Does the Sun's atmosphere have a scale height? Much of the Earth's atmosphere has a scale height $h$ of roughly 7 to 8 kilometers such that the local density varies as $\exp(-(r-r_0)/h)$ where r is the radius vector and $r_0$ would be some reference radius above the surface of the Earth.

This works approximately, at least up to 50 km (six or seven scale heights).

The derivation of this approximation is based on a simple gas at constant temperature, and those probably don't apply to the Sun's atmosphere very well.

Nonetheless, are there regions of the Sun's atmosphere where density varies approximately exponentially in such a way that a scale height characterization would work over the range of at least a few scale heights?

The exponential decrease in density comes out naturally whenever you have a gas in hydrostatic equilibrium. The scale height $H$ is then given by the balance between the kinetic energy of the particles due to thermal motion, $kT$, and the gravitational energy of the particles, $mg$. This is often a good approximation, both in planetary and stellar atmospheres, and even in galaxies.
That is,
$$
H = \frac{kT}{mg}
$$
where $k$ is Boltszmann's constant, $T$ is the temperature, $m$ is the average mass of the particles, and
$$
g = \frac{GM}{r^2}
$$
with $G$ the gravitational constant, and $M$ the mass inside the radius $r$.

On the surface of our Sun, $g$ works out to $274\,\mathrm{m}\,\mathrm{s}^{-2}$, 27 times as high as on Earth.

The average particle mass depends weakly on metallicity, and mostly on the ionization state of the gas, since the small mass of free electrons compared to that of atoms pulls the average down. For a fully ionized gas, the mean molecular mass — i.e. the mass in terms of the hydrogen mass — is $\mu \simeq 0.6$, whereas for a fully neutral gas it is (e.g. Carroll & Ostlie 1996)
$$
\mu \simeq \frac{1}{X + Y/4 + Z/15.5} \simeq 1.25,
$$
where $X$, $Y$, and $Z$ are the mass fractions of hydrogen, helium, and metals, respectively. I mistakenly wrote initially that the gas is fully ionized, but that's not true; it is only partially ionized, and the hydrogen is largely neutral.

Taking the average mass of a particle to be roughly equal to the proton mass $m_p$ (i.e. setting $\mu=1$), and taking the temperature to be $T = 5770\,\mathrm{K}$, the scale height is thus
$$
H_\odot = \frac{kTR_\odot}{GM_\odot m_p} \simeq 170\,\mathrm{km}.
$$
With $\mu=0.6$ you'd get $H\simeq290\,\mathrm{km}$, while $\mu=1.25$ yields $H = 140\,\mathrm{km}$.

Realistic density profiles

The above calculation are quite basic, assuming a completely isotropic Sun. But observations and more realistic models, both 1D and 3D, do indeed predict exponential density profiles, although with quite large variations across the surface (according to a Solar physicist colleague down the hall). I found this model from these lecture notes where the yellow curve shows the number density profile in the Sun's atmosphere.

Extracting the data and plotting on a log-linear scale shows reasonable agreement with an exponential decrease of scale height $H = 140\,\mathrm{km}\,\mathrm{s}^{-1}$:


What is the difference between albedo, absolute magnitude or aparent magnitude? What is the difference between albedo, absolute magnitude or aparent magnitude? I can’t really tell what’s the difference between them.

Albedo is a measurement of reflected light. It varies between 0 and 1, with 0 being completely black (reflecting no light) and 1 being completely white (reflecting all light). The albedo of coal is about 0.04, which is actually very similar to the albedo of our moon. The Albedo of ice is about 0.7. Only planets and other small objects are described using Albedo because they cannot make their own light, and so its apparent brightness is due to reflecting sunlight.

Absolute magnitude is a measure of how much light a star is emitting overall. Absolute magnitude is related to a bunch of other properties of the star, mainly how massive it is and at what stage of its lifetime it's in.

Apparent magnitude can be used to describe both stars and planets. It is simply a measure of how bright the object appears to us. An illustrative example is Sirius and Canopus. Sirius is the brightest star in the sky, but it is bright because it is close, only being 8 light years away. Canopus is the second brightest star in the sky, and appears about half as bright as Sirius, but is about 300 light years away, which means the star itself must much brighter than Sirius. If we check the absolute magnitudes and do some maths (Sirius absolute magnitude = +1.42, Canopus = -5.71, the more negative the number the brighter the star, brightness difference = 2.5^(5.71+1.42)) we get a brightness difference of 680 times between Canopus and Sirius.


Why are certain elements so common? I understand why hydrogen and helium are the most common elements in the universe. I also understand iron, because that's where stellar fusion stops. But why are (specifically) oxygen, carbon and neon more common than iron? I would expect iron to either outrank them, or some steady flow of atomic nnumbers to lead up to iron. These three, and the scattershot of seemingly random elements trailing behind iron, seem a very odd mix!

Source: https://en.wikipedia.org/wiki/Abundance_of_the_chemical_elements

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

I'll limit myself to the following observations:

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

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


How bright would the sun appear from the hypothetical Planet Nine proposed by Caltech? Theoretical Planet Nine, proposed by Caltech's Mike Brown and Konstantin Batygin, is said to have a 15-20 thousand year orbit. Approximately how bright would the sun appear from Planet Nine aphelion and perihelion?

Between $1/40,000$ and $1/8,000,000$ of the brightness as seen from Earth, depending on what the actual orbit would turn out to be, and where the planet is in its $15,000$ year orbit period.

Brightness drops as $\dfrac{1}{r^2}$ with distance from the light source. Earth is at $1~\textrm{AU}.$ The theoretical planet is at $200~\textrm{AU}$ when it's closest to the sun, and up to $2800~\textrm{AU}$ at the point furthest from the sun at the upper end of the estimated orbit.

So e.g. $1/200^2 = 1/40,000$ of the brightness (luminance) as seen from Earth.

For how a human would experience it, we can convert to exposure value as used in photography: The difference in exposure value (photographic 'stops') is $\log 2$ of the luminance ratio, so we would have $15$ to $23$ stops less light than on Earth.
Sunny noon on Earth is $15~\textrm{EV}.$

So the brightness at noon on the planet surface would be:

At $200~\textrm{AU},$ planet orbit is closest to the sun: About $0~\textrm{EV},$ roughly the same as a dimly lit interior
$400~\textrm{AU},$ lower bound on semi-major axis: $-2 ~\textrm{EV},$ similar to a landscape lit by the full moon
$1500~\textrm{AU},$ upper bound on semi-major axis: $-6~\textrm{EV},$ similar to landscape lit by a quarter moon
$2800~\textrm{AU},$ upper bound on aphelion (the point on orbit most distant from the sun): $-8~\textrm{EV}.$ This would be dark, but you would probably still see enough to avoid running into things.


Will Hubble's law always be accurate? We have Hubble's law:

$$v=Hd$$

where H is the Hubble parameter, which is decreasing in value, but it will be constant in the distant future. So, assuming the Standard Model of Cosmology holds true, will Hubble's law be accurate when the Hubble parameter becomes a constant?

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


Is dark energy evenly spaced throughout Universe? Are there certain parts of the Universe that are expanding faster than others? Is there anyway to definitively measure this?

If dark energy varied by location, then plots of 1a supernova brightness vs redshift should vary depending on which direction you look in the sky. AFAIK, that's not the case. For example, although coords are not accounted for, there's not a lot of scatter in this plot:

  In the relationship between the distance and redshift of Type 1a supernovae, the data (points) agree with the equation in which light propagates through the expanding universe on the least-time path (solid line). Image credit: Annila. ©2011 Royal Astronomical Society

Read more at: http://phys.org/news/2011-10-supernovae-universe-expansion-understood-dark.html#jCp


Where do the photons come from at the edge of an Event Horizon? I was reading about Professor Finkelstein's speech at Cambridge University back in the 50's. He was explaining the dual reference frame of a black hole's event horizon.  (A person falling in has one reference frame, and a person watching has a second.).... This I understand.  My question is about the person watching reference frame...
It is my understanding that a person watching someone fall into the black hole would never actually see the person disappear.  My question is how is this possible?  THe photons coming from the person at the edge of the event horizon are only given off while the person is on this side of the EH.  Once the person crosses the horizon, then the photons should not be there anymore for me to see. I know that time is slowed... but if I stared for 1000 years would I ever see the person cross the threshold?  If not, where are these photons coming from?

From the distant observer's perspective the person falling in becomes more and more time dilated and redshifted. As a result you see fewer and fewer photons from them and eventually none at all (although you never have any way of knowing that one more, very redshifted photon isn't going to arrive some time). So there are no extra photons.


Why is there no concavity in the orbit of the moon around the Sun? Such questions have been asked here and I have read the answers some of which are quite informative. However, I want an answer that's straightforward.

I have gone through the following paper by A B Turner titled The Moon's Orbit Around The Sun.

In this, the author concludes that the curvature of the Moon's orbit around the Sun in the new moon positions, $k_n$ is approximately 1.5 times of the Earth's orbit around the Sun, which is approximately the same as that of the Moon's orbit around the Sun in the full moon positions, $k_f$.

$k_n$$\approx 1.5k_f$, which is intuitive considering the fact that in the new moon positions, the moon experiences gravity from both the Sun and the Earth in almost the same direction.

My question is, then, how does the Moon's orbit manage to not have any concavity, no matter how minute, while transitioning from the full moon to the new moon positions? Isn't that not a mathematical impossibility?

Turner's expression for the radius of curvature ρ is correct, but in case (3), ρ = -0.988 a where the Moon is in the first or last quarter (θ = 7.5° or 22.5°).
Where the Moon is full (θ = 15°), ρ should be -0.749 a.

Case (1) models a spacecraft stationed at L1 (b = a / 100, ρ = -0.99 a) or L2 (b = -a / 100, ρ = -1.01 a), roughly 4 times the lunar distance from the Earth.

The case (2) example b value is too large for a stable orbit around the Earth.
Instead we can model a geosynchronous satellite with n = 366 and b = a / 3550.
Then ρ is +0.0218 a on the day side and -0.0314 a on the night side.
This trajectory is alternately concave toward and away from the Sun, resembling Turner's figure 1 but with shorter waves.

