How to detect emission lines in optical spectra?

The programs I am aware of that you could use do require some programming expertise to operate. I would recommend looking at either IRAF (or PyRAF, which uses Python to interface with IRAF), or SPECTRE. Unfortunately, I'm not aware of a "black box" solution where you can just press a button and go.

SPECTRE is written in FORTRAN77 and is quite easy to use once you get the hang of it, though it has a somewhat limited feature set and the documentation is brief. It will help you to manually fit Gaussians to spectral lines, giving their central wavelengths and equivalent widths. A caveat for SPECTRE is that it requires the plotting package SuperMongo, which is not free software. If you are affiliated with a university, most Physics and Astronomy departments should have a license.

IRAF is a quite fully-featured, free, cross-platform application purposefully built for astronomical image analysis. It can help you measure the spectral lines, and can analyse the spectrum to determine the object's radial velocity.


How is the mass and semimajor axis of a directly imaged exoplanet found?

A single direct image of an exoplanets can give an estimate of the mass and orbital period of an exoplanet.

For the former, you need a model for how the luminosity of a giant exoplanet (which all the directly imaged ones are) evolves with time, as a function of mass, and an estimate of the age. One simply looks up what mass of planet would have cooled to its current luminosity at the age of the star that hosts the exoplanet. The luminosity measurement needs an accurate distance to the system and benefits from observations at multiple wavelengths. Realistically, these masses are uncertain by a factor of two at least, because of cooling model uncertainties.

Here is an example of this applied to various directly imaged exoplanets. Evolutionary tracks for various masses are shown in the luminosity versus age plane. Note the size of the age error bars and that is before you factor in uncertainty in the models (two flavours, "hot start" and "cold start" are shown as solid and dashed lines respectively - from Bonnefoy et al. 2013).

The semi-major axis can only be estimated (a lower limit), from a single image, and only then if you have a good distance to the star/planet system. The angular separation is converted to a projected linear separation.

A perfect example of the above is the estimate of the mass and orbit for the exoplanet GJ 504b (Kuzuhara et al. 2013).

With more images spread over an appreciable fraction of the planet's period one can be more accurate. The projection of the elliptical orbit can be mapped, yielding the orbital inclination and thus a better semi-major axis measurement. The orbital period can then also be roughly estimated and checked for consistency with Kepler's third law and an estimate of the stellar mass. Note though that currently imaged planets have orbital periods of at least decades.

Wang et al. (2016) give an example of this approach for beta Pic b. A dozen precise images constrain the semi-major axis to a few percent and give the total system (star plus planet, but dominated by the star) mass.


How often do comets survive passage by the Sun?

Key factors:

How close is perihelion? Too close and it may be destroyed on its first pass. We know Halley's Comet, which has a perihelion of about 0.6AU, has been orbiting for over 2000 years, passing the sun every 74-76 years and is still going strong.
How big is it? Every pass loses material, so a bigger comet could last longer.
What is its composition? Darker comets can absorb more energy when close to the sun, which may accelerate outgassing and breakup.

And of course some comets hit large objects - Shoemaker-Levy hit Jupiter and was destroyed.


If the sun is a second or third generation star, where is the neutron star?

Stars are not like animals, with well-defined parents. They form from interstellar gas clouds that in turn originate from stellar winds and supernova ejecta from other stars in a mixture with primordial gas. It is unlikely that the sun is "descended" to a great degree from any particular past star.
(Even if it had been, the sun formed about 18 galactic rotations ago and we are nowhere near the bodies in that past neighbourhood now, due to mixing)


How is the mass of solar system objects calculated?

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

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

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

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


just check what 'median' means. the median stellar mass is such that half of all stars have lower and the other half hihger mass. it has nothing to do with the logarithm. I havn't seen the term (median stellar mass) in the scientific literature.

Looking at the paper you're referring to, they never use the expression "median stellar mass". Moreover, this study is not concerned about the masses of individual stars, but with the total stellar mass of a galaxy, which they donote by $M$. They then consider the typical or characteristic value for $M$ (as obtained by fitting a Schechter function to the observed distribution of $M$), and denote it by $M^\ast$. Since this is a rather large number, of the order of $\sim10^{10}$M$_\odot$, they prefer to use its logarithm to base ten: $$\log\left(\frac{M^\ast}{\mathrm{M}_\odot}\right).$$
I think the paper explains all that quite clearly and am puzzled by your difficulty.


Would a satellite in geosynchronous orbit between the earth and moon track across the sky together?

The answer to the headline question is clearly: No.

The full moon has an apparent diameter of about 0.5°. Its apparent motion is roughly 360°/24h, neglecting the orbital motion of Earth and Moon. Hence it apparently moves about 24h*0.5°/360° = 0.033h = 2 min.
Hence the moon moves one moon diameter in about 2 min.
(The neglected orbital motion induces an error of about 3.65% (0.997d/27.3d), divide the siderial Earth day by a siderial moon orbit, hence an error of just a few seconds for the apparent motion of 0.5°.)
In 35 minutes, Moon's apparent motion would be about 17 moon diameters.

The geosynchronous satellite, in contrast, remains staying at the same apparent position, by definition.

Planets or stars move apparently about the above error estimate different from the moon, hence roughly 27.3-times slower relative to the apparent moon than a geosynchronous satellite.


Is it possible to see a moonrise or moonset twice in a day?

This can only happen if the moonrise at a certain date is earlier than the moonrise at the previous day. There are two reasons why this could happen:

The body (Moon, Mars,...) moves in the opposite direction of the Sun viewed along the right ascension (or ecliptical longitude) axis. The sidereal day is 4 minutes shorter than 24 hours which means that a static celestial body (a star) will rise 4 minutes earlier than the day before. However planetary bodies move (more or less) along the ecliptic, which is the Sun's path during the year along the celestial sphere. The Sun moves on average 4 minutes in ecliptical longitude (and hence in right ascension with a bit more variation) every day. So for a body to rise earlier, the body would have to move slower along the celestial sphere than the Sun (or move in a retrograde direction). This is true for the outer planets (including mars) but not for the Moon, which moves in the same direction as the Sun and much faster (one orbit per month).
If the observer is at a sufficient high latitude on Earth (either north or south of the equator), the inclination of the celestial equator will be small enough so that a body that increases its declination rapidly enough will be able to rise earlier on the second day.

See for instance the figure below where each drawing is at the same time of day. The moon will increase its right ascension during the day (along the celestial equator downwards, the red arrow in the figure). If the observer is on the Earth's equator, moonrise will be later the next day. It does not matter whether the declination (the position above or below the celestial equator (to the left or right in the figure)) will increase or decrease. All objects with the same right ascension will rise at the same time on the equator. If the observer is not on the equator but near one of the poles (for instance at 65° latitude north), the increase in declination may be sufficient to counteract the increase in right ascension (see bottom panel). Note that even though  the right ascension increases (the moon moves left along the celestial equator axis), the Moon will rise earlier during the day.
This only works (in the Northern hemisphere) when the Moon increases its declination (the height above the celestial equator) very rapidly. This happen only when the Moon is near the Vernal Equinox (i.e. in Aries or Pisces). And the observer needs to be at a high latitude (e.g. near the North pole such as Thule as mentioned by @barrycarter)


It depends what you mean by "real position".    Also, just to clarify something that should be gotten out of the way, the Earth rotates, obviously, and the night sky appears to rotate around the Earth a complete 360 degrees in a 24 hours.  That's why the geocentric model lasted as long as it did.  That's what it looks like, and that's what everyone believed prior to Copernicus, but that's not really relevant to your question.   I mention it, because what is relevant is that the stars are so far away that they appear fixed.   They rotate together, they don't move past each other so the stars operate as a fixed background against which the sun, moon and planets move.   So, when we talk about "movement" in the sky or position, we're talking about against the fixed stars in the background, not the rotation of the Earth.

After the fixed stars, the Sun is the simplest object in the sky, it moves a little bit against the background every day and it returns to the same place every year, the Sun's position against the fixed background stars was known long before Kepler, so Kepler always knew the position or direction of the Sun, even at night because the Sun held it's position against the background stars, moving only a little bit each day.   He didn't know the distance to the sun (more on that later), but he knew the position.

Kepler also knew that Mars' orbit around the sun was 687 days.   How did he know that?   Well (these words aren't my own, but it's discussed here)

  It's a simple matter to measure that the synodic period of Mars is
  about 780 days. This is the time between successive oppositions of
  Mars, meaning that Mars is due South at midnight. Given the
  heliocentric model, it is easy to see, as shown in the Wikipedia
  article, that 1/Synodic Period = 1/Earth's orbital period - 1/Mars
  orbital period. So, once the heliocentric model was understood, it all
  becomes easy.

Kepler was a brilliant Mathematician.   Brahe wasn't so much.   Now, I'm not saying Brahe was dumb, cause that's not accurate either, but he wasn't skilled in math the way Kepler was.   Brahe was very patient and dedicated and he made the most accurate observations to date, some 10 times more accurate than anyone before him.   He had the best equipment for making accurate observations in history, so to a person like Kepler, Brahe's notes were pure gold, containing decades of great accuracy.    There was even a rumor that Kepler killed Brahe to get access to his observatory and his work, but most people think that's just a rumor, that it didn't actually happen, but there was enough uncertainty that Brahe's body was exhumed and tested for poison.  (none was found).    But I digress.

A 3rd bit of information is that by using parallax and the maximum angle that Mars varies in the sky, you can get a rough estimate of the distance to the planets relative to each other.   Copernicus did that so Kepler didn't have to.   A bit more on that here.  and here.

So Kepler had a lot of information at his fingertips when he got access to Brahe's notes.   He knew that Mars was about 1.5 times as far from the Sun as the Earth (from Copernicus) and he knew that Mars orbit was 687 days (not difficult given the heliocentric model), which means, every year and 322 days, Mars is in the same spot relative to the Sun, but the Earth has moved. - So you have 2 still objects and one moving object that's some "43 days" earlier in it's orbit each time Mars is in the same place and by taking that measurement every 687 days, you can learn a lot about the relative position of Mars, Earth and the Sun.   And that's what Kepler did.

