Is there a limit to how fast spacetime can bend/warp around an accelerating object?

Changes in the bending of spacetime propagate through spacetime at the speed of light. If I wave my hand thus the (very, very, very tiny) changes to spacetime will propagate outwards as gravitational waves at the speed of light and won't reach the Sun for 8 minutes.

No matter how massive an object or how great the acceleration, spacetime elsewhere only "feels" that acceleration after it has passed into the acceleration's lightcone.


Real images (not "artist concept") by NASA

NASA mission data is archived at a handful of NASA archives.  If you do not know which archive a data set is in, then you can search the Master Catalog at the NASA Space Science Data Coordinated Archive (NSSDC).
The main astrophysics archives are:

HEASARC (High Energy Astrophysics Science Archive Research Center)

IRSA (Infrared Science Archives)

LAMBDA (Legacy Archive for Microwave Background Data Analysis)

and MAST (Mikulski Archive for Space Telescopes)

Planetary/Solar System Data are at the various Nodes of the PDS.
NSSDC holds data from many of the older missions that predate these archives.

New Horizons data is being archived at the Small Bodies Node of PDS. But, it usually takes several months for data to get into these archives.  So, if you are looking for very current images, you have to rely on what the mission team has put up on its web site.  That is at their discretion.


Does the Abell 1689 supercluster have angular momentum?

I have not studied this galaxy cluster specifically. However, I can speak about systems like galaxy clusters that have a large number of bodies that are in close enough proximity to have a non-negligible gravitational effect on each other. For any such n-body system, where n is large (a solar system, a galaxy, a galaxy cluster,...), it can be assumed that, 1. there is a net angular momentum of the system, and 2. it is reasonable to assume that this angular momentum is non-negligible. I state 1 and 2 based on the following (very general model of solar system and galaxy cluster formation).

Such n-body systems form by the collapse of a very large amount of gas, dust, and other minerals. When fully extended, these clouds are cold, the the velocities of any given molecule or atom in them is comparatively slow. However, it is very improbable that the total angular momentum of the cloud of gas is zero, and more importantly, that it is close to zero. From this point, it is easy to reason that, so long as no angular impulse is applied to the system and no mass is added or removed from the system, the gas cloud can collapse or disperse in whatever way is consistent with the conservation of angular momentum, and the angular momentum of the system will not change. Thus, a galaxy cluster like Abel 1689 should have a non-negligible angular momentum.

But I have skipped over the most important logical step in the above argument. I simply assumed again that the angular momentum of the gas cloud would not be even close to zero, effectively shifting the level of the explanation up one level. Unfortunately, the best answer to "why do the gas clouds have non-negligible angular momentum?" requires explaining where the gas clouds came from. The extent of my understanding here is that the gas clouds that formed galaxy clusters only existed during the very early universe. Once they collapsed into galaxies, there is no process that could have caused a galaxy cluster (or even a galaxy) to disperse so completely that it forms a new very low density gas cloud that can re-collapse. Thus, the gas that made the galaxy cluster must have had an angular momentum if the galaxy cluster system is to have angular momentum.

I do not know if the sum total of the angular momentum in the universe is said to be zero or some finite value. However, I do know that even if it is zero, localized regions of matter in a system that has zero angular momentum do not also need to have zero angular momentum. Thus, the Abell 1689 system could be one such local region with non-zero angular momentum.

Continuation: According to The Cosmic Perspective: Seventh Edition, when considering galaxies that have not yet collided with another galaxy, the initial angular momentum of the protogalactic cloud is a large contributor in determining if a galaxy will be spiral or elliptical galaxies.

Regarding elliptical galaxy clusters, TCP has this to say: "Observations of galaxies in clusters support the idea that at least some elliptical galaxies result from collisions and subsequent mergers. Elliptical galaxies dominate the galaxy populations at the cores of dense clusters of galaxies, where collisions should be most frequent. This fact may mean that any spirals once present became ellipticals through collisions."

Thus, TCP suggests in this paragraph that at least certain clusters contain dense cores of elliptical galaxies that resulted from collisions and mergers of galaxies that had markedly different angular momenta, resulting in a new galaxy that has much less angular momentum.

But this doesn't really answer your question either. I have looked also for scholarly articles to find some consensus on what the typical angular momentum characteristics are, and found very little. Most discussions of galaxy cluster formation discuss filament formation and collapse. The one article of note I found was from Tom Crawford at the University Of Chicago:
askanastronomer.org

"The most obvious and uncontroversial class of structures beyond the galaxy is the galaxy group or cluster. These structures consist of tens to thousands of individual galaxies that are bound by gravity to one other and orbit a common center. That center is not defined by some huge central object, as the Sun defines the center of our solar system, but instead by the center of all the mass in the cluster, including dark matter. In fact, the dark matter dominates the mass budget of galaxy clusters, outweighing the normal matter by about a five-to-one ratio."

This description suggests more than just "random" motions of gravitationally bounded objects (galaxies), but does not definitively say anything about the net angular momenta of these systems.


If you are in the Northern Hemisphere, it is currently close to midsummer. The Sun is never very far below the horizon in the North, even in the middle of the night. There are plenty of satellites that are visible even late into the night. Heavens Above

Most are not nearly as bright as the ISS can be. However, against a dark sky a satellite may appear brighter. Lots of the objects that are visible then are actually rocket bodies. They often have a higher orbit than functioning satellites.


Have caged molecules ever been observed in space?

Fullerenes have been detected in space, including C60, C60+ and C70 and appear to be responsible for the diffuse interstellar bands (DIBs), which are absorption features caused by the interstellar medium. The question of endohedral fullerenes is an interesting one, it is discussed in the paper "Interstellar fullerene compounds and diffuse interstellar bands" by Omont (2016).

In section 3.1.2, the situation for metallofullerenes M@C60 is discussed:

  Analysis of the optical spectra of metallofullerenes thus
  presents the same difficulties as do C60 or C60-. It is therefore not surprising that even theoretical studies of their
  energy levels and excitation are lacking, incomplete or ambiguous, and we are still waiting for laboratory spectra.
  The accuracies of wavelength determinations are still far
  from being useful in constraining possible associations of
  visible DIB features with metallofullerenes. Estimates of
  line strengths are practically always lacking.

From this it would appear that more lab work is necessary before we are able to identify metallofullerenes from astronomical spectra.

In section 7, Omont describes the situation for hydrogen molecules in fullerenes (H2@C60), with section 7.2 discussing the prospects for finding interstellar H2@C60:

  It seems that the presence of endohedral H2 would not
  change the basic properties of interstellar fullerenes much,
  making it difficult to reveal its presence, except thanks to
  specific spectroscopic features. However, even if it were
  significant, the presence of endohedral H2 should be difficult
  to confirm because the abundance of H2@C60+, for instance,
  could hardly be expected to exceed or even reach 10−9.
  There is thus practically no chance to ever detect its near-IR H2 2 µm absorption. UV spectroscopy is more sensitive
  by orders of magnitude than absorption of weak IR lines.
  It can detect H2 column densities lower than ∼1014 cm−2
  (e.g. Gillmon et al. 2006). But even so, the UV detection
  of H2@C60 seems out of reach.

A quick check of the papers citing Omont (2016) doesn't turn up any detections of endohedral fullerenes, so it looks like the situation remains as described.


Can you provide a visual description of the sun rise?

Elaborating on the answer from James K, the azimuth of the sunrise depends on the declination of the Sun and the observer's latitude. The declination ranges between approximately +23.4 to -23.4 and depends on the date.
$$\cos(\theta_R)=-\frac{\sin(declination)}{\cos(latitude)}$$

where $\theta_R$ is the angle measured around the horizon from due south to the location where the object rises or sets. From this formula, the azimuth of rising is due east when the declination is 0 which occurs on the spring and autumnal equinoxes.

The calculation of the Sun's declination is not too involved, but it is longer than I want to type at the moment. Since the declination does not change very much from year to year, it would be relatively easy to pick a few dates throughout the year, get the declination for those dates, and calculate the azimuth.

Also see the following question: Solar azimuth for celtic fire festivals


Would a telescope that uses Fresnel lenses be more practical than using regular lenses?

The main advantage of a Fresnel lens is its reduced mass compared to a normal lens. Its disadvantages include difficulty of manufacture and diffraction from the discontinuities in the aperture, these I suspect are killers given the practicality of the alternatives, reflecting telescopes, to a Fresnel lens refactor.

Diffraction will give reduced contrast in the image, already noticeable in a reflector with partially obstructed aperture, but may well be worse with a Fresnel lens. But the difficulty in manufacture may well be the real limit given the relative ease of manufacturing large mirrors.

While I'm at it; it seems as thought making a achromatic Fresnel multiplet might be quite fiddly.

We might also note where there is no alternative, systems analogous to Fresnel lenses are used. The example I have in mind are grazing incidence x-ray telescopes.


Is it possible to explain the difference between synodic month and sidereal month by degrees?

A better diagram is something like this:

Day 0. The Moon is aligned with the Sun (New Moon) and some star X behind the Sun (of course :-), as shown by the arrow at 1. (Neither the Sun nor the star X are shown in the figure.)
Day 27.32. One sidereal month after figure 1. The Moon has travelled through 360 degrees orbiting around the Earth. (During that time, an arrow from the Earth through the Moon is pointing at different stars.) After one sidereal month, the Moon is again aligned with star X, as shown by the solid arrow in figure 2. (The arrow in 1 and 2 are parallel because the star is very far away.) Because the Earth has moved in its orbit around the Sun, the position of the Sun (shown by the dashed arrow in figure 2) has changed relative to figure 1. The Moon is still a few days prior to New Moon.
Day 29.53. One synodic month after figure 1. The Moon is again aligned with the Sun (New Moon), as shown by the solid arrow in figure 3. The direction to star X is shown by the dashed arrow which is parallel to the solid arrows in figures 1 and 2.

Note how the arrows pointing to the Sun all point to the center of the Earth's orbit while the arrows pointing to star X are parallel.

You are probably interested in the angle between the two arrows in position 3; that is, how far has the Moon moved in its orbit from the position of the sidereal month to the position of the synodic month. The average angle = (360 degrees)/(27.32 days) * (29.53-27.32 days) = 29.12 degrees.


Cosmological redshift - How do we know it's not caused by the observer's time dilation?

The reason we think that the cosmological redshift is caused by the metric expansion of space is 1) that there is a well-known, physical mechanism that can cause this effect, and 2) that this mechanism is a prediction of a well-established and thoroughly tested theory, namely the theory of general relativity.

