Are harmonic relationships among the rotation periods of the planets well known? Are they factors in other planetary phenomena?

I'm very sorry for you, but this is just (mathematical) probability at work and no (astronomical) pattern. I'm on mobile right now so I can't do a full analysis, but note that if you divide Mercury's period (3520.2) by 89 instead of 88, the result is 39.55. Still pretty close to 39.88. In other words, the larger the numbers get, the easier is to find a 'divisor' close to 39.88.


At Mars' equator how long does twilight last?

According to https://mars.jpl.nasa.gov/MPF/science/clouds.html twilight lasts up to two hours because of dust high up in the atmosphere scattering the sunlight. Mars Pathfinder is in Ares Vallis, 3°North.


Difference in energy released in stellar mass black hole merger and supermassive black hole merger

Too long for a comment, but it's an incomplete answer.

The gravity waves detected were from a stellar mass binary black hole merger, sometimes abbreviated bbh for binary black hole.

The two black holes are thought to be about 36 and 29 solar masses with a final combined mass of about 62.  So, roughly 5% of the black hole mass turned into into gravitational wave energy.   Source.

and from Wikipedia article (which can be taken with a grain of salt)

  As the orbiting black holes give off these waves, the orbit decays,
  and the orbital period decreases. This stage is called binary black
  hole inspiral. The black holes will merge once they are close enough.
  Once merged, the single hole settles down to a stable form, via a
  stage called ringdown, where any distortion in the shape is dissipated
  as more gravitational waves.3 In the final fraction of a second the
  black holes can reach extremely high velocity, and the gravitational
  wave amplitude reaches its peak.

  The existence of stellar-mass binary black holes (and gravitational
  waves themselves) were finally confirmed when LIGO detected GW150914
  (detected September 2015, announced February 2016), a distinctive
  gravitational wave signature of two merging stellar-mass black holes
  of around 30 solar masses each, occurring about 1.3 billion light
  years away. In its final moments of spiraling inward and merging,
  GW150914 released around 3 solar masses as gravitational energy,
  peaking at a rate of 3.6×1049 watts — more than the combined power of
  all light radiated by all the stars in the observable universe put
  together.3[4][5] Supermassive binary black hole candidates have been
  found but as yet, not categorically proven

Supermassive bbh are thought to have been observed, see here and we may see one in our lifetime, not sure how often they happen.  But beyond that, I have no way of knowing how much more energetic a supermassive BBH merger would be other than to say, I would think, a whole lot, as it would have to scale upwards by some formula.

With the case of Andromeda and the Milky Way, we're talking Andromeda's 100 million solar mass black hole and the Milky way's 4-6 million solar masses.  Maybe someone here can estimate, but if the energy output is even 5%-10% of the smaller of the two, that's still a few hundred thousand solar masses of energy, an absolutely crazy energy output.  Even if it's far smaller than a few hundred thousand solar masses, it's still enormous energy.

I'm also not sure if it's a dangerous form of energy output as it basically squashes things back and forth maybe a couple times before returning them to normal.  I have no idea what the safe distance is from that kind of merger and gravity wave creation.    Interesting question though.  Fun to think about.


How come the northern and southern lights are the same?

As most people are aware when they play with bar magnets, magnetic poles repel magnets of the same polarity and attract opposites. Magnets have two polarities: north and south.

Electric charges are classified as positive and negative, and also attract opposites and repel charges of the same sign.

However, magnetic poles and electric charges do not match up in the way your question suggests - it's not like North is positive and South is negative and they attract electric charges. Electric and magnetic fields are connected and interact (in fact, both are produced by moving electric charges) but the signs/polarity do not interact in the way you are imagining..

Aurorae (northern and southern lights) are produced because charged particles are streaming toward the Earth all the time in the solar wind. Charged particles (positive or negative) are blocked by magnetic fields - the easiest way to imagine this is to draw the Earth's magnetic field like a bar magnet and the charged particles cannot cross the field lines. At the poles the field is not parallel to the Earth's surface (blocking charged particles) but perpendicular to it, so the charged particles can move into the atmosphere without crossing magnetic field lines. In the atmosphere they interact with particles in the air and create the beautiful lights that we see.

So, in short, the lights are the same at the north and south poles because it is the same kind of particles interacting with out atmosphere.


In the book Galactic Dynamics by Binney and Tremaine (second edition) there is a whole section explaining the difference between the Jacobi radius and the tidal radius (page 677-chapter 8).

Here, $r_J$ is defined as:

  $$r_J=  R_0\left(\frac{m}{3M}\right)^{1/3}$$
  The Jacobi radius $r_J$ (also, Roche or Hill radius) of an orbiting stellar system is expected to correspond to the observational tidal radius, the maximum extent of the satellite system. However this correspondence is only approximate.

Five reasons are subsequently given to explain why the correspondence is only approximate.


How can we tell that the Milky Way is a spiral galaxy?

There are several clues.
The Milky Way is a flat disk
The first one, and the simplest one, is that we live in a disk. As you can see on images like the ones from the 2MASS survey in the infrared range:

We clearly see that we are inside a disk, since what we see when we look around us is a disk seen edge-on. It does not look like an ellipsoid or any other shape; it looks like a disk.
The Milky Way has arms
The other strong clue we have is the velocity of stars of the Milky Way, we can measure. First, we see an overall rotational motions, and, when you look closely to the patterns (like in the following image, that is observations in the radio range of the Milky Way at Green Bank Observatory), you can deduce that stars are even ordered into spirals:


What caused the comet ISON to disintegrate?

ISON slingshot past the Sun close enough (around one Solar diameter) for its surface temperature pointed towards it to reach 6,000 °C and it wasn't a fast spinner either to help this temperature average out along its total surface. In a nutshell, it melted to its nucleus into a hot gaseous plasma of materials it was composed of long before it reached perihelion. These materials still followed more or less same ballistic trajectory prior to perihelion, it merely elongated substantially due to strong gravitational influence of the Sun.

As it was approaching perihelion, the Sun's tidal forces broke this plasma into a wave-like pattern, increasing its surface area and exposing its materials to solar winds (and possibly hit by one CME, Coronal Mass Ejection that was detected near comet ISON on the perihelion day), changing orbit of different mass and size particles differently.

This was apparent past perihelion when the comet failed to create tails pointing away from the direction of solar winds like comets with a nucleus do, but instead started splitting into two  masses inclined roughly 90° to each other, suggesting different mass particles separated and the comet was indeed torn apart by Sun's gravity.

Since most of its materials thus gained angular momentum different enough to negate its own gravity, and they weren't bound to each other as they would if it remained solid, it entered a spin of death, slowly evaporated and dissipated entirely. No nucleus was observed to survive the perihelion, end even Hubble wasn't successful in finding any of its remains, further proving this fate of the comet ISON.

This is also additionally described with the help of ISON perihelion videos from various spacecraft in this Emily Lakdawalla's blog post, and since a picture tells a thousand words, here's my capture of a frame from SDO (Solar Dynamics Observatory) that matches the time, estimated size and distance, suggesting it could indeed be comet ISON being torn apart by tidal forces on its perihelion:

   A frame from SDO/AIA - 4500 203/11/28 18:00:07 (ISON pre-perihelion time). Click on image to download full frame JPeG.

This SDO frame (SDO/AIA - 4500 203/11/28 18:00:07) seems to have been missed by all other astronomers that were following the demise of ISON, likely because SDO was just positioning itself for ISON's perihelion at the time and this isn't a complete frame. So, in a sense, this is Astronomy Stack Exchange exclusive. :)

Update: Turns out it was a reflection / filter defect after all. I would like to point out that I never claimed this SDO/AIA frame shows comet ISON, merely that it seemed to me that it might. Position did seem odd, but not being equipped with all required SDO data, I had no way of confirming that. Clearly, I'm not experienced in interpreting SDO data, let alone a qualified expert, but my perseverance did pay off regardless and Dean Pesnell of the SDO team kindly looked into this and wrote an eye-opening explanation for this SDO/AIA frame in his SDO blog post Was Comet ISON Seen in AIA 4500? A polite debunking of my suspicions, an interesting read, and another win for science over one layman's speculation. Kudos to Dean Pesnell and the NASA SDO on Twitter!


What may be the period and speed of revolution of the star that is closes to galactic centre?

Star S2 reaches maximum velocities of 5,000 km/s according to this ESO page.

The orbital period is given as a little over 15 years and this paper gives a peri-center (closest approach to the black hole) of 17 light hours.

There's another star in the area called S0-102 which has a shorter period, but its orbit is less eccentric - so its closest approach to the black hole is further out at 36 light hours. Its maximum velocity will probably be less than S2 but I haven't been able to find a figure.


In traditional constellation shapes / outlines, yes. But the International Astronomical Union (IAU) standardised on a set of 88 constellations in 1922, and from 1924 to 1930 formalised the constellation boundaries, splitting the sky into separarate areas (that cover the complete sky) so there's no ambiguity now.

For example, before the constellations were standardised in 1930, the same star (on the boundary of Auriga and Taurus) was known both as Gamma Aurigae and Beta Tauri.

And the traditional constellations of Ophiuchus (The serpent bearer)  and Serpens (the snake) (Where Serpens ran across Ophiucus) was handled by splitting them into Ophiucus and Serpens Caput (snakes head) and Serpens Cauda (snakes tail).

see http://www.ianridpath.com/boundaries.htm for more details.

Different cultures have different traditional constellations. If you have a look at the free Stellarium program (see http://stellarium.org/), that has a set of overlays showing constellations from a number of different cultures. (I don't remember if that's part of the standard installation, or an optional extra you can download).


No. One explanation for the equatorial ridge on Iapetus involves a now-lost subsatellite, so perhaps such objects existed in our Solar System's past.

Subsatellites tend to get rapidly removed by tidal forces causing changes to the orbit. Depending on the parameters of the system, the orbit may decay until the subsatellite collides with the moon or breaks up at its Roche limit. Alternatively it may migrate outwards until the subsatellite becomes unbound: at this point it would become a moon of the planet, though likely on an unstable orbit that would lead to scattering or a collision. The orbital evolution tends to be faster for more massive subsatellites. See Kollmeier & Raymond (2018) for details, who note that any subsatellites orbiting the candidate Neptune-sized exomoon in the Kepler-1625 system would need to be smaller than Ceres to survive.


