Are all models of ocean planets theoretically cloud covered?

Having given this some thought, it's a bit of a monster question with a few variables that you don't mention.  Planetary tilt, which causes seasons is one.   Length of day, for example, a rotation like Venus' 116 days would be quite different than an earth rotation of 24 hours and a tidally locked planet would also be different.   Atmospheric pressure is another.   I'm assuming Earth like tilt, 24 hour day since you didn't mention, and 1 ATM, but I think, if and when we finally get a good look at earth-like exo-planets, which, granted, might be several decades away, I think we'll see a few unexpected things.

OK, moving on.   Earth likely got much of it's atmosphere from comets which suggests healthy amounts of CO2, CH4 and NH3, maybe N2, some from the NO family, maybe some Helium, Argon and Neon, those 3 I think we can mostly ignore, some hydrogen perhaps from solar flares, which might get stripped along with the helium from the planet over time.

The total amount of comet and asteroid impacts and ratio of gasses add some variability and a really important factor is whether or not the planet has life and undergoes photosynthesis.   Photosynthesis would put O2 into the atmosphere and reduce CO2 and would also have the effect of, over time, reacting with and eliminating the virtually all of the CH4 from the air and any dissolved Iron in the oceans, turning the oceans from a muddy brownish/red to clear blue.  Photosynthesis likely played a key role in turning the Earth from a hot planet to a snowball.

So, lots of variables, but it's a fun speculative question, so I'll give it a shot.

How much smaller than Earth?   I'm going to go with Venus size but if you have different in mind, let me know.   .815 Earth masses, 6,052 km in Radius and gravity of .905 of Earth.  Source.

10 miles deep oceans (16.09 km), surface area, 4 Pi R^2 about 7.4 billion cubic km of water.  (Earth, by comparison, about 1.37 billion cubic km)  Source.   Your planet has about 5.4 times as much water as earth (not counting water in the crust, but you didn't mention that, so lets not go there) - and planet size doesn't affect the predictions much.

  I wonder how much outgassed CO2 etc. would stay dissolved in the ocean
  and how much would accumulate in the atmosphere? Nitrogen?

I'm going to assume that the theory is correct that Earth (and your water world) get most of it's atmosphere and water from asteroids and comets Source and Source, though the 2nd source suggests some could have come from the primordial material of the Earth.   If your water-world has a lot of gas trapped below it's crust in it's primordial material, then out-gassing from volcanism becomes a bigger factor and the volcanoes, not the atmosphere above would keep the oceans saturated with gas.   That's possible, but I'm going to assume that most of the gas is already in the atmosphere following a late heavy bombardment period.   I'll touch a bit more on tectonics later.

Titan Probably got it's thick atmosphere from outgassing Source and perhaps Venus from a recent large outgassing event too - just throwing that out there.

I'm also assuming that your planet wouldn't have permanent ice over it's oceanic poles due to the warmish temperature you implied and likely oceanic circulation not blocked by land-masses would prevent ice formation.

  Would it necessarily have a runaway greenhouse?

Greenhouse Gases, CO2, CH4, NO family, (H20, more indirectly)

If your planet has a lot of CO2/CH4, comparable amounts to the amount of H20, then it's unlikely the oceans could begin to dissolve enough gas and this would lead to a run-away Greenhouse, almost without question.

If your planet has more earth like CO2/CH4 levels, perhaps driven by photosynthesis capturing carbon and released oxygen chemically reacting with the CH4, then the runaway greenhouse can be avoided and CH4 be very low concentration.   CH4 on Earth is currently about 1.8 PPM and prior to farming and livestock and oil drilling & Fracking which can release some CH4, it was probably less than half that much, less than 1 PPM.  In a water world CH4 would probably be even less as there's no biodegrading or digestion of plant-mass releasing CH4.

CO2 is more complicated and the amount dissolved in the ocean depends on the total amount available.  If there's not too much CO2 to saturate the oceans, then you have an ocean/air equilibrium which is also affected by temperature.    For a more detailed answer on this, look into Henry's law, but I'm going to cheat and do a quick and dirty calculation rather than use Henry's.

on Earth, the oceans contain about 50 times the CO2 as the Atmosphere:   Source.  And the oceans weigh about 265-270 times more than the atmosphere.   (ocean mass given above, atmosphere mass here.)

So if we estimate this as parts per million (ppm), the CO2 concentration in Earth's atmosphere is about 5 times the concentration in Earth's oceans, and at higher temperatures, that ratio goes up, lower temperatures it goes down, but it never gets to the point that there's no CO2 in the air, cause not possible in an equilibrium.   With 10 mile deep oceans, we can assume a more CO2 in the oceans, maybe 99% vs about 98% ocean to air ratio currently on earth but the 1% in the air would depend on the total mass of CO2 in the water world's ecosystem, ignoring anything permanently trapped deep on the ocean floor, such as sea-shells.

  Would such a world without land always be covered with clouds?

Lets say you have an Earth like Oxygen/Nitrogen atmosphere, 1 bar, a planet very similar to earth but just oceans.  Similar weather (as a starting point), this planet should have more clouds as clouds are more common over the oceans, source, and you would have no low humidity/dry pockets of land, but I would think, probably not 100% clouds, just more clouds.   The cloud effect on temperature has some uncertainty to it, as clouds both both reflect sunlight making the planet colder during the day and they trap heat at night, so they play for both the warming team and the cooling team, but what I've read, the overall effect is pretty small.   Days could be colder, nights warmer.

Also, evaporation tends to cool the air at the surface, which is why islands surrounded by ocean don't ever get the scorching hot temperatures you get in death valley for example, even if they're closer to the equator, so you wouldn't get heatwaves, but you wouldn't get freezing cold either.  (I would think).

Finally, oceans have lower albedo and a water-world earth would likely have no ice-caps, so, Overall, I think it's likely a water-world earth would be warmer primarily due to lower albedo and this would increase atmospheric water-vapor (not clouds but transparent water vapor, which is a greenhouse gas), so a water-earth would probably several degrees warmer on average with no ice-caps.    How much warmer - I have no idea but I don't think a water-earth would be run-away greenhouse.

A water world (Earth) could also have less CO2 in the atmosphere due to more dissolved in water and probably less CH4 from decaying matter on land, so, it might actually be colder, but this would more likely depend on the life cycle's O2 to CO2 ratio rather than oceanic absorption of CO2.   Too many unknowns to say for sure.

Another curious effect to a water world earth is you might get some hurricanes that last for weeks, maybe even months or years in the right conditions, kind of like mini versions of Jupiter's great red spot.   All you need for a hurricane to gain strength is warm water and cold air and with no land for the Hurricane to lose strength on, You could get some real jumbos.   Maybe category 7s, perhaps 8s.  That would be awesome.

Now if (and I think this is the gist of your question), what happens if you decrease the atmospheric pressure by significantly lowering the amount of O2/N2 in the atmosphere (or raising it).  Lower atmospheric pressure lowers the boiling point of water and if you lower the atmospheric pressure enough, then water starts to boil and you get a partial water vapor atmosphere as a result of low atmospheric pressure and lots of water.   I suspect this is unlikely to actually happen as a planet should always have enough other types of gas to prevent this unlikely outcome but a water vapor atmosphere can be approached logically, even if it's not going to really happen.  Now, this isn't just high humidity, this is an actual water vapor atmosphere.

I imagine a water-vapor rich atmosphere would form droplets and rain fairly regularly and be permanent or nearly permanent cloud cover but that's just a guess, though it would probably appear more like low to the ground smog than cloud cover, the clouds might hover much closer to the earth.

If you increase atmospheric pressure, then water has a higher boiling point and takes more energy to vaporize.   A denser atmosphere might hold heat better and be a bit warmer.  It might also have fewer clouds due to more air and perhaps less variation in temperature and circulation, as it's circulation and warm air cooling off that's the primary driver in cloud formation.

Water vapor would still form under a higher pressure atmosphere by wind and by photons.  Sunlight plays an important role in evaporation, perhaps even more than temperature based on pan evaporation studies.

If you have a very thin atmosphere made almost entirely of water (getting back to our improbable example).  You'd drown if you tried to breath it, but as far as temperature of such a planet, at least with earth temperatures, water stays liquid even at quite quite low atmospheric pressure.

At 1/2 PSI (1/29th of 1 atm) the boiling temperature of water is 79.6 degrees C.  Source.   I'm not sure how much heat 1/29th of an atmosphere could trap even if it was mostly water which is a greenhouse gas, so this one is hard to predict.  You'd likely see wild night to daytime temperature swings, frequent boiling of the oceans during the day and very heavy rainfall at night, much more rain than we ever see on earth, as 1/29th of an ATM of water vapor is several times more water than is in the earth's atmosphere at any given time.

If a water-vapor atmosphere planet (unlikely), got enough heat from it's sun, it could certainly turn into a run-away greenhouse, but I'm not sure if that happens at the radiance earth gets.   But a planet like this would probably be subject to bigger temperature swings than we see on Earth and very fast wind.

  How about such a world with no vulcanism?

(Vulcanism - as in, spock, . . . sorry).

We think of plate tectonics and volcanism as a process by which gas is given to the atmosphere, and that's true to an extent.  Certainly any gas trapped inside a planet at formation can be outgassed through Volcanism, but perhaps a more important aspect of plate tectonics isn't the release of gas but the absorption of gas.   Oxygen is very reactive and it combines with basaltic rock to make lighter/stronger rock like granite and bedrock which leads to continents and mountain ranges and all that good stuff.  Nitrogen can bind with basalt too, but Oxygen I think, binds more readily.   Without photosynthesis and the creation of oxygen, not just life on earth but the continents would look very different and they'd probably be much less permanent.

So if all the volcanoes are under water, that limits the atmospheric absorption of oxygen and other gases to just what's disolved in the oceans which could be a lot of NH3, but relatively lower amounts of other gasses.

Another effect of all tectonic activity and volcanoes being under water is that volcanic gas like sulfates and tiny dust particles which can cool the planet after a large eruption, are unlikely to reach the atmosphere in any significant volume, so you're likely to avoid any volcanic cooling like you get on occasion on earth.

Finally, you might think that volcanoes under oceans would warm the oceans and they would locally, but not much overall.   A significant percentage of Earth's volcanoes are under water, but because ocean water circulates, most of the deep oceans stay at a chilly 4 degrees C, as opposed to land which gets warmer as you dig into the ground.  Convection circulates heat much faster than conduction.

Undersea volcanoes in your water world could be very helpful in the continuation and maintenance of extremophile life due to repleneshment of nutrients but not much in the way of temperature change or cloud formation.

