This is a simple explanation to give a basic understanding as to why the solar system rotates.

It's caused by gravity and velocity. If you put two objects with mass in space a distance apart and at rest, they will accelerate directly toward each other, until kerpow! they hit.

But, if they have some small, random relative velocity to begin with that's not directly toward or away from each other, then no amount of acceleration toward each other will zero that out. It means they will pass each other at some finite distance and keep going. If the random velocity is low enough, this dance will repeat and they will move in elliptical orbits around their center of mass.

This kind of motion can be expresses in terms of its angular momentum and in this example (as in most) it is going to be a conserved quantity, so it's always going to be there.

So when things in the universe like solar systems or galaxies initially form, they have some random, residual, collective motion, and this angular momentum in the center of mass frame of whatever blob is coalescing doesn't go away. Objects in the solar system keep rotating because there's no other option, at least in the short term.

If there was a large amount of gas in the solar system, that might slow things down a bit, and there is a teeny tiny loss of energy due to gravitational waves, although for planets this is so extremely small it can be ignored. But basically the "wheels" of the celestial sphere are well lubricated and solar system bodies will continue in their orbits for millions and perhaps billions of years because there's nothing we know about right now to stop them.


Magnetars (and neutron stars in general) don't need a dynamo to create their magnetic fields. Their magnetic fields are "frozen in" at the time of their formation. To really see why this is, you have to understand a lot about electromagnetism, but I can boil it down to the basics. Keep in mind that neutron stars are highly mysterious objects and we don't have a huge amount of good, observational evidence behind our theories since they're so hard to find and observe.

Neutron stars (and magnetars) are superconductors (or at least most theory and objective evidence suggests that is the case). What this means is that there is literally no resistance to the motion of electric charges moving throughout the star. To point out, even if the entire star was all neutrons (which zibadawa timmy points out is not the case) you'd still have charges since neutrons themselves are composed of charged particles. In any case, since there's no resistance, there cannot exist any electric fields inside the star. Any field which arose would induce a force on the charged particles which would be able to immediately move to cancel said field out. All electric fields destroy themselves immediately in a superconductor.

If we venture into electromagnetic theory equations, you'll find the very useful equation:

$$\nabla \times \textbf{E} = -\frac{\partial \textbf{B}}{\partial t}$$

Without getting into the details of this equation, the general idea behind it is that electric fields result in time varying magnetic fields and vice versa. But we just said above that our neutron star was superconducting and thus has no electric fields. This means the left hand side of that equation is zero. The right side represents the change in magnetic field over time, but we know that has to be zero now so we're forced to conclude magnetic fields in neutron stars (and magnetars) cannot change. They're frozen in at the point of creation (when the star becomes superconducting).

We can venture one step further and say that the magnetic fields of neutron stars and magnetars is so strong simply because of magnetic flux conservation. That is, you have a huge star of a few solar masses which has a massive field in its own right. That star then collapses into a neutron star but in the process, the magnetic field flux through its surface has to stay conserved. Magnetic field flux is a function of both the field strength and the radius of the star. Since the radius is decreasing so immensely, the magnetic field strength has to increase in proportion to the radius decrease (squared) to compensate and keep the overall flux the same, at which point that magnetic field is locked into place due to the superconducting nature of the star.


Delete key does not always work in IRAF's xgterm on Mac

This sounds like a problem with your terminal settings.  Quoting the Setting up IRAF section this page, "We assume the user has defined this correctly when issuing the MKIRAF and no longer key off the unix TERM to set a default."  The page goes on to say that one should add stty xgterm the to your login.cl file (directly above the final keep line might be a good place).  The page linked above goes into more detail.


Day/Night Cycles and Seasons if the Earth's Rotational Axis and Orbit Changed

The Earth's rotational axis is essentially the line connecting the North and South poles.  This axis is slightly tilted, making the Earth's equator 23.5 degrees offset from the orbital plane (the ecliptic).

From Wikipedia

The first part of the question is asking what would happen if we were to make the rotational axis line up with the orbital plane.  That would essentially mean tilting it so that the Earth's equator would be 90 degrees offset from the ecliptic.  One way to do this would be to point the North or South pole directly at the sun, or to point it in the same direction as Earth's orbit.

The second part of the question is asking what would happen if the rotation axis was perfectly up and down.  This would mean that the equator would not be offset from the ecliptic at all.  However, for this part of the question, the Earth's orbit would also be highly eccentric, meaning that our distance to the sun would change a great amount.

I won't post the answer here, since this looks like it is probably a homework question, but to find the answer, first read up about what causes the seasons.  To talk about the day/night cycle, think about how the Earth's rotation causes day and night, by moving the sun across the sky.  If we changed how the Earth was tilted, how would that affect the seasons?  How would that affect where the sun appears in the sky?


Theoretical limits for natural satellites having natural satellites

Wikipedia says there may be a reason (emphasis mine):

  No "moons of moons" (natural satellites that orbit a natural satellite of another body) are currently known as of 2014. In most cases, the tidal effects of the primary would make such a system unstable.

This seems to be because the secondary would have to be very close to the primary - close enough that it doesn't become simply another satellite of the actual primary (i.e. the planet).

The page then goes on to talk about Rhea, a moon of Saturn that may or may not have a ring system. This gives some more information.


Does coronal mass ejection include radiation at optical wavelengths?

Coronal mass ejections consist of a very hot, but thin, plasma. Their very weak intrinsic emission would be dominated by ultraviolet and X-ray lines and bremsstrahlung continuum. There is very little optical radiation.

However, CMEs can and are monitored at optical wavelengths using the light that they scatter from the Sun. The process is Thomson scattering from the free electrons in the plasma. This is done by spacecraft because the Earth's atmosphere makes it very difficult to achieve the contrast necessary to observe the scattered light close to the Sun (except during total solar eclipses).

A CME takes a couple of days to travel from the Sun to the Earth and such monitoring can give "space weather forecasts" associated with magnetospheric phenomena triggered by the impact of CMEs.

An example picture taken by the LASCO white light imager on board the SOHO spacecraft is shown below. Everything here is viewed in scattered light. The Sun is obscured by an occulting disk.I have picked an example showing the eruption of a CME.


Where does Love's number come from in $\delta m = \frac{1}{2}k_2M_2\lbrace\frac{R_1}{r_\star}\rbrace^3$, the formula for the mass of a tidal bulge?

Let us begin with the basics. If an exterior perturber generates the potential W, then an originally spherical planet gets distorted. Owing to this distortion, its potential acquires an increment U which is usually called the additional tidal potential or, simply, the tidal potential.

Now let us fix a point on the surface of the planet. Let $\gamma$ be the angular separation between the planetocentric vectors aimed at this point and the perturber.

The perturbing potential W in that point can be expanded into a series over the Legendre polynomials $P_l(\cos\gamma)$. This is an almost trivial fact, see equations (9 - 11) in
https://link.springer.com/article/10.1007%2Fs10569-009-9204-7

https://arxiv.org/abs/0803.3299

In those equations, $r^*$ is the position of the perturber, while R is the surface point.

As it turns out, a degree-$l$ term in that equation contains a factor of $(R/r^*)^{l}$.

The additional tidal potential U in that point can be expanded over the same $P_l(\cos\gamma)$.

Moreover, you can write down the potential U in a point r located right above R (i.e., having the same angular separation from $r^*$ as R has).  This expansion is rendered by equation (19). In that equation, you need only the first line. Ignore the rest for now.

As we see from this formula, the degree-$l$ term of U is equal to the degree-$l$ term of W multiplied by $(R/r)^{l + 1}$ and by an additional factor $k_{l}$ named the degree-$l$ Love number.

Derivation of this formula is boring. It is a solution to a boundary-value problem for an elastic medium, and the factors $k_{l}$ emerge as some functions of the shear elastic modulus of the material of which the planet is composed.

This is how it is for static tides. In real life we, of course, have tides evolving in time, and all this machinery has to be generalised. We have to expand both W and U into Fourier harmonics and to write down all this formalism for each harmonic separately.
Then you get different lags for different harmonics involved. Also, the dynamical analogues of the Love numbers will be functions of other rheological parameters (like the viscosity).  However, the central idea will be more or less the same.  I can give you more references, if you are interested.

Jérémy Leconte in his work tried to substitute the degree-2 part of U with an equivalent potential generated by two equal masses located oppositely. You can think of this as two tidal bulges approximated with point masses. This was done for illustrative purposes solely and, in my opinion, bears no practical value.


Wouldn't be "dark matter" better called "massless gravity"?

I don't think the title of your question and your question are really in sync here and I'm not sure I will correctly answer your question. However, I will give it  a try.

To give you a hint of the conclusion. My answer would be, "NO".

There are several clues / reasons why I come to this answer.

Gravity is not necessarily a real thing. When you read the literature concerning general relativity about gravity you will learn that acceleration can be experienced as gravity and in the fabric of space/time gravity is "merely" a curvature in this fabric that. This means that gravity is highly debatable as something that could be seen as an independent force / object in the universe.
When gravity exists as a force, whether it is an exchange of gravitons between two objects with a mass, or whether it is a curvature in space / time due to the mass of an object, the phenomenon "gravity" is always related to the mass of an object and doesn't stand on its own as an independent force of nature.
Dark matter is not a "mysterious" thing that passes through itself. Dark matter is just the container concept of matter out there that doesn't interact a lot with other matter, but just leaves its marks in the universe by showing a gravitational force as interaction with other matter. Since we have only little understanding of gravity and we still have difficulty measuring gravity as a force in the universe, we have difficulty understanding and measuring dark matter directly. This doesn't mean it is not out there.

