Does the universe curve in on itself? I've read a little bit about the universe, but its not clear to me if there is a place in the universe beyond which there are no stars.

For example, is there anywhere in the universe that one could go where we would not generally see stars in every direction?  Its seems uncertain to me that there is.

If I were to go to the farthest star from earth magically, rather the via relatively (for argument sake).  I"m guessing it would have stars surrounding it in every direction.  But perhaps I'm wrong, perhaps there is some edge to the explosion of matter from the big bang, beyond which there is nothing.

Do we know if the universe curves in on itself? or is there some other dominant theory?

In the currently dominant theories, the Universe is basically the same everywhere, if you look on a large enough scale. There may not be a furthest star from Earth at all (the universe may be infinite) or it may be like "the furthest point on Earth from London" which exists, but is not a specially interesting place.  The distinction between those two possibilities is exactly whether the universe "curves in on itself" (in which case it is finite) or doesn't in which case it is infinite. We don't know for sure, but any curvature is definitely pretty slight.

To see the universe as uniform, though, you do have to look on quite a large scale (billions of light years). On smaller scales than this the distribution of galaxy clusters (and so galaxies and so stars) is quite uneven, and there are places where there are no stars close enough to see with the naked eye. Nevertheless, with a big enough telescope in one of those places you would still see stars scattered in all directions.

You mention the big bang, and what you write suggests a common misconception -- that the Big Bang happened in the middle of a big empty space into which matter expands. That is not what the theory suggests. Instead all points in space, when you trace things back far enough, seem to have contained a very hot, very dense, sea of matter, which, as far back as we can extrapolate, was all spreading apart, creating new space uniformly spread throughout the old space in the process.


Why is there a gap in this image of supernova discoveries? I came across this gif showing supernova discoveries from the late 19th century to 2010. Here's the data in 2010:

Notice that there's a prominent region shaped like an inverted U in which there are few detection points, which I have crudely enclosed:

Why is this? Is it an artifact of detectors simple not observing that part of the sky, or is there some underlying astronomical reason?

The coordinate system in this image is RA and Dec. It is a coordinate system which uses the Earth's equator (projected onto the sky) as its midline.

The inverted U is the Milky Way. The Milky Way is full of dust and gas, and blocks our view of galaxies (and supernovae) behind it. There is enough dust in the plane of the galaxy to block our view in that direction. For example the galaxy IC 342 is one of the nearest galaxies, and would be brilliant if it were not close to the galactic plane. There may be other galaxies that are completely hidden.

Our galaxy's bulk not only hides supernovæ that are in other galaxies, it also hides most of the supernovæ that occur in the Milky Way


How is the bar formed in a Barred-Spiral galaxy? How does the bar in a barred spiral galaxy form? What prevents it from being spirals all the way down like water spiralling down a plughole?

"Galactic bars develop when stellar orbits in a spiral galaxy become unstable and deviate from a circular path. The tiny elongations
in the stars’ orbits grow and get locked into place, forming a bar. The bar becomes even more pronounced as it collects more and
more stars in elliptical orbits. Eventually, a high fraction of the stars in the galaxy’s inner region join the bar. This process has been
demonstrated repeatedly with computer-based simulations."

Source:
http://hubblesite.org/hubble_discoveries/science_year_in_review/pdf/2008/barred_spiral_galaxies_and_galactic_evolution.pdf


How much mass would have to be added to the Sun to significantly alter any of its characteristics? While this might at first sound like an XKCD What If? question, let's proceed with it because astronomical objects do occasionally merge.
How much mass would have to be added to the Sun to significantly alter its characteristics (lifetime, spectral type, size, etc.)?

As an example, from Wikipedia's Rogue planet:

A rogue planet (also termed an interstellar planet, nomad planet, free-floating planet, unbound planet, orphan planet, wandering planet, starless planet, or sunless planet) is a planetary-mass object that does not orbit a star directly. Such objects have been ejected from the planetary system in which they formed or have never been gravitationally bound to any star or brown dwarf. The Milky Way alone may have billions to trillions of rogue planets, a range which the upcoming Nancy Grace Roman Space Telescope will likely be able to narrow down.

...if one doesn't fall into the Sun first!
Just in the last few years alone we've seen both a comet and an asteroid with extra-solar velocities enter the inner solar system, compare that to the 4.6 billion year age of the Sun.

2I/Borisov 2.01 AU perihelion
ʻOumuamua 0.255 AU perihelion

How much mass would have to be added to the Sun to significantly alter
its characteristics

Asking how much would be significant is inexact.  The Sun is classified as a G2V main sequence star.   Though the chart lists one solar mass as G4V, so there's some variation in there.   The classifications seem to relate to temperature.    To go 1 step up (and using the sun or any G star), the increase of one ranking (decrease in 1 number after G) alternates between a 2% and 5% increase in mass.   I'm going to round that to 3%-4%, so for our sun, that's about 30-40 Jupiters (31.4-41.9 if you care about significant digits), but I'm going with 30-40 Jupiter masses added for each step up the classification ladder, ignoring the effects of impact, which would be considerable.
That increase of one in classification level viewed from Earth would be quite significant.   3% more mass corresponds to about the 4th power of that or 12.5% more luminosity.   Earth would be very different if the Sun was 12.5% more luminous.  I would think that would be enough to turn Earth at least into a steam-bath, perhaps a full blown runaway greenhouse effect due to the abundance of water and feedback mechanisms.   Adding mass would also change Earth's orbit, drawing it in slightly and shortening the year, so the increase from Earth would be a bit more than 12.5%.
The stellar mass math goes something like this.
Lifespan decreases by about the 3rd power of an increase in mass.
Energy output or luminosity increases by about the 4th power of the mass, around 1 solar mass.  This increase slows down to about the 3.5th power with more massive main sequence stars.
Surface temperature increases slightly with mass, perhaps 10-20 K per 1% increase in mass, or about a 1% increase in temperature for every 3% increase in mass would be about ballpark.
Radius increases with added mass too, so the heavier sun would be both brighter and larger.   The radius increases enough that the surface gravity decreases.  A 3% increase in mass corresponds to around a 2% increase in radius, so the size wouldn't appear very different on these scales.
If the Sun became just 0.5% to 1% more luminous  (about 1.5-3 Jupiters in mass) that would be in the ballpark to man made climate change so far.  Enough to melt surface ice over time and cause oceans to rise, so, significant depends a lot on point of view.
These numbers should be taken with a grain of salt because other factors like metallicity and age matter too.   Stars like our Sun also change in size and luminosity during their lifetime, growing both larger and more luminous over time.  There are more massive main sequence stars than our Sun that are less luminous because they're younger and the reverse as well, though off hand, I can't name any.
Errors fixed.  Thankyou for pointing out.
I want to add, since you mention XKCD in the question, if Randall Munroe was answering this question, he would take it to the extreme, answering what happens if you add million Jupiters, or a billion.   Fun times.  Without his clever diagrams, it wouldn't be the same, so I'll stop here.


Does the Earth ever cross the Moon's path with respect to the CMB? Title says it, really. As I understand, our total peculiar motion with respect to the CMB is 631 (+/-20) km/h, in the direction of 10.5h RA; -24 deg Dec (or rather away from 22.5h RA; +24 deg Dec, right? That's close to the direction of Lambda Pegasi).

So, if we draw a line from the Moon to Lambda Pegasi, does Earth ever cross it?

I think the answer is no, but don't have enough of a model to prove it to myself.

The inclination of the Moon's orbit is 5.14 degrees, and the Earth subtends about 1.2 degrees from the moon, but the Moon's path extends off at 24 degrees above the ecliptic. So the Earth won't cross it, right?

This is all assuming that the vector info above is the total, and already accounts for the motion of the sun around the milky way. If that's not the case, please let me know.

Not within plus/minus a few hundred years at least! The Moon's orbital plane around the Earth (viewed in an inertial frame) rocks back and forth a bit (Lunar Nodal Precession with a period of about 18.6 years), but that's not going to help here.

You can play with the numbers in this program.

import numpy as np
import matplotlib.pyplot as plt
from skyfield.api import Loader

degs = 180./np.pi

load = Loader('~/Documents/yournamehere/SkyData')
data = load('de405.bsp')
ts   = load.timescale()

timez = []
dayz  = np.arange(1, 31, 0.1)
dayz  = np.arange(1, 366, 0.1)

#for year in range(1700, 2200):
for year in range(2017, 2028):
#for year in range(2000, 2030):

    #time    = ts.utc(year, 7,  dayz)  # July
    time    = ts.utc(year, 1, dayz)   # January
    timez.append(time)

earth = data['earth']
moon  = data['moon']

RAdegs, Decdegs = [], []

for time in timez:

    RA, Dec, dist = earth.at(time).observe(moon).radec()
    RAdeg, Decdeg = degs*RA.radians, degs*Dec.radians

    # lazy way to block the wrapparound
    dRAdeg = RAdeg[1:] - RAdeg[:-1]
    dropit = np.abs(dRAdeg) > 100
    RAdeg[:-1][dropit]  = np.nan
    Decdeg[:-1][dropit] = np.nan

    RAdegs.append(RAdeg)
    Decdegs.append(Decdeg)

RAdegpts  = [157.5, 157.5+180]  # coordinates from the question
Decdegpts = [-24.0, +24.0]

if 1 == 1:
    plt.figure()
    for RAdeg, Decdeg in zip(RAdegs, Decdegs): #[::5]:  # plot every 5th year
        plt.plot(RAdeg, Decdeg)
    plt.scatter(RAdegpts, Decdegpts)
    plt.title("2017 throuth 2027 continuously", fontsize=20)
    plt.show()


What's this red stuff in some astronomical photos, e.g. Centaurus A? Sorry for the rookie question, but I've got no idea about that red "belt" crossing over center of the galaxy, in this photo. I've actually tried all sorts of google searches but I couldn't get anywhere with keywords like "The red stuff" or so. Years ago I had a color book in which I saw this photo for the first time, and I can remember my attempt to remove the red dirt and clean the image with my shirt!