Finally - a fun bit worth mentioning.   NH3, which is a very common gas/ice in the solar system.  It's in all the outer planets and comets and outer moons.   The neat thing about NH3 is it's water soluble and it's an energy source to primitive life, so simply by having oceans, stinky but useful NH3 is fairly quickly, mostly removed from the atmosphere and dissolved in liquid oceans and if there's life, it can be consumed, some released as N2 or NO-family and some can be built into amino acids.  Some nitrogen can bind with basaltic rock and magma from undersea volcanoes forming different types of rock, but I'm no geologist, so I'm not sure how much that would happen.

There's no one simple answer to this, but that's my best guess as a layman.   I enjoy thinking about stuff like this.


There was no single designer of the Hubble telescope. It was designed and build by NASA, with contributions from the European Space Agency (ESA).

The advantages of a telescope in space have been clear for over 100 years. These ideas were crystallized in Lyman Spitzer's 1946 paper. Spitzer continued to push for a telescope through the '50s and '60s.

When Congress eventually funded the telescope in '78, the task of design was given to the Marshal Space Flight Center. The head of the space telescope project was Dr. Bob O'Dell, chair of the Astronomy dept. at the University of Chicago. In a video from NASA, Dr O'Dell talks about the early years of HST development.

Over the years others have worked as Project Manager. There never was a single "inventor". Hundreds of scientists have been involved in its design and planning. Something as complex as the HST could never be designed by a solo inventor.


Why did Venus not lose its atmosphere without a magnetic field?

There is an interesting article on the magnetosphere of Venus on the ESA Science and Technology site. You can find the article here and it will probably answer your question.
The article states, like you did, that some planets, like Earth, Mercury, Jupiter and Saturn, have magnetic fields internally induced by their iron core. These magnetic fields shield the atmosphere from particles coming from solar winds. It also confirms your statement that Venus lacks this intrinsic magnetosphere to shield its atmosphere from the solar winds.
The interesting thing, however, is that spacecraft observations, like the ones made by ESA's Venus Express, have shown that the Venusian ionosphere's direct interaction with the solar winds causes an externally induced magnetic field, which deflects the particles from the solar winds and protects the atmosphere from being blown away from the planet.
However, the article also explains that the Venusian magnetosphere is not as protective as Earth's magnetosphere. Measurements of the Venusian magnetic field show several similarities, such as deflection of the solar winds and the reconnections in the tail of the magnetosphere, causing plasma circulations in the magnetosphere. The differences might explain the fact that some gasses and water are lost from the Venus atmosphere. The magnetic field of Venus is about 10 times smaller than the earth's magnetic field. The shape of the magnetic field is also different. Earth has a more sharp magnetotail facing away from the Sun and Venus has a more comet-shaped magnetotail. During the reconnections most of the plasma is lost in the atmosphere.
The article explains therefore that although Venus does not have an intrinsic magnetic field, the interaction of the thick atmosphere with the solar winds causes an externally induced magnetic field, that deflects the particles of the solar winds. The article suggests, however, that the different magnetic field may mean that lighter gasses are not as protected and therefore are lost into space.
I hope this sufficiently answers the question.


Is there a general term used to identify the high flat features within Noctis Labyrinthus on Mars?

The geological term mesa has been used to describe these structures.

  A transitional zone on Mars, known as the fretted terrain, lies between highly cratered highlands and less cratered lowlands. The younger lowland exhibits steep walled mesas and knobs. The mesa and knobs are separated by flat lying lowlands. They are thought to form from ice-facilitated mass wasting processes from ground or atmospheric sources. The mesas and knobs decrease in size with increasing distance from the highland escarpment. The relief of the mesas range from nearly 2 km to 100 m depending on the distance they are from the escarpment.

Mesa is a term used to describe flat topped isolated geological structures that are composed of horizontal layered rocks (usually sedimentary).

If the flat topped structure is not composed of horizontal layered rocks it can be called a plateau or tableland.


What is the probability of seeing a neutron star as a pulsar?

Out of 100 pulsars, how many will have a beam that crosses the Earth?

About twelve.

"The beaming fraction f , that is the mean value of the fraction of observable pulsars or the mean probability of observing a normal pulsar, is 0.124 ± 0.004." M. Kolonko et al.: On the pulse-width statistics in radio pulsars


Yes, minerals can be observed using spectroscopy on a telescope, typically from their thermal signature. We can determine what type of mineral based on the elements that make up the mineral as well as the type of crystal structure, which will give a different spectral signature


How do you determine the apparent magnitude of the stars leaving main sequence with CMD diagrams?

There is no clearly-defined "turn-off" magnitude. Stars move from the zero age main sequence (roughly speaking, where core hydrogen burning begins to dominate the luminosity) to the terminal age main sequence (where the core hydrogen runs out) relatively slowly, increasing in luminosity as they do so. Their evolution then accelerates redward in the colour magnitude diagram. The turn-off is roughly the most luminous star for which this acceleration in evolution is yet to occur, but this is confused/blurred by observational uncertainties, binarity, differential reddening and spreads in age and rotation rate.

In practice what you do if you really want to do a proper job of estimating the age, with uncertainties, is you do a population synthesis (including binaries and observational uncertainties) for clusters of stars with a variety of ages. You then compare these with the observations using maximum likelihood or Bayesian techniques (e.g. Naylor & Jeffries 2006).


As mentioned in this comment if you were using low power, say 40x with a 50 degree apparent FOV eyepiece that would be about 0.4 degrees per second which is in the ballpark of the apparent motion of an artificial satellite in low Earth orbit (LEO).

A geostationary satellite (GEO) or even a geosynchronous one would move much more slowly relative to Saturn, and if your telescope wasn't motorized, it would appear in a fixed position even as Saturn slowly moved across your field of view due to Earth's rotation.

In the bottom plot below, taken from this answer it shows that 0.4 degrees per second is quite reasonable for satellites with a wide variety of altitudes. For satellites in LEO between 300 and 1000 km and viewing angles from zenith to near-horizon apparent motion can range from below 0.1 to over 1.4 degrees per second.

As pointed out in this comment it could perhaps be an airplane as well if there was some source of light. It could have had some lights on, or you were watching at dawn/dusk and it received some sunlight, or I suppose it might be possible that it was reflecting Moonlight.

But I think a comet (also mentioned in that comment) or a tiny near-Earth asteroid couldn't possibly appear to move a substantial fraction of a degree per second unless it was closer than the Moon, and we would likely (but not for certain) hear about that, though there are surprises as discussed in the (currently unanswered) question What is the “interesting story on the limitations of (NASA's) current (near Earth object) survey network”?

For further reading on artificial satellites in low Earth orbit, see What actually happened here with a satellite, the ISS and the moon? and also this answer to Why is taking a picture of the moon surface from earth possible but taking one of an artificial satellite is difficult? and while you're at it, answers to Are commercial communications satellites in GEO being constantly monitored by telescopes?


The newly discovered comet NEOWISE has been given the name C/2020 F3. What is the meaning of the F3 and the C in this name?

C means "comet": it has a coma which means that volatiles are being released due to solar heating.  Other possible letters are "A" for asteroid,  "P" for (short) periodic comet, "D" for disappeared comet, "I" for interstellar object and "SN" for supernova.
2020 is the year of discovery
F represent the half-month of discovery, so "A" is the first half of January, and so on. "F" means "the second half of March".
3 means the third comet identified in that half month, and "NEOWISE" is the name of the discoverer, in this case it was discovered by a space telescope that scans the sky looking for Near Earth Objects.
Such designations are considered provisional and can be replaced with a permanent number if the object is found to be in a permanent orbit. So a full designation of Halley's comet is "1P/1682 Q1 (Halley)" as it has permanent number 1, it is periodic, it's orbit was the first published in the second half of August 1682 by Edmund Halley.
See https://minorplanetcenter.net//iau/lists/CometResolution.html


If I have amateur astrophotography image data that might be of scientific value, what should I do?

It depends, as you say, on the subject matter. For instance, images useful in variable star study can be provided to the AAVSO http://www.aavso.org/ and images relating to asteroids can be provided to the IAU Minor Planet Center. http://www.minorplanetcenter.net/ Other subject matter will be accepted elsewhere.

However, for the images to be useful in scientific fields, you will need to comply with the submission instructions, which a casual snap probably will not comply with. On the MPC guide for beginners there is a list of 44 technical suggestions for submitting scientific data: http://www.minorplanetcenter.net/iau/info/Astrometry.html The AAVSO has a tutorial in six chapters on photometry using a DSLR camera (and another for astro CCD cameras) http://www.citizensky.org/content/dslr-documentation-and-reduction

If you just want to get it out there and have people comment on what is in it, post it to Google+.


Where did this famous Planetary Precession Formula come from?

I do not know where that formula was first published in full, but Oppenheim at least does something very close to it. First, keep in mind some of the relevant symbols in Oppenheim, though they're rather standard:
$$\def\anode{☊}\begin{eqnarray*}
k &=& \small\sqrt{G} = \small\text{Gaussian gravitational constant}\\
\anode &=& \small\text{longitude of the ascending node}\\
\omega &=& \small\text{argument of perihelion}\\
\varpi &=& \omega+\anode = \small\text{longitude of perihelion}\\
n &=& \small{k\sqrt{(m_0+m_1)/a^3}} = \small\text{mean motion}
\end{eqnarray*}$$
We can see that if we massage the notation, the proportionality we're looking for is equivalently stated as:
$$\frac{\pi m}{Lc^2} \leadsto \frac{\pi GM}{a(1-e^2)c^2} \leadsto \frac{\pi n^2a^2}{(1-e^2)c^2}\text{.}$$
Since the mean motion is $n\equiv 2\pi/T$, where $T$ is the orbital period,
$$\frac{n^3a^2}{c^2} = 2\pi\frac{n^2a^2}{c^2}\frac{1}{T}\text{,}$$
which means that if we want to talk about a perihelion advance of $\delta\varpi \propto \frac{\pi n^2a^2}{(1-e^2)c^2}$ per orbit, it is equivalent to talk about a term in the form
$$\frac{\mathrm{d}\varpi}{\mathrm{d}t} \propto \frac{n^3a^2}{(1-e^2)c^2} = \frac{n^3a^2}{c^2}\left(1+e^2+\mathcal{O}(e^4)\right)\text{.}$$
I cannot find where, if indeed anywhere, Oppenheim considers the missing factor of $(1-e^2)$ as belonging to the anomalous precession, but otherwise the formula is definitely there. I suspect that he only uses a circular orbit approximation, $e^2\approx 0$, because it's nowhere to be found in Sec. IV (Weber's 1846 theory) and performing his calculation (sans $1-e^2$) gives me $\delta\varpi = 13.72''$ per century for Mercury, in good agreement with his stated result of $\delta\varpi = 13.65''$, whereas putting in a factor of $(1-e^2)$ by hand gives $\delta\varpi = 14.32''$ instead.