So, to get back to your second part of the question, "massless gravity", this is not a viable alternative. Gravity exists either by massive objects exchanging gravitons and therefore exercizing gravity upon each other, or a massive object "denting" the fabric of space / time and therefore creating a gravity field. In either case gravity cannot be seen as a farce that exists without the existence of mass.

Secondly, to the first part of your question. Whether we can observe objects directly or indirectly, this doesn't make it more or less realistic. Dark matter is generally accepted to be there. We just haven't found a way to measure it. We can only see that the visible matter in the universe can't explain the gravitational pull on the visible matter and therefore we assume there is some other matter out there we haven't been able to "see". Hence, dark matter.

However, dark matter does interact through gravity and our current successes in measuring gravitational waves might also result in a better understanding of gravity and how to measure the effect of dark matter on the gravitational pull in the universe. So, to answer your question: "No, we call dark matter because we can only observe it through its influence on other matter due to gravity. There is no such thing as gravity without mass."


MOOG will search for the libraries through a .sm file. To solve this error, download an example sm file like this one, and specify where your libraries are in this file. Then use mv .sm ~/.sm to place this in your home folder, then MOOG should be able to create plots with SM.


The fundamental assumptions of LCDM cosmology are:

General relativity is valid on cosmological scales.
Universe is dominated by "cold" dark matter (origin, composition unknown).
The metric of the universe is given by the  Friedmann–Lemaître–Robertson–Walker metric.

Reference: https://en.wikipedia.org/wiki/Lambda-CDM_model
 A combination of 1 and 2 helped explain large scaled structure formation, which is smaller masses merging to become bigger systems while 3 helped explain the expansion of space depending on the matter density.

The lambda parameter comes from invoking general relativity. Although Einstein added it to support his own view of a static universe (disproved a decade later by Hubble's observations), it was later re-interpreted as a negative energy density of space (see Harvey, Alex 2012 https://arxiv.org/pdf/1211.6338.pdf and/or Carroll, Sean 2001 https://arxiv.org/abs/astro-ph/0004075).

Hope this helped.


Can a donut-shape planet or star be formed?

In principle, yes. In practice, no.
The question has been studied for a long time. The classic treatment is Dyson's papers on "anchor rings" in 1893 (paper I, paper II) but it goes back to the study of "figures of equilibrium" starting with Newton's considerations of the oblateness of the Earth and then continuing with Maclaurin and Jacobi's analysis of rotating ellipsoids of rotating homogeneous self-gravitating fluids (see Chandrasekhar's book Ellipsoidal Figures of Equilibrium 1969 for everything on the topic; a shorter overview).
The basic story for ellipsoids is that as you add angular momentum they become more oblate (the Maclaurin ellipsoids), undergoing an instability beyond a critical point to become general three-axis ellipsoids (the Jacobi ellipsoids), that may also contain fluid motions (the Dedekind ellipsoids). Eventually there is too much angular momentum and it breaks up, but Cartan showed that the mode that grows fastest is the low-frequency harmonic making it just split into two. So this way you will not get a torus-shaped planet or star.
Dyson considered the instabilities of a ring already existing. He found that for nearly circular rings:

The annular form of equilibrium of rotating gravitating fluid is stable for disturbances symmetrical about the axis, and for disturbances  which alter the shape o,f  the central curve,  but is unstable for long beaded disturbances.This result  was, perhaps,  to  be  expected, as  by  means  of  beaded  waves, the mass would naturally be broken up into spheroidal masses.

Similar results seem to hold for more ellipsoidal rings.
The condition for instability is $R>3r$, that is, thin hoops are unstable while really thick doughnuts may be stable.
On the other hand, an accreting disk may somehow shed angular momentum. Could that produce a fluid or solid torus (albeit with some body in the middle)? This is the problem of why Saturn retains its rings. Tisserand showed that $N$ satellites could orbit in the same orbit if $m/M < 2.3/N^3$ -- the more satellites, the more they would tend to clump unless they were very tiny. (Sheeres & Vinh 1993) tighten the bound a bit. So unless there are other forces than gravity keeping things in place a dust or gas ring will tend to clump.
Since then others have studied the question, such as (Wong 1974). They found that in principle it looks like there is a pathway of unstable intermediate shapes from the Maclaurin ellipsoids to the torii (see Ansorg, Kleinwächter & Meinel 2003), and if there is enough dissipation of the right kind it might counteract the beaded instability splitting it. But it all looks unlikely to happen unless everything is set up perfectly.
Wong's paper speculates in various astrophysical cases where a torus may be possible. The closest to reality is likely ring galaxies, where the stellar ring remains for a relatively long time. There is also a fair bit of interest in toroidal clouds around black holes (with or without accretion disks), since they make sense as models for active galactic nuclei. But here there are gas pressure and luminosity from disks complicating things: it is far from a ring-shaped star or planet.
(Which is a shame, since I had so much fun modelling them!)


Can I sense a bright star pointing an eight foot antenna towards it?

Stars are too dim for amateur radio equipment. There are two possible radio sources that you can detect: the sun and Jupiter.

Jupiter is particularly interesting as interactions between Io and its magnetic field produce beams of radio waves that sweep past earth every 10 hours.  These are detectable in the amateur range, at about 20 MHz.

Nasa make a kit for detecting these radio signals, or it is possible to use a ham antenna, but of course it must be cut for the frequency of operation. The Nasa kit uses a phased dipole antenna which must be set up in a field or similar as the antenna is about 7m long.

Stars are not very good radio sources. Supernovae remnants such as Cassiopeia A or the Crab nebula are much brighter at radio wavelengths. Most supernovae are too distant to be powerful radio sources; radio supernovae are rare. A local supernova would be a radio source but we haven't observed a supernova in the milky way for several hundred years.


Can white dwarfs have accretion discs too?

Yes. They are called cataclysmic variables.


Using parallax to find distance to the moon

I am sure that measuring the angle compared with distant stars is the best approach, but you can in fact work out the angle even without looking at distant stars, if you assume the Earth is spherical and you know its radius, and you know both the length and the direction of the great circle between you & your friend.
Here's a terrible drawing:

So here, the radius of the Earth is $R$, and the great-circle distance between you and your friend is $l$.  So the angle between you and your friend is $\theta = l/R$ (radians).
Now, you need to agree both to measure the Moon's angle above the horizon at a particularly good moment: that moment is when it crosses the plane defined by you, your friend, and the centre of the Earth.  In other words, you need to measure its angle above the horizon as it crosses the great circle between you.  This is annoying, because you have to know the direction of the great circle, but it's also nice because you don't need a clock.
So, when you measure its angle above the horizon you get two angles, $\psi$ and $\phi$.  And now you have the quadrilateral I drew below the main diagram, and you know three of its angles and two of its sides.  The total angle in a quadrilateral is $2\pi$, so this tells you the fourth angle, which is $\alpha = \pi - (\theta + \psi + \phi)$.  And you can then go on to solve the quadrilateral for the distance to the Moon, $d$ (I am leaving this out because I can't remember how to do it!)
Note that this will only work if the Moon is above the horizon for both of you when it crosses the great circle: it may not be I think.


How to align equatorial mount in the Southern Hemisphere?

I'm going to post answers for both hemispheres, because there are good points from both sets of directions I've found.

For the northern hemisphere:

  The "declination drift method" is the most accurate way to accomplish this. The method is straightforward, but it does require some time and patience.

  First, aim the mount's polar axis roughly at Polaris. Now point the telescope at a star that's somewhat above the celestial equator and as close to south as you can judge by looking opposite Polaris. Put in a high-power eyepiece. If the eyepiece has cross hairs, center the star on them. Otherwise put the star on the north or south edge of the field and defocus it a little. Turn on the clock drive, and ignore any east-west drift.
  If the star drifts south in the eyepiece, the polar axis is pointing too far east.
  If the star drifts north, the polar axis is too far west.
  Shift the polar axis left or right accordingly, until there is no more drift.
  Now aim at a star that's near the celestial equator low in the eastern sky.
  If the star drifts south, the polar axis points too low.
  If the star drifts north, the polar axis points too high.
  Again, shift the polar axis accordingly.
  Now go back and repeat from the beginning, because each adjustment throws the previous one slightly off. When all visible drift is eliminated the telescope is very accurately aligned, and you can take long deep-sky exposures.

Via Sky & Telescope

For the southern hemisphere:

  Part 1

  Direct your telescope to a bright star that's low-ish (about 20 degrees) on the Eastern Horizon, near the Celestial Equator (i.e. 0 degrees DEC). The Orion nebula would be good to look at in mean time (given the right time of the year - as it may be too high or not visible).
  At this point, the east/west rotation of your mount has little effect, leaving you to correct the angle of elevation of the mount.
  Now, if the star drifts NORTH, your polar axis is too low - change the axis to angle further up in the sky (i.e. the latitude knob on your eq mount)
  But if the star drifts SOUTH, your plar axis is too high - change the axis to angle further down toward land (i.e. the latitude knob on your eq mount).
  You can tell if the star is drifting north or south by allowing the star to drift for a little bit - now, move your telescope to "catch up with the star" using the DEC control. If you see it heading north, then it's drifting north and vice-versa!

  Part 2

  Now, point the telescope to a bright star on the Meridian and on the celestial equator i.e. right above your head somewhere. This is so the elevation of the mount (as corrected in the steps above) will have minumal effect, allowing you to accurately correct the east/west rotation of the mount.
  Now, if the star drifts NORTH, your polar axis is too far EAST - so rotate your mount further West (i.e. rotate your mount clockwise while looking down at it).
  Now, if the star drifts SOUTH, your polar axis is too far WEST - so rotate your mount further East (i.e. rotate your mount anti-clockwise while looking down at it).

  Part 3

  Repeat these two steps a few times to increase the accuracy of your plar alignment - repeating Part 2 will allow you to get Part 1 more accurate.