"Clean the image", how apt, because most of what you can see is caused by dust.

It is a bit difficult to comment specifically because I think you have chosen an example which is a combination of images and not all the images are in the visible part of the spectrum and so have had a "false colour" applied. In fact it is a combination of optical, X-ray and sub-mm images.

In general terms red is present in false colour astronomical images, usually used to representing the coolest or longest wavelength light in the image.

Red is also present in true-colour images of nebulae and is caused by light emitted due to transitions of atoms and ions between excited states separated by particular energies. Red light is predominantly due to Hydrogen alpha and ionised nitrogen emission.

The coolest stars are also red in true colour images, simply because their pseudo-blackbody radiation spectra peak in the red part of the spectrum for temperatures lower than 5000 K. The integrated light from old stellar populations (predominantly cool main sequence stars and red giants) would therefore have a reddish tinge - for example in elliptical galaxies.

As I said, it is hard to say in the case of this particular composite image, but the band across the centre of the galaxy is dust. It obscures (makes dark), the optical image, but warm dust glows in the sub-mm image. The structures coming out of the galaxy at right angles are jets from the active galactic nucleus traced in the X-ray.


Where can I find good places to watch stars? Where can I find good places to watch stars near me? I live in Leicestershire, England and am looking for good places to watch the sky nearby.

It would have been better to ask about "which conditions must have a good starsighting place".

It must be:

Dark
Far from cities
Not under main flight paths (this depends on which altitude is the object you want to observe)
High altitude (or any special place with low cloud formation and low turbulence)
Far from big water bodies that can cause local humidity


Why most planets don't spin retrograde? Edit: just found duplicate

Since orbital velocity is greater on small orbits, and smaller on large orbits, why don't planets spin retrograde around their own axis?

One portion of planet's mass is closer to its star, and should move prograde relative to center of mass.

Part of the planet's mass far from the star orbits higher and should move retrograde relative to center of mass,

which should create a resulting retrograde spin, with rotation speed depending on planet's diameter.

What's wrong with this? ( since fact is : most planets spin prograde )

In this not to scale second illustration, planet is replaced by two masses linked by a beam. Kepler's law of equal surfaces swept per unit of time is used
to display the retrograde rotational force that should be applied relative to center of mass, which orbits black path.

EDIT : Ok I think I found where my mistake is(?) Differential gravitational pull (yellow arrows) counteracts retrograde rotational force (green arrow) Which results in no rotational force at all. Anyway I still don't get how conservation of momentum of everything orbiting the sun, also applies to spin direction of orbiting objects.

Basically, go back to the pre-planet particle clouds, and as explained in

geek.com,

  To answer the question we have to go back to the beginning, to the
  formation of a solar system like ours. Before a star and its planets
  exist, there’s just a cloud of disorganized gas and small molecules.
  This is often called a molecular cloud or “stellar nursery.” The Eagle
  and Orion Nebulae are some of the more famous stellar nurseries we
  have been able to observe with the Hubble Space Telescope. These
  clouds are composed mostly of molecular hydrogen, which wouldn’t be
  able to congregate outside the dense molecular clouds. These clouds
  can be of any size — not just massive structures like the Orion
  Nebula. Over time the energy of the molecules in the cloud pushing
  outward can be overcome by slower molecules collapsing together
  farther in. As long as there is sufficient mass in the molecular
  cloud, it continues collapsing in toward the center until it reaches a
  high enough mass to fuse hydrogen and become a new star. The spin that
  we see quite clearly now is related to the process of the molecular
  cloud collapsing. The original cloud was very, very large and made up
  of many individual molecules and small clumps of matter. On that
  scale, there is some small amount of rotation within the cloud. It
  could be caused by the gravity of nearby stellar objects, local
  differences in mass as the cloud churns, or even the impact of a
  distant supernova. The point is, most molecular clouds have at least a
  little rotation.

  As the cloud collapses to form a star, it has what physicists call
  angular momentum. This is the movement an object has as it rotates
  around a central point. In a large system like a molecular cloud, each
  particle has some angular momentum, and it all adds together across a
  very wide area. That’s a lot of momentum, and it is conserved as the
  cloud continues to collapse. But how does that get us to objects that
  spin and orbit?

  Imagine a figure skater spinning around with arms outstretched. That
  is a model of angular momentum just like a collapsing cloud of gas.
  When the arms are drawn inward, the rotational velocity goes up
  because the total angular momentum is conserved unless there is some
  external force acting on it. There are such forces acting on the
  figure skater, but less so on a collapsing molecular cloud. So if a
  molecular cloud was maybe a light year across, then collapsed down to
  be just a fraction of that, it would be a huge change in size. Just
  like the figure skater pulling her arms in, the velocity must increase
  to conserve angular momentum and therefore form a spinning
  protostellar disc. It is from this orbiting matter that all the
  planets form, and of course, they are also spinning and orbiting in
  the same direction because of the conservation of angular momentum.

Quoting from wikipedia as to the mystery of  the misbehaving planets:

  All eight planets in the Solar System orbit the Sun in the direction
  that the Sun is rotating, which is counterclockwise when viewed from
  above the Sun's north pole. Six of the planets also rotate about their
  axis in this same direction. The exceptions—the planets with
  retrograde rotation—are Venus and Uranus. Venus's axial tilt is 177
  degrees, which means it is spinning almost exactly in the opposite
  direction to its orbit. Uranus has an axial tilt of 97.77 degrees, so
  its axis of rotation is approximately parallel with the plane of the
  Solar System. The reason for Uranus's unusual axial tilt is not known
  with certainty, but the usual speculation is that during the formation
  of the Solar System, an Earth-sized protoplanet collided with Uranus,
  causing the skewed orientation.[6]

  It is unlikely that Venus was formed with its present slow retrograde
  rotation, which takes 243 days. Venus probably began with a fast
  prograde rotation with a period of several hours much like most of the
  planets in the solar system. Venus is close enough to the Sun to
  experience significant gravitational tidal dissipation, and also has a
  thick enough atmosphere to create thermally driven atmospheric tides
  that create a retrograde torque.


How young could the asteroid belt be? I heard a theory (possibly from a crackpot) that there was an event in our solar system where two objects collided and created the asteroid belt, the orbits of some of the planets were altered, and that this event took place late enough for homo sapiens to have referenced observations made during the previous orbit in their oral traditions. (Mars parsing close enough to Earth for people to observe its moons)

Is any of this possible given readily observable evidence?
Are asteroid belts ever created in this way?

Dating the age of the asteroid belt is a problem that has to date no general solution.

Rocks are usually dated in the laboratory via isotopical analysis. This is possible for meteorites, but what to do about those rocks that remain in space?

Several methods to date the asteroid belt are known to me, apart from ad-hoc assumptions that it has the same age as the planets:

Creating links between dated meteorites and asteroid families: The meteorite family of L chondrites have possible origin families in the asteroid belt. Those 'families' are small blobs in the eccentricity-inclination diagrams, that probably originated in one common parent-body breakup event. Dating meteorites originating from those families, gives minimum ages for the parent bodies of i.e. 500 Myrs for the Ordovician meteor event.
Dynamical age modeling: After a breakup of a parent body, the cloud of fragments experience different strengths of YORP and tidal forces, which causes their semimajor axis distribution to spread. Taking a given distribution, one can calculate back to an age of when the cloud was one particle, i.e. the moment of breakup. This again is a minimum age and found for the Flora family to be more than $10^8$, but less than $10^9$ years. This does exclude a very young belt, but does not say much about the upper age limit.
Crater ages: For the largest bodies, i.e. Ceres, Vesta, etc. we can assign relative ages to the lunar cataclysm. Those usually give ages in excess than 3-4 Gyrs. For bodies like Vesta, where the surface is saturated with craters, this age is a lower limit.
Dating of crustal ejecta: There is another family of meteorites that can be linked specifically to Vesta via spectroscopy, called the HED meteorites. Those meteorites look like they're coming from a differentiated parent body (melted, low siderophile element content), and they containt CAIs. Thus, we can date Vesta at least back to the origin of the solar system, as CAIs give us the effective $t0$ for the formation of the solar system.


When did "resilvering" large telescope mirrors actually refer to aluminization, and why was it necessary? This answer nicely summarizes the general picture of historical use of silver for mid to large size telescope mirrors, and the shift to aluminum once good quality vacuum pumps and chambers suitable for this process were available. A freshly evaporated, atomically clean aluminum surface will can sometimes violently react with oxygen, so this has to be done under high vacuum, and vented carefully with inert gas first.

When did the term "resilvering" generally start to actually refer to aluminum coating, and not silver? Was that actually the case by the 1960's or 1970's?

From what I remember, evaporated aluminum will form a protective oxide layer ("native oxide") of the order of 10A thick within seconds or minutes that is self limiting - no further oxygen can penetrate this layer.

If this is so, why was re-silvering actually necessary? The cost, risk, and downtime associated with removing the primary and other mirrors every few years sounds like something you wouldn't do unless absolutely necessary.

Amateur telescope and mirror maker here. Not sure if I qualify as a "citable source" but anyway, here it is:

All metals will eventually tarnish. It may take a long time, but it will happen. The process is not entirely chemical always. Sometimes it's purely mechanical (abrasion). Other times it's in between. Surface phenomena are complex.

Even gold-coated mirrors will tarnish. You could use gold for infrared observations. But no matter how much care you take of it, eventually you will have to re-coat it. Sure, gold lasts a pretty long time, but not even gold is forever.

Aluminum is covered with a natural layer of colorless sapphire (aluminum oxide), which protects the metal from further destruction - sapphire is very hard and very inert chemically. But oxygen still diffuses through sapphire (albeit extremely slowly), and so do other corrosion agents like sulphur oxides, marine salt, etc.

