Based on the math in the Wikipedia article about the Airy Disk:

The intensity of the starlight as a function of angular distance $\theta$ from the center is reasonably approximated by a Gaussian $$ I(\theta) = I_0 \exp(-\theta^2/(2\sigma^2)) $$ where $\sigma$ is a measure for the width of the intensity profile, which for an ideal optical system and diffraction-limited images is $$ \sigma \approx 0.42 \lambda N $$ (where $\lambda$ is the wavelength and $N$ is the f-number) and which is independent of the peak intensity $I_0$.  For a non-ideal or not-diffraction-limited system, the Gaussian should still provide a reasonable approximation, but then $\sigma$ is (much) larger than for the ideal system.

The apparent edge of a star in an image occurs where the intensity $I(\theta)$ drops below a certain minimum $I_\text{limit}$, which (from the first equation) corresponds to an angular distance $$ \theta_\text{edge} = \sigma\sqrt{2\log(I_0/I_\text{limit})}$$

We can rewrite this in terms of visual magnitude $V$.  For fixed $\sigma$, $I_0$ is a measure for the brightness of the star.  Brightness is related to magnitude $V$ through $$ V = -2.5 \log_{10}(I/I_\text{ref}) $$ (with $I_\text{ref}$ the reference intensity for the magnitude scale) so $$\log(I_0/I_\text{limit}) = 0.4 \log(10) (V_\text{limit} - V_0)$$ and
\begin{eqnarray}
   \theta_\text{edge} & = & \sigma \sqrt{0.8 \log(10) (V_\text{limit} - V_0)} \approx 1.36 \sigma \sqrt{V_\text{limit} - V_0}
\\ V_\text{limit} - V_0 & = & \frac{\theta_\text{edge}^2}{0.8 \log(10) \sigma^2} \approx 0.543 \left( \frac{\theta_\text{edge}}{\sigma} \right)^2
\end{eqnarray}

So, the difference between the limiting magnitude and the star's magnitude is proportional to the apparent angular area of the star in the image.


Could the planets in the Solar System have been captured?

I'll address the second question first, Venus uniquely has retrograde rotation; how would this have come about based on current theories? The answer is that there are a number of hypotheses related to Venus's rotation. See the related question What is the current accepted theory as to why Venus has a slow retrograde rotation?. None of them involve capture of a rogue planet. Note that this also applies to Uranus; it too has an odd orientation.

So why not? Note that I am now addressing your first question, is it possible that the planets could have been captured by the Sun's gravitational force after drifting through space while retaining their axis of spin and speed?

The answer is simple. Capture is extremely unlikely (and that's putting it mildly), and it is not needed to explain the odd orientations of Venus and Uranus.


Is the newly discovered exoplanet 'Teegarden b' habitable?

TLDR version: at this point there are too many unknowns to determine whether this planet candidate is habitable.

The discovery paper for this system, Zechmeister et al. (2019) "The CARMENES search for exoplanets around M dwarfs. Two temperate Earth-mass planet candidates around Teegarden's Star" is available on the arXiv. Table 1 lists the properties of the star and table 2 lists the properties of the planets.

The planetary quantities we have reliable estimates for are the minimum masses, the orbital periods and the distances. The eccentricities are consistent with zero so I'm going to assume for the moment that the orbits are circular. The reason that minimum mass is relevant is that this is an example of a radial velocity detection, which results in a degeneracy between the inclination of the planet's orbit to the line-of-sight and the mass of the planet.

Habitable zone

According to figure 13 in the discovery paper, Teegarden b is too close to the star to be in the conservative estimate of the habitable zone, though it is in the optimistic habitable zone. The Earth Similarity Index (ESI) appears to be a fairly misleading metric (I am of the opinion that ESI is a bad metric that should be abandoned), from the discovery paper:

  In addition, planet b receives nearly the same stellar flux
  as the Earth and therefore has a nearly identical equilibrium temperature, but is outside the conservative HZ. As discussed by
  Kopparapu et al. (2014), this is caused by a runaway greenhouse
  effect that is due to water vapour starting at lower insolation for
  low-mass stars. This effect is neglected in the ESI calculation
  and leads to the curious fact that planet b has a higher ESI value
  than planets inside the conservative HZ.

The position of Teegarden b therefore suggests it may have undergone a runaway greenhouse effect, though this might be avoided if it can build up enough reflective cloud cover on the daylight hemisphere, or if it is a desert planet without enough water to destabilise the climate.

Even if a planet is located in the habitable zone, it is not necessarily habitable. An obvious counterexample would be the Moon.

Terrestrial or not?

The next question: are these planets terrestrial? Unfortunately we don't know: Teegarden b does not transit the star, and there has so far been insufficient coverage of Teegarden c to say whether it transits. As a result, we do not know the radii of the planets so we cannot tell whether or not the planets have densities compatible with being terrestrial. The minimum masses are low enough that these planets do likely fall below the transition region between rocky and non-rocky planets suggested by Kepler data, see for example Rogers (2015) "Most 1.6 Earth-Radius Planets are not Rocky" - using the mass-radius relationships of Zeng et al. (2016) "Mass-Radius Relation for Rocky Planets based on PREM" the transition is somewhere around 3–5 Earth masses.

Another point of comparison here is the other known system of low-mass planets around an ultracool dwarf star: TRAPPIST-1. In this case, despite the small size of the planets, the inferred masses indicate the presence of substantial amounts of volatiles, e.g. Dorn et al. (2018) "Interior characterization in multiplanetary systems: TRAPPIST-1" where the volatile contents of the planets are estimated to correspond to a few percent by mass, compared with 0.02% by mass for the Earth. A planet like this would be very different to the Earth, in an optimistic scenario it might result in an ocean planet. The habitable zone for ocean planets would be different to that for Earthlike planets, since the negative feedback of continental weathering would not occur.

Red dwarf stars

Red dwarf stars have a variety of problems when it comes to habitability. One issue is that they have an extended high-luminosity pre-main sequence phase which might result in runaway greenhouse effects and/or severe water-loss on close-in planets, see Ramirez & Kaltenegger (2014) "The Habitable Zones of Pre-Main-Sequence Stars".

An additional problem is that red dwarf stars have severe stellar activity that would erode an Earth-like atmosphere in a very short amount of time, as an example Garcia-Sage et al. (2017) "On the Magnetic Protection of the Atmosphere of Proxima Centauri b" estimate an Earthlike atmosphere at Proxima Centauri would be lost in less than 400 million years. Teegarden's Star is fairly quiet for its spectral type, but it is an old star and was likely far more active in its youth.

These issues are not necessarily fatal, outgassing from volcanism might be able to replenish the atmosphere to some extent after the stellar activity dies down, or if the planet was born with a large amount of volatiles it might be able to emerge as a habitable planet. Unfortunately we don't have sufficient data on habitable zone planets around red dwarf stars to know for sure.

A further issue is that the atmospheric chemistry on planets around stars like Teegarden's Star would likely result in substantial quantities of carbon monoxide accumulating in the atmosphere, see Schwieterman et al. (2019) "A Limited Habitable Zone for Complex Life". This would be lethally unpleasant for any human visitors (or other organisms that use haemoglobin as their oxygen carrier) caught outside without an air supply. Note that haemoglobin is not the only possible oxygen carrier: some groups of life on Earth use alternatives (haemerythrin, haemocyanin) that are less vulnerable to carbon monoxide poisoning, as noted by Howell (2019) "Comment on "A Limited Habitable Zone for Complex Life" (Schwartzman et al. 2019 ApJ 878, 19)":

  In short, there is a notable diversity of extant invertebrates on this planet that would be quite comfortable outside of the range of CO concentrations proposed by Schwieterman et al. to constrain the habitable zone. There is no a priori reason to rule out the possibility that these lineages could have given rise to intelligent life during the history of life on this planet. A significantly higher atmospheric CO concentration would have prevented the evolution of hemoglobin, but not necessarily prevented astronomically observable biosignatures from Earth.

Summary

It is too early to be able to say whether the planets around Teegarden's Star are habitable. There are plenty of reasons to be sceptical but worlds like Teegarden b are still promising enough to be worth investigating in more detail. It is unfortunate that it is not a transiting planet, but similar systems like TRAPPIST-1 should be helpful in constraining what the possibilities are.


Collision of asteroid and gaseous planet - what would happen

A very well documented example of a small celestial object with a gas giant is the collision between Comet Shoemaker-Levy 9 with Jupiter.  As it was a comet that was tracked, the whole process of the collision was well documented and observed.

Though the collision occurred in July, 1994 - the first effects occurred 2 years earlier, according to the NASA Technical Report "Shoemaker-Levy 9 and the tidal disruption of comets", when the

  break-up of Periodic Comet Shoemaker-Levy 9 into multiple pieces following its grazing encounter with Jupiter

This represents one of the first stages of the collision that could occur as the celestial object approached the large planet.  In this case, this breakup resulted in Comet Shoemaker-Levy 9 to form a train of fragments to eventually collide into Jupiter - the full process at this stage is shown in the diagram below:

Image Source: Tufts University

The train of fragments was captured by the Hubble Space Telescope (shown below):

Image Source: NASA

Each fragment was designated a letter (e.g. Fragment A etc).  According to the article "The collision of Jupiter and Comet Shoemaker-Levy 9." (Zahnle and Mac Low, 1994), these fragments were in the order of about 100 m to 5 km in size, heading straight towards Jupiter.

Then to the main event - the collision of the comet fragments with Jupiter.

An example was from Fragment G, one of the larger fragments, when it collided with Jupiter (image below):

Image source: NASA

As you can see, the explosion was spectacular - about 21 fragments impacted into Jupiter over a number of days, given Jupiter's relatively fast rotation, the impacts became quite distributed on the top of the clouds of Jupiter, as shown in the ultraviolet image below (the circle shadow below is the moon Io):

Image Source Air and Space Museum

There have been several impacts into Jupiter (and presumably into the other gas/ice giants as well), another one that underwent scientific scrutiny was an asteroid 'the size of the Titanic' colliding with Jupiter in 2009, reported in the article "New evidence that asteroid, not comet, struck Jupiter in 2009", describes a comparison between the 2009 asteroid and Comet Shoemaker-Levy 9 are described as:

  The dark debris, the heated atmosphere and upwelling of ammonia were similar for this impact and Shoemaker-Levy, but the debris plume in this case didn't reach such high altitudes, didn't heat the high stratosphere, and contained signatures for hydrocarbons, silicates and silicas that weren't seen before. The presence of hydrocarbons, and the absence of carbon monoxide, provide strong evidence for a water-depleted impactor in 2009.