His two major breakthroughs (in my opinion) were observing that Mars wasn't on the same plane as the Earth (He did that observing Copernicus' observations, not Brahe's - I read that, I can't find the notation right now).    And his other breakthrough was to figure out that the Sun wasn't at the center of the Earth's orbit.   He did that by measuring the angle between the Earth and Sun and Earth and Mars every 687 days.   More on that here.

His measurements telling him that the sun wasn't at the center of the Earth's orbit led him to his ellipse model.   After that it was all clarification and more detailed measurement that led him to his 3 laws, but I think, recognizing that the Sun wasn't at the center of the Earth's orbit was his big "aha" moment that and his discovery of Mars being on a different orbital plane.   Kepler's measurement was quite accurate.   Mars' inclination to Earth is 1.87 degrees.   Kepler had it measured at 1 degree, 50 minutes, or 1.83 degrees.


What is period doubling in a variable star?

I think part of the confusion comes from the fact that "period doubling" can mean two rather different things, which are probably related but this is not completely known.  The first is a purely observational meaning, which happens when you have a clear periodic pulsation in a star, but the amplitude shows a higher-lower-higher-lower pattern that implies a "double" period to see the full cycle.  Then there's the theoretical meaning, which is a step in the approach to chaos.

The way the approach to deterministic chaos often works is, you have a simple system that shows a simple period.  But then you increase the degree of nonlinearity in the system, and you see bifurcations into something that requires double the number of iterations to be cyclic.  If each iteration represents a cycle time, like the time to go around an orbit or the time for a pulsation, then that means double the period.  Maybe you started with an interative mapping that converges to a fixed point, but then you increased the nonlinearity and it turned into a limit cycle where the iterations go back and forth between two points.

Now, it's not immediately obvious the connection between that and the amplitude modulation of a simple cycle, and I think many of the sources are purposefully ambiguous about that connection because nonlinear dynamics is still a mysterious field.  But the basic idea seems to be, if you have a linear system, it has uncoupled pulsation modes that are higher harmonics of the lowest frequency "fundamental" mode.  But if you have a nonlinear system, you can have coupling between those modes, which end up looking like amplitude modulation in a lower harmonic due to resonant coupling with a higher harmonic.  If the coupling has an odd-integer relationship, like a frequency of 2 coupling to a frequency of 3, then if the first oscillation in the frequency 2 mode couples resonantly with the frequency 3 harmonic, then the second oscillation will not couple well, it will hit the harmonic when it is 3/2 of the way through a cycle, so is out of phase.  Then the third cycle of the frequency 2 mode will hit the frequency 3 mode when it has gone through 3 full cycles and is back into resonance.  So that gives the "period doubling" where the amplitude of the frequency 2 harmonic gets a boost every second time around, owing to its ability to receive energy from the higher mode as well.

The connection between that and period doubling bifurcations is not entirely obvious either, but it seems that the idea is that if you have nonlinearity, the period doubling you find on the path to chaos might be related to this kind of resonant interaction between the simple behavior you see, and higher frequency behaviors that could exist but only make their presence known as a kind of residue from how they interact with the behavior you do see.  In other words, if the linearity is weak, some simple oscillation gets excited, but if you crank up the nonlinearity, that simple oscillation starts to overlap with more complicated behaviors that require more cycles to return to the original resonance relationship.  Crank up the nonlinearity even more, and all you see are the complicated behaviors, you have complete chaos.  The period doubling in variable stars represents only a small amount of nonlinearity, it's not chaotic and the relatively simple nature of the resonances allow for a large amplitude to build up and allow the stars to be classified as variable in the first place.

To do more research into this, you might look up the "Blazhko effect," or how period doubling also applies to other variable stars.  It sounds like an active area of research where there are still more questions than answers, but Kepler data has the promise of producing the necessary information to solve it.


How large could a second moon of Earth be?

Your question doesn't work because it's based on a false premise - that you can have a stable 2nd moon in orbit. You can't. If you accept the fact that the 2nd moon won't be stable, the size it becomes dangerous is the size that you wouldn't want to fall to Earth - if it's garbage and loosely held together, 100-200 feet in diameter would probably be safe enough and make a big flash and a lot of noise if it hits the Earth, but not cause too much damage.

Space garbage disposal is pretty easy - because there's so much empty space. In densely packed regions like low Earth orbit with lots of satellites, you'd want to avoid dumping garbage, but in the vastness of the Earth's Hill sphere, letting garbage just float in empty space wouldn't be all that dangerous, unless there was a lot of it. If there's enough manufacturing that you can't let garbage float, or if there's toxic minerals like lead or plutonium that you don't want to let fall to Earth, then creating a garbage dump in space makes sense, but it'd be way smarter to just use the Moon as a garbage dump than to build a 2nd one.

If the Earth-Moon system could hold an additional moon in a stable orbit, then your question on how massive could be answered, but there's no practical reason and no benefit towards creating a garbage moon, it would have an unstable orbit and require periodic re-adjusting or it would be dangerous. Just make a garbage dump on the moon.  It's way smarter.

Would the L4 and L5 Lagrange points be unstable?

Doing a bit more research, it seems that Earth-Moon L4 and L5 are stable. NASA did some research into it. Perhaps that makes my answer not correct to this question.  An L5 garbage moon would be possible. - See @Andy256 answer here.

As noted in the comments, a moon-sized object is a lot of material. The Moon is 20 times the mass of the asteroid belt. The number of comets and asteroids that would need to be captured, mined and discarded into the L5 orbit to create a 2nd moon is huge. We're not likely to create anything but a tiny moon out of captured objects because the Moon is comparatively enormous. I rather like the idea of using L5 as a place to capture, place and mine asteroids.

There's still a size limit though. If it grows to a few times bigger than the Moon, it's no longer the Moon's trojan, the Moon becomes its trojan and there's a mass limit for a stable trojan orbit, perhaps about 2 lunar masses, after which, the Moon's orbit would become unstable - and could be a world-ending scenario.

(old update below)

That's a tough call. The Earth-Moon system is far from ideal for stable L4/L5 Lagrange points because the Moon's orbit wobbles due to solar tugging. The mass ratio of 81 to 1 is large enough, with about 26 to 1 being the minimum required for stable L4/L5 saddle points, but the eccentricity and lunar apsidal precession could be problematic for long term stability.
I'm not smart enough to say for certain whether they would be stable or not.  Perhaps another question, or it also fits here if someone else wants to answer.


Is there any real chance of getting hit with a meteorite if we build a base on the moon?

Being struck by a piece of rock that is pebble sized or bigger could do damage to any base. Fortunately such objects are rare. Their danger could not be entirely avoided, but would be just part of the overall risk of such a mission.

Micrometeorites would hit any exposed base, just as they have hit the ISS  and the space shuttle in the past. The base would be make strong enough to survive very small impacts. The base could be built partially underground, which would also reduce cosmic radiation. And in would be possible to isolate parts of the base, so a breach doesn't cause destruction of the entire base.

Finally, astronauts have to accept dangers that most people would find intolerable. The ISS was estimated to have a 5% chance of catastrophic failure over its lifespan. 4% of astronauts have died in a spacecraft. Breach due to micrometeorite impact would be one of the many risks.


We don't have theories that can describe the big bang (or whatever happened) properly so we can't say anything definite about the big bang itself, let alone if/what was before it.  We would need (at a minimum) something like a quantum field theory for gravity and we don't have one yet.  We've no reason to think that's all we'd need.

It's not even possible (as I understand it) to define "before" the big bang.  Time, in essence, started at the big bang.  It has no common sense meaning beyond that.


Jezero crater is roughly 50 km across, and it is emplaced within the wall of Isidis crater ("basin").  Isidis itself has seen a lot of erosion, but it has been dated by lots of different people through the use of superposed crater counts (craters on top of it, with the number-age relationship tied to the Moon) to somewhere around 3.85–4.05 billion years old, including by me (Robbins et al. 2013).  Jezero cannot be older than Isidis, so it is capped at 3.85–4.05 billion years.
Jezero must have been formed when there was liquid water on the surface of Mars in enough quantity to form a delta deposit.  That would mean that it should have formed by the Late Hesperian chronostratigraphic epoch, which dates, in various chronologies, to about 3.00–3.65 billion years ago.  So, Jezero formed at least 3.00* billion years ago, but no later than 4.05* billion years ago.
Beyond that, no one that I know of has done superposed crater counts on Jezero's rim to try to model its age.  Jezero is highly degraded, and it has only one larger crater emplaced on its remaining rim, making any such crater counts highly suspect.  This is why the work I linked to above, I only tried to date craters >150 km, and even that is somewhat suspect.  Jezero certainly has deposits in the floor (such as the delta), so any dating of the floor from craters would only be a minimum age because it can't be younger than its floor deposits.  Therefore, even if someone does try to do superposed crater counts to get an age for the crater, I don't think they will be successful in getting a very good one.  For example, this paper found that the age of the floor deposits is 2.6±0.5 billion years, so again Jezero is at least 2.6 billion years old, but that's a minimum, and the floor deposits must be younger than the crater.
*The Mars crater chonology has an uncertainty of a factor of 2-ish, so even though there are two decimal points on my numbers, treat that with a grain of salt.
And yes, an impactor formed the crater, either an asteroid or a comet.


Open source library to calculate image from signals of a distributed set of radio dishes?