So, yes, replacing silver with aluminum was an improvement. But not even aluminum is forever. In practice, you will maybe double the lifetime of the reflective layer when you migrate from silver to aluminum.

It should be noted that silver is still used. Aluminum is most popular nowadays, but you'll still find silver-coated mirrors (its reflectivity is higher than aluminum), gold-coated (for infrared), etc.

There are further coating techniques, like layering silicon oxide on top of metal, which further improve the lifetime of any kind of metal.


Magnetars and the Dynamo Effect How do magnetars create/sustain such strong magnetic fields? If a dynamo effect did create a magnetar, then what creates the magnetic field? The dynamo effect requires an electrically conductive fluid, but magnetars are purely composed of neutrons. In Earth, the dynamo effect is supported with evidence, but how does it apply to magnetars?

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

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

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

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

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

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


Could photons orbit a black hole if fired at the correct angle? My experience with Kerbal Space Program gives me the impression that it's impossible to transfer from a stable orbit of one body to the stable orbit of another without deceleration. But is this true for a photon? Could light enter a stable orbit of a black hole if fired at the correct angle from a distance?

There are no stable photon orbits around black holes (see this answer).
Black holes do have an unstable photon orbit though. "Unstable" here means that any tiny deviation from this orbit, will case the photon to be either scattered to infinity or sent into the black hole. It should be emphasized that there such orbits are always circular an can exist at only one radius (1.5 times the Schwarzschild radius for a Schwarzschild black hole). There are no elliptic photon orbits. This orbit is often called the "light ring" (or sometimes "photonsphere")
As long as we treat the photon as a perfect massless point (test) particle, we can find a highly fine-tuned trajectory that will asymptote to the light-ring. Of course, in reality, a photon always needs to be a wave packet with a finite size, making this fine tuning impossible in practice.


Fingers of God effect for galaxy clusters Galaxies that reside in galaxy clusters are know to have peculiar velocities that cause the Fingers of God effect.

But what about galaxy clusters themselves?

Do they also have peculiar velocites by interacting with other galaxy clusters?

If so, are they high enough to be taken into account or low enough to be ignored?

I'm going to break your question down into two sections.

  Do galaxy clusters have measurable peculiar velocities?

The answer to this depends slightly on what is meant by "peculiar velocity". In the most general sense, the peculiar velocity of some object is the velocity it has with respect to some standard of rest. In the cosmological sense, the peculiar velocity of a galaxy in a cluster is simply any motion that galaxy may have which is not attributable to the Hubble flow. Primarily this peculiar velocity is a result of that galaxy's orbit around the cluster barycenter.

With that in mind, it should hopefully be easy to reason out that clusters themselves will have peculiar velocities. It is highly unlikely that a cluster will be "fixed" in space and only move according to the Hubble flow. A quick google search revealed Bahcall & Oh 1996 which cites an rms one-dimensional galaxy cluster peculiar velocity of $298\pm28\:\mathrm{km/s}$ (based on 22 clusters). While this is smaller than a typical galaxy's peculiar velocity, it is certainly measurable and capable of being distinguished from the Hubble flow (at least for reasonable redshifts).

The real difficulty here is not whether the cluster has a non-negligible peculiar velocity, but rather, is that peculiar velocity even measurable due to many of the potential limitations. While the value of the peculiar velocity of a cluster is certainly "high enough to be taken into account" it can often be difficult to measure such a velocity with any reasonable error (a point that Bahcill and Oh discuss).

  Can these galaxy cluster peculiar velocities produce the "Fingers of God" effect?

To answer this you have to consider a very particular point I made in the last section which was that for galaxies the primary contribution to their peculiar velocity is their orbital motion around the cluster barycenter which can result in peculiar velocities on the order of $10^3\:\mathrm{km/s}$.

When trying to map out a particular cluster in redshift space, you can often get "Fingers of God" in your maps which are a result of the cluster appearing elongated towards the observer and due entirely to the orbital peculiar motion. An example of such an effect is below, with the red circles added by me to indicate these fingers of god.

Galaxy clusters on the other hand are not gravitationally bound to other galaxy clusters! This means any peculiar motion a cluster has is simply its general motion through space. Since the fingers of god are a result of orbital peculiar motion, it stands to reason that galaxy clusters will not have such a motion and as such clusters won't produce these fingers of god. You'll find that more or less, the peculiar motion of clusters will simply move cause them to be randomly re-positioned without any real "structure" that you see in the fingers of god for galaxies.


How long for a complete solar system cycle Considering the solar system as a closed system, how long until all the planets pass twice by the same position relative to each other? I would say that you'd have to get the length of the year in each planet and find the lowest number that can be divided by them, is that correct or are there more things to take into account?

Furthermore, can one do this also considering the moons of the planets?

Thank you!

It depends how precisely you mean the same position.

Take 2 since my previous attempt was ... optimistic.

In the late 1970's, the outer planets were moving into a configuration that made a Planetary Grand Tour much cheaper and quicker. They will be in a similar configuration in about 175 years.

But when will the planets be in the exact configuration? Never, well, not before the Sun turns into a red giant and swallows the inner planets.

But from time to time the planets will all be on the same side of the Sun. If the planets were all in a line, then after 495 years they would all be on the same side of the Sun, then again at 500 years, 522 years, 523 years (these last two are, of course 88 days apart, the period of Mercury), 995 years, 997 years, 1006, 1183 years, etc.


Is Optical VLBI theoretically feasible? If not why not? There are plenty of optical interferometers in use with baselines of up to maybe 1km. As far as I can find out, they all work by directly collecting the light at all the telescopes, using mirrors to bring them together and then actually interfering them and observing the fringes with photodetectors.

Meanwhile, radio astronomers do interferometry over distances of thousands of km, but they do it quite differently. Essentially, they record the baseband electromagnetic waveform at each telescope, with an accurate timestamp, transport them all to a central location and combine them computationally.

So my question is whether in principle you could use the radio astronomers methods with an optical (or near IR) signal. Obviously the engineering and computational challenges would be formidable -- you would need to record data at speeds close to 1 Pb/s, timestamp it with femtosecond accuracy and do a ludicrously huge computation to recover an image, but that's not my question here.

On the one hand, it seems like the answer should be "yes" -- electromagnetic waves are electromagnetic waves. On the other hand, why should the phase of a photon observed in one telescope have anything to do with the phase of one observed in another? If you are digitising the data at each telescope, you are ruling out interference  fringes on a quantum level, just as happens if you find a way to measure which slit a photon went through in the double-slit interferometer. This suggests that the answer should be "no".

If the answer is "no" then what wavelength does the cutoff happen at? We have radio-style VLBI at 1mm now and optical-style interferometry at 1 $\small\mathsf{\mu m}$, what is possible in between?

This question amounts to: is optical interferometry possible when detecting photons? The answer to this is yes. Many experiments have been done using interferometers where one photon at a time is passed through the apparatus and still reveals an interference pattern when detected on the other side.
The problems with optical interferometry are manifold: optical interferometry at the moment relies on bringing the signals together in hardware; there is an immediate problem with applying the radio interferometry technique of "taping" the data and doing the correlation offline, which is simply that the bit rate of the higher frequency optical light is far too high to do this at the moment. I have to admit ignorance as to whether this will be possible in the forseeable future. There is an obvious issue in recording the time of arrival of photons more accurately than the uncertainty principle allows for even the fastest optical transitions of $\sim 10^{-8}$ s, which is well below the resolution needed to sample frequencies of $\sim 10^{15}$ Hz. (Edit: the fastest detectors don't use transitions in atoms and can achieve timing of $10^{-10}$ s; still too slow by orders of magnitude.)
However, even if you could "tape" the data, there are many other problems to be overcome. These difficulties mostly relate to the coherence length and coherence time of phase changes introduced in the atmosphere above each telescope and the inability to use phase-reference sources as a result; and the inability to use phase-closure techniques unless the sources are very bright. These difficulties are explained in more detail in https://astronomy.stackexchange.com/a/35215/2531
To summarise, telescopes in different places observe the same object through different atmosphere. This introduces phase errors. These can be calibrated out in radio observations by using a nearby reference source. This is because the phase errors persist over a reasonably long time and are similar over a reasonably large ("isoplanatic") patch of the sky. For optical observations, the isoplanatic patch may only be a few arcseconds across, so there is rarely a bright reference source, and the atmosphere changes in any case on timescales of tens of ms.
Optical interferometers get round this by using phase closure techniques. The phase error can be eliminated by comparing the phase differences from at least three stations. But again, you have to do this on very short timescales and this means optical interferometry is limited to bright objects in order to get enough signal.
Where is the cut-off? Well certainly beyond the near-infrared, because those interferometers also use similar techniques (delay lines, phase closure) in their techniques.


Could some bacteria theoretically evolve to survive in space? There is no edge to the atmosphere, and therefor I expect there would be a gradual reduction of concentrations of bacteria as we ascend into space. Logically anaerobic bacteria should be able to obtain its energy from the sun and nutrients from whatever scarce but available organic floating materials. Cosmic rays and high energy particles could damage DNA, but there might be someway life could protect itself against them. Could it be that bacteria is evolving/ has already evolved at the edge of the atmosphere to live in space?

Currently there is no life that we know of that evolved in, or can survive for prolonged time in a vacuum.

To say 'Life in space is possible because there is anaerobic life' is a completely wrong point, that ignores the effects of vacuum on life.

Sure, there are extremophiles that can survive for a limited time in space (see also this list) when going into hibernation modes. However all life as we know it needs liquids and/or gases to survive. Those would evaporate once the cell/water bear would become active again. Activity is necessary to sustain and multiply life, so no, nothing that we know of can stay active and alive in space.

