ and $\ell$ such that you're constrained by the fact that $\tau=\sigma N\ell=1$.

  How large is this refraction - like 0.1 degrees or 10 degrees?

This is even harder to answer. The only way I know to determine something like this is to run a full ray tracing model of light through Saturn's atmosphere. I can hazard a guess that the most refraction you're seeing there is on the order of 10 degrees. This is supported by Dalba et al. 2015 who actually did a ray tracing analysis of light through Saturn (in regards to planetary transits) and found refraction angles on that order (ref. fig. 2).


Focusing light through fiber optic cable to extend telescope viewer

In principle a fiberscope eyepiece could work but not very well.
By putting one end of the cable flush with the field stop of an eyepiece (few allow this) and the other end exactly in the telescope's focal plane, you'd get a fiber-pixelated image.
A binoviewer would require two cables, difficult to align.


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.


Could there be life beneath the surface of Mars or moons?

That's one of the big questions. ESA scientists, at least, think it's worth looking for underground life. See ExoMars mission.

More likely than earth worms are microbes, since some microbes on Earth live under similar conditions as presumed for underground Mars.


Can New Horizons be used to measure the distance to Betelgeuse (despite its fickle photocenter)?

In principle yes, in practice no. The telescope is good enough, but the CCD camera will saturate, preventing a good positional measurement.
Salient facts. The parallax to Betelgeuse as seen between Earth and New Horizons will be about 250 mas. The current distance uncertainty is ~20%, so to better that, the position of Betelegeuse measured by New Horizons would need to be more precise than ~25 mas.
Looking at the LORRI camera, it has a plate scale of 4.95 microradians/pixel and a point spread function with FWHM of 1.8 pixels, which equals 1.84 arcsec (Cheng et al. 2007).
A useful rule of thumb is that your positional accuracy in a resolved image is given approximately by
$$ \Delta \theta \sim \frac{ {\rm FWHM}}{{\rm SNR}}\ ,$$
where SNR is the signal-to-noise ratio of the image. Using this, it suggests that if one could get an image of Betelgeuse with a SNR of around 100, then you should be able to centroid the position to better than 25 mas using the New Horizons data. Combining this with a simultaneous Earth-based image (which one presumes would be much more precise), then the parallax could be determined to better than 10%.
The problem appears to be the brightness of Betelgeuse. Cheng et al. also give a photometric calibration formula for the number of analogue to digital units per second, $S$, recorded by the camera for a point source with a given $V$ magnitude.
$$V = -2.5 \log S + 18.94 + 0.4\ ,$$
where the extra $0.4$ is probably appropriate for a star of Betelgeuse's red colour.
Using this formulae we see that $S = 2.16\times 10^8$ ADU/s, assuming $V\sim 1.5$ for Betelgeuse.
The CCD camera on LORRI only returns 12 bits and has a frame transfer time of 13 ms, which means that 13 ms is about the smallest exposure time you could use and hope to get accurate positional information (Cheng et al. 2007), though in principle a 1 ms exposure time could be done. Even in 1 ms you would get $2.16\times 10^5$ ADU from Betelgeuse spread over perhaps 4 pixels. Given that a 12 bit ADC saturates at 4096 ADU, then Betelgeuse is more than 10 times too bright and will produce a completely saturated image whose centroid cannot be determined with any precision. If it were possible to remotely reset the gain on the CCD (currently 22 electrons/ADU) to something much larger, then maybe it would be possible.


Why do objects all orbit in the same direction?

This video provides a visual answer: https://youtu.be/MTY1Kje0yLg?t=3m24s
objects orbiting in different directions are more likely to collide with each other. By the process of elimination, eventually one direction of orbit dominates.


Did uncertainty principle triggered big bang?

I think there is a real question in there, just not the one that you think.

Gerald isn't saying that the uncertainty principle played a role in the big bang.   He's saying it played a role in and shortly after the Planck era which came after the big bang.

On the big bang - nobody knows.   Really, that's the answer.  "Was the big bang this" answer "We don't know".   Period.   End of discussion.