The physical mechanism in question can be realized by considering the propagation of a light ray traveling along a null geodesic in the Friedmann-Robertson-Walker metric describing the expanding space (see e.g. Watson 2000 for a derivation). It is a mathematical result, but it is founded on physics, and verified through experimentation.

On the other hand, there is no known mechanism that would cause time to speed up as the Universe evolves. Moreover, even if there were, this would only make photons arrive less often in our reference frame (compared to photons arriving from a local object), but it wouldn't make the individual photons have less energy (unless they somehow kept oscillating at their original frequency, even if entering a reference frame where time goes faster).

But in a sense you're right that our time goes faster than the time of distant objects: In our reference frame, their time is slower by a factor $(1+z)$, where $z$ is the observed redshift. For instance, the brightness of a supernova observed at $z=1$ declines twice as slow as a "local", $z=0$ supernova. This effect is well-known and "in addition" to the redshift, in the sense that the flux received on Earth is diluted by a $(1=z)$ factor from the redshifting and by a $(1=z)$ factor due photons arriving at a slower rate.


To several decimal places, how many days are in one year?

A mathematician would probably find the imprecision of precise orbital time a little frustrating.
A second can be defined as

"the duration of 9 192 631 770 periods of the radiation corresponding
to the transition between the two hyperfine levels of the ground state
of the caesium 133 atom"

A 24 hour day is 86 400 of those, well, more or less.
A problem that early time-keepers ran into was that the length of the day varied based on the time of year.   Earth does a complete rotation in about 23 hours, 56 minutes, so, the rotation really doesn't factor into it directly, but because Earth orbits the sun about 1 degree of arc every 24 hours, an apparent solar day is has been defined as 24 hours long for centuries.    Days are shorter than that when the earth is closer to the sun, at perihelion, (by about 20 seconds) and days are longer than the 24 hour average by about 30 seconds at aphelion.   Not enough for anyone to notice unless they were measuring solar time very precisely.
So a day has to be averaged out, called a mean solar day, and that's what 24 hours was based on.   The problem is, Earth's rotation is slowing down.  The Moon is very gradually slowing down the rotation of the Earth by tidal interaction, so an average day is currently about 86 400.002 seconds.   Leap seconds are added almost every year and in the future, leap seconds will be added more often.   Officially a day is still 86 400 seconds, and when a leap-second is added, the "leap-day" is 86 401 seconds.   But you could use either number for number of days in a year, average day (currently 86 400.002) or SI day* (86 400).  As I understand it, a year is usually measured in 86 400 second days or SI days, not "average" days, but it's important to be specific about which one you're using.  I've seen the occasional article that says that 200 million years ago a year was about 400 days long.  That's clearly using average says, not SI days.
And if you live in high altitude, your clock will run even faster and a day will run a tiny bit longer thanks to relativity but . . . lets not even go there for the calculations).
A solar year, based on the position of the sun and the precise time of the solstice of your choice, which is a precise and momentary alignment.   The length of a solar year changes a bit on a year-to-year basis.   The position of the moon and gravitational perturbations from other planets, affect Earth's period of rotation a little bit, so like the day, the year should be averaged out also.   Modern leap-year calendars adjust days to years quote well, but not to the accuracy you're asking about.
Looking at the time of recent summer solstices (I've copied them down below for the June solstice).
2010-06-21 11:28
2011-06-21 17:16
2012-06-20 23:09
2013-06-21 05:04
2014-06-21 10:51
2015-06-21 16:38
2016-06-20 22:34
2017-06-21 04:24
2018-06-21 10:07
2019-06-21 15:54
2020-06-20 21:44
There's as much as 15 minutes variation in the length of a year over this 11 year sample.  That's not because Earth's orbit is permanently changing.  These are mostly fluctuations due to the Moon and nearby planets.
An "average year", or mean tropical year, is currently about 365.2422 86 400 second days and if you want to get more accurate, 365.2421897 days on 1 January 2000.  Formula in the link for adjusting to different years.
I suspect, but I'm not 100% certain that the Earth's mean tropical year moves back and forth with some of Earth's Milankovich cycles.   It's not slowing consistently in the sense that a day is slowing due to the Moon's gravity and inverse tidal force.
According to Kepler's laws, the orbital period is based mostly on Earth's semi-major axis.  The problem is that Kepler's laws aren't 100% accurate.   Relativity is a factor when you try to pinpoint a year precisely to the second and variation in eccentricity, which affects Earth's orbital speed and the precise timing of an average year.
What I know about orbital stability, is that Earth's semi-major axis changes very little, even during the Milankovich cycles, and orbital projections say, that it will probably change little even over time-frames of millions of years.
The Sun is also losing mass, which should slow the Earth's orbit over time, but that effect is also very tiny.   Over millions of years, it's thought that the length of a year doesn't change very much.  The length of a day, over time-periods that long, changes quite a bit more rapidly.  An unanticipated orbital change is always possible, but not thought to be likely.   Orbits don't leave much in the way of track-able footprints, so they're hard to pinpoint over long periods of time and are mostly estimated by mathematical models so, take with a grain of salt, but models do suggest that a year is pretty much constant even over periods of millions of years.
For further reading, maybe start here:   https://en.wikipedia.org/wiki/Equation_of_time
Hope that helps.  Some of the math gets kinda tricky.

Thanks zephyr.  I wasn't familiar with that term.


Why is the color of the moon so similar to the clouds?

The Moon is a dark gray in color -- it reflects light poorly but uniformly. (Its color is about the same  as fresh asphalt!  It's albedo is about 12%)  Clouds are (usually) light gray with an albedo ranging from 10% to 90%.

Both the Moon and the clouds shine by reflected sunlight, so they're both the same color. (Clouds usually have a greater surface brightness owing to their lower albedo.)


What is the force that changes a satellite’s inclination when orbiting a planet with an equatorial bulge

I can explain it for rings.

The phenomenon you're looking for is called nodal precession. Meaning, when the planet has an equatorial bulge, all non-equatorial orbits will precess, will wobble around the spin axis.

https://en.wikipedia.org/wiki/Nodal_precession

  Because of the bulge the gravitational force on the satellite is not
  directly toward the center of the central body, but is offset toward
  the equator. Whichever hemisphere the satellite is in it is
  preferentially pulled slightly toward the equator. This creates a
  torque on the orbit. This torque does not reduce the inclination;
  rather, it causes a torque-induced gyroscopic precession, which causes
  the orbital nodes to drift with time.

For rings, any such precessing orbits would be unstable - ring particles would collide with each other. The only non-precessing, non-colliding orbits are the equatorial ones. So that's where the rings persist.

As for satellites, I would venture a guess that their orbits actually do not migrate towards the equatorial plane, they stay where they are, just slowly precessing. Someone please correct me if your orbital mechanics is in better shape than mine.

EDIT: There's a ton of perturbations from an oblate planet, and some do apply to the inclination of the orbit. So it's not just precession. Some terms are periodic. Google "inclination of orbit oblate" without the quotes. It's... complex. I'll edit this answer tomorrow if I find something relevant.

EDIT2: This is far more complex than I thought. A thorough treatment is given in Kozai, Y. The motion of a close earth satellite:

http://adsabs.harvard.edu/abs/1959AJ.....64..367K

Chapter 11.15 in J.M.A. Danby, Fundamentals Of Celestial Mechanics, is entirely dedicated to this topic. He talks about critical values for initial parameters where his results are not applicable, so obviously there isn't a simple conclusion that applies in all cases.

I am not going to interpret the math, sorry. That's all I had to say on this topic.


Pictures of galaxies, nebulae: which way is up?

Images are often presented with (Earth's) North at the top but there is no correct orientation. For example, in this image from the VLT north is 0.1 degrees from the vertical, the sidebar on the image page has details on the position, field of view and orientation.

However, this image from the same telescope has been rotated left 9 degrees, which is likely an artistic decision.

Amateurs also regularly align the cameras on their scopes so that north is up as it can make it easier to diagnose tracking problems. However, it's fairly common to change the camera alignment on the telescope to frame particular targets or groups of targets.

Many classes as astronomical objects, such as elliptical galaxies, irregular galaxies or nebulae don't have a well defined or known axis of rotation, which is why Earth's poles are used as a reference.


I figured I might as well expand my earlier comment into an answer, as I've cobbled together some more information.

You're talking about the cyclic model, which states that the universe goes through cycles of Big Bangs and Big Crunches. We live in the middle of one of those cycles. Albert Einstein was one of the first to investigate it, but his efforts came to nothing.

The theory has been reincarnated, most recently in 2002 by Paul Steinhardt and Neil Turok. They propose that the cosmological constant has changed over time, which could make the accelerating expansion of the universe fit into the theory. Another good article on their idea can be found here. Roger Penrose has also come up with a theory - Conformal Cyclic Cosmology - along similar lines as Einstein's. You can use this as a starting point for more reading.

A related topic is the Big Bounce, which attempts to address the Big Bangs and Big Crunches.


Public Data Set on Atmospheric Absorbtion / Extinction

One source with absorption data worth mentioning is hosted by the Gemini Observatory at https://www.gemini.edu/observing/telescopes-and-sites/sites#Transmission
There are plain-text files with e.g. for the range 0.9-5.6 microns as measured on Mauna Kea.


Does red shift need to compensate for space expansion

Redshift is the "stretching" of the wavelength of light as it travels, over finite time, towards us, due to the ongoing expansion of space.

The effects of this redshift are indistinguishable from that caused by the doppler effect, so that is why some elementary treatments choose to discuss it in that way.

Most galaxies only have peculiar velocities with respect to thei comoving volume of a few hundred km/s, and this does perturb the cosmological redshift by that (small) amount. That is why Hubble's law only applies at distances such that cosmological expansion dominates.


With the exception of the Andromeda galaxy and the Magellanic clouds1, every star, star cluster and nebula that is visible to the naked eye is part of the Milky Way. The Pleiades is a star cluster in the Milky Way.

Objects in other galaxies are too far and too dim to be visible with the naked eye.

1The Triangulum galaxy (M33) may also be visible to the naked eye from a dark site to trained observers with excellent vision


Where are heavier elements than iron produced?

The question of "where" the heavy elements are produced is slightly ambiguous. I gather from your mention of the Milky Way that you are interested in "where in the Milky Way" are the heavy elements produced, rather than what type of stars are they produced in.

However, to answer the former, you need to understand the latter. So I point you to my answer on Physics SE for a summary of that.