As per the above comments, there will still be v=Hd, if v is the rate of change of current distance, and d is the current distance-- as long as the universe continues to obey the cosmological principle.  The nearby version of that will still look like correlating Doppler-shift velocity to distance indicators, and the faraway version will require a dynamical model, albeit a very simple one due to dark energy.  It will lead to a constant H that will not be anything like the inverse of the age of the universe.  And if you go too far into the future, you won't be able to see the CMB, or quasars, so alien cosmology will really struggle.


What you're asking about is not known at the moment, but I can briefly comment on how flawed our knowledge is.

1.) Theoretical predictions:

As one comment mentioned, the question "where do planets end up?" is dependent on the answer to "where do planets form?" and the answer to "how fast do planets migrate?" which is not well understood at the moment.
There are of course predictions for both those processes, but those essentially don't agree on anything and especially not with the observations.

A recent overview and a very short video by the authors of that paper summarizing it.

2.) Observations:

We have only very few multi-planet systems, and not a single one where we can claim we know all planets down to a certain size or mass limit.
I suggest you take a look at the data on http://exoplanets.org/table and use the filter "NCOMP >= 5" (also sort by name), this way you get current candidate systems with equal to or more than 5 planets.
You will see there are not many systems that we can base any systematic understanding on.

Plots like the first one in the answer to this question that seem to indicate that we know quite many systems, hide the fact that those systems are all single or double planet-systems.

This prohibits basing your generated systems based on data or planet formation theory. But I suggest that you can use a different method:

3.) Long-term stability:

We don't know all systems that are possible, but we certainly know that all systems under certain conditions become unstable. There is an overview article or also a long video outlining that there should be a simple stability criterion for planetary systems:
If any two planets come closer to each other than ~10 mutual Hill-radii, their orbits will become unstable and possible send the whole system into chaos.
The mutual Hill radius of two planets with masses $m_1$ and $m_2$, and central star $m_s$ and semi-major axis distances $a_1$ and $a_2$ from the star is

$$R_{hill,mutual} = \left( \frac{m_1+m_2}{3 \cdot m_s} \right)^{1/3} \cdot \frac{\left( a_1 + a_2 \right)}{2} $$

So my idea of a simple solar-system generator would be to choose some first planetary mass, place it somewhere.
Then build up the star system from the inside out planet by planet, by chosing the next planetary mass from one of the above mentioned 1-or-2-planet distributions (this one could be a realistic one, like in Mayor et al. 2011, if you know how to re-shape random distributions).
You calculate $a_2$ for them such that their mutual distance is $> 10 \cdot R_{Hill,mutual}$.
In this way you'll probably get pretty packed systems, but you can of course take their mutual distances to be larger than $10 \cdot R_{mutual, Hill}$.

So in this way the properties of your generated star systems would be:

Mutual distances: Stable physical foundation
Planetary mass distribution: Would be on shaky feet, as we don't know enough about multi-planet systems, and our data on low-multi systems is heavily biased towards the methods used to detect them.
System architecture / order of the planets: So far we've seen many unexpected things and there is no data or reason to assume there should be any strong constraints. I think here you can let your fantasy or the random masses generator roam free.

You could even go one step further and include one of the (half-way) realistic mass distributions for stars, to generate your $m_s$ values.

Your code (e.e. if you code in C) will only be able to give you uniformly distributed random numbers. If you want them to follow a specific distribution, you need to transform them. Google "transform probability density function" or ask someone with math education that you know, if you don't know how that works.


Is there an objective difference between space expansion and reduction in speed of light

In physics the "speed" of anything depends on the coordinate system you choose, since speed is measured as change in coordinate position in some interval of coordinate time. Even in the special theory of relativity, which doesn't take into account gravity and hence involves no spacetime curvature, the notion that the speed of light is always equal to the same constant (labeled $c$ in physics and astronomy) would only true in a special class of coordinate systems known as inertial frames, it is quite possible to define a "non-inertial" coordinate system in special relativity such as Rindler coordinates in which the speed of light does not have the same value $c$. In the general theory of relativity, which models gravity in terms of mass/energy curving spacetime, you can only have "local inertial frames" defined on very small patches of spacetime (specifically, the limit as the size approaches zero)--see this article on the "equivalence principle" for the conceptual details of how local inertial frames can be defined by observers in freefall measuring events in their immediate neighborhood (like an observer looking at events within an elevator in freefall). Such observers will always measure the local speed of light within a vacuum in their local region to be equal to $c$, regardless of the larger-scale properties of the spacetime they're embedded in, like the "expansion of space".

But if you try to define a global coordinate system on a large region of curved spacetime, this coordinate system is always a non-inertial one, so there is no guarantee that the coordinate speed of light in this coordinate system will be equal to $c$, and indeed the coordinate speed of light may vary from one region of spacetime to another depending on what coordinate system you choose (the equations of general relativity work in all smooth coordinate systems, as long as you define the metric correctly relative to your chosen coordinate system). In the basic model of curved spacetime in cosmology (the FLRW model), the simplifying assumption is made that matter is a sort of uniform fluid filling all of space, so that if you pick the right definition of simultaneity (multiple definitions are always possible in relativity due to the relativity of simultaneity), you will find that the density of this fluid is identical at every point in space at any given moment of cosmic time. This obviously isn't completely true to life, but it's expected that on large scales the density of matter is close to uniform at any given cosmic time, so it's seen as a reasonable approximation. The expansion of space basically means that the density of the fluid gets lower as time passes, and that if two objects are at rest relative to the local fluid in their immediate neighborhood, then the proper distance between them will grow with time (proper distance corresponds to what you would measure if you laid a bunch of short rulers end-to-end between the two objects at a particular moment in time, and then added up the distances).

As it happens, this cosmological model has a further nice feature (as discussed in the 'proper distance' link above which is based on this paper, see p. 99 of the 'Full Refereed Journal' link). The most "natural" coordinate system to use in this model is one in which the time coordinate corresponds to the proper time measured by a set of observers who have been at rest relative to the cosmic fluid since the big bang, and the spatial coordinate is such that the coordinate distance between any such observers at a given time corresponds to their proper distance at that time. If you use such a system, it works out that the overall coordinate velocity of any object can be broken down into a sum of two velocities:

The "recession velocity" at any given space, which is the velocity that an observer at rest relative to the cosmic fluid would be moving (the rate that their proper distance from the origin of the coordinate system is growing as a function of time, where we can assume the origin corresponds to our own location in space).
The "peculiar velocity" of any object which is not at rest relative to the cosmic fluid, which is just the same as the velocity of that object as measured in the local inertial frame of an observer at the same location who is at rest relative to the cosmic fluid. So, the peculiar velocity of a light ray must always be $c$.

So, if we know the recession velocity $v_{rec}$ at some distant location in space, then a light ray emitted directly towards us from that location will have an overall velocity $v_{rec} - c$ in this coordinate system, and a light ray emitted directly away from us will have an overall velocity $v_{rec} + c$. So from the perspective of this coordinate system, it makes sense to say as a shorthand that the light itself always travels at $c$, but space is also expanding away from us and this accounts for why the light is redshifted, and also why light originally emitted at distance $d$ won't necessarily take a time of $d/c$ to reach our own location. But this neat way of describing things is specific to both the cosmological model being assumed and the coordinate system used, things may not work out so neatly in other choices of spacetime or other coordinate systems. The only really general statement you can make about the speed of light is the one I mentioned earlier, that regardless of what global coordinate system you use and what the speed of a light ray works out to be in that system, it's always true that in a local inertial frame defined on a small patch of spacetime, light traveling through that patch always has a speed of $c$ as measured in that local frame.


Does conservation of energy make black holes impossible?

For simplicity, let's consider a Schwarzschild black hole, so that the spacetime is spherically symmetric and static. In particular, the Schwarzschild time $t$ coordinate gives a direction in which the geometry 'stays the same' (a Killing vector field), and its inner product with the orbital four-velocity $u$ is conserved:
$$\epsilon = -g(u,\partial_t) = \left(1-\frac{2M}{r}\right)\frac{\mathrm{d}t}{\mathrm{d}\tau}\text{,}$$
where $\tau$ is the proper time of the orbiting particle. One can think of this as the specific (per-mass) energy, including mass-energy: a particle escaping to infinity that becomes asymptotically at rest with respect to stationary observers would have $\epsilon = 1$.

  If we consider escape velocity for a particle at the event horizon, it has an escape velocity of the speed of light. The energy required to achieve this escape velocity would therefore be infinite.

The energy as measured by whom? Imagine a family of stationary observers everywhere surrounding the black hole, or at least along an infalling particle's trajectory. Those observers measure the speed and energy of the particle as it falls past them. As the particle nears the horizon, they will report speeds arbitrarily close to the speed of light and arbitrarily high energies.

But to a far-away stationary observer, those observers near the horizon are experiencing increasingly divergent gravitational time dilation, and the orbital energy of the particle stays constant. Flinging a particle of mass $m$ into the black hole will increase the mass of the black hole by $m\epsilon$.

  What's the critical flaw in this argument?

All energy gravitates, so I would say the main flaw is forgetting to add the gravitational potential energy of the particle to the mass of the black hole as well. Or better put, you should be concerned with the orbital energy, not just its mass+kinetic parts ('relativistic mass').

In this situation, you have a well-defined conserved orbital energy. If you insist on measuring the Lorentz factor according to close-by stationary observers, then yes, it diverges to $+\infty$, but then you would have to admit a gravitational potential energy term that diverges to $-\infty$, because their sum must resolve to be the orbital energy.


What symbol is used for the moon when doing calculations?

There is no authorative standard - you can use what you want, as long as it is clear what you mean. Besides that: yes, ☽ is reasonably standard. Often also written indices like $M_{moon}$ or $M_m$ are used.

Also see Wikipedia with its many references.