  Cosmic rays and high energy particles could damage DNA, but there might be someway life could protect itself against them

Until life comes up with such mechanisms, we just don't know. The problems with liquids and gases evaporating would need some tougher solutions even.

Bacteria are also pretty heavy 'particles' and therefore cannot float anywhere near the 'edge of space'. There are reports of bacteria in the stratosphere, but that requires already strong upwinds to put them there.


How accurate are rederings of something entering Earth's atmosphere? Supposedly how accurate is it when you see a movies showing something entering Earth's atmosphere traveling 1 mile to 10 miles an hour or barely at all but skinning the atmosphere they show it heating up and burning just by entering the atmosphere no speed or barely just entering?

As a second question, something coming in - say, a vehicle with rockets - to ensure that the speed does not exceed that would cause  any heating at all. What speed would that be?

In the movies you reference, I doubt that it was ever the intention of a movie maker to suggest that the craft was moving at a crawl when it began to enter the atmosphere. If the craft doesn't look like it is moving very fast, the idea is probably that the point of reference (the "camera") is moving at nearly the same speed as the craft, or perhaps that the craft is so huge, that the reference point is miles away. Everything is relative.

There is essentially no realistic scenario where any object would enter the atmosphere at such a slow speed. Imagine, for example that you have an object, like Felix Baumgartner or something, which is standing approximately stationary in space above the Earth's atmosphere. Gravity will immediately begin pulling on that object and because there is no atmosphere to slow the object down yet, it just keeps accelerating as it falls. Even as it enters the edge of the atmosphere, it doesn't slow down much, because there isn't much atmosphere yet. Felix was moving more than 800 mph before he started to slow down.

The International Space Station is another object that's sort of "stationary", in a manner of speaking. It's not going anywhere because it's in a stable orbit (sort of). But in order to stay up there, it's actually moving more than 17,000 mph, relative to the atmosphere. If something is "stationary" with respect to the sun, just sitting there in Earth's orbit as Earth comes at it, the Earth's atmosphere would smack it at 67,000 mph. That's how fast Earth orbits the sun. So there's really isn't anything that would just gradually drift into the atmosphere at 1 to 10 mph.

If a craft were to use rockets to halt itself before entry, as you suggest, remember Felix again; it can become stationary with respect to the Earth, but then gravity begins its work, so if the craft wants to remain at a very slow speed through the entire decent, it has to keep firing those rockets all the way down. And certainly, nothing would burn up at that pace, other than a lot of rocket fuel. What causes the burning is not something special and hot about the atmosphere, but the fact that the object is smashing into the air (thin as it may be at that altitude) at thousands of mile per hour. It is even theoretically possible for meteors to enter at such speeds that the atoms are literally forced to hit each other's nuclei, and a part of the fire generated would actually a brief nuclear reaction.

Response to Question Edit:
(Thank you HDE. The initial question was very unclear.)
The speed at which friction completely stops is 0. Any movement at all, even in that 1 - 10 mph range, produces a small amount of heat from friction, even if the medium you are moving through is very thin air (though certainly nothing one would characterize as "burning"). You seem to be looking for the speed at which that friction would be low enough to not cause a problem, but there's no magic answer. It depends on the situation. How delecate is the craft? How much wear is acceptable? Essentially, this question becomes something very similar to asking "How fast can I drive with something sticking out the window?" You are trying to keep the entry speed low enough that the wind is not a problem for you as the density of that wind gradually increases from practically nothing to surface air density. If your craft has a heavy, durable heat sheild, properly oriented, then extraordinary speeds are still not a real problem. But if you are trying to create a sand mandala on the roof of your craft as you descend, then you are going to want to control that wind level very carefully.


From CMB anisotropy data observed in 1992, did astronomers figure out that the universe should be accelerating before its discovery in 1998? CMB anisotropy was measured in 1992. I assume that astronomers, then, like now, would have been able to deduce the cosmological constant and things like that from the CMB anisotropy data. Then, from the cosmological constant and things like that, they would have been able to find that the current universe was accelerating. Did they indeed know that the current universe should be accelerating before its observation in 1998? Or, was the discovery in 1998 a great surprise? What was the exact situation between 1992 and 1998?

The only CMB anisotropies that were well determined in 1992 (and by the date, I assume you refer to the results from the COBE satellite) were those at small angular frequencies. This revealed the overall scale of the CMB structure (which tells you about a need for dark matter in order to get the large scale structure seen at the present epoch) and the dipole anisotropy, which tells you about the Galactic motion with respect to the CMB rest frame. There were no constraints on any possible dark energy - or rather, the effect of dark energy on the low frequency anisotropies is degenerate (i.e. cannot be distinguished from) the effects of the overall curvature of the universe.

To measure the curvature of the universe and constrain the mix of dark matter and dark energy (and constrain other cosmological parameters) needs information from higher angular frequencies - in particular information from the location and amplitudes of the first three "acoustic peaks" at angular scales of . This really only became available with the WMAP data (launched in 2001) and with improvements in ground based CMB measurements that took place at the end of the 1990s. The results and conclusions from this emerged over several years from 2000 onwards (e.g. Durrer et al. 2003; Spergel et al. 2003).

In actual fact, the evidence for $\Lambda$ from the CMB alone is rather weak. As explained in one of Wayne Hu's cosmology tutorials, the strongest evidence from the CMB is that the total density of the universe $\Omega = 1$ from the CMB and this can then be combined with other data (e.g. primordial abundances of D, He; galaxy cluster dynamics; gravitational lensing) to suggest that the matter contribution to this is only $\simeq 0.3$.


Why is cosmological event horizon closer to us than the particle horizon? From what I learned, cosmological event horizon is the horizon that moves away from us at/more than the speed of light. And particle horizon is the horizon that particles inside the horizon can have a causal connection with us. It is the maximum distance from which particles could have traveled to the observer in the age of the universe. This simply indicates a contradiction to me. How is it possible that the cosmological event horizon is closer?

The particle horizon marks the region from within which we may have received light. It started out at zero, because light from nowhere had had the time to reach us, and increased as time went by because light from increasingly larger distance reached us. Due to the expansion it will always increase in physical size, but due to the accelerated expansion, there is a limit to the region of the Universe that we will ever be able to receive light from. Today, the particle horizon is 46 Glyr (billion lightyears) away, and the limit — in the far future — is the part that today is 63 Glyr away (but in the far future will be much farther away).

The event horizon can be thought of as sort of the opposite. At any time $t$, it is the distance from which we will be able to receive light. If the Universe contained no dark energy, there would be no limit to how distant a galaxy could be and we would still be able to see it — just wait long enough. However, the accelerated expansion means that an increasingly smaller part of the Universe will be able to emit light toward us with us receiving it, before the expansion carries the photons away faster than they can travel toward us.

At Big Bang, the event horizon consisted of particles that today are 63 Glyr away — that's what I meant by "the opposite of the particle horizon".

The following figure from Davis & Lineweaver (2004) may help explain. It show basically three versions of the same thing, namely how the sizes of various horizons evolve with time. The top panel shows time as a function of physical size, while in the middle panel, the $x$ axis show comoving coordinates, i.e. the coordinates that expand along with the Universe. The lower panel shows not time on the $y$ axis, but conformal time, which can be thought of as the time it would take a photon to travel a distance if we "froze" the Universe.


Sunrise and sunset times when compared between the vernal and autumnal equinoxes Why is it that sunrise and sunset occur at different times when compared between the vernal and autumnal equinoxes?

Consider Pontianak, Indonesia, 0 deg N, for simplicity. Timeanddate.com gives sunrise and sunset times 5:46 AM, 5:53 PM at 3/21 (vernal equinox) and 5:31 AM, 5:38 PM at 9/23 (autumnal equinox). Given the symmetry of the problem, I would expect that sunrise would occur at the exact same time during the two events. Same for sunset. What is the reason for the 15 minute difference between the time sunrise occurs on the vernal equinox and the time sunrise occurs on the autumnal equinox?

Instead of looking at sunset/sunrise times, it's easier to look at noon, which is right in the middle between sunset and sunrise: 11:50am in March and 11:35am in September.

The fact that noon doesn't always happen at the same time of the day (when measured with steady clocks) is expressed by the Equation of Time. Two factors play a role here:

The plane in which Earth rotates around the Sun (the ecliptic) is tilted with respect to the one it rotates around its axis (the equator)
The Earth's orbit around the Sun is an ellipse, not a circle, and it moves faster than average in its orbit from (roughly) October up to and including March, and slower in the other months


What are shooting stars and how are they formed? What are shooting stars? How are they formed and how often do they occur during the night?

Also, why are there more shooting stars on some nights than on others?

A shooting star is simply the visible path a meteriod takes burning up in the atmosphere. As it flies through the atmosphere, it leaves a trail of fire, creating the streak of light you see.

Before they enter the atmosphere, they are simply chunks of rock. They could form in one of many ways - be broken off of a planet, etc. After they do this:

they look like this:

As for the reasons for meteor frequency varying, it's because of where the Earth is in it's orbit around the sun. Since almost all meteors are in orbit around the sun, and there tend to be clumps of them, this makes a 'donut' of meteors in orbit around the sun.

When the earth intersects one of these donuts, there is a spike in meteor activity. Ultimately, it's about how many space-rocks happen to hit Earth.

One annual 'high', for example, is the Perseid peak, which tends to occur mid-late summer each year. Sometimes, this even delays launches of space vehicles.


Are new stars less pure as generations go by? If stars are primarily made of hydrogen, which is then burned to helium, and then on and on and on down the chain until you either reach iron, or in extreme cases much heavier metals, this then implies that as time goes on, more and more heavy elements will disperse into the universe.
Now, the universe is huge, and there's still a lot of hydrogen left to make stars out of.
That being said, in denser star-forming regions, much of this gas is available for recycling as the remnants of old stars get turned into new ones.
Doesn't this all this imply that as time goes on, more and more heavy metals will be included within young stars from the start?
It may be a very small percentile, sure, but it I assume it would increase slowly as the generations go by.
Furthermore, I assume many rocks get swallowed up by a star during the formation of the planetary disc, which likely adds to the impurity.
Does any of this have any adverse effect on the lifespan of the sun? Or are we talking quantities so small in objects so huge in time-spans so large, that even a dip of a hundred years is completely meaningless?