Between cases (2) and (3) we can find parameters such that, on the sunny side of the Earth, the ρ expression denominator approaches zero and the trajectory is momentarily straight.
This is nearly the case if n = 24 and b = a / 575, corresponding to a 15.2 day orbit at 0.677 lunar distance.

Any curvature in an object's trajectory is due to acceleration in the direction of the net force acting on it.
We know from Newton that
$$ F = \frac{G m_1 m_2}{r^2} $$

For the Moon, the Sun is ~390 times as far as the Earth but 330,000 times as massive, so the Sun pulls roughly twice as strongly: $F_\oplus$ ≈ 0.45 $F_\odot$.
At full moon, both Sun and Earth pull the Moon in the same direction, and the combined force is 1.45 $F_\odot$ toward the Sun.
At new moon they pull in opposite directions; the resultant force is only 0.55 $F_\odot$ but still toward the Sun.
While the radius of curvature fluctuates, the direction of curvature never reverses.

For a geosynchronous satellite at 0.11 lunar distance, $F_\oplus$ ≈ 38 $F_\odot$.
On the night side of the Earth, the resultant force is 39 $F_\odot$ toward the Sun.
On the day side, it is 37 $F_\odot$ away from the Sun.


Need help in creating telescope I have 100cm F.L objective and 10cm F.L eyepiece 10cm in diameter and double convex both.
what are the planetary objects I can see with telescope made with these lenses.

I also have a double convex and double concave lens of 20cm F.L both having 5cm diameter. will any of the above said lens help the telescope I'm going to make if included somewhere in the design?

If F is the focal length of the objective, and f the focal length of the ocular, then the magnification of the instrument is:

M = F / f

So you'll get 10x with the first combination. This is enough to see craters, mountains and valleys on the Moon. You will probably not see planetary disks or the rings of Saturn.

The large diameter of the ocular might be awkward, but it should work. Make sure both lenses are installed on a rigid tube - plastic tubes from water pipes should work. Make sure the lenses are perpendicular to the axis of the tube - small errors there reduce overall performance.

You will need a way to finely adjust the distance between objective and ocular - the ideal distance is F + f but don't just glue them in place; instead, see if you can make two concentric sliding tubes, you know, like the spyglasses from pirate movies. Then you'll be able to look through the instrument and adjust the distance until the image is sharp.

If there's any way you could find a used ocular from an old microscope or binoculars, use that as the ocular in your instrument. It will provide higher magnification (smaller focal length) and its performance should be much better than a simple biconvex lens - fewer aberrations, since true oculars have internal corrections (well, most of them). With the higher magnification you will probably see the rings of Saturn, the moons of Jupiter, or even some equatorial belts on Jupiter.

Take it slowly and do it right. This is totally doable with simple materials. This is the design of the first telescope I've built many years ago. You'll be very proud when something you made with your own hands actually works.

EDIT: Yes, you can stack lenses for extra power in the ocular. If you stack two convergent lenses, so close that they touch each other, having the same focal length, the resulting system has about half the focal length of one lens. If there's some distance in between them, things are a bit more complex.

You will notice that power is not everything, and sometimes more power is beneficial while other times it creates problems. It depends on many factors, which you can discover through experimentation.

Also read about the various eyepiece designs, which are made for specific purposes - not just magnifying power, but also lack of aberrations and a high quality image. But those require special components to make.

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

By all means go ahead and try various combinations yourself. There's a lot to discover. It's a lot of fun.

EDIT2:

Telescopes for astronomy cannot be used hand-held. You must use some kind of mount. You could make one, doesn't matter how simple, as long as it's stable and it can turn in both directions (up/down and left/right). E.g. you could make a mount out of a cardboard box:

https://aquillam.wordpress.com/2010/01/04/a-cheap-tripod-alternative/

Google "diy refractor mount" or "galileoscope mount" for more ideas.

EDIT3:

When the eyepiece is convergent, the image is reversed. This is normal for usage in astronomy, where the orientation of the image doesn't matter.

There are special devices called prism diagonals that can be inserted between objective and ocular, that fix the reverse image problem. Star diagonals show the image correct-side-up, but left and right are reversed. Correct-image diagonals show the image correct-side-up and also left and right are correct. They achieve this by doing one or multiple reflections in prisms.


How massive does a planet need to be to create gravitational lensing? Is it ever possible for starlight to be distorted in a noticeable way from something as small as a planet?

I'm guessing that the chances of this working are better on a dense terrestrial planet than a gas giant.

How do you calculate this?

Edit: I was unclear, sorry. I'm specifically thinking about lensing that would distort the shape of a star, not just move its position.

You could google "Maccone focal" for the most ardent proponent of this project (and I love it too, the most furthest meaningful mission possible within a life time, and what a view!) But others may have variations of his idea. One should be a bit wary about enthusiasts.

Here's a link to some article where Maccone is cited to explain that planets need 6-17 greater distance to become gravitaional lenses than the Sun needs. Neptune surprisingly happens to be next best after Jupiter: http://www.centauri-dreams.org/?p=15290

Paper (might not be available for free):
Maccone, “A New Belt Beyond Kuiper’s: A Belt of Focal Spheres Between 550 and 17,000 AU for SETI and Science.”

Here are slides from one of his presentations where he explains his calculations more generally, but I think only for the Sun:
http://www.spaceroutes.com/astrocon/AstroconVTalks/Maccone-AstroconV.pdf

And then there's micro gravity lenses! A very different thing in terms of our observation. I now think my answer above misunderstood what you were looking for.


Why is there a Phillips relation? I want to know the explanation of Phillips relation, the relation seen in type Ia supernovae(SN Ia with broader light curve is brighter). More Ni-56 implies a larger luminosity at peak but what about broader light curve?

Check this paper: http://adsabs.harvard.edu/abs/2001ASPC..229..287M

LargerNi-56 directly implies brighter at peak, since it is the dominant source of radiation energy at early times.

The broadness of a light curve depends on the light curve timescale, which depends on ejecta mass, opacity, and kinetic energy. Since SNe Ia are from WD progenitors, there are not much different in ejecta mass. From simulations, kinetic energy is roughly the same across broad parameters. This leaves opacity to be the origin of the broadness. Since Fe-group elements have higher opacity than intermediate mass elements (IME), and more Ni-56 implies more Fe-group/IME, it is to be expected that high Ni-56 implies high opacity that implies broader light curve.

The slide, that someone provided in your comment, explains the same idea here (with more references).


Given a star's mass, age, and composition, how do I calculate the rest of its properties? If I'm given a star's starting mass, starting composition (metallicity), and its current age, How do I calculate the star's luminosity, current mass, and radius?
I want to know how to do this for any star, from the minimum size of 0.08 solar masses all the way up to stars 100-150 solar masses for a personal project.

I know that there are equations that relate these things, but my desperate googling has only found bits and pieces of the answer I'm seeking. Also, many of the equations I've found only apply to stars within a certain mass range (usually near the sun.)
I would seriously appreciate some guidance from an expert in the field.

In astronomy research, models of stars are computed using large software packages that solve a set of differential equations that describe the structure of a star as it evolves, mostly because of nuclear fusion transforming elements, under a number of assumptions.  (I once wrote a more detailed answer about this on Physics.SE.) So I'd say the first option is to find and operate one of these programs. One of the easiest to access and use is MESA but there are many others that exist, although they often aren't freely available.

Another option is to use (probably by interpolating in) large tables of models that have been produced using those codes.  MESA is used to produce the MESA Isochrones and Stellar Tracks for which the isochrones Python package provides an interface, but there's also the Dartmouth Stellar Evolution Database and the PARSEC grid, among others.  (Apologies to all the isochrone grids I've missed!)

You can also try your hand with the program SSE, which uses analytic fits to a grid of models from the late 1990s. While SSE implements them, the analytic fits themselves are described in Hurley, Pols & Tout (2000).


Moving-Cluster method for determination of the distance of Hyades. A starter problem I am currently following a class of observational astronomy lab. I will present a brief description of the method in first and then proceed to the question, so anyone is welcomed to read the entire post or move directly to the problem. The current question is about the Moving-Cluster method, for which I quote from Wikipedia

  In astrometry, the moving cluster method and the closely related convergent point method are means, primarily of historical interest, for determining the distance to star clusters. They were used on several nearby clusters in the first half of the 1900s to determine distance. The method is now largely superseded by other, usually more accurate distance measures.*

One can find more here http://pages.uoregon.edu/soper/Stars/movingcluster.html . A brief description of the method is like that:

From photographs taken, say, 10 years apart, we can see that a star of the cluster has moved. (It has proper motion.) Let us suppose that the proper motion is an angle μ. Then the angle θ is the angle between the convergent point and the star or equally the angle between the radial and tangential speed of the moving star

If one knows the angle θ, he/she will be able to find the movement of the star and in the end, calculate it' s distance, through deduced relations beginning with $ v_t =v_r \tan θ.$

Question
I know the declination and the right ascension of both the star and the convergent point.

My problem as is to find the angle θ. In the exercise I have it is requested to find the angle between the star and the convergent point, that is θ , using spherical trigonometry. My mind goes on to using the cosine relationship:

$$ \cos a = \cos b \cos c + \sin b \sin c \cos A ,$$ for the triangle

But, although it may be simple, I cannot understand where this triangle is to be implemented. What are it' s angles in my problem. Wherever I have looked, the angle θ is taken as granted.

So, should I use something like the triangle? If yes, what the triangle should be? Should the Sun be on one angle as the image above with the convergent point? Should I define another triangle? In the end, for whatever triangle I have thought, how will I calculate the needed sides? Note that I know the right ascension and the declination of both the convergent point and the star's.

The angle you are looking for
$$\cos \theta = \sin \delta_1 \sin \delta_2 + \cos \delta_1 \cos \delta_2 \cos (a_1-a_2),$$
where $(a_1, \delta_1)$ are the RA and Dec of the star and $(a_2,\delta_2)$ the RA and Dec of the convergent point.


3C273 jet mass estimate In this picture of 3C273, we can see a jet—estimated to be around 200,000 light-years long—being emitted.

I'm trying to find an estimate for the mass of the jet. It's a stream of electrons, accelerated through synchrotron radiation (see here). There seems to be lots of information on its spectroscopic properties and the mass of the quasar (886 million solar masses), but I can't find an estimate for the mass of the jet itself. Does anyone know of a source I could look at to give me some idea?