We don't know what came before the big bang, We don't know if space was empty or not empty.  We don't know what physical laws may have applied.  Nobody knows.   The initial, basic theory was that the big bang happened at a point or a singularity, which if you think about it, doesn't make much sense cause gravity wins at a point that dense, but the big bang at a point idea does resemble, kinda/sorta a really crazy particle that, by something kinda like quantum tunneling, could become a universe - er, maybe.   String Theory brings other dimensions into this particle becomes a universe - idea, and again, maybe, but nobody knows.

A more modern model, but still, we don't know, is that the universe was infinite at the time of the big bang expansion, and it grew, not from a point but it expanded from infinite into infinite, which works mathematically and it makes (I think) a bit more sense than the super-particle model.  Still the answer is "We don't know".   The answer is always "We don't know".   There's also 4 dimensional models or two time dimension models (those are my favorite when you have x-axis and y-axis time), which are creative ways to avoid the singularity but . . . I'm repeating myself.  Nobody knows.  The answer to what happened at the big bang is always nobody knows.  We don't know if it was empty or not, we don't know if there was some 4th dimensional crossing of membranes which caused the 3d universe we live in.

Now, on modeling back to the big bang, the experts can run the model backwards and apply known laws and have a theory that works to a time shortly after the big bang, which still isn't certainty, but it's a reasonable hypothesis using the laws of physics as they work today.  But it doesn't work all the way back to the big bang.

So, the paragraph that you quote, I think there's a solid question in there, but the quote doesn't mean what you think.

  You may think of the pre-Planck and the Planck epoch as a state of the universe, which is dominated by quantum fluctuations. They don't need a reason or a cause to occur. They are a consequence of Heisenberg's uncertainty principle.

  So,Did any law like uncertainty principle caused this to be happened or any somethings?

Gerald is talking about a period AFTER the big bang and what the models tell us about that period when the universe was, (theoretically), very dense and very hot and governed by quantum mechanics.

Perhaps a better question, which I'm not qualified to answer but might be worth asking, is, "how did the uncertainty principle apply during and shortly after the Planck Era" - at least, that's the question I take away from the paragraph you quoted.   Gerald didn't say the uncertainty principal applied to the big bang, only to the Planck Era.


Are Sun-Earth L1 & L2 points generally considered outside the Earth's Hill Sphere?

I'm going to answer this in a definition-based manner.

Here is the definition of a Hill Sphere according to Wikipedia:

  The Hill sphere or Roche sphere of an astronomical body is the region
  in which it dominates the attraction of satellites.

Here is the definition of a Lagrange point according to Wikipedia:

  In celestial mechanics, the Lagrangian points are the points near
  two large bodies in orbit where a smaller object will maintain its
  position relative to the large orbiting bodies.

So, if you think about it - L1 and L2 are on the edges of Earth's Hill Sphere - technically, neither inside nor outside, because neither of the two objects dominate gravitational attraction. This is stated on the Wikipedia page for the Hill sphere - and userTLK alluded to this in his comment as well:

  In the example to the right, Earth's Hill sphere extends between the
  Lagrangian points L1 and L2, which lie along the line of centers of
  the two bodies.

The Lagrange points are where the Sun's and Earth's gravitational influences have no net effect on an object. So, if an object at L1 moved ever so slightly towards the Sun, then the Sun would dominate the attraction of that object, meaning it would be inside the Hill sphere of the Sun. The same goes for the Earth, just in the opposite direction. (The conditions for L2 are different, but detailing them likely isn't relevant to the answer.)


Is it possible to filter radio-waves using another radio telescope?

Coherent interfering signals are usually referred to as RFI (or Radio Frequency Interference).  One could certainly digitally filter out a coherent interfering signal from a telescope by observing from an additional telescope focused on the source of the interfering signal, given the proper geometry.  Here is a research paper on Radio Frequency Interference Mitigation that outlines some of these very techniques in section 3.4: Post correlation studies.
Incoherent interference is typically referred to as noise, and won't be the same at the two telescopes. This means an astronomer can't use one telescope to filter out the noise at the other. Some examples of noise sources are:

Noise that comes from thermal from the telescope's electronics
Ground backscatter into the antenna receiver
Inaccuracies introduced by analog to digital conversion and other signal processing.


Adaptive Optics Shack-Hartmann Sensor and Deformable Mirror

Speaking as one who has dealt with those equations (and those problems) since 1984.