This is an interesting question!
I'll post a partial, preliminary answer and let's see where this takes us.
The NRAO block quote mentions two methods, interferometry and intensity mapping (like how a camera works). There may be hybrid options and other esoteric options, but lets look at these first.
See also this answer to Which techniques are used to convert radio signals received by antenna to images? and since there are so many good answers to Radio telescope targeting those might be worth a good read as well.
Interferometry
This is where all dishes point in the same general direction and you have:

...the exact location(s) on the globe, knowing their exact time-stamp(s) of observation...

and those time stamps must be made to nanosecond relative accuracy; you use the phase information of the radio waves. You either hook them up with coax or fiber optic cable in real time, or write at up to gigabit/second speeds to hard drives and then cart them all to one location and run a very very big and special computer called a correlator to analyze the data.
Intensity mapping
This is just making an intensity map by sampling many spots in the sky and building up a histogram; basically exactly like how a camera works but kilopixels rather than megapixels. Each telescope is an independent sensor, and it contributes to the the big map. You might start with a grid of pixels and as the antennas drift into each box you measure the intensity of radio power, and write that number down in that box. Earth's rotation scans in RA already. so no matter where they are, the dishes should be evenly spaced in declination.
See answers to How did single dish (or single receiver) radio telescopes originally generate images?
Hybrid and other esoteric options
There is intensity interferometry but I don't understand it well enough to write authoritatively on it. It looks at fluctuations in statistical noise, and so you can probably get away with less critical timing; each dish would record the instantaneous power level at perhaps microseconds or even millisecond intervals. I haven't a clue how this works with arrays; I've only heard it used for two receivers, but you could certainly analyze it pair-wise. (see also Narrabri Stellar Intensity Interferometer)
You could also do something fun/interesting with pulsars, but I don't know what exactly.

Intensity plus Doppler frequency can sometimes give you distance if you  have a model; this is an early radio map of neutral hydrogen in our galaxy from  Why the blank wedges in this very early 21 cm map of the Milky Way? (Oort et al. 1958)
 click for larger


The evaporation occurs through Hawking radiation. This is a very slow and low energy process. So low that the cosmic microwave background radiation, which is just a few degrees above absolute zero, pours far more energy into the black hole than Hawking radiation takes away. So in principle a black hole cannot evaporate.

With the exception of conjectured atomic sized black holes, that is, as Hawking radiation would be more pronounced then.

Edit: See this answer on the physics SE: https://physics.stackexchange.com/a/26607/55483
The black hole would need to be less massive than the moon to radiate more energy than it absorbs. Bigger than atomic size, as I suggested.


What percentage of the hydrogen today has never been in a star

About 70% of the baryonic matter in the universe is hydrogen, with a mean density of about $4\times 10^{-29}$ kg/m$^3$.

Most of the stars that have ever been born are still alive, since an average star is only about $0.25 M_{\odot}$ and has a lifetime much longer than the age of the universe (so very little material has actually been recycled).

If we assume there are $10^{22}$ stars of $0.25 M_{\odot}$ in an observable universe of radius 47 billion light years, that are 70% H by mass, the "stellar" hydrogen is only one part in 73.

So, there is only one hydrogen nucleus (a proton) in a star for every 73 in the universe. This ratio would have been smaller in the past (E.g. when theSun was born). But as I mentioned, most of this hydrogen (about 90%) is in stars that are longer lived than the universe. Therefore my very rough estimate is that around 1 hydrogen atom in every 1000 on Earth has been inside a star. This contrasts markedly with say that 100% of carbon and oxygen atoms have been inside a star.

EDIT: To be fair, this calculation hinges a lot on how many stars there are in the observable universe. This number is very uncertain and could be higher - perhaps $10^{23}$ (see here), in which case my numbers are somewhat pessimistic and it might be more like 1 H atom in 7 is inside a star and 1 H atom in about 100 on Earth were inside a star. However, I don't think there is any argument that the majority of hydrogen in the universe is not, and has never been, in a star, but whether that is 90% or 99% is still moot.


When the astronaut is far outside the black hole we can see the astronaut normally.
As the astronaut falls towards the black hole, things get strange. We don't see the astronaut pass the event horizon, instead as the astronaut falls towards the event horizon, he will seem to be time dialated.  The clock on his wrist will seem (from our perspective) to run slower (he won't notice this from his perspective, the watch is normal).  As he gets closer to the event horizon, light reflected off him will be more and more redshifted.  As he gets closer to the event horizon the light will become redshifted to infra-red (at which point he will appear invisible) then microwaves then longer and longer wavelengths carrying less and less energy.  He will turn black and invisible to us.
So you can't see the objects that have fallen to the black hole.
From the point of view of the astronaut, he falls down and in a short amount there is a singularity in the equations. It seems that his body is compacted into a single point of infinite density, it seems that general relativity can't yet describe this state, which will require a quantum theory of gravity.


Why don't they make an image of the black hole in the M33 galaxy?

M33 does not appear to contain a supermassive black hole: in fact there's no evidence that it contains a central black hole at all. The upper limit on the mass of a central black hole based on the dynamics of the core region is a few thousand solar masses.

Merritt et al. (2001) "No Supermassive Black Hole in M33?" derive an upper limit of 3000 solar masses on a central compact object in M33, noting that this is still consistent with the M-σ relationship between the mass of a supermassive black hole and the velocity dispersion in the stellar bulge, using which they obtain a predicted mass of 2600–26300 solar masses.

Gebhardt et al. (2001) "M33: A Galaxy with No Supermassive Black Hole" obtain an even smaller upper limit on the mass of a black hole of only 1500 solar masses (their best fit mass is zero, i.e. no central black hole), which they state is significantly lower than the predicted mass from the M-σ relationship.


Did NASA/JPL get "waning" and "waxing" backwards in this video?

I think they did make a mistake.  The moon waxes after a new moon and wanes before a new moon. To wax means to grow bigger or stronger (and is cognate with "waist"!). To wane means to become smaller or weaker (cognate with "waste" and "vaccum").

The moon that sets shortly after sunset in the evening is a waxing moon. The moon that rises shortly before sunrise in the morning is a waning moon.


To expand a little on James's answer: The pole angle method doesn't care what time of year it is, the celestial pole isn't going anywhere. ;) However, if you're in the tropics, the altitude of the celestial pole is rather low, which can make accurate observation difficult.

Of course, if you're doing this on another planet, (or even in Earth's southern hemisphere) you may not have a convenient pole star, and doing this sort of thing without decent equipment isn't going to be particularly accurate.

To determine longitude you need to have a good way of determining the time relative to a fixed reference, like the Greenwich meridian. The Earth makes one rotation on its axis (relative to the stars) in a (sidereal) day, so one day is equivalent to 360° of longitude, one hour corresponds to 15°, and one minute of time corresponds to 15 minutes of longitude, which is 15 nautical miles at the equator, and around 11.5 nautical miles at 40° latitude. In other words, if your time calculations are off by one minute, your position calculations can be off by 10 to 15 miles.

Newton described a method of determining the time from the moon, you can read about it in Wikipedia's Lunar distance (navigation) article. From the Earth, the Moon's angular diameter is roughly 0.5°, and it takes roughly an hour for it to travel through that distance relative to the stars. So if you want time measurements accurate to the minute you need very good observations. However, that's the easy part.

You also need very accurate calculations that tell you where the Moon's supposed to be at that point in time. And that's not easy to do - the Moon's motion is tricky! I won't go into the details here, but Wikipedia has a good introduction to lunar theory. If you're curious about the kind of formulas that get used, take a look at the links in this question.

So it wasn't just the difficulty of making sufficiently accurate shipboard observations that prevented Newton's method from being adopted, it was also the difficulty of preparing sufficiently accurate lunar tables. Actually, Newton's method of lunar distances was used, principally from 1763 (when the necessary tables and method were first published) until about 1850, when it was superseded by the marine chronometer. However, as Wikipedia mentions,

  The method saw usage all the way up to the beginning of the 20th
  century on smaller vessels that could not afford a chronometer or had
  to rely on the this technique for correction of the chronometer.
  [...]
  Captain Joshua Slocum, in making the first solo circumnavigation in
  1895–1898, somewhat anachronistically used the lunar method along with
  dead reckoning in his navigation.

To quote Captain Slocum:

  Even expert lunarians are considered as doing clever work when they
  average within eight miles of the truth.

Of course, on another planet you may have a better-behaved moon, or no moon at all. Or, as James mentioned, you may be very lucky and have a nearby large planet with fast-moving moons that are visible to the naked eye.


Has the Kepler data been examined for light curves due to phase changes of non-transitting hot Jupiters and other?

Here's an example paper looking for phase curves of non-transiting planets in the Kepler data: Millholland & Laughlin (2017) "Supervised Learning Detection of Sixty Non-transiting Hot Jupiter Candidates".

The candidates themselves can be found on this webpage.

Unfortunately I'm not aware of any RV followup on them to determine which of these candidates are actually planets.


What are the differences between TESS and PLATO exoplanet telescopes?

Here is what one commentator has said about TESS in comparison to PLATO:

The Transiting Exoplanet Survey Satellite (TESS), to launch in 2017, seems superficially to be a similar mission to Plato. It will potentially discover hundreds of planets before Plato even gets off the ground in 2024. However, the limited sensitivity of its cameras mean it is completely blind to Earth-like worlds around sun-like stars. Astroseismology is also off-limits for TESS, meaning the size of any worlds it does discover will be highly uncertain. Unlike Plato, it will also move between patches of sky every 30 days, allowing only hot, short-period planets to be found.
With all other new telescopes, both in space and on the ground, limited to finding super-Earths around small stars, Plato is the only mission on the table truly capable of discovering an Earth-like world around a star like our Sun. And by targeting bright stars that allow atmospheric follow-up, it is not impossible to think that, as well as the first truly habitable planet, Plato could find the first inhabited one too.

Credit: Hugh Osborn, PhD student, University of Warwick

http://lostintransits.wordpress.com/2014/01/30/what-can-plato-do-for-exoplanet-astronomy/


Would studying massive stars that are experiencing time dilation due to proximity to a black hole be useful?