Trigger Warning:  Major hacking going on here.
[edited from 1 ly to 1000 ly diameter]
Scaling from the image, I'll approximate the jet diameter as 1000 ly . So there's a volume of roughly 2e4*1e6 * pi/4 cubic ly.   For absolutely no good reason, I'll propose an electron ejection density of 1E6 per cubic km -- at least those are round numbers and easily scalable if anyone can find references to actual calculations/measurements --, so using 1 cubic ly = 8.468e+38 cubic km, I get 1.33e+56 electrons in the jet.


Why was StDr56 discovered only now? StDr56 is a newly discovered planetary nebula (maybe). link1 link2
It was found by amateur astronomers Marcel Drechsler and Xavier Strottner.
According to the above articles, it's pretty big:

With an extension of 44 x 36 arc minutes, StDr56 is not only the
largest probable PN in the region, but in the night sky it covers more
than half of the area of the famous Triangulum Galaxy M33, which is
located only a few degrees from Strottner-Drechsler 56.

Or

It’s about the same size as the full Moon on the sky.

So, if it's so big, why was it only discovered now? Wouldn't it make sense if all large objects were already discovered a long time ago?

Large objects can be very faint if they are far enough away. So large objects wouldn't necessarily be discovered a long time ago.  The object you mentioned is very faint and required a long view time to acquire enough photons to "see" it. As stated in a comment by @Pierre Paquette, the object was viewed for over 59 hours.


Why are some planets so close to their star? "Kepler-10b orbits its parent star once every 0.84 days, which means
  it is more than 20 times closer to its star than Mercury is to our
  sun" space.com

Are there any accepted theories that successfully predicted that so many planets would be found so close to their star?

Edit: This is especially a question, since it seems that planets are moving away from their star. newscientist.com

Big picture:

The reason why there are planets so close to their host star is probably planetary migration. These planets are difficult to explain by in situ formation models (first of all there is not enough material to form that kind of massive objects in these regions). There are different theories to explain planetary migration:

disk-planet interactions: planets form in a protoplanetary disk that still contains lots of gas after planet formation, and that can interact with the formed planets. There are different interactions that can explain migration, mostly depending on the mass of the planet you consider. We distinguish 3 main types of migration processes: Type I (migration of low-mass embedded planets), Type II (migration and gap formation by massive planets) and Type III migration (rapid migration of intermediate mass planet in relatively massive disk).
planet-planet interactions: gravitational scattering can be also a very efficient process to modify planet orbits.
Kozai mechanism: this mechanism happens in a three body problem. One object can be perturbated by an other object orbiting farther out; the object then starts to oscillate around its orbit, and can therefore migrates.
Interaction with planetesimals (also refers as the Nice model for the Solar System): interaction with a remnant planetesimal disk (as the Kuiper disk for the Solar System) can cause migration. Planets can exchange angular momentum with planetesimals in the outer edge of the disk by gravitational interaction, and thus migrating.

Disk-planet interaction:

Some more details about the Type I, II and III migration mechanisms:

Type I: planets are generating density waves propagating in the disk. Due to Keplerian rotation, this density waves become spiral waves, generating a Lindblad torque. The inner spiral arm is pulling the planet forward, the planet gaining angular momentum and migrating outward, whereas the outer spiral arm is is pulling the planet back and drives inward migration. In most cases, the outer Lindblad resonances take over the inner Lindblad resonances because they lie closer to the planet, and therefore drive an inward migration.
Type II: massive planets open gaps in the disk in which they form. The gravitational interactions become stronger and thus the angular momentum transfer to the disk. If angular momentum is deposit locally in the disk, material inside the planet loses angular momentum and material outside gains angular momentum, and therefore it recedes from the planet, opening a gap in the disk. This gap reduces the importance of Lindblad torques, and the planet is then coupled to the viscous evolution of the disk. Since material from the disk is accreted on the star and angular momentum transported away, massive planets are migrating inward.
Type III: if the surface density distribution in the co-orbital region of the planet is asymmetric, it can lead to a large torque that can cause the planet to fall inwards on a short timescale.

Sources:

Kley & Nelson 2012


Why is the Solar Helical (Vortex) model wrong? Could someone give a lay description of why the Solar System's depiction of motion as a vortex is wrong?

ref: Youtube video The helical model - our solar system is a vortex

Please don't take for granted that because of my previous question about why planets or asteroids don't have cometary tails, that I believe the 'tails' shown in the video are real. I do not. I know that the tails as shown are simply a graphic representation of the path 'flown' through space by the planets. It seems that the solar system we inhabit has to be moving through interstellar space in some fashion. If the paths of the planets could be shown in a video, it seems to me (a lay person) that there would have to be motion 'forward' in addition to the eliptical path around Sol.

It isn't correct, because a vortex is not a helix, and so while the planets do trace a helical path as they move through the galaxy, this is not evidence of a vortex.

Yes, the sun actually is moving through space, as it traces a path around the centre of the galaxy. The whole mass of the solar system moves with it, so the planets are not left behind as the sun moves.

Rhys Taylor and Phil Plait have comprehensive smackdowns debunking this vortex idea and other misunderstandings/delusions by the author.


How to find Greenwich Mean Sideral Time? I'm writing a program that converts position in the ECI (CIS, EPOCH JD2000.0) coordinate system to WGS 84 (CTS, ECEF) coordinates. To do so, I'm following what's described in this document and implementing C++ code.

My knowledge of astrophysics is very limited, so I find it hard to understand some terms. Especially, I can't seem to understand properly how the Greenwich Mean Sideral Time is computed, and why I always get different values.

For instance (assuming time is 2019-01-01 08:00:00 and Julian Date is 2458484.833333):

If I use this on-line GMST calculator, I find that GMST is 14:42:45.3766.
If I use the expression used in this MATLAB script (showing the value in time, not degrees) I also get 14:42:45.350.
However, if I calculate it myself with the expression given in the above appendix, I get 06:42:45.379163 (with a clear offset of

According to the equations in page A-26, GMST depends upon the number of Julian centuries (Tu). H0 expression is exactly the same than that of CelesTrak for θ(0h). However, I can't really understand what the du truly stands for, or more specifically, why the document shows du → ±0.5, 1.5, 2.5.... Moreover, to find Λ (which is stated to be in seconds of time) the expressions use values in radians and seconds of time indistinctly.

Could somebody shed light into this?

To do so, I'm following what's described in this document and implementing C++ code.

That is a very old document you are using. It's using the 1982 IAU precession model. There have been multiple updates since then. The latest set of changes are working toward eliminating the concept of Greenwich Sidereal Time. (In software terms, GMST and GAST are deprecated concepts. While still currently supported, support will be dropped in some future release of the Earth rotation model.)

Other than using an outdated version of a deprecated concept, what did you do wrong? The expression you are using to compute GMST is at the top of page A-26:
$$\Lambda = H_0 + \Delta H + \omega^{\ast}\,(t-\Delta t)$$
The $\Delta H$ term takes one from GMST to GAST. Dropping that term, the expression for GMST is
$$H = H_0 + \omega^{\ast}\,(t-\Delta t)\tag1$$
where
$$\begin{align}
H_0 = &\,24110.54841 + 8640184.812866 T_u + 0.093104 {T_u}^2 - 6.2\times10^{-6} {T_u}^3\tag 2 \\
&\,\text{is the Greenwich sidereal time at midnight, in seconds of time} \\
\omega^{\ast} = &\,7.2921158553\times10^{-5} + 4.3\times10^{-15} T_u\tag3 \\
&\,\text{is the Earth's sidereal rotation rate, in radians}/\text{second} \\
t\phantom{\Delta}\quad&\text{is UTC time of day, in seconds, and} \\
\Delta t\quad&\text{is the difference between UT1 and UTC.}
\end{align}$$

Note that there's a units mismatch in equation (1). Equation (2) yields a result in units of sidereal seconds while equation (3), when multiplied by time of day, yields a result in units of radians. The latter needs to be converted to sidereal seconds. To do this, simply multiple both terms in equation (3) by $\frac{86400}{2\pi}$:

$$\begin{align}
\omega^{\ast} = &\,1.00273790935 + 5.9\times10^{-11} T_u\tag4 \\
&\,\text{is the Earth's sidereal rotation rate, in sidereal seconds}/\text{UT second}
\end{align}$$

Applying this to 2019 Jan 01 08:00:00 UTC yields the following:

The Julian date at midnight UTC 2019 Jan 01 is $2458484.5$, or $2451545+6939.5$. See footnote 1 for an explanation of this convention.
Dividing the residual 6939.5 by 36525 yields $T_u$: $T_u = 6939.5 / 36525 = 0.1899931553730322$.
Applying equation (2) yields the Greenwich Mean Sidereal Time at midnight: $H_0 = 1665686.527373333$ sidereal seconds.
Applying equation (4) and multiplying by $t=8*3600$ seconds yields another 28878.85178960284 sidereal seconds to the above. What about $\Delta t$, aka dUT1? I've ignored it (i.e., treated it as if it was zero.) See footnote 2.
1665686.527373333 + 28878.85178960284 = 52965.37916293584 (modulo 86400). This is the mean sidereal time, in seconds. Converting to hours, minutes and seconds results in 14:42:45.4. The 0.4 seconds is bogus because I ignored dUT1.

Footnotes:

There's a problem with using Julian dates to express time when high precision is of concern. The problem is that adding 2.45 million and change to the time since noon on 1 Jan 2000 results in potential truncation problems. The next representable Julian date after midnight 1 Jan 2000 is 40 microseconds after midnight. I myself try to stay a couple of orders of magnitude away from the quantization level, which means problems for dates at the millisecond level.

Almost every scientific package that supports Julian dates either uses a pair of double precision numbers or uses some sort of modified Julian date, such as time since 12 Noon 16 Nov 1858 (omits the leading 2.4 million in Julian date), time since 12 noon 23 May 1968 (omits the leading 2.44 million), or time since J2000.
The term $\Delta t$ in equations (2) and (4) is better known as dUT1. This is the difference between UT1 (time using the mean rotating Earth as a clock) and UTC (time that runs at the same rate as an atomic clock, but with leap seconds occasionally inserted to keep UT1 and UTC in sync). Those leap seconds keep $|\text{UT1}-\text{UTC}| < 0.9\,\text{seconds}$. Simply ignore the $\Delta t$ term if that's the case -- but remember that fractions of a second are meaningless.