The heavy elements can be divided in a number of ways. Most are produced by neutron capture in either the slow s-process or the more rapid r-process. Different elements are produced in different quantities by these two mechanisms and are created in different types of star. s-process elements are created inside massive stars before they become supernovae, but also in stars of more modest mass that never go through the supernova stage. The r-process is now thought to occur predominantly in neutron star mergers, but there is probably also a contribution from type II supernovae, particularly for the lighter r-process elements. A summary, which still has plenty of uncertainties, but probably gives a reasonable summary of current thing is shown below (diagram produced by Jennifer Johnson).

Your question then amounts to where these stars were when they created these heavy elements towards the ends of their lives, and as a function of time since the Milky Way was born. The key facts here are that the progenitors of type II supernovae are massive stars with short lives. They are born in gas-rich regions and do not have time to travel far from this location before they explode. Thus most of the elements they produce are produced in the bulge and disc regions of the Milky Way. This is especially true for the bulge, which seems to have formed quite quickly at the start of the life of the Milky Way and there should (and indeed does seem to be) a radial gradient in metallicity increasing towards the disc plane and towards the Galactic centre.

Type Ia supernovae result from exploding white dwarfs can produce elemts at the iron-peak and somewhat beyond, mostly by "standard" fusion than by neutron capture. These events may have been produced by comparatively lower mass and long-lived progenitors, and may also have had long lives as white dwarfs before the supernova event. This means that type Ia supernova progenitors are spread more widely. They have had plenty of time to migrate from where they were actually born, so we expect their heavy element production to be more dispersed - still concentrated towards the disc, but much less so than the progenitors of type II supernovae.

The progenitors of the asymptotic giant branch stars that can produce lots of s-process elements can also be quite old. The elements produced by these stars would be distributed in a similar way in space and time to those produced by type Ia supernovae.

Finally we have merging neutron stars. These are quite problematic and not much is known about their progenitors or evolutionary paths and histories. They must originate from massive stars, with short lives that are born close to the Galactic plane or bulge. However, after both stars in the systems have exploded as supernovae and left behind neutron star remnants, the final neutron star merger can take place much later, due to a gradual inspiral triggered by the emission of gravitational waves. This gives the neutron star pair a long time (hundreds of millions or billions of years) to move about in the Galaxy before the merger takes place. In addition, the supernova explosions can give a "kick" to the neutron stars that can displace them well out of the Galactic plane with fast velocities. I expect therefore that the products of neutron star mergers could have been produced over a very large volume in our Galaxy.


cosmic-microwave-background - 2.7 degrees but does the Big Bang care?

The expansion of the universe is described by a scale factor that we normally call $a(t)$. We take the scale factor to be equal to one right now, so if $a(t)$ is less than one that means the universe has contracted while if $a(t)$ is greater than one the universe has expanded.

General relativity allows us to calculate how the scale factor depends on time. If you're interested I do this in the question How does the Hubble parameter change with the age of the universe? on the Physics SE. This is what the results look like:

The point of this is that the CMB temperature is directly related to the scale factor so if we can calculate how the scale factor has changed with time then we can calculate the current CMB temperature.

But the way the scale factor changes with time depends on the density of matter and energy in the universe. Specifically it depends on how much light, matter (normal and dark) and dark energy is present. The matter/energy density could in principle have any values, that is there is no fundamental principle in physics that specifies what the densities of the of the various types of matter and energy have to be.

Our current best measurements of the matter/energy density come from the Planck experiment, and they suggest the total density has a value close to the critical density. This density gives us the form of the scale factor that I drew above, and it gives us the CMB temperature of 2.7K. If the density were lower than critical we would get a lower CMB temperature and if the density were higher than critical we get a higher CMB temperature. So the answer to your question is that the CMB temperature is not uniquely determined and it could be higher or lower. Ultimately the CMB temperature is an experimental parameter that we have to determine by making experimental measurements.

But we should note that there are excellent reasons why we would expect the matter/energy density to be equal to the critical density, so we would expect the CMB temperature to be what we observe. If the CMB temperature were different by a factor of ten that would be hard to explain theoretically.

Finally we should note that there are other independent constraints of the form of the scale factor function. Rob Jeffries describes some of these in his answer so I won't go through them again here, but for example the ratio of hydrogen to helium depends on how quickly $a(t)$ increased after the Big Bang. If the scale factor had increased significantly faster or more slowly it would have changed the hydrogen to helium ratio and we'd be able to measure this change.


The Sloan Digital Sky Survey has saturation limits at around 13th or 14th magnitude. You are looking at $\phi$ 1 Cancri, which is a 6th magnitude star. It is hardly surprising that SDSS fails to morphologically classify it correctly.

No idea what your second question is about. You have given a link which has a latitude and longitude in the url. Longitude and latitude are not valid ways of expressing the coordinates of a celestial object. When I open this link, I see an image of your star and in the bottom left of the screen there is a display which gives the RA and Dec of where the cursor is.


Does any asteroid we have ever seen have a hot core like earth.?

See Could a planet form without a molten state?, where Anders Sandberg calculates that a body with radius ~ 1000 km would melt from the heat of formation.

The largest asteroid, Ceres, has a radius of 473 km, so it's a bit too small to have totally melted just from gravitational accretion, but it may have picked up a little more energy from impacts that were faster than freefall. It could also generate some internal heat from radioactive decay, but it's not expected to contain a significant percentage of radioactive material.

Presumably, Ceres was plastic to some degree during formation, since it's (almost) spherical (and it's the only asteroid large enough for gravity to make it spherical). But that wouldn't have required a very high temperature, since Ceres contains a lot of water ice.

But even if it did totally melt when it formed, it's had over 4 billion years to cool down, so it's extremely unlikely to still have a hot core. However, it does emit more water vapour than expected, and it may be warm enough to contain some liquid water below the surface.


Does a flat universe mean not finite and circular?

Yes, a "flat" universe is infinitely large, and hence "non-circular". I think perhaps you are confusing a flat and "closed" geometry.

Flat universes

A flat universe is a universe where your good old high-school math works as expected: the angles of a triangle sum up to $\pi$ radians (180 degrees), the circumference of a circle is $\pi$ times its diameter, parallel lines never meet, etc. In a 2D analogy, this corresponds to the geometry of a table; hence the name "flat".

Closed and open universes

But "flat" isn't the only possible geometry of space. You may, in principle, imagine any crazy geometry you like. However, we like to believe that the Universe is homogeneous and isotropic, i.e. that it looks the same everywhere and in all directions. Indeed, it seems to be on sufficiently large scales. If this is true, then there are only two other possibilities, namely "closed" (or "spherical") and "open" (or "hyperbolic").

Non-flat geometries are difficult to visualize in 3D, but the 2D analogy of a closed universe is that of a sphere, like the surface of a ball. On the 2D spherical surface, the angles of a triangle sum up to more than 180º, the circumference of a circle is less than $\pi$ times its diameter, and lines that start out parallel eventually meet. Moreover, if you travel far enough, you will get back to where you started.

The same is true in 3D spherical space. You may even, in principle, see your own back, although that would require either 1) that you start looking exactly at the Big Bang, in which case the light from your back will reach exactly when the universe collapses in a Big Crunch, or 2) that you have a somewhat fine-tuned amount of dark energy to slow down the collapse, without making it expand too fast so the photons never reach you.

Measuring the curvature in theory

Visualizing the 2D sphere is easy when it's embedded in a 3D space. But to describe how large its curvature is, you actually don't need the embedding 3D. You can just do a geometric measurement and compare your result to what you'd expect in flat space.

The quantity usually used to describe the magnitude of the curvature is $K$, which for a 2D sphere equals $1/R^2$ where $R$ is the radius of the sphere. If you draw a triangle of area $A$, and measure its three angles $\theta_{1,2,3}$, then the curvature is given by
$$
K = \frac{\sum_i \! \theta_i - \pi}{A}.
$$
This formula is valid for spaces of any numbers of dimensions (higher than 1D).

For instance, consider a triangle going from Ecuador, to Gabon, to the North Pole, and back to Ecuador. In this case, the sum of the angles is 270º, or $3\pi/2$, and the area of the triangle is $1/8$ of Earth's surface area, i.e. $4\pi R_\mathrm{Earth}^2/8$. Hence, the curvature is
$$
K = \frac{3\pi/2 - \pi}{4\pi R_\mathrm{Earth}^2 / 8} = \frac{1}{R_\mathrm{Earth}^2}.
$$

There are several other mathematical ways to measure the curvature. For instance you can parallel transport a vector along two difference routes; only in a flat space will they end up pointing in the same direction. In the example above, transporting a vector pointing from Ecuador to Gabon along the shortest path (red path in the figure below), and transporting it via the North Pole (green path), gives a difference of 90º.

So, to see the curvature of a 2D space in a higher-dimensional, flat space requires that theis space be 3D. But as described above, seeing the curvature is not required in order to measure it.

Furthermore, to see the curvature of a 3D space in a higher-dimensional, flat space, actually 4D wouldn't be enough — you would need 6 dimensions. And to see a 4D space (e.g. spacetime), you'd need 10 dimensions. But just as in the 2D case, there's no need for it, and to my knowledge (but please correct me if I'm wrong) no-one claims the existence of such higher dimensions.

Measuring the curvature in practice

In practice, this it not how we measure the curvature of our Universe. Rather, we use the Friedmann equation which gives the relation between the expansion rate $H$ of the Universe, the density parameters $\Omega_i$ of its various components (dark matter, normal matter, dark energy, and radiation), and its curvature.

The density parameters are defined as the densities $\rho_i$, divided by the critical density $\rho_c$, which is the density needed for a flat universe. If the total density $\rho_\mathrm{tot}$ is greater than $\rho_c$ — i.e. if $\Omega_\mathrm{tot} \equiv \sum_i\Omega_i > 1$ — the universe is closed.
Although "curvature" is not really a component of the Universe (such as e.g. matter), it may be parameterized as a corresponding density, and thus, in the Friedmann equation, appear as a term $\Omega_K \equiv 1 - \Omega_\mathrm{tot}$:
$$
\frac{H^2(a)}{H_0^2} =
\frac{\Omega_\mathrm{r}}{a^4} +
\frac{\Omega_\mathrm{M}}{a^3} +
\frac{\Omega_K}{a^2} +
      \Omega_\Lambda.
$$
Here, subscripts $0$ refer to the values measured today, whereas in general the values are measured at a time when the scale factor — the "size" of the Universe — is $a$. Subscripts $r$, $M$, and $\Lambda$ denote radiation, matter (dark+normal), and dark energy.

