How can the universe be expanding faster than speed of light? So the story goes like this:

A long time ago, 13.799±0.021 billion years to be exact, something happened. It was a big bang, loud explosion and universe came to existence. It grew and grew, and now we know that the observable universe is about 93 billion light years in diameter. So it means that

  objects that were once close ... are now up to around 45.7 billion
  light years away rather than up to 13.799 billion light years away ...

Hence my question is that everything is traveling faster than the speed of light?

If it is, then what does it mean in terms of special relativity?

I did read:

Is the universe expanding faster than speed of light?
How fast is the universe expanding?

It was a loud explosion

First, we have to clarify what the Big Bang was. The name is a misnomer, as it was neither loud nor an explosion. All we know is that the Universe is currently expanding — that is, space is literally being created between all matter. We have plenty of answers here explaining this more intuitively, such as mine (which also answers your question about its implications on special relativity).

Anyway, if we were to look further back in time, we'd expect there to be less and less space between matter. Our equations tell us that at 13.8 billion years in the past, the amount of space in the Universe should have been zero.

This is a problem. In general relativity, you can't have a metric with zero space. Thus, we know general relativity is incomplete. It cannot explain what happened at the moment of the Big Bang.

  [the] universe came to existence

No such claim is made. All we know is that something must have happened back then. Our current physical models do not accurately represent the extremely early Universe.

  Hence my question is that everything is traveling faster than the speed of light?

The expansion of the Universe is not happening at the same rate as it was back then. It is slow enough that everything within our Hubble sphere is receding at a lower velocity.


How does the Sun's gravity have so much force and pull on the solar system? How does it scale? Watched this video: https://fb.watch/1U0vCFBr0L/
Ok, so the distances are huge. And the sizes are much different. But I wonder how then the sun keeps the planets etc. orbiting it. If we swap out all those objects in the video (soccer ball for the sun, grapes, pin heads, etc.) for equivalently sized balls of iron, equally dense objects, they would have a negligible pull on each other.
A ball of iron the size of a football could never keep a grape 4 football fields away orbiting it. Or could it? Is gravity proportionate? At larger star and planet sizes, it pulls harder?
What is the explanation here?

Many questions can be answered using the vis-viva equation:
$$v^2 = GM\left(\frac{2}{r} -  \frac{1}{a} \right)$$
which gives the velocity of an object in a Keplerian orbit at distance $r$ from a body of mass $M$ and with a semi-major axis $a$. $G$ is the gravitational constant. And for convenience and accuracy, the product $GM$ or standard gravitational parameter for the Sun and for the Earth are 1.327×1020 and 3.986×1014 m3/s2.
For circular orbits set $r=a$ and get
$$v^2 = GM/a.$$
The orbit's circumference $C=2\pi a$ and the time for one orbit (period) is $T=C/v = C=2\pi a / v$ so
$$T^2 = 4 \pi^2 \frac{a^3}{GM}$$
The mass of a sphere is
$$M = \frac{4}{3} \pi R^3 \rho$$
and we're going to keep the density of the Sun fixed at 1408 kg/m3 which is only 41% higher than water! (see At what depth below the Sun's surface does the density reach that of water?) So:
$$T^2 = 4 \pi^2 \frac{a^3}{GM}$$
$$T^2 = \frac{3 \pi}{G \rho} \left(\frac{a}{R}\right)^3$$
or
$$T = \sqrt{\frac{3 \pi}{G \rho}} \ \ \left(\frac{a}{R}\right)^{3/2}$$

punch line: So the period is going to be one year, i.e. about 365 days whether we use the current values for $a$ and $R$ or scale them up or down by any factor!!
In other words, while:

in plain English, in fact a sun-density sized soccer ball will keep a grape-sized object (of the same density as the planet it represents) in the same scale orbit with the same orbital period. This does in fact all scale down.

...is nearly correct. If the Sun were a 22 cm diameter ball with the same 1.4 g/cm^3 average density, and the sesame seed-sized Earth was 47.4 meters away with a diameter of 2 millimeters and the same average density of 5.5 g/cm^2, then it would orbit the soccer ball-sized Sun once a year, unless there were external forces pulling on it from other astronomical object.
Alternatively you could keep the Sun and Earth and all the planets the same sizes and distances but make them a hundred times less dense, and the orbital periods would be $\sqrt{\text{100}} = $ 10 times longer.

This is actually variant of the rule of thumb that the period of a low orbit around a spherical body is just inversely related to the square root of density. So a dust particle orbiting a 1 meter diameter spherical chunk of "average Earth" will orbit in about 90 minutes just like the ISS orbits around the whole Earth in about 90 minutes.
But you can always replace a spherically symmetric mass distribution with a smaller spherically symmetric mass distribution, (even a point).
Not the same, but similar to what's discussed in this answer to Delta-V required for lift-off from a planet/ asteroid


B-V to U-B colour index Is there a formula to convert the B-V index to U-B index or are they totally different and separate observations and can't be convert like you can with Fahrenheit->Celsius?  (Updated)

I know F->C is not astronomy related directly but there is a formula to convert between the two measurements.

Both $B-V$ and $U-B$ are a reflection of how hot the surface of a star is. To first order (inhomogeneous surface temperatures do exist) this has a single value for a given stellar photosphere and so there ought to be a direct relationship between $B-V$ and $U-B$. Here it is, shown below as a graph appropriate for main sequence stars with a composition similar to the Sun. You can reproduce this yourself using data found (for example) in the Handbook of space Astronomy and Astrophysics (pp68-69) compiled by M.V. Zombeck.

You can see that the relationship is complicated; this is because stellar surfaces are not blackbodies and the absorption by various chemical species cuts out chunks of the spectrum, particularly in the blue and ultraviolet. However, given a $B-V$ one could look up the appropriate $U-B$. Note that this is not straightforward the other way around, where there may be two values of $B-V$ possible for a given $U-B$.

In practice things are much more complicated. The plot above assumes the stars are on the main sequence. Stellar atmospheres change a bit as stars evolve and get bigger, with lower surface gravity. Thus changes the relationship a little. The relationship also changes if the composition is not similar to the Sun, especially around the bump at $B-V \simeq 0.4$. Finally, this relationship is intrinsic, it assumes you are not viewing the star through extinction caused by dust. If that is not the case then the whole relationship slides redward in both colours (the vector direction of which is shown  in the plot above for about 2 magnitudes of visible extinction) but also changes shape slightly, depending on how much extinction there is.


2017 Eclipse intersection with I-95 I am excited to see the 2017 eclipse in South Carolina and would like to understand where exactly the eclipse-path intersects I-95, so that I can plan a day-trip.

I have approximated at roughly Santee,SC, but would really like to overlay and pinpoint the eclipse-path intersection with I-95 in Google maps (blue line in .kml file).  Unfortunately, the KML file does not show I-95

Hasn't someone already done this with the web version with Google maps?
Is there a simple process to either embed I-95 in the Google Earth .kml or
Command Google Earth to overlay the blue eclipse path over the internet connected map data?