Why is the statement "black holes make up 1% of the mass of the Milky Way Galaxy" unknown?

(a) We can't detect ALL of anything, but we can certainly makes estimates of their integral properties. For example, we estimate that there are about 100 billion stars in the galaxy, but we can observe only a tiny fraction of these. We have also observed a tiny fraction of the stellar black holes in our galaxy - those few examples that have revealed themselves through their binary companions.

(b) Yes we do. It is just over 4 million solar masses and thus a negligible fraction of the Galaxy's total mass.

(c) and (d) It is not just stellar evolution that is involved, it is the distribution of stellar birth masses (the initial mass function) that needs to be known, possibly as a function of time.

The stellar evolution part has a major uncertainty in that although it is reasonably certain that all stars above about 10 solar masses end their lives in supernovae, it is not certain what fraction will leave black hole remnants. This is probably a strong function of mass - higher masses favour a black hole and progenitor metallicity; low metallicity stars lose less mass during their lives and are more likely to directly collapse to a black hole - see Heger et al. (2003).

You can do a back of the envelope calculation. Suppose the IMF has the Saltpeter form $N(M)\propto M^{-2.35}$ between 0.1 and 100 solar masses. Further assume that: because a 1 solar mass star has a lifetime of 10 billion years, the age of the Galaxy is 10 billion years and the stellar lifetime is a very strong inverse function of mass $(\propto M^{-2.5})$, that the 100 billion stars that we deduce exist in our Galaxy (by counting up the local stars and measuring stellar densities elsewhere and extrapolating) are predominantly all the 0.1 to 1 solar mass stars that have ever been born. This allows us to then estimate how many stars have been born in any mass range - even if they have long-since died.

So making the above assumptions, I will make the further assumption that all stars from 20-100 solar masses die and leave a black hole remnant.Then if my maths is correct then there are about $10^{11}/1400 \sim 7\times 10^{7}$ black holes in our Galaxy. That is there is 1 black hole for every 1400 stars of 0.1-1 solar mass.

The remaining parts of the puzzle are what is the mass of a typical black hole and what is the mass of the Galaxy? A good number for the former is about 10 solar masses, since the black hole binaries appear to cluster at this value or a little lower (Ozel et al. 2010). The latter is still a topic of investigation. It appears to be an order of magnitude bigger than all the stars, gas and dust due to some form of dark matter. Let's use the number $7\times 10^{11}$ solar masses (Eadie & Harris 2016).

So my final number is that black holes form $7\times 10^{7}\times 10/7\times 10^{11} = 0.1$ percent of the Milky Way mass.

You could push this number up by a factor of a few by assuming that the IMF contained more higher mass stars in the past, or that the threshold mass for black hole production was lower in the past for low metallicity stars. Both seem theoretically possible and are active topics of investigation.


Explaining Dark Matter and Dark Energy to layman

Dark Matter
Your understanding of dark matter isn't bad, but here's a few clarifying details.
Orbits: The speed of an object's orbit is related to 2 things: the radius of its orbit and the mass inside of it. In the solar system, over 99% of the mass is concentrated at the centre, so radius is the dominant effect on orbital speed. As we look at planets further away from the sun, their orbital speed decreases. In the galaxy, it's a similar story, but with increasing orbital radius, you're also getting more and more mass inside of your orbit since the galaxy is full of other stars. Even so, we would expect orbital speeds to drop as you look at objects toward the edge of the galaxy, since the stars get more diffuse as you move out and not enough mass is inside the orbit to compensate for the increasing radius. Instead, we see that after a certain point, the orbital speeds stay the same, suggesting that there is more mass there than we can account for with stars and other visible objects.

Here is an example of a so-called rotation curve of the spiral galaxy NGC 3198:

Without dark matter, we would expect the speed of objects as a function of radius (from the center of mass) to follow the curve labeled 'disk'. However, what we see appears to be the sum of the two contributions (the disk and the 'dark matter halo' which surrounds this disk), which is overplotted with actual data.

We also have evidence of this extra mass from gravitational lenses. As light passes by a large object like a galaxy cluster, its path can get bent by that object's gravity. We can see this effect when a galaxy cluster is in front of an even more distant galaxy, the light from the background object gets magnified and distorted. We can calculate the path the light must have taken to appear as we see it, and the mass needed to bend it that way can't be accounted for with just the stars and gas we see in the lensing cluster. This again suggests that there is extra mass that we can't see.

Dark Energy
You're not too far off with this one, with one major distinction: Dark energy is the 'force' that's causing the universe to accelerate its expansion. We expect the universe to be expanding from the Big Bang cosmology model, but we'd expect it to be slowing down as the mutual gravity of everything in the universe acted upon each other. Dark energy appears to be a "pressure" on every part of space to expand, but it's very weak and has been dominated by gravity and other factors throughout most of the universe's history. It's only in the past few billion years it's started to take over.

It works something like this. Imagine you have a 1-D region of spacetime 1 unit long (just for convenience's sake). Here it is: |--------|
Now this region has this dark energy 'expansion pressure' acting on it. Let's say it's 0.07 units per year per unit. This means that every year, this region of spacetime gets bigger by 7%. In 10 years, the length will be double:
|--------|--------|
Now the thing is, each of these regions has the same expansion pressure as the original one! So both will double in 10 years, and then those will double, and so on. So what happens is you get a small expansion locally, but the further away something gets the faster it's accelerating away from you. The real effect of dark energy is orders of magnitude smaller, something like 67.15 ± 1.2 (km/s)/Mpc (Wikipedia), but it still means that any galaxies more than 4.5 Gigaparsecs away from us are currently being carried away from us faster than the speed of light. (We can still see them right now because the light we see was emitted long before the expansion rate got that high.) The expansion adds up over the huge distances we see in the universe.

Dark energy doesn't affect things like planets, solar systems, and galaxies because the effect is so weak on small scales that gravity more than counteracts it. It's effect can mostly be seen over the vast empty stretches of space between galaxy clusters.

Hope that helps!


Does the accelerating expansion of the Universe contradict Hubble's law?

The Hubble law gives the velocity of a distant galaxy right now. A galaxy at a distance $d$ recedes at a velocity $v = H_0\,d$ right now$^\dagger$.

However, the relation between $d$ and the redshift — which is the quantity that we observe — is a non-trivial function of the expansion history of the Universe, obtained by integrating the (inverse) scale factor over time, which in turn depends on the densities of matter, radiation, and dark energy.

Analogy

Paint dots on a rubber band, hold onto one end, and let your friend walk away with the other end. It doesn't matter whether your friend changes pace, maybe even stops and walks toward you for a while, then starts sprinting away from you. At any given time, the velocity of a given dot with respect to you is simply proportional to its distance from you.

$^\dagger$Strictly speaking, this is only true if our assumption is true that the expansion rate of the Universe is the same at every location, which is a corollary of the Cosmological Principle (i.e. that the Universe is homogeneous and isotropic).


What are the current observational limits on the existence of Dyson spheres/swarms/rings?

Reading Dyson's original argument gives some useful information. He says that such a sphere would have a surface temperature of 200-300 Kelvin, as it would reradiate some of the energy it absorbs at infrared wavelengths. That's a temperature comparable to cool Y-type brown dwarfs. If we can assume a black body approximation, and a radius of about 1 AU, we get a luminosity for a 200 K sphere of
$$L_{\text{DS}}\simeq4\pi\sigma R^2T^4\sim10^{25}\text{ Watts}\simeq0.1L_{\odot}$$
That's substantially more luminous than a cool brown dwarf of the same temperature. Additionally, Planck's law gives a peak black body wavelength of about 14,500 nm, placing the emissions in the mid-infrared range, as expected. Relatively hot brown dwarfs should be expected to have high infrared emissions. In other words, a Dyson sphere should look something like, from a distance, a large, luminous brown dwarf.

Dyson spheres, in the classic picture of a solid shell enveloping a star, are actually unstable, so it's unlikely that they would ever be built. We should expect other variants - Dyson swarms, rings, bubbles, etc. - that are composed of large arrays of smaller objects are much more likely to be created.

Fermilab used data from the Infrared Astronomical Satellite (IRAS) to look for Dyson spheres as a secondary mission, searching a band in infrared wavelengths centered around 300 K (see Carrigan (2009)), covering 100 K to 600 K. The study started with 250,000 sources over 96% of the sky, then whittled them down to about 6,000 based on temperature range and flux. 17 "ambiguous" candidates were finally left. This is, to my knowledge, the most complete survey yet undertaken.

There have, of course, been others, most using IRAS data. WISE and 2MASS also contributed data. Very few candidates have been found, and it has been estimated that any within about 1 kpc in reasonable temperature ranges (less than 400 K) should have been among these.

There are indeed natural sources that could prove to be false positives. Carrigan lists several:

Dust shells around stars possibly undergoing mass loss
Stars embedded in dusty areas, including dense nebulae
Mira variables, which undergo mass loss towards the end of their lives
Planetary nebulae
AGB stars, which may create circumstellar envelopes, as well as post-AGB stars


Why is this video showing radio waves transmitted from a radio telescope?

I guess it's a mistake in the animation. At radio wavelengths turbulence doesn't affect the measurments anymore and above 200 MHz the ionosphere doesn't play a role yet, so I guess the notion of a guide star is unnecessary here.


Why do gas giants have clearly delineated surfaces, whereas the Earth's atmosphere fades into space?

In an isothermal atmosphere, the exponential scale height of the atmosphere is
$$ h \sim \frac{k_\mathrm B T}{\mu g},$$
where $g$ is the gravitational field, $\mu$ is the mean mass of a particle and $T$ is the temperature (in kelvin).

i.e. The pressure/density of the atmosphere falls exponentially, with an e-folding height given by the above expression.

I suppose what matters when you look at a photo, is how this height compares with the radius of the planet.
$$ \frac{h}{R} \sim \frac{k_\mathrm B T}{\mu g R}$$

Jupiter is half the temperature, 11 times the radius and with 3 times the gravity of Earth. However $\mu$ is about ten times smaller (hydrogen vs nitrogen/oxygen). Overall that means $h/R$ for Jupiter is of order 5–10 times smaller than for Earth and so it will appear "sharper".