Hence, measuring the densities and expansion rates (preferably at various epochs given by $a$) will tell you $K$. This can be done in several different ways, e.g. by observing supernova brightnesses, the cosmic microwave background, the large-scale structure of the Universe, gravitationally lensed galaxy clusters, etc.

Luckily, the various more or less independent methods return roughy the same answer. And this answer seems to imply that our Universe is very close to flat; within the margins of the errors, it is consistent with being exactly flat (Planck Collaboration et al. 2018). Hence, for most practical purposes we may regard the Universe as flat. There was, however, recently a paper by di Valentino et al. (2019) claiming evidence for a positively curved (i.e. closed) Universe. But they considered only one out of Planck's many data sets, so it seems to me a bit like cherry-picking.


Do we know a star that is similar to the Sun when it would be a red giant?

Models for the future behaviour of the Sun do vary, mainly as a result of uncertainty of mass loss during the red giant (H shell burning) and asymptotic red giant (H+He shell burning) phases.

A highly cited paper by Schroeder & Smith 2008 claims that the Sun will reach its maximum size of about $256 R_{\odot}$ (1.18 au) at the very tip of the red giant branch (and not at the end of the asymptotic giant branch phase, which is suggested by some other models). This maximum size will occur 7.6 billion years in the future (not the 5 billion years of popular literature), when the Sun's mass will be reduced to about $0.7M_{\odot}$ and have a surface temperature of 2600 K (or about 2300 Celsius). This would have a spectral type of M6III, or perhaps even M6I.

Do we know of a star similar to this? It depends what you mean by similar, but there are unlikely to be any stars like this in our Galaxy. The reason for this is that star formation began in our Galaxy about 12 billion years ago. But a 1 solar mass star like the Sun requires about 12 billion years (or a bit more) to reach the tip of the red giant branch. Even were this just about possible time-wise and we were to find a red (super)giant star of about $0.7 M_{\odot}$, it would be highly unlikely that the star would have a similar chemical composition to the Sun. That is because stars born early in the life of our Galaxy would have very low concentrations of metals like iron and nickel that are only produced inside stars and only present in the interstellar medium when a generation of stars have lived and died.

So my perhaps pedantic answer, is that there aren't any big red giants we can see now that started life as 1 solar-mass stars and have a similar chemical composition to the Sun.


Origin of the magnetic field of neutron stars

The strong magnetic fields in neutron stars are supposed to come from magnetic flux conservation. If we have:

$$\Phi_B = \int B\ \mathrm{d}S = \text{const}$$

where $\Phi_B$ is the magnetic field flux, $B$ is the magnetic field strength, and $\mathrm{d}S$ is the elemental closed surface; then, this integral is constant through the surface.

If we consider the star surface over which take the integral, than

$$S = 4\pi R^2$$

where $R$ is the star radius. This can be translated, altogether with the magnetic flux conservation law,  as:

$$B_f = B_i \left(\frac{R_i}{R_f} \right)^2$$

where $i$ and $f$ are the indices for initial and final stages.
We know that the star implodes from a whatever star size to $\sim10 \; \mathrm{km}$. So the radii ratio is huge. You just need a starting magnetic field of $10-100 \ \mathrm{G}$, to get a final magnetic field of the order of $10^{12}  \ \mathrm{G}$, that is typical in neutron stars.


Long term effect on rocks of high pressure and temperature?

Consider not only the physical attributes of the Venusian atmosphere, but its composition, from the NASA Venus fact sheet:

Major:       96.5% Carbon Dioxide (CO2), 3.5% Nitrogen (N2)
Minor (ppm): Sulfur Dioxide (SO2) - 150; Argon (Ar) - 70; Water (H2O) - 20;
             Carbon Monoxide (CO) - 17; Helium (He) - 12; Neon (Ne) - 7

The 1st and 3rd most abundant components - carbon dioxide and and sulphur dioxide, at the pressures and temperatures of the surface would result in:

  The reactions probably include the deterioration of silicates by carbon dioxide to produce carbonates and quartz, as well as the deterioration of silicates by sulfur dioxide to produce anhydrate calcium sulfate and carbon dioxide.

From the Venus Wikipedia page

A study reported in the paper Venus Geochemistry: Progress, Prospects, and New Missions (2009), focusing on the basalt-atmosphere interaction, determined that the conditions on Venus is roughly the equivalent of:

  a metamorphic environment of ‘amphibolite
  grade,’ ‘hornfels facies;’ the paucity of H2O
  in Venus’ atmosphere (30±15 ppm) prevents formation
  of amphibole

The metamorphic facies for Earth that are used in the above quote are shown below:

Image source

The implications of the surface-atmosphere chemistry, pressure and temperature would, according to the Venus geochemistry article (linked above), result in the formation of:

  anhydrite from essentially
  all the Ca in basaltic minerals and rocks. Iron in the
  rocks and minerals will likely react to either oxides or
  sulfides, depending on the oxidation state of the atmosphere.
  The fate of alkalis is not clear – our calculations
  yield only alkali feldspar (albite), but alkalis
  could be partitioned into ionic (carbonate-sulfate)
  melts. Some Venus basalts are so rich in potassium
  that phases besides feldspar may be present (e.g.,
  leucite, nepheline-group, or sodalite-group).


Why is the letter *z* customarily used to denote redshift?

It may be that someone will come along with a better answer, but the first example I can find for the use of "z" explicitly for redshift is is William de Sitter's 1934 paper.

I'm tempted to think this is one of the earliest uses if not the first use, certainly looking at de Sitter's 1933 book where he touches on this briefly, he does not use the letter "z". However he did use "z" in a 1930 paper to represent the ratio between the radius of the Universe and initial radius, which in this context is actually the same quantity as 1 + z and so why he would've used "z" in his 1934 paper.

If I had to hazard a guess, there is no particular reason for the choice of "z" for redshift, it was just a convenient letter to use for the variable.


Fate of the Spiral Arms of the Milky Way and Andromeda

It indeed appears the Andromeda galaxy (M31) and The Milky Way (MW) are en route to a collision. This will lead to a merger of the two galaxies forming an elliptical galaxy. The flattened disc structure of M31 and MW comes about because their gas dissipated energy but conserved momentum: a disc is the minimum energy state at given angular momentum. The gas in the disc then forms stars, which are more or less on the same orbit as the gas, i.e. circling the respective galaxies. A disc galaxy is dynamically cold, i.e. the speed of the mean motion of the star around the galaxy is much larger than their random motions.

The merger destroys all these structures and replaces them with a much smoother and dynamically hot structure, when the density is roughly constant on (concentric) triaxial ellipsoids. There are several types of stellar orbits in such galaxies, but most important are the so-called box orbits, when the stars oscillate in a large box-like volume and have no preferred mean direction of motion. In the early phases, the elliptical galaxy will have some temporary structures (so-called shells), which are remnants of the merger process itself.

The gas in M31 and MW will most likely be swept into the inner parts of the elliptical where it may form many new stars and contribute to the feeding of the AGN which emerges from the coalescence of the two supermassive black holes of M31 and MW. But energy fed back from the new formed stars via supernovae and stellar winds as well as from the AGN will drive the remaining gas out of the elliptical galaxy, leaving it "red and dead", i.e. without star formation and young blue stars.

The merger of M31 and MW will hardly increase the already neglible rate of stellar collisions. Given the vastness of galaxies compared to the size of stars, such collisions are extremely unlikely. The only places where such collisions occur are the very dense cores of globular clusters.


Latitude, longitude and altitude of Ckoirama Observatory; where can things like this be looked-up?

Near the pinned location, at 24.0893°S 69.9306°W,
Bing Maps shows a building like the one in the University of Antofagasta photos,  with signs of recent construction.

It's probably a matter of time until Google updates their satellite imagery in that region.

The Minor Planet Center maintains a list of observatory codes with positional information.
A related page there says:

  The KML file of observatory-code locations has been removed, following a request from some Spanish observers concerned about security at remote, unattended sites.


If you take a look at the Hertzsprung-Russell Diagram that helps us to classify stars

(c) Wikimedia commons

(while this one uses data gathered from the Hipparcos satellice, roughly 100.000 stars) we see, that there are 2 main types of red stars: Giants and M-dwarves. The latter are stars on the main sqeuence (they have stable hydrogen-burning in their cores), while the giants are stars that have left the main sequence due to aging. They appear red as they bloat up to bigger radii compared to their main-sequence life which also makes them cooler, and thus redder.

Betelgeuse is now one of those giants, and he's actually more massive than the sun (somewhere between 7 and 20 solar masses, according to wikipedia), so there is no contradiction with M-Dwarves also being red.

Please let me know if you desire to know more details on this.


What causes the dimensions of a star increase when its hydrogen fuel is exhausted?

The Sun will never run out of Hydrogen. This is a common misconception.

At this moment the Sun is fusing Hydrogen into Helium. This fused Helium remains at the core until it will reach a critical mass. At this point the core will begin to collapse. This collapse increases the temperature and pressure around the core where Hydrogen is being fused causing the increase the Hydrogen fusion which also causes an increase of the radiation pressure making the outer layers to expand and cool.

Once the core reaches a temperature of about 100 million Kelvin Helium fusion starts dramatically (Helium Flash) with high radiation pressure in the core. Here is where the Sun will reach 250 times it's current radius.

When the Helium fusion stabilizes the radiation pressure at the cores decreases making the Sun's radius to shrink.

The temperature will only drop at the surface because as it's expands it's further from the core.


tl;dr: Perfectly full moon lasts, never more than, approximately 2 minutes.

Estimating an answer from the point of view of Plain Geometry:

If we approximate, that the sun is at an infinite distance, then exactly half of the moon is illuminated.

Next, how much of the Moon do I see when I look at it from Earth? (This has nothing to do with illumination of the Moon.) Estimate: Radius of moon (Rm), 1740km. Distance to the moon (Dm), 384,000km. Doodle this into a diagram that is VERY much not to scale...

arctan(Rm / Dm) gives us 0.260 degrees. So as long as the observer is within that angle from the Moon-Sun-centerline, the portion of the Moon which is visible, would be a circular disc fully illuminated.

Here, I suspect that the variation in the Moon's distance is the biggest thing I've estimated away.

The question now is: combining the Moon's motion of revolution (aka orbit about the Earth) with the Earth's rotation, how long does it take an observer to perceive the moon moving half a degree (1/4 degree either side of the Moon-Sun-centerline)? Rotation of the Earth gives 15 degrees-per-hour, minus the Moon's 1/2 degree per hour... 14.5 degrees per hour apparent motion.