Perhaps Oppenheim did not consider it explicitly and the author of the MathPages took it as obvious that the eccentricity factor should be there. Or perhaps there's a side comment in the text that I don't see; unfortunately, I'm nowhere fluent enough in German to understand much of what's going on.


Is there a term to describe all stars that are not neutron stars or white dwarfs?

Every so often someone will note that a "star" is a ball of plasma that is supported by nuclear reactions at its core, and so technically neutron stars and white dwarfs are stellar remnants, and not stars.

The technical term is therefore "star". And if you think that could be ambiguous then just spell it out: "Stars with supported by fusion reactions"/"Stars, excluding stellar remnants"


Determining moonset time from Nautical Almanac

It means that there is a moonset at that time on the third of February.

At the equator, it is certain that the moon that set at 0041, rose the previous day (Feb 2nd), at about midday.  (the moonset is about 50 minutes later each day, which means that there can be days with no moonset, but you can't have a day with two moonsets)

On days with no moonset the time of moonset in your almanac might be indicated with a time that is more than 23:59

For example:

Date of month
1    2    3    4
h m  h m  h m  h m
2052|2250|2500|0100

Here you see on the first day of the month the moon sets at 20:52, on the second it sets at 22:50, on the third day there is no moonset. The moon sets one hour after midnight. The moonset 25:00 and 01:00 refer to the same event.


They must be referring to two different filters:

the (Bessel) I filter which has a central wavelength of $\lambda \simeq 800\,\mathrm{nm}$  and a width of $\Delta\lambda\simeq 150\,\mathrm{nm}$, and
the (Sloan) z' filter, which has $\lambda,\Delta\lambda \simeq 970,255\,\mathrm{nm}$.

Hence their transmission is in the near infrared ("NIR"), and they're both above 700 Å.

The exact transmission curves vary a little depending on the producer, so if you want to calculate exact magnitudes, for instance, you need to find the transmission curves for the telescope in question. From the TRAPPIST equipment website, it seems they use filters from Astrodon. Astrodon don't seem to provide filter curves for these exact filter, but if you're okay with a good approximation, you can find the curves here:
Bessel I and
Sloan z'.


What should I look for in an astronomy tripod?

I have the exact same binocular, and in fact I also have a very similar tripod and head. I'll take a photograph this evening to demonstrate my setup.

I am very happy with the setup, and I find it very stable. I usually sit the tripod on grass, but I do not recall ever having a problem with it on concrete or stone. I do try to keep the neck as short as possible, and use the legs for height. Also, the binocular can have it's tripod mount moved fore or aft, I think that mine is in the full aft position.

You might be able to hang a large mass (bag, perhaps) from the underside of the tripod; you should find a hook already there made specifically for this purpose.


Will Earth lose the Moon before the Sun goes into supernova?

As HDE 226868 noted in his answer, the Sun is not going to go supernova. That's something only large stars experience at the end of their main sequence life. Our Sun is a dwarf star. It's not big enough to do that. It will instead expand to be a red giant when it burns out the hydrogen at the very core of the Sun. It will continue burning hydrogen as a red giant, but in a shell around a sphere of waste helium. The Sun will start burning helium when it reaches the tip of the red giant phase. At that point it will shrink a bit; a slight reprieve. It will expand to a red giant once again on the asymptotic red giant branch when it burns all the helium at the very core. It will then burn helium in a shell surrounding a sphere of waste carbon and oxygen. Larger stars proceed beyond helium burning. Our Sun is too small. Helium burning is where things stop.

The Sun has two chances as a red giant to consume the Earth. Some scientists say the Sun will consume the Earth, others that it won't. It's all a bit academic because the Earth will be dead long, long before the Sun turns into a red giant. I'll have more to say on this in the third part of my answer.

The current lunar recession rate is 3.82 cm/year, which is outside your one to three centimeters per year window. This rate is anomalously high. In fact, it is extremely high considering that dynamics says that
$$\frac {da}{dt} = (\text{some boring constant})\frac k Q \frac 1 {a^{11/2}}$$
Here, $a$ is the semi major axis length of the Moon's orbit, $k$ is the Earth-Moon tidal Love number, and $Q$ is the tidal dissipation quality factor. Qualitatively, a higher Love number means higher tides, and a higher quality factor means less tidal friction.

That inverse $a^{5.5}$ factor indicates something seriously funky must be happening to make the tidal recession rate so very high right now, and this is exactly the case. There are two huge north-south barriers to the flow of the tides right now, the Americas in the western hemisphere and Afro-Eurasia in the eastern hemisphere. This alone increases $k/Q$ by a considerable amount. The oceans are also nicely shaped so as to cause some nice resonances that increase $k/Q$ even further.

If something even funkier happens and the Moon recedes at any average rate of four centimeters per year over the next billion years, the Moon will be at a distance of 425,000 km from the Earth (center to center). That's less than 1/3 of the Earth's Hill sphere. Nearly circular prograde orbits at 1/3 or less of the Hill sphere radius should be stable. Even with that over-the-top recession rate, the Moon will not escape in the next billion years.

What about after a billion years? I chose a billion years because that's about when the Moon's recession should more or less come to a standstill. If the Earth hasn't already died before this billion year mark, this is when the Earth dies.

Dwarf stars such as our Sun get progressively more luminous throughout their life on the main sequence. The Sun will be about 10% more luminous than it is now a billion years into the future. That should be enough to trigger a moist greenhouse, which in turn will trigger a runaway greenhouse. The Earth will become Venus II. All of the Earth's oceans will evaporate. Water vapor will reach well up into what is now the stratosphere. Ultraviolet radiation will photodissociate that water vapor into hydrogen and oxygen. The hydrogen will escape. Eventually the Earth will not only be bare of liquid water on the surface, it will be bare of water vapor in the atmosphere.

Almost all of the Moon's recession is a consequence of ocean tides. Without oceans, that tunar recession will more or less come to a standstill.


Solar / lunar positions in ECEF using SOFA C libraries

A few responses to my own question:

(a) TDB time isn't critical because it only varies from TT by a few milliseconds.

(b) The resulting vector was not ECI but the vector from the earth to the sun in the celestial frame.  I needed to build the celestial to terrestrial rotation matrix using iauC2t00a() (assuming zero polar motion since it's generally less than 20 meters), and then rotate the resulting pv (position, velocity) vector-pair into the terrestrial frame using iauRxpv().

(c) (Still no clue on this one.)


I made a graph trying to show the Sun's movement relative to the Galactic Plane. It doesn't seem to be correct

Unfortunately, what you are trying to do is not possible. We do not know the details of the Sun's motion or the Galactic potential precisely enough to make a prediction of exactly where the Sun was 4.5 billion years ago.
The amplitude and period of the oscillatory motion that I specified in my answer to the previous question is known now to of order 10%. If you extrapolate backwards for more than a few cycles then the phase could be anything.
In addition, the Sun has likely changed its radial position in the Galaxy during its life. If so, then the amplitude and period of the vertical oscillatory motion will also have evolved.


Is it possible that the Universe is expanding due to additional Big Bang events?

You may need to revisit the theory behind the Big Bang. The Big Bang is more or less a description of what the early universe looks like if you view expansion in reverse. The single "Big Bang" at the beginning of our universe is the only "Big Bang" event predicted by expansion. Also, the Big Bang did not happen in a single place, it happened everywhere in the universe. Finally, the Big Bang did not create space-time, and the expansion of the universe is not the creation of new matter. The Big Bang describes a time when the metric of space-time was extremely dense, and expansion is the metric of space-time increasing as below:

The distance between points in space increases, but the same mass-energy is there as before.


How to find how many degrees from the beginning to the end of the sunset?

The easiest way to measure the sun would be to use a pin-hole camera.

Use a piece of card with a pin hole. Hold it up so the sun shines through the pin hole and onto a piece of paper.  (Don't look at the sun through the hole - eye damage)

You will see a circle of light on the paper, this is the image of the sun. If the paper is 100cm from the card and the size of the image of the sun is $x$ cm, then you can work out the angular size of the sun.  It is about $\tan^{-1}(x/100)$ which is approximately $\frac{180x}{100\pi}$ You should get an angle of about 0.5 degrees.


When the asteroid hit the earth 65 mya did the earth's gravity pull change? By how much?

The gravity of Earth did not change substantially.

The strength of gravity on Earth is about 9.8, but varies between 9.76 and 9.83 ms-2(due to Earth not being a perfect sphere)

This can be approximated from the mass of the Earth using $$g=GM_e/r_e^2$$

Where $M_e = 5.97219\times 10^{24}kg$ and $r_e= 6371000m$.

The mass of the asteroid was about $10^{16}kg$ That changes the mass from $5.97219\times 10^{24}$ to $5.97219001\times 10^{24}$. It changes the gravity from 9.8 to 9.800000001

In other words. The change in gravity was utterly insignificant. You probably get a greater change in gravity by walking downstairs.


How can there be anything "beyond" the CMB?

Yet we also have a third axiom:

  There are some parts of the universe we can never observe because they are receding away from us at a superluminary speed.

This is simply false, so of course it gets you in trouble if you insist on taking it as axiomatic. The average recession velocity goes luminal at redshift $z\approx 1.4$, while there are observed galaxies and objects at $z\sim 8$, e.g., Z8 GND 5296 dwarf galaxy and the GRB 090423 gamma ray burst. Additionally, there are at least some candidates for objects much farther than even that, possibly $z\sim 12$.

That means that other than the CMB itself, the most distant and most ancient object we might have observed hails from when the universe as around $370\,\mathrm{Myr}$ old, which is around a thousand times older than the recombination epoch when the universe first became transparent and hence when the cosmic background radiation was emitted. In terms of redshift, the CMB has $z\gtrsim 1100$ or so.

  But this surely also implies that we can see beyond the CMB if we see anything which has a red shift indicating an expansion speed very close to $c$.

Something at redshift $z\gtrsim 1100$ has a recession velocity of $v\gtrsim 3.2c$.

  This suggests axiom 2 is incorrect - so what should it be instead?

Since (3) says incorrect things, it's best to throw it out instead.


Can air pollution affect the observed positions of stars?

The air pollution has a negligible effect on the refraction, because pollution barely affects the air density.
However, the refraction is proportional to the air pressure and inversely proportional to the air temperature in Kelvin. So different weather conditions or ground elevations will result in different refraction values.