I was able to plot Santee, SC (on I-95) on the .kml file:

Any edits to sharpen or clarify questions are appreciated.  I pose this question to the astronomy community because I think that someone has already done this.  See you at the eclipse!

NASA has a very good interactive online map which also shows highways.  It looks like I-95 intersects with the path by a wildlife refuge.

A note on planning your trip: there will most likely be a very large number of people who also want to go out to see the eclipse.  So be prepared for lots of traffic.  Also, depending on where you plan on going, you will probably have to leave very early in order to get a good spot.


Does our sun/solar system orbit around any other celestial objects? Does our solar system orbit around another massive celestial object? Like a massive star or black hole for instance? From my understanding our sun rotates around the centre of our galaxy, thus our solar system follows. But I'm curious to know if there is/are other(s) gravitational force(s) that affect our suns? Other than the rotation around the centre of the milky way.

It would make sense to me that the gravity of massive objects would affect our sun, which in turn would affect our solar system, but I cannot find any answers other than ones concerning the "galactic year".

The orbit of the Sun around the Galaxy is quite complicated, because unlike the solar system, the mass is not completely concentrated at the centre. So, in addition to the roughly circular 230 million year orbit in the plane of the Galaxy, there are superimposed motions in and out of the plane and towards and away from the Galactic centre. These roughly sinusoidal additional motions, called epicycles, do not have large amplitudes - a few hundred light years - and take about 70 million and 160 million years respectively.

The Sun does not systematically orbit any other Galactic structures or stars (see the flagged potential duplicates for more details - there is no evidence for any binary companion to the Sun larger than a few Jupiter masses)  and is unlikely to do so in the forseeable future. The space between the stars in our Galaxy is large enough that they are essentially non-interacting.

The Galaxy itself is in motion with respect to the galaxies around it. The nearest tens of galaxies form the local group and probably have complicated orbits within their summed gravitational potential. These orbits cannot be precisely determined, because although we can measure line of sight velocities using the Doppler effect, the tangential motions require extraordinarily accurate position data over many years. These are now becoming available - for instance we know our Galaxy and M31 will collide in around 4 billion years.

Further afield, the local group is part of a larger galaxy aggregate, called the Virgo supercluster, but here, determining an orbit is impossible.

Even further, large galaxy clusters are arranged into a web of larger superclusters, voids and filaments that are all in motion with respect to each other.


What theories are there in explaining the formation of the Pluto-Charon system? After reading the question and answer to Effect of Charon on Pluto, a further question is this: how is the Pluto-Charon relationship explained?

Specifically, what theories are there in how did the Pluto-Charon system was formed?

Additionally, does this explain the presence of other moons (Nix, Hydra etc)?

The favored theory is a giant impact, at the end of the final stage of planetary formation (Canup, 2005). A "giant impact" is a collision with an object of a comparable size of the planet. That is also the most probable scenario for lunar formation (Canup, 2004).


How did Johannes Hevelius' long telescope work? Why all the round holes? The drawing below, found in Wikimedia and at lib.harvard.edu is of a very long tubed aerial telescope. I believe it is taken from his 1673 work Machinae coelestis.

I've always wondered about the purposes of all those (20+) square blocks with round holes, spaced evenly from one end to the other.

I've included two cropped views from Harvard University, Houghton Library, pga_typ_620_73_451_fig_aa (found here) the first of which I've zoomed and enhanced the contrast both to better appreciate the cute bird, and to try to read the annotations; there seems to be a 772 or 722 next to one of the blocks near the top left.

Question: What was the function of all of these blocks? Is it optical, or mechanical, both, neither? When I was young I thought there were lenses in each one, but I'm assuming now that at least the central circular holes are completely open.

I think it is pretty easy to guess either one, but presumably with such detailed annotation there is text to go along with this image somewhere and it is likely that we can know Hevelius's intention. I just don't know where to read the text.

I believe that the text is here but it's 1,000 pages long and in a language that I can't understand.

In general,  the less the support structure weighs, the easier it is to move the whole thing around.  Home-user telescopes usually have a solid body because it's by far the cheapest and simplest way to get a small rigid structure.

Large observatory 'scopes often have a minimal framework supporting the secondary (the far element in a reflector);  here the primary refractor is supported with a similar structure.

Most certainly the support blocks do not contain lenses.  For one thing, just imagine trying to align and focus a 12 or 15-element system!  For another, producing the lenses for telescopes of that size was incredibly tedious back then (and now, really).

There are a couple great chapters in "The Age Of Wonder" by Richard Holmes about a couple of the early astronomers & scope builders in Great Britain for those who are interested.


If gaseous planets have solid core, can they theoretically be considered as rocky planets? I read that gaseous planets such as Jupiter and Saturn, have a solid core deep inside. So, I wondered whats the difference between them and rocky planets apart from the fact that they have - what can supposedly be considered, a very thick atmosphere, and that their atmospheric pressure would probably be too high.

Is there any other difference which I am missing? I mean, if we go by definition, couldn't all their gaseous content, be considered as an atmosphere?

As a general rule, a gas giant planet is large enough that it retains a lot of hydrogen and helium, such as Jupiter and Saturn.  Their chemical makeup is similar to the universe, nearly 90% hydrogen.

Ice giants, like Uranus and Neptune are made up of material more similar to comets, lots of ices like H20, CO2, CH4, NH3 are the primary 4 ices that form into bodies beyond the frost line, but neither Uranus or Neptune have much free hydrogen or helium, either because they're too small or because of where they formed, to close to the sun where it was too hot for them to retain or keep the lightest 2 gases.

As a general rule, in addition to your definition, one possible definition, of a gas giant is an abundance of hydrogen.

When we get a bit of better look at planets in other solar systems, some of the terminology might need to change.    Super-Earths with a lot of water would be similar to hot Neptunes.   Any very massive planet that is able to collect and retain massive amounts of hydrogen will look like a gas giant, whether it has a super-earth origin or not.    Perfect definitions aren't possible.


What known comets are binary? Many asteroids are today known to be binary, to have "moons".

But what comets are known to be binary? Or what fraction of them are theoretically thought to be binary? And if possible, then separate Oort cloud comets from comets with more local asteroid like orbits in this respect.

None. and that doesn't surprise me much.

Comets are thought to be Oort-cloud objects which by interaction either with other Oort-cloud objects or with objects passing the Solar system have been perturbed to venture into the gravitational reach of the outer planets, whence they are
flung into the inner Solar system via close encounters (gravitational slingshots) with one of the outer planets, in particular Jupiter and Saturn. For such a slingshot to be efficient, the encounter must be close when any binary (proto-)comet will most likly be disassociated by the gravitational tides of the sling-shooting planet.


Do other G-type stars follow an 11-year cycle of activity? Our Sun follows an 11-year cycle of low-to-high activity. During high activity, there are much more sunspots and much higher chance of flares and other disruptive behavior.