I get approximately 2 minutes duration when the Moon's shifting of apparent position would only hide/show illuminated portions of it's disc.

But the lunar orbit is inclined

So it's only 2 minutes if the Moon is very close to the plane of the Equator. As the Moon's orbit carries it out of the plane, the time would be reduced. Beyond half a degree out of the plane, you would actually never (pure geometry) seen a fully illuminated disc.

So I think the best answer is probably, "Never more than 2 minutes."


Locating a star cluster on a Hertzsprung-Russell diagram with other color index

With an absolute magnitude of zero and a colour (any colour) of zero, your stars are around spectral type A0 (by definition). The only thing that can confuse this is reddening/extinction.

If there is appreciable reddening/extinction, then your stars could be more luminous and hotter (O/B stars), but there isn't any way to tell from your data.


"A little" (compared to how much visible light there is)

(from a question on physics se

The sun produces relatively little UV radiation, most of its light is visible, or infrared, and anything shorter than about 290nm is completely blocked by the atmosphere (remember to thank the atmosphere today)

UVa (in the range 400-320) and UVb (320-290nm) do get through. The sun is so very bright, that it is by far the brightest natural source of UV on the Earth.

The moon is about 400,000 times dimmer than the sun. Ultraviolet reflected off the moon could reach the Earth. It would be very dim.

At night UV camera systems use active illumination rather than depending on the moon.


I see two real questions here. First, whether it's possible to have a black dwarf with a companion object. For a given black dwarf, this is unlikely, since the orbits would likely be unstable at the time scale required to produce a black dwarf. Given the size of the universe, however, it's not out of the question. A black dwarf could even capture a companion star after cooling down, although this would be extremely rare.
Assuming it's possible, the next question would be whether the supernova produced would be the same as a normal type Ia supernova. The supernova is generally triggered based on the critical (Chandrasekhar) mass which triggers carbon fusion. Since it's dependent on mass, it would likely be a "normal" type 1A supernova. Keep in mind not all 1A supernovae are identical to begin with, although they can still be "normalized" to function as a fairly reliable standard candle.
One thing to note is, the accretion process would cause the black dwarf to heat up, meaning it would not remain black.


What might be the cause of the hot summer on Earth in 2018?

Seen globally, the Earth is 0.75—1.0 degrees warmer than the average. This is what is modelled by warming due to raised levels of greenhouse gases.

Currently the sun is close to a solar minimum. Perturbations of the Earth's orbit by other planets do not significantly change the level of solar insolation. Also the ENSO is currently close to neutral. The hot temperatures across western Europe and East Asia are a combination of the effects of greenhouse gases, and the random influence of weather.


Are our 'observations' of or near a black hole affected by gravity induced time dilation?

The Schwarzschild black hole is the simplest, being just an isolated, nonrotated, uncharged black hole. If $\tau$ is the proper time measured by a clock moving along an arbitrary path in this spacetime, then in Schwarzschild coordinates
$$-\mathrm{d}\tau^2 = -\left(1-\frac{R}{r}\right)\mathrm{d}t^2 + \left(1-\frac{R}{r}\right)^{-1}\mathrm{d}r^2 + r^2\left(\mathrm{d}\theta^2+\sin^2\theta\,\mathrm{d}\varphi^2\right)\text{,}$$
where $R = 2GM/c^2$ is the Schwarzschild radius. If the observer is stationary, then neither the radial nor angular coordinates change ($\mathrm{d}r = \mathrm{d}\theta = \mathrm{d}\varphi = 0$). So, for example, relative the rate of time of stationary observers at some radial coordinates $r_1$ and $r_2$ is just
$$\frac{\mathrm{d}\tau_1}{\mathrm{d}\tau_2} = \frac{\mathrm{d}\tau_1/\mathrm{d}t}{\mathrm{d}\tau_2/\mathrm{d}t} = \sqrt{\frac{1-R/r_1}{1-R/r_2}}\text{.}$$

  how might that affect our measurements?

In the sense of the measurement process, it doesn't, since a measurement apparatus is not time-dilated relative to itself. In the sense of what results we get, then of course it does.

  So it would logically follow that events happening near a black hole (a very large gravitational field) would seem to us to happen much more slowly than if they happened outside that field.

Yes. If the first stationary observer is shifted closer to the horizon ($r_1\to R$), then $(1-R/r_1)\to 0$, so the above time dilation goes to $0$ as well. So, for example, if that observer shines a laser beam toward the other, it will be redshifted when it arrives to the other.

This is really the same thing as time dilation. The laser beam has a certain frequency, so it takes a certain amount of time for one oscillation. Thus a frequency shift is just another way of talking about time dilation, except that time dilation means that this happens to every process, not just light: e.g, if the first observer taps his foot at a certain frequency (say, a tap per second), the second observer will see him tapping at a lower frequency (say, a tap per two seconds). Slowed. We can can still call this "redshift" in the sense that the frequency was shifted in the same way, although the term is usually used specifically for electromagnetic radiation.

  On a related note, does light being 'bent' around a black hole experience time dilation or red/blue shift from our viewing perspective?

Light itself experiences only trivial time dilation in the sense that the proper time along any lightlike trajectory is exactly $0$. Above, we were talking about stationary observers, and light is not stationary. Rather, $\mathrm{d}\tau = 0$ for light beams, always. So strictly speaking, light doesn't experience anything at all.

However, if you mean what the stationary observers would get when measuring light bent around the black hole, then it is blueshifted when coming closer to the black hole and redshifted when moving away--in the sense that this would be what stationary observers closer or farther away would measure. The details of the path it took between the observers do not matter; the ratio of measured frequencies will just be the inverse of the time dilation factor between the observers.


Why do medium stars collapse to form supernovae while big stars collapse to form black holes?

A supernova is caused by the sudden (less than 1 second) collapse of the core of a massive (initially more than 8 solar masses in total) star. Less massive stars do not undergo core collapse and do not produce core-collapse supernovae.

The thing that determines whether a supernova remnant is a neutron star or black hole is largely the mass of the core at the onset of core collapse, but may also depend on how effectively the rest of the star is blown into space by the supernova explosion and how much material falls back onto the collapsed core.

If the collapsed core has a mass of less than about 2 solar masses (and perhaps up to 3, though there is no observational evidence for this), then the collapsed core, consisting mainly of neutrons, can be supported against its gravitational weight and stabilised as a "neutron star". The support is provided by neutron degeneracy pressure and the extreme repulsion between neutrons when they are squeezed together, which is caused by the strong nuclear force.

If the collapsed core is more massive, or if more material from the exploding star falls back onto the collapsed core, then even this strong nuclear repulsion will be unable to support its weight. In fact, above about 3 solar masses, General Relativity dictates that no force can possibly support the weight of the compact remnant and it will collapse further to become a black hole. No light (or anything else) can emerge from inside a black hole; so-called Hawking radiation will be negligible; so the only light received from such objects (after any supernova has faded), will be due to material being accreted into the black hole that is heated as it spirals towards it.

A third possibility exists, which is direct collapse to a black hole. It is possible that for very massive stars (at least 25-30 solar masses), that the collapsing core is never stabilised even as a proto neutron star, that no bright supernova is produced and that much of the mass of the star directly collapses into a black hole. This is a very active research topic since the discovery of 20-30 solar mass black holes by the LIGO gravitational wave experiment.

To summarise, I can do no better than quote from an excellent review paper by Fryer (1999):
"Thus for core collapse models we can define three regimes of compact object formation: (1) low-mass, core-collapse stars drive strong explosions with little fallback and produce neutron stars; (2) moderate-mass stars produce explosions, but the fallback is sufficient to form black holes; and (3) high-mass stars are unable to launch shocks and collapse directly to black holes. The question for core-collapse theorists, then, is to determine the limits for these regimes.",
where the separation between these regimes is (1) $\geq 8$ up to perhaps 20 solar masses; (2) 20-40 solar masses and (3) more than about 30-40 solar masses, but these boundaries depend on metallicity, rotation and a correct understanding of supernova and neutrino physics.


Orbital path projections onto a 2D plane

Yes, the 2D projection of a 3D ellipse (or circle) is always an ellipse (or circle). However, in general, the various elements of the 3D ellipse do not project to the corresponding elements of the 2D ellipse.
A simple parameterization of the 2D ellipse uses the eccentric anomaly. For an ellipse centred on the origin, let $a$ be the semimajor axis, aligned with the X axis, $b$ the semiminor axis, aligned with the Y axis, and $\theta$ the eccentric anomaly. Then
$$x = a\cos(\theta)$$
$$y = b\sin(\theta)$$
Note that $\theta$ is not the central angle of the ellipse, it's the central angle of the auxiliary circles associated with the ellipse. Here's a diagram, courtesy of Wikipedia (click the image for an SVG version):

The eccentric anomaly of point P is the angle E. The center of the ellipse is point C, and the focus is point F.

