EDIT: http://wgc.jpl.nasa.gov:8080/webgeocalc/#AngularSeparationFinder lets you find planetary conjunctions online using NASA's data. It's still iterative, but fairly fast (since NASA uses fairly powerful servers, even for their websites).

Summary: I'm still researching, but there appears to be no well-known,
reliable non-iterative method to find conjunctions. Using the
iterative method and the C SPICE libraries, I created a table of
conjunctions for the visible planets (Mercury, Venus, Mars, Jupiter,
Saturn, Uranus) here:

http://search.astro.barrycarter.info/

Full "answer":

I am still researching a general answer to this question ("How to
calculate conjunctions of 2 planets"), but here's what I have so far.

The iterative method:

Compute the positions of the planets at regular intervals (eg,
daily). The "daily" works for planets (but not some asteroids and
definitely not the Moon) because the planets move through the sky
relatively slowly.
Find local minima in the daily lists.
For efficiency, carefully discard local minima that are too
large. For example, Mercury and Venus may approach each other, reach
a minimal distance of 20 degrees, and then drift apart. The 20
degrees is a local minima, but not a conjunction.
However, be careful when discarding minima. If you are searching
for 5-degree conjunctions, two planets may be 5.1 degrees apart one
day, and 5.2 degrees apart the next day, but less than 5 degrees
apart sometime in the interim.
For 5-degree conjunctions, you only need daily minima less than 8
degrees, and even that is overkill. The fastest a planet can move in
the sky is 1.32 degrees per day (Mercury), and the second fastest is
1.19 degrees per day (Venus). In theory, these movements could be in
opposite directions, so the fastest two planets can separate is 2.51
degrees per day. So, if two planets are more than 8 degrees apart
two days in a row, there is no way they could be closer than 5
degrees between the days.
In reality, planets maximum retrograde angular speed is slower
than the prograde maximum speed, so the 2.51 degree limit above is
never actually reached.
After finding local minima, use minimization techniques (eg, the
ternary method) to find the instant of closest approach.
I ended up using the C SPICE libraries, and found 32,962
six-degrees-or-less conjunctions between -13201 and 17190, averaging
about 1 conjunction per year. Of these, 2,185 occur between the "star
of Bethlehem" and the 2015 conjunctions:

http://12d4dc067e0d836f1541e50125c24a03.astro.db.mysql.94y.info/

This iterative process works, but can be tedious. Since planetary
positions are semi-well-behaved, you'd think there would be a faster,
non-iterative method. Well...

As https://math.stackexchange.com/questions/95250/ notes, people
have been interested in finding planetary conjunctions for at least
2000 years, probably longer. The first such known tool is the
https://en.wikipedia.org/wiki/Antikythera_mechanism. The "Operation"
section of the wikipedia page suggests this device was hand-cranked,
and thus iterative.
SPICE
(https://en.wikipedia.org/wiki/Spacecraft_Planet_Instrument_C-matrix_Events)
is one of the oldest digital computer attempts to compute planetary
positions, and it includes a routine to compute conjunctions:

http://emfisis.physics.uiowa.edu/Software/C/cspice/doc/html/cspice/gfsep_c.html

However, this also uses an iterative method, as the long description
of the "step" parameter indicates:

"step must be short enough for a search using step to locate the time
intervals where the specified angular separation function is monotone
increasing or decreasing. However, step must not be too short, or
the search will take an unreasonable amount of time"

By experimentation, I found that a step size of 6 days does not find
all Mercury/Venus minimal separations, although a step size of 1 day
does find these. In other words, reducing the step size from 6 days to
1 day found additional conjunctions, but reducing the step size to
well below 1 day did not produce any additional conjunctions.

Akkana Peck (akkana@shallowsky.com), creator of a well known script to find
conjunctions (https://github.com/akkana/scripts/blob/master/conjunctions.py)
tells me:

M[y script] iterates through dates.

[...]

In Astronomical Algorithms, Meeus has a chapter (17) on Planetary
Conjunctions, but the chapter is less than three pages and lacks
detail except that he's pretty clearly iterating and looking for
the smallest separation, same as my Python program does.

He does a little better in Astronomical Formulae for Calculators:
sounds like he's advocating iterating over big steps to find the
place where the separation stops decreasing and starts increasing,
then using "inverse interpolation" to find the exact time.

"Mathematical Astronomy Morsels" (also Meeus) talks starting on p.
246 about triple conjunctions (defined as two planets having several
conjunctions in a short time, due to retrograde motion, not three
planets all in conjunction at once) and gives some examples for very
specific pairs, like Jupiter and Saturn, with ways you could look for
those conjunctions.

It doesn't look like any of these books would be very helpful in
finding a non-iterative solution.

I haven't had a chance to read the books above, but did find:

http://adsabs.harvard.edu/abs/1981JRASC..75...94M

where Meeus confirms the standard iterative method, but also provides
a different, less accurate method. Unfortunately, Meeus only uses this
method to compute solar conjunctions, elongations and oppositions, not
interplanet conjunctions.

Jon Giorgini of NASA (jdg@tycho.jpl.nasa.gov) tells me:

As far as NASA computed planetary/stellar conjunctions/occultations, I
don't know of anyone within NASA that does that routinely. There is an
external network of volunteers that does that under the umbrella of
the International Occultation Timing Association (IOTA), and they have
developed pretty refined internal software for that purpose.

I contacted IOTA (http://occultations.org/) and they confirmed
they use an iterative method as well:

[...] the software package Occult does generate planetary conjunction
predictions - based on a two-body solution. The approach used is a
crude brute-force method of generating the planetary ephemerides on a
daily basis.  It is not particularly efficient - but it is sufficient
for its intended purpose. [...]

As a note, IOTA focuses on asteroid occultations, so computing
positions daily doesn't always work. Especially for near-Earth
asteroids, IOTA must iterate considerably more frequently.
I also tried contacting Fred Espenak, the creator of
http://eclipse.gsfc.nasa.gov/SKYCAL/SKYCAL.html, but was unable to
do so. Jon Giorgini tells me that Fred has retired.
I'm still looking, but my current conclusion is that there is no
good well-known non-iterative way to find planetary conjunctions. As
the image in my
https://mathematica.stackexchange.com/questions/92774 shows,
planetary separations aren't really as well-behaved as I had hoped.
I just got a reply from Arnold Barmettler, who runs calsky.com:

I'm using a very time consuming iterative approach to pre-calculate
Bessel Elements for conjunctions. This allows fast online calculation
for any place on earth.
Initial calculations are only done every few years, so CPU time does not
matter. This would change if I'd enter the business to calculate
asteroidal occultations.

re-iterating (pun intended) the same theme.

Miscellaneous:

I used planetary system barycenters (the center of mass of a
planet and its moons) for an entirely different reason. If you ask
HORIZONS (http://ssd.jpl.nasa.gov/?horizons) for the position of
Mars and set the date, you'll see this notice:

Available time span for currently selected target body:
1900-Jan-04 to 2500-Jan-04 CT.

However, if you use Mars' barycenter, this becomes:

Available time span for currently selected target body:
BC 9998-Mar-20 to AD 9999-Dec-31 CT.

In other words, NASA computes the position of Mars' planetary system
barycenter for a much longer interval than they compute Mars' actual
position. Since I wanted to compute conjunctions for a long period of
time, I went with the barycenters (DE431 computes barycenters even
beyond 9998 BC and 9999 AD).

I've complained that this is silly, especially for Mars, since the
distance between Mars' center and Mars' planetary system barycenter is
only about 20cm (yes, centimeters, since Phobos and Deimos have very
little mass) per
http://naif.jpl.nasa.gov/pub/naif/generic_kernels/spk/planets/aareadme_de430-de431.txt. However,
NASA apparently plans to keep it this way.

I ignore light travel time, which introduces a small error. Most
planetarium programs, like Stellarium, ignore light travel time by
default, although Stellarium has an option to turn it on.
I also ignore refraction, aberration, and similar minor effects.

More details on how I generated these tables (in highly fragmented form):

https://github.com/barrycarter/bcapps/tree/master/ASTRO

(README.conjuncts in the above is a good starting point)

Some of the "cooler" conjunctions I found are at:
http://search.astro.barrycarter.info/table.html. Here's a screenshot:


Is the Tombaugh Regio on Pluto visible from Earth?

Your First Question - Can it be done from the ground?

Unfortunately making maps that have this level of detail from the ground is not possible with current telescopes. There are 2 problems; first, it takes a lot of magnification to resolve Pluto and even more to resolve surface features. Second, from the ground you need more than just magnification, you need EXTREMELY stable skies or that will wash out the surface features. Getting just one super sharp image is not enough, you need several to ensure that you have high enough signal to noise to believe the surface features are real.

This is why Marc Buie, who used the High Resolution Channel on the Advanced Camera for Surveys on HST to create the image you put in your question. Unfortunately, even that level of resolution is not possible anymore as HRC is no longer functional. These kind of maps can be very hard to make.

There have been other maps of pluto, both from the ground and using HST. Here is a compilation of several using a compilation of ground based photometry, the Fine Optical Camera on HST and the ACS HRC. As you can see from the ground based photometry you can make a map, but it is very poor resolution compared to what HST can do.

There have been other HST press releases about those other images

Second Question - How well do the maps match?

There is a good blog post about how they match up from the museum of applied arts and sciences at the Sydney Observatory. But the bottom line is that they match up pretty well, but not perfectly. This is expected as we expected volitle transport on the surface so the surface features will change. As you can see in the image below, that there were a lot of changes from the map made in 1994 to the one made in 2003. So one would expect changes going out to 2015.

So, yes your intuition is valid, but Pluto's surface is dynamic so it will be changing on you!


Could information be transmitted from inside a black hole using gravitational waves?

A black hole is defined as a region of spacetime where the gravitational effects are so strong that the escape velocity within the event horizon is greater than the speed of light.  According to relativity, nothing can propagate faster than the speed of light, so at most gravitational waves would propagate at this speed.  According to particle physics, mediation of gravity would be controlled by massless particles called gravitons, which, by virtue of being massless, would travel at the same speed as photons, i.e. the speed of light.  Suffice to say gravitational waves (analogous to electromagnetic waves) would not be able to escape from inside the event horizon.