Horizon
The horizon is Earth's horizon ... absent any obstructions such as buildings, trees, etc.  Or another way to think of it is the horizon you would observe looking out across the ocean while standing on a beach.
The notion that it is "10° above the horizon" means that from the horizon line, it would be 10° up.  You closed fist (as if you are about to punch someone) measure about 10° (from your thumb to your pinky finger) if held at arm's-length.  Close your fist and extend your arm ... placing the bottom (the edge along the side with your pinky finger) on the horizon.  10° up is roughly where the knuckle of your thumb is located.
Caveats:  Time and Date
But a couple of cautions should be mentioned regarding the article...
Objects in our Solar System move from night to night and this motion can sometimes be noticed even from hour to hour.   Tonight (July 21, 2020) it is still below the "bowl" section of the Big Dipper asterism.  But in a few days that will no longer be true.  One week from today it will no longer appear "below" any part of the Big Dipper asterism.  It is heading through the Coma Berenices constellation.
Also, the 10° measurement... this assumes both a time in the evening AND a latitude on Earth.
I'm located at roughly 42° North latitude.  At 10:30pm for me and tonight (because time and date both matter) the comet is 22° above the horizon.  But if I wait an hour and check 11:30pm ... comet is only 14° above the horizon.  This is because the comet is setting ... sort of.  Technically if you are far enough North the comet doesn't "set" below the horizon.  Since I am located at roughly 42° North, anything with 42° of the celestial pole (near the North Star) doesn't actually "set" below the horizon.  It will be up all day and all night.  But as dawn approaches and the twilight brightens, the sky glow will wash out the comet until it is no longer possible to view from any "daytime" part of the planet.
Cheating
Rather than trying to imagine where the comet is, there's a much easier way to just nail it.
Technology has come a LONG way over the years.  These days, you can get an app for your smartphone that will tie into the GPS, compass, and level/tilt/gyro sensors on your phone.  It's amazing how well this stuff works.  You hold the phone up to the sky, and it knows everything in the sky that should be visible from your current location and in that direction.  It will literally draw the sky and show you the precise location of the comet.


Why do pictures of the Milky Way look like a spiral?

The objects in a galaxy orbit it's center as a planet does with it's star, they keep the same distance. They look like a spiral for something that is called density waves that make some parts of the galaxy (the arms) more dense.

You have to imagine a star system like ours orbiting the galaxy in your picture following those radial lines, so sometime the system will be inside of and arm and sometimes outside of it.

Thanks to this extra density in the arms is where most of the star formation takes place.


Finding Quintuple Planetary Alignments with SkyField

This is not an answer.

I computed all major planet conjunctions in DE431 to answer How to calculate conjunctions of 2 planets and you might be able to port what I did (using CSPICE) to skyfield.

Several of the more interesting conjunctions I found are here:

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

including 5 and 6 planet conjunctions:

http://search.astro.barrycarter.info/STELLARIUM/stellarium-191.ann.png
http://search.astro.barrycarter.info/STELLARIUM/stellarium-092.ann.png

I note that the 6 planet conjunction is unique in DE431, but I refer to the 5 planet conjunction as "rare", so it's likely there's at least one other one.

I'm too lazy to look through my own results, but this might give you a start.


What would be the apparent magnitude of Betelgeuse if it were as close to Earth as Sirius?

Given a distance of $d_1 = 640~\mathrm{ly}$ for Betelgeuze and $d_2 = 8.6~\mathrm{ly}$ for Sirius the difference in the apparent magnitude for Betelgeuze at distances $d_1$ and $d_2$ that is given by

$$m_1 - m_2 = 5~\mathrm{mag} \cdot log_{10}(\frac{d_1}{d_2})$$

is $9.4~\mathrm{mag}$. With magnitude $m_1 = 0.45~\mathrm{mag}$ we get $m_2 = -8.9~\mathrm{mag}$ for Betelgeuze at the distance of Sirius.

This is about $860$ times brighter than Sirius ($-1.46~\mathrm{mag}$). It would be visible during daytime


Specific word for the point of the terrestrial surface "under" a star or other object in the sky

Positions are usually given relative to the observer, not relative to the object. So instead of saying "I am under the star X", you would say "star X is overhead". The word for an object directly overhead is zenith: "the star X is at the zenith".


Would a human body float in the dense atmosphere of Venus?

No. According to the Nasa Venus fact sheet the density of the atmosphere at the surface is ~65 kg/m^3.

For comparison, water is 1000 kg/m^3. We only just float in water.

So if you were there, in a suit that could somehow withstand the heat (464 C), the "air" would feel thick.

Maybe a person could strap on some wings and fly ...


Are there any known low-inclination trans-Neptunian objects?

Queries to the JPL Small Body Database or the ESA Asteroids Dynamic Site for such bodies return several results.
Some of the bigger ones are:
Centaurs with H < 10 and i < 4°

386968 (2012 BR61), q = 11.7 au, Q = 31.6 au
527328 (2007 TK422), q = 17.0 au, Q = 25.2 au

TNOs with H < 5 and i < 2°

119951 (2002 KX14), q = 37.3 au, Q = 40.9 au
42301 (2001 UR163), q = 36.8 au, Q = 66.4 au
48639 (1995 TL8), q = 39.9 au, Q = 65.0 au


Yes, there probably are more.

There is a minor problem of definition. Unlike the case for "planet", the IAU has not produced a definition of "moon". There is no requirement to be round, or have cleared an orbit. So any chunk of rock or ice that we can see orbiting a planet can reasonably be called a moon, and Jupiter has, over the years, picked up a lot of stuff: Small asteroids that it has captured from solar orbit.

It is likely that there are other, sub-1km sized objects in orbit around Jupiter, that we haven't noticed yet.

Juno is, as of 2016, in orbit around Jupiter. However her main focus of study is the planet itself, not hunting for moons. The recent moon discoveries have been made by terrestrial observations.


time on earth compared to completely stationary in space

Time dilation is a relative phenomenon, not a local phenomenon.

As adrianmcmenamin mentioned, Earth's motion around the sun is cause for no more time dilation than any other point in the universe. Consider that the sun is travelling around the galaxy at 200 km/s, and the Milky Way is approaching Andromeda at 130 km/s. We do not know at what speed we are approaching the great attractor, and I hope that at this point it becomes obvious that there is no difference in "us approaching the great attractor" to "the great attractor approaching us". Nothing is considered "stationary to the background" in an infinite, expanding universe as there is no "background" to measure against.

Consider the gedankenexperiment of comparing a clock on Earth to a clock orbiting the galaxy. Now compare those clocks to clocks outside the galaxy. You might consider the clock that ticks the most time away to be the "most at rest clock". Now I ask you, how will you compare the clocks? You must move at a significant portion of the speed of light for a very long time to get information from one clock to the next. Thus, the messenger would suffer the effects of Special Relativity, rendering the comparison meaningless.

To further the point, consider the development of International Atomic Time. I'll quote directly from wikipedia:

  In the 1970s, it became clear that the clocks participating in TAI were ticking at different rates due to gravitational time dilation

Consider, those were clocks on Earth suffering from measurable time dilation due only to their latitude and altitude differences and the effect of moving information across the distances between them!


How much mass will the Sun have when it becomes a white dwarf?

Short answer:

The Sun will lose about half of its mass on the way to becoming a white dwarf. Most of this mass loss will occur in the last few million years of its life, during the Asymptotic Giant Branch (AGB) phase. At the same time the orbital radius of the Earth around the Sun will grow by a factor of two (as will the outer planets). Unfortunately for the Earth, the radius of the Sun will also reach to about 2 au, so it will be toasted.

There is the possibility that the decreased binding energy and increased eccentricity of the Earth and the outer planets will lead to dynamical instabilities that could lead to planetary ejection. This is highly dependent on the exact time dependence of the late, heavy mass loss and the alignment or otherwise of the planets at the time.

Long answer:

Stars with mass less than about 8 solar masses will end their lives as white dwarfs on a timescale which increases as their main sequence initial mass decreases. The white dwarfs that are formed are of lower mass than their progenitor main sequence stars, because much of the initial mass of a star is lost through stellar winds (particularly during the thermally pulsating asymptotic giant branch phase) and final ejection of a planetary nebula.  Thus, the current distribution of white dwarf masses, that peaks between $0.6$ and $0.7 M_{\odot}$ and with a dispersion of $\sim 0.2 M_{\odot}$, reflects the final states of all main sequence stars with $0.9 <M/M_{\odot}<8 M_{\odot}$, that have had time to evolve and die during our Galaxy's lifetime.

The most reliable information we have about the relationship between the initial main sequence mass and final white dwarf mass (the initial-final mass relation or IFMR) comes from measuring the properties of white dwarfs in star clusters of known age. Spectroscopy leads to a mass estimate for the white dwarf. The initial mass is estimated by calculating a main sequence plus giant branch lifetime from the difference between the age of the star cluster and the cooling age of the white dwarf. Stellar models then tell us the relationship between main sequence plus giant lifetime and the initial main sequence mass, hence leading to an IFMR.

A recent compilation from Kalirai (2013) is shown below. This shows that a star like the Sun, born with an initial mass of $1M_{\odot}$ (or maybe a per cent or two more, since the Sun has already lost some mass), ends its life as a white dwarf with $M = 0.53 \pm 0.03\
M_{\odot}$. i.e. The Sun should lose approximately 50% of its initial mass in stellar winds and (possibly) planetary nebula ejection.

A comprehensive treatment of what happens to solar systems when the central star loses mass in a time-dependent way is given in Adams et al. (2013). The simplest cases are initially circular orbits where the mass loss takes place on much longer timescales than the orbital period. As mass loss proceeds, the gravitational potential energy increases (becomes less negative) and thus the total orbital energy increases and the orbit gets wider. Roughly speaking, $aM$ is a constant, where $a$ is the orbital radius, which is a simple consequence of conservation of angular momentum: so the Earth would end up in a 2 au orbit.

However, in the presence of a non-zero eccentricity in the initial orbit, or in the case of rapid mass loss, such as that which occurs towards the end of the AGB phase, then things become altogether more unpredictable, with the eccentricity also growing as mass loss proceeds. This has a knock-on effect when considering the dynamical stability of the whole (evolved) solar system and may result in planetary ejection. The faster the mass loss, the more unpredictable things get.