Yes, the simplest approach is to use matrix methods to force the phases at the edges of every subaperture to be the same.  The assumption is that there is no phase tearing , and if there were, adaptive optics pretty much would be unable to handle such disruptions.   Once the full "map" is constructed, the mean value is removed (aka the "piston" term, zeroth Zernike) to reduce the overall stroke requirement on the deformable mirror (DM).  In most systems, the mean tip & tilt values (first Zernike terms) are removed as well and passed off to a full aperture tip/tilt correction mirror.  Again, that reduces overall stroke requirements on the DM.

There's lots more in practice, because the subaperture actuators on the DM (other that segmented ones, which have a host of other problems) have what's known as the "influence function".  That is, moving an actuator essentially creates a "hill" or "valley" over a decent region of the mirror.  These influence functions may or may not overlap neighboring actuators, depending on the mirror design, but in any case the final correction matrix has to take into account the difference between an idealized subaperture tip/tilt correction and what the mirror actually can do.

Follow that up with spatial and temporal (history) noise reduction filters, and off you go.

Now, back to the start: the constant of proportionality between Hartmann spot motion and phase tilt (or tip) is a simple function of the lens focal length. Draw the triangle consisting of the optic axis distance from lenslet to image plane (i.e. lens focal length), the distance from the origin to the centroid of the spot image, and the hypotenuse from the center of the lens to the centroid of the spot.  You can thus easily see that the tilt is the difference between the length of the hypotenuse and the focal length.  Convert that from meters (or microns) to wavelengths of the sensing light, and there you are.


Shouldn't tidal locking be impossible for a satellite that has a considerably eccentric orbit?

You are right, that would be weird if the Moon speeds up and slows down this way to always show the exact same side to the Earth. That's why it doesn't.
At some point in the orbit the Moon's rotation (its phase) lags behind and at some point it is too fast. It is (pretty much) constant and does not adjust to the varying orbital velocity. That's (one reason) why it wobbles. And this wobble is what's called libration.
But the Moon is still tidally locked as its rotation period is still equal to its orbital period.


How does gravity affect a physical clock?

If it has to do something with gravity affecting the inner working of wind up and pendulum clock will this also affect digital clock ?

As astrosnapper noted in his comment, gravitational time dilation affects all time-dependent processes. This includes not only clocks (including digital clocks and atomic clocks), but also the rates at which chemical reactions and radioactive decay occur, measured relative to the equivalent process far removed from the Earth's gravitational well.

General relativity is a descriptive theory. It describes what happens, but not why. (Newton's theory of gravitation similarly was a descriptive theory.) What happens is that time flows slowly in the vicinity of a gravitational source compared to how time flows well away from that gravitational source. The time dilation is a function of gravitational potential energy, not gravitational acceleration.

If you want an explanation regarding why this happens, the answer is no one knows. As noted above, general relativity is a descriptive theory. It explains what will happen but it does not explain why.


You're right; the effect of solar tides on the Earth would be to increase the Earth's (or Earth+Moon's) orbital angular momentum around the Sun. But it would be difficult to compute this properly, and probably impossible to measure the effect, because of all the other small alterations to the Earth's orbit going on.

A simple way to see the effect (which I gather you've already worked out) is to imagine two equal-mass objects in circular orbit about their common center of mass (midway between the two), one effectively a point mass (e.g., black hole) and the other being the Earth. If all motion is prograde (Earth rotates in same direction as orbit) and the Earth rotates faster than the orbital period, then the Earth's tidal bulge will be slightly ahead of the Earth-Sun direction. Then the near-side bulge will experience a retarding force from the point mass (slowing the Earth's rotation) while simultaneously pulling on the point mass in approximately the same direction as the latter's orbital motion. So the point mass is accelerated, adding to the angular momentum and making the orbit larger.

I believe that most attempts to estimate the solar tidal torque on the Earth end up with it begin about 5 to 10 times weaker than the lunar torque, so most of the slowdown in the Earth's rotation is still due to the Moon.

Of course, in the real Solar System the Sun also experiences a tidal bulge raised by the Earth, which (because the Sun rotates faster than the Earth's orbit) will accelerate the Earth forward in its orbit, also adding to the orbital angular momentum. But then you have remember that Venus and Mercury will raise larger tidal bulges on the Sun with different periods (relative to the Earth's orbit), and so the combined effect gets complicated.