The question of information, in regards to black holes is open however. The No-Hair Theorem would seem to imply the destruction of information as the black hole consumes matter.  However, the holographic principle, if correct would imply that information regarding the matter that formed the black hole would be encoded on the 2-dimensional boundary of the event horizon.  This information could very well be exhausted by Hawking radiation or gravitational waves.  Since this information was outside of the event horizon this would be fully consistent with the formulation of black holes.


Do the stars in irregular galaxies orbit anything?

Although the answer you link to doesn't mention irregulars, the answer applies to those as well: Star move around in the common gravitational potential created by everything in the galaxy, i.e. gas, stars, and, in particular, dark matter.
This potential has a center, but there isn't necessarily anything exactly at this center. The stars then move on elliptical orbits around the center, but are perturbed by local irregularities.
In spirals and elliptical, the potential is rather symmetric, whereas in irregulars it is quite… irregular. That means that, if you were to take two images of a galaxy, separated by a period of order the dynamical time scale ($t_\mathrm{dyn} \sim \sqrt{R^3/GM} \sim 100$ million years), a spiral and an elliptical would look more or less the same, while the irregular would probably have changed its shape notably.


From the Report of the IAU/IAG Working Group on Cartographic Coordinates and Rotational Elements of the Planets and Satellites (2000), section 3:

  The rotational pole of a planet or satellite which lies on the north side of the invariable plane will be called north, and northern latitudes will be designated as positive.
  The planetographic longitude of the central meridian, as observed from a direction fixed with respect to an inertial system, will increase with time. The range of longitudes shall extend from 0° to 360°.

  Thus, west longitudes (i.e., longitudes measured positively to the west) will be used when the rotation is prograde and east longitudes (i.e., longitudes measured positively to the east) when the rotation is retrograde.

The "invariable plane" in question is the invariable plane of the Solar System.

By contrast, the usual convention for quoting obliquity is to use the right-hand rule (fingers curl in direction of rotation, thumb points north). This is inconsistent with the definition used to determine the north pole but people seem to prefer having the rotation period always be a positive number.


I recommend you to take a look at Wikipedia about the ISS's orbit: Wikiepdia - ISS: Orbit

There is also a very interesting video that I think will answer to your question: ISS Orbit animation


Is there a web service providing orbital data of solar bodies (planets, comets, asteroids, dwarf planets)?

Not quite the webservice, but there is a JavaScript library Orb2.js with examples including X-Y(-Z) in ecliptic coordinates. So maybe you can compute it directly, not relying on some external service:

http://www.lizard-tail.com/isana/lab/orbjs/

As far as I do not use it actively, it just a pointer to site to explore, not a definitive solution for you.


Can someone explain this diagram showing the spectral type distribution of bright stars

It's all to do with the relationships between mass, spectral-type and luminosity and the initial mass function of stars.

I think your explanation of points 1 and 2 are completely correct. O and B stars are rarely born and short-lived; so even though they have enormous luminosities relatively few make it into a list of stars ordered by apparent brightness. A further important reason is that although they are immensely luminous, the more distant examples that might have made it into the list, cannot make it into the list because they are in the Galactic plane (they are young and still close to where they were born) and thus severely affected by extinction.

M-dwarfs are indeed the most common type of star, but not common enough to overcome the fact that their luminosities are much, much lower than hotter stars.

The dip in G stars is actually a peak in the K-stars caused by giants. Most low(ish) mass stars will evolve towards more or less the same point in the HR diagram forming the red clump. These stars are luminous, older and not concentrated in the galactic plane, can therefore be seen to great distances and produce a peak in the number counts.

So you can see this using actual data rather than a cartoon, I producedthe HR diagram for the brightest 10,000 stars in the revised Hipparcos catalogue (about V<6.7). You can clearly see the red clump of giants. Now collapse this distribution onto the B-V axis and you can see there is a local minimum at $0.6<B-V<0.8$ corresponding to G-stars.

Absolute colour-magnitude diagram for the brightest 10,000 stars

The corresponding B-V distribution for these stars. G-stars have $0.6<B-V<0.8$


Was Guth inflation propelled by what we now call "Dark Energy"?

One flaw is saying that the big bang occurred at a location. It didn't: the whole universe was once part of the big bang. So the stuff about deceleration being seen as acceleration isn't true.

Secondly, shortly after the big bang, the universe behaves "ballistically" (although following the rules of General Relativity, not Newtonian gravity). It was initially decelerating and the influence of dark energy was negligible.

But in the popular $\Lambda$CDM model (the prevailing model at the moment), the dark energy density is constant, so as the universe gets bigger it becomes more influential and the universe began to accelerate a few billion years ago.

None of this has much to do with inflation, which, if it happened, happened in the first tiny fraction of a second, when the universe was dominated by some energy density, similar in effect to dark energy (in that makes space expand), but much, much stronger. Exactly what this energy density was is still speculative, and inflation remains a "bolt-on" to the big bang model to explain why space is flat and why different parts of the universe appear very similar.


Orbital eccentricities and inclinations are damped to zero during the protoplanetary disc phase. This is because the gaseous parts of the protoplanetary disc are a great sink for excess angular momentum.
After the gas disc dissipates, the formed planets start to interact dynamically as they do not necessarily form in stable configurations. This can lead to the exchange of angular momentum. A giant planet with small initial eccentricity can hereby pump a lot of eccentricity into less massive planets, as the angular momentum $L=rmv$ is conserved.
So 'classical' thinking involves the exchange of angular momentum between planets after the gas disc dissipates as the origin of exoplanetary eccentricities.
A modern caveat to this is that there are mechanisms to incline parts of the inner gas disc during the planet formation phase w.r.t. the rest of the disc. As eccentricity and inclination are intrinsically linked, those inclined sub-discs may form inclined planets, which later lead to strong eccentricity pumping.


The systematic review of past observations in the hope of finding things that may have been missed

The first step for most of these projects is to make the old data available. Just about all of the old data is on photographic plates, so there are projects at many of the great observatories to scan their archives.

A good example is the project at Smithsonian Astrophysical Observatory (located at Harvard) to digitize the entire plate collection, ranging from ultra-early photographs to sky surveys that Harvard Observatory did during a large part of the 1900s. The plates are scanned, and they are matched with star atlases to determine their exact coordinates and their limiting magnitude and the filters used, if any.

This data -- and lots more from all over, including the epochal Palomar Sky Survey -- is available (or soon will be) using special software such as Microsoft's Worldwide Telescope.

So the old data is being collected and curated and being made available to everyone.

(Note that the old data, being photographic, is limited in how deep they go, so many objects are simply not visible in them, including most minor planets.)

And lots of projects make use of it.  The ones I'm familiar with tend to use the data to look for prediscovery images or to extend light curves into the past.

New surveys like the Large Synoptic Telescope will generate cubic shittons of data and are being deliberately designed to separate the collection of data from the use -- it'll collect deep full-sky images every few days and then people will mine that data for things that the people who planned the survey hadn't thought of.  It will revolutionize astronomy.  Key point: All this old data is sure to be made available in the same way and the very bright people developing ways to find interesting stuff in the LSST will also find good stuff in old photographic plates from the 20s.


Is nuclear pasta or neutron star crust iron stable outside of neutron stars?

"made of a form of highly compressed iron called nuclear pasta,"

This isn't the case. The crust of a neutron star consists of (starting at the surface going down).
(Picture from Watanabe & Maruyama 2012)

Iron-peak nuclei plus degenerate electrons.

Increasingly neutron-rich elements with higher atomic numbers (i.e. not iron) plus degenerate electrons.

Very, very neutron rich heavy nuclei (mass numbers of several hundred) plus degenerate electrons and free neutrons.

Nuclear pasta, where the identity of individual nuclei disappears and the nucleons (mainly neutrons) form long chains, or sheets, surrounded by degenerate free neutrons and electrons.

None of these compositions is stable at low pressures. Specifically, you need to have them surrounded by a high density gas of degenerate electrons to suppress the beta decay of neutrons and neutron-rich nuclei. Putting this stuff into a low density environment would be explosive (but somewhat less) just as in the case of "neutronium" (neutron matter inside neutron stars contains protons and electrons too). This is not because of the beta decay, which is a slow process, but because the constituents have a huge amount of kinetic energy (that is what pressure is).


What is the composition of the Solar Wind?

We have very good data on the heavy metals in the solar wind from the Charge, Element, Isotope Analysis System (CELIAS) on SOHO:

Some of these elements were previously known; others were observed for the first time. The new ones are phosphorus, chlorine, potassium, titanium, chromium, and nickel. They were detected in smaller amounts than, say, carbon, oxygen or calcium.

Now, the seeming overabundance of heavier elements, with the peak at iron, might seem surprising. One would think that the lighter metals - carbon, oxygen, etc. - might appear in higher quantities, given their substantially lower masses, which make it easier for them to escape the Sun's atmosphere. However, this is due to experimental bias. The instruments on SOHO were designed to search for these particular heavier elements, which of course led to the detection of several new isotopes. We now have a much better picture of what heavy metals can be found in the solar wind, but the chart above should not be taken as evidence of the relative quantities, as a whole.

Additionally, there actually is some neutral hydrogen, both in the solar wind and the solar atmosphere (see Olsene et al. (2004) and Collier et al. (2001)). It should be present in the corona, and can be detected and modeled by Lyman $\alpha$ scattering. It was also detected and studied by the Low Energy Neutral Atom (LENA) imager on IMAGE.


Is it possible for the year to have 11 full moons (i.e. the opposite of a blue moon)

Since a full moon happens once every 29 days (and that's not exact, but close enough), the longest period of time where you can have 11 full moons but not 12 is 11 x 29 + 28, or 347 days, so in a normal year, it's not possible.  There are slight variations in the time between full moons but not nearly enough to make up the extra 18 days needed.

So you'd need a special circumstance.   Great Brittan's short year in 1752 wasn't sufficiently short, it had 11 days removed.