EDIT: If you put some reasonable numbers in for Jupiter ($T \sim 130$ K, $\mu=2$, $R=7\times 10^7$ m), then $h/R \sim 3 \times 10^{-4}$. This means even if Jupiter fills a photo that is 3000 pixels across, the atmosphere will be about 1 pixel high.


Which lens and magnification is best to see planets?

Eyepieces

The highest useful magnification of the Celestron Travel Scope 70 is about 165x. However, you are unlikely to find that much magnification to be usable except under very ideal conditions ("good seeing" i.e. clear atmospheric conditions, dark skies with no light pollution, viewing on a night with a new moon, etc). Even then the image at the eyepiece is going to be dim at that magnification.

Any telescope can in theory achieve an unlimited magnification. But in practice, its magnifying power is limited by the amount of light the telescope is taking in, which is to say, its aperture. The telescope is only collecting so much light. Every increase in magnification is zooming in on a smaller and smaller area, which means it is looking at a smaller and smaller fraction of the original amount of light.

On your telescope, the aperture is 70mm. This is the main thing that dictates how bright of an object you can hope to see at the eyepiece. There is another stat that is important to figuring out what eyepieces are likely to show you, though, and that is the telescope's focal length. Your scope has a 400mm focal length. Total magnification is the focal length of the telescope divided by the focal length of the eyepiece.

Your telescope came with two eyepieces:

400mm (telescope focal length) / 20mm (eyepiece focal length) = 20x magnification
400mm (telescope focal length) / 10mm (eyepiece focal length) = 40x magnification

As you can see, the smaller the eyepiece focal length, the greater the final magnification. If you tried an 8mm eyepiece, you'd get a 50x magnification. To get a 65x magnification, you'd need somewhere around a 6.2mm eyepiece.

(Interesting aside: Since the magnification also depends on the telescope's focal length, note that each eyepiece will have a different magnification depending on the telescope it is in.)

Before you go buying any extra eyepieces, though, I'd strongly recommend you go see what you can see with the equipment you already have! You should certainly be able to see Jupiter, Saturn, and Mars.

Jupiter will appear as a disc. You can see its four Galilean moons, Io, Europa, Ganymede, and Callisto, so-named after the first person to see them, Galileo. They appear as points of light around the disc of Jupiter.
For Saturn, you should be able to make out the rings poking out either side of Saturn's disc. They look like tiny little teacup handles. You can also see Saturn's largest moon, Titan.
Mars is a little bit less exciting, but it should be visible as a reddish disc.

If you do opt to buy another eyepiece, one thing to pay attention to is eye relief. Eye relief is how far away from the eyepiece you can be to still see a decent image. Eyepieces with shorter eye relief are cheaper, but they are also harder to use. An eye relief around 6mm to 8mm is much more comfortable for me.

As for me, I can never help myself: for every telescope I've owned, I always buy an eyepiece that pushes the limits of my telescope, just to see what it looks like. But from experience, I can tell you it is not a great view. Still, it can be fun to see what your equipment can do. But before you jump all the way to an eyepiece that gives you 130x magnification (or whatever), I'd say to shoot solidly for the middle range of your scope (in your case, maybe 60x - 80x). If you can afford it, buy a mid range eyepiece with decent eye relief and field of view (FOV). You'll be much happier with the view through this eyepiece than you will an eyepiece that maxes out the magnification.

Filters

Do yourself a favor and at least pick up a lunar filter. They filter a certain percentage of the light and allow the rest through without modifying the color. The moon is so bright, it will make your eyes water trying to look at it through even a modest telescope. Lunar filters make looking at the moon a much more enjoyable experience.


Would an extremely bright star produce same kind of shadow band effect as seen in solar eclipses just before totality?

Yes, the nonuniform illumination pattern would be visible for extremely bright stars.

Venus (mag -4.6) is bright enough to cast visible shadows.
Let's move Sirius, the brightest star in the northern hemisphere closer, to match the Venus brightness:

currently: mag -1.46, 8.6 light years away, 0.006" angular diameter
relative brightness difference: $\sqrt[5]{100}^{4.6-1.46} \approx 18$
required distance factor: $\sqrt{18} \approx 4.24$
at $8.6/4.24 \approx 2$ light years away, Sirius apparent diameter would be $0.006'' \cdot 4.24 \approx 0.026''$ which is still much smaller than planets

Moreover, we can even see planets twinkling sometimes (depending on the seeing), so the "point light source" is not a strict requirement (well, strictly speaking, point light sources do not even exist ;))

Interestingly, there are no mentions of shadowband-like phenomena in the supernova observation history, even though some of them were bright enough to be visible in daylight.

I guess we have to wait for the next Milky Way supernova to be sure...


The National Solar Observatory historical archive lists at least one source of sunspot data. Under 'Sacramento Peak Telescope'->'Evans Solar Facility' there is a Dropbox link to Sunspot numbers data which seems to have spot numbers against time in ASCII format; this may be a suitable resource to get you going.


How can the Kuiper belt coexist with the 9th planet?

The recent evidence for the 9th planet is an alignment in various parameters of known Kuiper belt objects.

These are not Kuiper belt objects. The Kuiper belt is taken to refer to the neighborhood of Neptune resonances. The scattered disc is another population, further out, and the objects relating to Planet 9 are further still.

Legend: grey = Kuiper belt, green = integer Neptune resonances, blue = scattered disc. The objects in the recent announcement are all off the chart, although their perihelions are <100 AU.

Source: Wikipedia.

The shape of the grey splotch at the left of this chart shows that the Kuiper belt is indeed a collection of populations dominated by Neptune. This isn't news, and it's still taken to be a belt. It goes all the way around the solar system and the orbital eccentricities are low.

This postulated planet might help explain the "Kuiper cliff", the otherwise unexpected cutoff before the 2:1 resonance. But, that's another question. Failing that, there's simply no relation between the Kuiper belt and Planet 9 except that they're both very far away.


This is an unofficial explanation, but I doubt you'll get an answer from official sources here, unless you ask them directly (e.g. their Twitter account is @NavyOceans), so I'll give it a shot. The small thumbs appear to be merely approximation, representation icons that aren't resized from the real-time image, but merely prepared in advance images, displayed depending on the current Moon's phase, calculated by the server.
If you try their Moon phase image generator (at the bottom of this page), you'll see why they did it like that. The prepared thumbs are numbered from m180.gif for Full Moon to m360.gif for New Moon, with a 2° step, and apparently the server calculates based on input parameters the Moon phase and displays prepared thumbnail with a minimum approximation precision of 2° (roughly 8 hour step). The same page also offers an explanation for how this thumbnail generation works:

These lunar phase images were created by R. Schmidt from ray-traced
images of the Moon. A Clementine spacecraft mosaic of the lunar
surface was mapped onto a sphere, and scenes were rendered as a
virtual Sun "orbited" the Moon. The depiction of lunar surface
features suffers geometric distortion but the terminator is correct
with respect to the spherical Moon.

The lunar phase calculator offers date selection that goes forward in time or before the time any images of actual observations were made and stored, so it's unreasonable to expect actual images displayed for, say, some date in 22nd or 19th century. Some examples off U.S. Naval Observatory's server:

(source: navy.mil)

(source: navy.mil)

(source: navy.mil)

(source: navy.mil)

(source: navy.mil)
            Full Moon                        Waning Gibbons                     Last Quarter                    Waning Crescent                     New Moon
The larger image of how the Moon looks like now seems to be generated by a lot more precise renderer (image is refreshed every minute by a JavaScript call) that takes into account not merely lunar phase, but also its libration in respect to the surface on the Earth. It would look more like any frame off this animation:

                         Fig. 2: Lunar librations in latitude and longitude over a period of one month (Source: Wikipedia)
And inclination is likely calculated for geolocation of the Astronomical Applications Department of the U.S. Naval Observatory in Washington, DC. The page however doesn't explain if this assumption of mine is correct, but it would certainly appear so.


How does the Large Binocular Telescope resolve so well in both orthogonal directions simultaneously?

The simple answer is that it isn't resolving in both orthogonal directions equally well. The horizontal dimension is the binocular dimension, and from looking at your animation, it looks to have about ~3 times the resolution. The horizontal banding, I'm pretty sure is not ringing, and is in fact, representative of additional information.

This article does a nice job of explaining what is being seen, and it has a picture, both of a simulated depiction of Io from one of the two telescope mirrors, and the combined image.

The thing you should notice from the above image is that the 8.4 m telescope picture is at the same resolution as the vertical axis of your animation.

What is going on with the interferometer image?

By increasing the separation between telescopes, we increase the angular resolution. Just about everyone with a casual interest in astronomy will have learned that fact in this past week.

But the other thing you do, is introduce interference patterns. The two telescopes essentially act like a double slit experiment.
In the vertical (non binocular) dimension, the diffraction can be treated like a single-slit experiment. The angular intensity formula for a single slit is as follows:
$$ I(\theta) = I_0 sinc^2(\frac{\pi b sin(\theta)}{\lambda})$$
where $b$ is the diameter of each mirror, and $\lambda$ is the wavelength of the light.

The horizontal (binocular) dimension acts as a double slit. The double slit formula is as follows:
$$ I(\theta) = I_0 cos^2(\frac{\pi d sin(\theta)}{\lambda})sinc^2(\frac{\pi b sin(\theta)}{\lambda})$$
where $d$ is the distance between the centres of the two telescopes.

When you combine these two together, using the telescopes dimensions that you quotes (b=8.4, d=14.4), you come up with a pattern remarkably close to what you actually see from the telescope.

On the left, screencapture from above animation, on the right, the predicted double slit intensity.

Fringe removal

It seems that the animation you saw is based off unprocessed images from the LBT. Obviously, they have methods for removing the fringe bands. As to how, I have no idea. I found a paper that discusses interferometry in depth and they mention:

  What we therefore have is a series of fringes, whose amplitude is given by the Fourier transform of the source intensity distribution.
  In practice, steps are usually taken to get rid of the fringes using a
  phase rotation whose rate is known (as both B and s are known). This
  is done in optical interferometers by use of accurate delay lines to
  compensate for the path difference, and in radio interferometers by
  the insertion of electronic delays.