I don't think this is feasible, because of two competing effects.
To have significant time dilation, the velocity needs to be large.  The velocity in a Keplerian orbit of radius $r$ scales as $r^{-1/2}$, so the star would need to be close to the black hole.
Tidal forces from the black hole (the difference in gravitational force on the near side vs. far side of the star) scale as $r^{-3}$, so they get much stronger for orbits closer to the black hole.  So there would be strong tidal forces trying to pull the star apart.  Massive stars have outer layers that are only loosely bound in the first place (they are close to the Eddington luminosity) and thus have significant mass loss due to radiation pressure pushing on gas in their outer layers.  Tidal forces would exacerbate this, so the star would quickly start shedding mass.
The mass loss would have (at least) two effects.  First, it would generate a lot of luminosity as it heats up and spirals into the black hole; these accretion disks are in fact one of the ways that black holes are observed.  So the star itself would probably be hard to observe separately.  Second, the star would feel a drag from the gas in the accretion disk, and would spiral in toward the black hole, which would make the tidal forces stronger.  It might not seem like the star should feel drag from the gas if both are orbiting at the same distance, and thus should have the same orbital velocity.  But the gas disk is hot, so its internal pressure (or more fundamentally, its pressure gradient with $r$) partially supports it, allowing the gas to have a stable orbit with a slightly sub-Keplerian velocity.  The star, as a macroscopic object, has to orbit with the regular Keplerian velocity, so it feels a "headwind" from the gas.
I would guess that complete tidal disruption of the star would happen before it inspirals all the way to the event horizon, but either way, I don't think it would last long.
EDIT:  As @ProfRob points out in the comments, this analysis doesn’t work for supermassive black holes.  They have very low densities inside the event horizon, and thus relatively weak tidal forces there.


Journal standard for plotting section of skymap?

I don't think there's a journal standard for this.  An easy and commonly used method is to plot latitude $\delta$ versus longitude $\alpha$, where the aspect ratio is scaled by $\mbox{cos}(\delta_0)$ so the projection is orthonormal at the image center $(\alpha_0, \delta_0)$.  Depending on your region size and on $|\delta_0|$, this may or may not come out looking distorted.   You might be able to pick a coordinate system (i.e. equatorial or galactic) where the latitude is never large enough to give major distortion.

Another common thing to do, arguably better when you're imaging extended objects, is to use a tangent-plane projection, also called a gnomonic projection.  This projection depends on the chosen center, which is usually the image center.  It is included in some plotting and analysis packages, but none that I use regularly.  I recently coded this up for the Astropy package but haven't really tested it yet.

Projections like Aitoff, Hammer, and Mollweide have been included in many plotting packages, but usually only for whole-sky projections.  You could try to write your own versions but I'm not sure it would be worth the trouble.

(Whatever you do, make sure longitude increases to the left since you're looking at the sky not the ground...)


Could undiscovered smaller black holes within galaxies be an explanation for dark matter?

MACHOs (massive compact halo objects) are thought to make up a portion of the dark mass and portion of these are thought to be black holes, but there are several very solid reasons why black holes not generally seen as a serious alternative explanation to non-baryonic dark matter:

1) Observed elemental abundances are not consistent with the amount of baryonic matter being produced in the early Universe to explain the dark mass

2) Microlensing searches for MACHOs of the size of black holes formed by stellar evolution rules them out as making a significant amount of dark matter in the Milky Way.

3) Black holes are observable through accretion. For black holes to make a significant amount of dark matter they would have to be in isolated locations within the galaxy so as not to be visible. However most dark matter must be distributed evenly throughout the Milk Way to explain the rotation curve.


How much has the apparent magnitude of the Sun changed?

No. The apparent magnitude of the sun is constant to within 0.2%  How long it has been that way is only speculation since measurements aren't available. 1000 years is a very short time period in the life cycle of a star. Its safe to say yes its been constant for 1,000 years. There is a small variation based on the 11 year sun spot cycle but this variation is reported at less than 0.1%. There is also a larger variation due to the difference in Earth-to-Sun distance caused by the earths elliptical orbit. This effect is about 7%. This is no due to the sun but rather due to the eccentricity of Earths orbit. There are also minor effects with a cycle time or much greater than 1,000 years. These also due to variations in Earths orbit.


Venus is a very bright "morning star" at the moment, but what your image shows is only a blur. The telescope is very far from focus and you can see no details of the the planet. You may also be getting some reflections from the glass of the window
If Venus was in focus you would see a very small "half moon" (or perhaps "waxing gibbous" shape) at the moment. Try to practice focusing on distant landmarks during the day. That makes it easier to get the focus right at night.


How many earth-like planets do we know at this moment?

Twelve plus Mars are considered as potentially habitable (as of February 10, 2014).


Star - Black Hole binary. Star core Collapses. Does the black hole immediately grow from neutrinos?

Yes. Neutrinos which hit the BH will add to its mass.

But you may be thinking that the neutrinos pass through the nascent supernova without interaction. In a dense enough environment, neutrinos do interact, and absorption of neutrinos by the layers beyond the core is a major source of the energy that disrupts the star.

In a bit more detail, the core collapse happens through inverse beta decay, where a proton and an electron react to form a neutron and a neutrino. This basically turns the star's iron core to neutrons. The energy is carried away by the neutrinos and the absorption of this enormous flux of neutrinos by the outer layers drives the SN.  The Wikipedia article has a good (but IMO somewhat incoherent) longer discussion.

Still the many neutrinos which do escape do so before the shock wave blows the star apart and the energy they carry will bulk up a nearby BH.

I doubt that it would easily be observable, though. Stellar-mass BHs are small in cross-section and would only absorb neutrinos from a patch of the sky proportional to their area. If a 10km BH was orbiting a million km star right at its surface, it would only absorb one part in 1010 of the neutrinos, producing roughly a 10-9 fractional increase in size.  Very difficult to measure!

If you're using this to provide an early warning of the core-collapse (perhaps so you can start you FTL drive and get outta there) it's fairly useless. First, the neutrino flux is probably high enough to fry people who are close enough to need the early warning.  And second, if you're far enough away not to be fried by the neutrinos, you have plenty of time to see the star rapidly swelling  and get moving before the comparatively slow shock wave gets near you.


Where is the exomoon? (Possible discovery of a companion to Kepler-1625b)

"Detrending" refers to removing systematic effects that are caused by the spacecraft, detector or the environment and which are not part of the particular star. The systematic trends can be caused by a large variety of things; the main ones affecting Kepler are drifts in position and focus, thruster firings to compensate for angular momentum build up in the reaction wheels and the roll of the spacecraft every quarter. Because these trends are systematic and affect all the stars that Kepler sees in a similar way, the effects can be modeled and the trends removed - detrending. There are some examples in slide 15 and 16 of this Kepler workshop presentation which shows the "raw" brightness measurements (labeled 'SAP') and the detrended data on the right (labeled 'PDC').

You can see examples of this in the Supplementary Materials. On page 47 in Figure S2, you can see the data before detrending and you can see the dips caused by the transiting exoplanet (Kepler-1625b) but these are superimposed on long-term upwards or downwards slopes or curves. Detrending removes these effects (which will be common to all the stars, not just to Kepler-1625b's star) allowing you to get corrected flat light curves, enabling the search for the much smaller exomoon signal.

The evidence for the exomoon is in two parts. The timing of the planet transit that was observed by HST in October 2017 was 77.8 minutes early compared to the predicted time from the Kepler transits. This is shown (but not very well) in Figure S12 in the Supplementary Materials. If there was no exomoon or anything else tugging on Kepler-1625b, we would have expected the HST transit (the data point at transit epoch 7) to be on the extended black prediction line at an O-C value (y axis) of +77.8 minutes, ( off the top of their plot as it scaled). The second piece of evidence is that is if the transit is early, then they should expect to see the shallower/smaller transit of the exomoon after the transit as it would be on the other side of the orbit with the exoplanet. This is what they show in the bottom half of Figure 4; no matter which detrending method they use for the HST data, they see a small shallow dip after the main transit is over, centered around a time (x value) of 3056.25. This gives more confidence that the exomoon interpretation is real and not an artifact of the detrending process.

What they are saying after Table 1 is that the Δχ2 (the measure of the change in the goodness of fit of the model as you add extra things to the model) is better with the moon models (models Z and M) over one that just accounts for the transit timing variations (model T). This is because the presence of the exomoon is predicted to cause changes in the length and depth of the exoplanet transit but these are harder to measure than the change in timing.


Does naked eye miss a lot between local stars and distant milkiness?

Even the Hubble telescope cannot "see" many of the stars and galaxies in the dark parts of space at first glance.  If they fix the Hubble on a dark patch for several days, the photons from extremely distant stars and galaxies will fill the image.

http://en.wikipedia.org/wiki/Hubble_Deep_Field

Also, the Earth's atmosphere will obscure and distort many stars that are visible from low Earth orbit and beyond.  Astronauts can see many more stars than we can.

Regarding your question, "how many are we missing", that's a tough one to answer. It depends on what level of detail you are talking about.


What exactly is a Hamiltonian telescope? Is this one?

Gennady Borisov wrote on the astronomy.ru forum:

The first observations of the object were made in the morning of August 30 with a 650 mm telescope, F / 1.5 Hamilton

Первые наблюдения объекта были сделаны под утро 30 августа на телескопе 650 мм, F/1.5 Гамильтоне

Yes, this is Hamilton. All surfaces are spheres. Has a 3-lens corrector.
Good image quality, high aperture ratio. Large field of view: on FLI ML16803  FOV:  2.1 х 2.1 deg 1.87"/pixel

Да, это Гамильтон. Все поверхности сферы. Имеет 3-х линзовый корректор.
Хорошее качество изображения, высокая светосила. Большое поле зрения: на матрице FLI ML16803  FOV:  2.1 х 2.1 deg 1.87"/pixel


What's the difference between apparent brightness and apparent magnitude? Are they the same?

Yes, $B$ and $I$ are the same things in this context. The $\Delta m$ you derived goes by the name "distance modulus" when one of the distances is set to $10\operatorname{pc}$. The thing to be careful of is that $I$ is also sometimes used for spectral radiance (or surface brightness) when working with radiative transfer. Spectral radiance doesn't obey the inverse square law, though, so there's no risk that we're confusing them here.


Calculation of Eccentricity of orbit from velocity and radius

If you have the relative position $\mathbf{r}$ and velocity $\mathbf{v}$ at the same point in time, compute the eccentricity vector:

$\mathbf{e} = \frac{\mathbf{v} \times \mathbf{h}}{\mu} - \frac{\mathbf{r}}{\|\mathbf{r}\|}$

Where $\mathbf{h} = \mathbf{r} \times \mathbf{v}$ is the specific relative angular momentum and $\mu = GM$ is the standard gravitational parameter.