Overall, the article mentioned the likely process during any collisions of a celestial body with a gaseous planet, using the example from Jupiter:

  Plunging through Jupiter's atmosphere, the object created a channel of super-heated atmospheric gases and debris. An explosion deep below the clouds – probably releasing at least around 200 trillion trillion ergs of energy, or more than 5 gigatons of TNT

and the celestial object itself - vaporised, totally destroyed in the resulting explosion.


Yes, if the planets are tide locked to one another, their orbital period and sidereal day are the same. That's assuming their orbital plane about each other isn't much inclined to their orbital plane about the star.

From the Wikipedia article on orbital period:

$T_d=2\pi*\sqrt{a_p^3/(G*(M_1+M_2))}$

Where G is gravitational constant, $M_1$ is mass of 1st body and $M_2$ mass of 2nd body. $a_p$ is semi major axis of elliptical orbit of the planets about each other.

For the planet's year you'd use a very similar equation.

$T_y=2\pi*\sqrt{a_s^3/(G*(M_0+M_1+M_2))}$

Where $a_s$ is the radius of their orbit about the sun and $M_0$ is the mass of the sun. If you like, you can leave off $M_1$ and $M_2$ as their masses are probably neglible in comparison to the sun's mass.

I am guessing by "conventional day" you want the time an inhabitant sitting on his front porch would measure between one high noon and the next.

Solar day = sidereal day * (1 + (sidereal day/year))

For example our solar day is about 1/365 longer than a sidereal day.


Why is it impossible to estimate the inclination of an exoplanet with the Radial Velocity Method?

Essentially the problem comes down to geometry: radial velocity provides a 1-dimensional view of a 3-dimensional system. See the diagram below:

The observer sees only the component of the velocity vectors along the line from the observer to the star. That means that they cannot distinguish between the two blue velocity vectors: they will see only the purple (radial) component. This means a larger velocity vector that is tilted further away from the line-of-sight can masquerade as a smaller vector in the line of sight.

No matter how you define your variables, you will always run into this issue because it is inherent to the geometry of the situation.

This does break down in the case of planet-planet interactions, which can break the degeneracy here (a good example of this is Gliese 876).


Is it possible to move a planet out of its orbit? At least a lighter planet?

Yes. Even with todays technology we can get started.
In this article by Korycansky et al 2001 it is suggested to use rockets to move an asteroid (like a larger version of the NASA ARM Asteroid Redirect Mission) to a cycling orbit between Earth and Jupiter. The asteroid would pass near infront of the Earth in order to give the Earth a slight gravity assist which moves it outwards. The asteroid then passes near behind Jupiter to itself get a gravity assist. This transfers energy from Jupiter's to Earth's orbit. I've also heard of a proposal to use Venus instead.


What is the detection threshold of gravitational waves for LIGO?

I'm afraid this is not straightforward

The amplitude of the gravitational wave strain signal from a merging compact binary (neutrons star or black hole) is
$$h \sim 10^{-22} \left(\frac{M}{2.8M_{\odot}}\right)^{5/3}\left(\frac{0.01{\rm s}}{P}\right)^{2/3}\left(\frac{100 {\rm Mpc}}{d}\right),$$
where $M$ is the total mass of the system in solar masses, $P$ is the instantaneous orbital period in seconds and $d$ is the distance in 100s of Mpc. $h \sim 10^{-22}$ is a reasonable number for the sensitivity of LIGO to gravitational wave strain where it is most sensitive (at frequencies of 30-300 Hz).

So you can see that to increase the observability you can increase the mass, decrease the period or decrease the distance.

But here are the complications. LIGO is only sensitive between about 30-300 Hz and the GW frequencies are twice the orbital frequency. Thus you cannot shorten the period to something very small because it would fall outside the LIGO frequency range and you also cannot increase the mass to something too much bigger than the black holes that have been already seen because they merge before they can attain high enough orbital frequencies to be seen. (The frequency at merger is $\propto M^{-1}$).

A further complication is that the evolution of the signals is more rapid at lower masses. That is - the rate of change of frequency and amplitude increase rapidly with total mass. That is why the recent neutron star merger was detectable for 100s by LIGO, whereas the more massive black hole mergers could only be seen for about 1 second. But what this means is that you have fewer cycles of the black hole signal that can be "added up" to improve the signal to noise, which means that higher mass sources are less detectable than a simple application of the formula I gave above would suggest. A further complication is that there is a geometric factor depending on how the source and detectors are orientated with respect to each other.

OK, these are complications, but the formula can still be used as an approximation. So if we take the GW170817 signal, the total mass was about $2.8M_{\odot}$, the source was at 40 Mpc, so at frequencies of 200 Hz (corresponding to a period of 0.01 s) you might have expected a strain signal of about $3\times 10^{-22}$.  This gave a very readily detectable signal. The discovery paper (Abbot et al. 2017) says the "horizon" for detection was approximately 218 Mpc for LIGO-Livingston and 107 Mic for LIGO-Hanford. As the source was much closer than these numbers then it is unsurprising that the detection was strong.

Taking the formula above and a fixed orbital period of 0.01 s, we can see that the horizon distance will scale as $\sim M^{5/3}$. So a $10 M_{\odot} + 10 M_{\odot}$ black hole binary might be seen out to $218 \times (20/2.8)^{5/3} = 5.7$ Gpc (this will be an overestimate by a factor of a few because of the issue of the rapidity of the evolution towards merger that I discussed above).

A more through and technical discussion can be read here, although this is a couple of years out of date and LIGO's reach has been extended by about a factor of five since these calculations were done.


The ecliptic is a plane that is inclined 23.43 degrees (approximately, insert your more accurate value as needed). One position on the ecliptic is 0 hours Right Ascension, 0 degrees Declination (0,0). If you plot that on the Earth at the date and time you have, the rest of the ecliptic's path on the Earth can be calculated as follows:
$$\tan(23.43) \sin(longitude-longitude_{equatorcrossing}) = \tan(latitude)$$

where $longitude_{equatorcrossing}$ is the longitude where the ecliptic at 0 hours Right Ascension crosses the equator.

As a check, the latitude should be 23.43 degrees when the longitude is 90 degrees east of the (0,0) longitude point, and the latitude should be -23.43 degrees when the longitude is 90 degrees west of the (0,0) longitude point. In other words, if (0 Right Ascension, 0 Declination) corresponds to 15 degrees east longitude, then 105 east longitude is where the ecliptic is at the maximum northern latitude, and 75 degrees west longitude is where the ecliptic is at the maximum southern latitude.


If earth were the size of a marble would it be smoother than a marble?

No.

This What-If page describes the case for 'smoother than a bowling ball"

  These scans (along with various measurements of ball roughness1 tell us that a high-end bowling ball is quite smooth. If blown up to the scale of the Earth, the ridges and bumps[2] would be between 10 and 200 meters high, and the peaks would be between one and three kilometers apart:

  By Earth standards, this is quite smooth; our highest mountains are 40 times higher.

And it is simple to continue the scaling process to show that the same holds true for marbles. (Although I have not found a good source of data from people who scan marbles...)


Will Neptune be visible with the naked eye if I am standing on its satellite

NASA's Neptune fact sheet states that Neptune's irradiance is $1.51 W/m^2$. This is roughly three orders of magnitude less than at Earth (at $1367.6 W/m^2$). This sounds like a lot, and it is quite an attenuation. So much that for example solar panels for interplanetary probes are not worthwile at these distances from the sun.

However, if you compare this with the sensitivity of the human vision system, this seems still pretty bright. A sunny day outside will be approximately $10^5 cd/m^2$. So equally on Neptune a sunny day will probably be $10^2 cd/m^2$. This is comparable to an overcast sky, and still well within human photopic vision!

The perceived brightness of Neptune, as viewed from one of its moons, will not change, so yes, it will appear 1000 times less bright than earth does, but since you don't have any comparisons to brighter light sources there, you will perceive it as pretty bright!

As for the other gas giants, their surface irradiance will be equally higher, so they will in turn appear much brighter than Neptune.


What factors determines the size of a galaxy?

Galaxies reside in dark matter halos, and the size of the galaxy is more or less proportional to the size if its halo. The distribution of the sizes of halos is given by the halo mass function (HMF), which is a declining function with mass — that is, the smaller the mass, the larger the number. At low masses, the HMF falls off as a power law, while at higher masses there is an exponential cut-off.

Hence, a galaxy such as IC 1011 is much rarer than a galaxy like M87, which in turn is rarer than Andromeda and the Milky Way.

An analogy is the size distribution of towns and cities. Capitals are rare, towns are common, and villages are even more common.

The reason the galaxy masses follow this distribution is first given by how dark matter in the early Universe fragments into clumps, and start collapsing and thus detach themselves from the Hubble flow. The first analytical attempt to solve the problem of structure formation assumed spherical collapse, and was presented by Press & Schechter (1974). Later, assuming ellipsoidal collapse, Sheth & Tormen (1999) gave an improved solution. More realistic forms cannot be calculated analytically, but must be estimated using large-scale $N$-body simulations (e.g. Tinker et al. 2008, Klypin et al. 2011).

As the Universe evolves, galaxies merge to form larger galaxies. Thus, the distribution depends on the time one considers. The figure below (from Klypin et al. 2011) shows how the HMF evolves from early times ($z=10$) to now ($z=0$). You'll see that the amplitude grows with time, because more galaxies are formed. From $z=2.5$ to $z=0$ you can also see that the number density of small halos actually decrease a little, because small galaxies are "eaten" by larger ones.

Halo mass function, i.e. number density of dark matter halos as a function of mass. Points are from an $N$-body simulation, while curves show the Sheth-Tormen approximation, which provides a very accurate fit at $z = 0$, but overpredicts the number of halos at higher redshifts.


The energy driving Io's volcanism is derived from its orbital energy and (to a lesser extent by the orbital energy of the other moons and the energy of Jupiter's rotation).  Similar mechanisms power the cryo-volcanism on Enceladus and the other icy moons.  The available orbital energy reservoir is huge compared to that required to drive the volcanic activity.

This does imply that the gravitational interactions on all the moons in the system are changing their orbits and rotations, albeit slowly.  Given unlimited time, the orbit of Io might decay to a point where it falls within Jupiter's Roche limit and is torn apart.

(Note: the answer to How can tidal heating lower Io's orbit describes this more precisely, and points out that tidal forces are transferring energy from Jupiter's rotation to the orbits of its moons, so Io's orbit in the long term gets larger (to the point of it leaving Jupiter orbit) rather than decaying.)