But I have no idea what that means. Perhaps some boffin from Physics.SE would be nice enough to answer.


Can Gravitational astronomy look beyond the CMB?

Yes. Gravitational wave observatories like the proposed eLISA laser interferometer may be able to detect gravitational waves that originate from the early moments of the big bang itself. If some part of the big bang energy goes into gravitational waves then those waves will be redshifted by expansion and waves produced in the first $<10^{-10}$ s should be present in eLISA's frequency sensitivity window (Amaro-Seaone et al. 2012; Bartolo et al. 2016; Ricciardone et al. 2017). This might probe exotic physics like inflation, extra dimensions, phase transitions and cosmic strings.


What happens when a star gets hit by a supernova?

There are certain cases where the remaining star survives; this happened to high-mass X-ray binary star systems, pairs which consist of a neutron star or black hole (the remnant of a supernova) and a massive star. The X-rays are produced when some of the material ejected by the star is captured by the supernova remnant.

The most famous of these is Cygnus X-1, the first object widely accepted to be a black hole. The massive star in this pair has the same name as one of our moderators.


Why did the big bang not just produce a big black hole?

Your problem essentially arises from trying to apply Schwarzschild black hole logic, the assumptions of which are pretty much maximally violated at the big bang.
The following were true at the big bang, and violate the usual black hole formation logic.
The event occurred everywhere in space, not actually a point. In particular, the energy was uniformly distributed everywhere. The net gravitational potential was therefore near zero, and there was no one point to which everything could collapse. Furthermore, since stuff was everywhere, there was no expanse of vacuum (in a flat spacetime, no less) outside the collapsing region. And furthermore, things were moving rapidly, were in a highly excited state, and were not in thermal equilibrium (until inflation hit, and then things were too diluted and causally disconnected to collapse en masse).
As for what we could describe the universe as at time $t=0$, we have no idea. General relativity has a spacelike singularity there, and it subsequently cannot say anything beyond that. It is expected that general relativity is not a correct description of spacetime in the pre-inflation era, in large part because quantum mechanics introduces significant effects in those conditions, and it is well known that the two theories are incompatible.
Edit:
This question has been asked on the physics SE many times.
https://physics.stackexchange.com/q/20394/55483
https://physics.stackexchange.com/q/3294/55483
https://physics.stackexchange.com/q/26435/55483
Perhaps the answers there will be illuminating.


How did VLT's adaptive optics obtain this resolution for Neptune? Is it really working in visible wavelengths?

It is an image taken with the new narrow field mode of the MUSE instrument using the GALACSI Adaptive optics module on a single (UT4) VLT telescope using laser guide stars.

I am having a great deal of difficulty (e.g. from this press release) in working out at what wavelength(s) this image was taken. I do not believe that the AO system is working at blue wavelengths and therefore the comparison with HST (which certainly does) is somewhat misleading. I think (from ESO's own instrumentation pages) that the MUSE narrow field mode has zero efficiency at blue wavelengths and the description of GALACSI only talks about AO correction approaching diffraction limited images at 650 nm (i.e. red wavelengths).

The captioning to the press release images found here suggest that the shortest wavelengths used in this false-colour image were 550nm and that most of the detail you see is coming from redder wavelengths (600-920nm).


Could any known, living organisms on Earth survive on Mars?

The most likely candidate would be the Tardigrade. These little guys handle vacuum and radiation just fine. So long as water is provided, according to tests done in LEO the Tardigrade would survive on Mars. Even if they do dehydrate, they spring back to life once water is provided again.


No, stars do not have a rocky core. They start as clouds of gas, become gravitationally unstable and collapse.

Some planets are thought to start this way too - though probably not the ones in our solar system, because the outer three three giant planets have solid cores and Jupiter is likely to also. The two modes of giant planet formation are reviewed in detail by D'Angelo et al. (2010).


Problem about finding distance with magnitude given

Short answer: no, you can't solve the question if all you know is its apparent magnitude. It could be a faint star that's nearby, or a very luminous star that's far away. (The Sun's parameters are irrelevant.)

(I'm ignoring the bit about it being "a star in the Pleiades", which implies that it's a physical member of the Pleiades cluster, in which you already know its distance: it's the same as the distance of the Pleiades cluster. Maybe you just meant "it's in the same part of the sky".)


What we call a black hole is a solution of Einstein's equations of gravity.

We know several exact solutions: Empty and flat space corresponds to special relativity. Other known solutions include that of a point mass, a charged point mass, and a solution, known as the Kerr solution in which spacetime around a point mass is being "dragged"

This last solution can be interpreted as corresponding to a black hole that would form if a rotating mass were to collapse. As you note, a black hole contains no matter: it is a vacuum solution of Einstein's equations. However a black hole does have properties like mass and angular momentum.

The shape of space around the Earth can be approximated the Kerr solution of a spinning black hole, with a radius of about 1cm. We would therefore expect some dragging of spacetime around the Earth. In 2008 the Gravity B probe had measured the gravity around the Earth and confirmed that the Frame Dragging effect predicted was present and its results were consistent with  General relativity. Frame dragging around the Earth is fantastically small and the Gravity B probe was working right at its limit of sensitivity. It causes a gyroscope to deviate from its spin by 0.00001 degrees per year.


How long does an over contact binary star system last?

Short answer: $t \lesssim 10^5\ \mathrm{years}$ (maybe)

An "overcontact binary" is just another way of saying "common envelope binary". The two phrases are exactly the same and it's frustrating that the authors on the VFTS 352 paper decided to create their own convention - as if astrophysical classifications weren't confusing enough!

A contact binary exists on timescales predominantly dependent on stellar evolution, so figuring out how long a contact binary will exist is heavily dependent on the mass, metallicity, and rotation of the primary star among other things.

Deriving the timescale:

Let's keep the scope to systems like VFTS 352, where the primary is massive and the binary has an orbital period less than 4 years (2.5 AU separation). In order to have a common envelope event, the stars must have overflown their Roche lobes. The radius for the Roche lobe of two point masses is
\begin{equation*}
r_L = \frac{0.49 q^{\frac{2}{3}}}{0.6q^{\frac{2}{3}}+\mathrm{ln}(1+q^{\frac{1}{3}})}a
\end{equation*}
where $a$ is the separation. For close binaries, the general observed trend is a high mass ratio $q=M_2/M_1$. So, if we assume $q=1$, then $r_L = 0.38a$. Hence, for a binary with $a<2.5$ AU,
\begin{align*}
r_L &\lesssim 1\ \mathrm{AU}\\
r_L &\lesssim 215\ R_{\odot}
\end{align*}
since $q=1$ is an upper bound on the Roche lobe radius. Now, performing some trivial rearrangement of the blackbody luminosity equation $L=4\pi\sigma_{SB}R^2T^4$, we find that
\begin{equation*}
R \approx 3.31\times10^{7} \bigg(\frac{L}{L_{\odot}}\bigg)^{\frac{1}{2}}\bigg(\frac{1\ \mathrm{K}}{T}\bigg)^2 \ R_{\odot}.
\end{equation*}
Massive stars typically have roughly constant luminosity, so we will choose $L\approx10^5\ L_{\odot}$. Hence,
\begin{equation*}
R\approx 1\times10^{10}\bigg(\frac{1\ \mathrm{K}}{T}\bigg)^2 \ R_{\odot}
\end{equation*}

The massive star needs to evolve until its radius is equal to that of the Roche lobe radius, so we find that the star reaches the common envelope phase for
\begin{equation*}
T \gtrsim 7000\ \mathrm{K}
\end{equation*}
Taking a peek at an HR diagram, this star varies from about $30000\ \mathrm{K}$ to $4000\ \mathrm{K}$ from ZAMS to end of main-sequence. Thus the primary spends roughly 3/4 of its time on the main-sequence not in the common envelope phase. Hence, this binary's common envelope phase lasts for, at most, 1/4 the primary's total lifetime, which is on the order of $10^6$ years. Thus, the upper bound for the timescale of a common envelope event with massive stars with negligible rotation is $\sim10^5$ years.

Please note that this derivation does not take into account the bulging effect that occurs as the separation decreases. This will certainly lower this upper bound, but by how much I'm not sure. It could lower it by 1 year, or $10^5\ \mathrm{years}$.

Lower bounds to this timescale are entirely ambiguous and not particularly helpful in any physical context. The stars could be spinning really fast, have high or low metallicity, the binary could have a different mass ratio, there could be another binary close by, and there may be magnetic interaction (?). The list goes on! I'm sure there's something I left out.


Maxwell stress contribution to $\nabla \cdot \mathbf{P}$ in the Navier-Stokes equation for fluid in stars

$\nabla B^2 \sim B^2/l,$ where $l$ is a length scale on which $B^2$ varies.

In SI units, $B^2 \sim 4\times 10^{-8}$ T$^2$ and $\mu_0 = 4\pi \times 10^{-7}$, so an order of magnitude for $\nabla \cdot P_B$ is $4\times 10^{-8}/(8\pi \times 10^{-7}\times 10^{6}) \sim 10^{-8}$ Pa/m.

The density of the solar photosphere (at optical depth unity in the visible spectrum) is around $\rho \sim 10^{-4}$ kg/m$^3$.

Thus $\nabla \cdot P_B/\rho \sim 10^{-4}$ m/s$^2$.

So no, I can't quite reproduce your number.