Via OZScopes

Edit to provide less-precise-but-more-prompt method

  Polar alignment for southern hemisphere observers is very similar to techniques used for the northern hemisphere. The major difference is the lack of a bright pole star like Polaris near the south celestial pole (SCP) to aid orientation used in several alignment procedures.

  Rough polar alignment using the latitude scale of your scope is identical to the procedure used in the northern hemisphere.

  Either look on a map, use Google Earth, or consult an almanac to find your observing site's latitude. Unlock any latitude adjustment screws on the sides of the mount and turn the latitude adjustment screws until the index on the polar axis reads your latitude. Tighten the adjustment screws if needed to secure the latitude setting. (You may also need to loosen the center pivot bolt by turning the hex nut to allow the equatorial mount head to be tilted.)

  Latitude adjustment of the mount

  Now complete the polar alignment by turning the entire mount (not either axis; both should be clamped tightly) to align the upwards end of the polar axis with south on the horizon.

  Another more accurate rough method is to point to Sigma Octantis. This star is located only one degree from the SCP, the point in the sky around which all the other stars appear to rotate, and where the polar axis of a properly aligned equatorial mount should point.

  First, set up the mount so that the polar axis is pointing south.

  Second, unlock the declination clamp and move the scope in declination so that the tube is parallel to the polar axis. Your declination setting circles should read 90° in this orientation. Clamp the declination lock.

  Note: These last steps involve moving the entire mount. Don't use either the RA or Dec motions to change the position of the tube.

  Move the mount in altitude and azimuth until Sigma Octantis is in your finder's field of view or centered in your finderscope. At magnitude 5.5, it is a fainter star--just slightly brighter than the unaided eye limit--so your finder will definitely be helpful here.

  Sigma Octantis and the SCP

  Tweak the position of the mount by again moving the mount, this time centering Sigma in the eyepiece field of view. Altitude can be adjusted using the latitude adjustment screw or shortening-lengthening tripod legs.

  For greater accuracy, you can point directly at the SCP using an offset from Sigma Octantis. (This can be done after aligning the optical axis of your finderscope with the polar axis.) Use the chart above or these patterns of southern circumpolar stars to find the SCP. Draw an imaginary line toward the SCP through the Southern Cross stars Gamma Crucis and Alpha Crucis (down the long axis of the cross). Draw another imaginary line toward the SCP at a right angle to a line connecting Alpha Centauri and Beta Centauri. The intersection of these two imaginary lines will point you close to the south celestial pole.

Via Celestron


How precisely we know the Gaia spacecraft baseline? [Question related to Planet 9 and the cosmic distance ladder]

This page of the documentation sort of answers the question.

An important part of Gaia data processing is the Gaia ephemeris allowing one to compute Gaia position and velocity in the BCRS for any moment of time covered by observations. Clearly, the accuracy of Gaia ephemeris is crucial for the project. The required accuracy of Gaia velocity is driven by the aberration of light: 1 μas in direction corresponds to about 1 mm s−1 in velocity. The requirements for the accuracy of Gaia position comes from the paralactic effect for the near-Earth objects (NEOs) and was assumed to be 150 m. The latter requirement is only important for data releases that include solar-system data.

So the position is known to 150m and the velocity to a few millimeters per second. This is probably relative to the Earth, but the Earth's orbit is known to comparable precision. In fact a far less precise determination of the position would do for observing the parallax of stars. The largest such parallax is about 1 second of arc and the Gaia's observations of the position of stars is accurate to a few micro arc seconds, so the errors there are at least a few parts per million. On the other hand an error of (say) 1km in the diameter of the orbit is a few parts per billion. The precision location data is only needed for observing objects within the solar system.


Configuration required to tidally heat Earth-like rogue planet

Do you need tidal heat?  What about core heat?

Ice is a fair insulator.  Put enough ice on top and the heat flux from the core can keep the bottom of the oceans liquid, no tides required.

If you need the water on the surface, I don't think it's possible. Even a huge amount of interior heat (whether from tides, radiation, or primordial) wouldn't be sufficient to hold the surface liquid.  The Earth possibly had periods where there was no surface water (snowball earth), and that was with the advantages of an atmosphere and solar input.

The lack of a sun is going to make having an Earth-like atmosphere difficult.  In an interstellar environment, any thin atmosphere will freeze out.  Without an atmospheric blanket, the surface temperature plummets.

So I think you could have an ocean, but not with surface water.

One more thing.  Assuming the current heat production of the earth, and its size, I calculate that something around 5.6km of ice is sufficient insulation.  Water below that depth with average heat flow would be liquid (ignoring pressure effects and only assuming temperature and heat flow issues).  That's a lot of ice, but not implausible.


Are there any examples where the transverse doppler effect is applied in astronomy?

Accounting for the transverse Doppler effect (and other relativistic effects) is essential in modelling the X-ray spectral emission lines from the accretion discs around black holes (e.g. Cadaz & Calvani 2005). In this case the transverse Doppler effect is "mixed up" with gravitational redshift and it is treated holistically in the Schwarzschild or Kerr metrics.

The transverse Doppler effect is also essential in interpreting the signals from binary pulsars and even in single pulsars because of the rotation of the Earth and its motion around the Sun.


The Chelyabinsk meteor didn't explode.

In fact meteors in general usually don't explode, this is a common misconception.
When an object at orbital speeds ($\sim11 km/s$) enters the atmosphere, this speed corresponds to extremely high mach-numbers (M$\sim 30$). This generates a strong shock-front in front of the meteor which causes the meteor to heat up enourmously.
To give a feeling of how important this heat-up is, you can look up on Wikipedia that already at Mach-numbers of 2-3 there are severe engineering problems in not letting aircraft overheat. At Mach-numbers of 30, the air in the shock is so hot that it locally ionizes the air and easily evaporates the solids away from the falling meteor. Upon reaching the ground one hears this shock as loud thunder.
The other thing that happened with the Chelyabinsk meteor is that it broke up in mid-flight. This caused the total solid surface of the meteor that is exposed to the shock to momentarily increase strongly. More surface exposure to the heat causes more surface to glow. This lasts until the smaller fragments that resulted from the breakup evaporate themselves.

Taken together, the mach-shock and the increase in glowing surface area upon break-up give the impression of an "explosion".
Also for an explosion a reservoir of energy would need to be suddenly tapped, which is just not there.


Why do photos of Earth always have a black background?

The photo of the Milky Way is an exposure that was 30 seconds or longer in order to capture the stars. The photo of the Earth was an exposure that was probably 1/100 of a second. (It is daytime!) The stars will not appear in such a short exposure.


The Earth will keep revolving in the same direction during a geomagnetic reversal.

There have been a few articles about this phenomenon published during 2018:

National Geographic - No, We're Not All Doomed by Earth's Magnetic Field Flip

Science Daily - Earth's magnetic field is not about to reverse

The Conversation - The Earth’s magnetic field reverses more often – now we know why


Why does the Sun track out a seemingly sinusoidal path on the celestial sphere?

Both the equator and the ecliptic are great circles on the celestial sphere.
The appearance of each on a map depends on the map projection.
In an equirectangular projection centered on the equator, the equator (brown) is a straight line, and the ecliptic (blue) is approximately sinusoidal.

If the same projection is centered on the ecliptic instead, the ecliptic is a straight line and the equator is approximately sinusoidal.
As uhoh's answer illustrates, neither curve is exactly a sinusoid.

Lucey provides both views too, but his ecliptic-centered map doesn't show the equator.
The axes are labeled in ecliptic rather than equatorial coordinates.

He also provides a stereographic projection centered on the midday zenith at 55°N.
Here both equator (red) and ecliptic (green) are mapped as circular arcs.


Lunisolar calendar: what phase was the moon Y years ago on the spring equinox?

This is a challenging problem and I can't give you a definitive answer, but here are some thoughts, mostly on the methodology rather than implementation.
First, you are right that fundamentally the way to do this is to integrate the equations of motion backward in time, taking into account the relevant forces.  You are also right that only considering the Sun, Moon, and Earth is inadequate.  The farther back in time you want to go, the more precise your model needs to be, so you need to include the influence of the other planets.
However, just considering the gravitational forces of the various bodies, but treating them as spherical masses, is also inadequate for your question.  For different reasons for the two parts, you have to take into account the fact that the Earth isn't a perfect sphere.
For the equinox, this is because of precession.  For determining the date of the equinox, it's not enough to know where the Earth is in its orbit; you also have to know where the pole is pointing.   There are detailed models of the Earth's precession available, but I don't know how far back they go before uncertainties become large.
For the moon phase, you have to take into account the fact that interactions between the Moon and the Earth's equatorial bulge are gradually transferring angular momentum from the Earth's rotation into the Moon's orbital motion, causing the Moon to slowly move away from the Earth (and thus causing the length of the lunar month to increase slowly). Again, this has been measured and could be incorporated into a model.  The rate of orbital period change is such that 10,000 years changes the Moon's period by several seconds.  That doesn't sound like much, but the cumulative effect is several full orbits' difference over that time period.  In other words, you'd be completely wrong about the Moon's position in its orbit on a given date if you didn't take it into account.
I suspect that there are other effects that would need to be taken into account as well to get an accurate prediction in the distant past, but those are the two that come to mind.
Regarding implementations, currently JPL Horizons is what I would consider the state of the art (though there may be others of comparable accuracy).  Their time horizon is 10,000 years (i.e. you can't request positions farther back than that).  I don't know if that limit is set by accuracy (which will certainly get worse as you go farther back) or by computational resources.


Why does in-the-sky.org say that Mercury is not visible from Taipei these days?

The logic that In-The-Sky.org uses to decide whether events are observable or not doesn't currently work very well for Mercury and Venus. I have some plans to fix this in the long term, but always have far too many projects on the go.