The eccentricity is then given by the norm of the eccentricity vector $\left\| \mathbf{e} \right\|$.


Does the holographic principle contradict eternal inflation?

There is no obvious implication to re-assess the large-scale topology of the universe.
You can still stereographically project infinity to a point on a sphere, if you need.
Implications to the holographic principle aren't immediate, at least.

When talking about objects beyond the cosmic horizon, it's a difference, whether we talk about observation or mathematical models. When talking about mathemetical models of the universe, it makes sense to talk about space and objects beyond the cosmic horizon, because we can get a higher symmetry of the theory. We can use mathematically simple structures like de Sitter spaces as simplified models of the universe, with no need to add an observer to the structure.

For an evidence-based science it's of course an extrapolation far beyond immediate observation, hence questionable. But there is also no evidence-based argument to assume, that the universe looks fundamentally different from a distant observer with a different cosmic horizon overlapping with our cosmic horizon, but reaching beyond.


As said, winds on Uranus are quite strong blowing upto 560 miles/hr (900 kmph). But, the question is what forces drives these winds?
Researchers has been digging for quite a while. They investigated the gravity fields of Uranus using data gathered by NASA's Voyager 2 spacecraft and ground-based telescopes. The strength of a planet's gravity field depends on its amount of mass, and this strength can vary over the surface of a planet depending on the amount of mass lying under it.  By analyzing the gravity fields of these worlds, the investigators could deduce how their atmospheres circulated. From the readings, they deduced that the winds blow in relatively thin weather layers no more than 600 miles (1,000 kilometers) but comparing to size of Uranus, the layers are proportionally very small (0.15 % of the mass of Uranus). So, the researchers propose that the winds are mostly driven by atmospheric and surface conditions (possibly moisture evaporating and condensing in atmosphere causing the winds to swirl).
Some suggest that since the outer planets are gaseous, there is virtually no surface roughness to act as a drag on winds like on Earth and hence the atmospheres is more fluid like. Moreover, Uranus is very far away from the sun and thus there is less solar energy to impact the turbulence in the atmosphere.
Some also suggests internal heat to be the reason. Uranus only puts out about 6 percent as much heat as it receives from the Sun. So, internal heat source might be involved on some level.
For more information, read this excellent paper: Atmospheric confinement of jet streams on Uranus and Neptune
Additional source:

https://www.space.com/21157-uranus-neptune-winds-revealed.html
https://phys.org/news/2013-05-uranus-neptune-confined-thin-atmosphere.html
https://www.csmonitor.com/Science/2013/0517/Why-do-planets-farthest-from-sun-have-highest-winds-Team-closes-in-on-answer
https://arstechnica.com/science/2013/05/powerful-winds-on-giant-planets-confined-to-a-thin-atmospheric-layer/


Are there ways to estimate size of the "whole universe"?

tl; dr The universe is probably infinite, but if that's the case it's impossible to verify. If the universe is finite, and small enough, and the global curvature is equal to the curvature of our observable universe, then we will be able to estimate its size.

If the global curvature of the universe isn't positive, then the size of the universe is infinite, (and it's always been infinite since the dawn of time, at the Big Bang), assuming the topology of the universe is trivial.
Measurements of the observable universe indicate that the curvature may be zero (giving a flat universe), or even negative; if the curvature is positive, then its value is very small. We assume that the observable universe is typical of the whole universe, but of course that's impossible to verify.
Wikipedia says that

experimental data from various independent sources (WMAP, BOOMERanG,
and Planck for example) confirm that the observable universe is flat
with only a 0.4% margin of error.
[...]
The latest research shows that even the most powerful future
experiments (like SKA, Planck..) will not be able to distinguish
between flat, open and closed universe if the true value of
cosmological curvature parameter is smaller than $10^{-4}$. If the true
value of the cosmological curvature parameter is larger than $10^{-3}$ we
will be able to distinguish between these three models even now.

So even if the total universe does have positive curvature, its size is much larger than the observable universe, assuming that our observable universe isn't a patch that's abnormally flat.
According to How Big is the Entire Universe? by Ethan Siegel, if the curvature is positive, then the diameter of the total universe is at least 14 trillion lightyears, but that article is from 2012, more recent calculations may give a larger value.


Definitely a hoax as other commenters have said.

Dark rifts exist in the galaxy - large clouds of dust, several can be made out by eye, overlapping the milky way, in ideal conditions.  See for example the Great Rift.  However these are so large it would take far longer than three days for a solar system to pass through them.  Probably thousands or millions of years.  So this is definitely incorrect information.


How many times has the Milky Way rotated?

The question doesn't have a single answer, because the Milky Way's rotation is differential rather than solid-body. This means that stars in the inner part complete their rotation about the center in a shorter period of time than stars at a larger radius. Stars orbiting with a radius of 1 kpc (about 3,000 light years) will take about 30 million years to go once around; stars like the Sun, with orbital radii of about 8 kpc, will take about 220 million years; stars orbiting with a radius of 20 kpc will take about 500 million years.

If you mean "how many times has the narrow region of the Milky Way currently containing the Sun's orbit rotated" -- ELNJ's answer is probably reasonable, given that the last major merger contributing to the Milky Way's growth was probably about 10 billion years ago, so changes in the amount of mass inside the Solar radius (which would change the orbital timescales) were probably fairly small after that.


Does the Sun impose back it's tidal forces onto the Earth (such as the Earth's to the Moon)?

Yes - the earth and sun do have tidal forces like the moon and earth. There are two main reasons this is happening.

The sun is always losing mass due to nuclear reactions, the sun is always converting a tiny amount of its mass into energy. This means the pull on the earth is weakened.
The tidal forces that happen between the earth and moon also happen between the sun and earth, except on a much smaller scale.

The sun is also losing its angular momentum at a rate of 0.00003 seconds per year


Is that possible?

Yes!

You can see bright stars and planets through a telescope in the evening and in some cases during the day. As magnification increases the brightness of the sky or any extended object decreases with the square of the magnification. For example, a patch of sky will have a certain brightness per square degree, but at 100x magnification it will be 10,000 times dimmer.
A star will (roughly speaking) be simultaneously brighter by the ratio of the square of the telescope's diameter to your pupil (which is 6 millimeters or less), so an 6-inch (15 cm) diameter telescope will make the star look 625 times brighter.
When it got dark, the sky naturally became far dimmer, and your pupils dilated to larger diameters, making the star brighter.

So you can think of the two methods (viewing through telescope versus waiting for night) as similar, in that they both make the sky dimmer and the star brighter.

It's possible to start guessing what that star might be, but I don't know. Here are two predictions of the sky on November 15 at 5:30 PM and 8:30 PM local time at roughly 42 degrees North latitude.

That far North there are no planets that could be very close to the zenith, and it would be necessary to know the year as well to see how close one might have gotten.

Images are from in-the-sky.org's planetarium feature.


Tips for a beginer who wants to learn more about astronomy

To me it sounds that you just want to expand your astronomical knowledge and are not actually looking for a career advice.

I would suggest to just start with reading popular science books.
https://en.wikipedia.org/wiki/Cosmic_Evolution_(book) ,
https://en.wikipedia.org/wiki/A_Brief_History_of_Time for example.


Why is Saturn invisible in this radar image of its rings?

The main issue is that there is relatively little material in Saturn's atmosphere that can efficiently scatter radar waves, so the radar basically just gets absorbed.

The key point is that it's much harder to get a radar return from the very small objects (aerosol droplets or tiny ice particles) that would make up clouds in Saturn's upper atmosphere than it is to detect the meter-sized ice chunks in the rings.

The figure below (from here) shows $\sigma / (\pi r^{2})$, which is the "effective" scattering cross section relative to the geometric cross section of a scattering object, plotted versus the size of the object $r$ relative to the wavelength of light $\lambda$ (e.g., the radio waves used in radar). When the size of the object is roughly the same size as wavelength of light, or larger, it efficiently scatters them: its effective cross section is roughly the same as its geometric cross section (ignoring other effects such as chemical composition, surface roughness, etc.). Sometimes it can even be several times larger!

But when the objects become smaller than the wavelength, you transition to the Rayleigh scattering regime, where the scattering depends on the size of the object to the sixth power.

Imagine a sphere 10 cm in radius, roughly the same as typical radar wavelengths (so $2 \pi r / \lambda \sim 1$). It will have an effective cross section similar to its geometric cross section: about 300 cm$^2$. Now imagine subdividing that object into spheres one-tenth the radius. This would mean about 1000 objects 1 cm in radius (so that the total volume is the same), with a total geometric cross section of about 3000 cm$^2$ (realistically a little less, because of shadowing). But the plot shows that the effective cross section goes down by a factor of 1000, so the total radar cross section of all the small objects -- and thus the backscattered energy of the radar waves -- would be ten times smaller than it was for the single big object of the same total volume. And that's just for a reduction in size of ten; in reality, we're talking about going from centimeter-to-meter-sized objects (the ring) to sub-mm and micron-sized cloud particles in Saturn's atmosphere.

This is why weather radar on Earth can show you precipitation (rain, snow, hail) -- because the raindrops/etc. are large (a few mm in size), and can scatter radar waves with some efficiency -- but usually can't show you clouds -- because the water droplets making up clouds are so small. (And obviously radar wouldn't be nearly as useful to the military if it was easily scattered by clouds.)

So -- absent floating chunks of centimeter- or large-sized ice in the upper atmosphere of Saturn -- you're not going to get much of a radar return from Saturn itself. The original discovery of radar returns from the rings surprised people, because the assumption had been that the rings were made up of very small (e.g., micron-sized) bits of ice, which wouldn't efficiently reflect radar waves either.