Okay, well, calcium is just an easily recognizable spectral feature, so I guess that's why they used it. The y-axis shows the relative intensity. I am assuming this is a spectrum that is normalized or scaled down so that the highest value of the intensity is 1, but this is not so important. The laboratory wavelengths of the absorption lines are shown in red in the graph. If you look carefully, you can see that the minimum of each spectral line does not exactly line up with these red tick marks. That is because the lines are Doppler shifted. M31 is moving and thus the light that we see has a different color (or frequency) from what we would observe in the lab. Let's focus on the left absorption line. The minimum is a bit to the left of the red mark. I conceive this to be approximately 4 angstrom off. The wavelength that you measure here on Earth will thus be 3630 angstrom and not 3934. If I take the formula from the previous page and rewrite I get
$$\frac{\Delta \lambda}{\lambda_0} = \frac{v}{c}$$
where $\Delta \lambda$ is the difference between the observed and the lab wavelength of the spectral line. In this case $\Delta \lambda = -4$ angstrom and $\lambda_0 = 3934$ angstrom. The speed of light is given in km/s, not in m/s, so the value is indeed correct. Now you can get the speed of M31 from
$$v = c\frac{\Delta \lambda}{\lambda_0} = 3 \times 10^5 \frac{4}{3934} = -305km/s$$
I entered this value and it was accepted.


How much of a difference do good lenses make?

...a very blurry, small view of Jupiter with the 4mm and Barlow...

Be aware that a 4mm eyepiece and a 3x barlow at the same time will give you a very high magnification - too high!  For regular Jupiter viewing I would suggest you stick to 100x or 200x at most, unless the air is exceptionally still. (After a few sessions you'll find out what "Still" air means.)

A lot of beginner scopes like this are sold with barlows that tend to give too much power, my advice is to keep it packed away for most sessions.

So I'd stick to using that 4mm on its own, jupiter will look quite small but with practice you can usually tease a bit more detail out of the image.

I don't have one of these scopes, but my experience of comparing cheap modern eyepieces with expensive ones is that the lower cost ones are generally OK these days. 4mm is quite a high power so the extra magnification will cause a lot of blurriness and it will take patience to get the focusing at its sharpest, so don't panic if everything is looking fuzzy at the moment.

For a first stage I would suggest a few things that are nothing to do with the eyepieces at all...

Collimation - look this up, it just means adjusting the two mirrors
so your eye is looking right down the tube in a straight line. For a
long focal length scope like yours, it's not likely to be a problem
unless one of them is wildly out of line.
Tube currents/thermal behaviour - on almost any cold night, when
the tube and main mirror are still warm from being indoors, rising
air currents in the tube will mess up your image and make it
shimmery, at high power. Low power will look OK. It might take an hour or so for the image to improve (a guess)

Anyway, in summary my guess is changing the eyepieces straight away won't make a dramatic difference. Eyepiece makers will say otherwise of course :)


Is it possible that some of the confirmed exoplanets are in fact primordial black holes?

The first you need to determine if you're looking for a zebra or a unicorn.  That is, if there's a herd of horses before you, you expect to see all horses.  Maybe there's a wayward zebra in there.  Unlikely but possible, but you're not going to find a unicorn because, as far as we know, unicorns don't exist.

We know how planets form and we expect to find mostly planets where planets should be, so if primordial black holes do exist, finding one orbiting a star in a near orbit wouldn't be impossible, but it would be a zebra in a herd of horses.

That said, primordial black holes, at least those with planet-mass, are probably unicorns. People have looked for them without success and some estimates have put a size limit at below the mass of a planet.    Does that prove they don't exist?   Well, no, but they probably don't exist, not at close to planet mass anyway.  They've gotten some attention because a planet-mass black hole is a cool idea, but cool idea or not, they're probably not real.  Quantum black holes may form by additional dimensions and gravity getting stronger at very short distances, but CERN hasn't found any evidence of that, so it remains a neat mathematical explanation without a speck of evidence.

We can only detect exo-planets reasonably close to their star, either by transit or wobble.   A captured primordial black hole would probably have a more distant orbit - like planet 9 is thought to have.   A close orbit that could be detected and pass as a planet would be unusual.   If it formed within the proto-solar-system, it would likely clump matter around it, becoming part of a large planet or the central star, so it would need to be captured later, after the system cleared out and that's harder to do.  Gravitational captures that close to a star are rare.

A primordial black hole, with a planet's mass would be tiny, so, as you noted, detection by the transit method would be unlikely.   An Earth-mass black hole would be about the size of a golf-ball.   A Jupiter-mass black hole would be about 10 feet across.   Even when you factor in that the observed light-halo is about 2.6 times the size of the event horizon and beyond that, the majority of the lensing is just another 5-10 times the size of that, you're still looking at a very small object with just a feet to maybe a few hundred feet of transit.  That's probably much too small to create the kind of shading needed to recognize a planet.  You'd need stellar mass to begin to detect a transit and that would be detectable in other ways.

A theoretical planet mass black hole could create a wobble and could be detected by the wobble method, if it was close enough to the star that a few wobbles could be detected, timed and the planetary mass object confirmed, but for reasons above, we don't expect that kind of black hole to exist and if they did, one getting captured that close to a star would be an unexpected.  The odds are approaching draw a straight flush odds.  Not impossible, but very unlikely, and coming telescopes should provide better insight on this.  Maybe some surprises await us when the next round of telescopes start getting images, but I would guess that planet mass black holes won't be one of them.


How long time since a typical asteroid collided with another asteroid?

How ancient they are, I'd have to say as old as the solar system.
Collision frequency:
Collision amongst Asteroid Belts
From ref 1  below: Based on the particle in a box model which assumes a homogeneous distribution in the interactive volume, the intrinsic collision probability is computed to be around $5 \times 10^{-18} km^{-2}yr^{-1}$.
However Wetherill(ref below) have some modifications to Opik's formulation for collision between two elliptical orbits, suggests that the impact is nearly independent of the orbital elements of the test body and is nearly half of the value obatined from PIAB model. It also seems to suggest that intrafamily collisions are higher than collision amongst non family objects.

Collision with the Earth Here it characterized in various classes:

bolides or upper atmospheric explosion,
land impacts leading to local or regional devastation,
ocean impacts leading to regional disasters,
global ecological catastrophe,
mass extinctions

Based on the scale, the impact events are categorized as harmless, Class-1, to devastating, Class-5.

The 440-KT Chelyabinsk event of 2013 belongs to impact hazard class 2, causing regional or local devastation. This was caused by a mere 20 meters wide meteor entering the atmosphere of Earth. For a cumulative number of impacts described for a given reference time interval of $T_{ref}$ = 1Ga ($10^9$ years)
which is long enough to statistically include a meaningful number of impacts from all the classesabove, the class 2 impacts cumulative will be $2\times 10^6$ , with small bodies falling in the size range of 20-100 m. Where the characteristic time scale is of $10^2$ a.

For class 3-5 events leading to regional disasters to global extinction events, the cumulative number of impacts within the reference interval is $10^4$ with characteristic time scale being $10^5$ a. These events can lead to large devastation ranging from an entire country or region (class 3, with an explosive energy of approximately 104 MT TNT), to mass extinctions (class 5 releasing an approximate energy 108 MT of TNT). An example from the history for such event is the classic K-T extinction events 65 Ma ago that wiped out more than 99% of the living species.

References:
1. http://articles.adsabs.harvard.edu//full/1992LPI....23..283D/0000283.000.html
2. http://www.springer.com/fr/book/9780792354666
3. http://onlinelibrary.wiley.com/doi/10.1029/JZ072i009p02429/abstract
(PS: If anyone has a link to the open access for final paper, or a pdf; I would really appreciate a copy.)


What measures are currently in place to detect asteroids on possible collision courses with earth?

This site is an asteroid tracking program started up by NASA. It seems to be fairly effective at detecting asteroids. Read up on this article for further information about NASA's plan.


After a quick google, This page pops up showing that Anaxagoras was the first documented person to make the link between the sun and stars being the same thing.


Chemical reactions in white dwarf and carbon allotropes

White dwarfs are objects the size of the Earth, but with a mass more similar to the Sun. Typical internal densities are $10^{9}$ to $10^{11}$ kg/m$^{3}$.

White dwarfs are born as the contracting core of asymptotic giant branch stars that do not quite get hot enough to initiate carbon fusion. They have initial central temperatures of $\sim 10^{8}$K, that swiftly (millions of years) drops to a few $10^{7}$K due to neutrino emission.

An important point to make is that after that, the interior of a white dwarf is almost isothermal. This is because the degenerate electrons that provide the pressure support also have extremely long mean free paths for scattering interactions and thus the thermal conductivity is extremely high.

The exterior of the white dwarf is a factor of 100 cooler than the interior. The temperature drop happens over a very thin shell (perhaps 1% of the outer part of the white dwarf), where the degenerate gas transitions to becoming non-degenerate at the surface.

This outer layer acts like an insulating blanket and makes the cooling timescales of white dwarfs very long. From interior temperatures of say $3\times 10^{7}$ K it takes a billion years or so to cool to $5\times 10^{6}$ K and then another 10 billion years to cool to around $10^{6}$K, and such white dwarfs, which must have arisen from the first stars that were born with progenitor masses of 5 to 8 solar masses, will be the coolest white dwarfs in the Galaxy.

At these temperatures there is no possibility of the carbon undergoing chemical reactions, it is completely ionised; the carbon and oxygen nuclei are in a crystalline lattice at these densities, surrounded by a degenerate electron gas. There is evidence that crystallisation does take place, via asteroseismology of some pulsating massive white dwarfs.

The details of the crystalline structure in these objects is unknown, and the subject of theoretical investigation. However, diamond is pure carbon and white dwarfs are expected to be a carbon/oxygen mixture. A further complication is that the process of crystallisation may be accompanied by gravitational separation of the carbon and oxygen, so that the inner core is more oxygen-rich than the outer core.

Original ideas were that the crystalline form would be body-centred cubic (bcc), but other more complex possibilites are opened up by the mixture of carbon and oxygen. bcc Carbon would be a new allotrope of carbon and not like diamond - it is a denser way of arranging the nuclei.

EDIT: To answer a point in the comments. Even if you were to wait trillions of years and allow white dwarfs to cool to the thousands or even hundreds of degrees that you might think would allow electrons to recombine and for chemistry to occur, that is not how it works. In the degenerate electron gas, the typical Fermi energy of the electrons is an MeV or so, compared with the eV-keV of bound electron states, and this is completely independent of the temperature. So the high electron number densities ensure that they will never recombine with the carbon nuclei (a theory first developed by Kothari 1938).