In my mind, none of these explanations really cover the actual reason that red giants expand.  Indeed, this subject seems like an area where people just make up anything that sounds plausible, but it's often quite wrong (Fraser Cain mentions both light pressure and a higher volume in the fusing shell, but light pressure plays no role at all, and the volume of the shell isn't much different than the core, which is way smaller than the Sun's core). So let's get the story straight.

Many of the descriptions do include some of the key elements, including the fact that you have a shell of hydrogen fusion going on atop an inert degenerate helium core.  But the key reason for the expansion is that the way this shell self-regulates its fusion rate is quite a bit different from the way the Sun's core is self-regulating its fusion now.

Now, the Sun's core self-regulates its fusion rate to match the rate that energy (in the form of light) is diffusing out through the mass of the Sun.  The way it does this is essentially the same in all main-sequence stars: they regulate their core temperature, which is why the core temperature of more luminous main-sequence stars is a little higher.  But this is not at all the way the shell fusing in a red giant self-regulates its fusion rate-- it cannot regulate its temperature, because the temperature is handed to it by the gravity of the degenerate core its sits upon.  (This sets the temperature via the virial theorem, that is the key way the degenerate core affects the shell-- it sets its temperature.)  Since the shell doesn't regulate its own temperature, the temperature tends to be quite high, especially as the core gains mass (this is why the luminosity rises with time).  Fusion is very sensitive to temperature, so being stuck with a very high temperature makes the fusion rate go berserk.  The rest of the star cannot support this spectacular fusion rate, so something else happens.

The star puffs out, and in so doing, we find the way the shell regulates its own fusion rate: it lifts weight off the shell.  That reduces the pressure in the shell, which compensates for the high temperature and brings the fusion rate down to what the rest of the star can manage (which is set by the rate light can diffuse through the shell).  So there's the real reason-- the star must find a way to lift weight off its crazy-high-temperature shell to keep the fusion rate from going nuts, but the upshot is that the fusion rate is still pretty high, and gets higher as the mass of the core rises, forcing the shell temperature up ever more and forcing the star to puff out even more.


Question on Friedman equation involving cosmological constant

When including a cosmological constant, the acceleration equation reads
$$\frac{\ddot{a}}{a}=-\frac{4\pi G}{3c^2}(\epsilon+3P)+\frac{\Lambda}{3}$$
In a pressureless universe, $P=0$, and in a static universe $\frac{\ddot{a}}{a}=0$. The equation becomes:
$$\frac{4\pi G}{3c^2}\epsilon=\frac{\Lambda}{3}$$Energy density is always positive and non-zero, so $\Lambda>0$.

The solution is unstable since after a slight perturbation, if $a$ grows, $\epsilon$ will go down ($\epsilon_m\propto a^{-3}$ for matter for example), which will result in an increase of the acceleration, making $a$ go up even more.


How to find a deep sky object 1° NE from a given star

Usually when the direction to an object is given relative to a star, the directions are related to equatorial coordinates (the coordinates on the sky). They are not related to the compass or cardinal points. It is best to avoid directions relative to the horizon/compass/cardinal points since those depend on the time and location.

In your example, 1 degree northeast of $\delta$ Cas means this direction:

north is toward Polaris along a line of right ascension, so you would go in the direction of increasing declination from $\delta$ Cas in the direction of Polaris.
east is along a line of declination in the direction of increasing right ascension.

When looking at the sky, and if north (the direction toward Polaris) is thought of as the 12 o'clock position, the direction north east would be approximately 10:30. (Likewise, east would be the 9 o'clock position, south would be the 6 o'clock position, and so on.)

Here is a view of Cassiopeia for an arbitrary time and date with the directions shown. Are you trying to locate M103?


Simple mathematical models of solar systems

I "grew up" with the old roleplaying game 2300AD's system creation rules, which were based on fairly hard science (and used the Gliese star catalog as a map!) They have stood the test of time fairly well (here is an astronomer friend's take on it).

Here is a rough sketch of how I typically make my system generators, which is based on the same logic:

Generate a number $N$ of planets in the system. At present there is no clear data constraining this, although we may imagine high metallicity stars could have more planets (but maybe not).
Generate orbits $a_i$: start with a random innermost orbit $a_1$, and then make $a_{n+1}=(1+r_n)a_n$ where $r_n \sim U(K_1,K_2)$ is a random number. This imitates the Titus-Bode law pattern and looks roughly like observed systems. I tend to use $K_1=0.1, K_2=2$ but this is guesswork. Note that this process has problems handling hot Jupiters really close - you may want to add them by hand. It also ignore resonances etc.
Generate a planetary seed size/mass. This can be done in an elaborate way by looking at exoplanet $(a,M)$ distributions and extrapolating to unobserved corners, or using some suitable skew random number distribution.
From the mass and distance to the star, estimate planetary and atmospheric composition. Basically, outside the snowline ($\approx 1.4\sqrt{L/L_\odot}$ where $L$ is star luminosity) planets will start accumulating volatiles, and having ice crusts become possible. Essentially, decide on a core density randomly.
Calculate the approximate minimum molecular weight retained based on the surface gravity. This will determine the atmospheric density. Especially, if the planet retains hydrogen it will become a gas giant: multiply up the mass a lot (the overall radius of a planet scales as $M^{1/3}$ for solid planets up to a few earth masses, and up to Jupiter size for gas giants - then degeneracy pressure will make the size level off and they become denser instead). Note that this is going to be the most fudgy step, since actually estimating a consistent atmosphere (especially exosphere) temperature, composition and pressure is really involved (and nontrivial even in reality).
Given the size, temperature zone and atmosphere classify it suitably ("ice ball", "gas giant", "venusian hothouse", "super-earth", ...).
Add random moons, rotation periods, eccentricities and whatnot, calculating their effects (like tide sizes, estimated magnetic fields, temperature variations, number of Hadley cells...) - lots of handwaving here, even when you base it on real astronomy and atmospheric physics papers.

This is the simple version. It does not try to actually simulate planet formation, where important processes like migrating gas giants can really affect systems.

You extra questions:

Right now it looks like planets are very common, so a system with no planets is likely pretty unusual.

Big moons might be uncommon, or they might be ordinary. In the solar system Pluto/Charon is also a pretty even pair. I personally think double-planets may be more common than people expect. We'll see.

Asteroid belts are likely pretty common. In a way we have two, the main belt and Kuiper belt. I would expect most systems to at least have a bit of debris that has not cohered in the outer regions.


When black holes such as those detected by LIGO merge, in what ways does the Schwarzschild Radius warp?

This is a better simulation SXS did for the LIGO detection (still frame below), the two event horizons (Schwarzschild Radii as you say) just simply come together and merge quite effortlessly like two bubbles in a bath coming together and merging into one. I don't know if you were expecting something more dramatic than this but there is no reason to suggest its a violent event.


It might depend on what qualifies as a plant.   A thin film of rugged mold for example . . . just maybe.

The tricky part is that asteroids are too small to retain an atmosphere and you need an atmosphere to have liquid water which plants and all life (so far as we know) needs.   I don't want to dismiss the possibility that there's underground water and some kind of very sturdy life that can eek out a living on the airless surface of an asteroid, but I wouldn't bet money on it.  I think it's unlikely, maybe unlikely multiplied by about 5 orders of magnitude.    It's very close to impossible.

That said, Ceres is an asteroid and Ceres may have a liquid underwater ocean.   Life in an underwater ocean is thought to be possible.


Why is SPT0418-47 ("the most distant Milky Way look-alike") expected to evolve into an elliptical galaxy?

The full paper can be downloaded from the ESO site. This contains the following:

Dusty starburst galaxies are believed to be the progenitors of early-type galaxies (ETGs),
which are the most massive galaxies observed today, dominated by old stellar populations. The
most popular evolutionary track for this transformation predicts that the dusty-starburst phase
is followed by a quenching phase, during which AGN feedback leads to gas consumption and
heating with the consequent formation of a population of compact quiescent galaxies at $z \approx 2$. In
the final phase, dry minor mergers are expected to be responsible for a growth in galaxy size and
the transformation into present-day ETGs.

(AGN = active galactic nucleus)
They then compare the stellar/baryonic mass with the likely influence of dry mergers:

The comparison between the ETGs and the stellar/baryonic quantities for SPT0418–47 in the size–mass plane (Fig. 4a) indicates that this starburst galaxy should increase its stellar mass by a factor
of 6 (3 for the red diamond) and its effective radius by a factor of 11 (3 for the red diamond), in
order to evolve into an average ETG (yellow cross). This is in agreement with a simple toy model
for mergers, in which a single dry major merger event would be responsible for an increase in both
the size and stellar mass of SPT0418–47 by a factor of 3.

They also compare the distribution of dark matter with local ETGs:

Finally,
we derived the fraction of dark-matter mass within the effective radius and found that, with a value
of $f_{\rm DM} (< R_{\rm e}) = 0.095^{+0.004}_{-0.004}$ (red diamond), the central regions of SPT0418–47 are dominated by
baryons. As shown in Fig. 4c, such a low fraction of dark matter is compatible with observations
of local ETGs, implying that the physical mechanisms responsible for the mass and size growth
of this galaxy with cosmic time should preserve the dark-matter contribution within the innermost
∼1 kpc.

Note that the term "early-type galaxy" refers to elliptical and lenticular galaxies, which are near the start of the Hubble sequence. It does not refer to evolutionary stage, as noted in Hubble (1927) "The classification of spiral nebulae":

The nomenclature, it is emphasized,
refers to position in the sequence, and temporal connotations are
made at one's peril. The entire classification is purely empirical
and without prejudice to theories of evolution—comparison with
theories will be the more significant for this very reason.


Finding the radius of an eccentric orbit at any point

If you know the apoapsis and periapsis, you can find the eccentricity, using $r_a = a (1+e), r_p = a (1-e)$ (https://en.wikipedia.org/wiki/Apsis#Mathematical_formulae).

Using the eccentricity, you can find the radius at any angle $\theta$ from the periapsis (this angle is also called the true anomaly).

$r(\theta) = \frac{a(1-e^2)}{1+e.\cos \theta}$(https://en.wikipedia.org/wiki/Kepler_orbit#Johannes_Kepler)

With time, it is little more difficult. The mean anomaly (M) is defined as

$M = \frac{2\pi t}{T}$ where t is the time since last periapsis, and T is the time period.

This mean anomaly relates to true anomaly through a variable called eccentric anomaly (E). From mean anomaly (which you get from the time), you can get eccentric anomaly using

$M=E-e.\sin E$

Now this equation isn't analytically solvable, but you can solve it graphically or numerically, or using a simple iteration which I would demonstrate here (technically included in the numerical methods, but it's easy enough to show here). It is easy enough to be done on a simple scientific calculator.