Can we observe the activity of other stars? I'm particularly interested in other G-types like the Sun. Do they also follow an 11-year cycle?

The answer is yes. Early work showed that just over half of sun-like stars exhibit magnetic activity cycles similar to the Sun. These cycles can be monitored by looking at chromospheric activity, measured in the emission line cores of their Ca II H and K lines; by looking at spot modulation of their light curves; or for a few stars, looking at their coronal X-ray emission over time.

Subsequent work has found cycles in most stars where there is good data (Olah et al. 2016). The cycles vary in length from star-to-star, where they have been found. Some stars show multiple cycle periods (e.g.  Olah et al. 2009). There is obviously an observational bias towards finding shorter periods, but the Sun's cycle is towards the longer period end of what has been found. There seems to be a rough correlation between cycle period and rotational period - the cycle period is 100-300 times as long as the rotation period.

An excellent short review is given by Korhonen (2015).


Implementation of formulae for Meeus for the Moon With Astronomical Algorithms by J. Meeus (1998)
I have obtained the relevant illuminated fraction of the disk and position angle.

But they're numbers (all Greek to me!) I know, Meeus explains that they're ratios of the lengths of the illuminated limb and areas.

Okay, so I have a ratio.  How do I implement that together with the position angle to work out the direction of the illuminated limb and the amount illuminated?

I'm assuming that the ratio 1.0 is totally illuminated (i.e. Full Moon)
and 0.0 is nothing (i.e. New Moon).

and then there is the position angle, I'm assuming that the direction in degrees is starting at 0 pointing upwards.  Is that the crescent shape direction i.e. the horns are pointing away from the directional angle?

I would like to be able plot this so that I can show visually the moon's illuminated fraction and direction for any day in any given year.

As I said, I have the numbers, I'm trying to translate that into a more visually appealling and meaningful result.

Meeus says this is the selenocentric elongation of the Earth from the Sun (Phase Angle) - i.

the formula for illuminated fraction k is:

k = ( 1 + cos i ) / 2

tan i = (R sin l) / (D - (R cos l))

R = distance Earth-Sun
D = distance Earth-Moon
l = geocentric elongation

Meeus calls k the ratio of the lengths BC:AC
the same as the ratio of the areas: NBSC:NASC

Where

Your understanding of the illumination is correct. k is the ratio of the illuminated length BC to the diameter AC. New Moon has an illumination of 0, Full Moon an illumination of 1, and the quarter phase an illumination of 0.5.

The position angle is measured counterclockwise from north (celestial north, along a line of right ascension) to the bright limb C. You need to calculate the parallactic angle, q, which is the angle from "up" to celestial north. From chapter 14 (in the 2009 version)
$$
\tan q = \frac{\sin H}{\tan \phi \; \cos \delta - \sin \delta \; \cos H}
$$
where $\phi$ is the geographical latitude, $\delta$ is the declination of the Moon, and H is the hour angle. The diagram below shows a positive value for both q and PA.


Opening VLA Export Format data on Windows I recently downloaded some data from the NRAO archive.
It came as a file in the "VLA export format" with the extension .xp1. From what I gather I can open this using some software called CASA or a Python library called casatools using the importvla command.
The trouble is that I can't seem to find any way to install CASA on Windows 10. Infact as far as I can tell CASA isn't supported at all for Windows.
How do I open this file on Windows? I just want a big matrix of data that I can process with my own tools and astropy. Is there some reason that the NRAO distributes data using this (as far as I can tell) arcane and undocumented (I couldn't find any documentation after about 1/2 an hour of looking) file format instead of something like NetCDF or hdf5?

The VLA data downloaded from the archive is raw data. It contains the output of the correlator in its own format, data on weather, water vapor in atmosphere, turnings of the antennas, wellbeing of each antenna, all calibrators, information which part of the data might not be healthy and why, comments by staff, if any, who examined the data, etc.
CASA allows the user to run the data through the data reduction pipeline, which is designed to process these outputs and deliver “the product”, i.e. calibrated data.
The product is a large file containing the UV data (Fourier transform data) on calibrators and targets. To get images this data has to be imaged and cleaned. Imaging implies doing the inverse Fourier transform. Cleaning implies application of CLEAN algorithm which takes care of the fact that the telescope beam has non-trivial shape.
Unless one knows everything about the telescope hardware and software, it is impossible to make any use of the raw data from the archive. The purpose of CASA is to eliminate necessity of the telescope’s inner working knowledge as well as necessity to write ones own imaging code.
CASA is essential tool. Try NRAO’s Helpdesk. They might know a workaround the Windows issue.
If you are determined to play with data yourself, ask Helpdesk to point you to the documentation about VLA’s correlator and its output format.


Why do stars become red giants? Disclaimer: I’m not a career astronomer.  I don’t own a telescope.  I have no professional credentials.  But I do find this stuff fascinating, and I consume all the astronomy documentaries I can.

So, I’ve watched lots of documentaries describing stellar evolution.  I understand that below a certain threshold, stellar death does not involve supernovae.  I understand that above that threshold, supernovae may create neutron stars, magnetars, or (if the supernova qualifies as a hypernova) black holes.

However, for a long time, I was curious about why stars below the supernova threshold—like our own Sun—become Red Giants.

From documentaries, I have been instructed that (for stars below the supernova threshold), when the star’s core’s fusion cannot continue…fusion ceases, and the star begins to collapse under gravity.

As gravity crushes the star, I understand that the star heats up as gravity crushes it.  As a result, although the stellar core remains “dead” (no fusion occurs), a “shell” of gas around the stellar core becomes hot enough to begin fusing helium.  Since the fusion occurs as a “shell” around the stellar core, the outward-push from the fusion is what pushes the star’s outer layers further.  The result is that the star grows into a Red Giant.

My question is this:  Why does fusion cease in the core?!  It seems to me that as gravity crushes the star, stellar fusion would reignite in the core itself—not in a sphere around the core.  Why does the stellar core remain “dead” while its “shell” begins fusion???

(This is somewhat simplified but I hope it gets the idea across.)

The reactions stop in the core because it runs out of fuel. During the main sequence, the star is supported by the fusion of hydrogen into helium. Eventually, the hydrogen runs out at the centre, so hydrogen fusion is no longer possible there.

Why doesn't it start fusing helium into carbon right away? That's because the core isn't hot or dense enough yet. Different reactions broadly rely on the presence of different resonant states in the nuclei and, in the case of helium, such a state cannot be reached often enough until the core temperature is about $10^8$ kelvin.

In order to get that hot, the core has to contract and heat up. It eventually does (if the star is massive enough) but it doesn't happen instantaneously. Remember that the gas is still hot and at high pressure, which it exerts on itself and its surroundings.

Meanwhile, at the edge of the core, the star (partly as a result of said contraction) is hot enough to turn hydrogen into helium, so it does so. This is exactly the nuclear-burning shell that distinguishes the internal structure of a red giant.