As you correctly say, it's blazingly obvious that Mercury is highest in the sky at lunchtime, and the code I wrote to generate these descriptions should realise that (but doesn't, currently).

The statement that Mercury is going to be very challenging to observe from Taipei is basically correct. At this time of year, Mercury doesn't venture very far from the Sun, even at greatest elongation. The best time to see it in the evening sky is around late September. This is true regardless of where you live in the world, so the statement "all around the world..." in the web page is basically wrong.

As I say, I have plans to improve all of this, but the website is a one-man effort, done entirely in my spare time, so things don't always get fixed quickly :-)


Are X-ray telescopes with glancing angle surfaces basically "funny-looking" Cassegrain telescopes mathematically?

Short answer, yes, the x-ray grazing incidence mirrors set up like this still function like a Cassegrain to minimize off-axis aberration.  However, it is still different in that a classical Cassegrain uses a fast positive paraboloid and a slow negative hyperboloid.
The Cassegrain design combines the paraboloid and hyperboloid to minimize off-axis aberrations effectively. In 1952, Hans Wolter outlined 3 layouts of paraboid-hyperboloid mirrors which could be used in a grazing-incidence for x-ray applications.  The type 2 layout is most equivalent to a Cassegrain, using a positive paraboloid followed by negative hyperboloid, the application used here (in the XMM-Newton observatory) is a Type 1.  As a result, the remaining aberrations are slightly different, though still minimized rather effectively. Also notable, is that the focal lengths of the paraboloid and hyperboloid in the Wolter designs are of similar focal lengths.
Type 1:

Type 2:

https://www.cosmos.esa.int/web/xmm-newton/technical-details-mirrors
https://en.wikipedia.org/wiki/Wolter_telescope


Difference in creation of solar flares and CME?

The point of my comment was that the connection between CMEs and flares would make for a good 20 page research paper in a college course.  The connections are still only understood at a kind of cartoon level, after decades of concerted research.  But I can tell you what the basic cartoon is, and you can think of that as a kind of operating hypothesis that helps organize many of the observations, but is still a long way from a predictive model.  The basic idea is that the energy for flares and for CMEs both seem to come from magnetic free energy stored in the corona due to the magnetic and turbulent nature of the convection zone.  The magnetic field is capable of trapping pockets of hot gas that has been energized by this magnetic activity, but then when the field lines get significantly twisted up they reach a point where they can release that energy and relax to a smoother configuration.  This relaxation does two things-- it releases energy, and it reconfigures the field such that it can release these pockets of trapped plasma.  The former is the solar flare, the latter is the CME, and that's why the two often come together.

However, there seems to be a full range of weak to strong flares, and weak ones would not be associated with CMEs.  On top of all that, there may be other processes that heat the corona, and other processes that lead to the solar wind, so one big question is if the heating of the corona is mostly from lots and lots of small flares, and the solar wind is lots and lots of small CMEs, or if the general heating of the corona and general driving of the wind are separate processes from flares and CMEs.

Also, you mention closed and open field regions, so these questions can be asked separately in those two domains.  But CMEs would normally come from closed field regions that open up when the field reconfiguration occurs, I believe open field regions tend to be rather sedate and not associated with either strong flares or CMEs, but rather a more steady wind.  In the closed field regions, CMEs are often associated with features called  "helmet streamers," which are regions of trapped plasma that are kind of combed out at their highest point by the solar wind, creating an interface between closed and open fields that is a good setup for getting field reconfiguration that could launch out some plasma.


The ratio $T/\Omega$ tells you about the acceleration of the system - or more specifically, the second derivative of its moment of inertia - it does not tell you about the velocity.
If the system collapses because it has $T/\Omega<0.5$, then when it reaches $T/\Omega=0.5$ it stops accelerating. That doesn't mean it stops collapsing. It overshoots in the same way that if you compress a spring it goes beyond the equilibrium position before bouncing back once the deceleration becomes large.
NB: I answered the question (of which your question is a subset) that you linked to in the same way.


Naming of the planets of the solar system

Despite of the explanation of your question, it is a valid question to ask why planets all have Roman names. First of all, the Romans could, like the Greeks and Sumerians, could only see Mercury, Venus, Mars, Jupiter and Saturn. These planets can be seen with the naked eye. However, the fact that the Romans could see these planets, didn't give all the planets latin names.

If the Names of the planets were given by the ones that discovered them first, the planets would have Sumerian names. The answer to the question why the planets have latin names is because they don't. The first western scientist took over the Roman names, because latin was they scientific language of the middle ages and the renaissance. After the discovery of Uranus and Neptune these planets were not given these names by their discoverers. Only after longer controversy western astronomers have standardized the names of planets and moons according to Greek and Roman mytholohy, but this was only around the second half of the 19th century.

Since the beginning of the 20th century the IAU is setting the standards for naming celestial bodies including planets. However, the IAU doesn't set the names, it only sets the standards. Newly discovered objects are now named according to these rules by their discoverers. Older planets are still living with their old names and for western astronomers those are still the names from Greek and Roman mythological figures. However, the IAU allows you to consistently name the planets by their Arabic or Chinese names. So, the question is valid, the text around your question is unfortunately not true, since the planets don't necessarily have Roman names. If you are An Asian person and correctly think the planets should have their Chinese names, because they were earlier in discovering these objects than the Romans (although not earlier than the Sumerians), you are allowed by the IAU (sic) to call them by their Chinese names.


By far the most particles in the visible universe aren't quantum entangled. That's obvious from observation, since if all electron spins e.g. would be entangled, all electrons would flip their spin at the same time, and we couldn't observe varying statistics of electron spins in a sample, resulting in varying degrees of magnetism, e.g. by applying gradually changing exterior magnetic fields.

A second example: All (bosonic) atoms of spin 0 would form a Bose-Einstein condensate not just near absolute zero, since they would be in the same quantum state, hence in the lowest.

The "single source" isn't a single quantum state. In the early phase of the big bang temperatures were very high, allowing for a huge amount of possible quantum states within a small volume. Without warranty you may think of the initial state of the universe as a superposition of all possible quantum states, the wave functions of which collapse relative to an observer, regarding the observer as a particular quantum state.

That way we get to the many world interpretation of quantum theory and a universal wave function. See also a proposed Hartle-Hawking state as a pre-Planck epoch of the universe.


Do celestial bodies have many asteroids orbiting them?

The short answer is not very often.   Article on that here:  can moons have moons?

Nothing can orbit the Moon long term cause the Earth affects a moon/moon-satellite orbit and it draws tidal energy from it so an object would crash into the moon.

A 2nd moon around earth might be more stable than one around the moon, but I suspect, not long term stable.   If it fell into orbital resonance with our moon, like the inner 3 moons of Jupiter, I don't know, maybe, but I suspect not, cause the Moon's tidal force on a 2nd moon would (I suspect) be too great.

From time to time, the Earth captures an asteroid, but they usually don't stay for long:  http://www.scientificamerican.com/podcast/episode/earth-usually-has-second-tiny-tempo-11-12-27/

Mars' 2 moons and many of the gas giant planet's moons are captured asteroids or Kupier belt objects and if the earth didn't have our large moon, we might have a few captured asteroids in our orbit as well, so planets absolutely can capture asteroids, but the closer to the sun (Mercury/Venus), the smaller the planet's gravitational sphere's of influence and the less stable long term.

Curiously, some asteroids even have asteroids, and Pluto in particular is a rather strange system with a comparatively enormous and very close moon, Charon and 4 outer moons that orbit around Pluto and Charon.   you can't get objects orbiting the Earth-Moon system, cause that would extend past the stable part of the Earth's Hill Sphere.   Pluto's Hill Sphere is 'about 3 times the radius of Earth's because it's so far away from any large bodies.

Generally, stable orbits only exist in the inner 1/2 to 1/3rd of a body's hill sphere, so those are the 2 factors to look at, tidal effects and hill sphere (or Sphere of influence, which is related to Hill Sphere).

Sphere or influence numbers for the 8 planets and Pluto:

https://en.wikipedia.org/wiki/Sphere_of_influence_%28astrodynamics%29

Related:  What is the difference between Sphere of Influence and Hill sphere?


Ceres is spherical, Vesta isn't – is it the largest non-spherical solid body in the solar system?

Among the asteroids 4 Vesta is the largest non-spherical object in the Solar system. But far out there in the Kuiper Belt is 136108 Haumea, and maybe yet undiscovered objects, that are very elongated. Haumea is 1,960 × 1,518 × 996 km. That's larger than the 960 km diameter Ceres and hardly spherical with 2:1 proportions. Its fast rotation and composition of malleable volatiles makes this possible. Maybe Haumea is "hydrostatically relaxed" without being spherical?

Haumea images below from Wikipedia.


Counter clockwise rotation of storms at Jupiter's north pole. What explanations have been proposed?

Let my try to stretch the analogy to anticyclonic tornados: The vast majority of all tornados move cyclonic, i.e. counterclockwise on the Northern Hemnisphere of our blue planet. There is e.g. a publication on anticyclonic tornados by Howard Bluestein et. al. which says:

It is also possible that the source of vorticity in the anticyclonic tornadoes is not from a parent mesoanticyclone (produced via tilting), but rather from (hypothesis 2) preexisting anticyclonic shear vorticity on the anticyclonic-shear side of the low-level jet associated with the rear-flank downdraft [...]

In simple words: It is not fully clear yet, how exactly these atypical atmospheric vortices form, but there are different possible explanations why tornados exist which rotate against the direction supported by Coriolis force. This reminds me somehow of the Coriolis Force Effect on Drains:

The notion that the Coriolis force determines which direction water spirals down drains is one of the most prominent scientific myths.