There is a phenomenon where not everywhere on Earth gets a full moon every 29 days.   In the North and South poles for example, where during summer they get 24 hours of sun and winter, no sun, a somewhat similar effect happens with the moon.   The Moon orbits the Earth at 5 degrees off the ecliptic or basically, the plane between the Earth and the Sun and that makes the Moon's apparent movement across the sky a little more complicated than the Sun's, but because full moons only happen when the moon is roughly opposite the sun, it's not hard to see that when the sun is in the sky for 24 hours during the Polar summer, they don't get a full moon for several months because the full moon is below the horizon when the sun is in the sky 24 hours.   You might only see 6-9 full moons a year at the poles (if I was to guess), and there's probably a range of latitudes close to the poles where 11 full moons might be about the yearly average.

If you want to split hairs even further, the Moon only takes an instant to pass the 180 degree point in relation to the Earth and Sun, so in a sense, a perfect 180 degree full moon is only visible to half the earth and by the time the other half of the earth turns to make that moon visible, the moon has already gone past 180 degrees and from a certain perspective, it's no longer full.   But, a full moon is generally regarded as a roughly 24 hour period, not an instantaneous moment, so I don't like that argument personally.


Why does the Moon never set in Svalbard, Norway?

You must have misheard it, or the documentary you watched wasn't presenting very precise information. It does set but it also stays on the night sky for several days during the polar winter (polar night) when the Moon if full. This is relatively simple to imagine, so I'll describe it;

So what's happening is that the Earth's axial tilt during the polar winters leans the whole Northern hemisphere towards the night side, away from the Sun. This tilt is big enough (~ 23.4°) that the night sky objects aligned with the Earth's equatorial plane stay visible relatively low on the horizon. With those regions being either relatively flat and/or with a view towards the sea, there's not many obstructions limiting the viewing angle, so the Moon (and analogous also the Sun during polar summers) stays "locked" low above the horizon. To help a bit with imagining this, here's an animation of the Earth's axial tilt, courtesy of Wikipedia:

If we imagine this animation of the Earth with the Sun in the distant left of the image, so during Northern hemisphere's winter (winter solstice to be precise), and the Moon to the distant right of the image (roughly 25 widths of the image away), so when it's either full or close to this lunar phase, it's not too difficult to appreciate that the northernmost polar regions have a direct line of sight of the Moon during Earth's full rotation on its axis, or a day. If you keep in mind that other celestials, including the Moon, are oblivious to the Earth's axial tilt (well, not exactly, but let's not nitpick about tidal effects that might take millions of years to make a difference), as the Moon moves farther in its orbit, in our case towards the viewer, this observation angle decreases further still and those northernmost latitudes hide to us for some part of the day. At lunar last quarter, it would be directly towards us relative to the image, so this direct line of sight relationship between the Earth and the Moon becomes reciprocal to how we're seeing places on the Earth on the animation.

Why when the Moon is full? Simply because that's when the Moon is also behind the Earth (but not in its shadow), so the relative angle between the observation point and the Moon would stay high enough to observe it. As it moves in lunar phase and in orbit around the Earth farther, this angle becomes lower and the Moon indeed does set also in the arctic region. For what is worth, this goes exactly the same for South pole, only with a half a year difference.

One other effect that plays a role here is the Earth's atmospheric refraction which also adds to the duration during which the Moon appears not to set. Meaning, that even when the Moon wouldn't be in direct line of sight, but only marginally so, it would still appear low on the skies due to optical effect (displacement) of the atmosphere. This effect would somewhat offset observing the Moon from lowlands with possibly shallower observation angle when compared to higher altitude observation points with less direct line of sight obstructions, due to denser atmosphere and thus higher refraction index.


Normally, the two main sources of cosmic rays are considered to be the Sun (solar energetic particles) and galactic or extragalactic sources, although extragalactic sources are more common. Surprisingly, though, there is a third source, a family of particles known as anomalous cosmic rays (see Hovestadt et al. (1973) and Garcia-Munoz et al. (1973)). These are particles - generally ionized nuclei of lighter elements - primarily originating from interstellar space with energies on the order of $\sim10$ to $\sim10^2\text{ MeV}$.

Anomalous cosmic rays are generally accelerated at the termination shock of the heliosphere, moving into the heliosheath. Some may then escape into the inner Solar System. Here's an image from Stone et al. (2005), which shows them as a distinct population from galactic cosmic rays:

Now, these particles are technically of interstellar origin, even though they are accelerated to great speeds only at the boundary of the Solar System. However, a related population may have been found in the Kuiper Belt (see Schwadron et al. (2002)), formed from grains there after collisions between objects. While these are not necessarily - or even likely - the dominant source of anomalous cosmic rays, they do exist, and thus constitute another source of cosmic rays inside the Solar System.


do rockets follow a straight path but is looks curved because of space time?

This is partly space exploration and partly physics.

Rockets do not normally go straight up. They start off going up, to get into thinner air, then they steer to turn themselves eventually to a horizontal path, until they are going fast enough to stay in orbit.

In general relativity, an object that is free to move in the Earth's gravitational field will appear to accelerate towards the ground as it is following a path (called a geodesic) in curved spacetime. This is not specific to rockets; a cricket ball will follow a curved path when thrown due to the curvature of spacetime around the Earth. And if one is far  from any gravitational field, then a cricket ball will travel in a straight line.

To directly answer your question: The path of a rocket going straight up will not look curved, it will look like it is going straight up.


Do Astronomers really never call Lagrangian points "libration points"?

A full-text search using the astronomical-literature collection on the Astrophysics Data System (more complete and more astronomy-specific than Google Scholar) yields 10,845 hits for "Lagrange point" and 2,302 for "libration point" (not sure how long that URL will be valid, but it's easy enough to redo the search). So it's clear the former is more common than the latter.

Nonetheless, my impression from looking at article titles is that use of the "Lagrangian" sense of "libration point" significantly outnumbers the "lunar" sense that Hale prefers.

And it's clear that actual astronomers and physicists have been using "libration point" to refer "Lagrange point" since at least the beginning of the 20th Century. E.g., in Trunpler (1923), "Search for Small Planets at the Triangle Points of Mercury and the Sun", the Lagrange points are referred to interchangeably as "triangle points", "libration points", and "Lagrange's libration points." And Plummer (1903), "On Oscillating Satellites", refers to "points of libration" when discussing what are clearly the Lagrange points in the 3-body problem (e.g., "the three collinear points of libration").

So, yes, astronomers often refer to Lagrange points as "libration points", and have for at least a century; Hale is clearly wrong.


Does the radius of your cosmic horizon decrease as velocity increases?

Your past and future lightcone surfaces are invariant: no matter what speed you have, at a certain point in space and time you can see the same things (although strongly Lorenz contracted and with aberration). So if you could see the object when at rest, you can see it near lightspeed.
The cosmic horizon is tricky since it is non-local: the distance where galaxies appear redshifted beyond detection changes with time because of the changing expansion rate of the universe. There are also visible galaxies that recede faster than light we can still see (!).
In some sense the horizon may seem to shift when you speed up because the redshift makes nearby parts of the universe undetectable, but this would be true even in a cosmology without any horizon. If you accelerate at a constant rate you can even get a true event horizon behind you where light cannot catch up to you (a Rindler wedge), but it does not have anything to do with the cosmology. Still, if the object is in that region it would be invisible. But that is because of acceleration, not speed.


Why aren't GPS clocks sufficient to synchronize VLBI observations?

There is a presentation that covers this subject which is a regular at the International VLBI Service workshops. The most recent version is at High-accuracy Time and
Frequency in VLBI and the relevant derivation of needed accuracy is on slide 5. Basically for interferometery, you are trying to match up the phase of the waveforms (position along a sine wave in the simplistic sense) of the two signals from each of the dishes.

If we use an observing frequency of 10 GHz (10e9 Hz) and want to keep the phase coherent to ~10 degrees (out of the 360 degrees) after 1000 seconds integration then you need 10 / (360 * 10 * 1e9 * 1e3) or 10 / 3.6e16 or a clock stability of about 2.8e-15. Higher frequencies such as the 230 GHz mentioned in the comment will be even more demanding.

If you look at Allan Deviation/Variance plots, which measure the stability of clocks, such as the plot on slide 4 (and copied below):  this level of performance at 1000 seconds is only reached by hydrogen masers. The typical GPS Disciplined Oscillator will be at about 1e-12@1000seconds, see this recent comparison of 17 GPSDOs. With careful correction of the sawtooth error caused by the internal oscillator of the GPS (discussed at the end of the presentation) you can improve the typical 50 ns jitter of GPS to ~5 ns. Beyond this atmospheric/ionospheric effects come in and you need to use a dual-frequency GPS to take those out.


Is there any planet or black hole where we could experience time dilation like in the movie Interstellar?

You can definitely scratch out any planets, as their gravitational field will not be strong enough. Even a very dense neutron star wouldn’t give much effect, as far as I know.
As for black holes, well, basically any black hole will give this effect. However, those are too far to reach with our current means.
Safe travels!


What are the azimuths of the planets' orbits?

As someone commented, you are actually short a term for fully defining the orbit.  The "azimuth" as you describe is commonly defined as a "longitude of perihelion", and you're forgetting the orientation of the inclination, commonly defined as "longitude of the ascending node." (Both of these values are included in the first link in the previous posted answer)

For a more detailed look at the math which uses the defining values, JPL made a whitepaper which goes into the math, and also allows you to account for changes in the orbital terms.  (Most of the math you're probably looking for is at the end of the document, beginning about section 8.10, on page 25)

ftp://ssd.jpl.nasa.gov/pub/eph/planets/ioms/ExplSupplChap8.pdf


Given the description, that light patch was probably Pleiades.

It is a star cluster in the constellation of Taurus which is inspired from a popular Greek mythology of Seven Sisters. You can read more about it here.

EDIT: One of the great softwares, free and open source, for amateur level observational astronomy, though professionals use it as frequently is Stellarium. It's easy to use and should clear all your doubts.


Can the theory of multiverse explain dark matter?

This is a bit of a speculative question, but I can answer it.

Dark matter has been observed in galaxies, and the distributions of dark matter in galaxies have also been measured. It seems clear that the matter is firmly in our universe - we just can't detect it with electromagnetic radiation.
*However, there are ideas (extremely speculative) that dark matter is from a different spatial dimension. It sort of explains why gravity is so weak (it "leaks off" through the other dimension(s)). Note, though, that this highly, highly speculative.