Was GRAVITY built to look at one star? GRAVITY (shown below) is a interferometric combiner of near infrared light from four very large telescopes called The Very Large Telescope in order to make careful astrometric measurements near the dusty galactic center where very exciting things happen near what we assume to be a supermassive black hole.
There have been publications of verification of general relativity by watching the star S2 pass through its periapsis around whatever the supermassive black hole-sized object is that it orbits. This was a major milestone in GRAVITY's contribution to science.

Is S2 still the fastest known star in the galaxy?

It may still be used for other things, but primarily...
Question: Was GRAVITY built to look at one star? Was the observation of S2 sufficient to justify this ESA effort? Or was it always also expected and required to do many more things?
From eso1825 — Science Release: First Successful Test of Einstein’s General Relativity Near Supermassive Black Hole; Culmination of 26 years of ESO observations of the heart of the Milky Way:

New infrared observations from the exquisitely sensitive GRAVITY [1], SINFONI and NACO instruments on ESO’s Very Large Telescope (VLT) have now allowed astronomers to follow one of these stars, called S2, as it passed very close to the black hole during May 2018. At the closest point this star was at a distance of less than 20 billion kilometres from the black hole and moving at a speed in excess of 25 million kilometres per hour — almost three percent of the speed of light [2].
The team compared the position and velocity measurements from GRAVITY and SINFONI respectively, along with previous observations of S2 using other instruments, with the predictions of Newtonian gravity, general relativity and other theories of gravity. The new results are inconsistent with Newtonian predictions and in excellent agreement with the predictions of general relativity.
[1] GRAVITY was developed by a collaboration consisting of the Max Planck Institute for Extraterrestrial Physics (Germany), LESIA of Paris Observatory–PSL / CNRS / Sorbonne Université / Univ. Paris Diderot and IPAG of Université Grenoble Alpes / CNRS (France), the Max Planck Institute for Astronomy (Germany), the University of Cologne (Germany), the CENTRA–Centro de Astrofisica e Gravitação (Portugal) and ESO.
[2] S2 orbits the black hole every 16 years in a highly eccentric orbit that brings it within twenty billion kilometres — 120 times the distance from Earth to the Sun, or about four times the distance from the Sun to Neptune — at its closest approach to the black hole. This distance corresponds to about 1500 times the Schwarzschild radius of the black hole itself.

Source

A new instrument called GRAVITY has been shipped to Chile and successfully assembled and tested at the Paranal Observatory. GRAVITY is a second generation instrument for the VLT Interferometer and will allow(s) the measurement of the positions and motions of astronomical objects on scales far smaller than is was currently previously possible. The picture shows the instrument under test at the Paranal Observatory in July 2015.

This link gives you all the papers that have cited the instrument description paper.
The list of papers shows that it has been used for studying: the centres of AGN, close binary systems, discs around young stars, the atmospheres of AGB stars and interferometric imaging of exoplanets at least.
Here is a paragraph from the instrument description paper

Inspired by the potential of phase-referenced interferometry to zoom in on the black hole in the Galactic center and to probe its physics down to the event horizon (Paumard et al. 2008), we proposed in 2005 a new instrument named GRAVITY as one of the second generation VLTI instruments (Eisenhauer et al. 2008). At its target accuracy and sensitivity, GRAVITY will also map with spectro-differential astrometry the broad line regions of active galactic nuclei (AGN), image circumstellar disks in young stellar objects and see their jets evolve in real time, and detect and characterize exo-planets especially around low mass stars and binaries – in short we will “Observe the Universe in motion” (Eisenhauer et al. 2011).


Can mechanical angular momentum be transferred by gravity? At PSE there are a few answers that explain that since the moon is responsible for tides and tides slow down spinning, L must be transferred to the moon.

This implies that if earth's L were reduced by an asteroid, this would not influence the moon's orbit.

Can you  clarify:

does the change of $\omega$ of the Earth alter the orbit of moon only when that change is attributable to the moon or in any case?
would there be tides if there were no moon?
what is the mechanism through which angular momenta can exchange magnitude? This is apparently impossible since there is no connection or, we we consider the obsolete Netwonian gravity, the connection is radial and never tangential?

does the change of $\omega$ alter orbit of moon only when that change is attributable to the moon or in any case?

Angular momentum is a conserved quantity (more on this later). What this means is that if some object is undergoing a change in angular momentum, that angular momentum must necessarily be transferred to some other object or objects. The Earth is losing rotational angular momentum. It's rotation rate is slowing down. That means some mechanism must exist for transferring that angular momentum.
The Earth's angular momentum would change if the Earth was struck by an asteroid, but that would be a very, very small change. The loss of gases from the top of the atmosphere is also represents a negligibly small transfer of angular momentum, as does the interaction of the Earth's magnetic field with the solar wind. The Earth is far too large to be subject to the YORP effect, and The only mechanism left is gravitation. Almost all of the Earth's loss of rotational angular momentum due to a slowing rotation rate is via gravitation. A small amount is transferred to the Sun and to the Earth's orbit about the Sun. All that is left is the Moon. That vast majority of the Earth's lost angular momentum goes into making the Moon recede from the Sun.
Angular momentum is a conserved quantity. The conservation laws are very important in higher level physics (Lagrangian and Hamiltonian physics); they are assumed rather than derived. Even higher level physics derives the conservation laws from Noether's theorem. However, angular momentum can also be derived from the strong form of Newton's third law. Proving this is standard fare in every undergraduate and graduate classical mechanics text. For example, it's the subject of chapter 2 in Classical Dynamics Of Particles And Systems by Marion et al. and it's the subject of chapter 1 (starting on page 5) in Classical Mechanics by Goldstein et al. There are some nice online derivations as well, such as the week 1 lecture notes for a class on Central Forces at Oregon State University.
Another name for a force that obeys the strong form of Newton's third law is a central force. A central force is one in which the force

Acts between pairs of particles that comprise a system of particles,

Obeys the weak form of Newton's third law (equal but opposite forces),

Is directed along or against the line connecting the two particles, and

Is a function of the radial distance between the particles only.

Examples include Hookean spring forces, Coulomb forces, and Newtonian gravitation. The gravitational force between a pair of particles is given by $\vec F = G M_1 M_2 / r^2 \hat r$, where $\hat r$ is a unit vector pointing along the line that connects the two particles.

would there be tides if there were no moon?

Yes. The Sun also raises tides, but smaller in magnitude than those raised by the Moon. The heights of the tides raised by the Sun are a bit less than half those raised by the Moon.

what is the mechanism through which angular momenta can exchange magnitude? This is apparently impossible since there is no connection or, we we consider the obsolete Netwonian gravity, the connection is radial and never tangential?

There are two issues with this thesis. One is that Newtonian gravity is not an obsolete theory; it remains widely use in space exploration and in astronomy. A better way to look at it is that Newtonian gravity has a limited domain of applicability.
The other issue is the claim that Newtonian is purely radial. This is only true in the case of point masses or objects with a spherically symmetric mass distribution. It is not the case for objects with an aspherical mass distribution. For example, the Earth has a significant equatorial bulge due to its rotation. This equatorial bulge exerts a torque on satellites orbiting the Earth that causes the satellites' orbits to precess. Scientists take advantage of this. There's a special class of satellites, the sun synchronous satellites, whose orbits by design precess by 360 degrees over the course of a year.
With regard to the Moon, the rotation of the Earth makes the tides raised by effectively lead the Moon by a small amount. While this isn't Newton's tidal bulge, it has a similar effect on the Moon. Averaged out over the course of time, the waters on the near side of the Earth pile up slightly in front of the Moon while the waters on the far side of the Earth pile up slightly behind the Moon. Because those far side waters are further from the Moon, the near side waters dominate gravitationally, causing the Moon to accelerate.


Could the Earth be used to cast an arago/poisson spot on something There's a potential design for a space-based telescope that uses a large opaque circular disk in place of a transparent lens to focus light for analysis. It uses the light that bends around the disk that focuses on the Poisson or Arago spot, but requires a circular object. It could theoretically produce much clearer images (1000x) than telescopes we have today, albeit with a narrower field of vision.

I understand that the Earth is not perfectly round, but I believe I've also heard it's very "smooth" despite mountains and such, so I wonder if instead of a spot (very small circle) on some point behind it, it might cast an oval or other oblong shape of some size that, using lenses or computer reconstruction, might produce an image that is usable, and presumably even higher resolution (millions, billions, trillions, or even more times as detailed?).

If that's not possible, what are the issues? Does the atmosphere get in the way? If we could imagine a sufficiently circular disk roughly the size of Earth, about how far would the image processing receiver have to be from the disk opposite the studied object in order to capture the convergence of bent light? Would the gravity of the Earth affect the distance at all, due to gravitational lensing?

One constraint is the recommendation $F=\frac{d^2}{b\lambda}\geq 1$, in this case with $d=12700 \mbox{ km}$ about the diameter of Earth, $\lambda=600 \mbox{ nm}$ some wavelength of visible light, and $b$ the distance between circular obstacle and observer.
The distance between Earth and the observer should hence be

$b\leq \frac{d^2}{\lambda}=\frac{(12700\cdot10^3\mbox{ m})^2}{600\cdot 10^{-9}\mbox{ m}}=481.67\cdot 10^{18}\mbox{ m}$

Another constraint is the surface roughness of the circular object:
$\Delta r < \sqrt{r^2 + \lambda\frac{gb}{g+b}}-r$, with $r=6350\mbox{ km}$ the radius of the circular obstacle (here Earth), $g$ the distance between the point light source and the circular obstacle, and $b$ the distance between the circular obstacle and the screen.

To simplify calculations, say $g\gg b$. Then approximately
$\Delta r < \sqrt{r^2 + \lambda\frac{gb}{g}}-r = \sqrt{r^2 + \lambda b}-r$.

After adding $r$ and squaring you get
$(\Delta r +r)^2<r^2+\lambda b$.
This simplifies to
$(\Delta r)^2+2r\Delta r<\lambda b$.
Assume $\Delta r\ll r$, and neglect the second order $(\Delta r)^2$ to get
$2r\Delta r<\lambda b$. Divide by $\lambda$ to get an approximate constraint for $b$ as

$b>\frac{2r\Delta r}{\lambda}$.
With $2r=12700 \mbox{ km}$ about the diameter of Earth, $\lambda=600 \mbox{ nm}$ some wavelength of visible light, we get

$b>\frac{12700\cdot 10^3\mbox{ m}\cdot\Delta r}{600\cdot 10^{-9}\mbox{ m}}
= 21.1667\cdot 10^{12}\Delta r$.