The radius of an AGB star can be calculated using $ L = 4\pi R^2 \sigma T_{eff}^{4}$. Stars at the tip of the AGB branch have luminosities of $\sim 10^{4} L_{\odot}$ and $T_{eff} \simeq 2500\ K$, leading to likely radii of $\sim 2$ au. So it is quite likely that unless the Earth is ejected or has its orbit significantly modified by some dynamical instability that, like the inner planets, it will end up engulfed in the outer enevelope of the AGB star and spiral inwards...

Even should it narrowly escape this immediate fate, it is then quite likely that tidal dissipation will rapidly extract energy out of the orbit and the Earth will spiral in towards the envelope of the giant Sun... with the same result.


Does the Sun orbit the Milky way in a (kind of) flat or inclined orbit or more of a sine wave?

It is correct that the sun bobs "up and down" (relative to the plane of the galaxy as it orbits the galaxy, and it takes about 64 million years  to complete a full cycle, so it does pass through the plane of the galaxy every 32 million years or so (there are quite large error bars on those numbers.)

This idea is uncontroversial, but it gets caught up in other theories that are less well established (for example the idea that there are dense clumps of dark matter in the galactic plane, that can have disrupt the Oort cloud, or otherwise cause mass extinction) or are simply nonsense (the idea that the solar system is a "vortex").

Phil Plait rubbishes this second idea on his blog, and includes some further discussion of the "bobbing" path of the sun. He notes that the disk is about 1000 ly thick, and we bob up and down by about 200 light years every 64 million years.

Lisa Randall's is a serious scientist, who would have no truck with "solar system vortices", and the ideas in her book are reasonable, but not uncontroversial. We have no evidence of dense clumps of dark matter, nor do we know exactly where the Earth was 65 million years ago. We last passed through the galactic plane about 3 million years ago, without serious consequences.


Difference of Gas Giant Planet and Ice Giant?

Gas Giants consist primarily of the gases hydrogen and helium, with a rocky/iron core.

The interiors of Ice Giants consist primarily of what astronomers call 'Ices'. These are compounds like water, methane and ammonia, and are so-named presumably because they are solid at cold (-200 C) temperatures, unlike hydrogen and helium, which remain gaseous up until very close to Absolute Zero.

Hydrogen and helium are very light, and it only takes a small amount of heat energy to make them fast enough to escape the gravitational field of most planets. Jupiter and Saturn are the only planets heavy enough to retain a significant fraction of hydrogen and helium (though the atmospheres of ice giants Uranus and Neptune consist mainly of hydrogen and helium, it's mainly ices further down).


I'm doing some palaeobiology-landscape rendering; so I'm thinking if I want to add the sun in such pictures; should I draw it just like present day ?

Although the Sun was less luminous 4.5 billion years ago (about 70% of it's current luminosity, it is thought), and while this would affect it's color, making it perhaps a little more orange than yellow, the effect is small.

The Sun is still going to be a blazing near-white object in the sky of roughly the same size.

Although there was probably a color shift to a slightly more orange tint, the composition of the Earth's atmosphere was radically different (see for example this page on Paleoclimatology) and this would have had a much more significant effect on the appearance of the Sun (and the lighting of the landscape) than any minor change in the Sun's appearance.

There are some issues relating to the development of life and this change in the Sun's luminosity (like the Faint Young Sun Problem) but these won't make much difference to your rendering, I think, compared to the atmospheric and climate issues.

However the climate changes are not fully understood and it's rather speculative to try and make specific claims about what the overall effect would have been on the illumination levels and color of light reaching the surface in these periods.


How do astronomers estimate the total mass of dust in clouds and galaxies?

Dust absorbs stellar light (primarily in the ultraviolet), and is heated up. Subsequently it cools by emitting infrared, "thermal" radiation. Assuming a dust composition and grain size distribution, the amount of emitted IR light per unit dust mass can be calculated as a function of temperature. Observing the object at several different IR wavelengths, a Planck curve can be fitted to the data points, yielding the dust temperature. The more UV light incident on the dust, the higher the temperature.

The result is somewhat sensitive to the assumptions, and thus the uncertainties are sometimes quite large. The more IR data points obtained, the better. If only one IR point is available, the temperature cannot be calculated. Then there's a degeneracy between incident UV light and the amount of dust, and the mass can only be estimated to within some orders of magnitude (I think).

If lines from various atomic or molecular transitions are seen as well, the composition can be better constrained. The size distribution can be determined from fitting the theoretical spectrum of a given distribution to observed dust spectra. This information is often not available in a given high-redshift galaxy, so here we can be forced to assume that the dust is similar in nature to "local" dust, i.e. in the Milky Way and our nearest neighbors.

If you're interested in the relevant equations, they can be found many places, e.g. here.

Another way to estimate the dust mass is to measure the metallicity of the gas with which the dust is is mixed, either from emission lines or absorption lines if a background source is available. The dust mass is then found from an assumed dust-to-metals ratio, which is pretty well-established in the local Universe, and to some extend also at higher redshifts.


Is it the paper David H pointed out, by Geringer-Sameth et al. (2015)? If so, then I should point out that they're not the first. Just using Wikipedia, I came across Weniger (2012) and Albert et al. (2008).

Geringer-Sameth et al.'s abstract reads (in part)

  We present a search for gamma-ray emission from the direction of the newly discovered dwarf galaxy Reticulum 2. . . . Reticulum 2 has the most significant gamma-ray signal of any known dwarf galaxy. If Reticulum 2 has a dark matter halo that is similar to those inferred for other nearby dwarfs, the signal is consistent with the s-wave relic abundance cross section for annihilation.

Albert et al. also studied a dwarf galaxy:

  The nearby dwarf spheroidal galaxy Draco, with its high mass to light ratio, is one of the most auspicious targets for indirect dark matter (DM) searches. Annihilation of hypothetical DM particles can result in high-energy γ-rays, e.g., from neutralino annihilation in the supersymmetric framework. A search for a possible DM signal originating from Draco was performed with the MAGIC telescope during 2007.

The point is, this has been done before. These folks aren't the first.

  Has someone actually detected gamma rays emitted by neutralino (or any other dark matter) annihilation?

More analysis is probably needed. I'll quote the paper itself:

  While Ret2's X-ray signal is tantalizing, it would be premature to conclude it has a dark matter origin. Among alternative explanations, perhaps the most
  mundane is the possibility that an extragalactic source lies in the same direction. Searching the BZCAT [54] and CRATES [55] catalogs reveals a CRATES quasar (J033553-543026) that is 0:46 from Ret2. Further work must be done to determine whether this particular source contributes to the emission, though we note that at spectrum radio quasars rarely have a spectral index less than 2 [56]. One of the much-discussed astrophysical explanations for the apparent Galactic Center excess is millisecond pulsars [24, 26, 57{61]. In the case of Ret2, it is the high-energy behavior which disfavors a pulsar model, as millisecond pulsars exhibit an exponential cut-offat around 2.5 to 4 GeV [26, 30, 61{64]. Alternatively, high-energy cosmic ray production could potentially arise in the vicinity of young massive stars. Upcoming photo-metric and spectroscopic analysis of Ret2 will check this possibility.

Don't jump to conclusions. They didn't. It stinks, but the answer to your question is, "Nobody's sure."


Expansion of the universe and visibility of distant galaxies

(My answer here isn't correct, as pointed out by the first comment on this post, please refer to it.)

Yes, it certainly is possible - and it already has happened. The age of the universe is 13.8 billion years, yet it is something like 90 to 100 billion light-years in diameter. This means that we definitely can't see all of it, not because of our own observational limitations, but because the light from those parts of the universe just hasn't reached us yet - and perhaps it never will.

As for the future, yes, if cosmological expansion overpowers gravity at close range (i.e. the Milky Way and Andromeda, and the world "close" is relative here) then eventually many galaxies and clusters will disappear out of view due to light not being able to "keep up" with the expansion.


How bright is the full Earth during the lunar midnight?

Let's take an average albedo for the Earth of 0.3 (it depends, which hemisphere is visible, how much cloud cover etc.). That means the Earth reflects 30% of the light incident upon it.

The flux $f$ falling on the Earth is given by
$$ f_{\odot} = \frac{L_{\odot}}{4\pi d^2} = 1.369\times10^{3}\ Wm^{-2}$$
where $L_{\odot}=3.85\times10^{26}\ W$ from the Sun and $d= 1$AU.

The integrated luminosity from the illuminated hemisphere will be
$$L_{earth} = 0.3\pi R^2 f = 5.2\times10^{16}\ W$$

So now we can compare this with the Sun. One hemisphere of the Sun radiates $1.93\times10^{26}\ W$, and produces a flux of $1.369\times10^{3}\ Wm^{-2}$ at 1 AU. Therefore the illuminated hemisphere of the Earth results in a flux of approximately $f_E=0.056\ Wm^{-2}$, assuming the average Earth-Moon distance of 384,400 km. This calculation assumes isotropic emission, but it is quite likely that the albedo is higher for light reflected through 180 degrees.

The Sun has an apparent magnitude of -26.74, so the magnitude of the "full-earth" at the moon is
$$ m_{Earth} = 2.5\log_{10}\left( \frac{f_{\odot}}{f_{E}}\right) - 26.74 = \underline{-15.77}$$

The answer will of course vary with the albedo of the visible hemisphere, which in turn depends on the time of year and how much of the polar regions can be seen (e.g. http://www.climatedata.info/Forcing/Forcing/albedo.html ). Variations of a few hundredths seem possible, which will lead to apparent magnitude variations in $m_{Earth}$ of $\sim \pm 0.1-0.2$ mag. The albedo may also vary in detail with the exact angle at which the sunlight hits the Earth - an "opposition surge" in brightness, when the Sun-Earth and Moon are almost aligned is possible. The Earth-Moon distance varies from 363,000 to 405,000 km. This will lead to magnitude variations of $\pm 0.12$ mag.

A further way to check this is that the albedo of the Moon is 0.12 and it has a radius of 0.273 times that of the Earth. Therefore the Earth seen from the Moon ought to be $(0.3/0.12)\times(1/0.273)^2 = 33.5$ times brighter. This is 3.81 magnitudes brighter. The mean magnitude of the full Moon is -12.74 (maximum is -12.92), so the brightness of the "full Earth" should be -16.55 on average.

I am not sure why these figures don't agree; I suspect it is that the albedo for reflection when the Sun's light is normally incident on the Moon is quite a bit larger than 0.12. The so-called "opposition surge". If the Earth's albedo behaves in the same way, then the latter figure may be more accurate than my first calculation. My gut instinct is that the answer is somewhere between the two.


Why do some solar eclipses' umbra cross arctic/antarctic regions?

I wanted to ask this question even though I realized the correct geometry in the middle of typing it. So I will answer this myself.