How is the first detonation in Supernove type Ia triggered?

Nobody really knows how type Ia supernovae detonate (or deflagrate) - there are a number of possibilities. The "vanilla" possibility is not what you state in your question, it is that the white dwarf accretes sufficient mass that it approaches the Chandrasekhar limit and becomes dense enough in its core to commence carbon burning.

However, the emerging diversity that is seen among type Ia supernovae, once thought to be a single population, suggests there may be other possibilities. There is some evidence that white dwarfs may explode at masses well below the Chandrasekhar limit. If a white dwarf in a binary accretes enough He-rich matter, this can become compressed enough to ignite He burning near the surface (this happens at a lower density threshold than Carbon burning). This then drives a shock wave into the white dwarf and the compression caused by this can ignite the carbon.

Why does the He "explode"? Well, the accreted He will form an electron-degenerate layer at the surface. A fundamental property of this degenerate gas is that the pressure is independent of the temperature. Thus, if the He ignites then at least initially, the temperature goes up but the pressure does not. Since the He fusion rate depends on something like $T^{40}$ this allows a runaway reaction to develop that could be characterised as an "explosion" in the surface layers.


Techniques for locating origin of gravitational waves

Short answer
Three detectors would improve the ability to pinpoint the direction of the gravitational waves drastically.
Long answer
Finding the direction of the source of a gravitational wave, as with most things in science, is harder than it might initially seem. I'd like to point out that the process used in determining direction is not precisely a form of triangulation (even if it had 3+ detectors) because triangulation involves determining precise position from knowing distances to the object from 3 or more locations. That's not really what's happening here.
I'll outline the basics for determining the direction and how that method might be improved with more detectors, but you can also read this excellent source.
Time of Arrival
The main way we can determine direction is time of arrival at the various detectors. The currently active detectors are located some 10 milli-lightseconds apart. That means one of them is going to detect the wave first, then sometime later, the other one detects it. Given that the wave has to be traveling at the speed of light, you can determine the direction it was traveling from such that it traversed the distance between the two detectors in the time difference measured.
But, this presents a problem because this method has degeneracies with only 2 detectors. Really, all you can calculate is the angle of the direction of the wave with respect to the line connecting your two detectors. That only gives you a cone on the sky of where the wave could've come from and you can't pinpoint the azimuthal angle without more information. This concept can be seen in the image below showing the probable direction. You can see they've limited it further, but their shape still takes part of a circle (the projection of that cone on the sky).

What if you added another detector?
As stated above, you have a degeneracy with only 2 detectors, but adding a third would break that degeneracy. You'd turn your cone into a point and you'd know much more precisely where the source came from. More detectors would allow you to pinpoint that location even more precisely, but only 3 are necessary to remove all degeneracies.
Now, of course I've been describing how it would work "in theory". In practice, observations are usually much more messy. Any measurement comes with error, so even with 3 detectors you're not going to be able to pinpoint the location exactly. You'll more likely be able to pinpoint it to a small patch in the sky. That's where other direction detection techniques can be used to refine that location.
Detector Sensitivity
The way in which the experiment works is that a mass is set up which can length contract under the influence of gravitational waves. This tiny contraction of space-time (and thus the mass) by the passing is measurable with precise lasers and interferometry. However, the lasers can only measure the contraction of the mass in one direction. The degree to which the mass contracts in that direction depends on the direction in which the wave is hitting it. This is an oversimplification, but if the modulation of the wave (that is the direction in which it goes "up" and "down") is aligned perfectly with the direction in which the contractions are being measured, you measure the full effects of the wave. If the wave comes in at some angle to that, you're still measuring the effects, but you'll only be measuring some part of the effect, reduced by the sine of the angle.
This process of wave direction detection requires at least 2 detectors, but more is better for beating down the noise.


How would latitude affect day/night lengths on a planet with no axial tilt?

The length of the day would be the same on (almost) every latitude, and just a bit longer than the night. The reason for that is that the planet's star (let's call it 'Sun') is a disk, not a point. So even if the middle of the Sun is below the horizon, part of it might still be above the horizon and it's technically 'day'. For the same reason, the length of an Earth day at Equinox is slightly more than 12 hours.

At the poles, and a small region around them (depending on the apparent size of the Sun), it will always be day. At the North Pole of the planet, you will always see exactly the northern half of the Sun; the southern half will always be under the horizon. It will still take the Sun exactly the same amount of time to go 'around' the sky 360 degrees as it will on the equator or any other place on the planet.

The diagrams in the Wikipedia article are a good visualization how every day on such a planet would look like (click to enlarge):


Why do Uranus' and Pluto's moons orbit the equator? What makes a moons' inclination tilt with the rotational axis of their planet?

I have managed to find two hypotheses for Uranus's tilt that explain why its moons are also orbiting on a tilted plane.

The multiple impacts hypotheses lies on the fact that, if Uranus had become tilted by the force of a single large impact (as was commonly believed), then the moons should have stayed on their original plane. A corrected version of this hypothesis said that, if the accretion disc from which Uranus was forming was still around, an impact would have destroyed it and recreated it on the same place as the new equator. The moons would then have accreted from the disc on that plane. This, however, would have produced moons with retrograde motion (which is not the case for most). So the latest idea is that Uranus suffered multiple impacts, which, according to the simulations, works to produce the planetary system as we see it today. (There's another problem with this hypothesis.)

The collision-free theory (or collissionless scenario as described in the paper) proposes that Uranus could have become tilted progressively as it formed without the need to postulate an impact (or series of impacts), if it had an additional satellite and started with a large inclination (which is allowed by current models); the extra moon could have been ejected later. The simulations worked for an initial inclination of more 17 degrees and a moon of 0.01 Uranian mass and at 50 Uranian radii.


Is the whole universe is rotating on an axis?

From what we observe it is extremely unlikely the universe is rotating, but it is a good question nonetheless.

Perhaps you thought that since we see galaxies moving away from us they could be being pulled outwards as a result of the rotation. All the galaxies we observe (outside the local group) are accelerating away from us, and this includes galaxies in all parts of the universe. If the universe were like a giant spinning cylinder (or whatever object you like) then the galaxies "above" and "below" us in the cylinder would appear stationary since they are rotating at the same radius.

This observation does not support the claim that the universe is spinning on some invisible axis.

As always we should remember that our local universe may not represent the entire cosmos, and we would in fact be spinning around some axis. As more observational evidence is gathered we understand more and more about the universe, and one would be foolish to rule out a theory based on current observational evidence.


When is the best time to observe M81 from the Northern Hemisphere?

You are correct that M81 transits the meridian at upper culmination just past mid-February in the eastern US. As long as the Moon is well below the horizon, as it will be in Feb. 2020, this will be the best time to observe M81, and its neighbor, M82. Since these galaxies are relatively bright, good views are available from late December through early May when they are 40+ degrees above the horizon, as long as, there is no interference from the Moon. I have taken some of my best photos of M81 & M82 in March and April, mainly because they are a bit farther to the W, avoiding some local light pollution here towards the NE.

It's likely that many think early Spring is the best time to observe M81, since they don't have to stay up so late; in the eastern US, M81 transits the meridian about 10 pm in mid-March and 9 pm in early April.

Edit:  Added answer to question from the comments.

The simplest and most accurate way to determine when an object transits the meridian at your location is to  use astronomy software such as, SkySafari, Stellarium, Cartes du Ceil, or any of the many other planetarium type programs. However, I started amateur astronomy many years before smartphones, tablets, gps, and even personal computers. So I learned to estimate when objects in the sky would be visible.

At my location, around 10 degrees of longitude to the east of yours, I remember that the meridian on January 15th corresponds to about 8 hours of right ascension. Since the sidereal day is about 4 minutes shorter than the solar day, RA at the same location and time increases about one hour in 15 days, or about 2 hours per month. Precession changes this RA slowly over time, however, it is a useful approximation that is roughly valid over decades.

Thus, M81 at 9h 56m RA, will transit my meridian near midnight around Feb 15th, and at your location a few days later. A month later M81 will be two hours farther to the west, so will cross the meridian earlier, around 10 pm.


When was it worked out/discovered that our sun can't go supernova?

I think the definitive work is that of Hoyle & Fowler (1960).
They argued that supernovae were produced by two possible mechanisms - what they called an implosion/explosion or an explosion within degenerate matter. Both of these mechanisms required very high internal temperatures ($>2\times 10^{9}$ K) and they argued that this could only be achieved for both mechanisms if the star is more massive than the Chandrasekhar limit of $5.8M_{\odot}/\mu_e^2$, where $\mu_e$ is the number of mass units per electron (1 for hydrogen, 2 for helium, carbon, oxygen, 2.25 for 56Fe etc.); otherwise it could achieve a degenerate state, supported by electron degeneracy pressure, and cool indefinitely without contracting any further.
In addition, they successfully identified these two types of supernovae with Type I and Type II supernovae and argued that very high mass stars accounted for Type II supernovae via the implosion/explosion mechanism, whereas the Type I supernovae arose from the ignition of degenerate material in less massive stars, but still larger than the Chandrasekhar mass, which is $>1.2M_{\odot}$ for all plausible compositions. They even then conclude that stars up to $10M_{\odot}$ may lose enough mass to avoid a Type I supernova and become stable white dwarfs.
Though it seems not to be explicitly stated in the paper, it is implicit that the Sun could not explode as either type of supernova.
As an aside - I reminded myself of the plot of "Songs of Distant Earth" (I read it loooong ago). Clarke's story posits that the "neutrino problem" (the lack of detected solar electron-neutrinos), which was a thing back in the 60s, was actually because something odd was going on in the Sun's core. So I don't think that Clarke was suggesting that under normal stellar evolutonary circumstances the Sun would imminently explode. Indeed it was well known at that time that the Sun should continue on its main sequence life for billions of years.


Is the discovery of Alien or Extra Terrestrial Life a function of the science of Astronomy?