According to the Case Western Reserve University webpage The Edge of the Solar System (2006) an important consideration is that

The whole concept of an "edge" is somewhat inaccurate as far as the solar system is concerned, for there is no physical boundary to it - there is no wall past which there's a sign that says, "Solar System Ends Here." There are, however, specific regions of space that include outlying members of our solar system, and a region beyond-which the Sun can no longer hold any influence.

The last part of that definition appears to be a viable definition of the edge of the solar system.  Specifically,

valid boundary region for the "edge" of the solar system is the heliopause. This is the region of space where the sun's solar wind meets that of other stars. It is a fluctuating boundary that is estimated to be approximately 17.6 billion miles (120 A.U.) away. Note that this is within the Oort Cloud.

Though the article above is a bit dated, the notion of the heliopause has been still of interest to scientists, particularly how far away it is - hence, the interest in the continuing Voyager missions, which states on the website, that it has 3 phases:

Termination Shock

Passage through the termination shock ended the termination shock phase and began the heliosheath exploration phase. Voyager 1 crossed the termination shock at 94 AU in December 2004 and Voyager 2 crossed at 84 AU in August 2007.

(AU = Astronomical Unit = mean Earth-sun distance = 150,000,000km)

Heliosheath

the spacecraft has been operating in the heliosheath environment which is still dominated by the Sun's magnetic field and particles contained in the solar wind.
As of September 2013, Voyager 1 was at a distance of 18.7 Billion Kilometers (125.3 AU) from the sun and Voyager 2 at a distance of 15.3 Billion kilometers (102.6 AU).

A very important thing to note from the Voyager page is that

The thickness of the heliosheath is uncertain and could be tens of AU thick taking several years to traverse.

Interstellar space, which NASA's Voyager page has defined as

Passage through the heliopause begins the interstellar exploration phase with the spacecraft operating in an interstellar wind dominated environment.

The Voyager mission page provide the follow diagram of the parameters listed above

It is a bit complicated as we do not know the full extent of what the dynamics are like out there, a recent observation reported in the article A big surprise from the edge of the Solar System, reveal that the edge may be blurred by

a strange realm of frothy magnetic bubbles,

Which is suggested in the article could be a mixing of solar and interstellar winds and magnetic fields, stating:

On one hand, the bubbles would seem to be a very porous shield, allowing many cosmic rays through the gaps. On the other hand, cosmic rays could get trapped inside the bubbles, which would make the froth a very good shield indeed.


The Oort cloud is more of a theoretical construct. It's a catch all, in effect, for small icy fragments anywhere from 2000 AU to over 100,000 AU.
From NASA:

The Oort Cloud lies far beyond Pluto and the most distant edges of the Kuiper Belt. While the planets of our solar system orbit in a flat plane, the Oort Cloud is believed to be a giant spherical shell surrounding the Sun, planets and Kuiper Belt Objects. It's like a big, thick bubble around our solar system, made of icy, comet-like objects. The Oort Cloud's icy bodies can be as large as mountains — and sometimes larger.

The inner edge of the Oort Cloud, however, is thought to be located between 2,000 and 5,000 AU from the Sun, with the outer edge being located somewhere between 10,000 and 100,000 AU from the Sun.

While the inner edge of such a cloud may have been closer or further in the distant past, as it has been disturbed by the planets, the idea of that region of space having icy planetesimals that may become comets doesn't really have any physical constraints other than those that Opik-Oort decided upon, unlike the Kuiper Belt, which does have measured boundaries.
Image of Oort Cloud from study.com:


Moon, center of mass & geometric center, tidal locking

Since the moon is tidally locked to earth, what about center of mass and geometric center? How far from each other are they?

They differ by about two kilometers.

  Can a celestial body be tidally locked to an other one if its mass distribution is perfectly homogeneous (center of mass at the very same place of geometric center)?

Yes, but that condition is not what I would call "perfectly homogenous". What's needed is a non-spherical mass distribution. For example, a uniform density ellipsoidal body would have its center of mass and center of figure of coinciding, but would still be subject to torques. It's the moment of inertia that is key rather than the offset between center of mass and center of figure. An object whose center of mass and center of figure do not coincide necessarily has a non-spherical moment of inertia tensor.


What would the Sun be like if nuclear reactions could not proceed via quantum tunneling?

Short answer: Without tunnelling, stars like the Sun would never reach nuclear fusion temperatures; stars less massive than around $5M_{\odot}$ would become "hydrogen white dwarfs" supported by electron degeneracy pressure. More massive objects would contract to around a tenth of a solar radius and commence nuclear fusion. They would be hotter than "normal" stars of a similar mass, but my best estimate is that they have similar luminosities. Thus it would not be possible to get a stable nuclear burning star with 1 solar luminosity. Stars of 1 solar luminosity could exist, but they would be on cooling tracks, much like brown dwarfs are in the real universe.

A very interesting hypothetical question. What would happen to a star if you "turned off" tunnelling. I think the answer to this is that the pre-main-sequence stage would become significantly longer. The star would continue to contract, releasing gravitational potential energy in the form of radiation and by heating the core of the star. The virial theorem tells us that the central temperature is roughly proportional to $M/R$ (mass/radius). So for a fixed mass, as the star contracts, its core gets hotter.

There are then (at least) two possibilities.

The core becomes hot enough for protons to overcome the Coulomb barrier and begin nuclear fusion. For this to happen, the protons need to get within about a nuclear radius of each other, let's say  $10^{-15}$ m. The potential energy is
$e^2/(4\pi \epsilon_0 r) = 1.44$ MeV or $2.3\times 10^{-13}$ J.

The protons in the core will have a mean kinetic energy of $3kT/2$, but some small fraction will have energies much higher than this according to a Maxwell-Boltzmann distribution. Let's say (and this is a weak point in my calculation that I may need to revisit when I have more time) that fusion will take place when protons with energies of $10 kT$ exceed the Coulomb potential energy barrier. There will be a small numerical uncertainty on this, but because the reaction rate would be highly temperature sensitive it will not be an order of magnitude out. This means that fusion would not begin until the core temperature reached about $1.5 \times 10^{9}$ K.

In the Sun, fusion happens at around $1.5\times 10^7$ K, so the virial theorem result tells us that stars would need to contract by about a factor of 100 for this to happen.

Because the gravity and density of such a star would be much higher than the Sun, hydrostatic equlibrium would demand a very high pressure gradient, but the temperature gradient would be limited by convection, so there would need to be an extremely centrally concentrated core with a fluffy envelope. Working through some simple proportionalities I think that the luminosity would be almost unchanged (see luminosity-mass relation but consider how luminosity depends on radius at a fixed mass), but that means the temperature would have to be hotter by a factor of the square root of the radius contraction factor. However, this could be academic, since we need to consider the second possibility.

(2) As the star shrinks, the electrons become degenerate and contribute degeneracy pressure. This becomes important when the phase space occupied by each electron approaches $h^3$. There is a standard bit of bookwork, which I am not going to repeat here - you can find it something like "The Physics of Stars" by Phillips - which shows that degeneracy sets in when
$$\frac{ 4\pi \mu_e}{3h^3}\left(\frac{6G R\mu m_e}{5}\right)^{3/2} m_u^{5/2} M^{1/2} = 1,$$
where $\mu_e$ is the number of mass units per electron, $\mu$ is the number of mass units per particle, $m_e$ is the electron mass and $m_u$ is an atomic mass unit. If I've done my sums right this means for a hydrogen gas (let's assume) with $\mu_e=1$ and $\mu = 0.5$ that degeneracy sets in when
$$ \left(\frac{R}{R_{\odot}}\right) \simeq 0.18 \left(\frac{M}{M_{\odot}}\right)^{-1/3}$$

In other words, when the star shrinks to the size of $\sim$ Jupiter, its interior will be governed by electron degeneracy pressure, not by perfect gas pressure. The significance of this is that electron degeneracy pressure is only weakly dependent (or independent for a completely degenerate gas) on temperature. This means that the star can cool whilst only decreasing its radius very slightly. The central temperature would never reach the high temperatures required for nuclear burning and the "star" would become a hydrogen white dwarf with a final radius of a few hundredths of a solar radius (or a bit smaller for more massive stars).

The second possibility must be the fate of something the mass of the Sun. However, there is a cross-over point in mass where the first possibility becomes viable. To see this, we note that the radius at which degeneracy sets in depends on $M^{-1/3}$, but the radius the star needs to shrink to in order to begin nuclear burning is proportional to $M$. The cross-over takes place somewhere in the range 5-10 $M_{\odot}$. So stars more massive than this could commence nuclear burning at radii of about a tenth of a solar radius, without their cores being degenerate. An interesting possibility is that at a few solar masses there should be a class of object that contracts sufficiently that nuclear ignition is reached when the core is substantially degenerate. This might lead to a runaway "hydrogen flash", depending on whether the temperature dependence of the reaction rate is extreme enough.

Best question of the year so far. I do hope that someone has run some simulations to test these ideas.

Edit: As a postscript it is of course anomalous to neglect a quantum effect like tunnelling, whilst at the same time relying on degeneracy pressure to support the star! If one were to neglect quantum effects entirely and allow a star like the Sun to collapse, then the end result would surely be a classical black hole.

A further point that would need further consideration is to what extent radiation pressure would offer support in stars that were smaller, but much hotter.


Tools for creating a multiwavelength view of the sky

Aladin is a good way to do visualization from multiple catalogs. MAST also offers a cross-mission search.