The blue circle provides the X coordinate of P, and the green circle provides its Y coordinate.
The vertices of an ellipse are the points where the major axis crosses the ellipse, the co-vertices are the points where the minor axis crosses the ellipse. The focal points lie on the major axis at a distance $f=ae$ from the origin, where $e$ is the eccentricity. Note that $a^2=b^2 + f^2$.
If we rotate the ellipse in 3D around the X axis by $\gamma$ and project it back onto the XY plane, the projected ellipse has the same semimajor axis, but its semiminor axis shrinks to $b\cos\gamma$. Similarly, if we rotate it around the Y axis by $\beta$, the semiminor axis is preserved, but the semimajor axis shrinks to $a\cos\beta$. With more complex rotations, things get complicated...
We can represent rotations using rotation matrices. This is useful because we can combine rotations by multiplying their matrices. Please see the Wikipedia article for details.
Using rotation matrices, we can produce parametric equations for the general 3D ellipse. I won't reproduce those equations here, since they're a bit messy. But using those equations we can show that the projected ellipse is, in fact, an ellipse.
Below is a Sage / Python script which creates an interactive 3D plot of a rotated ellipse and its projection onto the XY plane. I use the same conventions as that Wikipedia article: I use a right-handed coordinate system, with angles $\alpha, \beta, \gamma$ representing anticlockwise rotations (in degrees) around the Z, Y, X axes, respectively. (If you "grab" an axis with your right hand, with your thumb pointing in the positive direction, then your fingers curl in the positive angle direction).
The $z=0$ XY plane is rendered in translucent grey. I plot the vertices, co-vertices, and foci of the 3D ellipse in primary colours, and use the same colours for the projections of those elements onto the 2D ellipse. The major axis is red, the minor axis is blue, the foci and the 3D ellipse itself are green. I also plot the actual vertices, co-vertices, and foci of the 2D ellipse: magenta for the major, cyan for the minor, and pale green for the foci and the 2D ellipse. When one of these points coincide with a projected point, the graphics system "decides" which one to render.
""" Plot an ellipse in 3D & its projection onto the XY plane
    Written by PM 2Ring 2012.04.11
"""

from math import radians
t = var('t')

ps = 10
# Plot a 3D line segment & its endpoints
def pline(pts, color):
    P = point3d(pts, size=ps, color=color)
    P += line3d(pts, color=color)
    return P

# Project a 3D vector to the z=0 plane
def proj(v):
    return vector((v[0], v[1], 0))

@interact
def main(ecc=slider(0, 1, step_size=0.05, default=0.6),
  alpha=slider(-180, 180, step_size=5, default=0),
  beta= slider(-180, 180, step_size=5, default=20),
  gamma=slider(-180, 180, step_size=5, default=20),
  frame=False, perspective=False, auto_update=False):
    a = 1
    f = a * ecc
    b = sqrt(a^2 - f^2)
    print("b =", b)

    # Build rotation matrices
    alpha = radians(alpha)
    beta = radians(beta)
    gamma = radians(gamma)
    Rx = matrix(RR, [
     [1, 0, 0],
     [0, cos(gamma), -sin(gamma)],
     [0, sin(gamma), cos(gamma)]])
    Ry = matrix(RR, [
     [cos(beta), 0, sin(beta)],
     [0, 1, 0],
     [-sin(beta), 0, cos(beta)]])
    Rz = matrix(RR, [
     [cos(alpha), -sin(alpha), 0],
     [sin(alpha), cos(alpha), 0],
     [0, 0, 1]])
    R = Rz * Ry * Rx

    # The z=0 plane, origin, & axes
    P = plot3d(lambda x,y: 0, (-a, a), (-a, a), color="#888", opacity=0.2)
    P += point3d([(0,0,0)], size=ps, color="#888")
    P += pline([(-a, 0, 0), (a, 0, 0)], "#aaa")
    P += pline([(0, -a, 0), (0, a, 0)], "#aaa")

    # The 3D ellipse
    fxyz = R * vector((a*cos(t), b*sin(t), 0))
    fxy0 = proj(fxyz)
    P += parametric_plot3d(fxyz, (t, 0, 2 * pi), color = "green")
    # The projected ellipse
    P += parametric_plot3d(fxy0, (t, 0, 2 * pi), color = "#7f7")

    # The vertices & co-vertices of the 3D ellipse and their projections
    # Major
    vx0, vx1 = fxyz(t=0), fxyz(t=pi)
    P += pline([vx0, vx1], "red")
    P += pline([proj(vx0), proj(vx1)], "red")

    # Minor
    vy0, vy1 = fxyz(t=pi/2), fxyz(t=3*pi/2)
    P += pline([vy0, vy1], "blue")
    P += pline([proj(vy0), proj(vy1)], "blue")

    # The foci of the 3D ellipse and their projections
    f0 = R * vector((-f, 0, 0))
    f1 = R * vector((f, 0, 0))
    P += pline([f0, f1], "green")
    P += pline([proj(f0), proj(f1)], "green")

    # Find the true vertices & co-vertices of projected ellipse
    # rf is the radius function of the projected ellipse
    rf = abs(fxy0)
    # Major
    ap, t0 = find_local_maximum(rf, 0, pi)
    _, t2 = find_local_maximum(rf, pi, 2*pi)
    vpx0, vpx1 = fxy0(t=t0), fxy0(t=t2)
    P += pline([vpx0, vpx1], "magenta")

    # Minor
    bp, t1 = find_local_minimum(rf, 0, pi)
    _, t3 = find_local_minimum(rf, pi, 2*pi)
    P += pline([fxy0(t=t1), fxy0(t=t3)], "cyan")

    # Find true foci of the projected ellipse
    eccp = sqrt(ap^2 - bp^2) / ap
    P += pline([vpx0*eccp, vpx1*eccp], "#7f7")

    P.show(frame=frame, projection="perspective"
      if perspective else "orthographic")

Here's a screenshot:

It's a bit hard to see what's going on in that screenshot. It's a lot easier in the interactive 3D view. This view is created using three.js, and responds to the usual OrbitControls: you can pan & rotate the camera using the mouse, use the mouse wheel to zoom. On a touch screen, one finger rotates, use two fingers to pan & zoom.

The script now allows you to easily set the frame off or on, and to choose between a perspective or orthographic camera.


Is there any correlation between the Cosmic Microwave Background (CMB) and the distribution of distant galaxies?

CMB fluctuations
The CMB fluctuations are often analyzed through their power spectrum $P(k)$, which is a measure of the extent to which it is "clumpy" on a given scale $\ell$, with corresponding wavenumber $k = 2\pi/\ell$. The origin of this power spectrum is laid in the very early early Universe, just after the Big Bang, and it is of utmost importance in cosmology, as it specifies how structure subsequently evolves.
Various observations help us constrain $P(k)$. For the early Universe, we use the CMB, while at later epochs, the large scale structure of galaxies and, even later, galaxy clusters are helpful. Moreover, absorption lines in the spectra of background quasars enable us to map out the structure via the so-called Lyman $\alpha$ forest.
The figure below (from Norman 2008) show how a single cosmological model (from Tegmark et al. 2004) fit all epochs.

The power spectrum evolves with time, so to compare difference epochs, the figure shows $P(k)$ for the different observations extrapolated to the current epoch (i.e. today).
CMB dipole
Observing the dipole in the CMB is relatively easy because, to within 1 in ~100,000, it is described by the same Planck temperature in all directions. This is not the case for any astrophysical sources, so measuring the dipole in galaxies is less straightforward. With large samples of galaxies it has nevertheless been attempted. For 1.2 million galaxies in the WISE catalogue, Rameez et al. (2018) find the dipole anisotropy to be consistent with that of the CMB.


Contradiction between Theory of Relativity and Newton's Law of Universal Gravitation?

A) Don't take Newton's law too seriously in the relativistic regime.

  According to Newton's Law, when it reaches the event horizon it must be traveling at $c$, because $c$ is the escape velocity at the event horizon.

This is true, assuming the particle freefalls from rest at infinity, but the details are quite different from Newtonian theory. For orbits in Schwarzschild spacetime, the effective potential is
$$V = -\frac{GM}{r} + \frac{l^2}{2r^2} - \frac{GMl^2}{c^2r^3}\text{,}$$
where $l\equiv L/m$ is the specific angular momentum of the orbit. The last term offers relativistic corrections that have no analogue in Newtonian gravity, so in general Newton's law cannot be counted on when near the horizon of the black hole.

But for radial freefall, $l = 0$, and we a total orbital specific energy:
$$\mathcal{E} = \frac{1}{2}\left(\frac{\mathrm{d}r}{\mathrm{d}\tau}\right)^2 - \frac{GM}{r}\text{,}$$
which has exactly Newtonian form. In particular, a particle freefalling from rest at infinity will have
$$\left|\frac{\mathrm{d}r}{\mathrm{d}\tau}\right| = \sqrt{\frac{2GM}{r}}\text{,}$$
so at as it nears the horizon ($r\to 2GM/c^2$), we have $|\mathrm{d}r/\mathrm{d}\tau|\to c$, just as Newtonian theory would predict.

However, the motion is only roughly analogous to what's prescribed by Newton's law, because there are very important differences in the interpretation of those quantities. Specifically, (1) the $\tau$ refers to the proper time of the orbiting particle, not coordinate time $t$, and certainly not any absolute time as Newtonian theory proper, and (2) the $r$ refers to the Schwarzschild radial coordinate, which is defined such that a sphere at this coordinate have surface area $4\pi r^2$, and this is not the radial distance to the center.

B) Coordinates have no intrinsic meaning.

  Therefore, after it passes the event horizon, it must continue to accelerate past $c$, yet theory of relativity states that it's never gonna happen.

Suppose I fling a particle along a straight ruler, and that ruler is marked in feet. So if, for example, my particle goes five marked units along the ruler in a nanosecond, that's clearly a problem, because the particle must be superluminal ($c\approx 1\,\mathrm{ft}/\mathrm{ns}$).

Now suppose the ruler is not marked in feet, and I don't tell you how it's marked. Is it a problem that the particle goes five units in a nanosecond? Without knowing what the marks are, the statement of this sort of "coordinate speed" is not even meaningful. Finally, suppose I tell you how the ruler is marked, but its marks do not actually measure the distance along the ruler. Then is it a problem that it goes through five of those marks in a nanosecond?

The moral is simply this: by themselves, coordinates are simply labels to identify events. One can pick coordinates such that the coordinate speed is arbitrarily large or small, and it doesn't matter, because coordinates are not physical things. The universe does not come with its own coordinates; they're just labels we put on things.

However, what can give coordinates meaning is the metric tensor, which relates coordinates to actual lengths or durations, so that, for example, the Schwarzschild radial coordinate has a set meaning in terms of surface areas. Finally, the Schwarzschild coordinates are pathological near the horizon. They are not defined across the horizon (technically, the Schwarzschild coordinates are actually two entirely different coordinate charts disconnected by the horizon). You can see this in the Schwarzschild metric (units of $G = c = 1$):
$$\mathrm{d}s^2 = -\left(1-\frac{2M}{r}\right)\mathrm{d}t^2 + \left(1-\frac{2M}{r}\right)^{-1}\,\mathrm{d}r^2 + r^2(\mathrm{d}\theta^2+\sin^2\theta\,\mathrm{d}\phi^2)\text{,}$$
where the metric coefficients in the Schwarzschild chart are undefined at the horizon $r = 2M$.

C) "So, in this condition, what will the object do?"

It will cross the event horizon and meet its end at the singularity.