The two constraints allow for reasonable values of $\Delta r$.
Assume a surface roughness of Earth of e.g. $\Delta r = 1\mbox{ km}$.
Then a valid range of distances of observers would be between
$0.00224$ and $50912$ lightyears of $9.4607\cdot 10^{15}\mbox{ m}$ from Earth.

In astronomcal units of $149597870700\mbox{ m}$ the closest distance of an observer would be $141.49 \mbox{ au}$ from Earth.

Due to Earth's oblateness, however, you would get a point spread function significantly different from a dot for this "short" distance from Earth. It might be possible to correct this by an appropriate telescope optics.

The effect of gravitational lensing is
$\theta=\frac{4GM}{rc^2}=2.969\cdot 10^{-27}\frac{\mbox{m}}{\mbox{kg}}\frac{M}{r}$, after applying the constant of gravitation $G$ and the speed of light $c$. With the mass $M=5.97237\cdot 10^{24}\mbox{ kg}$ and a radius of $r=6350000\mbox{ m}$ of Earth, we get an angle of
$\theta=2.793\cdot 10^{-9}$ by gravitational lensing at the surface of Earth.

This would focus parallel rays of light to a point near a distance of
$b=\frac{r}{\tan \theta}=130.27\cdot 10^{15}\mbox{ m}$, or $13.77$ lightyears, hence well beyond the minimum distance where an Arago spot could form. But, of course, the innermost peak of the point spread function would be closer to a circular disc at this larger distance with relevant gravitational lensing.


Number of lenslets in wavefront sensor array I have to do a short calculation, but, quite frankly, I have no idea how to even start...
Suppose you have a 10-m telescope (f/10) with an infrared camera that observes at 10 micrometer. The seeing is 2 arcsec (FWHM of the PSF,  2”= 1/100000 rad) and you would like to implement an adaptive optics system.
The question is: How many lenslets do you need in the Shack-Hartmann wavefront sensor array? Below is a picture of the working principle of such an array. The telescope would also include a coronagraph, but I think (might be wrong) you don't need that information for this calculation.

Generally speaking as many as possible because the number of lenslets determines the lateral wavefront resolution. But in reality there are a few factors to be considered.
Wavefront sensor
Say your lenslet array is of focal length $f$ and subaperture size $D$, given a maximum desired detection angle $\alpha$ (in your case is $2'' \approx 1\times10^{-5}\, \text{rad}$), there is the following relationship:
$$
2'' = \alpha \approx \tan \alpha = \frac{\Delta x}{f} \leq \frac{D}{2f}. \tag{1}
$$

To ensure a good resolution and reduce cross-talk effect, $D$ is chosen to be at least a few image sensor pixel size. In other word, for a given image sensor to measure the seeing angle, the number of lenslets are limited to a maximum value.

Usually you do not design the lenslet arrays; those are pre-designed, e.g. https://www.thorlabs.com/newgrouppage9.cfm?objectgroup_id=2861. So Eq. (1) serves as a metric to select between different microlens array designs.

For AO applications, the primary goal of wavefront sensors is the measurement speed rather than the lateral resolution. The sensors need to be fast. The smaller the number of pixels, the smaller the sensor latency is, resulting in a low number of pixels, and hence low number of lenslets.

Adaptive optics (AO) system
$w \in \mathbb{R}^m$: Measurement wavefront of a lenslet array of number $m$.
$c \in \mathbb{R}^n$: Correction wavefront of the deformable mirror that has $n$ number of actuators.
An adaptive optics (AO) system works by solving a linear system to output $c$ given $w$:
$$
w = A c
$$
where interaction matrix $A \in \mathbb{R}^{m \times n}$ is pre-calibrated before usage.

To compute $c$ in a least squares sense, it requires $m \geq n$, meaning an over-determined system, mentioned also by @nflemming2004.

As for the aperture sizes of the two, it does not really matter that much because the conjugate lenses/mirrors in your optical system can scale down/up the apertures. That said, the sub-aperture of the wavefront sensor (i.e. $D$) and the actuator cover area $D_{\text{deformable_mirror}}$ do not have to be equal.

If $m \gg n$ (as in most cases), as you can imagine, the extra number of lenslets only serves to make a robust estimate, and is wasted from AO's perspective because of insufficient correction degree-of-freedom.

Example
Using WFS30-7AR as our wavefront sensor, that $f = 5.6 \,\text{mm}$ and $D  = 150\,\text{um}$, given Eq. (1) we know:
$$
2'' \leq \frac{D}{2 f} = 1.34 \times 10^{-2} \, \text{rad} \tag{satisfied}
$$
The number of lenslet is $m = 80 \times 80$, far larger than $n = 140$ the actuator number of deformable mirrors (say https://www.thorlabs.com/newgrouppage9.cfm?objectgroup_id=3208).
So for this design, we conclude it satisfies the requirement of this question.


What is the size of the largest asteroid orbiting the Earth, other than the Moon? It seems odd to me that no other asteroid around the Earth is even close to the size of the Moon. Anyone have any idea about the size and orbit radius of Earth's "second moon" ?

There are no other known natural satellites orbiting the earth. Most of the objects orbiting the earth are man-made and largest object other than the Moon orbiting the earth is the International Space Station which orbits the earth at a height of about 400km from the surface and is about 100m wide.

It may be odd to you that there are no other natural satellites orbiting the Earth. We are not sure but it maybe that the probability of an object ending up in a stable orbit around the Earth is not that high. There are roughly 3 different categories of natural satellites - Large moons in roughly circular orbit like the Moon, small rocks in low orbits formed as a result of collision and nearby asteroids that a planet catches in its gravitational well.

It maybe that there are Earth is not massive enough to capture fast moving asteroids in orbit. Or the Moon which is quite large relative to the earth destabilizes a satellite's orbit around the Earth and there are not many stable orbits that a random natural object can get itself in.


Can matter leave the cosmic horizon? Cosmic horizon in the de Sitter space is a sphere, centered at the observer with finite radius where the red shift due to cosmic expansion becomes infinite.

Given that no information can be transmitted from behind the cosmic horizon, I wonder whether any matter can ever pass behind the event horizon, because this will give rise to an information paradox, similar to the black hole information paradox.

The matter approaching the cosmic horizon seems to experience time dilation, so will that time dilation prevent it from crossing the horizon?

To work around the paradox, one have to postulate that the information about outgoing matter should be encoded in the de Sitter radiation.

A little bit of caution is needed when approaching this subject as there are several surfaces in cosmological solutions that can be interpreted (or misinterpreted in some cases) as horizons. Not all of them appear in all comsological solutions and in some cosmological solutions some of these surfaces are only present for part of the evolution of the solution.

If you were to mean the particle horizon, which is the limit of the observable Universe. The cosmological red shift at the particle horizon is always infinite, when there is a particle horizon. No matter can ever leave the particle horizon (regardless of the cosmological model, as long as there is a particle horizon) as it expands radially outwards at a local speed of c relative to any matter on the horizon (of course to us it expands even faster than c as it is also receding due to expansion).

An ever-expanding Universe with a cosmological constant approaches a state similiar to de Sitter space in late times, however de Sitter space itself is not a realistic cosmological solution as it is devoid of matter. De Sitter space does not have a particle horizon and for that matter it doesn't have any singularities either.

In de Sitter space the cosmic event horizon, which is the surface beyond which events happening now cannot affect us in the future providing we stay at the same spot (this should not be confused with the particle horizon), sits at a constant distant to the observer that is inversely proportional to the square root of the cosmological constant.

In answer to your question then, de Sitter space is symmetric in time and space which actually gives rise to seemingly different physical interpretations of it. However the black hole information paradox is specifically the loss of unitariness in the time evolution of a black hole and you have not justified why you think an equivalent thing should happen in de Sitter space. The loss of unitariness in actually arises because  the event horizon disappears, but (naively at least) the information behind it also disappears. In de Sitter space the event horizon remains and even if it were to disappear by the cosmological constant miraculously going to zero we would still find the information behind it, so there is no analog to the information paradox in de Sitter space.

To take another example, consider Rindler coordinates in Minkowski spacetime, there is a horizon (the Rindler horizon) and there is radiation (Unruh radiation), however there is no information paradox as we know that unitariness is preserved in Minkowski spacetime and when the Rindler horizon is made to disappear by selecting a different coordinate system, all the information that was behind the horizon is still there.


Transformation from Geocentric Coordinates into Equatorial Coordinates? I have the moon's position in equatorial coordinates (RA and Dec) and now I want to calculate same moon's position in terms of earth centered coordinates i.e. Geocentric (latitude & longitude). My main goal is to establish a c++-code that gives me this transformation from RA,Dec to Lat,Lon and vice versa.

EDIT: More precisely, I have moon’s position in equatorial geocentric coordinates with following details:

Origin: Center of Earth
Lat   : Declination With fundamental Plane: Celestial Equator
Lon   : R.A.        With Primary direction: Vernal Equinox

And I am able to obtain the moon’s position corresponding to a specific Modified Julian Date (MJD). I am also able to calculate moon’s position in xyz-coordinates by following calculations

$X= EarthMoon*sin(PI/2.-dec_moon*deg2rad)*cos(ra_moon*deg2rad);$
$Y= EarthMoon*sin(PI/2.-dec_moon*deg2rad)*sin(ra_moon*deg2rad);$
$Z= EarthMoon*cos(PI/2.-dec_moon*deg2rad);$

where EarthMoon is the distance between Earth and Moon. Further, I am converting them into Horizon Coordinates (topocentric) w.r.t to an observer at a specific location by using my subroutine and this part done.