So maybe think of it this way. Imagine a star at the end of the main sequence. Where is it hot enough to fuse hydrogen into helium? Everywhere up to the edge of the core! Does it fuse in the core? No, because it's out of fuel. So where does it fuse? At the edge of the core, which we recognize as the shell.


How to find period of a pulsar from a given time sampled voltage data? I've a data file with 2 columns, each containing time sampled voltage output (a sequence of real numbers) from a radio telescope, so there is a pair of telescope which are measuring the signal here. The source of the data is a pulsar and I've to find its period and dispersion measure.

I'm an amateur trying to understand what is really happening in the method. The data is time sampled at the Nyquist sampling frequency (2B) and the signal is measured at central frequency f and broadband B.

I studied the cross correlation between the two voltages V1 and V2 from essential radio astronomy, but didn't quite understand the method. Do we have to evaluate the cross correlation function of each reading and then plot it with time to get the pulse profile? Or do we have to take fourier transform of each pair of voltage(how?) and then plot it to get the pulse profile? We find period of a pulsar from its average pulse profile, right?

It is also written that to find pulsar timing, we've to fold data from many pulses modulo instantaneous pulse period (how can we do that if we don't know the pulse period in the first place?). This sentence is everywhere and I'm not able to understand it each time.

I'm confused about the appropriate series of steps to be done here to get the pulsar period and dispersion measure. Could you give me some insight?

Thanks.

To understand the basics of algorithms like that, I find it's often best to look at what amateur practicioners are writing (as opposed to diving directly into scientific papers). The description of the folding algorithm at joataman.net is thorough and pragmatic.

A great in-depth book on the topic is Lorimer/Kramer: Handbook of Pulsar Astronomy, Oxford University Press.

For the actual work, I would suggest the sigproc package by Duncan Lorimer. Its home is at Sourceforge, but there are forks all over the place because the original software is a bit quirky when run on modern Linuxes.


Was the Milky Way ever a quasar? Is there any evidence that the Milky Way could have been a quasar in it's early history? Is it thought that most galaxies come from quasars?

A quasar is simply an active galactic nucleus (AGN) that is viewed from a particular angle; see the picture below, in which quasars are labeled "QSO". This is really a remarkable figure because historically all of the names in the figure were thought to correspond to different types of objects, when really they all refer to the same thing!

Your question really shouldn't be "Was there ever a quasar in the Milky Way?", since the dotted line in the figure would correspond to the Galactic plane and we would not see Sagittarius A* (the Milky Way's super-massive black hole) from the correct angle. A better question might be, "Has Sagittarius (Sgr) A* ever been active?" The answer to that question is yes; according to this page it was probably active (very bright with a jet) about 10,000 years ago. However, at the moment, it isn't really doing anything, since it isn't currently accreting anything (to put it plainly, it isn't eating anything, so it doesn't have enough energy to be active). However, many astronomers (myself included!) are anxiously waiting for a cloud of gas called G2 to fall into Sgr A*. We are hoping that Sgr A* will burp or do something interesting.


Reshaping black holes to "push" the event horizon I watched this simulation made by LIGO (Laser Interferometer Gravitational-Wave Observatory) where 2 black holes merge into one.

https://www.youtube.com/watch?v=I_88S8DWbcU

I was therefore thinking that theoretically nothing can escape a black hole given the features of the space and the black hole remain the same due to the strong gravity.

However, I was thinking if 2 black holes stay together close enough, the "event horizon" would reshape as well.

We all know the gravity force is F=G(Mm/r^2).

Given a point next to the event horizon but "inside" the black hole the gravity there will be so strong that the Escape velocity will be higher than C (speed of light).

If we place another massive object (like another blackhole in the proximities of this point, but far enough, the gravity there will be counteracted and therefore this point will be able to escape its black hole, reshaping the event horizon and making it smaller. (We will have created a "concavity" in the event horizon)

My question is:

Given a hypothetical scenario where we create thousands of micro blacholes orbiting the main blackhole, in a gravitational equlibrium. Would we be able to reshape and "push" the event horizon strong enough that objects inside could now use the chance and escape?

Could we even pull the matter of the black hole and make it less dense along the time even transforming it into a start destroying the singularity?

The only places around one black hole where gravity would be strong enough to pull an object out from inside the horizon of another black hole are all inside the horizon of the first hole.

Remember that at minimum an object just inside the horizon of a black hole is traveling toward the center at the speed of light (as its speed would be measured from outside the hole, if we could see it) and getting faster. For a second black hole to catch up to the object to pull it out, it would itself have to be moving faster than the object - faster than light - which is obviously impossible but anyway would simply merge the two black holes, without rescuing anything.


Does the Milky Way orbit around anything? We know most of the objects in the Universe have a spherical or elliptical shape. The object which has less mass and gravitational pull orbits around the nearest object with more mass and gravitational pull. For example:

Moon orbits around Earth
Earth orbits around Sun
Sun orbits around Sagittarius A* which is the center of Milky Way.

Thus, is the Milky Way orbiting around some object or perhaps Black Hole?

I know that the Milky Way is going towards Andromeda as they are attracting each other and they will collide with each other after 3 billion years to 6 billion years. But it is possible that the Milky way is orbiting around some object at the same time? Perhaps both galaxies are present in a group of galaxies which is orbiting around some object.

If the Milky Way is not orbiting around some object then is there any proof found by the scientists for that?

The object which has less mass and gravitational pull orbits around the nearest object with more mass and gravitational pull.

Actually, both the heavier and the lighter object orbit around their common center of mass. It's just that the heavier object doesn't move much (has a tiny orbit), while the lighter object moves a lot (has a wide orbit).

E.g. our Sun actually orbits the center of mass of the whole solar system, but that motion is tiny, it barely budges.

In the case of a double star, where both partners have about the same mass, you can clearly see how both are making similar orbits around their common mass center.

  Sun Orbits Around Sagittarius A* which us center of Milky Way.

With galaxies, including ours, it's a little different.

There is no super-heavy thing at the center, around which everything else is orbiting. Not even the very large black hole at the center of our galaxy is heavy enough for that.

Rather, galaxies are clumps of matter that create a common gravitational field. Stars, and everything else, are trapped in this common field and orbit around the common center of mass.

  So the question is that is that is Milky Way is orbiting around some object or perhaps Black Hole.

Same idea. There is no single point-object nearby massive enough for our galaxy to "orbit" around it.

Our galaxy, along with Andromeda, and a handful of other galaxies, are bound together in what is known as the Local Group. Each galaxy is moving within the common gravitational field of the whole group. The Local Group has a diameter of about 10 million light-years.

The Local Group is part of a larger structure, the Virgo Supercluster, which is about 100 million light-years in diameter and has at least 100 galaxies. However, the Virgo Supercluster is more "loose" - it is not gravitationally bound together.