There are ideas like this in string theory, too. Some say that we are trapped on a brane, a section of space-time with three spatial dimensions. However, there are one or more additional dimensions. All of these dimensions are known as the bulk. Gravity can "leak off" through these extra dimensions. Some theories take this to the extreme, and say that there exist many universes (of a certain number of dimensions each, but with less dimensions that those of the bulk) that are branes, just like ours. These have been considered the source of phenomena such as dark energy - in which case your scenario is within the realm of plausibility.

All of these ideas, however, are speculative and have no evidence in their favor.


Were radio spectroscopic observations ever used to first measure a distance successfully?

Yes, radio spectra have been used extensively to find the distances and locations of HI regions and molecular clouds inside the Milky Way. Observations of the 21 cm hydrogen line and/or several carbon monoxide lines (in particular, $\text{CO}(1\to0)$) enable us to make radial velocity measurements of clouds within the galaxy. From there, some geometry (see Roman-Duval et al. 2009) allows us to determine the distance to the cloud, if the cloud lies farther from the galactic center than the Sun does.
Unfortunately, if the cloud is closer to the galactic center, there are two possible solutions for the distance between the cloud and the Sun, something called the kinematic distance ambiguity. It occurs because at two possible points along the line of sight - if the cloud's orbit is smaller than the Sun's - the projected cloud velocities are identical. There are a few ways we can try to do this, including HI absorption (Kolpak et al. 2003).
Here's a diagram to make the distinction clearer, from Roman Duval et al.:

Figure 1. Sketch of the HISA method to resolve the KDA. In the inner Galaxy, a single Galactocentric radius (determined by the radial velocity of the cloud)
corresponds to two distances along the line of sight, a near (in blue) and a far (in red) kinematic distance. The near and far kinematic distances correspond to the same
radial velocity $V_r$, which is the projection of the orbital velocity $V_0$ of a cloud around the GC onto the line of sight. At the tangent point, the orbital velocity of a cloud is parallel to the line of sight. In this case, the radial velocity is maximal and the near and far kinematic distances are identical. The cold HI embedded in a cloud
located at the near kinematic distance absorbs the 21 cm radiation emitted by a warm HI background located at the far distance. Consequently, the HI 21 cm spectrum
toward a near cloud exhibits an absorption line that is coincident with a 13CO emission line from the cloud. A cloud located at the far distance does not lie in front of
a warm HI background emitting at the same velocity as that of the cloud. Therefore, there is no absorption feature in the HI 21 cm spectrum toward a cloud located at
the far kinematic distance.

The technique has also been applied to extragalactic sources (though often lines in the optical region are used, even for radio galaxies). One example is the Leo Ring discovered by Schneider et al. in 1983. It's mostly neutral hydrogen, so the 21 cm line was initially used to derive a redshift, recessional velocity (a system velocity of $V=960\;\text{km s}^{_1}$) and distance. That velocity measurement corresponds to $d\simeq14\;\text{Mpc}$, fairly close to that of the Leo Group. I wouldn't be surprised if other extragalactic blobs of neutral hydrogen have also had their distances measured via the 21 cm line.


Sort of... There is a system called the International Celestial Reference System (ICRS) which has center at the Solar System Barycenter (normally inside the Sun but not the same as the Sun's center) and which has the x and y axis in the plane of the Earth's equator and the z-axis ("North" if you will) pointing towards the Celestial Pole. The x-axis points towards the equinox where the plane of the Earth's orbit and the plane of the Solar system (the ecliptic) cross (on average; this gets complicated quickly as everything related to the Earth is wobbling about on different timescales).

Note although this reference system is close to the older mean equator and equinox J2000 or FK5 system it's not actually the same as it is now defined by a set of several hundred very distant extragalactic radio sources which are called the International Celestial Reference Frame (ICRF) and which we assume don't move on the sky. The latest version of this frame is called ICRF3 and was formally adopted at the IAU General Assembly last year in Vienna. More details on how the reference systems and frames work at the USNO page on the ICRS


How does the Moon move in the "night" sky as seen from the poles?

The moon is quite interesting as you get near the poles. It still moves round due to the Earth's rotation once a day, but it also orbits with a period of a month so what you get is two weeks of moon above the horizon followed by two weeks where you cannot see it.

What changes through the year is what phase you see when the moon is visible, and in the middle of winter you have two weeks around full moon, whereas in summer you have two weeks around the new moon - not so exciting. Spring and autumn obviously have two weeks around first and last quarter, respectively.

This simulation using Stellarium shows the effect quite well.

From astro.cornell.edu - this wording may help:

  The second movement caused by the Moon's orbit around the Earth is analogous to the movement of the Sun over the course of a year only it repeats over the course of a lunar month. Near the new Moon phase, the Moon is near the Sun and therefore never rises during the winter. As the Moon approaches full, it will start to pop up above the horizon. Eventually near the full Moon phase it will be high enough in the sky to stay up all day and circle like the Sun in the video above. The elevation of the circle will rise as the Moon becomes completely full and then start to decrease until it begins to dip below the horizon. Eventually the Moon will stop rising at all as it gets close enough to the new phase. The cycle then repeats.

And another useful reference at planetarium.madison.k12.wi.us:

  At the beginning of Winter, when it's nighttime all of the time, the moon would be in the sky for the 2 weeks closest to Full Moon, and then below the horizon for the next 2 weeks. And at the beginning of Spring (click on the graphic), when the sun is at sunrise all of the time, the moon would be up in the sky for the 2 weeks closest to First Quarter (waxing), and then below the horizon for the next 2 weeks.

  The animated graphic above shows what we would see from the North Pole if we went out every day at noon, for 14 days in a row, from March 1st to March 14, 2006. We start with a thin crescent moon near the horizon, and end with a full moon near the horizon. Halfway through, the First Quarter moon would be when the moon is highest above the horizon.

  Keep in mind, that if you were observing the moon constantly, throughout a 24 hour period, the moon would seem to move to the right in the sky along with the sun, stars, and planets due to the Earth's rotation. Nothing would seem to rise and set: they would just seem to circle around you.


Central temperature of a stellar model with combination of gas and degeneracy

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

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

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


What happens to the momentum of a star as its mass is converted to energy?

Momentum is conserved. But it is the momentum of the system as a whole. The energy released by nuclear fusion escapes as a result of neutrinos and radiation. If these emissions are isotropic (the same in all directions), or at least with no hemispheric asymmetry, then NO net momentum is removed from the star, it is not accelerated and therefore its velocity with respect to any reference frame remains unchanged. It's momentum in that rest frame is then reduced by the percentage of mass that it has lost. The lost momentum exists in the escaping radiation and neutrinos which will appear anisotropic in any frame of reference other than the rest frame of the star.

In practice most radiation loss is fairly symmetric, if not spherically, then at least not in the sense that more radiation does not emerge from one hemisphere. I am not sure about neutrinos, but I see no reason for asymmetric emission and these are a fairly minor component in any case.


Relatable comparison of VY Canis Majoris to the Sun?

After some playing around with wolfram alpha and google my best comparison has been

The sun compared to VY Canis Majoris is like a donut compared to the London Eye.

The London Eye is about 120m in diameter, this divided by 1500 is about 8cm which is roughly the diameter of a ring donut.


Is it possible for a septuple eclipse to happen on Saturn considering there are 7 moons capable of eclipsing the Sun?

The short answer is no, septuple eclipses can’t happen.
Using a method described by Meeus in Mathematical Astronomy Morsels (p. 190), we find that while it is, in theory, possible, for the seven satellites to align, it happens only once per ~ 20 million years—and that’s among themselves only; I didn’t factor in the Sun’s position yet!
The possibility that the Sun is along the same line is $ \displaystyle \frac { \text {(angular size of the Sun as seen from Saturn)}°} {360°} = \frac {0.05} {360} $; let’s be generous and round it off to 1:10,000.
Multiplying 20 million years by 10,000 shows us that the seven satellites would line up with the Sun once in 200 billion years—longer than the life of the Universe!
Finally, the above was estimated by using circular orbits, which is not the case in practice. Plus, orbital parameters change (“chaotically”) over long timespans, so it would (at least currently) be impossible to compute the actual positions of satellites in the remote future on timescales such as the ones above.


Why are tides biggest during the equinoxes?

Only a syzygial tide during/near an equinox is the strongest. This means, there must be either new or full moon. In general, syzygial tides are strong because three bodies (Earth, Moon, Sun) align near one line, and tidal effects of Moon and Sun on Earth become (nearly) collinear and sum to the maximal possible magnitude. The Moon’s orbit is inclined to the ecliptic by about 5° only (cos 5° ≈ 0.996), so two tides are aligned almost perfectly on a Moon’s syzygy during whichever season. The difference is that, during/near an equinox, this line also lies in the equatorial plane and rotational motion of Earth’s surface/hydrosphere/crust can direct the stuff to move along this tidal line to a maximal range possible. At least, on the equator.

As for variations in terrestrial latitudes, namely mid-latitudes vs equator, comparison between syzygial tides during solstice and equinox far from the equator is a mechanical, not astronomical problem. Tidal bulge should be stronger during solstice, but projection of velocity to the tidal line will be greater on equinox. A detailed analysis of particular body of liquid (and its shore) is required to compare.

In contrast, a quadratural (where Earth–Moon and Earth–Sun lines are perpendicular) tide during an equinox is as weak as any quadratural tide. The solar tide basically cancels a part of the lunar tide. A quadratural tide during an equinox might even be the weakest, since lunar tide outmatches the solar one, but on an equinox and Moon’s quadrature simultaneously the Moon definitely lies away of the equatorial plane.


How is the redshift of starlight precisely measured?

I don't really think I fully understand the first part of your question.

Redshift is measured by comparing the wavelengths of a redshifted pattern of absorption or emission lines with the wavelengths they would have in an object at rest. All lines are shifted by the same factor of $1+z$, where $z$ is the redshift.

Gravitational waves cannot be used to obtain a redshift because the intrinsic wavelength of the waves is not known. An inspiralling binary emits waves with a frequency equal to twice the orbital frequency. This in turn depends on the mass, but the observed frequency is also redshifted. There is thus a "degeneracy" between the mass and redshift.