Further, I am initiating my simulated CR-particles from observer site. E.g. vCR[3]= {R,lat,lon}

vCR[3]={Rearth+2300, 31.30, 55.74} //Geocentric CS (earth-based)

And converting them into Geocentric Cartesian coordinates(x,y,z). Finally, I have

vM[3]={X,Y,Z}; in Equatorial geocentric coordinate system
vCR[3]={x,y,z}; in Geocentric Coordinate System i.e. Earth based

That is my particular question in which I want to make vM[moon position] and vCR[CR position] to lie in same coordinates to get my final result. i.e. transformation from Geocentric (R,lat,lon) into Equatorial Geocentric (h,R.A,Dec).
Note: For vCR[3]: zero-longitude is the Prime Meridian, Greenwich whereas for vM[3] it’s vernal equinox in celestial sphere there is the difference.

If I understand your question correctly than your problem is that the primary direction of vCR[3] is the intersection between the prime meridian and the Earth's equator, instead of the vernal equinox. The geographical latitude is the same as the declination, but to get the right ascension you will need to rotate your coordinates around the z-axis by an amount that is equal to the sidereal time at Greenwich ($\theta_G$). This is done, simply by replacing the the longitude (lon) with the sidereal time at your location $\theta = \theta_G-lon$.

x =  R* cos(lat) * cos(theta);
y=   R* cos(lat) * sin(theta);
z =  R* sin(lat);

vCR[3]={x,y,z};

where theta is the sidereal time $\theta$.

The sidereal time is not a time but actually an angle that gives the right ascension of objects that are at that moment crossing the local meridian (i.e. the line between South and the Celestial north pole, for observers on the northern hemisphere). It is in a sense the longitude of your position on Earth if the geographical coordinate system were to rotate along with the celestial sphere.

EDIT: To calculate the Sidereal time, you can use the method explained in Astronomical Algorithms by Jean Meeus:

$$T = \frac{JD-2451545.0}{3625}$$
$$\theta_0 = 100.46061837 + 36000.770053608 T + 0.000387933 T^2 - \frac{T^3}{38710000}$$

The sidereal time is expressed in degrees.

Then express the UTC time as an angle:

$$t = 15h+0.25m+\frac{0.25}{60}s$$

where $h$ is the number of hours, $m$, the number of minutes, and $s$ the number of seconds.

The sidereal time at Greenwich at the specified time is then:

$$\theta_G = \theta + 1.00273790935*t $$

and the local sidereal time is:

$$\theta = \theta_G - L$$

Where $L$ is the geographical longitude (positive west from Greenwich).


Is there any proof of space being created? Is there any proof for space being created?

Because I guess an explanation for the dark energy (dark = our understanding of it) could be that there is a gigantic amount of matter/dark matter that we cannot see (the gray clumps in the picture) that are attracting our matter (the known universe in the middle) => of course the universe is "expanding" with an acceleration.

Am I talking non-sense?

This is an intriguing proposition, but I would ask how your hypothesis explains that the universe appears to be flat? That is with $\Omega_M + \Omega_\Lambda = 1$. The evidence for this comes from measurements of the cosmic microwave background, yet if we sum up all the matter (including dark matter), we only arrive at $\Omega_M \sim 0.3$.
I do not think that your proposed arrangement both allows an accelerating expansion of the visible part of the universe (which note, can only be gravitationally influenced by "outside regions" that are within causal contact with it) and for it to be geometrically flat.
In addition, I note you are attempting to explain the expansion in some sort of Newtonian sense - I suppose your clumps of far-away dark matter are meant to be "stretching" the observable universe? But that isn't what happens in GR and any Newtonian interpretation would have to counter the problem that if the matter outside the observable universe was uniformly distributed (i.e. in a spherical shell), then it would exert exactly zero force on everything inside it (the shell theorem); but if it were not symmetric then you have to explain why the expansion of the universe appears to be (very close to) isotropic observationally.


Proof that Parallel Universes exist Today I heard on the radio that the Super Collider has discovered mini black holes. They said that this discovery is proof of the existence of Parallel Universes and that the "Big Bang" never happened.I don't understand this and why this discovery means that Parallel Universe exist ? Does anyone know ?

Your radio information source is wrong.
The Large Hadron Collider has not discovered mini black holes.
When it comes back on line this spring, the LHC will begin looking for mini black holes: Large Hadron Collider Could Prove the Existence of Star Trek's Parallel Universe

Journalists do like their headlines, but the gist is this:

  When the Large Hadron Collider is brought back online in the spring, researchers will be looking for the existence of mini black holes. These mini black holes would lend support to string theory, which posits that different dimensions and parallel universes are possible.

No miniholes yet, maybe never.


Is the three-body system "unique"? Given a state of an ideal 3 body system (i.e., without external interference) in time $t$: the velocity $v_{i,t}$, mass $m_{i,t}$ and position $x_{i,t}$ for $i\in \{1,2,3\}$, using numerical method it is possible to determine any state in time $\hat{t}$, where $\hat{t}>t$. But is it possible to determine a unique state in time $\bar{t}$ where $\bar{t}<t$?

In another word, is it sufficient to know a single state of the system for any given time to deduce the states of the system for all time?

Equivalently, will two different states of the 3-body system results in the same state (maybe at different time) in the future?

What you're really asking about is less to do with Astronomy and more to do with mathematics. You're basically asking if, given a system of differential equations, will a unique solution exist for all time? For an answer, you should check out the Existence and Uniqueness theorems of differential equations. You'd be better to ask questions like this on the Mathematics stack exchange.

However, to discuss the particular, astronomical case you've asked about, the answer is yes, you can run that system both forwards and backwards if you know some initial state. Newtonian mechanics is completely deterministic in that if you know all the equations of motion involved, as well as the entire state of the system at a given time, you can figure out the state of that system at any other time, both in the past and the future.

To speak to your particular 3-body orbiting problem though, I'll say that the system of equations is not solvable in closed form - that is, you can't write down an analytic solution to the equations like you could for the 2-body case. As userLTK states, you can write down approximate solutions in the restricted 3-body problem, where one mass is significantly less than the other two and orbits under specific conditions.

To get a solution at any time $\hat{t}$, you need to use numerical methods. Of course numerical methods are inherently flawed. Numerical errors build up the longer you simulate due to time steps which are not infinitesimal, from errors within the numerical algorithms used, and general floating point errors. In theory, if you had a computer with infinite precision and infinite computing power, you could solve a 3-body (or n-body) system perfectly, but we live in the real world where such things are impossible.

To prove though, that you can figure out the state at any time in the past or present for a three body body, I've written a basic simulation in Python 3. It can run both forwards and backwards from a given start condition and start time. Essentially it puts three nearly identical masses in contrived starting positions and velocities. Below the code are plots of results.

import numpy as np
from numpy.linalg import norm
from matplotlib.pyplot import *
from time import time

# Define physical constants
G = 6.67408E-11          # Gravitational Constant, m^3 kg^-1 s^-2

# Define body 1 parameters
m1 = 2.2E30                # Mass, kg
x1 = np.array([0,1E11])  # Position, m
v1 = np.array([-3.5E4,0])  # Velocity, m/s

# Define body 2 parameters
m2 = 1.9E30
x2 = np.array([1E11*np.cos(210*np.pi/180),
               1E11*np.sin(210*np.pi/180)])
v2 = np.array([3E4*np.cos(300*np.pi/180),
               3E4*np.sin(300*np.pi/180)])

# Define body 3 parameters
m3 = 2E30
x3 = np.array([1E11*np.cos(330*np.pi/180),
               1E11*np.sin(330*np.pi/180)])
v3 = np.array([3E4*np.cos(60*np.pi/180),
               3E4*np.sin(60*np.pi/180)])

# Define simulation parameters
n = 3                    # Number of bodies, unitless
t = 0                    # Simulation time, s
dt = 1E4                 # Simulation time step, s
tEnd = 1E8               # Simulation end time, s
m = np.array((m1,m2,m3)) # All masses
x = np.vstack((x1,x2,x3))# All positions
v = np.vstack((v1,v2,v3))# All velocities
xHist = [[list(x1)],[list(x2)],[list(x3)]]
vHist = [[list(v1)],[list(v2)],[list(v3)]]

# Simulate until end time is reached
start = time()
while True:
    # Calculate acceleration
    a = []
    for i in range(n):
        a.append(0)
        for j in range(n):
            if i == j: continue
            a[-1] += - G * m[j] / norm(x[i]-x[j])**3 * (x[i] - x[j])

    # Update velocities
    for i,vi,ai in zip(range(n),v,a):
        vi += ai * dt
        vHist[i].append(list(vi))

    # Update positions
    for i,xi,vi in zip(range(n),x,v):
        xi += vi * dt
        xHist[i].append(list(xi))

    # Update time and end simulation if past tEnd
    t += dt
    if dt > 0 and t > tEnd: break
    if dt < 0 and t < tEnd: break

end = time()
print('Simulation finished in {:.4f} seconds.'.format(end-start))

# Convert xHist and vHist to np arrays
xHist = np.array(xHist)
vHist = np.array(vHist)

# Plot everything up
figure()
for i,c in enumerate(['or','ob','oc']):
    # Plot starting positions
    plot(xHist[i,0,0], xHist[i,0,1], c)
for i,c in enumerate(['-r','-b','-c']):
    # Plot path of star
    plot(xHist[i,:,0], xHist[i,:,1], c)
gca().set_aspect(1)
gca().set_xticks([])
gca().set_yticks([])
show(block = False)

Note, the plots show the initial positions of the stars as the points and then trace out their paths over time.

Path of three masses for $t<t_0$

In this scenario, the three masses end up in the contrived scenario. I ran the simulation backwards by setting the timestep dt and the end time tEnd to be negative.

Path of three masses for $t_0<t$

From here, the simulation is run forwards with positive dt and tEnd, starting from the same contrived scenario as above.

Note how chaotic and unstable this system is. The entire system only remains a 3-body system for less than 6 years. Before that the three masses are separate and doing their own thing. They "coincidentally" meet (because I set it up so they should), orbit around each other for a little less than 6 years, and one gets ejected, resulting in the other two continuing to orbit one another.

Contrived Scenario with $m_1=m_2=m_3$, $|v_1|=|v_2|=|v_3|$, and all positions $120^o$ from each other

Just for fun, a "stable" 3-body problem with all stars of equal mass and orbits. This is really an unstable equilibrium orbit and any perturbations will screw it up and you'll see what you saw in the above two graphs. In fact, if you run this long enough, the numerical instabilities of my code will result in the orbits breaking down. This numerical instability, as I said above, is inherent in any numerical solution and cannot be overcome, only minimized. I find that, using the numerical method I did, my system is resistant to numerical instabilities for about 7 years. If I want to run this any longer (and continue to be accurate), I need more robust numerical methods.