Converting orbital elements from equatorial reference frame to ecliptic I have a bunch of orbital elements (inclination, longitude of the ascending node, mean anomaly, argument of periapsis) that are all referenced from an equatorial reference frame. But the program I am trying to use requires the orbital elements in an ecliptic frame. Is there an easy/convenient way to convert between the two frames?
I also have the RA and Dec for the orbit pole, also in the equatorial frame. I am aware of some rotation matrices that allow me to convert between equatorial and ecliptic frames—should I just apply the matrix to the RA and Dec of the orbit pole? Is there a way to turn that into orbital elements?
Or should I turn the equatorial elements into a state vector (Cartesian position and velocity) and then apply the matrix to that, and then convert the ecliptic vector into orbital elements?
What is the easiest way to go about this?

The orbital elements $\omega$, $i$ and $\Omega$ are Euler angles in the sequence $(3, 1, 3)$.
The easiest way to transform them is to convert them to a representation that's easier to manipulate, e.g. unit quaternions or a matrix, apply the required transformation then convert back to Euler angles.
A useful guide to converting between the systems is given by James Diebel (2006) "Representing Attitude: Euler Angles, Unit Quaternions, and Rotation Vectors".
For example, using the formulae in §8.10, the rotation matrix for the orbital elements is given by:
$$R_{313}(\Omega,i,\omega)=R_3(\Omega) R_1(i) R_3(\omega) =
\begin{bmatrix}
c_\Omega c_\omega - s_\Omega c_i s_\omega & c_\Omega s_\omega + s_\Omega c_i c_\omega & s_\Omega s_i \\
-s_\Omega c_\omega - c_\Omega c_i s_\omega & -s_\Omega s_\omega + c_\Omega c_i c_\omega & c_\Omega s_i \\
s_i s_\omega & -s_i c_\omega & c_i
\end{bmatrix}$$
Where $s_\omega = \sin \omega$ and $c_\omega = \cos \omega$, etc.
You can then apply the transformation from equatorial to ecliptic coordinates (or whatever other transformation you want to do) to this matrix. The equatorial and ecliptic coordinates are related by a rotation about the x-axis by $\epsilon$, the obliquity of the ecliptic, so combine the rotation matrix using matrix multiplication. Depending on the precise details of the coordinate systems of the input elements and the ones your software expects, you may also need to invert the axes, which is another matrix multiplication.
This yields a new matrix $R'$ with elements $r_{ij}'$. You then convert this back to Euler angles:
$$
\begin{bmatrix}
\Omega' \\ i' \\ \omega'
\end{bmatrix}
=
\begin{bmatrix}
\operatorname{arctan2}(r_{13}', r_{23}') \\
\arccos(r_{33}') \\
\operatorname{arctan2}(r_{31}', -r_{32}')
\end{bmatrix}
$$
If $i' \in \{0, \pi\}$ then $r_{13}' = r_{23}' = r_{31}' = r_{32}' = 0$ and the above formula for $\Omega'$ and $\omega'$ no longer works as it requires evaluating $\operatorname{arctan2}(0, 0)$. In this case, the angles $\Omega'$ and $\omega'$ are in the same plane and therefore cannot be uniquely separated. Choosing $\omega' = 0$, the transformation from the rotation matrix to Euler angles becomes:
$$
\begin{bmatrix}
\Omega' \\ i' \\ \omega'
\end{bmatrix}
=
\begin{bmatrix}
\operatorname{arctan2}(-r_{21}', r_{11}') \\
\arccos(r_{33}') \\
0
\end{bmatrix}
$$

Note that the two-argument arctangent function in the above uses the convention $\operatorname{arctan2}(y, x)$ giving the angle between the positive x-axis and the point $(x, y)$. Some systems such as Microsoft Excel, which provides the function as ATAN2(x, y) have the arguments reversed, consult the manual for the software you are using.
Depending on whether you pre- or post-multiply the rotation matrix with a vector, the rotation matrix elements may be transposed from the ones given above, in which case you need to switch the element indices in the reverse transformation.


If I know apparent planetary positions how do I calculate the date? I have planetary position in retrograde  how do I calculate the date?
For eg, I know Mars was in retrograde in Scorpion (Sidereal 210 to 240 deg), how do I calculate the date and year of this?
As of now the only other info I have is that just days before this was a lunar eclipse where Sun was in Libra.
Just a year or two before this
Mars was in retrograde in first half of Leo and at this time Jupiter was placed at Capricorn. These are all sidereal observations.
P.S: This is regarding an ancient scripture citation where it was said Mars was in retrograde in Scorpion, current proposed dates range from 500 BC to 6000 BC. This is all the information I have, that Mars was is in retrograde in sidereal scorpion and I want to find possible date of this happening. The iterative approach would take too long, because the span is about thousands of years

Mars's ~72 day retrograde loop occurs every ~2.1 years and intersects a given constellation once or twice every 15 or 17 years.
The more information you have about the positions of other planets, e.g. Jupiter or Saturn, the more you can narrow the set of possible dates.


Is there a standard mapping of symbols to terms for celestial and orbital mechanics Is there a standard mapping of symbols to terms for celestial and orbital mechanics (and physics in general)?

For example here are some ambiguities

Angular Momentum: L, H, J, $\Gamma$,

Orbital period: T or P,

M: mean anomaly or mass of the Sun,

E: energy or eccentric anomaly,

p: linear momentum or length of semi-latus rectum,

a: acceleration or length of semi-major axis,

e: eccentricity or base of natural logarithms,

There are a lot of instances (such as those you pointed out) in all of science and mathematics where a symbol can mean two different things. In mathematics, there's actually a good deal of this (and the article says that the list is incomplete!). Many are based on letters, although they may have deviated from the original shape in order to make a distinction regarding just what the heck the thing actually means.

Eventually, you run out of letters. You can opt to use symbols from another language, the first choice for many being Greek. That roughly doubles the amount of letter-based symbols available to you. But there are still a lot of duplicates, as you can see by browsing through the lists here and here.

To my knowledge, this is the most comprehensive list easily available. And yes, there are loads of duplicates. But the great thing is that you almost certainly won't have to use the same variable to mean multiple things in one equation, or set of equations. For example, I doubt you'll need to use $V$ for voltage, volume and shear force all in the same equation!


Solar Spectral Types and Dwarf Stars On the following page,

http://www.uni.edu/morgans/astro/course/Notes/section2/spectraltemps.html

It mentions Main, Giants and SuperGiants.

However for Dwarf Stars (e.g. D, Sd) , do I apply Main Sequence to the solar type?

For other types, IV, II, do I apply Main Sequence / Subgiant / Giant to the type?

If you look at the linked webpage, you'll see it mentions Types (V,III,I) but for instance where would Procyon (F5IV-V), Sargas (F1II) fit or should there be a different table for types IV, II?

Would groups (D, sd, IV,m II) have a different table?  Rigil Kentaurus (G2V) would clear fit into Main Sequence.