The chemical enrichment of the Universe over time is indeed a thing. The plot below (source) shows observational measurements of the cosmic density of ionised Carbon in the Universe against redshift (higher redshift -> further back in time). The abundance of other heavy elements over time shows a similar trend.

Stars are explicitly classified based on their metallicity, and its direct relation to which generation of stars they are part of. This is known as the Baade chemical classification. The first stars to form from primordial gas are confusingly known as Population III. These stars have yet to be observed, but it's hoped that the James Webb Space Telescope may glimpse them. They are theorised to be very massive and short lived, hence the difficulty in detecting them. The next generation, formed from the enriched material from this first generation of stars are known as Population II. And the subsequent generation, of which our own Sun is a member, are Population I stars. Population I stars have the highest metallicity, due to the fact they formed from material that has already participated in Population III and II star formation.
The metallicity of a star is almost entirely determined by the metallicity of its birth cloud.  Stars do swallow up small amounts of material over their lifetimes, but this does not significantly affect the overall metallicity of a star. A stars lifetime is mostly set by it's mass, and whilst metallicity does have a small effect, it doesn't change significantly due to outside effects over a stars lifetime.


Why is space and time spoken of as one thing I've never understood why it's spacetime and not space and time. In other words why aren't space and time spoken of as two different things. Can someone explain to me why this is

There are different words for different aspects of space. For example, consider: length, width, and height. Other words include depth and breadth. We can speak of them as different things if we choose to, but we generally consider them to be part of unified concept of space. Why?

It's because we understand that these words just pick out measurements along specific directions in space set by context or the speaker's viewpoint, and that there is no intrinsic difference between those directions. Depending on which way they are facing, what's length to one person is width to another.

More formally, there is a symmetry between directions in space. We can rotate freely and the intrinsic properties of space look just the same, e.g. the Euclidean metric (infinitesimal distance formula)
$$\mathrm{d}s^2 = \mathrm{d}x^2 + \mathrm{d}y^2 + \mathrm{d}z^2$$
is invariant under the orthogonal group of rotations, $\mathrm{O}(3)$. Together with spatial translations (which are like rotations about a point at infinity), this forms the Euclidean group of isometries, $\mathrm{ISO}(3)$. Because of this symmetry that "intermixes" directions of space, we think of those different directions as being unified one into one thing: space.

In special relativity, spacetime has a Minkowski metric (in units of $c=1$)
$$\mathrm{d}s^2 = -\mathrm{d}t^2 + \mathrm{d}x^2 + \mathrm{d}y^2 + \mathrm{d}z^2\text{,}$$
and it is invariant under the Lorentz group $\mathrm{O}(1,3)$ that act like rotations in spacetime; together with the translations (in space and in time), the full isometry group is the Poincaré group. In fact, the usual Lorentz transformation (say, boosting along the $x$-axis) is exactly a rotation with by a hyperbolic angle in the $tx$ plane.

That's why people began to think of spacetime as one unified whole. The reasons are essentially the same as why people think of space as one unified whole despite it having different directions: there is a symmetry that "intermixes" those directions, albeit slightly differently in spacetime because of the different sign of the temporal and spatial directions in the metric.

In general relativity, this is generalized further, but the details aren't immediately relevant here.


Meaning of critical density in radiative transfer I do not understand completely the meaning of critical density when talking about atomic or molecular line emission in astrophysical gas clouds: I know that it is way more complex than this, but just taking the two levels approximation (i.e. an atom/molecule with two quantum levels, 1 and 2, where 2 is the upper one) the critical density would be
$n_c = A_{12}/\gamma_{12}$, where $A_{12}$ is the Einstein coefficient for spontaneous radiative emission (i.e. de-excitation) and $\gamma_{12}$ the collisional de-excitation probability.

It seems to me that if the gas density is $n < n_c$ the radiative de-excitations dominate, so I would expect to see the photons; conversely if $n>n_c$ collisions dominate, so the particles will lose energy through collisions before having enough time to spontaneously decay and emit photons.

This argument of mine, however, seems to be wrong, because everybody use $n_c$ exactly in the opposite way: I see the photons only if $n>n_c$. I would expect that for $n \ll n_c$ the particles are experiencing almost no collisions, so neither collisional excitation, and they do not decay because they are not excited in the first place; but I feel that something wrong has been written in the previous lines, so I would like to read someone's opinion about it before to make even more pie in the sky conjectures.

Your misunderstanding is in the second paragraph. If $n<n_c$ then there are insufficient collisions to excite the molecule into the upper level. Thus in equilibrium there are very few excited molecules and since the luminosity is proportional to the number of molecules in the excited level then little radiation is seen.

The critical density is defined as that (temperature-dependent) density at which the rate of collisional excitation to the upper level of a molecular (or atomic, I suppose) transition, equals the rate of radiative de-excitation.

The rate of de-excitation is given by $n_2\ A_{21}$, where $n_2$ is the number density of the molecular species in the excited state and $A_{21}$ is the relevant spontaneous emission coefficient.

The rate of excitation is given by $\gamma_{12} n_1 n$,
where $\gamma_{12}$ is the collision probability (which depends on the speed of molecules in the gas and hence temperature), $n_1$ is the number density of the molecule of interest in the lower energy state and $n$ is the number density of all molecules in the gas.

There is also a rate of collisional de-excitation given by $\gamma_{21}\ n_2 n$. However there must be a relationship between $\gamma_{12}$ and $\gamma_{21}$ because for a gas in thermodynamic equilibrium, we know that
$$n_2/n_1 = \exp(-E_{21}/k_BT),$$ and so if collisions are in detailed balance then
$$\gamma_{21}\ n_1 \exp(-E_{21}/k_BT) n = \gamma_{12} n_1 n$$
$$ \gamma_{21} = \gamma_{12} \exp(E_{21}/k_BT)$$

OK, so now consider a gas where there are radiative losses from state 2 and allow it to come into equilibrium such that
$$\frac{dn_2}{dt} = 0$$
$$ -A_{21}n_2 - \gamma_{21}n_2 n + \gamma_{12} n_1 n = 0$$
$$ \frac{n_2}{n_1} = \exp(-E_{21}/k_BT)\ \frac{1}{1 + A_{21}/\gamma_{21}n}$$

If now we say that $A_{21} \ll \gamma_{21}n$, then radiative de-excitation is (relatively) negligible compared with collisional de-excitation then the $n_2/n_1$ ratio approaches the level expected by the Boltzmann ratio.

If on the other hand $A_{21} \gg \gamma_{21}n$, then the ratio of $n_2/n_1$ becomes much lower and radiation removes excited molecules faster than collisions create them.

Now we ask how much radiation we are going to see from this gas.

(1) If $A_{21} \ll \gamma_{21}n$, which we can restate as $n \gg n_c$, the "critical density".

The luminosity from the molecules will be
$$L \propto n_2 A_{21} \simeq n_1 A_{21} \exp(-E_{21}/k_BT)$$
If $k_BT < E_{21}$ this means that $n_1 \gg n_2$ and the luminosity per molecule in the gas has a fixed value that is independent of density $n$. There is just a fixed amount of radiative losses per molecule of the species in question.

(2) If $A_{21} \gg \gamma_{21}n$, which we can restate as $n \ll n_c$, The luminosity from the molecules will be
$$L \propto n_2 A_{21} \simeq n_1 A_{21} \exp(-E_{21}/k_BT) \frac{\gamma_{21} n}{A_{21}}$$
and so this is smaller than for the first case by a factor of $n/n_c$. So now there is less radiation per molecule of the interesting species because fewer of those molecules are in an excited state.

NB This is a different thing to the "quenching" density for forbidden lines that I refer to in my comment.


Inflation cosmology: slow-roll inflation versus tunneling between two vacua In slow-roll inflation models, the early inflation of the universe is driven by the flat non-zero part of the inflaton potential, and it ends as the ball rolls down the cliff and the potential energy released leads then to the formation of matter.

However, as I have heard such slow-roll models are no longer deemed realistic today in inflationary cosmology, instead other models where the end of the inflation is explained by quantum tunneling of the universe from a state with higher vacuum energy density to another state with lower vacuum energy density are considered for example.

Why (from a physical point of view) are slow-roll inflation models no longer considered to be realistic, what are their disadvantages? What can models that explain (the end of) the early inflation by a tunneling transition between to vacua explain or describe, that slow roll models can not do?

I...think you've got it backwards? Slow-roll inflation is alive and well, unlike models which involve tunneling between two vacua. The model of inflation first proposed by Alan Guth back in 1980 was a tunneling model. But it had a serious problem: it didn't reheat. Tunneling from a false vacuum to a true one (or a lower-energy false one) wouldn't release any radiation. And we need that radiation, because we are that radiation, cooled down and condensed after nearly 14 billion years of expansion. The solution was slow-roll inflation: as you mentioned in your question, rolling down the potential well releases a great deal of energy, which in turn leads to the wide variety of matter and radiation in the universe today. Slow-roll inflation also does a nice job of predicting the tiny density fluctuations in the early universe that seeded the formation of all the structures in the universe. Wikipedia has more on all of this.

A small disclaimer: there's a lot of inflationary models out there in the literature (there's a hilarious partial list in footnote 3 of this excellent paper by Yadav and Wandelt), and for all I know, one or more of them involve vacuum decay. But the dominant paradigm for inflation today is definitely slow-roll.