What you can do with gravitational waves is estimate the physical distance to the source using the wave amplitude and how quickly the amplitude changes with time. If the host galaxy of the source can then be identified, then you can use the redshift of that galaxy to get an estimate of the Hubble parameter (e.g. see Abbott et al. 2017).


Primoridial black holes can be found (hypothetically; there is no experimental evidence yet) of any small size above the Planck mass. Stellar black holes, however, cannot have a mass below the TOV limit (1.5 to 3 solar masses)

There does seem to be an upper limit1 of 50 billion solar masses. However, I suspect2 that this takes into consideration formative constraints (i.e. the constraints posed on the formation of such a BH); and does not prohibit such a black hole from existing. After all, the Schwarzschild metric certainly does not impose limits on the size of a black hole.

Note that talking about the limiting dimensions of a black hole is slightly meaningless as the dimensions change in different reference frames. It is far easier to talk about the mass of a black hole; the radius can be calculated in various frames from that information.

1. Natarajan, P. and Treister, E. (2009), Is there an upper limit to black hole masses?. Monthly Notices of the Royal Astronomical Society, 393: 838–845. doi: 10.1111/j.1365-2966.2008.13864.x

2. but cannot confirm yet; I will have to read the paper more thoroughly


Wouldn't the rings of Saturn experience tidal effect?

Most of the 60 moons in the Saturn system are far away from the rings and very small, so their effect on the rings is negligible. But larger ones that are closer in (Enceladus) do have a rather significant effect on the rings, but as the gravitational pull of these moons is radially outward, it is hardly visible. On the other hand, small moons inside the rings that are a few kilometers in diameter cause significant changes in the ring system.

(source: magicgate.eu)
The orbits of those giant rocks are slightly inclined relative to the rings, which causes the rings to create a bulge in the direction of the small moon as seen in the picture. And these effects are fairly significant as the rings are mostly just a few hundred meters thick or less without a moon present, but with one, they are possibly hundreds of kilometers spread out.


What is the closest one could get to a black hole and still stand a realistic chance of escape?

The simplest case is for a rocket that does not fire its engine when it approaches the black hole from afar, and hence follows a geodesic trajectory. In this case the closest you can get is $3r_s$, but you can make an arbitrary number of turns around the black hole.

The reason for this is that the counterpart of a hyperbolic orbit corresponds to moving in the effective potential $V(r)=-GMm/r+L^2/2\mu r^2 - G(M+m)L^2/c^2\mu r^3$ (this is the Wikipedia version of the formula; different authors use different measures). For a given energy and angular momentum this means that either you can come in from infinity and fall into the black hole (for low energy and angular momentum) or the potential has a maximum greater than your energy and you will turn back. The closest you can get the maximum to go to the hole is $r=3r_s$ (as your angular momentum increases).

However, the angular velocity behaves as $d\theta / d\tau = (L/\mu)/r^2$: you will swirl around a lot as you get close to the black hole. Since a carefully selected trajectory will have energy and angular momentum so that you turn around almost exactly at the potential maximum where the derivative $V'(r)\approx 0$ the proper time spent there will be long and you can get as many turns as you want. The orbital frequency as seen from the outside is 2200 Hz $(M_\odot/M)$ (!) (source).

Note that all of this is freefall: the only thing to worry about is tidal forces, but that is not much of an issue near a SMBH. The velocity needed is of course in principle zero at infinity if you come in along a parabolic orbit.

The second question is if one can "slingshot". I assume this refers to an Oberth manoeuvre rather than a gravitational slingshot, which is unpowered. Gravitational slingshots from black holes can give you any change in angle, but the velocity change is just like the classical case. Rotation does not affect this, except by making the angular shift weirder. Note that for prograde orbits around an extremal spinning black hole the ISCO approaches the horizon radius, so the above unpowered flybys can in principle get right up  to the horizon.

For the powered manoeuvre I don't have a full answer; I would actually like to see it answered (see this question and my half-answer to it). The basic story looks like that for velocity-at-infinity increasing manoeuvres you should not get closer than $3r_s$. But for a sightseeing trip close to the horizon spending some of your fuel (and mass) ought to work.

What you need to do is to fire the engines powerfully enough to change the sign of your radial velocity and retain more total kinetic energy than the potential maximum. The potential maximum is due to angular momentum.

For the freefall radial trajectory with no angular momentum you will have $(dr/d\tau)^2=(E/mc^2)^2-c^2(1-r_s/r)$, so if you start with zero velocity at $r=\infty$ (hence $E=mc^2$) the inward radial velocity will be $dr/d\tau = c\sqrt{r_s/r}$. So you need $\Delta v =2c\sqrt{r_s/r}$ to escape again this way. We can rearrange the formula into $$r_{escape}= \frac{r_s}{\left(\Delta v / 2c\right)^2}$$

In your case $m_0/m_1=100$ and I assume $I_{sp}\approx 0.6c$,  $\Delta v = c \tanh \left( \frac{I_{sp}}{c}\log\left(\frac{m_0}{m_1}\right)\right) = 0.9921c$, and $r_{escape}=4.0642 r_s $. That demonstrates that the purely radial approach is not a good choice; adding a bit of lateral thrust would allow a closer approach.

I don't have the full answer for the lateral case. If you come in along a flyby orbit that approaches the ISCO you can slow down a bit so you now are on an inspiralling trajectory. To escape you need to pay roughly twice the potential difference between your final radial distance orbit and the maxmimum of the new potential.

Again, rotating black holes allow you to use the Penrose process to gain a lot of angular momentum by dumping reaction mass in the right direction. However, the efficiency is proportional to the total momentum of the reaction mass and is rather negligible for non-relativistic rockets. Hence rockets not powered by antimatter will likely not benefit much from this possibility.


Retrograde motion of planets: reason and time?

The Wikipedia article on apparent retrograde motion seems to have a table containing exactly what you are looking for:

The apparent motion is observed when the projection of a planet is compared to the star background. There are no real change in the planetary orbits, but when for example the Earth catches up with Mars, it seems to move retrograde relative to the background because of our own change of view. (Like how it appears that you are scrolling further down on this page when the background moves upwards.)


Are the Trappist-1 planets close enough to be tidally heated?

This is a complicated question that would really require a full physics simulation and better knowledge of the system to accurately answer. But let's try a few back of the envelope calculations to see what we get.

Calculating tidal forces from TRAPPIST-1c on TRAPPIST-1b

I'm going to calculate the tidal effects of TRAPPIST-1c on TRAPPIST-1b (simply because, a priori, this seems likely to be where the strongest tidal heating will be induced). See the figure below which describes the parameters.

The tidal force of 1c on 1b is defined as the differential force of gravity across 1b, that is, the difference of the force of gravity on the side of 1b facing towards 1c and the force of gravity on the side of 1b facing away from 1c. Mathematically, we get.

$$F_{tide,c-b} = F_{g,-R_b} - F_{g,+R_b} = \frac{GM_bM_c}{(D-R_b)^2} - \frac{GM_bM_c}{(D+R_b)^2} = 4GM_bM_c\frac{DR_b}{(D^2-R_b^2)^2}$$

We can presume that $R_b << D$ (for this case $R_b/D = 1\%$) and reduce this to

$$F_{tide,c-b}(D) \approx 4GM_bM_c\frac{R_b}{D^3}$$

But this isn't enough to determine the amount of tidal heating that may occur. Tidal heating only occurs when the tidal force changes. It is this constantly changing tidal force which results in tidal flexing of the planet and thus creating heat through tidal friction. Fortunately, for these two planets, the tidal force will be changing since $D$ will be constantly changing. So let's calculate $F_{tide}$ for the two extremes where these planets are as close as possible and as far as possible and difference them.

$$\Delta F_{tide,c-b} = F_{tide}(0.004\:\mathrm{AU}) - F_{tide}(0.026\:\mathrm{AU})$$

If I plug in numbers to this, I find that

$$\Delta F_{tide,c-b} \approx 3.7\times10^{20}\:\mathrm{N}$$

Okay, but what do we do with this number? It's somehow a metric of the change in tidal forcing that 1c imparts on 1b, but is it negligible? To determine this, we have to compare it to something. Let's compare this to the tidal forcing that TRAPPIST-1b would receive from the star.

Calculating tidal forces from TRAPPIST-1 on TRAPPIST-1b

I've already set up the math, so we don't need to go over that again. But first, let me discuss where this tidal forcing is actually coming from. Quoting an article from space.com, the paper's author, Gillon, states:

  Because the seven alien worlds orbit so tightly, they're probably all tidally locked, Gillon said. That is, they likely always show the same face to their host star, just as Earth's moon only shows the "near side" to us.

As I said above, the only way to produce tidal heating is to have changing tidal forces. These planets are likely synchronized and always present the same side to the star. It's marginally possible these planets aren't perfectly tidally locked, but rather have some higher spin-orbit resonance. That is, their spin orbit resonance might not be 1:1 (as it would be if they were tidally locked) but instead could be something like 3:2 (which is what Mercury has). I'll ignore that complication though and just assume 1:1 resonance. So if they're tidally locked, they can't experience differing tidal forces through their own rotation. Instead, the differential tidal forcing comes from the orbit's ellipticity. Sometimes the planet will be closer and sometimes it will be farther, causing a differential tidal force on TRAPPIST-1b from the star as it orbits. This is exactly what occurs in the tidal heating of Io. Let's calculate $\Delta F_{tide,*-b}$ by using the varying distances TRAPPIST-1b will have from the star. I found that TRAPPIST-1b will orbit between $0.0101\:\mathrm{AU}$ and $0.0119\:\mathrm{AU}$1. This means the differential tidal force is:

$$\Delta F_{tide,*-b} \approx 4GM_bM_*R_b\left(\frac{1}{(0.0119\:\mathrm{AU})^3} - \frac{1}{(0.0101\:\mathrm{AU})^3}\right) = 1.8\times10^{23}\:\mathrm{N}$$

Is planetary tidal heating non-negligible?