You can stick a thermometer in space, and if it is a super-high-tech one, it might show you the temperature of the gas. But since the interstellar medium (ISM) is so dilute, a normal thermometer will radiate energy away faster than it can absorb it, and thus it won't reach thermal equilibrium with the gas.
It won't cool all the way to 0 K, though, since the cosmic microwave background radiation won't allow it to cool further than 2.7 K, as described by David Hammen.
The term "temperature" is a measure of the average energy of the particles of a gas (other definitions exist e.g. for a radiation field). If the gas is very thin, but particles move at the same average speed as, say, at the surface of Earth, the gas is still said to have a temperature of, say, 27º C, or $ 300\,\mathrm{K}$.
The ISM consists of several different phases, each with their own physical characteristics and origins. Arguably, the three most important phases are (see e.g. Ferrière 2001):
Molecular clouds
Stars are born in dense molecular clouds with temperatures of just 10-20 K. In order for a star to form, the gas must be able to collapse gravitationally, which is impossible if the atoms move too fast.
The warm neutral medium
The molecular clouds themselves form from gas that is neutral, i.e. not ionized. Since most of the gas is hydrogen, this means that it has a temperature of roughly $10^4\,\mathrm{K}$, above which hydrogen tends to get ionized.
The hot ionized medium
Gas that accretes onto the galaxy in its early phases tend to have much larger temperature, of roughly $10^6\,\mathrm{K}$. Additionally, the radiative feedback from the hot stars (O and B), and the kinetic and radiative energy injected by supernova explosions ionize and heat gas bubbles that expand. This gas comprises the hot ionized medium.
Cooling
The reason that the ISM is so sharply divided into phases, as opposed to just being a smooth mixture of particles of all sorts of energies, is that gas cools by various physical processes that have a rather temperature-specific efficiency.
"Cooling" means converting the kinetic energy of particles into radiation that is able to leave the system.
Hot gas
Very hot gas is fully collisionally ionized and thus cools mainly through free electron emitting Bremsstrahlung. This mechanism becomes inefficent below $\sim10^6\,\mathrm{K}$.
Warm gas
Between $10^4\,\mathrm{K}$ and $10^6\,\mathrm{K}$, recombinations (i.e. electrons being caught by ions) and collisonal excitation and subsequent de-excitation lead to emission, removing energy from the system.
Here the metallicity$^\dagger$ of the gas is important, since various elements have different energy levels.
Cool gas
At lower temperatures, the gas is almost fully neutral, so recombinations cease to have any influence. Collisions between hydrogen atom become too weak to excite the atoms, but if molecules or metals are present, it is possible through fine/hyperfine lines, and rotational/vibrational lines, respectively.
The total cooling is the sum of all these processes, but will be dominated by one or a few processes at a given temperature. The figures below from Sutherland & Dopita (1993) shows the main cooling processes (left) and the main cooling elements (right), as a function of temperature:

The thick line show the total cooling rate. The figure below, from the same paper, shows the total cooling rate for different metallicities. The metallicity is a logarithmic scale, so [Fe/H] = 0 means Solar metallicity, and [Fe/H] = –1 means 0.1 times Solar metallicity, while "nil" is zero metallicity.

Since these processes don't cover equally the full temperature range, the gas will tend to reach certain "plateaus" in temperatures, i.e. it will tend to occupy certain specific temperatures. When gas cools, it contracts. From the ideal gas law, we know that the pressure $P$ is proportional to the product of the density $n$ and the temperature $T$. If there's pressure equilibrium in the ISM (which there isn't always, but in many cases is a good assumption), then $nT$ is constant, and thus if a parcel of hot ionized gas cools from $10^7\,\mathrm{K}$ to $10^4\,\mathrm{K}$, it must contract to increases its density by a factor $10^3$. Thus, cooler clouds are smaller and denser, and in this way the ISM is divided up in its various phases.
So, to conclude, interstellar space is not as cold as you may think. However, being extremely dilute, it is difficult to transfer heat, so if you leave your spaceship, you will radiate away energy faster than you can absorb it from the gas.

$^\dagger$In astronomy, the term "metal", refers to all elements that are not hydrogen or helium, and "metallicity" is the fraction of gas that consists of metals.


Can smaller habitable planets (or moons) orbit a larger habitable planet?

First off, let me point out that if you have a central planet which is orbited by other bodies, those other bodies must necessarily be considered moons and cannot be planets. This is due to the official definition of a planet stating that it must orbit the central star. The objects orbiting your central planet are not primarily orbiting the central star and thus would be considered moons or satellites.

Orbital Dynamics

What you're proposing is that your central planet is orbited by two moons that are on opposite sides of the planet. This condition provides some strong constraints on your system as dictated by the laws of physics.

If we are to consider that the moons are orbiting around the central planet, then the planet must be fairly massive. If all three of your bodies are of a similar or comparable mass, you'd have a much different (and not necessarily stable) type of orbit than what you're describing (see this link for example). The fact that you have two moons orbiting your planet, without your planet having a noticeable orbit about the moons, implies the planet has most of the mass of the system. This results in one of two scenarios, either

your central planet is very large (e.g, the size of Jupiter or Saturn) with moons the size of Mercury or Ganymede, or
your central planet is relatively small (e.g., the size of Earth or Venus) with moons the size of asteroids (e.g., Phobos and Deimos, the moons of Mars)

Of course you could have any gradation between the two possibilities I just listed. The main point is, your planet must be much more massive than your moons in order to be the dominant gravitational player in its system.
You also state your moons are orbiting such that they're on opposite sides of the planet. Such an orbital configuration is known as a 1:1 resonance. It necessarily implies that your moons are orbiting at the exact same distance from the planet and take the same amount of time to complete a single orbit. While such a system is technically stable, I can't say with certainty that it is stable for a long period of time. More than likely, small perturbations will occur that will build up over time and cause your moons to leave their 1:1 orbital resonance. This will result in unpredictable outcomes, but my best guess is that the moons may start gravitationally interacting if they get close enough after no longer being on the same side of the planet. Eventually this may result in them exchanging energy and one getting kicked to a closer orbit while another getting kicked to a further orbit. But that's just a guess.

So short answer, yes your system is possible but it comes with some constraints and is likely not going to be stable for a long time period.

Habitability

You also ask about the habitability of such a system. This is a bit tricky and there's no single answer. There are so many conditions that affect habitability. To make matters worse, we only have a single example of a habitable planet. However, there are general conditions we believe must be true for a planet to be habitable (for humans at least).

The big condition is that your system must be at a good temperature such that water can exist as a liquid. If your central planet is a Jupiter/Saturn analog, this is not going to happen since such planets have large, mostly hydrogen gaseous envelopes. There's not really any pools of water for humans. However, your moons could still host water. It really depends on how far out your planet is from the star. These types of planets form very far from the star, where water doesn't easily exist as a liquid so either your planet must migrate in to the inner stellar system early on (and stop at a respectable distance), or else some sort of tidal effect has to keep your moons much warmer than they would be other wise (see Enceladus, for example). The former means your chosen moon orbits are likely to be less stable and the latter means you're going to need other moons to help provide the heating, again making your orbits less stable. If your central planet is Earth-like, it can certainly host water and be as Earth-like as you want, but the moons are unlikely to be big enough to be habitable.

There are various other conditions for habitability such as nice atmospheres or magnetospheric protection from the star but one could write a whole book on this subject. I'll leave off by pointing you to this question on habitability.


Is the amount of energy that dark energy applies to push objects apart equal to the amount of energy lost because light from distant galaxies is red shifted?

No, dark energy existed prior to the expansion and the shift towards blue or red is based on the direction of movement of the emitter relative to the observer, so the amount isn't equal.

Sources:

Wikipedia: Dark Energy - "Change in expansion over time":

  "Recent results from the Hubble Space Telescope Higher-Z Team indicate that dark energy has been present for at least 9 billion years and during the period preceding cosmic acceleration.".

Wikipedia: "Integrated Sachs–Wolfe effect":

  "The integrated Sachs–Wolfe (ISW) effect is also caused by gravitational redshift, but it occurs between the surface of last scattering and the Earth, so it is not part of the primordial CMB. It occurs when the Universe is dominated in its energy density by something other than matter. If the Universe is dominated by matter, then large-scale gravitational potential energy wells and hills do not evolve significantly. If the Universe is dominated by radiation, or by dark energy, though, those potentials do evolve, subtly changing the energy of photons passing through them.

  There are two contributions to the ISW effect. The "early-time" ISW occurs immediately after the (non-integrated) Sachs–Wolfe effect produces the primordial CMB, as photons course through density fluctuations while there is still enough radiation around to affect the Universe's expansion. Although it is physically the same as the late-time ISW, for observational purposes it is usually lumped in with the primordial CMB, since the matter fluctuations that cause it are in practice undetectable.".

Wikipedia: "Theories of Dark Energy":

  The Equation of State (EoS) of Dark Energy for 4 common models by Redshift:

              A: CPL Model   B: Jassal Model   C: Barboza & Alcaniz Model   D: Wetterich Model

This question was closed on our Physics.SE site: "Could some Red and Blue shifts be the result of light passing through “dark matter”? [closed]", while this: "Redshifted Photon Energy" was not.

See also:

"New Aspects of Photon Propagation in Expanding Universes" (Oct 6 2016), by H.-J. Fahr and M. Heyl.
Wikipedia: "Friedmann equations":

  "The Friedmann equations are a set of equations in physical cosmology that govern the expansion of space in homogeneous and isotropic models of the universe within the context of general relativity.".

Wikipedia: "Cosmic Expansion History":

  Since the densities of various species scale as different powers of $a$, e.g. $a^{-3}$ for matter etc., the Friedmann equation can be conveniently rewritten in terms of the various density parameters as

  $$H(a)\equiv {\frac {\dot {a}}{a}}=H_{0}{\sqrt {(\Omega _{c}+\Omega _{b})a^{-3}+\Omega _{\text{rad}}a^{-4}+\Omega _{k}a^{-2}+\Omega _{DE}a^{-3(1+w)}}}$$

  where $w$ is the equation of state of dark energy, and assuming negligible neutrino mass (significant neutrino mass requires a more complex equation). The various $\Omega$ parameters add up to 1 by construction.

As you can see the contribution to shift (either way) is very small compared to the amount of dark energy present. The dark matter particles interact with each other and other particles only through gravity and possibly the weak force.


Good source for the relationship between absolute magnitude, diameter, and albedo?

The JPL CNEOS
asteroid size estimator
and various asteroid albedo papers cite
Harris and Harris 1997.
That paper is behind a paywall, but
Stuart and Binzel 2004
attribute this formula to it:
$$ H = C - 5 \log_{10} D - 2.5 \log_{10} p_V$$
where $H$ is absolute magnitude, $p_V$ is albedo, and $C$ = 15.618.