The modern system of stellar classification is 2-dimensional, with one axis conveying spectral class, and the other conveying luminosity class. The spectral types are as follows:

O-type, "blue", $\geq 30000$ $\mathrm{K}$
B-type, "blue-white", $10000-30000$ $\mathrm{K}$
A-type, "white", $7500-10000$ $\mathrm{K}$
F-type, "yellow-white", $6000-7500$ $\mathrm{K}$
G-type, "yellow", $5200-6000$ $\mathrm{K}$
K-type, "orange", $3700-5200$ $\mathrm{K}$
M-type, "red", $2400-3700$ $\mathrm{K}$

There are also some extended spectral types, including D for white dwarfs, C for carbon stars, and several more for brown drawfs. The above are the predominant.

The magnitude axis consists of the following groupings:

0-type, "hypergiants"
I-type, "supergiants"
II-type, "bright giants"
III-type, "giants"
IV-type, "subgiants"
V-type, "dwarfs" (these are the main-sequence stars)
VI-type, "subdwarfs"
VII-type, "white dwarfs"

When these systems are used, a spectral class is combined with a number ranging from 0-9 which symbolizes the location between adjacent spectral classes, and the luminosity class is then appended.

For example, the Sun is a type G2V star. This means it is a G-type star on the main sequence. It could then informally also be referred to as a "yellow dwarf".

The system is more complex than this, and there are many additional suffixes and other symbols to further explain the nuances of individual stars. This does cover the basics quite well, however.


What are possible methods of discerning a starburst AGN composite galaxy? I'm currently working on an independent project that involves discerning starburst galaxies that are themselves within active galaxies.

I would assume radio observations would be best to discern star formation rates, and X-ray to understand the AGN in greater detail. However, assuming we have only optical data could I use the relative strength of the OII flux to the OIII flux to better understand the star formation rate of an active galaxy? Spectroscopic surveys of distant galaxies, particularly for redshifts between $z$ ≈ $0.4$ to $1$, routinely use [O II] $λ3727$, a prominent nebular emission line in H II regions, to estimate SFRs. If (high-ionisation) AGNs experience substantial levels of ongoing star formation, the integrated contribution from H II regions will boost the strength of the [O II] line (compared to, say, [O III] $λ5007$, which can be largely ascribed to the AGN itself).

For starbursts, their star formation rates are best estimated by a combination of their ultraviolet and infrared emission. [O II] and [O III] as well as Hydrogen recombination lines can be used to estimate star formation, but like the UV, these suffer from extinction.


What's the temperature of Pluto's core? The surface is freezing-nitrogen cold, but it's typical for the temperature to increase towards the core. What's the temperature of the core of Pluto? Is the planet icy and rocky all the way through? I wonder at the possibility if that frozen ice ball could have any internal heat at all?

TL;DR: 1000 K (according to differentiated model of Pluto)

According to the density value of Pluto, astronomers proposed three types of structural models:

Undifferentiated or "cold" model: rocks mixed with water-ice
Differentiated or "hot" model: rocky core and water-ice mantle
Rocky core only (temperature high enough to boil off water-ice.

There are two types of rocks to consider i.e. CI rock- carbonaceous chondrite($\rho$ = 2.66g/cm3) for undifferentiated models which is $\approx$13% by wt. eposomite(MgSO4) and Prinn-Frigley rock($\rho$ = 3.262 g/cm3) for differentiated models which is sulfide-rich rocks like troilite(FeS).
McKinnon and Parmentier(1986); McKinnon and Mueller(1988) correctly predicted the structural model of Pluto to be differentiated one. During initial formation of Pluto, radiogenic heating was the only source of heat, hot enough to melt the outside part of the undifferentiated core consisting of hydrated CI-rocks to form liquid matter of what is known today as water-ice in the mantle (water-ice melting temperature was 251 K at 0.21 GPa). During that time, 80% of core was at above 500 K . The anhydrous silicate core at the center was still rigid resisting the outside flow (the temperature of core was 500 K). Convection through heating allowed steady-state heat flow to form water-ice in the mantle, some of which erupted into the surface. Although the gravitational potential energy was modest, the process become self-accelerating.
This process was also accelerated by the "accretion process" due to a large-body impact that resulted in the formation of Charon. This collision heated up the core by 100 K and also resulted in several differentiation of Pluto that made the surface level ice layer mixed with organic compounds.
Present day scenario: Pluto is a rock-rich world with rock mass fraction $\gtrsim$ 60%. Core temperature is 1000 K, surrounded by 250-300 km thick mantle shell of predominately water-ice and mixed liquid rock. Surface temperature is 40 K. Core pressure is 1.1-1.4 GPa. Rock fraction (rock/rock+H2O+ice) for differentiated Pluto is 0.69-0.79(mainly PF-rock). There is also an extra thermal boundary layer between the ice-mantle and the ice-rock layer just outside the core. There might also be an organic layer in between the core and mantle but it hasn't been proven yet. The water-ice mantle is differentiated into several layers based on its transitions. You can check the details and calculation in ref.2

Fig. 1: Temperature of Pluto at several Pressure value

Fig. 2: Temperature of the mantle and outside layer

Fig.3: Differentiated model of Pluto
References:

http://articles.adsabs.harvard.edu//full/1988LPI....19..764M/0000764.000.html
Pluto and Charon, Alan Stern, David James Tholen, University of Arizona Press, 1997
https://arxiv.org/ftp/arxiv/papers/1510/1510.06604.pdf (This paper discusses the possibility that Pluto might have an iron core)


Assuming a hypothetical system without gas but only solid rocks, how big of a planet can form through the coalescing of these rocks? I have heard that gas giants are primarily huge solid bodies like regular rocky planets that exponentially gained more and more gas in their atmosphere through their increase in mass which  they use to  pick up even more mass ,gaining even more gas, and practically any planet we know today above a certain size threshold is a gas giant
That being said, assuming an area in space with very little or no gas but a bountiful supply of rocks, how big can a solid planet consisting of these coalesced rocks get?

That very much depends on the structure of the solids, i.e. whether they exist in the form of small dust, or ready-to-smash planetesimals.
In any case, the available median dust mass for planet formation is about 158 $\rm m_{earth}$ (see Tychoniec et al., (2020)). If you leave all this mass as dust, without any gas interactions, the dust will not coalesce into planetesimals, as a hydrodynamic instability is required to jump over the meter-sized barrier (Johansen et al., (2014)).
However, if you somehow allow all this dust to be converted into planetesimals, then the size of the planet you can form will be given by how narrow you can pack the planetesimals. The absolute upper mass of the formed planet will be given by the 158 $\rm m_{earth}$, but realistically , that is going to be lower, as planetesimals and collision ejecta are lost during the smashing phase of planet formation.
As you were asking about the size of this hypothetical planet, is we assume no compressional effects and hence the same mean density as earth, you would get a solid ball of the size of $\rm 158^{1/3} \; r_{earth} \approx 5.4\; r_{earth}$.


What happens when a star gets hit by a supernova? This may sound like a strange question, but say that two stars are relatively next to one another in a binary star system, what would happen to one of the stars if the other went supernova? Would it explode as well, or would it survive the blast?