How to know that the 21cm Hydrogen line is the actual emission and not any other redshifted line? Is it possible that we mistake some other wavelength emission to be 21 cm because of redshift?

No, not really.  The first thing is that we know that  ${H}$ is far more abundant than other elements or simple molecules in the universe.  The next thing is that the 21 cm line comes from a relatively unusual hyperfine splitting, and there just aren't any other sources near that wavelength and in intensity levels that can easily be detected.

In most cases, astronomers have other methods of estimating objects' speeds, so any measurement of 21cm apparent redshift that was inconsistent with those estimates would be an indication that the wrong line was being observed.


Inclination of planets Why do the planets have roughly the same inclination with respect to Earth?

Looking at the inclinations, which I got from http://www.astronomynotes.com/tables/tablesb.htm, we see that Mercury has the highest inclination at about 7 degrees with respect to Earth.  Why are there no inclinations at approximately 90 degrees, or even slightly less, like 45 or 30?  My guess is after some long time, gravitational influence causes all the planets to line up.

Question two is then: Do we know of any planetary systems where the planets have high degrees of inclination with respect to each other?

While the other answer here and the link provided by Jeremy give excellent explanations, I believe a bit more nuanced reasoning is required.

Although the theory of planet formation is currently still incomplete, it is generally accepted that planets form in a so-called proto-planetary disk as a part of stellar formation process. This is backed up by several observation of such disk and even by
directly observed planets in systems where remnants of the disk are still present (most notably Fomalhaut-b).

However, exactly how these planets form is still unclear. A popular theory is that these planets form in place and therefore keep the orientation and inclination of the disk. This beautifully explains the Solar System and specifically why all planets have roughly the same orbital inclination. But that's not surprising as the model was specifically designed to do just that! Yet the Solar System does not have to be the only possible type of planetary  system, or even the most common for that matter. Indeed the study of exoplanets has confirmed that very different planetary systems are possible.

An alternative theory is that planets are created far away from the star and migrate inwards. During this migration multiple planets can interact with one-another and be pushed out of the disk plane to higher inclinations. They could even flip over completely and become retrograde.

Now to come back to the actual question; the Solar planets probably have about the same inclination simply because they were created that way.

You are also right in saying that gravity will tend to align the planets. How this works physically is that the star can exerts a torque on the planet, which produces a tidal effect (like how the Moon creates tides on Earth). Over time this aligns the angular momentum of the planet to that of the star by pushing the planet towards the equatorial plane of the star.

As for your second question, yes, there are probably planetary systems which are misaligned with respect to the star. At least there are certainly planets that are not aligned with the equatorial plane of the star. The well named HAT-P-11b is a known example, but there are a lot more.

Note that for exoplanets the important parameter is not the inclination (which for exoplanets is the angle between the normal of the orbit and the line of sight), but rather the obliquity or tilt of the planet rotation with respect to the rotation of the star.

As for a third twist to this story, very recently there was a discovery of an entire proto-planetary disk that is misaligned with respect to the star, which opens the door to even stranger configurations.


Given a date obtain latitude and longitude where is the sun zenith Searching is easy to find terminator line (frontier between day and night) or the position of the sun in the sky given a position on the earth and a time; but I can't find how to obtain where is the zenith of the sun given a date (and time).

I need to obtain the center of the illuminated zone of the earth (latitude and longitude) at a given time. (Well, actually I need the opposite, the Nadir, but with one you can calculate easily the other).

Any knows the function?

Thank you.

What you're looking for is the latitude and longitude of the subsolar point.

Calculating it with accuracy is somewhat complicated.  Here is a python library that has routines to do it for you:
http://rhodesmill.org/pyephem/

You could also possibly query this information from other sites.

If you really want to work it out on your own, I suggest "Spherical Astronomy" by Robin M. Green published by Cambridge University Press as a starting point.


How to complete the Hipparcos Catalog? I've downloaded hipparcos catalog from the Strasbourg Universe but whilst there is a lot of information, I can't find the names of the stars that the HipID relates to.  There is the wikipedia page which lists some but not all the stars, for instance Zosma (54872) is missing from the list and don't fancy searching for each and every star page to find out the missing details such as what the HIP's name is, how far away it is, etc.  Is there a downloadable list which lists the HipID, the name, how far away it is, the constellation etc?

Most of the stars in the Hipparcos catalogue do not have a common name.

In the main catalogue file you will also find the Henry Draper (HD) number of the star (if it has one) at columns 391-396. You can use this ID to find the name in the Bright Star Catalogue.

The Bright Star Catalogue contains all stars brighter than magnitude 6.5 (naked eye stars) together with the HD number (columns 26-31) and a Name (columns 5-14), which in fact is the Bayer (greek letter - constellation) or Flamsteed (number - constellation) name.

Using both catalogues you can get the Bayer or Flamsteed name from the HIP identifier. I do not know of any catalogue to get the common name.

The distance of the star can of course be found in the Hipparcos catalogue, which is the reason the Hipparcos catalogue exists. Columns 80-86 give the trigonometric parallax $\varpi$ in milliarcseconds. You can get the distance from: $$d=\frac{1}{\varpi}$$ where the distance $d$ is in parsec and the parallax $\varpi$ is in arc seconds. For the Hipparcos catalogue you should divide the parallax by 1000 as the parallax is given in milliarcseconds.

You can get the constellation for any position in the sky from the algorithm and data in http://cdsarc.u-strasbg.fr/viz-bin/Cat?VI/42


What will the PRECISE time of the Summer Solstice 2020 be? I'm trying to explain to my daughter that the Summer Solstice is not a day of the year, but an exact instant in time.  My search shows that this will be about 21:43 or 21:44 UTC tonight.  To make my point I searched for a more precise time.  I figured I could find a time accurate to at least the second, or hopefully several places after the decimal , but I couldn't find anything.
Can anyone show me a source for the precise time of the Solstice to at least the second?  To the millisecond?  The microsecond?

TL;DR:
In a couple of weeks it might be possible to say precisely (within a second, perhaps a fraction of a second) when the summer solstice did occur. But until then, sub-minute estimates should be treated as fraudulent.

The reason you are seeing different times for when the summer solstice will occur is that different websites use different models of the orbits of the bodies that comprise the solar system and different models of the Earth's orientation in space.
It's very important to remember that "all models are wrong, but some are useful".
There are several issues with finding the exact time of when solstices occur. One is that the solstices are extrema of the Sun's declination. Without a model of derivatives, finding extrema (e.g., the solstices) mathematically is a much harder problem than is finding zero crossings (e.g., the equinoxes).
More importantly, a very precise prediction would require multiple very precise models. While many models of the orbits of the bodies in the solar system are very, very good, none are good enough to get the timing of the solstices down to the millisecond. Even more importantly, the tilt of the Earth's orientation in space with respect to its orbit about the Sun needs to be modeled. The best model, the IAU 2006/2000A precession-nutation model has over a thousand terms (1365 terms, to be precise).
And even then, "all models are wrong." The people at the International Earth Rotation and Reference Systems Service (IERS) know this very well. The IERS is the organization that is responsible for modeling time as measured by the Earth's rotation and for modeling how the Earth is oriented in space. There are some aspects of the Earth's orientation in space that are not quite predictable. The US Naval Observatory, one of the key members of the IERS, publishes daily updates of observed deviations from model predictions. It take a couple of weeks to fully digest measurements made by astronomers worldwide.


How close can planets form to one another? With the NASA announcement today regarding the discovery of a system containing seven earth-sized planets (3 within the habitable zone), I wondered about the seemingly crowded conditions.  What principles guide the formation of planets and the distances between them?  Are there laws governing this that are well established?  How close have we observed two planets form?

If you're thinking about how close planets can be, you should probably consider each planet's Hill sphere, the region in which it can retain satellites. Fang & Margot (2013) did an analysis of Kepler data and found that planets had mean values of $\Delta = 21.7$, where $\Delta$ is a parameter given for two adjacent planets by
$$\Delta=\frac{a_2-a_1}{R_{H1,2}}$$
where the $a$s are the semi-major axes and $R_{H1,2}$ is the mutual Hill radius.

One system the authors consider is Kepler-11, which has 6 planets, all with semi-major axes $\leq0.466\text{ AU}$ and with only one semi-major axis greater than $0.25\text{ AU}$. The smallest $\Delta$ there is approximately $5.7$, although all the other $\Delta$s are quite small. Kepler-36, with only two planets, still has a $\Delta$ of $4.7$.

According to the Nature paper about TRAPPIST-1, all seven planets have semi-major axes within $\sim0.063\text{ AU}$. They have mean $\Delta$s of $10.5\pm1.9$ - not much different from the Kepler-11 planets, because they have smaller Hill spheres. They may be closer together, but they can be much closer together without having stability problems. Additionally, they are in a "near-resonant" configuration.

How close planets can be depends strongly on their masses, then, which in turn determines their mutual Hill radii, which determines stability.

All that said, the authors believe that the TRAPPIST-1 planets may have migrated in from further out, thus entering the resonances. Without more information, we can't know whether this is the case, but if so, it is not, then, an example of planets forming close to each other.


What would be the dynamics of a double-planet system, similar to Earth / Moon, but with both bodies nearly Earth-sized? Researching a sci-fi story involving a "twin-planet" system. Is such an arrangement (however unlikely) physically possible? And if so, must they be tidally locked with each other, similar to the moon, month-long days, etc?