Here's the same Schwarzschild geometry in Gullstrand-Painlevé coordinates:
$$\mathrm{d}s^2 = -\mathrm{d}t^2 + \left(\mathrm{d}r + \sqrt{\frac{2M}{r}}\,\mathrm{d}t\right)^2 + r^2(\mathrm{d}\theta^2+\sin^2\theta\,\mathrm{d}\phi^2)\text{.}$$
In this coordinate chart, space is moving towards the singularity at the estate velocity $\sqrt{2GM/r}$, and the particles are carried along with it. In a related coordinate chart, Lemaître coordinates, particles freefalling from rest at infinity have zero coordinate speed, with $\mathrm{d}r/\mathrm{d}t = 0$, where here $r$ is the Lemaître radial coordinate. (Edit: corrected example mistakenly attributing this property to the GP chart.)

You can pick any coordinate chart that's well-behaved at the horizon, and any such chart will show it moving across the horizon with no trouble. But the Schwarzschild coordinates do not give such a chart, because they're undefined at the horizon.


Are stars really "burning balls of gas"?

While I agree that it boils down to semantics, I actually disagree on the scientific use of the term in the comments. In astronomy, we know there's a difference between gas and plasma, but we almost always use the term "gas" when talking about what's in stars. E.g. "the fraction of gas locked up in stars" (as opposed to in the interstellar medium). We also talk about the metallicity of gas, meaning both gas and plasma. I think we only use the term plasma when specifically discussing properties that are unique to plasma, like being "frozen" in magnetic fields. And googling ["ionized gas" -plasma] (the '-plasma' is to get rid of articles explaining what plasma is) returns 243,000 hits, of which by far the most seem to be scientific papers and websites.

So I think the answer is: "No, stars aren't made of gas, they're made of plasma, but go ahead and call it gas, you won't offend any astronomer."

However, I agree with Rob Jeffries that we don't like when you call it burning gas. It's nuclear power! It's a million$^\dagger$ times more powerful than fire!

$^\dagger$Why "a million"? Because nuclear reactions are measured in MeV, while chemical reactions are measured in eV.


A basic answer:

To us on Earth, it looks like everything falls mostly straight down and hitting the Earth.   We almost never see something falling, trapped in Earth's gravity and missing the Earth but in space that happens a LOT.

If we lived in space our observation would be very different.   Most of the stuff we see flies around, in orbits, around the Sun or Earth without colliding into them.  The reason for this is tangential velocity.  It's actually hard for something to hit something else in space, even with gravity, because space is (shockingly) mostly empty space.

Planets, the Sun, black holes, etc, are very small relative to their gravitationally dominant regions or hill spheres.

Why is it hard to hit the sun?  (this applies to black holes too, even more so)

Why is the solar system flat?

Mostly, objects caught in the gravitationally dominant region around a black hole will just orbit the black hole.

When an object passes inside a black hole's Roche limit, it will begin to break apart into smaller parts, and those smaller parts into smaller parts, and close enough, pretty much all matter just becomes a bunch of debris, kind of similar to Saturn's rings but much faster moving and more dense.   Ring systems can form inside the Roche limit around any massive body, large moon, planet, sun or star, though stars which emit lots of light and heat aren't good for the stability of ring systems.

Ring systems, kind of like Saturn's, can form around black holes when there is a collision or when an object passes too close and ring systems tend to flatten out over time due to collisions.  The angular momentum remains constant but the up or down motion of objects cancels out.  The flat accretion disk is actually conservation of momentum, it forms naturally, even if it seems like a lot of wasted space, the matter doesn't know where else to go but into a flattened, quickly rotating disk.

Because of the laws of relativity and gravitational waves, inside about 3 Schwarzschild radii, the ring material much more quickly spirals in towards the black hole, moving faster into a smaller region, getting very hot and extremely magnetic and that's generally what's called the accretion disk.  (Ring systems are generally further out and not hot or spiraling in, though they are loosely similar)

All happens outside the black hole.   Nothing that happens inside the black hole gets out, but what happens outside the black hole in the accretion disk can be surprisingly bright and energetic.  Brighter than the brightest stars.

It might seem like stuff should just fall into a black hole like a vacuum sucking things in, but in gravity in space doesn't work like that.  Anything with zero tangential velocity will fall straight in (kinda-sorta, there's also the dragging space around it problem for rotating black holes), but for simple non rotating black holes, Only an object with close to zero tangential velocity will fall straight, but almost nothing that moves through space has zero tangential velocity relative to it's nearest large gravitational body.  This applies to dust, gas, comets, meteors, planets, stars, anything that passes close to or orbits a black hole.

That's the overall gist of it.  Corrections welcome.


Is Mars visible from naked-eye from Michigan (47N, 88W) these days?

On 16 June 2014 at 11:45pm, it was up not far from 220az 30alt. It was actually 232az, 27alt, almost level and to the right of Spica. It would have been easily seen without binoculars being -0.2 visual magnitude.

You can answer this sort of question yourself by getting a planetarium app like Stellarium, The Sky X, Sky Safari, or one of many others.


Questions about amateur astrophotographer Nik Szymanek's telescope

Szymanek says he took some of his Flickr images with a GSO 10" Ritchey-Chrétien telescope.
At least one third party offers these with Altair Astro branding and says the tube is made of carbon fiber.

The Ritchey-Chrétien design is a Cassegrain with hyperbolic primary and secondary mirrors.
It minimizes coma and spherical aberration at the expense of field curvature.
The baffle tube forward of the hole in the primary is required to shade the image plane from direct skylight.
Many Cassegrain telescopes also have a short baffle tube aft of the secondary.

The multiple baffle rings inside the main tube just darken the tube more completely than black paint alone.
They are more likely separate rings than a continuous helix.
The alternative would be to roughen the surface e.g. with flocking paper.

The mount looks like an older Software Bisque Paramount, a robotic German equatorial mount.
On current models the declination axis housing is cylindrical rather than rectangular.


Why is there a difference between the cosmic event horizon and the age of the universe?

From the link you provide:

  The particle horizon differs from the cosmic event horizon, in that the particle horizon represents the largest comoving distance from which light could have reached the observer by a specific time, while the event horizon is the largest comoving distance from which light emitted now can ever reach the observer in the future. The current distance to our cosmic event horizon is about 5 Gpc, well within our observable range given by the particle horizon.

Your confusion somes from the fact that you are mixing to different horizons:

The particle horizon is the sphere centered on us that has a radius equal to the distance that light can travel in 13.8 Gyr (the age of the Universe). That is, light that was emitted when the Universe was born at a point on that horizon, reaches us today. Note that, because the Universe is expanding, the distance is not 13.8 Glyr, or 4.2 Gpc, as one might naively expect, but in fact 46.3 Glyr.

The cosmic event horizon is also a sphere centered on us, which is the boundary inside which light, if it is emitted today, may still reach us sometime in the future. If it is emitted outside this horizon, the expansion of the Universe ensures that the light will never reach us.

As time goes, we will see light that came from farther and farther away, and thus the distance to the particle horizon is increasing. Meanwhile, the accelerated expansion of the Universe ensures that light emitted from distant galaxies in the future are able to reach us only if they are increasingly closer. Thus fewer and fewer galaxies will be inside the event horizon. But since space is at the same time expanding, the boundary will asymptotically reach a finite size (~17 Glyr).

You can see the particle and the event horizons as blue and red lines in the plot in @Pulsar's great answer here.

The 16.5 Glyr that the distance to the event horizon is today is sort of a coincidence. It has nothing to do with the age of the Universe. It only depends on the future expansion of the Universe, which in turn depends on the densities of the components of the Universe ($\Omega_\mathrm{b},\Omega_\mathrm{DM},\Omega_\Lambda$, etc.). If the Universe had been dominated by matter (or radiation), then there would be no event horizon: No galaxy, ever-so far away would not be visible to us, if we just had the patience to wait. A galaxy is 10,000 billion lightyears away? Just wait long enough (exactly how long depends on the actual density).

However, our Universe happens to be dominated by dark energy, which accelerates the expansion without boundaries. This unfortunately means that the light leaving today from a galaxy 17 Glyr away will be carried away by the expansion faster than it can travel toward us. In contrast, the light emitted today from a galaxy 15 Glyr away will travel in our direction, but will nonetheless initially move away from us due to the expansion. However, its journey toward us makes this expansion rate smaller and smaller (since the expansion rate increases with distance from us), and after a period of time it will have traveled so far that it has overcome expansion and starts decreasing its distance from us and eventually reach us after 100 Gyr or so.


Line of lights moving in a straight line, with a few following

Given the date and timing, this could be most like the Starlink satellites in their "stacked" configuration. They are currently in a line, but they will later move to separate orbits. Dr Marco Langbroek has a video of a line of 60 satellites.

They won't stay in a line; they have ion engines that they will use to spread out into separate orbits.


Could the rotation of a black hole cause other planets to rotate?

A rotating black hole (or indeed any other rotating object) can cause nearby objects to rotate through an effect called frame dragging. However this effect is really small! Detecting it requires either a a body to be in a pretty close orbit around a black hole, or extremely sensitive instruments.

The fact that the galaxy and almost everything in it rotates in one way or another is not considered to have anything to do with this effect. Instead it is basically the rotation left over as clouds of gas, dark matter and later dust collapse under their own gravity. Inevitably they do not fall exactly straight into the middle of the cloud, but end up aiming off to one side or another and so orbiting around the centre, and the final spin of the collapsed object comes from the more or less random overall total of these spins. Objects spin so fast because very large clouds of gas and dust have collaposed to form fairly small objects like planets and stars, and just like an ice skater speeding up their spin by pulling in their arms, this speeds up the spin of the central object.

If you removed all the black holes from a galaxy there would be no major change. A shift in the orbits of stars, especially near the centre of the galaxy and some changes in the behaviour of the clouds of gas between stars as the radiation from the central black hole stops acting on them.


A reference frame that has its Z axis fixed with respect to the Earth's rotation axis is not inertial. The Earth's rotation axis undergoes precession and nutation, and on a shorter term, undergoes polar motion as well.


This is a very broad question: its answer involves the full details of stellar evolution, Galactic chemical evolution and nuclear physics.