Some educational German television show has been able to actively influence the rotation of water spiraling down a drain with appropriate boundary conditions like little wings in the drain (but the clip is not online). In the metereology of planets with solid surface, the effect of the ground on the atmospheric motion is usually termed orographic effect. I suppose the different rotations of the Juno IR image of Jupiter's north pole have similiar "guide wings" in the lower atmosphere, most probably density fluctuations.
Another important factor is that eddies in fluids live really long, plus these eddies can interact with each other. I remember listening to a scientific talk on some kind of "algebra of eddies", but I did not manage to figure out the reference. However, a similar idea can be found within the references of a preprint on the network of solar flares. The idea of this kind of meta-modelling of Navier-Stokes equation is to save computer-time and nevertheless predict eddy dynamics. What that means for your questions: It is not too unprobable that a single anticyclonic eddy formed, and due its interactions with the other eddies in the storm system, other anticyclonic eddies could have been branched of this one. Let me conclude with the words of @AtmosphericPrisonEscape

One can speculate much and say it should have been A or it should have been B, but unless those scenarios are tested in a 3D global circulation model, nobody will know for sure.

Related

Simulating oceans and ice sheets on other planets or moons?


Hmm - good question, Michael; and thanks for sharing your images with us.

For me, it's about eliminating possible causes:

Collimation - OK (the star test should be fine for your purposes)
Focus - probably OK (suggest you try the same kit - scope, Barlow, DSLR - during the day and focus on a far distance object if possible to double-check)
Chromatic Aberration - Newtonians are less prone certainly, but it can happen. The most common cause is atmospheric dispersion caused by looking at an object very low in the sky. Your profile doesn't say where you're observing but Saturn is very low in the Northern Hemisphere sky at the moment
Magnification - one problem for you is that your kit is probably not optimal for small objects. I have a 1000mm FL Newtonian and need to use a x3 Barlow and a specialist CMOS/CCD planetary camera for Saturn as it looks too small on my DSLR (which gets used for larger objects)

You said "The thing that I am really puzzled about is that through the lens the image did not look like either of these pictures" - please could you clarify what you meant by lens here; were you talking about using the scope, Barlow and a separate eyepiece, or were you using scope, Barlow and DSLR with live preview lens?

It is much more common for people to image the planets using a video stacked approach. The idea behind this is that they are very bright (so exposure times can be very short) and the limiting factor is random blurring caused by atmospheric seeing.

If you take a video file of say 1000 frames at 20fps then there is software available that can combine these into good shots (e.g. AutoStakkert, Registax). What DSLR are you using? For what it's worth, the human eye is very good at picking details and resolving out the seeing on bright objects, so my first photo attempts were dreadful compared to what I could see in the eyepiece.

I'd recommend you try reading up on this approach for inspiration. Some (but not all) DSLRs do allow video recording modes, so it might be worth a try.

Saturn is definitely a tricky object to go for - at the time of writing Jupiter isn't in a good Northern Hemisphere space (I'd normally suggest trying it out as a somewhat easier target). Have you tried practising by imaging parts of the Moon - it's a much easier target to build confidence and practical experience?

Please don't be disheartened ... it took me a couple of years to get semi-decent pictures of Saturn


Giant planets have an abundance of H and He. Hydrogen burning occurs at relatively low temperatures. Rocky planets, on the other hand, don't have much hydrogen and even less helium. For them, the first relevant process would be C burning, which occurs at about 8 solar masses.

But this is valid only as a thought experiment. Rocky planets loose or don't accrete H/He mainly because of insufficient mass. If they were massive, they would consist of H/He and therefore "ignite" at 17 Jupiter masses.


What is the cycle for when the maximum amount of the Moon's apparent area is sunlit?

I asked the same question on the Usenet group sci.astro and received the following helpful reply from Barry Schwarz that goes some way towards an answer, reposted here with permission. But the question remains hard. I have not seen this cycle addressed in any astronomical textbook, although it may be in one that I haven't looked at.

  Since the Sun's diameter exceeds the Moon's, the Sun always
  illuminates slightly more than 50% of the Moon's surface.  (To
  visualize, draw the external common tangents from a larger circle to a
  smaller one.)  Using 98M miles as the distance, 432K miles as the
  Sun's radius, and 1K miles for the Moon's, the Sun illuminates almost
  7 miles into the opposite hemisphere.

  Any observer a finite distance from the Moon can only ever see less
  than 50% of the Moon's surface. The further away the observer is, the
  closer the observable percentage is to 50.  (To visualize, draw the
  two tangents from an external point to a circle.)  However, when the
  Moon is in the plane of the ecliptic, an observer at the north pole
  can see approximately 12 miles over the Moon's north pole into the
  opposite hemisphere.  So he can see some 5 miles beyond the
  terminator.  Similar geometry causes his view of the Moon's south pole
  to fall short of the terminator.

  However, for a person standing in line with the ecliptic, his view in
  any direction falls 4 miles short of the terminator and that is the
  maximum you seek.

  When the Moon is within 4K miles of the ecliptic, an observer on Earth
  at the same offset from the ecliptic will have the view described
  above.

  When the Moon is more than 4K miles away from the ecliptic, all
  Earth-bound observers see some non-illuminated portion and therefore
  do not see the maximum you want.

  Not only is the Moon's orbit tilted with respect to the ecliptic, but
  the direction of the tilt wobbles like a child's top.  I have no idea
  how to factor that into an equation that would compute the  time
  between one visible maximum and another.


Were parts of the Universe deflating during Inflation?

That sounds... confused, or else just wrong. Negative pressure is what causes inflation: something (inflaton field, cosmological constant, or "dark energy" in general) with an energy density that stays approximately constant as space expands has negative pressure and will drive the exponential expansion of space.

The transition to "true" vacuum is what shuts off inflation, after which you get  the standard Big Bang, during which the expansion continues but slows down.

(The cosmological constant -- or "dark energy" -- also has negative pressure, and so the universe is gradually moving towards a state where the space between galaxy groups and galaxy clusters will again be expanding exponentially.)

The "eternal inflation" model has our visible universe -- almost certainly a tiny part of a much larger region of true vacuum -- as one of many such pockets of true vacuum embedded within a vastly larger false vacuum that's still expanding exponentially.


How big is the universe. How does it works?

There are features in the CMB, the distance and size of which can be derived from cosmological models (6-parameter Lambda CDM model). The expected size of the features can be predicted in terms of angles in the sky, depending on how curved large-scale space-time is. (The features are a statistical property of tiny temperature fluctuations in the CMB indicating the largest scale of causal connection - kind of sound waves - in the baryon-photon plasma before recombination 380,000 years after the Big Bang, called first acoustic peak of the CMB power spectrum. The angle should be about 1 degree for a flat space-time, in accordance with observation.) Things aren't quite straightforward, e.g. due to the way the universe has been expanding, therefore it's done with simulations.

Angles look larger in a positively curved space-time (e.g. a sphere) than in a flat space-time (e.g. a plane) than in a negatively curved space-time (e.g. a saddle).

The curvature of space-time follows the same principles as curvature of 2-dimensional surfaces. A triangle in an Euclidean plane has an angle sum of 180 degrees, while a triangle on a sphere has an angle sum of more than 180 degrees. Hence an object of a fixed size and a fixed distance looks larger (takes a larger angle) on a sphere than on a plane.

Here a sketch of a triangle on a sphere (may be interpreted as an observer looking to an object) and a triangle of the same size - meaning the same lengths of the sides in this case - in a plane:

(More on non-Euclidean geometry.)

If the observable universe is positively curved, it could be a fragment of the surface of a 3-sphere resp. 3+1-de Sitter space-time. A 3-sphere is finite. If the observable universe is flat (curvature 0), it could be a fragment of an infinite Euclidean space resp. 4-dimensional Minkowski space-time. That's plausible, but not the only possible extrapolation beyond the observable universe. An almost flat observable universe could e.g. be a fragment of a huge 3-spherical universe or of a huge 3-torus. Planck data don't indicate (in a significant way) that kind of structure outside the observable universe, but extrapolations become less valid the further we go beyond the observable universe.

According to the current (Lambda-CDM) model of the universe together with the best-fit parameters, the universe is flat within the limits of measurement. Hence it's plausible (by extrapolation), but not necessary, that the universe is either huge or even infinite.


Is there a far away probe that has recorded the light spectrum from Earth as if it were an exoplanet?

TL;DR: (1) we don't need to go very far to measure the spectrum of Earth's reflected light: a satellite in orbit around the Earth could easily do that; however, (2) detecting the reflected light of an Earth-size planet is extremely difficult: we currently use other techniques with much greater success.

The Wikipedia article on how scientists detect exoplanets provides a good introduction and overview - well worth a quick read. In summary:

Most of the thousands of exoplanets detected so far have been found using the transit photometry method, where the measured brightness of the star temporarily dips because a planet is blocking some of the light as it transits across the face of the star.

The next most successful technique is to measure changes in the radial velocity of the star to deduce the presence of a planet (by the "wobble" its gravity causes). This was the most productive method until 2012, after which transit photometry became the dominant detection method.

However, neither of these methods are looking for reflected light from the planet.

It is possible to deduce the presence of a planet from periodic changes to a star's brightness that would correspond with a nearby planet going through its phases (like how the Moon is bright when full and "dark" when new), but we're typically talking about a Jupiter-sized exoplanet in a close orbit - definitely not a relatively "tiny" Earth-like planet further out from the star.

It's also possible to directly detect the reflected light, or the infrared light emitted by a "hot" planet, or light that has been polarised by the planet's atmosphere – but all these methods rely on a much-bigger-than-Jupiter planet orbiting a relatively small/dim star, and preferably in our local stellar neighbourhood, since the light we're trying to detect is exceedingly faint. Note that currently the smallest directly-imaged exoplanet is still more massive than Jupiter, and Jupiter itself is over 300 times more massive than the Earth (and ten times the Earth's radius, which is even more relevant).