$E_{n+1}=M+e.\sin E_{n}$ where $E_0=M$

Using this E, you can find the true anomaly $\theta$ using

$\theta=2 \tan^{-1}[ \sqrt{\frac{1+e}{1-e}} \tan \frac {E}{2}]$

Then you can find the radius as described above.


How large (that is, radius) could a planet be?

Puffy planets tend to be Jupiter or Saturn like, probably lower mass than Jupiter, perhaps lower metalicity but the most important factor is heat.  Either close to the sun or recently formed.   Heat expands gas planets.  You're correct that as you add more mass the planet of Jupiter mass tends not to grow larger, but if there's enough internal heat, gas giant planets can get a bit larger than Jupiter.   Planets as much as 2 Jupiter Radii have been observed (though there's some inaccuracy in those estimates), but growing larger than Jupiter is largely a factor of high temperature.

Rocky worlds will never grow as large as Jupiter.  The greater mass will prevent it, and past a certain mass, a rocky world is unlikely to remain what we consider rocky.  Above a certain mass it holds onto hydrogen which is the most abundant gas in the universe, and that would give the massive rocky world an appearance more like a gas giant.

But to answer your question in theory, if you had a rocky world with Jupiter mass and a negligible atmosphere, it could never get close to Jupiter diameter.  The mass crushes the inside of the planet.  Mercury for example is made of denser material than the earth.  Higher Iron content, but it's less dense than the Earth because the Earth's mass crushes its rocky mantle and metallic core.  When you start to get a Jupiter-mass rocky world, the crushing becomes significant and you could never build a rocky world close to as big as Jupiter.  As with hydrogen, it would reach a certain maximum size, then it would begin to get smaller under the crush of gravity.   Even so called "non compactible" material, does compact at the pressure inside large planets.   Iron has a density of 7.874 g/cm3 and slightly less than that at high temperature, but the Earth's metallic core has a density of 12.6–13.0, and that's primarily due to crushing.  When you've reached Jupiter mass, the crushing and density is significantly greater.

At about the mass of the sun the planet would be smaller than the Earth, and it would basically resemble a white dwarf star, and no longer be distinguishable as a rocky world.

I can add a few links to back this up if needed and if someone wants to run the math or give an answer with more detail, feel free.


Why Helium could be produced in first minutes of big bang but not heavier elements

The triple alpha process requires very high densities, very high (but not too high) temperatures, and a good amount of time. Large stars have all three. In particular, they have millions of years during which the very rare triple alpha process can proceed.

The temperatures in the very early universe were extremely high, too high to create anything more complex than protium. By the time the universe had cooled enough so that it could produce deuterium (and thence helium), its density and temperature had already dropped too low for the triple alpha process to proceed. All of this happened in a few hundred seconds. Compare that with the millions of years during which large stars burn helium.


You have plenty of choices!

I am specialized in X-ray astronomy, then I can suggest you many facilities: they usually give at least some of their data publicly available. Tipically, you have data for light curves (photon count rates versus time), and/or spectra (that is flux versus energy).

I suggest you to work on the light curves: they are easily available, already reduced (you don't have to play with the data to make them usable, you can directly use them), and they fit your request of conforming to a regression model. Also spectra are enjoying, still I know some theoreticians who are not familiar with spectra, plus they usually need some more knowledge, because of response files, spectral fitting packages, etc. It is up to you (and your available time)!

Here a list of some facilities with available data:

Rossi X-ray Time Explorer

Swift/BAT

INTEGRAL

Chandra

MAXI

Also here you can find a list of many facilities with NASA participation, so the list is much larger.

Of course you can also use other kind of data (optical, infrared, radio) by the MAST project, but I have no experience on those. I guess, in the optical band, the Hubble Space Telescope will have a huge amount of data!

If you need a hand, beyond the SE vote up/vote down neurotic fashion, let me know. Good luck!


What could be a star-like object that shines at daytime for few seconds? What's the probability to see it?

Several satellites (but most notably Iridium) have large reflective panels.  If aligned with the sun properly, it can shine with sufficient brightness to be visible in the daytime.

An Iridium Flare usually lasts only a few seconds.  It sounds quite consistent with your description.

It can be hard to tell in a blue sky with no nearby references, but an Iridium satellite would also be moving while fading out.

There are sites that can show you upcoming flares for your area.  If you knew the exact date in the past, they could probably even "post-dict" what flares were available then and see if it matched.


Gravitational acceleration inside a planet

You can use Gauss's law for gravitation to work out the gravity as a function of (interior) radius.
$$ \oint \vec{g} \cdot d\vec{A} = -4\pi G \int \rho\ dV\ .$$
What this means is that the flux of gravitational field $\vec{g}$ out of a closed surface is proportional to the mass enclosed within that surface.

Let us assume that the density is a function of (interior) radius, such that $\rho = Ar^{\alpha}$ and that the planet is spherically sysmmetric, so that the left hand side becomes $4 \pi r^2 g(r)$ and the volume element $dV = 4\pi r^2\ dr$. Then
$$ 4 \pi r^2 g(r) = -4\pi G \int_{r=0}^{r} Ar^{\alpha} 4\pi r^2\ dr$$
as long as $\alpha \neq -3$.
$$ r^2 g(r) = - 4\pi G \frac{Ar^{3+\alpha}}{3+\alpha}$$
$$ g(r) = -\frac{4\pi GA }{3+\alpha} r^{1+\alpha}$$
What this shows, is that if $\alpha < -1$, then gravity will increase with decreasing radius.

Now we are in a position to answer your question. In the Earth's deep interior $\alpha \simeq 0$ and the density does not increase greatly as we head towards the centre. This means that $g(r) \propto r$ as is seen in the plot you reference. In the outer part of the crust $\alpha \simeq -1$ and $g(r)$ is roughly constant.

Below is a model density profile for Jupiter (solid line).

Just looking at this I would say that $\alpha$ is just a bit $<-1$ for most of the interior and so the gravity should be increasing slowly as $r$ decreases, but there is a sharp uptick in density at the boundary to the core which would see $\alpha \ll -1$ and the gravity becoming proportional to $r$ to the power of a negative number (i.e. increasing strongly with decreasing $r$), but then must drop to $g=0$ when $r=0$ shortly afterwards.


Should a station in orbit around the Sun travel in the same or the opposite direction to the planets?

It must orbit the sun in the same direction as the Earth and other planets.

The main difficulty in sending a transport to the station will be the the velocity change needed to put the transport into the station's orbit. If the orbit is retrograde with respect to the Earth, then the transport needs to make a massive change of velocity to reach the station, and this needs a massive amount of fuel.

If, on the other hand, the station is prograde with respect to the earth, then the transfer can be made with an economical Hohmann transfer. This is much more important than any apparent benefit of a retrograde orbit passing close to the Earth more often.


Where is a Hollow Object's Center of Gravity?

If by "a hollow object" you mean a spherical shell -- a theoretically perfect sphere Moon, with a perfect sphere of material removed from the center: The center of mass, and therefore the center of gravity remains at the geometric center. Ref Newton's Law; Bodies with spatial extent.

The body of the moon is plastic; Meaning the gravitational forces are strong enough to reshape the material... it basically flows into a sphere. So if you magically removed a small amount of the center, it would collapse into a smaller sphere. If you removed most of the center, then the "thin" shell may or may not be strong enough to support it's shape.

Inside a hollow spherical shell, there is no net gravitational force. So you would be in a zero-G environment everywhere within the shell. From outside the shell, the gravity is the same as if all the remaining mass was located at a point at the geometric center.


What does the Sun look like from Eris at its aphelion?

Light flux decreases as one over distance squared. So if Eris is 100 times further away from the Sun than the Earth is, then the amount of light that reaches Eris is 10 000 less than that on Earth.

Since a difference of five magnitudes is a factor of 100, The Sun would have an apparent magnitude -16.7 seen from Eris' aphelion, as opposed to -26.7 magnitude at Earth's 1 AU. For reference, this is still brighter than a full Moon, which is listed by Wikipedia's Apparent_magnitude; table of notable celestial objects at a magnitude of -12.9.

So you would still tell the difference between night and day, but the day would be rather dim and the Sun would appear point-like.

Solar eclipse glasses reduce the Sun's brightness by a factor of 500 000, so it might still be dangerous to look at the Sun with average binoculars (or even with the naked eye).


If a pulsar's rotational and magnetic axes were aligned and both aimed at us, would we see a steady radio source?

Two things would be required. First, your line of sight would have to be close to looking along the magnetic axis of the neutron star. Second, that magnetic axis would have to be closely aligned with the rotation axis of the pulsar.
If both of these are pseudo-randomly distributed and the pulsar beam is narrow, then this is inherently a very unlikely scenario. The cumulative probability of a randomly oriented axis in space having an inclination $<i$, where $i=0$, means aligned, is $1 - \cos i$. And if both line of sight and magnetic-rotation axes require close alignment, you are dealing with the product of two of these factors.
One factor that may work in favour of seeing such objects is that it seems likely that pulsars drift towards alignment of the rotation and field axes as they get older (e.g. Young et al. 2010), but on the other hand, the pulsar beam is expected to get narrower and the pulsar activity weakens and eventually switches off on a similar timescale. It is also possible that young pulsars are more likely to have misaligned, almost orthogonal rotation and field axes, in which case the scenario is impossible for young pulsars.
Still, this is an interesting question. If pulsars are born with random alignments, you might expect a small number to be born with aligned rotation and field axes, and for a small number of those to have beams aligned with our line of sight. They would be characterised by very little modulation of their emission.


Expected nature of LISA's data; will it be more like a forest of static peaks, or a series of individual events?

LISA's data will be very different from LIGO's. It will typically "see" many sources at the same time. Most prominently:

Mergers of pairs of supermassive blackholes. These will be very much like a scaled version of the events that LIGO sees (probably more variation in the mass-ratio and ecccentricity though). They will be much louder (SNR>1000) than any event seen by LIGO though. They will (potentially) enter LISA's sensitivity band months before merger. Event rates are uncertain with estimates ranging from a few to hundreds per year. Depending on where this rate is in nature, we may get many overlapping signals.
Wide binaries These are binaries of compact objects (mostly white dwarfs) that are far from merger. They evolve very slowly, and will therefore mostly show up as continuous almost monochromatic signals. There will be thousands of such signals "visible" at any one time.
There has been some suggestion that LISA may be able to see the earlier stages of GW150914 like events months to years before merger. The current design limitations make this somewhat unlikely though.
Extreme mass ratio inspirals These are mergers of stellar mass compact object with a supermassive black hole. Due to the very small mass-ratio their evolution is very small, meaning that they will stat in the LISA sensitivity band years before merger. Their event rate is even more uncertain than that of supermassive BH mergers with estimates ranging from 1 to 1000s per year. In all likelihood these signal will overlap.