The back of the envelope calculations show that the differential tidal force on TRAPPIST-1b from TRAPPIST-1c is about $0.2\%$ of the differential tidal force due to the star. Whether you consider this negligible or not is up to you. I'd personally consider it a pretty small effect and say that most of the tidal heating these planets experience comes from the star itself.

Could the inter-planetary tidal heating still contribute to the tidal heating of the planets enough to heat the interior?

This is a remarkably hard question to answer and I can't even really do a back of the envelope calculation without making wild, unjustifiable assumptions. The calculations above simply determined the maximum tidal force variation over time. That doesn't tell us anything about how much tidal heating this may induce though. That requires knowing more about the planet itself, in particular the planet's Love numbers which define the rigidity of the body and thus how easy it is to stretch via differential tidal forces. You can vary your tidal forcing as much as you want, but if your planet is pure iron (and thus very rigid) you're unlikely to have as much of an effect as if it were primarily silicate (and thus much less rigid). The paper produces the plot below which defines the potential constituents of each planet. This would be a first step in determining the planet rigidities, but as you can see from the error bars, it would be highly uncertain.

Overall, and this is entirely opinion based and from my calculations above, but I'd say the chances of the inter-planetary tidal heating having significant effects on the interior heat of of these planets is negligible. More than likely the biggest contributing factor is radioactive decay, followed by tidal heating from the star (but this is amplified by the eccentric orbits induced by planetary gravitational perturbations)

1 Note that this calculation involves using the eccentricity and the paper only provides an upper limit. These distances then represent an upper limit as well and the final answer will also be an upper limit. It may be less.

Values used in calculations:

$G = 6.67\times10^{-11}\:\mathrm{m^3 kg^{-1} s^{-2}}$
$M_b = 5.075\times10^{24}\:\mathrm{kg}$
$M_c = 8.239\times10^{24}\:\mathrm{kg}$
$M_* = 1.604\times10^{29}\:\mathrm{kg}$
$R_b = 7.34\times10^{6}\:\mathrm{m}$


Finding equation for energy density of matter

Energy density $\rho \sim nmc^2$ if kinetic energy is unimportant (and you are ignoring dark energy and other forms of energy density).

But the number density of particles varies as $a^{-3}$.

Thus
$$ \frac{\rho}{\rho_0} = a^{-3},$$
where $\rho_0 = n(0)mc^2$ is the present day energy density and the present day scale factor is 1.


As of this moment (2016 May 18, 13:15 UTC), the Earth is 1.0116 astronomical units from the Sun. WA is smart enough to know that "Sun Earth distance in AU" is a time-dependent question.

Update: The stricken text that follows from my original answer is incorrect. I am leaving it present (but stricken) for the sake of humility.

It interpreted your query to mean Sun Earth distance today in AU, and because today is 2016 May 18 and because you didn't specify a time of day, it picked noon (UTC), it in turn interpreted your query to mean Sun Earth distance on 2016 May 18 at noon UTC in AU.

The correct answer: I happened to ask WA the very question posed in the OP a couple of minutes apart and got two different answers (1.014 and 1.015 AU), and this did not occur across a day boundary. WA apparently interpreted your query to mean the current distance between the Sun and the Earth.

You have to be very careful of what you ask WA.

A better query is to ask WA What is the mean distance between the earth and the sun in AU? That query will give you an answer of 1.0000010178 au. An even better question is What is the semimajor axis of Earth's orbit? That will give you an answer of 1.00000011 au. Note the extra zero, but also note that for some reason, the precision is reduced.


What is the term for a star swallowing another star?

The term you're searching for is probably accretion (I guess it might sound similar to a kerzan if the speaker had a cold or the listener couldn't hear it well enough for other reasons). This term is not limited to close binary stars (accreting binaries) cannibalizing each other because they orbit each other at such proximity that the L1 Lagrange point between them extends into their own Roche lobe, they form accretion disks connected through an accretion stream, and they feed off each other (thus why this is often referred to as cannibalism):

   Artist's conception shows an accreting binary at the center of the globular cluster NGC 6624 (Source: U. of Tennessee AstroWiki)

Some examples of accreting binaries are symbiotic novae, X-ray binaries, microquasars, X-ray pulsars, and so on. See links for more examples and explanations. Other examples of accretion include stellar accretion of planetary materials of a planet in a low altitude orbit around its parent star, formation of planets via accretion of materials of the protoplanetary disk, active galactic nucleus (AGN) like quasars (could also sound similar), novae, and e.g. the Earth still accretes on average ~ 110 metric tons of new materials per day, too.


Is the moon's orbit circularizing? Why does tidal heating circularize orbits?

Short answer: YES. I will assume you mean the eccentricity in the orbit of the Moon around the Earth.

In general, tidal forces in binary systems (like the Earth-Moon system, or a binary star, etc...) effect the binary in three main ways: in order of longest timescale to shortest timescale

1) circularization of the orbit (eccentricity goes to zero, binary separation goes to minimum).

2) alignment of the binary components' spin angular momenta with the orbital angular momentum (the directions of $S$ and $L$ are the same).

3) synchronization of the binary components' rotational frequency with the orbital frequency.

  but why?

There's different ways of answering "why," and here is a great conceptual answer given by the God Father of astrophysical tides himself, Z. P. Zahn:

  A fundamental property of closed mechanical systems is that they conserve their
  total momentum. This is true in particular for binary stars, star-planet(s) systems,
  whether they possess or not a circumstellar disc, if one can ignore the angular
  momentum that is carried away by winds and by gravitational waves. Through
  tidal interaction, kinetic energy and angular momentum are exchanged between
  the rotation of the components their orbital motion and the disc. In the absence
  of such a disc, which is the case that we shall consider here, they evolve due
  to viscous and radiative dissipation to the state of minimum kinetic energy, in
  which the orbit is circular, the rotation of both stars is synchronized with the
  orbital motion, and their spin axis are perpendicular to the orbital plane. How
  rapidly the system tends to that state is determined by the strength of the tidal
  interaction, and thus by the separation of the two components...

So, basically, the tidal torque drives dissipation, and this dissipation brings the binary to a minimum kinetic energy state, i.e. circular orbit, synchronized spins with the orbit, aligned spins with the orbit.

Here's a contour plot of the timescales provided by equations #$9 - 13$ from Hut's 1981 seminal paper, assuming the separation does not change much relative to the other quantities, for a binary composed of a black hole and a Wolf-Rayet star, which is a similar system to a planet-satellite system, where the timescales are parameterized in terms of the mass of the WR star and the separation of the binary:

The dotted black line is the merger timescale for the binary due to gravitational waves, meaning below that line you are binary that mergers within the lifetime of the universe. The synchronization timescale is independent of the initial spin of the WR star, which is why there is only one synchronization line in the plot, but the alignemnt timescale does depend on the initial spin of the component feeling the tides. Points below the contours achieve that process (below the red dashed line are synchronized). The quantity $f_{B}$ is the break-up fraction parameter, is between $0$ and $1$, and the choice of which determines the initial WR star spin as a fraction of its break-up spin. (the tidal timescale contours were made by equating the WR lifetime with the tidal timescale). I did not include the circularization timescale because it is essentially always less than the sync. timescale (seen simply from the fact that generally, $S << L$).

Lastly, it must be said that a WR star is quite different from a moon/satellite, however the effects of tides on each is similar enough to make a conceptual comparison (since a WR star is much less massive than the BH, as is the moon than the Earth, the hierarchy of tidal timescales is the same, but you must keep in mind that the WR's apsidal response to the tidal gradient is very different than a rocky moon's).

As you pointed out, the Moon's orbit is slightly eccentric, but the moon is also tidally synchronized with the Earth's rotation (which are both sync'ed to the orbital rotation). This should make sense since it takes longer for tides to circularize than to synchronize.

I'm sure somebody can post an example of using tides in an actual planet-moon system, rather than my black hole-star system. :)


Millennia ago, when only a few thousand stars, a handful of planets and nebulae, and some transient objects like supernovae and comets, were known, people usually named these objects after gods and heros, but also after everyday objects. Stellar constellations that vaguely resembled something known, was named after this. Mars is red, so it was named after the god of war; Mercury, being the fastest to complete one cycle around the Sun, was named the god of messaging and traveling.
With the advent of telescopes, so many objects are now known that we had to come up with a more pragmatic convention, so usually an object is named according to its place in some catalogue, its type, and/or it coordianates on the sky. For instance, a quasar detected at the "J2000" coordinates $\{\mathrm{dec.},\mathrm{RA}\} = \{22^\circ\, 22'\, 56'', -09^\mathrm{h}\, 46^\mathrm{m}\, 36^\mathrm{s}\}$ may be called J2222–0946. An x-ray source observed in the constellation Cygnus may be called Cyg X-1, while its binary counterpart is number 226,868 in the Henry Draper Extension Catalogue and is thus named "HDE 226868" (not to be confused with one of our site moderators).
Oftentimes, an object appears in multiple catalogues, and will thus have multiple names. For instance, a Lyman $\alpha$-emitting galaxy appears in a catalogue of Ly$\alpha$ emitters, but due to its infrared properties, it will also appear in some catalogue over dusty galaxies (though Ly$\alpha$ emitters tend to contain little dust).


The paper you linked is specifically trying to understand massive $(M\gtrsim 8\:\mathrm{M_\odot})$ star formation. It is still an open question as to exactly how very massive stars form. Do they form all at once in a giant accretion event ("top-down" formation) or do they form by having a bunch of low mass stars merge ("bottom-up" formation) or by some other process altogether? The authors of the paper state that it is particularly tricky to study high mass star formation.

  Evolution in low mass systems is often characterized by changes in the spectral energy distributions (SEDs). This is not as easily done in high mass systems (Molinari et al. 2008)1, which is why studying the gas dynamics is very important.

Basically, they can't easily figure out what's going on by looking at the SED of the high mass star forming region, so instead, they're going to look at how the gas is flowing and moving during the formation. Klassen et al. break their observations into observing "infall" and "outflow" (via two different molecular transitions). What they mean by this is that one molecular transition (that of the $J=4-3$ transition in $\mathrm{HCO^+}$ and $\mathrm{H^{13}CO^+}$ molecules) will show evidence of gas flowing into the core, i.e. infall, while another molecular transition (that of the $J=8-7$ transition in the $\mathrm{SiO}$ molecule) will show evidence of gas flowing out of the core, i.e., outflow.