Note also that Proxima b (an Earth-size planet orbiting our very closest neighbour, the red dwarf Proxima Centauri) was discovered using the radial velocity method, and while efforts are under way to observe it directly, no direct image has yet been obtained. This might give you an indication of just how hard it is to directly image such a (relatively) tiny planet!

Also, we don't need to send a probe to the edge of the solar system to work out what we're looking for, as we already know exactly what the reflected light from Earth looks like – the iconic Earthrise being the most famous example. And in 1990 Voyager 1 took a famous photo of the Earth from 6 billion kilometres (41 AU) away: see the Pale Blue Dot.

By Sept 2017, Voyager 1 was 21 billion km (140 AU) from the Sun - by far the most distant probe we've ever sent. Nonetheless, that's a long way from the "border of the solar system". Voyager has entered the interstellar medium but it hasn't yet left the Kuiper Belt, and it will be another 300 years before it reaches the Oort Cloud, and tens of thousands of years before it reaches the real edge of the Solar System.


Average amount of annual daylight at any place on earth

Wikipedia strikes again:

  The naive expectation is that, for every place on Earth, the Sun will
  appear to be above the horizon for exactly half the time. Thus, for a
  standard year consisting of 8760 hours, apparent maximal daytime
  duration would be 4380 hours. However, there are physical and
  astronomical effects which change that picture. Namely, atmospheric
  refraction allows the Sun to be still visible even when it physically
  sets below the horizon line. For that reason, average daytime
  (disregarding cloud effects) is longest in polar areas, where the
  apparent Sun spends the most time around the horizon. Places on the
  Arctic Circle have the longest total annual daytime of 4647 hours,
  while the North Pole receives 4575. Because of elliptic nature of the
  Earth's orbit, the Southern Hemisphere is not symmetrical: Antarctic
  Circle at 4530 hours receives 5 days less of sunshine than its
  antipodes. The Equator has the total daytime of 4422 hours per
  year.

Further details here on astronomical causes of average daytime variation, and here on Insolation, the solar radiation received at the top of the atmosphere and its effects on the energy received at ground level.


What are the apparent visual magnitudes of Spica A and Spica B?

The Wikipedia article says that Spica A has 8 times the luminosity of Spica B (presumably that's at maximum.)  Magnitude is -2.5*log10(I/I0). So if the combined magnitude is 0.97, the magnitude of component A is ~1.1 and the magnitude of component B is ~3.3.


This particular object I tracked down by comparing the visual with the infrared image of the sky. It was a little tricky because no very bright star appears at that location. But after making comparisons, the star in the original image is HD 155161, also known as V* AH Sco, indicating that it is a variable.

In fact, Simbad lists it as a red supergiant at coordinates [17 11 17.02114,  -32 19 30.7132], and visual magnitude of about 7.5 Red supergiants are often unstable and are semi-regular varible stars.

That explains why it is much brighter in the VISTA infrared image than in the optical. Red supergiants have cool outer layers that give off much of their radiation as infrared, not as visible light.

Looking at the star on Aladin Lite, even in the infrared, there is not a hint of nebulosity around the star. The conclusion is that this is an optical effect, lens flare or similar.


This is most likely not noise.

Photographic noise is generally going to be limited to roughly single pixels. This is because much of it is generated at the individual pixel level (often a single color, so not even an entire pixel) on the sensor. There is also some relationship to this and overheating - I'm not familiar with the cooling systems of the Hubble, but presumably it's sophisticated enough to prevent serious noise from overheating.

When examining the photo, you can see that these are not individual pixels but rounded seemingly point-source light. It's hard to say whether this is an accurate representation of their actual angular diameter, as they may have acquired some of that rounding by diffraction. Regardless, this suggests that these are not artifacts from the sensor but are actual light sources. For a long exposure photograph, this also means that they are a persistent light source. The colors that they are composed of also fits the colors we would expect to see from stars.

Unless there has been some very heavy editing done with the intention of representing noise as stars, I think it's fairly safe to say that these are all stars or other star-like light sources.


Calculating the date and year when a lunar eclipse will fall on the Winter Solstice

A university library in 1947 might have a copy of T. von Oppolzer,
Canon der Finsternisse (1887),
which lists those lunar eclipses on pages 368 and 374.
A reader with a working knowledge of astronomy could make sense of the tables even if not proficient in mathematics or German.

If that reference were not available, a text such as R. Buchanan,
The Mathematical Theory of Eclipses (1904),
would equip the professor to work it out from scratch.
As lunar eclipses are visible from the entire night hemisphere,
they are simpler to compute than solar eclipses,
which are total only along a narrow path.

Eclipses recur in
saros cycles
of 18 years and 11.3 days.
Given dates of some contemporary eclipses,
one could trace
saros 119
back: 1945-06-25, 1927-06-15, ..., 1638-12-21; or trace
saros 125
forward: 1938-11-07, 1956-11-18, ..., 2010-12-21.
The saros numbers were not assigned until 1955,
but the cycles were known to ancient astronomers.


Does Dark Matter affect the motion of the Solar System?

The dark matter model that is used to explain the "missing mass" problem relating to our Galactic rotation curve, consists of a pseudo-spherical distribution that is much more extended than the visible stars and gas. Even though this "halo" contains more than ten times the mass of the visible matter, when you work out what it's density should be in the solar neighbourhood, it turns out to be around 100 times less than the "normal" interplanetary medium.

This means, that inside the Earth's orbit, there could be as much as $10^{10}$ kg of dark matter. This sounds a lot and in principle, yes, this would speed up the orbits of planets in a way that depended on their distance from the Sun. However, the predicted effects of this are still about 6 orders of magnitude below the current precision of measurement.


How many photons does it take to determine the existence of a distant object?

What matters is how many photons you collect versus how many you would expect to see if the object wasn't there.
Photons would be present, without a source, for a variety of astrophysical (e.g. diffuse background) or non-astrophysical (night sky brightness, dark current) reasons, so you need to be able to rule out the null hypothesis that what you have seen is consistent with no object at all. In principle, if you expect no background counts, then the detection of one photon is significant.
The lowest backgrounds tend to occur in space-based X-ray and gamma ray observations, where the detection of a few photons is often taken to be evidence of the presence of an X-ray source.
The backgrounds tend to be higher in optical observations. More photons from the source are needed to get a significant detection, because the variance in the background is proportional to the expected background.
Determining anything beyond detection typically needs an order of magnitude more photons because you are slicing your data into several positional or wavelength bins and you need significant detections in those individually. For rotation you need both position and wavelength, so I would say you need yet another order of magnitude increase to be able to say anything.
It is hard to write anything but a vague, general answer, because the details depend exactly on the telescope, the detector and the type of object being observed.
For example, one way of looking for very distant objects is to look for mere detections of high redshift Lyman alpha emitters. To minimise the background you use a set of narrow-band filters which would select the Lyman alpha line over a narrow range of redshifts. Detection of a source through one filter combined with non-detection in adjacent filters tells you it is an emission line object, and assuming it is Lyman alpha you then also know the redshift.


Client-side JavaScript astronomy libraries which are not based on node.js?

I found "Astronomy" from "cosinekitty" on Github:
https://github.com/cosinekitty/astronomy
It supports both node.js and browser.
Browser demos:
https://github.com/cosinekitty/astronomy/tree/master/demo/browser
Testing the library without downloading whole repository:
Save this file locally:
https://github.com/cosinekitty/astronomy/blob/master/demo/browser/positions.html
Into the file, replace line
<script src="astronomy.browser.js"></script>

by
<script src="https://cdn.jsdelivr.net/gh/cosinekitty/astronomy@1.0.1/source/js/astronomy.js"></script>

Reference: https://cdn.jsdelivr.net/gh/cosinekitty/astronomy/
Change this line:
let observer = new Astronomy.Observer(latitude, longitude, elevation);

to
let observer = new Astronomy.MakeObserver(latitude, longitude, elevation);

Remove line:
 <link rel="stylesheet" href="astro_demo.css" />

You should get the table in the file filled up with data calculated by the library.

Another one is astronomy-bundle-js:
https://cdn.jsdelivr.net/gh/andrmoel/astronomy-bundle-js@6.0.4/src


When was the last time a great conjunction coincided with a solstice?

It depends how you define “conjunction.” For example, this year, Jupiter and Saturn will be at the same ecliptic longitude and the same right ascension within hours of each other on December 21, 2020. However, in 2000, they were at the same right ascension on May 30, but at the same ecliptic longitude on May 28, 2000.
I have done calculations (using the VSOP87 planetary theory for positions and Meeus’ Astronomical Algorithms, Second Edition, 1998, for dates of solstices), for conjunctions measured by ecliptic longitude. The result is at https://astronomie.quebec/conjsols.html
This doesn’t give you the precise moment of conjunction; it just lists solstices where Jupiter and Saturn were at less than 5° from each other, between the years −1000 and +3000. As you can see for 2020, the spacing between the two in ecliptic longitude (i.e. neglecting latitude difference) is 0.0387°. The last time these planets were so close in ecliptic longitude at the solstice was in −204 (i.e. 205 BCE).
The only other times in the time span calculated where Jupiter and Saturn were at less than 0.1° in ecliptic longitude (again, neglecting latitude) at the solstice are in −860 and in +2159.
Interestingly, it is in 2020 that the planets are so close in latitude as well at the solstice; their spacing was slightly more that 56′ in 861 BCE and slightly more than 19′ in 205 BCE; and will be more than 1° 12′ in 2159 CE. (Again, these are spacings at the solstices; they may come closer at other dates than solstices, but your question was solstices.)


Which came first: black holes or galaxies?