Consequently, LISA will be observing thousands of sources at any one time, which will be a significant challenge for the data analysis.


Explanation for the first spinning neutron star detected in the Andromeda galaxy?

In the article it notes that the partner has less mass than the sun, or possibly up to about twice the mass of the sun (quite a range of uncertainty there!) It's not clear, from the article how they made these calculations, measurement of doppler shifts to get orbital velocity I suppose.

The mass suggested there is not especially small, there are pulsar binary systems in which the secondary is one hundredth the mass of the sun, and could be considered to be planets.

However it does rule out a black hole companion, which would have a mass of 3 solar masses at the least, (and more likely more than 5). There is no need to hypothesis a black hole to explain the spin.

The last few sentences in your question don't make any sense to me.


How does an accreting black hole acquire magnetic fields?

Addressing (2) and (3) first:

In a case like this, your best option is to go straight to the source. In this case, that would be Blandford & Znajek (1977), the originators of the proposed mechanism. they have several figures regarding different magnetospheric structures depending on different magnetic fields. However, the most important one is Figure 1, the general case:

$D$ represents the accretion disk, while $T$ is the transition zone where matter is rapidly moving radially towards the black hole. The lines represent magnetic field lines, along which charged particles. Electrons (e-) travel in the direction of the medium arrows on each field line; positrons (e+) travel in the other direction. The field lines are closer together in the disk area closer to the black hole, indicating the stronger magnetic field.

If you want a good overview of the process, this is the paper to look at. That said, it is far from an easy read.

Regarding (1), I'm not an astronomer and not really qualified to tell you what the mainstream opinion among astronomers is. However, as far as I'm aware, the Blandford-Znajek process is currently accepted as a strong possibility for generating the magnetic fields observed near most black holes (including stellar-mass black holes).


Is there an established distinction between a geyser and a cryovolcano in solar system bodies?

An educated guess: on Earth, the working fluid in a "geyser" is water, which remains a fluid more or less permanently after each eruption.  This means that geyser eruptions are not associated with buildup of new geological material around the geyser (apart from the much slower process of dissolved minerals accumulating as the water evaporates).

Contrariwise, on Earth, the working fluid in a "volcano" is molten rock, which solidifies after reaching the surface.  Each new eruption permanently deposits new solid material in the vicinity of the volcano.

If I were writing definitions, I would define a "cryovolcano" as a geological vent system where the working fluid is water, but where the environment is cold enough that the ejecta from eruptions accumulates in a permanent mountain of ice around the vent.


If GRB 080319B happened next to nearest star: Proxima Centauri, what effect would that have had on us?

Faster than light's answer is fine if the gamma ray burst is "pointing" towards us. The emission from rapidly moving material in a GRB is thought to be beamed in the direction of that motion, along the rotation axis of the progenitor object in fact.
The opening angle of the beam may only be a few degrees (e.g. Frail et al. 2001), which would mean that for random orientations of the GRB source there would only be about a 1 in 500 chance of the solar system being "in the beam". Thus the highly destructive conclusion of Faster than Light's answer is a remote possibility.
As the fireball phase of the GRB develops then it is thought that the beam opening angle will widen over days and weeks, but of course the source brightness also fades drastically on these timescales too. It should also be noted that most of the highly destructive X-ray and Gamma ray emission arises during the tightly beamed phase within minutes or hours of the burst.
Thus although I'm not saying we're not "screwed", I am saying that the issue is more complex than scaling the apparent magnitude of a GRB and comparing it to the Sun. I suspect that in most cases the effects would be similar to the explosion of a supernova at the distance of Proxima Cen.
The conclusions I can garner from such as https://physics.stackexchange.com/questions/26578/near-earth-supernova?rq=1 , suggest that would still likely remove the ozone layer for some time and therefore could have serious and deadly consequences for life on Earth.


How well do planetary orbits fit with Johannes Kepler's in- & circumscribed Platonic solids?

It is easy enough to do the calculations, formulae for the in and circum radii of the Platonic solids can be found here which give ratios of circ to in radii of (note formulea for the radii have dropped common factor of the side length, which we don't need as we are interested in the ratios):

>ri4=sqrt(6)/12,rc4=sqrt(6)/4
     0.204124
     0.612372
>
>ri6=1/2,rc6=sqrt(3)/2
          0.5
     0.866025
>
>ri8=sqrt(6)/6,rc8=sqrt(2)/2
     0.408248
     0.707107
>
>ri12=sqrt(250+110*sqrt(5))/20,rc12=(sqrt(15)+sqrt(3))/4
      1.11352
      1.40126
>
>ri20=(3*sqrt(3)+sqrt(15))/12,rc20=sqrt(10+2*sqrt(5))/4
     0.755761
     0.951057
>
>rho4=rc4/ri4
            3
>rho6=rc6/ri6
      1.73205
>rho8=rc8/ri8
      1.73205
>rho12=rc12/ri12
      1.25841
>rho20=rc20/ri20
      1.25841

Which may be compared with the orbital radius ratios from here (radii in km)

>RMecury=57.9e6;
>RVenus=108.2e6;
>REarth=149.6e6;
>RMars=227.9e6;
>RJupiter=778.3e6;
>RSaturn=1426.7e6;

Now we can compare the corresponding radii ratios:

>[RVenus/RMecury,rho8]
      1.86874       1.73205
>[REarth/RVenus,rho20]
      1.38262       1.25841
>[RMars/REarth,rho12]
       1.5234       1.25841
>[RJupiter/RMars,rho4]
      3.41509             3
>[RSaturn/RJupiter,rho6]
       1.8331       1.73205

Which, as these things go, is not bad.


The image is about a merger of two black holes, before the actual merger takes place.
The two largest of the dark regions are slightly distorted images of two the black holes. Distortion is due to light being bended asymmetrically by the combined gravity of the two black holes.

The two smaller dark areas show the respective other black hole in a strongly distorted version, since the light - or missing light - of the other black hole is bended around the respective less distorted appearing black hole by about 180 degrees, somewhat similar to an arrangement of uneven mirrors.

You may interprete the phenomenon of the smaller dark areas as an extreme version of Einstein rings.


What does observational evidence say about the number of protoplanets believed to exist in the early solar system?

Earth and moon are thought to be formed by the collision of two planets. So we have at least these two protoplants. The cores probably ended mainly in the core of Earth; that's inferred by the much higher mean density of Earth in relation to Moon.

Iron-nickel meteorites strongly indicate at least one pre-existing planet large enough to form an iron-nickel core.

Comets of the Kuiper belt are hypothized to be a collision product.

As you go further back in time the building blocks of planets get ever smaller and more numerous till the interstellar dust and supernova remnants our solar system is formed of.

Some of the early material will most likely have been ejected from the solar system; this may include protoplanets.

By the chaotic nature of multi-body systems like our solar system it's getting more and more difficult to track single bodies back into the past.

The NICE models try to describe the early solar system.


Very bright star in the east at northern hemisphere. What is it?

It will be very difficult to tell you without knowing where you are located.  Particularly, if you are in the northern or southern hemisphere.  At this time of year, most of the brightest stars are in the sky.  Also where in the sky, the particular star that you are wondering about is.  And any other constellations that you can identify.

For stars rising this time of year, particularly bright ones you might be thinking of Sirius.  Which would be near the constellation Orion below the hourglass shape of the it (http://starchart.polaris.net/chart.php?constel=ori&layoutType=Key&layoutOri=Portrait&project=SANSONS&type=PNGC).

Another possibility would be Aldebaran, it is located in a "V" shape of stars that make up the constellation Taurus.
http://starchart.polaris.net/chart.php?constel=tau&layoutType=Key&layoutOri=Portrait&project=SANSONS&type=PNGC

To find out I would take a look at some of the online star charts that are available for your time and location.  That will give you a good idea of what you are seeing.  Here is a list of some online:

http://www.skyandtelescope.com/resources/internet/3305011.html?page=1&c=y


Moving from computerised to normal amateur telescope?

A Celestron Nexstar 130 SLT is not suitable for spectroscopy and the only photography that it will be mush use for would be afocal or DSLR imaging of the Moon (or Sun with a suitable filter).

For more serious work you will need a equatorial mounted scope with a better focuser.

Exactly which one to choose will depend of your budget and what you want to do.

When you say the computer does not work can you be more specific.


The magnetic activity is driven by a dynamo that relies on a combination of rotation and convection.

M dwarfs have large convection zones or are fully convective beyond spectral type M4. They also have much longer spin-down timescales than G- and K-dwarfs; billions of years rather than tens of millions of years.

This means that a large fraction of cool M dwarfs are rotating fast and have deep convection zones.

This is the basic reason, but it is still a very busy research area.


How did the orbits of Phobos and Deimos circularize?

Astronomers are not still completely sure that these satellites are captured asteroids (see for example this paper). However, in case they are, it's true that their eccentricity should be much higher. There are several theories that provide scenarios for the circularization of the orbits (the most recent I found is this one, which invokes impacts on the moons causing them to start tumbling and exciting the spin), but more data is needed to test these hypothesis (see this last paper for a review of this).


Do galaxies change shape and size over time?

The size of a galaxy is dependent on how the matter was distributed in the early universe, and how it collapsed under its own gravity to form clumps of (dark) matter that the galaxies would form around.

The very earliest galaxies are irregular, but as matter fell towards them, and they developed a consistent orbital direction they formed into a disc. Gravity waves in the disc then show up as spiral arms. Large galaxies that are actively forming stars tend to have this structure. Small galaxies may never reach the point of forming a disc and remain irregular.

As galaxies collide and interact, the orbits of the stars are disrupted. The result of the merger of two large spiral galaxies is often a galaxy in which the stars orbit in all directions, which looks like an elliptical galaxy. Star formation in elliptical galaxies is often very low. Elliptical galaxies are the most evolved form of galaxies.

The time scale of this evolution is slow. It takes billions of years for galaxies to evolve from one form to another.

The details of the process of galaxy formation and evolution are still uncertain. The exact nature of dark matter will be important in understanding how galaxies form. The role (if any) that black holes play in galaxy formation is uncertain.