During star formation, you tend to get both types of flows. Infall will occur because the gas is simply accreting onto the core due to gravity. Outflow will occur because the accreting matter and central core get very hot, causing an energy release outwards.

You can see precisely how they track infall flows in Section 3.2. They have a paragraph discussing precisely how they determine that the gas is falling into the core from the observations of the $\mathrm{HCO^+}$ and $\mathrm{H^{13}CO^+}$ molecules.

  Double peaked or asymmetric H$\mathrm{HCO^+}$ line profiles in which the brightest emission is blueward of the source rest velocity can be interpreted as due to infall if the $\mathrm{H^{13}CO^+}$ is single peaked at the rest velocity. If the optically thin $\mathrm{H^{13}CO^+}$ also has a double peaked profile, the double peak in both lines is likely due to multiple components along the line of sight. Churchwell et al. (2010)2 describe the different mechanisms that could produce a double peak in an $\mathrm{HCO^+}$ line profile towards massive star forming regions, and conclude that self absorption best fits observed line profiles. The blue or red skew of the absorption profile can then be used to distinguish infall (blue) or outflow (red).

I won't go into detail to precisely explain the spectroscopic interpretation of the above quote, but suffice it to say, that is their method of detecting matter infalling towards the core of the high mass star.

1 Molinari, S., Pezzuto, S., Cesaroni, R., et al. 2008, Astronomy and Astrophysics, 481, 345

2 Churchwell,  E., Sievers,  A., & Thum, C. 2010, Astronomy and Astrophysics, 513, 9


The short answer: Atmospheric gases never left Earth, they're in it!

The long answer to this question isn't just about the planets' current states, but rather the processes that led them there. Let's start at the very beginning (a very good place to start).

The very early years

When our solar system began forming 4.6 billion years ago, most of the mass from the collapsed portion of a molecular cloud (see Nebular hypothesis) collected in the center to form the Sun. The mass that didn't collapse into the Sun left a protoplanetary disk—a cloud of dust and gas—surrounding the new star. Gradually, dust particles began to come together through accretion, attracting more and more particles into fledgling planets.

Close to the Sun, where Venus and Earth both reside, it was too hot for many particles to condense, so the planets in this region formed with metals and silicates, which have high melting points. This is why the four planets in the inner Solar System are called the "rocky" or "terrestrial" planets. The earliest atmospheres on these planets began forming with the gradual collection of gases from the solar nebula, mostly hydrogen.

Goldilocks and the two planets

At this point in the evolution of the two planets, they looked pretty similar, but there is one major difference: the distance to the Sun. Earth, it seems, got lucky to be in the "Goldilocks zone", where the temperature is just right to support life. Being in this zone has two major implications: liquid water and, as a consequence, active plate tectonics. (See this article for an in-depth view into why the two are related.)

Carbon sinks

On Earth, there is considerable water contained in liquid oceans. On Venus, this is not the case. It is simply too hot that close to the Sun, so all the water evaporated into the atmosphere. (Venus likely contained liquid water in its early stages, but it all evaporated after a billion years or so.) It is also likely that the young Earth once had a dense, punishing atmosphere like Venus's today. However, surface oceans and plate tectonics both provided ample pathways for gases to be absorbed into the surface of the Earth. Oceans and plate tectonics offer huge amounts of carbonate storage, allowing the transfer and balance of carbon compounds to and from the atmosphere.

So now we have two things reinforcing the differences between Earth's and Venus's atmospheres:

Evaporation of liquid water: It is too hot on Venus for liquid water to exist. All water has evaporated, resulting in a denser atmosphere. On Earth, water can reside on the surface, lessening the amount in the atmosphere.
Carbon sinks: Liquid water and plate tectonics allow Earth to absorb considerable amounts of gas, allowing the atmosphere to be thinned of certain compounds like carbon dioxide. There is no such pathway on Venus, forcing all gas to remain in the atmosphere.

With no major mechanisms for gases to be absorbed by the planet, Venus is experiencing a runaway greenhouse effect.

Atmospheric escape

You mention Jeans escape. It is true that this effect is greater at higher temperatures; however, it is much easier for smaller molecules to escape than for larger ones. Hydrogen and helium, being the two smallest elements, are the most greatly effected by this phenomenon. In comparison, carbon dioxide, which makes up the majority of Venus's atmosphere, is not much affected by Jeans escape.

You also mention solar winds. While these play an effect, especially on planets without a magnetic field, this phenomenon is not as powerful as you might think. Ultraviolet light (i.e. photoionizing radiation) causes ionization in the uppermost region of the atmosphere. These charged particles now form a shell (called the ionosphere) that deflect solar winds, much like a magnetic field would. On Venus, the thick atmosphere provides more particles for ionization, resulting in a more powerful deflection. (Compare this to Mars, where solar wind is the primary non-thermal escape mechanism due to the thin atmosphere with few ionized particles.)

The primary atmospheric escape mechanism for Venus is actually a bit more complicated. In the absence of a magnetic field, it is easier for charged particles to escape. In particular, electrons are most susceptible due to their small mass. As electrons escape, the net charge of the ionosphere leans positive, causing the ejection of positive ions, mostly H+.

Conclusion

While Earth and Venus formed similarly, Earth got lucky. It has pathways to remove gases from the atmosphere, while Venus does not. Additionally, the two planets do not experience significantly different rates of atmospheric escape. This results in the atmospheric densities where know today: 66 kg/m3 for Venus and just 1.2 kg/m3 for Earth.


I think the reason you're suffering from conflicting sources is that you're mixing both new and old, out-of-date pieces of information. First off, the book you cited was published in 2001 - 15 years ago - and the other article you cite was published in 1999 - 17 years ago. There's been a lot of work done in the past 15 years, often under the term "precision cosmology", in an attempt to really nail down the precise content, shape, size, etc. of our Universe. By the early 2000's we pretty much knew the science behind everything (we knew about dark matter, dark energy, had well-developed theories on the Big Bang, etc.) but what we didn't have, were good, solid, believable numbers to put into these theories, explaining why the flatness of the universe was still contested in your sources.

I'll direct you to two incredibly important observatories which have been paramount in achieving our goal of having "good numbers". The first is the Wilkinson Microwave Anistropy Probe (WMAP), launched in 2001, and the second is the Planck satellite, launched in 2009. Both missions were designed to stare intently at the Cosmic Microwave Background (CMB) radiation and try to sort out the treasure trove of information which can be gleaned from it. In this vein, you might also come upon the Cosmic Background Explorer (COBE), launched in 1989. This satellite had a similar purpose as the other two, but was not nearly as precise as the later two missions as to provide us with good numbers and definitive statements by the early 2000's. For that reason I'll mostly focus on what WMAP and Planck have told us.

WMAP was a hugely successful mission which stared at the CMB for 9 years and created the most detailed and comprehensive map of its day. With 9 years of data, scientists were really able to reduce the observational errors on various cosmological quantities, including the flatness of the universe. You can see a table of their final cosmological parameters here. For the flatness, what you want to do is add up $\Omega_b$ (the baryonic matter density), $\Omega_d$ (the dark matter density), and $\Omega_\Lambda$ (the dark energy density). This will give you the overall density parameter, $\Omega_0$, which tells you the flatness of our universe. As I'm sure you know from your sources, if $\Omega_0 < 1$ we have a hyperbolic universe, if $\Omega_0 = 1$ our universe is flat, and $\Omega_0 > 1$ implies a spherical universe. From the results of WMAP, we have that $\Omega_0 = 1.000 \pm 0.049$ (someone can check my math) which is very close to one, indicating a flat universe. As far as I know, WMAP was the first instrument to give a truly precise measurement of $\Omega_0$, allowing us to say definitively that our universe appears flat. As you say, the BOOMERanG experiment also provided good evidence for this, but I don't think the results were nearly as powerful as WMAP's was.

The other important satellite here is Planck. Launched in 2009, this satellite has provided us with the best high precision measurements of the CMB to-date. I'll let you dig through their results in their paper, but the punchline is that they measure the flatness of our universe to be $\Omega_0 = 0.9986 \pm 0.0314$ (calculated from this result table), again extremely close to one.

In conclusion, recent results (within the past 15 years) allow us to definitively state that our Universe appears flat. I don't think, at this time, anyone contests that or believes it is still uncertain. As it usually goes with science, answering one question has only resulted in more questions. Now that we know $\Omega_0 \simeq 1$, we have to ask why is it one? Current theory suggests it shouldn't be - that it should be either enormously small or enormously large. This is known as the Flatness Problem. That in turn delves into the Anthropic Principle as an attempted answer, but then, I'm getting out of the scope of this question.


Where on Earth could dinosaurs feel the impact of their extinction asteroid?

The whole Earth "rippled" after the Chicxulub impact. By ripple, I specifically mean surface displacement.  Current models indicate that it rippled enough that most of the dinosaurs and other species with brains certainly would have felt the ripple.
How much did the Earth ripple? Well, that depends on the location. Meschede et al. [2011] in their paper Antipodal focusing of seismic waves due to large meteorite impacts on Earth, visualize a computer model of the surface wave displacement due to the Chicxulub impact in the following figure:

Note that the displacement doesn't get much smaller than 1 centimeter, which is certainly large enough to feel over the order of tens of minutes.  As another point of interest is that their model shows large displacement near the antipodal point, which some geoscientists correlate to the increased volcanic activity at that time in the Deccan Traps. Also note that these are just the seismic waves.  We know from geological evidence at least from the Southeastern United States, that the resulting tsunami was much bigger.
As far as a "wobble", or change in angular momentum, the Chicxulub impact was too small to have a noticeable effect for the dinosaurs.  JamesK has a nice example here: If an Asteroid was to strike the Earth, would it affect the Earth's rotation?.


how to calculate distance between galaxies in the distant past

I'm assuming you're talking about physical distances (as opposed to any of the other distance measures in cosmology).