This is a duplicated question but I think it fits better into astronomy than into physics. I'll try to answer all aspects of this question including the consequences.
First, is it possible? As already pointed out it is, like Pluto and Charon. But is it possible for bigger planets, with the size of Earth? Well we didn't find this planet system yet, but the universe is big. There is no physical law that would forbid this system, it might be unlikely and very rare but it would be possible.
Let's assume somewhere is a planet system with two Earth-sized twins. What would be the dynamics of this systems on the consequences for possible living beings on one of this planets?
One consequence is that the gravity center of this system would be between both planets, if they exactly are the same size (same mass) the gravity center would be in the exact middle of both. That means both planets are rotating around this point (while they are rotating around a sun). This would lead to some effects like in a centrifuge and lead to some weird appearances. The gravity point of our Earth – Moon system is already shifted from the geometrical center point of our earth. And this leads into tides. The tides on our double Earth system would be vast (assuming at least one of this planets is just like our earth and has water on it). Complete areas would be periodically covered with water. Most living beings would probably be amphibians. This of course means the planet would rotate. In the beginning, the planets would rotate but they would slow each other down until they show the same side to each other. Just like our Earth did to the Moon. The Moon continually slows down the rotation of our Earth, but with his minor mass this takes a while. This processes would be much quicker on the double earth system. But even this would take some time. So it would be a matter of time when you look at this system. If it's a young system they would still rotate. If it's an older system they would show the same side each other.
The last point would fix the tides problem, because the water level would be stationary. But this would lead into new problems. The day length would be as long as they need to rotated around their common gravity center. I'll be honest. I am not sure how long this would be. For a stable double Earth system the distance between both planets would be further then the Earth-Moon distance. Let's just assume a day in that system would take 2 Earth months for simplification. This would mean on daylight it would be very hot and at night it would be very cold. One side of the planet would be heat up for hundreds of hours (our earth hours) and the other half would cool down for the same time. Temperatures over a period would jump up and down in big steps. There is no living being known on earth that can handle both extreme hot and extreme cold in such a short period of time. Also this would lead into very heavy and constant storms in the area where the sun sets or rises, making living there even harder.
Also your weight (stepping on the bath scale) would vary depend on where on the planet you are. If you on the side facing the other planet you would be heavier than on the other side of the planet.
Another point that would be possible, but I will point out I am not completely sure about this. The planets would struggle keeping their atmosphere for a long period of time because of the rotation. The side facing away from each other would have thicker atmosphere than on the sides (see my sketch).

(source: directupload.net)
At this point the gravitational pull of the planet is also lower then everywhere else. This could help the molecules to escape the Earth, in a long period of time.


When using adaptive optics, what is the shortest timescale of atmospheric changes in refractive index that astronomers have to deal with? I am interested in the astronomer's need for speed in wavefront correction, especially when measuring in NIR and visible domain.

What is the fastest abberation coming from atmospheric turbulence in any measurement setting (sun observation during day, night-time observations)? Does any astronomer need correction in the visible spectrum beyond e.g. 100 Hz?
Are fast systems practically limited by spatial resolution of the sensor?
Is there any way to get a quantitative overview how widespread adaptive optics is in astronomy today and which setups are currently in use?

Thank you for your thoughts!

I apologise, I'm posting an answer because I can't comment.

What is the fastest abberation coming from atmospheric turbulence in any measurement setting (sun observation during day, night-time observations)?

It depends on the astronomical seeing conditions at the location of the telescope, and also varies quite a lot. It could be anywhere from $1-100~ms$.

Does any astronomer need correction in the visible spectrum beyond e.g. 100 Hz?

It depends on the what you're after. In solar physics, for example, observations are generally interested in resolving features on the Sun. You can also find polarimeters that operate beyond $100~Hz$, so an Adaptive Optics system that operates in the $kHz$ is a fairly standard requirement for most modern Solar telescopes. However, if you're only interested in collecting unresolved spectra (say, a star), you may not really benefit from having a really fast Adaptive Optics system.

Are fast systems practically limited by spatial resolution of the sensor?

I'm not sure what you mean by limited and sensor. Are you talking about the sampling of the wavefront by the wavefront sensor?

Is there any way to get a quantitative overview how widespread adaptive optics is in astronomy today and which setups are currently in use?

I think you'd find this reference helpful: https://link.springer.com/article/10.12942/lrsp-2011-2


Hohmann transfer orbit (Earth-Jupiter system) Consider the Hohmann transfer orbit of a probe launched from Earth to Jupiter. Assume the orbits of Earth and Jupiter are circular.

I want to calculate the following:

The semi-major axis and eccentricity of Hohmann transfer orbit from Earth to Jupiter.
The speed of the probe at Earth.
The launch speed of the probe from Earth to reach Jupiter.
The duration of the flight to Jupiter.
The elongation of Jupiter at launch to ensure that the probe arrives at Jupiter.

Here my attempt:

$a = \frac{1}{2}(r_E + r_J) = \frac{1}{2}(1+ 5.203) = 3.102 \text{ AU}, \ \ e = 1 − \frac{r_E}{a} = 0.677$
I am thinking of using the vis-viva equation but I don't know what I should use for $r$? Should I use $r_E$ = 1 AU for $r$ where $r_E$ is the average distance from the Earth to the Sun? Am I right? Please correct me if I am wrong.
I am thinking of using the escape velocity formula where $M$ is the mass of the Earth and $R$ is the radius of the Earth. Please correct me if I am wrong.
$t = \frac{1}{2}P = \frac{1}{2} \sqrt{a^3} = 2.73 \text{ years}$.
I really don't know how to do this question. Please help.

I want to calculate the following:

  The semi-major axis and eccentricity of Hohmann transfer orbit from Earth to Jupiter.
  The speed of the probe at Earth.
  The launch speed of the probe from Earth to reach Jupiter.
  The duration of the flight to Jupiter.
  The elongation of Jupiter at launch to ensure that the probe arrives at Jupiter.

$a_E$, $a_J$ are 1 and 5.2 AU, respectively. The Hohmann transfer ellipse touches both, so the major axis is 6.2 AU and the semi-major axis is half of that, or 3.1 AU as you wrote.
Definitely use the vis-viva equation whenever you can! The standard gravitational parameter $GM$ of the Sun is 1.327E+20 m^3/s^2. You can use that if you convert your distances to meters to get the velocity in m/s,

$$v^2 = GM\left( \frac{2}{r} - \frac{1}{a} \right)$$

or as long as you are orbiting the Sun you can stick with AU and years and use:

$$v^2 = \frac{2}{r} - \frac{1}{a}$$

which will give the velocity in AU per year (once you take the square root).
I'm not exactly sure what "The launch speed of the probe from Earth" means and how it is different from the speed of the probe at 1 AU. If it means the speed you'd fire it from the surface of the Earth from a gun if there was no air, then just add the Earth's escape velocity to the velocity from #2.
this answer should help you calculate the period of an orbit for a given semi-major axis . For the Hohmann transfer ellipse, use its semi-major axis to calculate its period, and then use half of the period for the duration of the flight from Earth to Jupiter.
I think you can figure this one out for yourself. Draw a picture of the 1 AU and 5.2 AU circles and the Hohmann ellipse that touches both. Put a dot where you want Jupiter to be when you get there, then move Jupiter backwards by a certain angle, which would be 360 degrees times the ratio of the flight time to Jupiter's period. If the flight time is 5 years (it's not) and Jupiter's period is 10 years (it's not) then you'd back Jupiter up by 180 degrees for its starting position when you launch from Earth. Then read how elongation is defined.


What is the formula to predict lunar and solar eclipses accurately? A number of ancient civilizations had devised methods to predict exact dates and times of such eclipses, marking them as important events.
Hence I assume the predictions were based on calculations, which should be quite easy to do now.
So what is the exact formula to predict the exact date and time(this is optional, but desirable) of lunar and solar eclipses?
Also how to calculate if the solar eclipse will be visible from a particular location or not?

The NASA sites have some very useful resources for this I will list them below:

Lunar Eclipses

This Link has an index for all lunar eclipses from -1999 to +3000, predominantly a statistics page but also has this page that contains how to calculate when lunar eclipses are.

There is more than one formula depending on which time frame you are trying to look in.

This is the formula for eclipses between the year 2005 and 2050:

  $$\Delta T = 62.92 + 0.32217 * t + 0.005589 * t^2$$

  Where:   $$y = year + (month - 0.5)/12$$  $$t = y - 2000$$

Solar Eclipses

This Link has an index like above but for all of the solar eclipses from -1999 to +3000.

This link has the formula for calculating solar eclipses. This is the formula for between 2005 and 2050:

  $$\Delta T = 62.92 + 0.32217 * t + 0.005589 * t^2$$

  Where:  $$y = year + (month - 0.5)/12$$  $$t = y - 2000$$


Hoag's object as an Einstein ring What are the main arguments against Hoag's object in particular and Hoag-like objects in general to be explained as Einstein rings formed by a farther edge-on galaxy?

It doesn't look like an Einstein ring.  An Einstein ring shows highly distorted shapes: stretched and multiple images of the distant galaxy, and a massive foreground object (with different redshift) which would have to be a galaxy itself as nothing else is big enough.
Hoag's object shows none of these distortions, and the central bulge is at exactly the same redshift (so the same distance by Hubble's law).  The areas of star formation in the ring look rather like those in more normal galaxies. It is just an unusual shape.


Synodic period of planets Many books explain the difference between synodic and sideral periods. The synodic period of planets are tabulated for centuries and it is defined as the time interval between two consecutive repetition of phase, phase in this case meaning the terrestrial measure of its angular position and speed relative to the Sun.

My question is, how do astronomers measure the angular position of a planet relative to the Sun if the major observations are made during the night?

I'm pretty far from an expert at ancient astronomy, but as I understand it, the ancient astronomers looked for the date when the planet was highest in the sky.   This is called being in opposition or opposite the sun.

Source of image

See here for further explanation.

I've also read that the synodic period can be calculated between periods of retrograde.   In neither case, the position of the sun is necessary.

The synodic period, because it was initially based entirely on observation, works for both Geocentric and Heliocentric models.

"Position of the sun" infers our current understanding of the heliocentric system.  The triangle, between the 3 points, Earth, Sun and observed planet and that only works for the heliocentric model.  Most of the ancient astronomers didn't use that model.  Their synodic period was based entirely on the motion of the planet relative to the fixed background stars.