First of all, the sub-lunar point never exceeds 28.545 N/S. If you're standing at the sub-lunar point, the Moon will be directly overhead. The Moon never wanders outside this latitude bracket.

But that does not mean the Moon's shadow can't wander outside it. Here's one way it could happen.

Edit: this is my attempt to draw something approximately to scale. I think I got the Earth/Moon sizes about right, but the Moon-Earth distance should around 4x greater. And the solar rays are of course not parallel but I don't think it's possible to draw or perceive tiny angles for this scale.

The green line is the ecliptic. The Sun and Moon are supposed to be aligned along a horizontal line, but for some reason I drew the solar rays half and half beside the green line which is deceptive. I'll try to edit the pic when I can.

In technical parlance, the Moon could be slightly above or slightly below its Node. If the Moon was exactly at its node, then the umbra would absolutely be centered somewhere between 28.5 N/S (actually 23.4 N/S because the Sun never goes beyond that). But as you can see, the farther it gets from its node, the steeper the shadow impacts Earth and the farther towards a pole it gets. Too far from the node, and the shadow will not intercept Earth at all and we won't have an eclipse.

If you draw a line from the center of the Moon to the center of the Earth, you can see that the sub-lunar point is definitely within 28.5 degrees N/S. But the shadow does not necessarily follow that imaginary line. The shadow always follows the line from the Sun to the Moon.

You could even imagine a very extreme case where the shadow falls on "the other side" of the pole. I think that may be the case here since that eclipse looks really short.


Star density in- versus out- of the arms of the galactic pattern?

I'm afraid there is no answer, yet... The problem is that you would have to know the star densities in the different regions (in the spiral arms and outside of the spiral arms). While it is certainly possible to have an accurate estimate of the density of massive stars because of their brightness, the density of low mass stars in different regions of the galaxy is much more difficult to estimate. And since low mass stars make up the vast majority of stars (see for instance the initial mass function which is a measure of the distribution of stars over the different masses), estimating the density of stars in different regions becomes very uncertain.

But, at the moment the Gaia satellite is measuring the position and distances of about a billion stars and other objects. A much more detailed three dimensional map will be created with that data, including distance data for dim stars whose distances we were as yet unable to measure. So there is hope ;-)

NB. The first data release of Gaia has been published in September. I haven't found any papers that were able to answer your question (my search was not extensive though), but they might be forthcoming. Of course if you have some programming skills, you might determine the density differences from the data (available here) yourself.

EDIT - Just took a look at the data, the parallax (and therefore the distance) is still lacking for many stars. Hopefully this will change with future data releases.

  = I understand that one of the reasons the arms "appear real" is just that there are more bright young stars forming in the arms (due to the
  higher pressure). Perhaps the actual density difference is very low?

I would expect that the density difference of stars is quite small.

  = Perhaps the density variation is dependent on the type of star, or other material - what about things like gasses?

The density variation is certainly different between different types of stars. High mass stars only occur near star forming regions (i.e. in spiral arms). Their lifespan is so short that they do not have the time to get very far from the region in which they were formed.

I'd expect that the density variation for gas and dust will be higher (but I have no knowledge to support this).


What does "located in the Hubble flow" mean?

The apparent motions of external galaxies can be broken down into two components and the redshift observed will be due to the sum of both.
The first is an apparent recession velocity that is given by Hubble's law. $v_H = H_0 d$, where $v_H$ is the apparent recession velocity, $d$ is the proper distance and $H_0$ is the current value of the Hubble parameter.
On top of that, a galaxy will have a peculiar velocity $v_P$ with respect to the cosmological rest frame (usually taken to be coincident with the frame in which the cosmic microwave background has no dipole anisotropy). This preculiar velocity is due to the gravitational influence of nearby other galaxies or due to a galaxy's motion within a group, cluster or supercluster of galaxies.
Since the first component grows with distance, whereas the second component is fixed and of order hundreds to perhaps a thousand or so km/s, then if you look far enough away from the Earth then $v_H \gg v_P$. This is what it means to be "part of the Hubble flow" - the velocity that you measure for that object is dominated by the cosmological expansion of the universe and local peculiarities and gravitational perturbations that the galaxy is influenced by play a small enough role that they become a negligible source of uncertainty.
Since $H_0 \simeq 70$ km/s per Mpc, and $v_P \sim 1000$ km/s, then a quick bit of maths tells you that in order for $v_H \gg v_P$, then $d \gg 14$ Mpc for an object to be considered part of the Hubble flow. There is no exact definition.


You are correct to say that all the heavier elements in the universe were formed in stars.

Stars like the sun fuse hydrogen into helium. When they get older they can fuse the helium into carbon (it actually takes 3 helium to make one carbon). Larger stars can fuse carbon into oxygen, and neon and elements in the first half of the periodic table.

When the star runs out of fuel, the outer layers are gently ejected in what is called a planetary nebula (though it has directly to do with actual planets). The ejected gas is enriched with the heavier elements that the star has fused. The heavier elements in the planetary nebula mix with the hydrogen and helium gas in outer space and can later go on to be found in new stars. Most of the carbon and oxygen and nitrogen on earth was formed by this process.

Very large stars will fuse all the way up to iron, and then collapse in a supernova. This releases a very large amount of energy, some of which is used to form elements heavier than iron. All the heavier elements (copper gold, uranium for example) are formed in supernovae.

The supernovae explosion distributes these elements back into space, mixed in with the debris from the star, as time goes on, and stars are born and die, the gas in space gets enriched with more and more heavier elements

So the heavier elements are formed in the cores of stars, and are stored there until the star dies, when some of the star's atoms are released into space. The heavier elements are not "radiated out" of the star, until the star dies. Some stars die in a supernova explosian, but most have a more peaceful death in a planetary nebula.


Are binary stars with only one visible sibling common enough to contribute noticeable to dark matter?

The luminosity of normal stars is a strongly increasing function of mass. e.g. $L \propto M^3$. If another star is "hidden" in a binary system, then it is of lower mass. So the amount of hidden mass is less than what is seen.
Of course this can (and is) be accounted for when estimating the mass present in luminous matter because we know typical binary frequencies and properties. The luminous matter is only about 1% of that required to explain the properties of the universe as a whole and less than 10% of the mass required to explain the dynamics of ours and other galaxies, so even a factor of 2 is not going to solve the problem.
Further, from big bang nucleoynthesis calculations and estimates of the primordial abundances of helium and deuterium, we know the missing mass is not "baryonic". It is not normal matter made if protons and neutrons.
So hidden binary companions are not an important contributor to the dark matter problem.
You ask for a quantitative answer. Well, a typical binary frequency for low-mass stars is 30-40% and they have a flat mass ratio distribution (i.e. the binary companions have a uniform distribution of mass). Given that stars make up less than 1% of the critical density of the universe then even if one ignored the "binary problem" (which astronomers don't), then the additional mass in hidden binary system is less than 0.15-0.2% of the critical density.
This is to be compared with the ~30% that needs to be there to satisfy the dark matter problem (and as I mentioned, the vast majority of that cannot be in the form of normal matter like binary star companions in any case).
To be even more specific. Hidden binary companions make no contribution to dark matter at all, since they are luminous. What they change is the average mass/luminosity ratios of stars that must be used in order to estimate a mass from a luminosity. The values that are used by astronomers to estimate the amount of luminous matter attempt to take account of a binary population. Are the effects of binarity noticeable? Yes of course we have many means to show that stars have a less massive companion. Are they noticeable on large scales - no. The precision with which the dark matter content of an individual galaxy is estimated (e.g. from a rotation curve) will have uncertainties of at least 10% and the contribution of hidden mass in stellar binary systems will be tiny compared with that.


Is it coincident that all planets with tight orbit are tidally locked to their parent body?

This is a physical result:

  The change in rotation rate necessary to tidally lock a body B to a larger body A is caused by the torque applied by A's gravity on bulges it has induced on B by tidal forces.

Stellar bodies get tidally locked with time. You even get graphs like there.

On how long it generally takes, from Wikipedia

$$t_{\text{lock}} \approx \frac{\omega a^6 I Q}{3 G m_p^2 k_2 R^5} $$

Where $Q$ and $k_2$ depend on the constituents and age of the satellite, which themselves evolve with time. Difficult to estimate. Moreover, you rarely know the age of your system.

But it explains that for long-time established system of relatively close bodies, tidal lock is generally the case. The more tide there is, the quicker it gets.


Suns Orbital speed and escape velocity question

It's asking exactly what it says- where would you expect this material to end up? Giving a distance from the center of the galaxy may be relevant, or it may not, depending on exactly what speed you end up with, and how it compares to galactic orbital / escape velocity. If the velocity is well above galactic escape velocity, giving a specific distance from the galactic core doesn't make much sense- there will be no fixed distance, the stuff will just keep going, achieving greater and greater distances.

So, the first step is indeed to solve and check if average speed > orbital velocity or not (because the question says to do so), and a good second step would indeed be to check if it's less or greater than escape velocity. And from that information, you can guess generally where the stuff will go- stay reasonably close to the original orbit of the star? or stay in galactic orbit, but spread out in some wildly different range of orbits than what the original star had? or completely ejected from the galaxy?


Number of stars & planets in Milky Way Galaxy?

A consensus number is that there roughly $10^{11}$ stars in our Galaxy (though this number is certainly uncertain by a factor of at least two, because it is based on extrapolating what we know about stellar populations in our vicinity). Most of these stars are of lower mass and are much less luminous than the Sun.

The number of planets is even more uncertain. It now seems probable that most stars like the Sun have at least one planet, but we really don't know that much about the diversity of planetary systems, how typical something like the solar system is, or how planetary systems might change as a function of position in the Galaxy. We also don't know that much about planetary systems around the dominant (by number) very low-mass stars. They certainly can have planetary systems, but the fraction that do is still a work in progress.


Going by those coordinates, the dark area you're seeing in Google Sky is the Chamaeleon I cloud. It's a nearby star-forming region that is part of the Chamaeleon complex (see also the "Notable features" section of this article on the Chamaeleon constellation). The star nearest to the coordinates you provided is the T Tauri star TYC 9414-787-1.


$t$ signifies time; see the Wikipedia article for spacetime, and then the subsection for 4-vectors.

The basics are pretty natural to understand. Suppose something happens, an event, like an apple falling off of a tree.

In order to tell someone else about it you need the three space coordinates $x, y, z$ and the time coordinate $t$. Without all four, you won't be able to see it happen.

update: The question was clarified to focus on the meaning of "negative time" or $-t$. That's no big deal. There's no problem with $-x$ for example. We don't think of that as negative distance or negative space. It's just a coordinate relative to some origin $(0, 0, 0)$, and we can put that anywhere, arbitrarily.