The comoving distance to a galaxy at redshift $z$ is
$$
d_C(z) = \frac{c}{H_0}\int_0^z
         \frac{dz}{\sqrt{
         \Omega_r(1+z)^4 +
         \Omega_m(1+z)^3 +
         \Omega_k(1+z)^2 +
         \Omega_\Lambda
         }},
$$
where $c$ and $H_0$ are the speed of light and the Hubble parameter, and
$\{\Omega_r,\Omega_m,\Omega_k,\Omega_\Lambda\}$ $\simeq$ $\{\lt 10^{-4},0.3,0,0.7\}$
are the radiation, matter, curvature, and dark energy density parameters, respectively.

By definition, the comoving and the physical distance $d_P$ coincide today, but at the time a galaxy at $z$ emitted the light we see today, it was at
$$
d_P = \frac{d_C}{1+z}.
$$

In your example, $z_A = 6$, and $z_B = 0.5$. The comoving distance to galaxy $B$ is, from the equation above and with a Planck Collaboration et al. (2016) cosmology, $1946\,\mathrm{Mpc}$ away. Hence, when $A$ emitted the light we see today, its physical distance was a factor $(1+6)$ smaller, i.e. $278\,\mathrm{Mpc}$.

In Python, with astropy:

from astropy.cosmology import Planck15
from astropy import units as u
zA,zB      = 6,.5
dC_B       = Planck15.comoving_distance(zB)
dP_B_at_zA = dC_B / (1+zA)
print(dC_B)
   -> 1945.5612969107992 Mpc
print(dP_B_at_zA)
   -> 277.93732813011417 Mpc
print(dP_B_at_zA.to(u.Mlyr)) # convert to light-years
   -> 906.5103217447773 Mlyr


Is it possible for a virus to come from a meteorite?

There is, believe it or not, a scientific theory regarding this: panspermia.

According to the panspermia hypothesis, microbes "hitch" a lift on bodies leaving a planet. They then travel through space and eventually land somewhere else.
Panspermia involves three difficult phases of travel:

Launch (as well as the impact event)
Travel in the harsh environment of space
Atmospheric entry and landing, with high temperatures

It's difficult to survive these. However, materials related to organic matter have been found that might be evidence of panspermia; see Bell et al. (2015) for one example.

Panspermia is traditionally applied to bacteria known as extremophiles, which are accustomed to living in harsh environments. Space contains temperatures ranging from hundreds of degrees Fahrenheit to hundreds of degrees below zero, and any organism would have to survive those. This may be possible - bacteria have survived in conditions like there - but difficult.

Viruses pose an additional challenge because they cannot survive long without a host organism on which to live. This means they would have to be carried along by an extremophilic bacteria. This also means that they might not be a huge harm to creatures on Earth. Viruses only attack certain bacteria because of proteins on their surfaces; extraterrestrial biology most likely differs from that on Earth, and these viruses might not be equipped to harm terrestrial organisms.


Why does Newtonian physics give the right radius for a black hole?

As explained in this answer, dimensional analysis indicates that this radius depends on $GM$ and on a velocity squared. However, the fact that the two quantities are exactly equal, rather than there being a constant factor between the two is a pure coincidence.


Why is CMB an evidence of hotter and denser early Universe?

By request:

Beyond the fact that the cosmic microwave background (CMB) is a direct prediction of the big bang model, there is the question of how you would produce it in any other way. It is remarkably close to being isotropic and remarkably close to being a blackbody spectrum - i.e. it is almost a perfect blackbody radiation field.

A blackbody radiation field is emitted by material in complete thermodynamic equilibrium (CTE). An example would be the interior of a star. A requirement for (CTE) is that the matter and radiation field are characterised by the same temperature and that the material is "optically thick" - meaning that it is opaque to that radiation at basically all wavelengths.

Given that the universe is mainly made up of hydrogen, helium and (presently) traces of heavier elements, we can ask how is it possible to produce a perfect blackbody radiation field? Cold hydrogen and helium are transparent to microwaves. To make them opaque they need to be ionised, so that the free electrons can be a source of opacity at all wavelengths via Thomson scattering. But this requires much higher temperatures - about 3000 K.

How do we uniformly raise the temperature of a gas (adiabatically)? By squeezing it. A smaller, denser universe would be hot enough to have ionised hydrogen and would be opaque to the radiation within it. As it expanded and cooled, the electrons combined with protons to form atoms and the universe becomes transparent, but filled with a perfect blackbody radiation spectrum. The light, originally at a temperature of 3000 K and mainly in the visible and infrared, has had its wavelengths stretched by a factor of 1100 by expansion of the universe, meaning we now see it mainly as microwaves.

Additional evidence for this model is that the radiation field is not absolutely isotropic. These small ripples encode information such as the expansion rate of the universe at the time of (re)combination and the density of matter. When inferred from measurements, these parameters agree very closely with other determinations that are independent of the CMB, such as  the Hubble redshift distance relationship and estimates of the primordial abundance of Deuterium and Helium.

There is now direct evidence that the CMB was hotter in the past and by exactly the amount predicted by an adiabatic expansion. The source of this evidence is measurements of the Sunyaev-Zel'dovich effect towards galaxy clusters (e.g. Luzzi et al. 2009); or more precisely by probing the excitation conditions in gas clouds at high redshift using even more distant quasars as probes (e.g. Srianand et al. 2008.


Will the Sun rotate slower in red giant form?

Indeed conservation of angular momentum dictates that in a single star like the the Sun, rotation should be much slower when it becomes a red giant. This is because at the present time the Sun does not rotate at vastly different rates with depth, thus when it expands, the moment of inertia increases drastically and convection in the outer envelope will ensure that slow rotation is enforced throughout most of the star.

The consequence for magnetic activity will be that the Sun will become magnetically inactive (in relative terms - magnetic activity will not be absent, but its signatures will be much reduced when expressed as a fraction of the stellar luminosity) as a red giant, because dynamo-generated magnetic activity is strongly correlated with rotation rate.

In general it is found that giant stars are magnetically inactive and slowly rotating, but there are exceptions. If a star considerably more massive ($\sim 3-5M_{\odot}$) than the Sun evolves to become a giant it may have preserved a considerable amount of angular momentum, since such stars are not magnetically active on the main sequence and are unable to lose their angular momentum through a magnetised wind in the same way that a solar-type star does. Secondly, it is possible that some rapidly rotating giants (e.g. the FK Com stars) could be the result of mergers in binary systems or possibly even giant planet engulfment. In which case the angular momentum is from the binary system and these stars can be very magnetically active. Finally it is also found that giants in close binary systems (the RS CVn stars) can be fast-rotating and magnetically active because tidal locking in the binary system enforces rapid rotation. The FK Com and RS CVn stars manifest many signs of extreme magnetic activity - hot X-ray coronae, chromospheres, coverage of large fractions of the surface by cool starspots. We thus deduce that the difference between these and the vast majority of magnetically-inactive giants is their rapid rotation.


Absolutely:
$$\theta = \frac{4GM}{c^2b},$$
where $\theta$ is the angle of bending, $G$ is the gravitational constant, $M$ is the mass of the star/blackhole/whatever, $c$ is the speed of light, and $b$ is the "impact parameter", the distance from the light ray to the center of the mass on its closest approach if it was NOT deflected (i.e., $b=0$ for an impact).


The speed of the blast front depends on the initial energy release and the density of the medium into which it is expanding, see here.

Theory suggests and measurements confirm expansion rates of the order of thousands of km/s or a few $\times 10^6\ \mbox{m/s}$ or  $\sim 1\% \mbox{c}$.


Try WinJupos. It can be used to measure Jupiter (as well as other planets)


Any cheap (<100 euro) tools for observing the lines in solar spectrum?

You need a diffraction grating in order to split the sunlight into a spectrum. It turns out even a normal CD or DVD because of the pits that they have to encode data work very well as high resolution diffraction gratings. You could make a simple enclosure with a CD and a slit like this:

http://coolcosmos.ipac.caltech.edu/cosmic_games/spectra/makeGrating.htm

This should allow you to observe the sun's spectrum, although I would also recommend that you be very careful when observing the sun directly. Preferably use a very narrow slit--this will both make it safer and make the absorption lines much clearer.


What is this rapidly twinkling red, blue, and white star I saw?

It's most probably Sirius.  At this time of year (at 1 am local time) it's low in the sky in the East, so there is a lot of atmosphere in the way, and as Sirius is a bright bluish star, it will show all the colours described as it twinkles.


One reason to prefer the space between the temperature and luminosity classes is that the hypergiant luminosity class is "0", which could be confused with a subtype. Usually this appears as part of a range with luminosity class "Ia" as opposed to on its own, but this range is often written as "0-Ia". Something like "G5 0-Ia" is more comprehensible than "G50-Ia".

If you're writing a parser, be prepared to handle both. In general, parsing spectral types is a bit tricky (I have written spectral type parsers before, it's "fun") and there are a surprising number of annoying edge cases. It would be nice if astronomical catalogues also included something like the IVOA spectral type encoding, but usually the furthest you get is some bespoke system that only encodes standard OBAFGKM types and provides no way to encode, say, carbon stars or Wolf-Rayet types, even though the catalogue contains examples of these.

Some other issues you may run into:

The ambiguity between "-" as a minus sign and "-" as a range symbol. Uusually the machine-readable files will be ASCII so none of the other dashes in Unicode are available, and given the general state of things I wouldn't trust them to be used reliably anyway!
The previously-mentioned "0-Ia" luminosity class range means that a parser will run into an ambiguous interpretation of "G0-Ia": either an open ended range "G0 or earlier" of luminosity class "Ia", or a "G" type star with no subtype and luminosity class range "0-Ia".
The fact that "+" may be an open-ended range symbol or a multi-spectrum separator "M5+" (star of type M5 or later) vs "M5+..." and "M5+K2" (M5 star with companion).
The peculiar A-type stars, e.g. "kA5hF0mA5 V".
The various two-parameter systems for carbon and S-type stars.
The fact that the entire spectral type or various subcomponents may be bracketed to indicate uncertainties...
...which must be distinguished from the use of square brackets to indicate that the star is the central star of a planetary nebula
If you're really lucky, your source will render the spectral types in ALL UPPERCASE (so that peculiar A-star spectrum from a few bullet points up becomes "KA5HF0MA5") because it's been compiled using some Fortran code that hasn't been updated since the 1960s or something...