Are galaxies moving away with constant acceleration? Suppose that we could measure the current acceleration with respect to us of a distant galaxy chosen at random in our observable universe (Ignore the peculiar motion of the galaxy). Is the acceleration constant, is it increasing or is it decreasing?

I think the correct answer is that nobody knows - it all depends on the behaviour, or equation of state, of the dark energy.

If the dark energy takes the form of a cosmological constant, then the energy density due to dark energy will end up completely dominant as the universe expands.

The acceleration of the scale factor $a$ is given by
$$\frac{\ddot{a}}{a} = -\frac{4\pi G}{3}\left(\rho + \frac{3P}{c^2}\right) $$
where $\rho$ is the energy density, and $P$ is the pressure. A cosmological constant has $P = -\rho c^2$, which leads to
$$ \ddot{a} = \frac{8\pi G \rho a}{3}$$

So, if dark energy dominates, then $\rho a$ increases with time, so the acceleration is positive and increasing with $a$ and hence time.

If matter dominates, which it did in the past, then $P \simeq 0$ and $\rho a \propto a^{-2}$, so the acceleration is negative, with a magnitude that becomes smaller.

The transition from deceleration to acceleration happens when $\rho_{\Lambda} + \rho_m + 3P/c^2= 0$. If we assume a cosmological constant with $P = -\rho_{\Lambda} c^2$, where $\rho_{\Lambda}$ is the dark energy density (which is constant), and $\rho_m$ is the matter density, where $\rho_m = \rho_0 a_0^{3}/a(t)^3$ and $\rho_0$ and $a_0$ are the present-day matter density and scale factor respectively. Thus
$$ \rho_{\Lambda} + \rho_m -3\rho_{\Lambda} = 0$$
and the transition point occurs when $\rho_m = 2 \rho_{\Lambda}$. But because $\rho_{\Lambda}/\rho_0 = 0.7/0.3$, then
$$ \rho_{\Lambda} = \frac{7\rho_0}{3} = \frac{7}{3}\left(\frac{a}{a_0}\right)^3 \rho_m$$
and so the scale factor at the time of transition from decleration to acceleration is found from
$$ \rho_m =  \frac{14}{3}\left(\frac{a}{a_0}\right)^3 \rho_m,$$
$$ a = \left(\frac{3}{14}\right)^{1/3} a_0$$
i.e. when the scale factor was 60% of the present day value. According to this cosmology calculator, this happened roughly 6 billion years ago.

Thus the universe was decelerating then accelerated. So clearly the acceleration now is increasing with time.

On the other hand, if for whatever reason, the dark energy started to "decay" then it might be that the acceleration would decrease.


The Three Elements in the Milankovitch Cycles According to the Serbian geophysicist Milutin Milankovitch, there are three elements that make an ice age possible:

Eccentricity (orbital shape):  Varying between 0.000055 and 0.0679 over the course of 100,000 years (1.0 being a perfect circle).
Obliquity (axial tilt):  Varying between 22.1 and 24.5 degrees over the course of 41,000 years.
Axial precession (change in the orientation of the rotational axis on a rotating body):  Polaris being the North Star for a total of 26,000 degrees.

The question is how connected those three elements are, in the event someone wants to change the numbers in a worldbuilding process to make either a longer or shorter ice age:

Does the duration and extent of eccentricity affect those of obliquity and precession?

Or does the duration and extent of obliquity affect those of eccentricity and precession?

Or does the duration and extent of precession affect those of eccentricity and obliquity?

The primary driver of ice ages over the history of the Earth is much more Earth science than astronomy.  The modern ice age cycle (last 2.6 million years or so) is only possible with relatively low levels of CO2 in the atmosphere and land masses near the North Pole where glaciers can form.  Other factors in the modern ice age period are thought to be the formation of the Isthmus of Panama, perhaps the relatively permanent antarctic ice sheet, and just maybe, the formation of the Himalayas as well.

The set of circumstances required on Earth that make milankovich driven Ice ages possible are pretty complicated and have to be set just about right.  When they are just right, then the Milankovich cycles, which are quite weak, can take over and be the drivers of ice ages.   Prior to 2.6 million years ago, the last time the Earth went through a period of rising and receding ice ages was 440 million years ago when most of the land was on the South Pole.  CO2 was much more abundant, but the single continent sitting over the south pole made the formation of ice sheets possible.   Over most of Earth's history, the Milankovich cycles haven't governed ice ages.

That said, onto the astronomy bits of this question:

  The question is how connected those three elements are, in the event
  someone wants to change the numbers in a worldbuilding process to make
  either a longer or shorter ice age:

There's a strong correlation between the Milankovich cycles and the length of recent ice ages, so the answer here is yes, if you were able to adjust the "numbers" then the length of ice ages would change accordingly.

The 3 Milankovich cycles combine into two primary patterns, the 40,000 year cycle and the 100,000 year cycle (more on that shortly).  These were observed first by geologists and later tied to Milankovich cycles.  What starts and stops ice ages is complicated and there's a few moving parts, so there's not 100% consensus on Milankovich cycles driving ice ages, but the correlation is strong and as far as I've read, this idea is largely agreed upon.

Ice age Cycles

From about 2.3 million years ago to about 800,000 years ago, the 40,000 year ice age cycle was dominant, and for the last 800,000 years, the 100,000 year cycle has been dominant.  The 40,000 year cycle is thought to be the stronger of the two, and one theory as to why the 100,000 year cycle became dominant 800,000 years ago is that the Earth grew slightly cooler, perhaps with the glaciation of Greenland.   See 100,000 year problem.  The 40,000 year cycle still plays a role even within the 100,000 year cycle currently the dominant one.  Article on that here.

Axial tilt.

Axial tilt is on a pretty consistent 41,040 year cycle and it's strongly tied to the arguably somewhat misnamed, 40,000 year cycle.

Axial tilt is the cause of seasons on Earth, so when the Axial tilt is high (over 24 degrees), the seasonal variation increases, meaning, warmer summers and colder winters on both hemispheres.  When it's low (below 23), it means more mild seasons.  While this doesn't change how much sunlight hits the earth, it does change climate on Earth for other reasons,  The overall result is that the more mild seasons lead to a cooler earth.

From Wikipedia

  This trend in forcing, by itself, tends to make winters warmer and
  summers colder (i.e. milder seasons), as well as cause an overall
  cooling trend.

The reason for this, briefly, is that it still snows in high latitudes in the Northern Hemisphere during mild winters and snow and Ice takes longer to melt during mild summers.  If this persists over enough years, Snow cover can build up, which causes Albedo to rise and reflect more sunlight back into space, and less summer melting of ice, over many years, can lead to an ice age.  Once an ice age gets going it's self perpetuating, sometimes referred to as feedback mechanisms.

Axial Tilt reached a maximum about 8,700 BCE and it will reach a minimum in about 11,800 CE.  The warming of the Earth from the Axial tilt maximum isn't sudden though, it begins several thousand years before the maximum and continues for several thousand years after.   This fits nicely with the ending of the last ice age around 10-11,000 BCE.  We're currently in a mid-point in the Axial tilt cycle and it's gradually approaching a cooling effect, though, at least for now, the effect of man made climate change is far bigger than the gradual Axial Tilt cooling.

Axial Precession

Precession is a bit more complicated.  The Earth is on a 26,000 year "wobble",  that moves the North Star and other stars around in a circle in the night sky, but what matters with this Milankovich cycle is what month the Earth is closest to the sun, or, at Perihelion.  The change in Perihelion over time works on roughly a 21,000 year cycle, because it combines Axial precession with Apsidal precession.

Currently the Earth is closest to the Sun in January, but that will gradually change and in about 10,500 years, the Earth will be closest to the Sun in July.  When the Sun is closer, the Earth gets slightly warmer, so the current position of this cycle leads to milder seasons in the Northern Hemisphere (which can lead to cooling) and more extreme seasons in the Southern Hemisphere.   Unlike Axial Tilt which affects both hemispheres in the same way, precession affects each hemisphere in the opposite way to the other.   This, in effect either adds to Axial Tilt's effect or subtracts from it.   As pointed out above, the simplest factor is that milder seasons in the Northern Hemisphere that are ice-age friendly and more extreme seasons are not.   So when perihelion is in January, that somewhat cools the Earth and when it's in July it somewhat warms it.  We're currently in a "cold" period of this cycle, we were in a warm one 10,500 years ago, which also times well with the end of the last ice age. This cycle is moving towards a gradual warming and more extreme Northern Hemisphere seasons currently working in opposition to the Axial tilt effect.  It's cycle is roughly twice as fast and so this also ties in nicely with the 40,000 year cycle too, as these 2 periods (21,000 years and 41,000 years) are close to in sync.  These two cycles are thought to be key factors in driving the greening and droughting of the Sahara desert.  Article here and here.

Eccentricity

The Earth's orbital period and semi-major axis of it's orbit remain roughly unchanged.   What changes with eccentricity is how far the focal point (the Sun is at one of the focal points of the ellipse), moves from the center of the elipse.   The effect of the Earth's orbital eccentricity changing is two-fold.  First, Axial precession only matters if the Earth's orbit is an ellipse.   If the Earth's orbit is circular, the precession Milankovich cycle has no effect.  Precession has maximum effect when the Earth's orbit is at it's most elliptical.

The 2nd effect is that a more elliptical orbit means that, because planets move more slowly in their orbit when they are further from the sun (Equal areas in equal time, per Kepler's law), a more elliptical orbit means that over the calendar year, the Earth gets less sunlight because it's average distance form the sun over time is slightly higher.  So, low eccentricity means slightly warmer, high eccentricity, slightly colder, even though the planet passes through an entire orbit every year, the total energy the Earth receives does drop slightly when the Earth's orbit is most eliptical.

In addition, greater eccentricity amplifies the effect of the Axial precession, which when it's at it's highest, tends towards ending ice ages, so this effect operates in both directions, which is part of the reason why the 100,000 year cycle is smaller.

In effect, eccentricity multiplies the effect of precession, but it has an additional effect that works in opposition to the precession ice age driver.  (Does that make sense? - I can try to clean up if not).