The search for extraterrestrial life is usually covered by "exobiology". But astronomy can provide clues of whether a planet or an exoplanet provides some of the conditions, which allow life or indicate life.

The habitable zone - in its various versions - defines, in which distance from a star a planet needs to be to allow for e.g. liquid surface water, a condition regarded as necessary for most life-forms we know from Earth. With the discovery of extremophiles the habitable zone for that kind of life has been extended.

Astronomy can provide information about the central star and the distance of a planet from the central star. From these data a decision is possible, whether the planet fulfills the surface temperature condition for habitability.

Astronomy can provide information about the stability of a star, necessary for life to evolve on its planet.

Astronomy can - in principle - provide information about the chemical composition of a potential atmosphere of a planet. Oxygen and methane at the same time would be regarded as an indicator of possible life on the planet.

Astronomy can determine size and density information about the planet. That way it can be decided, whether it's a rocky planet or a gas planet, and infer the gravitational acceleration at the surface. The gravitational acceleration is relevant, e.g. for constraints of the atmospheric composition, or for possible properties of life, if extant.


Can we apply Kepler's law of planetary motion to the moons motion around the earth?

Yes. Kepler's laws apply to orbits of the Earth, and the moon's orbit is elliptical, with an eccentricity of 0.055

Or at least nearly elliptical. In Newton's theory of gravity, if there are two particles in orbit around each other then they will have an elliptical orbit. If there are more than two bodies then the gravity of other bodies will perturb the orbits from exactly elliptical. A more accurate description of the moon's orbit is "Elliptical + perturbations"

The most significant perturbations come from the sun, and from the fact that the Earth is not a perfect sphere. Other planets and tides also have effects on the Moon's orbit.

There are two wiki pages you should read: Orbit of the moon and Lunar Theory


Consequences of strong wind on an alien Planet on the possibility of life

We know that life can start and evolve quite well on a planet with 250 mph winds: You're living on one. The jet stream can reach this speed in the Earth's atmosphere.

Beyond that it is speculation. While not much is going to survive on the surface if there are surface winds of 1500mph at a static pressure of anything like 1atm, I can see no problem of life developing in the depth of the ocean no matter what the wind speed is.

However I futher suggest that 1500mph winds cannot happen at the surface of a terrestrial planet that is anything like the Earth. The extreme conditions needed to create a surface wind of 1500mph on a daily basis may prevent any kind of life as we know it. But then the winds are not the acutal problem but the conditions needed to produce them.


There are two issues: Would there be gravitational waves to be detected and would LIGO detect them.

On the first issue, gravitational waves are quadrupolar, and a cylindrically symmetric system will not produce any. (Specifically, the second time derivative of the quadrupole moment of an isolated system's stress–energy tensor must be non-zero in order for it to emit gravitational radiation.) So a head-on collision of equally massive black holes would not produce significant gravitational waves.  If the BHs were of different masses, or the collision not quite head-on, gravitational waves would be produced.

On the second issue there are at least three ways LIGO might might miss them.  First, they might simply be too faint.  A merger through rotation is one of the best ways to produce gravitational waves, and a head-on collision is one of the worst (during a BH merger, of course).

A second way it might miss is if the merger is in one of its blind spots. Each LIGO detector has blind spots and there's no rule that says a BH merger can't be in one of them. (In fact, the neutron star merger was located by use of the blind spots. The third gravitational wave (GW) detector, Virgo, in Italy, didn't pick up the neutron star merger.  It was bright/loud enough that it should have, which meant that the merger was in one of Virgo's blind spots, which helped to narrow down its location.  A GW detected in only one detector would be listed as a possible GW, but not as a detection, because it might simply be terrestrial interference.

But another way it might miss is if LIGO had not been looking for the waveform of that kind of merger. General Relativity allows us to compute in very good detail the GW signal from any type of collision that we think likely. Because the GW signals are so faint, the first step of detection is matching computed GW waveforms to the data and looking for matches. Once a match is found, the signal is analyzed in more detail, but they have to know where to look. If no one expected the waveform and it was not strong enough to pop out of the data and yell "Here I am, dummy!" it might be missed. Since head-on collisions are expected to be very rare, it's possible that they are not in the library of waveforms that LIGO checks.


How do scientists know that a rock found on Earth came from Mars?

Mars is the only planet providing rocks with a similar chemical composition and age. So the origin of the meteorite is evident.

This doesn't mean, that it's absolutely waterproof. There might have existed protoplanets similar to Mars 4 billion years ago, which since then have been swallowed by Jupiter, the Sun, or have been ejected out of the solar system. But this would be much of coincidence without leaving convincing evidence. (Here a paper about a couple of simulations.)


Is there another habitable planet in our galaxy?

what'll happen when we destroy our planet or it becomes none-habitual for human life?

Very pessimistic idea.

We lack the technology to tell if a planet orbiting another star could be lived on by humans.  We barely have the technology to detect some planets orbiting a small number of other stars (compared to the number of stars in the galaxy).

The term for these planets is exoplanet and Wikipedia says we have detected about 3750 to date orbiting about 2800 stars.  As there are somewhere between 100 and 400 billion stars in the galaxy, we only know a tiny fraction of what's out there.

We have a very limited basis for listing some planets as possible candidates for supporting some form of life, but, with all respect to the serious scientists doing this work, this is nowhere near knowing if they really can support life at all, let alone humans.

So your question's answer is : we don't know.

We cannot travel fast or far enough at present to reach another star in a reasonable time frame and we also could not do it with enough people.  We do not currently have any reasonable expectation that we will be able to do that in the future.

So even if we knew of a suitable planet, it would be unlikely we could ever travel there to e.g. colonize it.

So we're stuck here so keep recycling. :-)


How would tidal effects behave in this setup: An icy moon orbiting a gas giant, as the gas giant orbits a red dwarf?

I'm not sure exactly what your question is, but Tidal locking is fairly common for moons.   If the Gas giant was in the habitable zone of the red dwarf star, then it would need to be fairly close and as a result the Moon would need to be fairly close to the Gas giant - it should be tidally locked to the gas giant, not the star.  (A moon being tidally locked to a star is probably impossible because the planet will always have the greater tidal effect on the moon).

So your scenario, you have a moon, tidally locked to a gas giant planet, which orbits a star.   Tides don't exist in a perfectly circular tidal lock because the pressure is constant, but there's no such thing as a perfectly circular orbit, so there will always be some tidal squeezing as the moon moves around the planet.

The real question is how eccentric the Moon's orbit is around the gas giant planet, because it's eccentricity will determine it's tidal squeezing and any libration.

As far as periodic habitable zones, that would depend on how eccentric the orbit of the gas giant was around the star.  You could have it moving in and out of the habitable zone, but it would presumably be a short-period orbit, so it would be a matter of days outside, then days inside, perhaps a week or two outside and a week or two inside, unless you went highly eccentric.

All these are general points.  There's lots of variables.

There's no one answer for where the Moon would receive most of it's heat - you'd need to work out the specifics or orbit, distance and solar temperature and then give the Moon a specific orbit and then you could work out an approximation of where more heat comes from.

and this feels like worldbuilding to me.   It's too hypothetical for astronomy.


As mentioned in the comments, in relativity, "all stop" can only be in reference to some other object, or set of objects.  Still, I would argue there remains some interesting potential for the "all stop" concept in physics theory, because Einstein was motivated by Mach's principle when creating general relativity (regardless of whatever version of that principle is formally supported in that theory), and so we might suspect that "all stop" can have the theoretical local meaning of "at rest with respect to the center of mass of everything around it that is affecting it."

That notion is trickier in an expanding universe, where the concept of a center of mass gets replaced with the concept of "at rest with respect to the cosmic microwave background", which is normally treated as a kind of accident of history, rather than a theoretical truth.  Still, since the Big Bang model is often taken as asserting that the accidental history of our universe is actually relevant to the origin of our universe and its theoretical properties, as well as to the origin of the laws that our universe obeys, it is not a great stretch to imagine that an "accident of history" kind of preferred frame could still, nevertheless, have a formal status in the laws as a kind of true "all stop" frame-- albeit only locally.


How deep and shaped is the depth of a black hole?

The first representation isn't depicting the black hole; it's depicting its accretion disk, which is indeed plate-shaped. The accretion disk is around the black hole, and - if it's a rotating (Kerr) black hole - it's perpendicular to the black hole's axis of rotation. Accretion disks are disk-shaped, and so the representation is disk-shaped.

The other is a common representation of a black hole, but it's misleading. It's an extension of the "rubber sheet" analogy, which describes the bending of spacetime. A heavy object, placed on a semi-taut sheet, will curve it a little bit. This represents ordinary matter curving spacetime. A black hole is represented in that image as bending spacetime by an infinite amount. The issue is that this representation embeds the two-dimensional representation of spacetime in three dimensional space, making it appear as if the black hole has this extra depth. Black holes are not embedded in an extra spatial dimension.

Note also that the small object moving near the edge of the black hole is also represented by a tiny funnel. The creator seems to have tried to take the analogy to the extreme, portraying the object's funnel as finite, and the black hole's as either extremely large or infinite, in an attempt to explain that light cannot escape from a black hole. It's not too accurate, but it works to get the point across.

By the way, black holes do have size (the central point singularities of non-rotating black holes are infinitely tiny, while the ring singularities of rotating black holes are in the shapes of infinitely thin rings). You might want to read up on event horizons. The radius of the event horizon of a non-rotating black hole is a finite number, the Schwarzschild radius. It is calculated as
$$r_s=\frac{2GM}{c^2}$$

In this case, the event horizon is spherical. In rotating black holes, it may be in the shape of an oblate spheroid.


What is the maximum Clear Sky Light Pollution rating to see the 2018 Perseid Meteor Shower?

The darker the better.

There are meteors of different brightnesses in the Perseid shower, including some fireballs that can be seen in any (cloudless) sky.

But many of the meteors are dimmer, and the quoted rate of 60 per hour assumes dark skies.  As there is no moon tonight, you would benefit if you can get to a dark sky location.  If you can see the milky way, you can see the Perseids.

I am under a blanket of cloud, but have seen a few while out running on Wednesday.