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

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


Is radar interferometry used, or feasible, for ground based astronomy? I've heard of "interferometric synthetic aperture radar" being used for Earth observation from satellites and airplanes. Is active radar used interferometrically for astronomical purposes too? Such as characterizing asteroids or space debris, or identifying volatiles in the polar craters of Mercury.

Unfortunately for this idea, Arecibo and FASTare located 173 longitudinal degrees apart and have limited fields of view, so they can't observe the same object simultaneously.

For obvious reasons you cannot use active radar to observe anything much further away than the Moon. There are plenty of phased-array microwave/radar passive sensing systems that map the energy in those wavelength bands that are emitted from stars, quasars and various other "hot" astronomical bodies.

To answer one of your examples, measurement of volatiles is usually done by measuring absorption spectra; which is tough to do with Mercury since we can't look at the solar-illuminated side (with special exceptions for certain short-lived orbital positions). Basically, that's why we send off probes to get close to various planets and asteroids, so that we can collect high-resolution visible and microwave images and, in some cases, do some active illumination (radar).  I am not aware of any interferometric active radar on such probes, but someone more knowledgeable may well correct me.


Math behind a uv plot in interferometry? I've seen standard uv plots used discussions of interferometric array designs regularly, and I have a vague idea that each arc may correspond to a pair of elements within the array, and the coordinates $u, v$ in the plot may be something like the path length difference between the source and the two elements projected into some cartesian coordinates, and that the length of the arcs corresponds to the total duration of the observation, but it's all still fuzzy.

Since the elements may be on topography or be so far apart (as in the example below) that the curvature of the Earth is even larger, the formal definition of the $uv$ plane should take this into consideration.

There are several more examples of uv plots and comparisons to point spread functions in the SlidePlayer presentation Array Design and Simulations.

Question: What is the math behind the construction of $uv$ plots in radio astronomy? A source that explains it and shows how one can construct them would be extremely helpful, as will any intuitive hints.

below: From the recent ArXiv preprint Detection of Intrinsic Source Structure at ~3 Schwarzschild Radii with Millimeter-VLBI Observations of Sagittarius A* (also ApJ Open Access.

below: From the 2016 ArXiv preprint High-Angular-Resolution and High-Sensitivity Science Enabled by Beamformed ALMA

Rick Perley gives a lecture about interferometry to a biannual NRAO Synthesis Imaging Workshop.
Part 2 of his 2016 talk discusses u,v plane coverage.
The 2016 IRAM Millimeter Interferometry School lectures by J. Pety and F. Gueth may also help.

Let $\hat{u}, \hat{v}, \hat{w}$ be orthogonal unit vectors with $\hat{w}$ pointing from Earth to the radio source, $\hat{v}$ oriented northward, and $\hat{u}$ oriented eastward.
Let $B_{ij}$ be the baseline vector between antennas $i$ and $j$ in the array.
Then $(B_{ij} \cdot \hat{u}, B_{ij} \cdot \hat{v})$
and $(-B_{ij} \cdot \hat{u}, -B_{ij} \cdot \hat{v})$
are points in the u,v plot, and
$B_{ij} \cdot \hat{w} / c$ is a time offset between the signals at $i$ and $j$.

If both antennas observe the source for several hours, $B_{ij}$ rotates with the Earth, and the two points in the u,v plot grow into elliptical arcs.
The shape of those ellipses depends on the declination of the source, so
the same array will have different u,v coverage for sources at different declinations.

The interfering signals from the array are combined in a visibility function $V(u, v)$.
The image of the source is an inverse Fourier transform of $V$, typically deconvolved using an algorithm called CLEAN.
To minimize image artifacts, a diverse set of baseline lengths and orientations, without large gaps in u,v coverage, is desirable.


What is the range of the zenith angle between velocity and position vectors? I'm coding on a space simulator and I want to draw a predictable orbit for a satellite. I have read about orbits on http://www.braeunig.us/space/ and under the section "Launch of a Space Vehicle" they describe how I can obtain the orbital elements like eccentricity, true anomaly etc. given that you have the position vector, velocity vector and the zenith angle between them.

I have two questions that I need to understand:

What is the range of the zenith angle (0-180 degrees or 0-360 degrees etc.) ?
Can the angle tell which "half side" of the orbit the satellite is? I.e. if it's onwards to periapsis or apoapsis?

Partial answer:
What Braeunig calls gamma or zenith angle is actually beta!

...and γ is the angle between the position and the velocity vectors, called the zenith angle (see Figure 4.7))

According to @TomSpilker's answer

This is a problem that has plagued groups of people very knowledgeable about orbital dynamics but who learned using different textbooks: there are two different definitions of "flight path angle"!!
In addition to $\gamma$, the angle between the tangential direction and the velocity vector, there is $\beta$, the angle between the radial direction and the velocity vector. People often say "flight path angle" without saying which definition they're using. Confusing! (I just noticed that the diagram in Julio's answer also shows $\beta$)
If you work with $\beta$ instead of $\gamma$, $\beta$ is given by
$$\arccos\left(\frac{\mathbf{r \centerdot v}}{|\mathbf{r}| \ |\mathbf{v}|} \right) \tag{1} $$

So Braeunig's gamma is NASA's beta. It's positive from peri to apo, and negative from apo to peri. Because I'm not smart enough to do it with math, I'll do it with Python.
I'm doing this in dimensionless units, so the period is 2π and the semi-major axis is unity.  You can see beta's behavior over one orbit, positive moving towards apoapsis, then negative moving towards periapsis.

def deriv (X, t):
    x, v = X.reshape(2, -1)
    acc = -x * ((x**2).sum())**-1.5
    return np.hstack((v, acc))

import numpy as np
import matplotlib.pyplot as plt
from scipy.integrate import odeint as ODEint

halfpi, pi, twopi = [f*np.pi for f in (0.5, 1, 2)]
degs, rads = 180/pi, pi/180

time = np.linspace(0, twopi, 361)

r0 = 1.8
v0 = np.sqrt(2./r0 - 1)  # vis-viva

X0 = np.hstack([r0, 0, 0, v0])

answer, info = ODEint(deriv, X0, time, full_output=True)

x,    v      = answer.T.reshape(2, 2, -1)
rn = x/np.sqrt((x**2).sum(axis=0))
vn = v/np.sqrt((v**2).sum(axis=0))

x, y, vx, vy = answer.T

beta = np.arccos((rn*vn).sum(axis=0))

if True:
    plt.figure()
    plt.plot(x, y)
    plt.plot([0], [0], 'ok', markersize=8)
    x0, y0 = 0.5*(x.min() + x.max()), 0.5*(y.min() + y.max())
    hwx, hwy = (np.abs(x-x0)).max(), (np.abs(y-y0)).max()
    hw = 1.05*max(hwx, hwy)
    plt.xlim(x0-hw, x0+hw)
    plt.ylim(y0-hw, y0+hw)
    plt.show()

if True:
    plt.figure()
    plt.subplot(3, 1, 1)
    plt.plot(time, x)
    plt.plot(time, y)
    plt.plot(time, np.zeros_like(time), '-k')
    plt.title('x and y vs time')
    plt.subplot(3, 1, 2)
    plt.plot(time, vx)
    plt.plot(time, vy)
    plt.plot(time, np.zeros_like(time), '-k')
    plt.title('vx and vy vs time')
    plt.subplot(3, 1, 3)
    plt.plot(time, beta)
    plt.plot(time, halfpi*np.ones_like(time), '-k')
    plt.title('beta vs time')
    plt.show()


Do we see planet venus everyday? Do we see planet Venus every day? (Just before sun sets or Just after sun sets) throughout year? or some days we don't see it?