The answer is right there in your screenshots. The 'dimensione' or angular diameter on the sky is 3 degrees for M31, whereas the moon has an angular diameter of only 0.5 degrees. In other words, M31 is about 6 times as wide as the moon.

As wikipedia correctly states, you need a large telescope to see the full extend of M31. With the naked eye of even a small telescope you simply cannot see the spiral arms.

I think this image shows it very neatly:
http://apod.nasa.gov/apod/ap061228.html


Are we sure solar cycle is not related to the orbit of Jupiter?

I think for people who want to believe the Jupiter link to solar cycles it will be impossible to disprove.  They'll cling to it forever.

I think in a practical sense the "link" is little more than numerology.  The force Jupiter can exert on the Sun's atmosphere and internal structure is a minute fraction of the magnetic forces at work generated by the Sun itself.

A simple model helps.  Place two magnets a foot apart.  Now place a magnet a thousand feet away from both.  The effect of the distant magnet is one millionth that of the two close together.  That gives you a feel for the relationship between magnetic forces within the Sun's atmosphere and Jupiter's distant influence.  And gravity is an even weaker force.

As I understand it the current daring of models is the Double Dynamo, but certainly the dynamo type models ( all relating to electromagnetic fields generated by the Sun ) are the main contenders.

With people who propose the Jupiter link ( often extended to include Saturn ), there is a tendency to do a lot of algebra to find ways to relate various periods without actually justifying the supposed link on the basis of forces at work ( i.e. doing a simulation or a model based on real physics ).  This is why I call it numerology.

The dynamo models actually are based on real physics and make predictions, both past and future, which can be tested.  This also means the details of these theories can be refined in a proper scientific manner to tease out the truth.


Under what situations can an aperture mask improve the resolution of a small/medium amateur telescope? Is this demonstrable mathematically?

The vast majority of people - almost everyone, really - should not rely on this on a daily basis. The chances are overwhelming that they will reduce the performance of their instruments. There are so many variables involved, and it depends on so many factors, a likely result is worse performance.

The link is to a site maintained by Mel Bartels. Mel is a very experienced observer and telescope builder, with hundreds if not thousands of hours behind the eyepiece. He has enough practical experience so that, just by looking at the overall image, and the shape and size of the Airy disk and diffraction figure, he can tell exactly in what way the telescope's behavior is affected by changes.

The effects of the secondary mirror and spider are a topic which is much misunderstood. You will encounter many strong opinions on this topic with not much basis in reality.

The spider vanes, if straight, will each create two spikes, diametrically opposed, perpendicular to the length of the vane. So a 3-vane spider makes 6 spikes. A 4-vane spider makes 8 spikes, but coincidental two by two (look like 4 brighter spikes). If the vanes are curved, then the diffraction figure is more like a faint, diffuse cloud.

So using an aperture mask can remove the spikes due to the spider vanes. But will it make the image "better"? Spikes don't matter from a practical perspective. If a spike is blocking one of the components of a double star, just rotate the scope. And the theoretical resolving power is diminished. You are, after all, using a smaller aperture.

The secondary mirror will increase the size of the diffraction figure. The Airy disk and the first minimum become larger in diameter - and so do the rest of the rings. This is a matter of much confusion, with some claiming that obstructed telescopes (scopes with a secondary mirror) are essentially inferior to unobstructed designs - this is demonstrably false.

Simple rule of thumb: if D is the diameter of the primary, and d is the diameter of the secondary, then the performance of an obstructed scope is equivalent, from the p.o.v. of resolving power (high frequency components), to an unobstructed instrument of aperture D - d. This is not an exact formula but it's close enough. A 300mm reflector with a 60mm secondary has about the same fine detail performance on planetary observations like an unobstructed 240mm instrument.

There are cases when you're observing close double stars, and it would be great if one of the stars could fall into the diffraction minimum of the other, brighter star, just outside the Airy disk. Sometimes you can do that with an aperture mask.

The article by Mel Bartels makes an important point, and I wish he emphasized it more: the aperture mask should only be used at low or medium magnification. In other words, if you're planning anyway to use low or medium magnification, then in some rare cases the aperture mask might be useful. It NEVER makes sense to use an aperture mask at high magnification - it's like pushing a telescope above its maximum useful magnification, and the result is just a blurry mess.

The linked article also mentions the issue of brightness. Some observers claim that the Moon is too bright in a telescope. Mel Bartels sees this as an opportunity to use an aperture mask (again, NOT at high magnification). Others use Moon filters - neutral density filters that reduce the incoming flux of light.

There is a third school of thought here. Do not become dark-adapted while observing the Moon. Do not observe the Moon from a dark place. Observe it with lights turned on around you, like from the sidewalk of a well-illumined street. This is what I, and others like me, do. Suddenly the Moon is not too bright anymore. The telescope is not performance-limited by the aperture mask, and you can zoom in at high magnification to observe small details of craters and mountains. And your eyes are working in their optimal regime, when the pupil is not fully dilated.

Surface brightness of extended objects never increases in a telescope, it just seems to increase. In fact, it always decreases. The Moon seems too bright because you're dark-adapted - but in reality it's quite pale.

There is another case when the image is demonstrably improved by an aperture mask. This is when the primary mirror has strong defects near the perimeter, like TDE (turned down edge), which is a defect both quite common (for mass-manufactured instruments) and very damaging to performance. A scope with strong TDE has poor resolving power and poor high frequency contrast. Sometimes it's so bad that even an aperture mask smaller than D/2 will improve performance.

The solution here is not the aperture mask, but instead masking off the perimeter. Make a ring with an inner diameter somewhat smaller than the aperture and place it permanently on the mirror, so as to block off the defective edge. Aperture is somewhat diminished, but now the optics are good across the whole surface.

One more case: let's assume seeing is so bad, the image is shaking like a giant, blurry pile of jelly. There's no way you're going to use high magnification anyway. You operate at low-to-medium magnification. Now apply the aperture mask. Is the image better?

Objectively the answer is no. If you're doing lucky imaging with a camera, it's still better to leave aperture alone, and by luck you will capture a few frames where detail resolution is pretty good.

But the human visual system is weird. Some people prefer to let go of the higher performance ceiling, instead opting for an image that's perceived as more smooth and stable. Personally I prefer to let low magnification do its thing, and leave aperture alone, but not everyone is the same.

Even visually, if you don't mask the aperture, sometimes seeing gets better briefly, and you can see small details for a while. With an aperture mask, you will miss those.

But this is a subjective thing, it's not something that can be objectively measured, except as a study of the psychology of visual perception.

If you want to experiment with the aperture mask, by all means go ahead and do it. But be aware of the many, many pitfalls here. It's not a silver bullet. It's not the solution to all of humanity's problems. It's just a trick that sometimes works, in a fairly narrow slice of use cases.


Low precision GMST formula clarification

The number 18.697374558 was the value of GMST (in hours) at noon UT on 1 January, 2000. The number 24.06570982441908 reflects the fact that it takes slightly less than 24 hours (23.934469591898 hours to be precise) for a star to the appear to rotate by 360 degrees.

This is a consequence of the fact that the Earth orbits the Sun. Our 24 hour day is based on the time between two consecutive solar noons. Because the Earth moves a bit along its orbit, the Earth has to rotate by more than 360 degrees (with respect to the stars) to see the Sun rotate by 360 degrees.

Another way to look at it: The 23.934469591898 hour long sidereal day means that a solar day is 24.06570982441908 hours long expressed in sidereal hours.


The idea of wormholes is completely abstract.  We have no evidence they actually exist and they are a result of following the strict mathematics of general relativity to and beyond it's intended limits.

When you push a theory past it's "design limits" you can't expect the results to be valid.  Sometimes they are, sometimes they aren't.

  for me curvature requires a higher dimension (paper bending or a stick bending, always +1 to its dimension)

The problem here is that you're thinking in a common, everyday sense about something that fundamentally is not understandable using common everyday sense.  One of the major problems people face learning even special relativity is that common sense ideas (like simultaneity) go out the window, and this is even more difficult in general relativity.

To understand how space-time is distorted in such extreme ways requires an understanding of the mathematics.  The simple explanations that involve things like rubber-sheet analogies just don't give you the right idea at all.

A typical tool to explore space-time and how it distorts is a Carter-Penrose diagram.  Another one is the Kruskal-Szekeres coordinates.

So really the problem here is you trying to use your common everyday sense in a situation that it does not work in.  The reason physicists do all that complex mathematics in the first place is because that is what works.


What is the Northernmost Latitude of Saturn?

Salem is at 42.5° north. At the equator, the position of the sun varies by 23.5 degrees on either side of the vertical over the course of each year, due to earth's 23.5° axial tilt with respect to the ecliptic.

Saturn, however is not on the ecliptic. It departs by about 2.5° over the course of its 29.45 year, 9.5 AU orbit. So once every 29.5 years Saturn's angle to the ecliptic will add 2.5° to the max latitude the planet can be seen overhead.
That adds up to 26° as the max latitude Saturn can be seen directly overhead.

There's one more factor to account for though. Earth being 1AU out from the sun, it's only 8.5 AU from Saturn at close approach. That will add another 0.2° (by trig) to the max latitude at that time of year when earth is nearest Saturn. That gets us to the possiblity of Saturn appearing directly overhead at up to latitude 26.2°: Miami, Jaipur, or Dubai, not Salem


Term for the moment when hydrogen fusion begins in a star

Ignition or more fully Stellar ignition is the term you want.


Why won't the Sun set for days at N66.2 which is below the arctic circle?

From the PyEphem Quick Reference Guide:

Rising and setting are sensitive to atmospheric refraction at the
horizon, and therefore to the observer’s temp and pressure; set the
pressure to zero to turn off refraction.

It seems likely that, if you're using the default settings, the result returned is including atmospheric refraction, giving the results you would expect to see at the location, where the sun is visible even though a direct line to the disc of the sun would put it below the horizon.

Image is from "What is Refraction of Light" by By Konstantin Bikos and Aparna Kher at timeanddate.com