Note: I am not an astrophysicist. If someone with the relevant expertise finds my analysis of the source paper to be invalid, I would appreciate your help with correcting my answer here.
This is a somewhat complicated question to answer. Black holes may precede galaxies in the form of primordial black holes which according to some models arose in the early Universe from quantum fluctuations. Primordial black holes are currently hypothetical. Supermassive black holes (that you find at the center of galaxies) may not come from these, however. There are a few possible ways that supermassive black holes may have formed:

Molecular cloud / dwarf galaxies collapsing to form supermassive stars (SMS) which later further collapsed into supermassive black holes (SMBH)
Dwarf galaxies forming massive stars which collapse into massive black holes, which later coalesce into SMBHs
Primordial black holes accreting sufficient matter to become massive black holes, which later coalesce into SMBHs

Depending on how strictly you define "galaxy" it does seem most likely that the supermassive black holes formed inside early galaxies, but a type of black hole (primordial black holes) may have preceded galaxies.
Source:

Formation of the First Supermassive Black Holes - Bromm and Loeb

UPDATE 10/21/2013:
This just in, recent observations have ruled out the possibility that SMBHs gain mass only through merging with other black holes. For more, see this Astronomy.com article.


About curvature, you can see it. Just look at a ship some hundred meters away: you'll notice if you are careful that you can not see where the ship touches water.

You can see at http://mathcentral.uregina.ca/qq/database/QQ.09.97/dyck2.html that the curvature is 8 inches per mile.

On the other hand, about rotation, planes do actually take advantage of earth rotation, but not directly. This was answered in https://physics.stackexchange.com/questions/16390/does-the-rotation-of-the-earth-dramatically-affect-airplane-flight-time


How did the solar system get its angular momentum? Why did Earth not fall into the Sun at the start?

A particle's angular momentum around a point is the particle's mass times its distance from the point times its velocity around the point.  So a small velocity at a large distance can still mean a lot of angular momentum.

The bit of a Giant Molecular Cloud which collapses to form one solar system is typically on the order of a light-year across or somewhat less. It normally possesses some angular momentum -- just from random motions if nothing else.

When it collapses to form a solar system, its size drops from maybe a light-year down to maybe 10-100 AUs. A light year is about 60,000 AUs (astronomical units), so it is collapsing to about 1/1000 the size. To conserve angular momentum, the circular velocity of the matter must increase by a factor of a thousand.  And the fraction of the matter which will form the star collapses by another factor of a thousand or so.

Not only does this give you plenty of speed, but for the star to form it actually needs to shed angular momentum.

See the Wikipedia article on angular momentum for a lot of good details.


How can we focus radio telescopes on a star when the earth is spinning?

Part of the answer that I suspect the original questioner needs is that although the Earth is indeed spinning very fast, the amount the surface of the Earth moves relative to an astronomical object is tiny.

So you put motors in the base of the telescope so that it slowly turns to look at the same patch of sky.   You don't need to refocus because telescopes are looking at objects so far away that focus doesn't matter.  You don't need to do anything else because the Earth's movement is smooth and continuous, and it's not about how fast you're moving, it's about how quickly you're turning.  In our case, one complete circle every 24 hours which is pretty slow.

Focus at infinity just means that you set the focus of the telescope so that an object which is infinitely far away would be in perfect focus.  It depends on the quality of the telescope, but the practical difference between focus at infinity and focus at the actual distance disappears after a few miles or so.  At the distance of stars, there is essentially no difference at all.


It's spread out in time. If a source emits a 1W pule of energy lasting 1 second and the receiver is receding so fast that it is Doppler shifted to a frequency which means the power is just 0.5W, then the pulse will take 2seconds to arrive (since the end of it had further to travel).


How does one measure vector magnetic field of astrophysical object?

Using spectropolarimeter one can measure vector Magnetic field.

A spectropolarimeter based on a ferroelectric liquid-crystal modulator is described. An optical system with spatial modulation of the positions of the components of Zeeman splitting is a specific feature of this instrument. In comparison to the familiar instruments, the developed spectropolarimeter utilizes the light flux more efficiently and contains only one photodetector array. An operating spectropolarimeter developed at the Sayan Solar Observatory is considered as an example. Comparative estimates of noises in different operating modes are presented.

(Ref: Kolobov, D.Y., Kobanov, N.I. & Grigoryev, V.M. Instrum Exp Tech (2008) 51: 124. https://doi.org/10.1134/S0020441208010156)


Gravitational field for oblate spheroid?

If you have a self gravitating sphere than all force vectors at the surface are normal to that surface (and other equipotentials interior to the surface). These force vectors only point towards the centre in the case of a sphere (or at the poles and equator of an arbitrary spheroid). However, even if you remove any centrifugal effects and just have a rigid, non-rotating spheroid, the gravity vector still does not point to the centre of the spheroid except at the poles or the equator and that is true either inside or outside the surface.

The gravitational potential outside of a uniform spheroid of mass $M$ can be expressed to high accuracy with
$$\Phi = -\frac{GM}{r} + \frac{kG}{2r^3}(3\cos^2\theta -1),$$
where $r$ and $\theta$ are the usual spherical coordinates and $k$ is a constant equal to the difference in the moments of inertia about axes parallel and perpendicular to the rotation (symmetry) axis (note that the object would have to be rotating in order to be a self gravitating spheroid, but that the potential above is just the gravitational potential and does not include any centrifugal, non-gravitational component and is equally appropriate for a rigid oblate spheroid).

Taking the gradient of this potential (since gravitational field  $\vec{g} = -\nabla \Phi$), you see there is a radial term in $\hat{r}$, but also a $\hat{\theta}$ component
$$g_{\theta} = \frac{3kG}{r^4}\sin \theta\cos\theta$$
that is not directed towards the centre of the spheroid.

In the limit of an almost disk-like spheroid, then at large distances from the centre and reasonably close to the disk plane, this term will dominate and the gravity will tend to act towards the disk midplane rather than towards the centre of the disk.

The above analysis only includes the monopolar and quadrupole gravity fields. There are (smaller) higher order terms that can be included in the potential, but the result is the same -- that the gravitational field outside an oblate spheroid is not a central force (Hofmeister et al. 2018).


Why the continued obsession in measuring the year as a multiple of days?

Why all the stunts with leap days and leap seconds, etc... when we now have atomic clocks?

Science exists to serve mankind, not to rule it. Calendars were one of the first concepts developed by humankind, arguably predating writing. Timekeeping is even more important to modern society than it was to the ancients. You're not going to get rid of calendars.

One key purpose of the Gregorian calendar is to keep the calendar in sync with the seasons. Every solar calendar requires some intercalation mechanism. The Gregorian calendar uses leap days. The Gregorian scheme (years divisible by 4 have an extra day, except for years that are multiples of 100 that aren't divisible by 400) will fall out of sync with the seasons at at the very slow rate of one day per 3030 years.

Leap seconds are a rather different beast. They are inserted, unpredictably, to keep the clock in sync with the day/night cycle. Without leap seconds, counting 86400 atomic clock seconds = one day would already have the clock out of sync by about half a minute. Leap seconds are problematic for business. There is an ongoing effort to remove leap seconds and just live with a day of 86400 seconds going out of sync with the day, at least until it truly becomes noticeable.

  Why not decouple the two and say that day X belongs to year Y if at 12:00 noon of that day is "in" year Y?

What's this "year Y" thing you're writing about? How do you know what year it is without a calendar?


Pinhole projector for the Transit of Mercury

I've just rewritten this answer - @MikeG caught a glaring error by pointing out a really basic handy relationship called the Rayleigh criterion.

\begin{align} {\theta}_R \approx1.22 \frac{\lambda}{D}. \end{align}

It's better to read the (or any) article, but very briefly, the angular resolution is roughly the ratio of the wavelength to the diameter of a circular aperture. You can apply this equally well to pinhole-only imaging, or to a system which images by focusing with curved mirrors or lenses.

Mercury's diameter is about 4900 km and since it will be on the line between the sun and the earth, the distance will be about 150,000,000 minus 58,000,000 or 92,000,000km. In that case the angular width of Mercury will be about:

\begin{align} {\theta}_{merc} \approx \frac{4.9 \times 10^3\ \mathrm{km}}{9.2 \times 10^7\ \mathrm{km}} \approx 0.000052 \ \mathrm{rad} \approx 0.0031°  \approx 11 \ \mathrm{arcsec}.  \end{align}

So to even poorly resolve Mercury as a dark fuzzy dot, you'd like the diffraction width to be equal or less than the angular width. If you set the two angles equal and let $\lambda$ = 580 nm, you get

\begin{align} D \approx 1.22\ \frac{5.8 \times 10^{-7}\ \mathrm{m}}{\theta_R}  \approx \frac{7.1 \times 10^{-7}}{5.2 \times 10^{-5}}\ \mathrm{m}   \approx 14\ \mathrm{mm}\ (minimum)   \end{align}

Since the light is essentially parallel, your geometrical resolution on the screen is about the diameter of the pinhole. To make a 5.2E-05 radian object 14mm on a screen the screen would have to be VERY far away:

\begin{align} L_{to screen} \approx \frac{D}{\theta_{merc}}   \approx  \frac{0.014}{0.000052} \approx 270 \ meters! \end{align}

You can try that, use a 14 or 20mm "pinhole" at 270 meters away, but I think the light will be far to faint to see. I once did something similar to see a solar eclipse. It may have been a 10mm "pinhole" but I'm sure I wasn't that far away. I used household mirrors to bring the light indoors to a very dark area, and it worked great.  But that was at most only abour 30 meters!

If you would really like to try it, here are some tests you can do ahead of time:

Search to see if someone has done it and given details
Try some simple experiments using the full sun. If the fuzziness at the edge of the sun's image really seems to be 300 times smaller than the sun's image, then that's a suggestion that it might work.
Check the internet for daily sun images and see if you can find a sunspot. Then test your setup and see if the sunspot suggests your resolution looks as good as 1/300.