I'll limit myself to the following observations:

The elements you mention (actually  are all "islands of nuclear stability". Whilst their binding energy per nucleon is not as high as that of iron, it is a little higher than elements immediately around them in the periodic table.
Nuclear fusion reactions are exothermic (up to iron), but they require energy to initiate them (in a similar way that you need to get coal hot before it will burn). This ignition becomes more difficult the higher the atomic number (the number of protons) in a nucleus. This means that it is not a given that nuclear reactions will simply proceed to the element with the highest binding energy. If there is a way to prevent the material in a star reaching this ignition temperature, then the reactions will not proceed through to iron.
This means that even in a very massive star, it is only the core where burning runs to completion in the iron peak elements. The majority of the star is still hydrogen and helium, with the layers outside the core containing the remnants of burning of lighter elements. It is these products that get blown into space in a supernova and mostly consists of products of helium burning (C/O) and alpha capture onto these elements (Ne, Mg, Si, S, Ar, Ca),  with each being slightly harder to produce than the last and accounting basically for their relative abundances. Most iron would end up trapped in the neutron star or black hole remnant.
Iron is mostly produced following thermonuclear detonation of C/O white dwarfs - a completely different production path. Hence the abundance of Fe is decoupled from things like Oxygen, and more dependent on the rate of type Ia supernovae, governed by binary interactions and statistics than the details of nuclear physics.
Carbon and Nitrogen are mainly produced in lower mass stars that never become supernovae and where conditions never reach ignition temperatures to produce heavier elements by fusion. There are many more of these than stars which end as supernovae.

So, the relative abundances of the elements is a very complicated issue that has no simple answer, but where the broad details are well understood in terms of the processes I mention above.


Is it possible to measure galactic red shift using consumer telescope equipment?

Looks like you can measure the cosmological redshift of quasars using that equipment and an 14" reflector:

http://www.rspec-astro.com/sample-projects/
(halfway down the page)

So the answer appears to be yes, you can do it, assuming that page isn't a total fabrication. Seems plausible to me: the redshift of that quasar is sizable at 0.15 (though that's small for a quasar), which means that the spectral lines will be shifted by something like 15%. So you  don't need to have a very fine diffraction grating to see the shift. And that quasar is pretty bright for a quasar, at magnitude 13, so you can easily spot it with a good scope. Sounds like a fun project!


When is it a good aproximation to consider a star to be an ideal gas?

There is no hard and fast answer. To be treated as an idea gas (your title question), the particles in your gas should be point-like and they should be non-interacting if you are to use the ideal gas approximation.

This means you can take the mean separation ($\sim n^{-1/3}$) and compare that with the size of the particles - it should be much bigger. You can also compare the particle kinetic energies with the potential energy of their interactions at typical separations. eg The ratio of $kT$ to the Coulomb potential energy should be large.

The rest of your question is confused. You are asking about an adiabatic index, but that is a completely separate question from whether the gas is ideal. It depends on the equation of state, on interactions, on internal degrees of freedom, on the way energy is transported.

Relativistically degenerate gases can be ideal gases and are often assumed to be so. Do not confuse an ideal gas with a perfect gas (even though many do so).


This is exactly what is done for template fitting. The shifted SED is one of the intermediate data products of template fitting codes such as Le Phare or Phosphoros.

For instance, with Phosphoros, you can shift your SEDs to any number of redshifts, then find the shifted SEDs (also called model grids) in the IntermediateProducts folder in the Phosphoros directory.

Based on your comment, you are also applying a filter to the shifted SEDs. This can also be done with Phosphoros, and you can directly get the flux through each filter for the shifted SEDs.

Phosphoros can be used from the command line, so you use the Python library subprocess to make calls to Phosphoros from your Python script.


How can gravitational lensing makes a quasar appear brighter?

The quasar gives out light in all directions. The light spreads out in space. Only a very small amount of that light would be pointed exactly in the direction of your telescope.

But if a large galaxy or galaxy cluster is between the quasar and us, it bends some of the light towards us, making the quasar brighter (it would also distort the shape, but quasars are too small for their shape to be seen) This makes it brighter for us (and dimmer in other parts of space which would have got the light that is now bent towards us.)


Which is really larger, Big Dipper or Small Dipper, in 3D

To calculate that you will have to build a 3D model based upon the distance of each of their stars and then calculate the resulting 3D object's volume. I know this may not answer the question, but I give the way to answering it. It's not an easy job.

Making a quick guess based upon the distances of the stars of both constellations in my opinion I guess the Small Dipper has a bigger volume since it's starts have bigger distance differences ranging from < 100 ly up to > 500 ly. In the case of the Big Dipper all distance are very similar since all stars come from the same open cluster. Their distance range between 78 and 124 ly.

Source:

Small Dipper stars
Big Dipper stars

Useful resources:

Big Dipper 3D model


Kepler's Law, focal points - Heliocentric or Barycentric?

Planetary orbits are normally described as heliocentric, but it is possible to describe them from a barycentric point of view. JPL Horizons (https://ssd.jpl.nasa.gov/horizons.cgi) provides for both possibilities.


How do CMB photons 'gain energy when they pass through normal regions of space with matter' and 'lose energy when they pass through voids'?

It is the late time integrated Sachs Wolfe effect.

As they travel towards us, apart from the general expansion, photons from the CMB gain energy when they fall into potential wells (where matter is). Of course, they lose it again as they emerge on the other side of the well, but the cosmic expansion means that the well isn't quite as deep by the time that happens. As a result, photons gain a little energy compared to the average CMB sightline.

The opposite happens for voids. Energy is required to enter a void (the potential is higher than average), but all this energy is not retrieved because the potential flattens due to expansion as the photon traverses it. Hence the directions of voids would be a bit cooler.


Elemental abundances in the crust of exoplanets and accessibility to them

No, it doesn't work like that. A rocky planet, to first order, will probably contain a similar mix of elements to the Earth.

The most likely reason that the star has a low metallicity is because it is old, and formed when the interstellar medium was less metal rich. But because rocky planets do not incorporate (much) hydrogen and helium, what you are really asking is how do the abundances of other elements scale to the iron abundance (which is often what is meant by metallicity)? The r-process events, such as type II supernovae that produce the heaviest radioactive elements (beyond lead), enrich the interstellar medium "promptly", because the progenitor lifetimes are short compared to the age of the Galaxy.

However, because a lot of the iron and nickel in the interstellar medium is not produced/disseminated by type II supernovae, but by type Ia supernovae with longer-lived, lower mass progenitors, then if anything I would expect a larger ratio of elements produced/disseminated by type II supernovae to iron/nickel. This will also be true of the "alpha" elements like Silicon, Oxygen and Magnesium.

The planet might be a bit short in those elements predominantly produced by the s-process in lower mass giant stars - so elements like carbon (!), nitrogen, barium, rubidium, strontium, zinc and many others may be scarce.

For a star of this age/metallicity I would expect these variations to be plus or minus a factor of two or so compared to the Earth.

Note, I am discussing bulk abundances. The accessibility of the material, where it is etc., is more of a speculative geology question, and I don't know the answer. A further complication is where exactly the planet forms compared with the condensation temperature of various elements and materials. The above discussion implicitly assumes that the planet forms in a protoplanetary disc material at a similar temperature and density to the conditions in which the Earth formed. However, a change in these conditions will alter the detailed compositional mix.


Calculate Earth's distance travelled in a year

You need to find the perimeter of an ellipse. This might be accurate enough for you (recalling that an ellipse is only an accurate description of an orbit for a two-body system.

An ellipse has no simple formula to find the perimeter. There appear to be several commonly used approximations based on infinite series (see for example "Maths is Fun").

e.g. Define
$$ h = \frac{(a-b)^2}{(a+b)^2},$$
where $a$ and $b$ are the semi-major and semi-minor axes and $b = a\sqrt{1-e^2}$, where $e$ is the eccentricity.

Then the result you want is
$$ p = \pi(a+b)\left[1 + \frac{h}{4} + \frac{h^2}{64} + \frac{h^3}{256} ...\right]$$

All this is general, but might be overkill for your problem involving the Earth. If my calculator skills are working, then for the Earth's orbit around the Sun, $a = 1.00000011$ au, $e = 0.01671022$, $b = 0.99986048$ au and $h = 7\times 10^{-5}$.

Using this and just the first 2 terms in the series, I get a perimeter of $1.999895\pi$ au. So hardly a correction from $2\pi a$.

However, if you were serious about accuracy then you would have to add to this about 12.5 orbits around the barycentre of the Earth-Moon system and a further correction for the motion of the Sun around the solar system barycentre.


How can we explain high redshift numbers?

There is evidence that this assumption is incorrect.

  Stars actually started forming before 500 million years post-Big Bang.

There is some evidence that Population III stars formed as early as 200 million years after the Big Bang. The oldest known star, the Methuselah Star, dates back to 200 million years after the Big Bang. It is a second generation star (judging from its metal content), so that means a generation of very large stars must have lived and died before this one was born.


Why isn't Ultima Thule spherical, while its two component parts are

The explanation is that as Ultima and Thule were forming, they collected out of small hunks of debris and piles of accumulated rubble.  The key thing is that once U and T became the biggest piles of rubble around, any new rock or pile of rubble that gently approached them would be much smaller then U and T and hit and gently bounce a bit and, on average, settles down in a low spot. The collisions are mostly pretty gentle and even rubble heaps probably mostly stay stuck together because the force of gravity isn't enough to overcome the cohesion of the material.

U and T grow and are -- roughly -- spherical since the things colliding with them and sticking to them are much smaller than they are, there are lots and lots of them, and they have a small tendency to end up in low spots. So a rough sphere emerges.

When U and T finally collide to produce Ultima Thule, it's another gentle collision, and the same thing happens: Gravitational forces are too small to overcome cohesion, so U and T stay pretty much the shape they were in and you have a body which is two spheres stuck together at the point of collision.

U and T formed spherical because they formed out of many much smaller pieces which are probably still pretty distinct, but which are too small to see -- yet.  UT formed out of two big pieces which are also still pretty distinct, but much easier to see..


Protocol for establishing longitudinal meridians on other heavenly bodies

As an example consider the prime meridian for Vesta.

Based on Hubble images, the IAU had established a prime meridian based on the observation of the Obler's regio (a dark region). The prime meridian was defined as passes through the middle of this region.

When the Dawn mission arrived at Vesta, it was discovered that the asteroid's pole was not in the same place that it had been assumed to be (based on Hubble data) and that Obler's regio was not a convenient basis for a prime meridian, as it was not even discernable from up close. Instead they chose a small crater, Claudia, that was close to the equator, and defined the prime meridian based on that (the prime meridian is defined as a passing 4 degrees from the centre of Claudia) This is now the generally accepted coordinate system, although it has not been "officially" adopted by the IAU.

In general, if you are the first to map a new world, you get to choose where do draw your meridian.


What is causing this strange shimmering pattern of light during a total solar eclipse?

These are called shadow bands.

As the sunlight is reduced to a very narrow strip, in the last few moments before totality, turbulence and refraction in the atmosphere will cause shimmering bands of light and shadow.

What you are seeing here is essentially the same as the twinkle that you see from stars.