I looked at the image and annotated it based on what it most likely is:

In black is the formation itself.
In blue is a depression underneath it.
In grey is part of the boundary of what you believed to be the disappeared half.
In green is a ridge to the north.
In red are several craters.

Here's a more likely hypothesis for what this actually is.
The depression
The blue oval contains a slight depression. It could be a number of things, including an old crater, a basin, or the result of the collapse of part of the surface. I'm actually a little bothered by the idea of it as an impact crater. We can see from the shadows on the western walls that it is only a little shallow, especially compared to other similarly-sized impact craters in the area. It's possible that it's a crater that was filled in with sediment.
If it's a natural basin, it could once have been a Martian lake - although this is quite speculative. It could also be the remains of a pit crater (there are many pit craters in the area), formed possibly from the cave-in of a series of lava tubes. Again, this is also speculative, and the depression is quite large, about 10 miles long on its major axis.
If one of these options is correct, the depression is not very mysterious. You've identified the grey part of it as the location of the other half of the formation, partly because it appears to be roughly the same size. I find this unlikely, for a couple reasons:

The ridges look similar in height to the edge of the depression on the other side of the formation, indicating that it is actually a part of that feature.
Half of an outcropping does not simply vanish by itself. I know of no mechanism for this to happen. Erosion could take its toll over time, and perhaps that caused weathering of the formation, but it couldn't move it.

I think it would be clearer that this is actually part of the depression if the ridge enclosed in green were not there.
The mesa
So if we assume that the outcropping is all that there even was of the object, we simply have to explain why it's so square. This really isn't that mysterious. There are plenty of other mesas in this cartographical area of Mars, designated the Elysium quadrangle by the USGS for mapping purposes. Here are some examples, which are similarly shaped:

The craters
Finally, we have the three large craters. They aren't too mysterious; craters are normal on Mars, and there are plenty, of varying sizes, in the Elysium quadrangle. The one reason the smallest stands out is because it is near the center of the northeast edge of the mesa. I'm reasonably certain this is just a coincidence.

Here's my hypothetical and likely not entirely correct timeline for this formation:

The large basin forms, from either a crater or some other process.
The mesa and ridge form, possibly as a result of the erosion of some larger feature.
Craters form through impacts. One happens to be on the mesa itself.

I think your identification is simply a case of pareidolia, which is the human tendency to find shapes and patterns where none really exist. A well-known Martian example is the Face of Mars.


Since every object in the known universe which has mass has an infinite gravitational field every object in the universe affects one another.
Every star in our surrounding affects Our solar system but the relative proximity of the sun masks these effects to us.
However if the sun would be a part of a binary system of stars

Our planet should have been affected by it because a star is a much larger mass when compared to the planets.
Since the star and our sun interact with each other they would produce a noticeable wobble.
Any such star would definitely be very close to the sun so as to be able to influence it and hence easily visible yet no evidence has been found for this.

So for now one can say that we and the sun are not part of any binary star system which has been (if it exists) undiscovered till recently


What makes small interferometers useful? Like NIRISS on JWST

A supplemental answer to probably someone's answer:

Why this can indeed be called interferometry:

Once one thinks in terms of physical optics (e.g. $\text{exp}(j(\omega t - \mathbf{k} \cdot \mathbf{r} ))$ ) instead of ray optics, imaging is always an interference problem, and the math behind correlating signals from an array of radio telescopes to produce an image is not so much different than the math behind calculating an intensity pattern at the focal plane of an optical telescope.

From the Aperture Mask Interferometry section on page 5 of the JWST Pocket Guide (see below) you can see that they have no problem using terms like "interferometry" and "non-redundant baselines" as one would when laying out a sparse pattern of dish antennas in an array. See for example a Google Maps image of the Meerkat Array Core below, which has already produced images with only 16 sites occupied with active dishes.

While the mask shown in the question is a pupil mask, the pupil plane is conjugate to the telescope's aperture. So it's very similar to having seven large hexagonal holes in front of the JWST, or just using seven of the telescopes primary mirror's hexagonal mirror elements (with additional masking, these hexagons to not cover the full size of the individual elements, as shown in the illustration).

An important distinction though is that these seven small apertures are of the same order in size as their separation, while in a large radio telescope array, the spacing between receivers is usually somewhere between much larger and much much larger than the aperture of individual elements. So the analogy breaks down at this point.

What Aperture Masking Interferometry is for:

The purpose of using the mask is to enhance the system's resolution by narrowing the central peak of the point spread function compared to what you would get from the full aperture.

By loosing contrast in the full field due to diffraction artifacts, as well as loosing ~90% of the transmitted intensity, you can enhance contrast near the center of the field by narrowing the central diffractive peak, in this case in the range of 70 to 400 milli-arc seconds, according to the JAM Team's (JWST Aperture Masking Team) slides linked below.

From the first link:

  Light admitted by 7 apertures in an otherwise opaque pupil mask interfere to produce an interferogram on the detector. This interfogram has a sharper core than is provided by normal "direct" imaging. The advantage is significant: while the ability to separate closely spaced objects with normal imaging is given by the familiar Rayleigh criterion (separation  δθ=1.22λ/Dδθ=1.22λ/D , where λ is the wavelength of light and D is the diameter of the telescope),  interferometry can resolve objects as close as δθ=0.5λ/Dδθ=0.5λ/D (the Michelson criterion).  AMI allows planetary or stellar companions that are up to ~9 magnitudes fainter than their host star and separated by 70–400 mas to be detected and characterized.  AMI  can also be used to reconstruct high-resolution maps of extended sources, such as active galactic nuclei.

You can read more about it in:

JWST's NIRISS Aperture Masking Interferometry
ArXiv NIRISS aperture masking interferometry: an overview of science opportunities
The JAM Team's: Non-Redundant Aperture Masking Interferometry (AMI) and Segment Phasing with JWST-NIRISS

Below: MeerKAT array core from Google Maps at (30.7136109S, 21.4399576E).


At last 80 nearby galaxies are believed to be part of a small cluster called the Local Group.
The three largest members of the Local Group, have their own system of satellite galaxies.
Over 50 galaxies are satellite galaxies of the Milky Way Galaxy and are closer to it than the Andromeda Galaxy is.
https://en.wikipedia.org/wiki/List_of_nearest_galaxies
The Andromeda Galaxy is actually number 86 in that list of nearest galaxies.
The Small Magellanic Cloud is number 23 and the large Magellanic Cloud is number 19.
The nearest galaxy on the list is the Canis Major Dwarf Galaxy, altough its status as a galaxy is soemewhat disputed.
If the Canis Major Dwarf Galaxy is not actually a galaxy, the next galaxy on the list would become the nearest galaxy. That is Draco II.


To answer the question

  Does spacetime return to being flat?

Yes it does. The picture you have calculated is not correct. It is not like that in gravity a solid body plow through spacetime and creates this valley like dragging a bowling ball throw snow would. The curvature of each point is affected by the mass and changes as the mass moves. So if the mass moves away far enough from a given point in space time the curvature becomes flat again.


How/where to check where the Sun is (constellation)?

There are plenty of online planetarium sites which should be able to do this for you. This one is the first hit for a Google search for "online planetarium software"


Is there any significant time dilation on other planets orbiting the sun?

It depends on what you mean by "significant."  None of the orbits of other planets exhibit relativistic effects at the level you specify.  You can estimate the overall size of relativistic corrections by squaring the ratio of v/c, where v is the speed of the orbit.  For the Earth, that number is .00000001, and it is not more than about a factor of 3 higher for any other orbit.  So it's pretty negligible, but that doesn't mean it has no consequences-- over many orbits, for example, Mercury precesses by a larger amount than can be understood without relativity.  Also, if one takes the relativistic perspective, the reason the planets follow the orbits they do is because of that tiny time dilation effect-- that small effect is enough to select the orbits we see over other possible ones.  So that's "significant," but not like a 59 minute hour.


How will "modern equipment" allow the Royal Observatory to now avoid some of the effects of light pollution at Greenwich?

It's a little hard to guess what research can be done there -- Today, Greenwich is close to the worst place on Earth to have an observatory: At sea level, in the light dome of one of the largest cities in the world, in a country famous for its clouds and fogs.
The Wikipedia article on the observatory mentions the new telescope and says:

In 2018 the Annie Maunder Astrographic Telescope (AMAT) was installed at the ROG in Greenwich. AMAT is a cluster of four separate instruments, to be used for astronomical research; it had achieved first light by June 2018:

A 14-inch reflector can take high resolution images of the sun, moon and planets.
An instrument dedicated to observing the sun.
An instrument with interchangeable filters to view distant nebulae at different optical wavelengths.
A general-purpose telescope.

At a guess, they hope to do some sort of star position measurements which is one sort of astronomy that is less affected by haze and light pollution (though it needs steady air -- I have no idea what the air steadiness is like there). But even that is still better done with larger instruments in better locations, and by satellites such as Gaia.
It really sounds like a public outreach facility or an undergraduate training site more than a research facility.  It's also worth noting that the pictures show equipment that is at the serious amateur level -- I certainly know amateurs who have better equipment.


What are the odds that the Sun hits another star?

TL;DR: Virtually zero.
The distances between stars are HUGE and stars are tiny compared to the astronomical scales of distance between neighboring stars. The sun, is about 0.0000001 or one ten-millionth of a light year. The probability of a star (to be generous, say a $10 R_\odot$ star) colliding with the Sun is tiny.
Every star has a different velocity vector and different positions in space-time. The probability that their positions will intersect in a relatively short timescale is almost zero. Imagine throwing a tiny dart, and hitting a bullseye the size of an atom from several miles away. That still is larger than the probability of the Sun and some other star colliding in the near future (5 billion years is a tiny moment of time when it comes to events like these).
According to Wikipedia, it takes about 30 trillion years for a star to undergo a close encounter with another star. Even if the encounter is less than 1 AU, the stars would probably just pass by and not do anything special. For a complete collision and merge, you'd need the distances to be much, much smaller, about $0.1 R_\odot$.

Years from now
Event

3×1013 (30 trillion)
Estimated time for stars (including the Sun) to undergo a close encounter with another star in local stellar neighborhoods. Whenever two stars (or stellar remnants) pass close to each other, their planets' orbits can be disrupted, potentially ejecting them from the system entirely. On average, the closer a planet's orbit to its parent star the longer it takes to be ejected in this manner, because it is gravitationally more tightly bound to the star.

Source: https://en.wikipedia.org/wiki/Timeline_of_the_far_future
Also, according to Wikipedia:

While stellar collisions may occur very frequently in certain parts of the galaxy, the likelihood of a collision involving the Sun is very small. A probability calculation predicts the rate of stellar collisions involving the Sun is 1 in $10^{28}$ years.