Does a star's rotation always match a planet's orbit? For a planet orbiting a star, is it ever possible for a star to be rotating in the opposite way on it's own axis compared to the planet that is orbiting the star?
Or do gravitational forces mean the planet will always orbit a star with the same rotational direction as the star's axis rotation ?

TL;DR: No, not always, but most of the time.
Well, most planets are in either an equatorial orbit (~0° in relation to the star's rotation vector / equator). Some are in polar orbit (~90° and almost perpendicular to the star's equator/rotation vector), and even fewer are in retrograde or strange (highly eccentric or inclined) orbits. For example, all the major planets in our Solar System orbit the Sun in an equatorial orbit. WASP-79b, a hot Jupiter, orbits in a near polar orbit (deviation of 5°). WASP-17b and HAT-P-7b are both in retrograde and highly inclined orbits to their parent stars. Neptune's moon, Triton, orbits in an inclined, retrograde  orbit, suggesting that Triton may be a captured body (but that is a long story that is unrelated). This means that a planet's orbit does not necessarily, have to align with the parent star's rotation vector.


Basic explanation of baryon acoustic oscillations; what if anything is actually oscillating? Sci-News' Sloan Digital Sky Survey Collaboration Releases New 3D Map of Universe links to the two videos

The eBOSS 3D map of the Universe
SDSS releases largest 3D map of the universe ever created

and says

The team’s results will be published in a series of 20 papers in the Monthly Notices of the Royal Astronomical Society.

After 01:30 in the second video Jiamin Hou from the Max Planck Institute for Extraterrestrial Physics says:

However the red shift has an additional component that is due to the galaxy’s own velocity, and which moves in response to the gravitational attraction of the surrounding matter, and these two components can not be separated from each other. But the statistical analysis of the eBOSS allows us to distinguish the effects of velocity from that of expansion.

I've also looked at:

Wikipedia: Baryon_acoustic_oscillations; Cosmic sound
SDSS.org's surveys; eBOSS

but I can't get a grasp of what the term "baryon acoustic oscillations" really means.
Question: Is it possible to write or quote a basic explanation of what these are? For example, does oscillation mean there are things moving back and forth due to some restoring force? Is there something analogous to a "sound wave" out there somewhere? Or is there just a plot of something that shows a wiggly line that goes up and down?

tl;dr nothing is oscillating now, but it was in the early universe. Now we just see the frozen relics
This is just an attempt at a simplified explanation of the Wikipedia article linked in the question.
There were pressure waves in the very early universe. They came from a balance of gravity and radiation pressure (the matter at the time was hot enough to be opaque). Whereever there happened to be a locally higher density (of matter, radiation and dark matter) the excess pressure would cause an expansion and then gravity would cause a contraction sending waves out at (according to the article) half the speed of light.
As the universe expanded and cooled, matter eventually became transparent (it was hydrogen and helium gas) letting the radiation pressure escape and leaving higher densities both in the location of the original concentration and at the crest of the wave a certain distance away (but the same certain distance for the first waves from each concentration). Those high densities are where the galaxies formed, so what we look for today is an excess of galaxies at a specific separation and that's what we see.


Solution for counting atmospheric layers of planets according to relational data This math states, that by multiplying the gravitational pull (we assume square of density of the planet), by its inclination (being how high and out it goes) and subtracting that loop from the velocity in a slope formula that measures to the borderline of being horizontal on the sun's x-axis, most parallel with the solar system's other planets, relegating the offset in order as to consign the preferred balance of the planet with the Sun. We can take then from the furthest point out once the Sun can lesser not take from the product and sum and difference of the clouds on the planet so that we may therefore continue on and calculate the priori aft dividing by 100 to delineate in a reciprocity to the true count, the atmospheric makeup of which the count of surface bars is.

This site is were to get the constants for the planetary facts you will need to use the formula.

https://nssdc.gsfc.nasa.gov/planetary/factsheet/

n = Orbital Velocity - root of density * inclination to sun
round((Apogee - n)/100) = Planet's Atmospheric layer count;

This is an easy formula to figure out how many layers of atmosphere your favorite planets have on them. It's preliminary because I don't have the satellite data from each planet to synthesize with my data from this formula.

If this is wrong, please tell me. If it is not, I suppose tell me. But I am only in search criticism and laud.

NASA Deg.   Real Deg.   Planet + Rank   Density Orbital Vel.    Apogee  Bars (Atmospheres)

17.2        15.3        Pluto+4         2095    4.7             7375.9  6.6 Bars

7           5.1         Mercury+3       5427    47.4            69.8    2.5 Bars

1.8         3.7         Neptune+2       1638    5.4             4545.7  44.01 Bars

1.3         3.2         Jupiter+1       1326    13.1            816.6   34.13 Bars

1.9         -           Mars -0         3933    24.1            2492.2  2.1 Bars

2.5         -0.6        Saturn -1       687     9.7             1514.5  15.07 Bars

3.4         -1.5        Venus -2        5243    35.0            108.9   2.52 Bars

0           -1.9        Earth -3        5514    29.8            152.1   3.22 Bars

0.8         -2.7        Uranus -4       1271    6.8             3003.6  31.06 bars

While the question is rather hard to follow, the formula simply doesn't work.

For Mercury the formula using a naive application of the numbers predicts:
four layers

Orbital Velocity (km/s)        47.4
Density (kg/m3)              5427
Orbital Inclination (degrees)   5.1
n                            -328.65368
Aphelion (10**6 km)            69.8
Layers                          4

That is clearly not true, since Mercury has no atmosphere. You can vary the formula, by choosing different units, or taking the inclination relative to a different plane, but no matter what I try, can't get the formula to give a sensible value for all the planets.

The formula is inconsistent, since you generally can't add velocity (km/s) to sqrt(density)*inclination (kg^0.5 degrees/m^(1.5)) as these quantities have different units. It rarely makes sense to add quantities with different units.

There is, moreover, no reason to suppose that the orbital characteristics of a planet, and its density would have any effect on the number of layers in the atmosphere, which is not a very well defined quantity anyway.

Sorry but this formula is entirely wrong.


In layman's terms, why do we put the "time" in space-time? What does "time" mean in the term space-time? I am trying to understand why anything that travels through space, from a proton to massive galaxies, is always referred to as traveling in space-time, not just space. Does this have to do with time being relative?

I understand that 3d space and time together constitute a space-time.

Can someone illustrate with a scenario to point out the need to describe things in terms of space-time.

To answer your point about why "spacetime" as opposed to space and then time - photons (and other massless particles) travel through spacetime on "null geodesics" - in other words they do not experience time at all. But they are not in some separate class as particles at approach ever closer to the speed of light travel on paths through spacetime that are ever closer to null geodesics.

In other words the four dimensional geometry is inextricably linked and "space" and "time" are a unified whole.

In theory humans moving also experience this effect but the speed at which we move is far too slow for this ever to manifest itself to our senses. But we do see its impact in high precision timing - such as with GPS satellites.


Do other supercluster like Laniakea also get pulled by the Great Attractor? I'm not sure, whether there is any other observable super cluster like our local one, Laniakea, and if they do exists, do they also get pulled by the Great Attractor?

I am a little confused about superclusters and The Great Attractor, how they do exists relative to each other and how we can see/know from our galaxy, specifically from the Earth.

In short, not much gravitational pull at all, not relative to the distance and relative velocity.

Everything in the observable universe is gravitationally attracted to everything else, at least within the appropriate horizon.  There should be some objects far enough apart, taking into account cosmic inflation, that gravity can't travel that far over the age of the Universe.  For example the farthest galaxy we can see in one direction and the farthest one we can see in the opposite direction, those two galaxies can't see each other and they likely experience no gravitational pull from each other, but that's not really relevant to your question.

The Great Attractor is a comparatively small but very dense part of Laniakea, with an estimated mass thousands of times the mass of the Milky Way.

The Laniakea supercluster has a mass of about 100,000 Milky Ways, so any gravitational attraction one supercluster experiences would primarily be from other entire superclusters, and the great attractor would be a small percentage of that.

Superclusters experience gravity from other superclusters but the Universe is very uniform, so that gravitational pull is largely balanced out because it's coming from all sides.

This touches on one of the modern puzzles in physics, called the Horizon problem. This question goes back to the big bang and why the young universe was very close to, but not precisely uniform.

It might be a fun exercise to calculate the gravitational pull neighboring superclusters have on each other.   I imagine it's pretty negligible compared to the expansion of space between them.


Supermoon Lunar Eclipse? I recently read this article: http://www.cbsnews.com/news/supermoon-lunar-eclipse-coming-this-month/ I'm having some trouble understanding how a "supermoon" lunar eclipse is possible.

Judging from this diagram, it would seem that the moon always reaches its apsides above or below the orbital plane of the earth around the sun, making any kind of eclipse at these points impossible. Can someone clarify for me?

Thank you.

Lunar Eclipses occur when a full moon occurs when the the moon crosses the ecliptic. The ecliptic is the plane in which the Earth orbits the sun, and so also the track which the sun appears to trace in the sky relative to the stars.

It appears from the diagram that there are two nodes each month (ascending and descending), but this is only approximately correct. The influence of the sun causes the nodes to move.It takes 29.5 days between two full moons, but only 27.2 days between two ascending and descending nodes). This means that a node can occur on any day of the year, and any day of the month.

A "supermoon" occurs when the moon when it is closest to the earth, the perigee. This point of perigee also moves, so the perigee can also occur at any time during the month, and the time of perigee is not related to the time of the node. Unlike terms like "node" which astro(nom|log)ers have used for many years, Supermoon is a neologism.

Thus is is possible, but rare for a full moon, a lunar node, and a perigee to occur at about the same time, and if it does, a supermoon eclipse will occur.