Precession and Axial Tilt can add together or cancel each other out.

and there's other effects.  How much sunlight is reflected off the high albedo south pole, how much sunlight warms the antarctic ocean, how strong the Atlantic current is.   The entire picture is hugely complicated, but from an astronomical point of view, eccentricity is the only one that can reduce or increase total solar energy hitting the Earth.   Axial tilt and precession have no effect on how much sunlight hits the earth, they only drive the variation between the seasons.

  Does the duration and extent of eccentricity affect those of obliquity
  and precession?

The extent of eccentricity directly increases the seasonal variation effect of precession.

  Or does the duration and extent of obliquity affect those of
  eccentricity and precession?

No, the obliquity doesn't affect either of those, but the effect of obliquity can add to, or cancel out the effect of precession.

  Or does the duration and extent of precession affect those of
  eccentricity and obliquity?

Precession doesn't effect eccentricity (which way the Earth is pointed obviously doesn't effect eccentricity), but precession can add to or subtract from Obliquity.

That's the gist of it.

You can run some numbers if you like, but that gets even longer.  The sunlight energy that reaches the earth and how it varies by axial tilt is longitude dependent, so any calculations on changes in energy by changes in axial tilt will vary based on longitude.  At many latitudes, the summer solar energy is more than double what it is for winter.  Even a one or two degree in the angle of axial tilt variation can make a measurable difference in the seasonal variation.

Precession energy variation isn't longitude dependent, but it varies every day with the earth's orbit.   Some bad math on it, with the current eccentricity of about 0.0167, what that means is that at perihelion, the Earth is about .9833 AU from the sun and at aphelion, about 1.0167 AU. Because solar energy diminishes with the square of the distance, the roughly 3.4% variation in distance from the sun works out to about 7% more solar energy hitting the Earth perihelion to anhelion.   When the Elliptical orbit peaks, that ratio exceeds 20% and when it's minimized, less than 2%, but those are ratios by season and location, the total energy the Earth receives from the sun doesn't change, only seasonal variability changes.

That's as far as I'll go with the numbers, as far as the timing, that's pretty straight forward, some of that covered above:

Axial Tilt, when at a maximum it tends to melt glaciers and leads to warming, this last happened 10,700 years ago and fits with the end of the last ice age.   Minimal tilt can lead to cooling and the earth will enter an Axial Tilt Minimum in about 9,800 years.   The entire cycle, high to low back to high lasts 41,040 years.

Precession is significantly amplified with high eccentricity and significantly reduced with low eccentricity.   Earth is currently in a low eccentricity so the precession effect for the last 10,000 years or so has been relatively small.  10,500 years from now precession will be at peak seasonal variation in the Northern Hemisphere, which will work against the Axial Tilt's cooling.   In about 21,000 years, Precession will again be working towards more mild seasons in the Northern Hemisphere, eccentricity will be high and the Axial Tilt will be past it's minimum, but still in a low point.   So, taken together, we might see an ice age begin to form somewhere around 18-20,000 years in the future, but we have little to worry about.  It's not difficult to generate enough atmospheric CO2 to prevent that ice age from happening.   We're already well over the necessary amount of CO2 to prevent a coming ice age.

Finally, Eccentricity was at a minimum about 9,000 years ago and will reach a maximum in about 27,000 years in the future.  Eccentricity is the least consistent of the cycles, it varies between about 90,000 years and 125,000 years and there's a larger, 400,000 year cycle on top of that.

If parts of that aren't clear, I can try to tidy up, or just ask.


Which is the farthest distance a particular distant known- identified object from the asteroid belt can get from the sun? According to

Space facts

  The Asteroid Belt is located in an area of space between the orbits of
  Mars and Jupiter. That places it between 2.2 and 3.2 astronomical
  units (AU) from the Sun

But asteroids from the asteroid Velt like Pallas and Hygiea have an aphelion of
3.4119 AU and 3.5024 AU respectively. So my question is, which is the farthest distance a particular distant known-identified object from the asteroid belt can get from the sun?

Updated since the question might seem unclear. I'm asking for a currently known object from the asteroid belt which gets closer to Jupiter.

tl;dr: The key part the question is the phrase

  ...from the asteroid belt...

There are lots of asteroids and only some of them are considered to be in "the asteroid belt" or to be main-belt asteroids. As @zephyr's answer to the question Do astronomers generally agree that the distinction between comets and astroids is not so clear? points out, even the definition of what is or isn't an asteroid is under debate and review. To the question of which is the farthest a main belt asteroid can get from the Sun, the answer depends on exactly how you draw your line.

There are also families and groups of asteroids, and while these have visually clear clumps and clusters when plotted in certain ways, they may not have absolute universally accepted definitions of who's in and who's out.

See this and this for some families within the main belt.

While the OP"s link says

  The Asteroid Belt is located in an area of space between the orbits of Mars and Jupiter. That places it between 2.2 and 3.2 astronomical units (AU) from the Sun. The belt is about 1 AU thick.

that wording seems awkward since "between the orbits of between the orbits of Mars and Jupiter" only really puts it between 1.4 and 5.5 AU.

The first sentence of Wikipedia's Asteroid belt says:

  The asteroid belt is the circumstellar disc in the Solar System located roughly between the orbits of the planets Mars and Jupiter

repeat, "roughly" and it goes on to say

  The asteroid belt is also termed the main asteroid belt or main belt to distinguish it from other asteroid populations in the Solar System such as near-Earth asteroids and trojan asteroids.

The article also includes the figure shown below, showing that there are plenty of other groups of asteroids besides those designated as "main belt". I am not sure if there is a 100% absolutely IAU-official definition that you can apply to any set of orbital parameters that will return Yes or No to the question "is this one main belt?" but in general it looks like asteroids with a semimajor axis beyond 3.3 AU, or a high eccentricity, or a high inclination would not be called "main belt" by most astronomers.

As @PM2Ring points out Kirkwood Gaps delineate statistical sub-groupings of the main belt.

  A Kirkwood gap is a gap or dip in the distribution of the semi-major axes (or equivalently of the orbital periods) of the orbits of main-belt asteroids. They correspond to the locations of orbital resonances with Jupiter.

You can see that while there is a minimum around 3.3 AU due to the 2:1 resonance, there are "green" stragglers extending to the right that continue on past the arbitrary 3.5 AU edge of this plot of "main belt asteroids".

Source

Source

Screen shot from Scott Manley's video What Rock Star Brian May Discovered About Interplanetary Dust

which gives a better idea that there are groups and stragglers than the image in the OP's link:

Source


Why are we building larger land-based telescopes instead of launching larger ones into space? This question is a follow-up to Do bigger telescopes equal better results?

How much bigger does a ground-based mirror have to be to match what a space-based one can do?  I guess I'm asking primarily for visible light, but I'm interested in general too.

I guess on the ground, you're safe from micrometeorites, so it will probably last longer.  At what point does it become cheaper to build a telescope on the moon or something?

It's cheaper.

(1) With adaptive optics you can get 0.1 arc second resolution on the ground (admittedly only on a mountain top with particularly good air flow, but still!). This eliminates one of the major advantages of space until you get above several meters mirror diameter.

(2) Rocket fairings are the shrouds which protect payloads during the supersonic atmospherics speeds reached during launch. A 5 meter fairing is about the largest that can be flown, which limits the size of the one-piece mirrors which can be launched.  (The Dreaded Webb Telescope's mirror is in pieces which will assemble themselves in space -- a very scary and very expensive piece of design.)

(3) Servicing a telescope on the top of Mauna Kea or in the high Chilean Andes is a difficult and expensive process. Servicing a telescope in orbit makes that look like small change.  (Cost comparable to the cost of building a new giant scope on Earth.) And in-orbit servicing can't even be done with current technology except in low earth orbit.

(4) While high resolution is one frontier in astronomy, going deep is another, and going deep requires big mirrors.  A 30 meter mirror on Earth gathers much more light than a 5 meter mirror in space.  The giant terrestrial telescopes simply do a better job of being light buckets for spectroscopy than anything we can yet put into space.

The bottom line is that with the development of adaptive optics, space-based telescopes of currently buildable and launchable size lost their main advantage over ground-based telescopes.  And since they're 10x to 100x the cost, they are simply not worth building for many purposes.

Space based telescopes still hold a significant edge in parts of the spectrum blocked by the atmosphere such as UV and IR (Webb), and for certain tasks involving long-term high accuracy photometry (Kepler) and astrometry (Gaia).  But for general purpose use, the balance seems firmly on the ground's side for large telescopes.

This will change if space flight becomes cheaper -- the SpaceX BFR, for example, with its 9 meter fairing and dramatically lower launch costs, offers great hope for space telescopes.


Bright pulsating light Was looking at the sky and I see a satellite , turned around and saw another light , but this light was pulsating like a strobe light , definitely not a plane , next thing I know the light goes out and was not seen again , could this be a dying star , I know what satellites and planes look like in the sky , but I have never seen anything like this , it just went dark after being super bright

This might be it:
Astronomers Say Giant Disco Ball in Space Is Bad for Science

Three foot diameter faceted mirror ball launched late January 2018.
I don't know either designation or orbit.


Average amount of annual daylight at any place on earth If this is the wrong group please direct me to the correct one.

It seems intuitively obvious that the amount of daylight per annum should be the same for any latitude on earth. For example, 12 hours per day at the equator. The poles have daylight for half the year and darkness for the other half (crudely).

Is there any way to get an answer to this apparently simple question - is the annual amount of daylight the same at any point on earth?

Wikipedia strikes again:

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

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


Are black holes expanding? We know that it is a commonly heard claim that in every black hole there is a universe. It is also believed that the universe is expanding. If we use both of these, can we conclude that every black hole is expanding?

Most$^\dagger$ black holes are indeed expanding, but not because the Universe is expanding. Rather, their size (more precisely their Schwarzschild radius) increases proportionally to their mass, so they grow as they accrete more matter.

It is a common misconception that everything expands along with the expanding Universe. It doesn't. On small scales, i.e. black holes, planets, stars, galaxies, and even galaxy groups, gravity ensures that these things don't increase in size. Only on large scales, i.e. galaxy clusters and beyond, does the expanding Universe pull everything apart.

As for the theory about a universe inside, I… I don't even wanna go there. I'll just refer to HDE 226868's comment.

$^\dagger$Very small black holes will decrease in size due to Hawking evaporation.