As for Mars and Ceres: yes, in principle the same effect would apply. But it would be weaker for Mars (smaller planetary radius, larger distance from Sun) and much weaker for Ceres (same reasons, except moreso).

In addition, all of the orbits are being affected by N-body interactions with the rest of the planets (especially Jupiter), which I strongly suspect means that any potential changes in orbital angular momentum due to solar tides would be washed out by the various N-body and resonant interactions.

(My understanding is that solar tidal braking could explain the current rotations of both Mercury and Venus, though you really can't be sure because you don't know what their initial rotations were.)


Are all the photos of the universe really so colorful or is it just visual effects that are added later?

There is no straight yes or no answer.

Some objects do have significant radiation in visible part of the spectrum which would make them colorful if we would be close enough while observing them (our eye needs large number of photons to distinguish color so that means, astronomy wise, that object needs to be either very bright or very close or both).

On the other hand lot of them are being adjusted for color later. Usually most beautiful images are mosaics of several images from different filters (even from parts of spectrum our eye cant see) where each image is given a specific color based on filter used and then later all of them get combined.

If you want to learn more check this very helpful paper, which has some tutorials as well:

http://www.stsci.edu/stsci/resources/imaging/papers/rector2007/files/Rector2005.pdf

Btw, as a curiosity - images from rovers on Mars are mainly black and white and are colorized later (thats why the rovers usually have a color wheel on them for calibration purposes as seen here). This lead to uncertainties of color of Mars sky which turned into a long lasting debate.


Why is it always planets orbiting stars?

"Why is a star by definition stationary as opposed to planets which are moving." This just isn't true. Both star and planets orbit a point in the system known as the barycenter (a.k.a. the center of mass of the system).

Because stars are much more massive, this barycenter is much closer to the center of the star than it is to the planets. Hence it appears to the casual observer that the less massive object orbits around the more massive object. In fact both objects orbit the barycenter. Like so:

The following diagram (seen on the wikipedia page above) shows the motion of the solar system barycenter with respect to the center of the Sun. Note that the barycenter is close to the center of the Sun, but often spends time outside the visible surface of the Sun. It follows this complicated path, mainly because of Jupiter but all the other planets make smaller contributions too.


Before the 1761 transit, what was our best estimate of the distance to the Sun?

Aristaurus gives a calculation which would result in a value for the Earth-Sun distance  of about 0.016-0.065AU, clearly that's a long way out. The true value is 1.00 AU
Archimedes gives a value that would correspond to 0.43AU, other Ancient authors gave a range of values, all less than 0.5AU and many were close to Aristaurus's value.  So among ancient writers, all we can really say is that "They had no idea how far the sun was".
Kepler and his laws of planetary motion could show that values of 0.02au were far too small (since then a very significant parallax would be visible in venus) but couldn't say with precision how far out.
In the 17th century, telescopic measurements of the Earth-Sun-Moon angle allowed for Aristaurus's method to be used with greater precision and yielded values of 0.5 or more
Huygens made a number of assumptions on the size of the planets and came up with a very good value of 1.068AU,  but his reasoning was faulty.  This value is more luck than good maths. However, Cassini used observations of Martian parallax (and a good estimate of the size of the Earth) to get a well-calcuated value of 0.925AU
This was the best estimate prior to the measurement using transit of Venus.


What causes the bright white area in this photo of the earth?

Taken on March 28 2012 by André Kuipers, the version here is labeled

  Amazing aurora lights and atmospheric glare with the countries Ireland, Netherlands, Scotland, England and Wales in the foreground.

Off to the north and east like that, it's likely refracted early sunrise. That's a lot of streetlights for that time of the day; likely a longish exposure to capture aurora. That'd increase sunrise glare as well.


Is 486958 Arrokoth (2014 MU69 aka Ultima Thule) the only solar-system object determined to be binary by occultation?