The rubber sheet only is not meant to be a qualitative model, it gives one concept and one concept only:  Mass causes curvature of spacetime.
You can't get any more than that from the rubber sheet.  If you have that idea in your head already then you are ready to drop the image because:

The sheet is 2d but spacetime is 4d
The 2d sheet is embedded in 3d space. Space-time isn't embedded in a 5d or higher dimension (or at least, if it is, it is irrelevant)
The sheet has two space-like directions with no time dimension. Space-time has a time-like dimension.
The basic way of finding distances on a flat sheet is $(x^2+y^2)^\frac12$. The basic way of finding distances in 4d spacetime is $(x^2+y^2+z^2-t^2)^\frac12$  (where units are chosen to make the speed of light = 1, eg time in seconds, distance in light-seconds)

You can't create a simple image of curved 4-d space-time. The rubber sheet analogy will hinder further understanding if you try to use it to understand why an object will travel in an apparently curved path in space when in a gravitational field. It's only purpose is to embed the notion that "space-time is curved" it can't tell you anything else about what that implies.
So to address your three points

In fact spacetime is 4 dimensional.
Yes.  The placing of balls is strictly metaphorical. There is no actual sheet.
Object don't move at the speed of light.  In their own frame of reference they are still (by defintion).  If you have two objects, they can move relative to each other, but always at less than the speed of light.


The universe is dying vs the universe is ever expanding

Both are true.

The universe is "dying" in the sense that stars eventually run out of hydrogen, and there aren't infinite amounts of hydrogen in galaxies to replace old stars. This will take a very long time, but eventually all stars will burn out and new stars will stop forming. This is what the study recently in the news is saying.

The universe is ever expanding in the sense that distances between galaxies which are not bound together gravitationally is expanding all the time. If you look at very distant galaxies, they are all receding into the distance.

Combine these two facts and the result is that the universe will become more and more "rarefied" (distances between galaxies will keep expanding), and also all existing stars in galaxies will burn out over trillions of years.


How close to a brown dwarf would a planet need to be to receive as much light as Earth?

Brown dwarfs are born moderately hot and luminous and then they contract and cool. Thanks to electron degeneracy in their cores, they never become hot enough to ignite hydrogen (though there is a brief deuterium burning phase) and as a result their fate is to cool and fade.

The plot below (from Burrows et al. 1997) shows how the luminosity behaves as a function of time since birth (age) for objects of various mass. The brown dwarfs are the green and red tracks (some would call the red tracks giant planets). The curves are labelled in Jupiter masses. The highest mass brown dwarf is 73 Jupiter masses, the next one down is 70$M_J$ and then the green curves count down in steps of 5$M_J$ to 15$M_J$, then the first red curve is at 13$M_J$ and the red curves count down in steps of 1$M_J$.

The axes are logarithmic (to base 10). Thus when the y-axis says -2, it means the brown dwarf is one hundredth the luminosity of the Sun, -3 = one thousandth etc. The x-axis is logarithmic in units of billions of years. Most brown dwarfs observed in the Galaxy will be in the range 1-10 billion years (i.e. between 0 and 1 on this x-axis).

The flux received by a planet in orbit around a star/brown dwarf will be proportional to the its luminosity divided by orbital radius squared. From this we can write that the orbital radius where a similar flux to that of the Earth around the Sun will be received is
$$ R = \left( \frac{L_{BD}}{L_{\odot}} \right)^{1/2}\ {\rm au} $$
$$ \log\left(\frac{R}{1\ {\rm au}}\right) = 0.5 \log \left(\frac{L}{L_{\odot}}\right)$$

This you can get you answer immediately from the graph below. Choose the mass of your brown dwarf and its age. Find the log luminosity on the y-axis and halve this value. That is the the log of the orbital radius in astronomical units where a planet would receive the same amount of flux as the Earth from the Sun.

e.g. Consider a 50$M_J$ brown dwarf that is a billion years old. Find the appropriate green curve and see that the y-axis value for x=0 (1 billion years) is -4.2. The log of the orbital radius (in au) at which a planet would receive the same amount of flux from the brown dwarf as does the Earth from the Sun would be -2.1. In linear units, this is $10^{-2.1} = 0.0079$ au, or 1.19 million km.

A further issue to consider is that the brown dwarf surface will be far cooler than the Sun and also changes (cools) with time. This means that the radiation received from the planet will push further and further into the infrared as the brown dwarfs gets older and hardly any of the received radiation will be in the visible part of the spectrum.


Why are accreting objects portrayed with a white flash at the point where the gas stream from the star enters the accretion disk?

When the stream of gas falls towards the disk it gains a significant fraction of the orbital kinetic energy (after all, it is falling from the top of the Roche lobe) which means that it is moving fast and then slows down sharply when it interacts with the disk. This produces a hot spot that in theory could reach $10^8$ K but in practice "merely" is very hot.


Milky Way mass fraction breakdown review

I should clear this up: the Milky Way isn't a completely well-defined body. Galaxies shouldn't be thought of as a single body — they are more like a collection of various material and stars. That said, I can only give you a wonky estimate of what might be the composition.

By "Milky Way", I'm assuming you mean only the Milky Way's disc and halo (and not everything that orbits the Galactic Center, like the satellite galaxies). That alone is hard to define, but let's get a rough estimate. The entire mass of the Milky Way is a bit uncertain, but it falls between $0.8–1.5×10^{12}M_☉$. Let's average that at about $1.2×10^{12}$.

Sagittarius A*, the black hole at the center of our galaxy, has a mass of roughly $4,000,000M_☉$. That would make it only ~0.0003% of the galaxy! A few sources state that dark matter composes about 90% of the Milky Way's mass.
So it has a mass of approximately $1.08×10^{12}M_☉$.

If we want to dissect the rest, we'll have to make generalizations. From some sources like this one, about 75% of MW stars are M type stars, 15% are F-K stars and the remaining 10% are mostly degenerate stars. Knowing that there are 100 billion stars in our galaxy, we can take their average masses and kind of conclude that 3% of the Milky Way's mass is from stellar objects.

Regarding the mass of gases in the Milky Way, this source states that both atomic gas and molecular gas would individually make up 0.25% of the Milky Way's mass. Combined, we get 0.5%. The rest is buried in a sea of contradictions and mystery.

That said, here's the best breakdown I could make:

Sagittarius A*: 0.0003%
Dark matter: 90%
Stellar objects: 3%
Gases and nebulae: 0.5%
Everything else: 6.5%

Again, it is extremely difficult to break down the mass of the components of a galaxy. This is the best you can really do without combining tremendous generalizations with shoddy math and sources.


How large will our Sun get during the red giant phase (ie which planets will it engulf) before the helium flash?

Low mass stars like the Sun do become very large prior to He ignition in the core. The exact value depends a bit on models for mass loss from the extended atmosphere (e.g. Guo et al. 2016), but estimates of 250 times its current size are possible (Schroeder & Smith 2008; Spiegel & Madhusudhan 2012).
At this radius both Mercury and Venus are engulfed. The mass loss from the Sun means Earth would orbit a bit further out than now. Nevertheless, it too would be engulfed in Schroeder and Smith's model at that point, because tidal drag dissipates energy from the orbit and it "falls" into the Sun very quickly.
Interestingly, the Sun achieves its maximum size at the tip of the RGB, rather than in the AGB phase.


You should definitely be able to see the 4 main moons of Jupiter. You won’t need additional magnification - they can be seen through 10 x binoculars. It’s possible that when you looked the moons were close to or behind Jupiter, so worth trying again. Or it may be that James is right and you were looking at a star.
As for the bands on Jupiter, these are quite subtle but at 60x magnification, 2 dark bands should be visible. Don’t expect to see the Great Red Spot, at least initially.


What kind of effects can maintain Uranus' rings' eccentricities?

A relevant paper is Papaloizou & Melita (2004) "Structuring eccentric-narrow planetary rings" which starts off promisingly with the following:

  The nature of the dynamical mechanism that maintains the apse alignment of narrow-eccentric planetary rings is one of the most interesting and challenging problems of Celestial
  Mechanics.

  According to the leading model (Goldreich and Tremaine 1979) the self-gravity of the
  ring counter-acts the differential precession induced by the oblateness of the central planet.
  Using this hypothesis, a prediction of the total mass of the ring can be made, which, in
  general, is not in good agreement with the inferred mass of the observed eccentric rings in
  the Uranus system

It then goes on to describe subsequent refinements taking account of additional effects including particle interactions and perturbations that bring the predictions more in line. The approach described is given in the introduction:

  In this work we build, from first principles, a simple general continuum or fluid like
  model of a narrow-eccentric ring. The eccentric pattern in the ring can be described as
  being generated by a normal mode of oscillation of wave-number m = 1 which may be
  considered to be a standing wave. Dissipation can be allowed to occur due to inter-particle
  collisions leading to a viscosity which would lead to damping of the mode. However, this
  global m = 1 mode can also be perturbed by neighboring-shepherd satellites which can
  inject energy and angular momentum through resonances. In this way losses due to particle
  collisions may be balanced. It is that process that is the focus of this paper. Other possible
  mechanisms, such as mode excitation through self-excitation through viscous overstability,
  that could arise with an appropriate dependence of viscosity on physical state variables (see
  Papaloizou and Lin 1988, Longaretti & Rappaport 1995), are beyond the scope of this paper
  and accordingly not investigated here.

The specific case of the ε-ring of Uranus is described in section 10.2, where the main satellite forcing considered is due to the 47:49 second-order mean-motion resonance between the ring and Cordelia.

The paper goes into quite a bit of mathematical detail on the various processes involved.