Source: https://en.wikipedia.org/wiki/Stellar_collision#Stellar_collisions_and_the_Solar_System
Back to crude math, assuming an uncertainty of 2 AU, the probability given these parameters is equal to:
$$\dfrac{5 \cdot 10^9 \text{ years}}{3 \cdot 10^{13} \text{ years} \cdot 10^{24} \text{ stars in the universe}} \cdot \dfrac{0.1 R_\odot}{450 R_\odot} = \dfrac{1}{27,000,000,000,000,000,000,000,000,000,000} = \dfrac{1}{2.7 \cdot 10^{31}}$$
Warning: This is only a very crude estimate, don't take this as a "real, super-mathy" answer. But I still hope this helps.
Edit: That equation above is still probably too big. Asked a question about this: If two stars collide, what is the probability that they merge to form a single star?


I don't know what could cause Upsilon Scorpii to appear red as it is not noted as being a highly variable star. It's spectral type is B2, which implies it would be blue/white in colour. If you know the spectral type of a star, which many stellar databases will note, you can work out its colour from this.


What day/night cycles, climate and seasons would experience Alpha Centauri Bb inhabitants?

This is just a partial answer (so far), but it will address at least part of your question.

  1.To what temperatures the dark surface of such planet could be heated? Is there possibility of liquid water?

Wikipedia gives a good estimate for this: 1,200 degrees Celsius. For a planet, that's pretty hot! Any water there would evaporate very quickly. In fact, it is expected that a lot of the planet's surface is molten. Now, does this apply to the back side of the planet? I would think so. While only the front side of the planet would bask in the warmth of Alpha Centauri B, the heat should dissipate throughout the entire planet.

  2.Will the radiation of the second star be enough to provide normal day-like illumination and heating?

The short answer here is no. In fact, when compared to our solar system, nothing about the radiation Alpha Centauri Bb gets is normal. It is situated extremely close to Alpha Centauri B - 0.04 AU. Given that both Alpha Centauri A and Alpha Centauri B are similar to the Sun, I think it's clear that this exoplanet will not receive a "normal" (normal here meaning similar to Earth) amount of radiation from Alpha Centauri B. Conversely, Alpha Centauri Bb is 11 AU away from Alpha Centauri A - much greater than the distance between it and Alpha Centauri B (a bit more than the distance between Saturn and the Sun). This means that the planet will receive only a tiny fraction of light from Alpha Centauri A relative to Alpha Centauri B.

  4.Will such planet experience seasons and how they would be arranged?

Seasons depend on the tilt of a planet's axis. Unfortunately, it is hard (if not impossible) to measure the tilt of the axis of Alpha Centauri Bb (I haven't been able to find any measurements). I would assume that it would experience some seasons, but I don't know how much they would extend.

As for question 3 - I'm not quite sure what you mean by "calendar." If you mean an artificial calendar, I would think that any beings on the planet (life there is extremely unlikely) would only use Alpha Centauri B to determine their years (just over three days and five hours).


What's the proper terminology for nebula clouds?

"nebula clouds" sound a little bit redundant, because, is there any nebula that is not a cloud?

I would stick to "nebulae". (And if you want to be consistent, you should be writing "supernovae" too).

I am not an expert on this topic, but I have never seen any use of "nebula cloud" before.


What happens for a "closed" universe without any content?

I see that you are using the convention that a has units of distance and is like the radius of curvature of the universe, and you are getting that either a or t would need to be imaginary.  So you can interpret that in two ways, either you say that is impossible, and conclude that it is impossible for an empty universe to be closed, or you attribute some physical meaning to imaginary radius of curvature or imaginary time.  I'm not sure what that significance would be, but what seems clear is there is something bizarre, or even bogus, about claiming that an empty universe can have positive curvature.

Another way to see this is to interpret your equation as an equation for k via H=Kc/a, where K^2 = -k.  If you further demand that nothing be imaginary, this requires that k be either 0 or -1.  Empty universes are flat or open, or else they are imaginary in some way.


Do all planets have a van Allen radiation belt?

Mars does not have a magnetic field of any strength and so does not have a van Allen belt (and this is a serious problem for potential human exploration of the planet). Similarly, neither Venus nor Mercury (nor the Moon, which is arguably part of a two-planet system with the Earth) have a van Allen belt.

The gaseous giants do, however, have radiation belts which are analogous to Earth's van Allen belt.

Spotting aurorae on distant exoplanets would be an indication of a strong magnetic field and thus of the likelihood of a radiation belt. I am not aware that such aurorae have been detected but distant aurorae have been seen around a brown dwarf - a substellar object somewhat bigger than a planet (see http://www.theguardian.com/science/2015/jul/29/astronomers-find-aurora-a-million-times-brighter-than-the-northern-lights).

Alternatively, the radio emissions from high energy electrons trapped in planetary radiation belts might be detectable - this was how Jupiter's radiation belt was discovered around 60 years ago. Work seems to be underway to detect such radiation: https://skaoffice.atlassian.net/wiki/download/attachments/22183971/PoS-exopla-AASKA14.pdf?version=1&modificationDate=1403003959989&api=v2


Saw a bright orange object in the sky. What was it?

If the bright orange object was low in the southeast around midnight, it may have been Mars.
In late July and early August 2018, it will be only 0.39 AU away from Earth and shine with apparent magnitude -2.7, slightly brighter than Jupiter.
This is not as bright as Venus but may stand out more strongly in a midnight sky than Venus does in twilight.
For most of Mars's 2.1 year synodic period, it is farther from Earth and appears fainter.

As the Earth rotates, celestial objects appear to move about 1/4 degree per minute from east to west.
For observers in the northern hemisphere, objects near the southern horizon appear to move horizontally left to right.


Are astronomers continuously monitoring exoplanetary systems?

Exoplanets is an almost over populated field of research now, so yes they are monitored. But maybe not too much beyond what is needed to confirm their existence.

I think you have to humble yourself in the enormity of time. There are some transient events in astronomy, but mostly, watching the sky is like watching a rock. Nothing much changes. Phoebes is crashing into Mars. But not until several million years. It will be a very slow process of it breaking up and parts of it landing one by one. Even in the midst of that event human intuition would think of it as a frozen constant non-event. Like the rings of Saturn. Look at a mountain and see how fast it grows or shrinks. Astronomers cannot see exoplanets crashing into each other in the way I think you mean.


Did comets 266P/Christensen or P/2008 Y2 (Gibbs) cause the Wow! signal?

This hypothesis was put forth by Paris Antonio in the paper "Hydrogen Clouds from Comets 266/P Christensen and P/2008 Y2
(Gibbs) are Candidates for the Source of the 1977 “WOW” Signal".

As you correctly suspected, he tested his theory in 2017 leading to the paper "HYDROGEN LINE OBSERVATIONS OF COMETARY SPECTRA AT 1420 MHZ". In this paper he claimed to have detected a radio signal from the comets and concluded that they were the source of the WOW-signal.

It seems like this paper has mostly been ignored by the community as it is, lets say, not very convincing.
It appeared only on arXiv and the "Washington Academy of Sciences". I am not familiar with the latter one, but judging from the paper, it has not been peer-reviewed.

I will let the following sentence of the paper speak for itself:

  The data collected was saved using the spreadsheet output format option of the SpectraCyber software and
  imported into Microsoft Excel as a text file. The data were then replotted and interpreted using the Chart
  Wizard feature in Microsoft Excel and converted into JPEG format

There is also this response to the paper, which makes very clear why it did not get any more attention:

  This is a statement regarding the claim that the “WOW!” signal was caused by hydrogen emission from an unknown comet or comets. It points out inaccuracies which are fatal to the theory that the “WOW!” signal was caused by a slow-moving comet. It also points out missing details from the author’s paper.

  The staff of the OSU Radio Observatory has examined the paper by Paris regarding the “WOW!” signal (Paris, Antonio. HYDROGEN LINE OBSERVATIONS OF COMETARY SPECTRA AT 1420 MHZ) and comet 266/P Christensen.

  We conclude that comet 266/P Christensen is not the source of the “WOW!” signal for a number of reasons.

  Using the ephemeris at:
  http://ssd.jpl.nasa.gov/horizons.cgi?find_body=1&body_group=sb&sstr=266P%20
  we see that Comet 266/P Christensen, at the time the “WOW!” signal was observed in 1977, was nowhere near the position of “WOW!”. (all positions are J2000).
  “WOW!” was at:        RA 19h 25m 31s  or  19h 28m 22s        Dec -26 deg 57’
  266/P was at:        RA 18h 32m 15s                        Dec -27deg 22’.
  The declination is near but the right ascension is nearly an hour off, placing 266P in a different part of the sky when the WOW! signal was observed.
  Another comet, P/2008 Y2 was closer, at:  RA 18h 39m 39s.        Dec -29 deg 38’
  The declination was almost 3 degrees off and the right ascension 47 minutes off, placing P/2008 Y2 in a different part of the sky when the WOW! signal was observed.
  It is important to point out that the OSU telescope beamwidth in right ascension was only about 3 minutes (of time), but the comets were 47 and 55 minutes (of time) away. The comets would have passed thru the OSU telescope beam roughly 55 and 47 minutes (of time) earlier.
  The observations of the comet made by the author in 2017 near the position of the “WOW!” object are irrelevant, because that is not where the comet was in 1977 when the “WOW!” signal was observed.
  The author does not cite any references regarding observations of hydrogen emission from comets. We have contacted a comet expert and a hydrogen expert and they are both unaware of any hydrogen emission ever having been observed from a comet.
  The author does not cite any references of variable emissions from comets. Such variability would have to be incredibly unusual to match the “WOW!” signal. The signal would have to be exactly constant at 30 sigma for 2 minutes since it matched the OSU antenna beam exactly. It would also need to be less than 0.5 sigma for 2 minutes, at a time 3 minutes earlier or later, since it did not appear in the other OSU beam.
  The author does not show any specific frequencies in the spectrum plots so it is impossible to compare them with the “WOW!” frequency, or to demonstrate that the comet signal bandwidth is less than the 10kHz bandwidth of the WOW! signal.

  Dr. Robert S. Dixon
  Director, Ohio State University SETI Program

So it seems that this hypothesis was already excluded before it was tested. This is already briefly mentioned in the article you linked to.
And while it is not clear (to me) what exactly he measured during his observation, it was most likely not the same thing that caused the WOW! signal.


Can black holes be a subset of quark stars? If not, what are black holes made of?