Actually, the very first binary asteroid was discovered by stellar occultation in 1980.  The evidence was rejected by the IAU, and wikipedia credits the first binary asteroid discovery to the Galileo spacecraft finding the moon of 243 Ida in 1993.
The first moon of the well known asteroid 216 Kleopatra was observed through stellar occultation in 1980.  From this article:

As occultation czar David Dunham would later report in Sky &
Telescope, nine different observing stations visually timed blinkouts
of the 9th-magnitude star SAO 128066. Collectively these timings
traced chords across an irregular, elongated silhouette measuring
roughly 60 by 80 miles (95 by 130 km). "But there was an unexpected
secondary occultation too!" Dunham wrote. "Gerald Rattley and Bill
Cooke, observing from sites 2,000 feet apart near Loma Prieta,
California, independently and nearly simultaneously saw the star
disappear for about one second."

Dunham concluded that they had discovered a satellite of 216 Kleopatra. At that time, however, it was unknown if asteroids had satellites, and their evidence was not accepted.  There was some stigma associated with the occultation methods.   In 2008, the Keck II telescope finally verified the moons of 216 Kleopatra:

In 2011, one of Kleopatra's moon's size and orbit was shown to be consistent with the original 1980 occultation observations, referenced in this paper, and reconstructed here with the original chords:

Even though there were two independent chords that occulted the star, the IAU's former Minor Planet Director Center called the evidence "shoddy", even decades later.  In 1980, he had refused to accept that asteroidal satellites existed.
Interestingly, even after vindication 30 years later, stellar occultation evidence for binary asteroid systems are still met with skepticism.  For example, the wikipedia page for the asteroid 911 Agamemnon says a 2012 occultation is merely 'suggestive' of the Asteroid having a satellite.  To me, the stellar occultation of the light intensity measurements from this paper, is far more than just suggestive, but instead clear evidence of a satellite of 911 Agamemnon:

The large steady dip is the main body of the asteroid passing over the V magnitude 8.0 star HIP 41337 = SAO 60804 = BD= +37° 1857 = HD 70920, spectral type K0.  The small dip is the satellite passing over the star.
Note:
It's strange to me that the discovery of binary asteroid systems is met with skepticism since the method is essentially the same method often used to detect extra-solar planets (the transit method explained nicely in this answer).


How is the Hubble constant determined from gravitational waves?

If you measure the gravitational waveform from an inspiralling binary, you can at any point measure the amplitude, instantaneous frequency and the rate of change of frequency. The last two give you the "chirp mass", which is related to the product and sum of the binary component masses.

The amplitude of the gravitational wave then depends on the chirp mass and the distance to the source. Thus we can estimate the (luminosity) distance.

If we then know the redshift of the source, or at least the redshift of the galaxy it is in, then an independent estimate of the Hubble parameter is possible. Therefore this can only be done for gravitational wave sources with identified host galaxies (so far this can only be done for GW sources that show an electromagnetic counterpart - merging neutron stars).

It is a bit more complicated than this, because the amplitude also depends on the inclination of the binary orbit to the line of sight. Fortunately, this can be constrained by measuring the relative amplitudes of the two possible gravitational wave polarisations (+ and ×) and this can be done if you have two interferometers with different orientation of their arms (like LIGO and VIRGO for example), which also helps to locate the GW source in the sky to look for electromagnetic counterparts.


Could we parallax measure stars just based on the Earth's size?

In principle, it's not impossible.

The Gaia spacecraft, designed primarily for measuring stellar positions, is able to measure parallaxes up to 10 kpc away with 20% uncertainty. Its baseline is 2 AU; $2.3\times10^{4}$ times larger than the diameter of Earth. Thus, placing two Gaias on each side of Earth would be able to measure parallaxes of stars up to a distance of $10\,\mathrm{kpc} \,/\, 2.3\times10^{4} \simeq 0.4\,\mathrm{pc}$, meaning that you would almost be able to measure the distance of our nearest neighboring star, $\alpha$ Centauri, which lies at 1.3 pc. So you would just need improve your Gaias a little bit.

This ignores small complications such as the atmosphere, but if you're willing to put them outside the atmosphere, you could do it. Of course, it would sort of be a waste of time, since we already know the distances to the nearest stars, but hey, go ahead.


There are a few details here that make this non-trivial. The year doesn't matter much, since the same stars are visible each year. Further, in Stellarium, you can set the date to 1889 so this is not an issue but we have to turn off atmospheric effects to make the stars visible.

More tricky is that the Brazilian flag is designed as if viewed "from above". We can flip the image from Stellarium to simulate this.