So you should really thing of the time coordinate as $t-t_0$ where $t_0$ is some arbitrary origin in time. It might be midnight yesterday, or midnight next Tuesday. That would make $t_{Now}$ positive in the first case and negative in the second case, but either way it's still Now.


Could the Earth survive a red giant Sun?

The answer is yes, there are many authors who do take mass loss from stars into account when trying to work out the fate of their planetary systems. Examples include Schroder & Smith 2008; Adams et al. (2013); Adams & Bloch (2013).

It is thought that (e.g. Kalirai et al. 2008) that the Sun will lose half its mass by the time it ends up as a white dwarf. Almost all of this mass loss occurs in the final hundred million years or so (in just over 7.5 billion years time) whilst the Sun is on the red giant branch and asymptotic giant branch. The mass loss that occurs before then is small in comparison - about $10^{-13}$ of a solar mass every year (see Noerdlinger 2008), so losing less than 1% of its mass prior to the RGB.

Consideration of conservation of angular momentum alone leads to the idea that $M(t)R$ is constant, where $M(t)$ is the mass of the Sun as a function of time and $R$ is the semi-major axis of a planetary orbit. Thus prior to the RGB phase, the Earth's orbit expands by maybe 1.4 cm per year, or by a total of $7\times 10^{-4}$ au prior to the RGB.

However, during and after the RGB we see that the Earth ends up with an orbital radius of $\sim 1.5$ au (or in solar units, about 290 $R_{\odot}$) after the RGB and maybe 430 $R_{\odot}$ after the AGB. So, whilst the Sun is in the RGB/AGB phase and losing mass, the Earth moves outwards from about 215 to 430$ R_{\odot}$. Meanwhile, as it ascends the RGB, the Sun will likely increase in size to about $200 R_{\odot}$. Therefore it is a close-run thing as to whether the Earth will be engulfed within the outer envelope of the evolved Sun.
Major uncertainties include just exactly how big the Sun will get, how much mass it will lose on the RGB and then on the AGB, and it turns out that the time-profile of the mass loss (which will happen in pulses, rather than continuously) may also play a role.

The plot below (from Schroder & Smith 2008) may help to make some sense of this. It shows the (log) of the radius of the Sun versus time in the RGB and AGB phases as it loses mass. Also shown with a dashed line is the orbital radius of the Earth. Fractions written on the plot show what fraction of the Sun's present mass remains.

However, even if the Earth escapes immediate engulfment, the Sun would be several thousand times more luminous than it is today, so the temperatures at the surface of the Earth could reach $>1000$ Celsius even if it stays an astronomical unit away from the Sun's surface.

A further issue is the effects of tidal dissipation. The Earth will be dragged back by tidal bulges induced in the Sun's extended envelope. The rate at which angular momentum is lost is however exquisitely sensitive to the exact radius of the Sun and the radius of the Earth's orbit, to the extent that their orders of magnitude uncertainty in whether the Earth will be dragged into the Sun's envelope even before it reaches the tip of the RGB (this is what Schroder & Smith 2008 claim) or whether the loss rates will be too low to have much effect. It all depends on how much mass loss there is (Schroder & Smith use a slightly lower estimate than above) and therefore what radius the Earth reaches and what the final radius of the Sun becomes.

At the moment I would say there is sufficient uncertainty that the fate of the Earth is undecided, but it will unarguably be uninhabitable.


At what latitude could Orion's belt and Aldebaran be ~10 degrees above the horizon at the same time?

This sky configuration is consistent with 40 degrees N, at about 6.30am (local time) at the end of November.  If you went to mid November, this would look the same at about 7.30am (ie 1 hour per half month).

The latitude of New York is about 40N, which makes sense for a this particular US TV show,  which is recorded live in New York City, "just down the street" from where Dr. Tyson works as directory of the Hayden Planetarium.

The way I guessed was from a combination of things. From where I live I'm familiar with the eastern sky, so it was obvious this wasn't looking east. So I played around with different times of day and directions, which confirmed this was a view from looking west, and south of where I live (50N).

This can be quickly confirmed with any on-line "planetarium" or sky viewer. For example here's a shot from in-the-sky.org in planetarium mode, set to New York City, 6:15 AM November 20. It puts Aldebaran and Orion's belt both at about 10 degrees above the Western horizon, with only the beginning of twilight. Obviously other combinations of nearby dates and times can work as well.

The point is that the light in the image in the question is not the sunset or sunrise, so it must be a distant city in the West.


How to collect photometric data for WW Aur

I did a bibliography search on the object via Simbad and there isn't a lot of data available since the Southworth et al. 2005 article you linked to. Looking at the paper it seems the only other source of photometric data that is mentioned is Kiyokawa and Kitamura, 1975 in Ann. Tokyo Astron. Obs., second series which seems to be a in-house publication of Tokyo Observatory and doesn't seem to be available online. The only recent references are a few times of eclipse, which would help any period change modelling, but not the general modelling of the system.

This situation is not uncommon; there are a lot of binary stars out there and at a period of 2.5 days, this one is quite "expensive" in terms of observing time. This is even more the case if you want to get lightcurves in several filters, to aid the modelling by constraining temperatures etc.


No, the blue and red shift of stars is not possible to detect with the naked eye. There are a couple of reasons for this. First the effect is slight: even for an object moving at thousands of km/s (way faster than stars in the milky way move), it is only a tiny change in hue. Next, since the light of stars is a mixture of many different colours (a black body spectrum) if the light from a star is red shifted, then ultra-violet is shifted down and takes the place of blue, the blue moves towards green and so on. The end result is a mix of all the colours, and so very little change in the perceived colour.

The colour of stars has relatively little to do with their composition, and a lot to do with their temperature. Cool stars (either small, or large, old and with puffed up cool outer layers) look red or orange. Hot stars (large and powerful stars) are white or blue-white. Stars have a spectrum that approximates a blackbody, and so the actual composition has relatively little effect on the colour.

The composition of the atmosphere of a star will absorb particular wavelengths of light, and this results in a dark line in the spectrum (not enough is absorbed to change the apparent colour). It is by measuring the position of these lines in the spectrum that red- and blueshift is observed.


No. If you only know the distance to your object, that tells you absolutely nothing about where it is. That's like saying, I know the grocery store is 5 miles from my house, how can use this information to figure out exactly where it is? The answer is that you just can't, you need more information. – zephyr

In 3D space, you need three coordinates to locate an object. Ra, Dec are two (position on a sphere), z (or some other radial distance measure for nearby objects) is the third. Astronomy usually tries to make do with just Ra, Dec because that's easy to measure. – Alex

(comments converted to answer under cc licence)


What would the Earth's peri- and aphelion have to be in order to have the same seasons due to its orbit?

TL;DR: about 5 times current eccentricity for a deviation by 5 degrees from the mean... but the devil is in the celestial mechanics and climate model details.
It would not quite work, because the length of seasons would be uneven. Currently Earth's north and south hemispheres get nearly exactly as much or little sunlight as the other. But on the eccentric Earth the hot "summer" period would be short while the cool "winter" period would be long since planets spend more time in the outer part of an elliptic orbit.
There is another complication: absorbed sunlight heats up air, water and soil over time causing a lag: the warmest part of summer and the coldest part of winter comes after the time of the highest solar influx.
To add to the modelling woes, the actual temperature of the planet is strongly affected by the greenhouse effect: while CO2 is not going to do anything weird, as the entire planet cools water vapour (a potent greenhouse gas) will decline as rain and snow, and the albedo (how much light is reflected by clouds and ice) will increase.
That said, here is a simplistic model ignoring the atmosphere inertia:
The distance to the star varies as $$r(\theta)=\frac{a(1-e^2)}{1+e\cos \theta}$$ where $a$ is the semimajor axis, $e$ the eccentricity and $\theta$ the "true anomaly" (annual winner for worst name of parameter since 1609). To convert it to and from time you need to solve Kepler's equation numerically.
The temperature in a zero-dimensional atmospheric model is $$T=\left( \frac{ I_0(1-a)}{4 \sigma \epsilon}\right )^{1/4}$$ where $I_0$ is the solar constant at the top of the atmosphere, $a=0.3$ the albedo, $\epsilon=0.78$ the effective emissivity in IR. If we let $I_0(\theta)=I_0/r(\theta)^2$ (measuring orbits in AU) we get a temperature dependency scaling like $T(\theta)=T_0/\sqrt{r(\theta)}$.
So, at perihelion the temperature is $$T(0)=T_0\sqrt{\frac{1+e}{1-e^2}}$$ and the aphelion temperature $$T(\pi)=T_0\sqrt{\frac{1-e}{1-e^2}}.$$ So given a desired $T_{high}$ and $T_{low}$ one can now solve for $e$ to get $$e=\frac{(T_{high}/T_{low})^2-1}{(T_{high}/T_{low})^2}.$$
If we want a 5 degree difference from a mean $T_0=288$ K this means $e\approx 0.067$, about 5 times the current eccentricity. This is probably small enough that the time asymmetry issues will be small enough to ignore (they matter more for more eccentric orbits).


Why is there a zig-zag in elemental abundances?

Nuclei are more stable if they have an even number of protons Z, and are also more stable if they have an even number of neutrons N. This is because the particles form pairs. (Almost all nuclei with both N and Z odd are unstable with respect to beta decay.)
If a nucleus is going to have significant cosmic abundance, it must be either absolutely stable or at least have an extremely long half-life. This is more likely to happen when Z is even (and also when N is even).
Stability also effects probabilities of different types of decay. E.g., if a nucleus can undergo both beta+ and beta- decay, then it is usually more likely to do the decay that results in the more stable daughter.


What was the "brilliant new star in Aquila" on June 8, 1918, just after the solar eclipse?

It was Nova Aquilae 1918.  Wikipedia has some detail. https://en.wikipedia.org/wiki/V603_Aquilae
Apparently it was discovered by Zygmunt Laskowski, an amateur astronomer.


Right ascension and sidereal time difference

Question 1: How do sidereal time and RA correlate together?

Think of the sky as a globe. The constellations and lines of right ascension (RA) are painted on the globe. Because of the Earth's rotation, it appears that the sky globe rotates east to west once every 23 hours 56 minutes 4 seconds.

The meridian is the line of RA that goes from due south to the zenith (the point overhead) to due north. The meridian splits the sky into an eastern half and a western half.
The sidereal time is just the RA that is on the meridian. Because of the rotation of the sky, the sidereal time increases by approximately 15 degrees (or 1 hour of RA) every hour of time.