Black holes are not made of anything... at least nothing in our current understanding of reality.
We understand black holes in terms of General Relativity, the theory of gravity.  In this theory there are some "vacuum solutions"  These are solutions that have no matter at any point in spacetime.  One solution is "flat space", but another solution is a solution with a singularity in time and space.  This solution was discovered by Schwatzchild, and for long was thought to be a mathematical curiosity but not physically real.
But it was later discovered that when a star collapses under gravity it reaches this state.  All the matter falls towards the singularity and we can't describe what happens later than the singularity in time. From the outside it seems as if all the mass of the star is concentrated at one point, but there is actually nothing inside, just the distorted spacetime.
In General relativity nothing can exist in a black hole. For anything falling beyond the event horizon, all directions lead to the singularity.  It is as unavoidable as "next Wednesday". No ball of matter could exist. In normal space, two particles can remain separate. Inside a black hole this isn't possible, since both particles end up at the singularity in a finite (and short) amount of time.
Now we aren't certain if this truly describes reality, because we lack a theory of gravity that takes account of quantum mechanics. It might be that the singularity is real. It might be that something happens (with strings or something) that prevents the ultimate collapse to a singularity. Whatever it would be outside our current models of reality.  Not "quarks".


The only light you would see, would be from the stars outside the black hole. Any light that is generated inside the event horizon would be refracted towards the singularity. You wouldn't see the object in front of you since the light would not reach your eye.


How to calculate the current distance between two planets

There are many different ways of doing it.

One way is to use the Keplerian elements and their rates obtained from the JPL Solar System Dynamics web site to determine the Cartesian coordinates of the planets in the ecliptic plane at a given time. Information and data to allow computation of approximate positions for the planets is provided in the document "Keplerian Elements for Approximate Positions of the Major Planets" by E.M. Standish.

Then for each of the planets, $i$, we have the position coordinates in the form
$$
\mathbf{r}_i =
\left[\begin{array}{l}
x_i \\ y_i \\ z_i
\end{array}\right]
$$
Since these coordinates are all given in the same reference frame, the displacement vector from planet $i$ to planet $j$ can be obtained from
$$
\mathbf{r}_{ij} =
\left[\begin{array}{l}
x_{ij} \\ y_{ij} \\ z_{ij}
\end{array}\right] =
\left[\begin{array}{l}
x_j-x_i \\ y_j-y_i \\ z_j-z_i
\end{array}\right]
$$
and the distance between planet $i$ and $j$ can be calculated using
$$
r_{ij} = \|\mathbf{r}_{ij}\|
= \sqrt{x_{ij}^2+y_{ij}^2+z_{ij}^2}
$$
The method described in the document to obtain the ecliptic coordinates are very easy to implement in software.


What does it mean when we say that information is potentially recoverable from a black hole

The first thing to understand is that we have no actual knowledge of how information and black holes interact.

What we do have are theories that are well supported by experiment (General Relativity, which predicts Black Holes but which says nothing about information) and Quantum Mechanics (which says a lot about information (in this technical sense) but nothing about BHs.)  We also have a variety of theories, none of which have any substantial experimental support, and all of which are incomplete (which means we don't know if there's actually a viable theory there -- all we have today is bits and pieces that look promising).  One of the goals of all of these theories is to -- somehow -- meld GR and QM into a single coherent theory with greater explanatory power than the two separately.

It is very fair to say that some of the work which has been done to date which tries to link information and BHs is plausible and (at least) does not contradict any experimental evidence we have.  And some of them (like what Hawking used in the prediction of Hawking radiation) are such a careful application of QM in the GR context that most physicists would be surprised if they turned out to be incorrect.

The bottom line is that GR and QM are both amazing theories which yet cannot be completely correct -- they must both be subsumed in some more encompassing theory. (Or at least everyone is convinced of that.  I am too, BTW.)

Due to the mathematical underpinnings of QM and GR, physicists vaguely can see how some generalization of QM might, possibly be able to include GR someday.  But virtually no one can see how GR could be extended to include QM.  So the smart money -- just about all the money, actually -- is on extending QM in some way.

One of the most basic features of QM is that it is linear and preserves information (in a certain well-defined, but technical sense).  If GR could be derived from some extension of QM, we'd expect that that extension of QM preserves information, and that GR also must preserve information.

On the face of it, GR by itself says that information which falls into a BH is lost. Zap. Gone. Never to return.  Therefor finding ways to get GR to preserve information may be an important clue to the ultimate combined theory.  And extensions of GR which preserve information seem more likely to be correct.

So the paradox is that BHs, which are predicted by our marvelously correct theory of gravity, GR, do not conserve information.  But we have reasons to strongly suspect that BH built from the hypothetical gravitational extension of QM would preserve information.  Something's wrong here!

One of the plausible theories that have been developed to resolve this says that Hawking radiation -- which seems very plausible, and which carries energy away from a BH in contradiction of vanilla GR -- might also carry the "lost" information away, eliminating the paradox.

Maybe.

But remember that there isn't any experimental evidence for any of this.  It's all based on smart hard-working, competent scientists trying to use their brains and their intuition to guess what the undiscovered theory which merges GR and QM will predict and to turn it into consistent math.  It's certainly better than a random-assed guess, but it is a far cry from established science.

Keep watching -- I hope that all this work will bear fruit and a real theory with experimental evidence behind it might emerge in our lifetimes.


Plotting Galactic Longitude from 180 to -180

Assuming

x, y = rrl_pm.l, rrl_pm.b

The transformation you want is

xx = [(q+180)%360 - 180 for q in x]

Add 180, do modulo 360, then subtract 180.

Then set your limits

ax.set_xlim(180, -180)

import matplotlib.pyplot as plt

x = (0, 10, 20, 40, 80, 160, 200, 280, 320, 340, 350)
y = (0, 10, 20, 30, 40, 50,  -50,  -40, -30, -20, -10)

fig = plt.figure(figsize=(8,5))
ax  = fig.add_subplot(1,1,1, aspect='equal')

ax.scatter(x, y)

ax.set_xlim(0, 360)
ax.set_ylim(-90., 90.)

plt.title('Old', fontsize=18)

plt.show()

xx = [(q+180)%360 - 180 for q in x]  # DO THIS!

fig = plt.figure(figsize=(8,5))
ax  = fig.add_subplot(1,1,1, aspect='equal')

ax.scatter(xx, y)

ax.set_xlim(180, -180)
ax.set_ylim(-90., 90.)

plt.title('New', fontsize=18)

plt.show()


We could define a distance $ds'$ in four-dimensional space-time as
$(ds')^2 = (dx)^2 + (dy)^2 + (dz)^2 + (cdt)^2$
This would be a Euclidean distance. The problem is that experimental observation shows that observers in different inertial reference frames disagree when they measure this $ds'$ distance between the same two points (events) in spacetime. This suggests it is not a physically meaningful value, since it depends on how the observer is moving.
However, if we use the Minkowski distance $ds$ defined by
$(ds)^2 = (dx)^2 + (dy)^2 + (dz)^2 - (cdt)^2$
we find that observers in different inertial frames agree on the value of $ds$ (at least if we restrict ourselves to flat spacetime), so this is an objective value (independent of how the observer is moving) and so more useful.
An everyday analogy is that when talking about locations in England (say) it is more useful to use the objective measure "distance from the centre of London" than the subjective measure "distance from where I am right now".


Two lights roughly 5 minutes apart about as bright as Mars or maybe Saturn

According to CalSky (https://www.calsky.com), there were two Iridium flares visible low in the south-east from Minneapolis at 21:18 (Iridium 60) and 22:28 (Iridium 55) on that evening. Both would have been about 15 degrees elevation and both in the right ballpark for brightness.

Iridium flares occur when the rising / setting Sun's rays bounce off the antennae or solar panels of one or other of the Iridium communications satellites. By their very nature they occur around dusk or dawn. They'll flare up, persist for a few seconds and fade away.


A little reading and effort is recommended and the Wikipedia page is straight forward enough, if you're not completely math-phobic.

But if you want an answer, Rob Jeffries 9,500 km is right, however, that's center point to center point.   Surface to surface, that would only be less than 2,000 km, which would be crazy close.   That close the Moon would take up about (guessing) close to 50% of the sky, maybe 70 degrees or so, when directly overhead.    But that 2000 km number doesn't take into account tidal stretching or rotation, as the Moon would almost certainly be tidally locked and rotating in sync with it's orbit like it does now and the Earth would be similarly tidally locked and have a tidal and gravitational bulge in the direction of the Moon.

Taking into account rotation, that pushes it out to the cube root of 3, not 2 or about 14% further, but that's center to center.  The surface to surface distance would be increased to about 3,000 KM.   Still insanely close.   Any orbital eccentricity would increase instability though, so for a safe distance, Earth to Moon where the Moon wouldn't break apart at all, but it would still stretch measurably, I wouldn't want it any closer than about 5,000 or 6,000 km from the Earth's surface.   That's a ballpark guess with tidal forces, tidal locking, solar perturbations and orbital eccentricity all in mind and even there, some loose moon dust and rocks might regularly blow off and shower onto the Earth.   6,000 KM (surface to surface) is less than twice the Moon's diameter, so it would still be enormous in the sky, but I think that's about the distance where the Moon would hold together in orbit.   The math says even closer, but that doesn't take into effect various issues like eccentricity, rigidity, etc.   It gets rather hard to calculate with rock solid accuracy taking into account all the details.  But, point is, the Moon would need to be stupid/gonzo/cray close before it begins to break apart.   Some 50-60 times closer (surface to surface) than it is now.

Now as far as escape, the math of the Hill Sphere or Sphere of Influence  is pretty straight forward, but the long-term stable zone is about 1/2 to 1/3rd of the Hill Sphere calculation, some 500,000 to 750,000 km.   Somewhere in there or perhaps just beyond that range the Moon probably becomes unstable and escapes the Earth's orbit becoming a very large and potentially very dangerous near earth object.   That doesn't mean that the Moon would just fly away past 750,000 km.  It would remain in orbit for multiple orbits, hundreds, maybe thousands even past that distance, but past the stable region it would probably destabilize over time and escape eventually.

The Moon is perhaps as much as 70% of the way towards where it could eventually escape the Earth, but that final 30% is a LOT.  The moon isn't going anywhere anytime soon.  It's still comfortably inside the long-term stable region and should remain orbiting the Earth for billions of years.

Hope that helps.


Does anybody know date of publishing of following spirit rover image in NASA apod?

It was posted on November 26, 2002, with the caption "Name This Martian Robot."
There was an essay contest to name the two Mars Exploration Rovers to be launched in 2003.
The winner was Sofi Collis, a third grader in Arizona.