Now, the stars are on a globe and the flag is flat, so a projection is needed to represent it. It's not clear what projection the flag designers used, but perhaps it is "orthographic". Stellarium can do that projection.

The projection also seems to be slanted, so "north-south" isn't vertical, although it's not clear by how much. We can estimate this slant from the position of the equator.

So here is an image of the sky from Rio. The Southern Cross is visible towards the bottom of the image.

If you want to do this yourself in Pyephem you can create a "fixed body" object for each star (Pyephem has a catalogue of brighter stars), create an observer in Rio, and call "body.compute(observer)" which will calculate the azimuth and altitude of the star. Their orthographic projection is then the application of the formulas found on Wikipedia, and you can plot that using any plotting package.


This might be it:
Astronomers Say Giant Disco Ball in Space Is Bad for Science

Three foot diameter faceted mirror ball launched late January 2018.
I don't know either designation or orbit.


Can stars be observed from space by x-rays, near infrared and radio wavelengths?

Yes, and not only from space but from the Earth surface too. Stars emit in almost all wavelengths depending on their surface temperatures. The hotter the star is the shorter (higher energy) wavelengths it'll emit.

You can try this simulator to check this:

http://astro.unl.edu/naap/blackbody/animations/blackbody.html


Abrevaya et al. (2020) in "The UV surface habitability of Proxima b: first experiments revealing probable life survival to stellar flares" (abstract, PDF) say:
"Our results show the impact that a typical flare and a superflare would have on life: when microorganisms receive very high fluences of UVC, such as those expected to reach the surface of Proxima b after a typical flare or a superflare, a fraction of the population is able to survive. Our study suggests that life could cope with highly UV irradiated environments in exoplanets under conditions that cannot be found on Earth. "
That suggests that total-extinction conditions aren't found on Earth, either.  The paper has a lot more detail of their assumptions and modeling, but that's my quick takeaway from the abstract.


Sky from Equator at 8,000 years Before Present?

You can equally well just type in a negative year and Stellarium will navigate you into the corresponding year B.C. Don't forget that the year 0 didn't exist. 1 A.D. follows directly after 1 B.C.

A while ago I've confirmed that Stellarium is very accurate down to 1500 B.C. using a babylonian list of lunar occultations of stars. Thus, it should be reasonably accurate for the year 6000 B.C. as well.


'Little h' usage in cosmological simulations

Partial answer; although the question is 17 days old, I hope that this guide to the dependence on h in cosmology can give you a good insight into this concept.

Quick guide to h dependence in cosmology

which links to Damn You, Little h! (Or, Real-World Applications of the Hubble Constant Using Observed and Simulated Data). The abstract says:

  The Hubble constant, H0, or its dimensionless equivalent, "little h", is a fundamental cosmological property that is now known to an accuracy better than a few per cent. Despite its cosmological nature, little h commonly appears in the measured properties of individual galaxies. This can pose unique challenges for users of such data, particularly with survey data. In this paper we show how little h arises in the measurement of galaxies, how to compare like-properties from different datasets that have assumed different little h cosmologies, and how to fairly compare theoretical data with observed data, where little h can manifest in vastly different ways. This last point is particularly important when observations are used to calibrate galaxy formation models, as calibrating with the wrong (or no) little h can lead to disastrous results when the model is later converted to the correct h cosmology. We argue that in this modern age little h is an anachronism, being one of least uncertain parameters in astrophysics, and we propose that observers and theorists instead treat this uncertainty like any other. We conclude with a `cheat sheet' of nine points that should be followed when dealing with little h in data analysis.


How do you estimate the error on the height/width of a Gaussian?

If you know the number of overlapping Gauss functions, you may write them as $\sum_{i=1}^n a_ie^{-(\frac{x-\mu_i}{\sigma_i})^2}$, then subtract your time series, with more than $3i$ items, to reduce the risk of an ill-posed problem.
Then apply a root mean square (RMS) minimization / least squares method to infer the $3i$ parameters.
If your FWHM, or $\sigma$ is known to be the same for all peaks, your number of parameters reduces to $2i+1$.

See also Rietveld refinement as a related approach.