Question 2: Why do the stars rise earlier each month (and each day!)?

The Sun (and Moon and planets) are not painted on the globe of the sky. They move on the globe at different speeds (depending on the distance from the Earth and Sun). The Sun moves west to east and goes through 360 degrees along the ecliptic in 1 year, or approximately 30 degrees per month.
Because we use the Sun to tell time, the Sun is on the meridian at local noon.
But one month after March 21, the Sun is at 2 hours RA (= 30 degrees east of 0 RA). Therefore, the star that is at 0 RA and was on the meridian at noon on March 21 is now 30 degrees west of the meridian at noon on April 21.

In other words, because the Sun moves west to east, at the "same time of day" the stars appear to move westward by 4 minutes per day, 32 minutes per week, approximately 120 minutes per month (2 hours per month). That's why Sirius appears 2 hours earlier each month.

Technically, the stars appear earlier because it only takes 23 hours 56 minutes for the Earth to rotate 360 degrees measured relative to the stars, but it takes 24 hours (on average) for the earth to rotate 360 degrees relative to the Sun. The 4 minute difference is due to the motion of the Earth around the Sun, or the apparent motion of the Sun on the globe of the sky.


How to calculate the altitude of the Celestial Equator from an observer's latitude? Exam question

Perhaps you have heard that the altitude of the celestial pole (3 in the figure below) is equal to your latitude. In other words, angle 5 is the same as your latitude. Of course, the angle between the celestial pole (3) and the celestial equator (4) is 90 degrees (angle 6). From this, you should be able to calculate the angle of the equator (angle 7).

Item 1 is the horizon, and 2 is the celestial sphere.


Could diatom blooms affect albedo of an exoplanet?

First off, I am rather certain that by diatom bloom you refer to algal bloom which is a seasonal change of sea color, at least here on Earth, the only system we have observed yet. Your question is indeed rather hypothetical, but from a astronomy/ physics points indeed an interesting one.
Next, let me summarize the assumptions I am reading out of your question:

You assume an exoplanet that is mainly covered by water. Such a waterworld is not only existing in movies, but also a common toy system in climate science (of Earth).

The exoplanet is not only habitable, but also has some kind of algae living in its ocean. A strong assumption indeed, but for our calculation, it gives us numbers for how the color and thus the emissivity of the planet changes.

Indirectly, you also assume our hypothetical waterworld has an atmosphere. Why? An waterworld exoplanet would otherwise have already quickly evaoprated evaporated into space. In terms of greenhouse calculation later on, we will assume that the atmosphere consists only of water vapor. This means that our hypothetical exo-algae on waterworld would not necessary be aerobic. After all, we are after the physics here.

Could algae bloom affect the albedo of an exoplanet? Definitivley yes, but the question is, how much. And if we could observe that from Earth would be a whole different question.
What do the assumptions mean in numbers?
The albedo of open ocean water (on Earth) is $\alpha_O = 0.06$. Concerning the algal albedo $\alpha_A$, I found e.g.

a poster "Algal blooms: how are they harming models used for climate management?" by Ian Jenkinson
an article John Raven: The possible roles of algae in restricting the increase in atmospheric CO2 and global temperature
another article by H Gordon, M Jacobs: Albedo of the ocean-atmosphere system: influence of sea foam.

In short: Algae do not only change the color of the ocean (which color depends on the involved species), but also may produce foam (with $\alpha_A \approx 1$). The latter seems to have the largger effect since $\alpha_A \gg \alpha_O$. One of the above mentioned papers suggests a 10% coverage of the ocean with foam, an assumption which sounds appealing to me.
Some calculations
For a planet with radius $r$ in thermal equilibrium, the incident radiation equals the black body radiation:
$$ \pi r^2 E (1- \alpha) = 4 \pi r^2 \sigma T_{\rm eqi}^4 \Longleftrightarrow
T_{\rm eqi} =\sqrt[4]{\frac{E (1-\alpha)}{4 \sigma}}$$
In this formula, $E$ is the solar constant ($E_{\rm Earth} = 1368 {\rm W}/{\rm m^2}$), $\alpha$ is the (overall averaged) albedo ($\alpha_{\rm Earth} = 0.3$) and $\sigma = 5.67 \cdot 10^{-8} {\rm W\cdot m^{-2}K^{-4}}$ the Stefan-Boltzmann constant. For Earth this leads to $T_{\rm Earth,eqi} = -18^\circ {\rm C}$ which shows that we have to include the Greenhouse effect, i.e. that the atmosphere absorbs part of the outgoing radiation. We do that by introducing a  factor $\beta$:
$$T =\sqrt[4]{\frac{E (1-\alpha)}{4 \sigma \beta}}$$
On Earth, all Greenhouse gases (${\rm H_2O, CO_2, CH_4}$) contribute to an average $\beta \approx 0.62$ which leads to a more realistic global average temperature of $T = 15^\circ {\rm C}$ for Earth.
For a water world (without algae) we may use $\alpha = 0.1$ which leads to $T_{\rm WW} = 32^\circ {\rm C}$ which compares to a snowball planet with $\alpha = 0.8$ and $T_{\rm SB} = -63^\circ {\rm C}$.


What is the specific term for time of the night when there is no moon?

You are right that the moon only rises as the sun sets when it is full. At other times there is some time each night when both the moon and the sun are both set.

The terms moonrise and moonset are well established, by analogy with sunrise and sunset. The time when the moon is at its highest point in the sky is known as lunar culmination. You can use "lunar day" for the 24.84 hour period between lunar culminations. I know of no specific term for the period when the moon is set, or for when it is risen, instead you can say "after moonrise" or "before moonset", as appropriate.


Time on Mars is measured in sols.

Venus could be a little trickier, depending on your application.

I'm not aware of any timekeeping system for any other planets. There hasn't been a need for one yet. So, again depending on your application, you could come up with various different possible schemes.

Speculation (based on the OP's input regarding application):

Mercury - because of the length of the day, I doubt would-be colonists would measure time in days. They would probably use Earth time (because of communications) and also keep track of Mercury years.
Jupiter, Saturn, Uranus, Neptune - no colonization, no timekeeping


Studies of isotropy from different ends of the visible universe?

The most distant objects which can be seen with telescopes are around 13.39 ly away, with a redshift $z=11$. Telescopes see a roughly uniform Universe, although some very large structures were already identified.

The Cosmical Microwave Background has a $z \approx 1000$.

It was measured for its possible minor anisotropies by the COBE, and the Wilkinson Microwave Anisotropy Probe (WMAP) and Planck satellites.

It measured the temperature of the 2.7K Cosmical Microwave Background with $\approx 10^{-5} K$ precision. The technology was not easy.

There were two results:

First, there is a global dipole anisotropy showing that we are moving with $\approx 368 \frac{km}{s}$ to the Hercules constellation. It is the nearest thing what could be seen, as if we had found an absolute reference frame.

Second, substracting this from the results, another, lesser anisotropy were found:

The distribution of the anisotropy, combined with the result of other experiments, helped to get

the most precise measurement of the age of the Universe (13.7 Gy).
the most precise measurement of the curvature of the Universe (0, below measurement precision).
that the expansion of the Universe is accelerating.

Also the polarity was measured, which has shown that the re-inozation happened earlier as it was previously believed.

This is the best result what we have until now.


If you project the orbits onto a plane, for example the plane of the ecliptic, the projections will cross. But that's only because you're looking at a 3D problem in 2D. If you look at the orbits in 3D, you'll see that Pluto's orbit is highly inclined (17º) from the ecliptic, so it never actually passes through Neptune's orbit. Each time it seems to cross (in the 2D view), it's actually well above or well below Neptune's orbit. Pluto and Neptune never come closer than about 17AU.

At least for now. Minor perturbations over tens of millions of years may change that. Or might not. Pluto and Neptune currently have a 3:2 resonance, and such resonances tend to be stable.

You might want to look at http://en.wikipedia.org/wiki/Pluto, especially the section "Relationship with Neptune".


Do today's professional astronomers use earth rotation angle or sidereal time?

Most professional astronomers tend to use whatever package they are used to or got from somewhere else and don't look "under the hood". Very few astronomers I've met tend to write the own code that needs to deal with the multiple time and co-ordinate systems needed.

Most tend to use AstroPy these days which hides a lot of details but uses the SOFA/ERFA routines underneath which do indeed use the modern IAU2006/2000 framework and Earth Rotation Angle. However most textbooks, webpages etc still use the old methods to compute an apparent place and then sidereal time to rotate and relate that to the terrestrial system. So a lot of people still use the old methods because that's what they learnt initially or read recently. Until there is an update of a readable book like Jean Meeus's Astronomical Algorithms to the new system, I suspect sidereal time etc will persist for a good while longer.

The two methods give the same answers for positions to more precision than most astronomers can measure for most objects; exceptions would be pulsars, quasars, spacecraft (all three mostly done in the radio and/or with VLBI techniques) and GAIA-related measurements.


Does the Milky Way's movement through space have spacetime implications?

Your question has a misunderstanding at its root:  You say, "So for simplicity's sake, let's say the galaxy is actually moving through space at 1,000km/s". The problem here is that the whole point of Relativity is that the idea of movement with respect to space is meaningless -- the only thing that has meaning is movement with respect to other bodies.

Time dilation is also relative. If you're moving at 1000kps relative to Fred, Fred sees your clocks as running slower, and in his "reference frame" they are.  But you, in your reference fame, see Fred's clocks as running slower! (This is the essence of the famous Twin Paradox.)

You're being led astray by your intuitions about space and time. (And don't be ashamed about that. Newton, who was probably the greatest physicist ever, even greater than Einstein, and Kant (one of the great philosophers) and many, many others had the same intuitions. Yet experiment has taught us that they were wrong.)  Neither space nor time is absolute, and an understanding of Special Relativity (which is much easier to understand than it was to discover in the first place!) is required to ask as well as to answer your question.


Iridium satellites are known for predictable flare events but are gradually being de-orbited and replaced.
In particular, Iridium 66 would have flared at 6:04 several degrees east of Orion,
but its orbit has almost certainly changed since its de-orbiting process started on August 2.
Other Iridium satellites have taken 3 to 5 weeks to finish this process.

Other satellites can flare too if a shiny surface catches the Sun just right.
Heavens-Above shows two satellites passing south of Orion at 6:12 on the morning of August 19:
USA 215, an American spy satellite;
and Cosmos 2084, a failed Russian early warning satellite.

Another possibility is a head-on sporadic meteor.

Update: Iridium 66 finished de-orbiting on August 23.