Possible occultation of HIP 99314 and Ganymede

For what it's worth, I just got a shot of what looked like a 5th Galilean moon near Ganymede tonight (from South Florida). Here's a quick, unprocessed image. I thought I was going crazy seeing an extra moon (in the correct plane) that kept showing up. But looks like it's HIP 99314 (2nd dot to the upper left of Jupiter).


Calculate NEO object position with nasa Near Earth Object data

The file you had linked as 'Detailed NEO Data' lists all the orbital parameters needed to determine the position of the object as a function of time. It lists these parameters for the objects orbiting the earth. This should be just what you need. They are located at bottom of file. There are routines to convert the orbit in orbital parameters to cartesian coordinates. This may be easier to use. You can google on orbital parameters conversion to find these routines. I think I have a copy in Fortran but not many people use Fortran now.


Finding the diameter of a molecular cloud using period of intensity variation of hydroxyl maser

The principle that you apply is that the properties of a body cannot change faster than it the time it takes for light to cross the body.  So pulsars can be very small, which we know from the high frequency of their variation.

A body which varies with a period of 5 minutes can't be larger than about 5 light-minutes across.


Note that for cosmic expansion, special relativity only applies when the spacetime region in question is approximately flat, which in our case happens when it is small.

  As the M31 galaxy is moving toward us at great speed it's "depth" should appear slightly flattened for us. A sphere moving toward us at the speed of light will appear "pancaked" in lack of a better word.

No, that's not correct. You have a worldline in spacetime, and its direction at an event is your four-velocity. The orthogonal complement of your four-velocity, i.e., all directions perpendicular to it, is your space. What Lorentz-FitzGerald length contraction says is that if you project a moving sphere onto your space, the result is contracted along the direction of motion.

The difference between this and what you say is that there is no mention of how the sphere will appear. The sphere is contracted in the inertial frame comoving with you at that instant. It doesn't appear contracted--in fact, if you took a picture of it with a hypothetical super-fast camera, the relativistically moving sphere will appear... still completely spherical. But not quite the same as a non-moving sphere if it has features on the surface, as Penrose-Terrell rotation makes some of its "back parts" visible.

  Does this also mean that electromagnetic radiation (light) from a flattened star is focused in the direction of motion relative to the observer?

Yes. Just as a vertically falling rain comes at an angle when you're riding in a car, the angles at which any signal comes are different in a moving frame compared to a stationary frame. This is described by relativistic aberration. Additionally, Doppler shift changes the intensities of a radiating source along different directions, making the overall effect relativistic beaming.


Why does hydrogen ionization happen in HII regions?

Stars are responsible.

HII regions$^\dagger$ can refer to several things, but usually I guess one thinks of the volumes around star-forming regions. The more massive a star is, the faster it burns its fuel, and at a higher temperature, meaning that the peak of their spectra are more toward the high frequencies. The most massive stars of a stellar population — the so-called O and B stars — produce enough photons with wavelengths below the hydrogen ionization threshold of $\lambda = 912$ Å that they carve out bubbles in their surrounding HI clouds, giving rise to HII regions.

Right: The HII region NGC 604 (from Wikipedia). Left: The spectra of three different stars. Only the B star has a significant portion of its spectrum above the hydrogen ionization threshold. Note the logarithmic scale on the intensity (from here).

Because of the high densities, the HII quickly recombines to HI. If the recombination goes directly to the ground state, a new ionizing photon is emitted, which again is absorbed by a hydrogen atom, but if it goes to one of the higher states, the emitted radiation is no longer capable of ionizing the gas. In this way the ionizing radiation is converted into photons of specific wavelengths, corresponding the energy differences between the excited level of hydrogen, most notably the Lyman $\alpha$ emission line with $\lambda = 1216$ Å.

Because hydrogen is the most abundant element in the Universe, and because Lyman $\alpha$ is the most common transition, the Lyman $\alpha$ line is an excellent probe of the most distant galaxies where other wavelengths are not observable. This is especially so because the most distant galaxies are also the earliest and hence still in the process of forming, meaning a high star formation rate which, in turn, means that the shortlived OB stars are still present.

In addition to this distinct regions of HII, ionized hydrogen also exist in a more diffuse component between the stars of a galaxy, in huge bubbles caused by stellar feedback and supernovae, and in the intergalactic medium.

$^\dagger$The terms HI and HII refers to neutral and ionized hydrogen, respectively.