No. Some days Venus is an evening star, visible for up to a few hours after the Sun sets. Some days it is a morning star, visible for a few hours before the Sun rises. Some days it is so close to the Sun in the sky that we can't see it at all (or even actually behind the Sun).

Venus can also be seen in daylight, if it is above the horizon, not too close to the Sun and the sky is clear. You need to know where to look for it, though, it's only a little brighter than the sky.


Can there be an object with planetary discriminant between Ceres and Neptune? The planetary discriminant is a measure of how dominant a body is within its region of the solar system. For (true) planets, it is $>10000$ and for dwarf planets it is $<1$. (See this answer to "How many planets are there in this solar system?" which explains the related details.)

Is it possible, given what has already been observed about our solar system, for there to be an object with  "planetary discriminant" µ between that of Ceres and Neptune?   In other words, could the clear cut line between planets and dwarf planets get fuzzier?

Obviously it would have to be a massive body pretty far out.  Is there a region in the solar system which could be devoid enough of many small objects, which also could be home to a large enough body that we also haven't observed yet?   How does the math work out?

I know this is an old question, but recent developments in astronomy have caused this to become relevant again. Early in 2016 there was a paper published by Konstantin Batygin and Michael E. Brown (here) which indicates the possible existence of a new planet, dubbed "Planet 9". Subsequent work by other authors have lent more evidence to its possible existence, e.g., a paper by Fienga et al. (here) which concludes that including Planet 9 in orbital dynamic models decreases residuals.

This being said, the original paper specifically states

  ...our calculations suggest that a perturber on a $a'\sim700\:AU$, $e'\sim0.6$ orbit would have to be somewhat more massive (e.g. a factor of a few) than $m'=10\:M_\oplus$ to produce the desired effect.

Such a planet is certainly massive enough to have cleared out its orbit under normal circumstances. For reference, Neptune has a mass of $17\:M_\oplus$ while Uranus has a mass of $14.5\:M_\oplus$. The special feature of this planet is that it is located $700\:AU$ distant from the Sun which is in a very ambiguous and unknown region of our solar system. It is not clear at all how much, if any, debris in the form of comets resides out there. There are models suggesting a disk (referred to as the Hills Cloud) exists at those distances, implying that this new planet could reside inside an extended cometary disk. Of course this is only speculation, but it is a distinct possibility.

If the above-mentioned disk exists, we have to consider a few scenarios.

Planet 9 was "inserted" into this disk by being expelled from the inner solar system early on in its formation. This is a possible cause for its current position and suggests one of two outcomes.

It has existed here long enough that it has cleared out its orbit, thus making its planetary discriminant well within the "planetary" side of the definition.
It has either not existed in the disk long enough to clear the orbit, or has achieved some sort of equilibrium such that it can't now clear the orbit (considering most of the clearing occurs during formation). Here is where you may run into an ambiguous case. For all intents and purposes we may consider this a planet, but there remains the possibility that, despite it being more massive than the Earth, it could formally be considered a dwarf planet or else exist in this fuzzy region as you suggest.

Planet 9 formed at its current orbit (plus or minus some migration).
In this scenario I think it is clear that it would have cleared its
orbit and resoundingly be considered a true planet.

Now these are all wild suppositions. I don't have any of the math to back it up, and I'm not sure the numbers to plug into the various equations are well known, or even exist. I'm merely remarking on the possibility that the situation you asked about now seems like it could exist.


Is the Milky Way seen as how it looked millions of years ago? The distant an interstellar object is, the more time is required for light to travel, which is known, measured as light years. So are the farthest stars and gas clusters in the Milky Way seen as how they looked millions of years ago & not how as they look today?

Most of the matter in the Milky Way is contained within a radius of 100,000 light-years from the galactic center. This encompasses the disk and most of the halo, including halo stars and globular clusters. It also includes quite a few (though not all) of the Milky Way's satellite galaxies. This is rather large - the Sun is only about 25,000 light-years from the galactic center - but not truly inter-galactic scales. Andromeda, the nearest large spiral galaxy, is 2.5 million light-years away, and so we see it as it was 2.5 million years ago. However, we see most of the objects in the Milky Way as they were no more than 100-200,000 years ago, usually much closer to the present day.

It's true that some galaxies are a couple million light-years across, but they are quite rare. In fact, the largest known spiral galaxy, Malin 1, is a mere 650,000 light-years in diameter. I should make it clear, though, that most of these galaxies are large because they have extremely large halos, consisting of gas, pockets of stars, and dark matter. Most matter is not contained in the halo. Additionally, it is not always clear where a galaxy ends and where the intergalactic medium begins.


Can all areas on Earth experience a total solar eclipse? A solar eclipse only puts part of Earth at a time in shadow, so I'm not talking about a solar eclipse that casts a shadow over all of Earth.

What I mean is: are there any areas on Earth where, no matter how Sol, Earth and Luna are aligned, a total solar eclipse is impossible? Can the penumbra cast a shade on any part of Earth, or are there certain areas where a penumbra due to a solar eclipse cannot occur?

Yes, total Solar eclipses can occur anywhere. On any given place, they occur roughly every 400 years, on average. If you take a look on the picture below (from Wikipedia), you'll see that within the last one thousand year, almost every place on Earth has experienced a total Solar eclipse. The places that haven't experienced one have no special latitude or longitude, so it's just a matter of coincidence.


Effect of particulates on the visibility of stars? I am looking for a (mathematical) relationship - either empirical or theoretical - which quantifies how the visibility of celestrial objects decreases with increasing amount of particulates in the air. I would be already happy if somebody knows such a relationship for the atmospheric light attenuation effects of black carbon or soot. Those two are actually not the same as Alfred Wiedensohler from the Leibniz Institute for Tropospheric Research points out in a presentation on the topic:

Soot is carbon particles resulting from the incomplete combustion of hydrocarbons. Soot contains polycyclic aromatic hydrocarbons (PAHs) and heavy metals. Black carbon (BC) itself is neither a toxic nor a carcinogen.

What I expect (but I am happy to be proven wrong) is that I am actually after the three constants in the following expression:
$$ f_{\rm attenuation}(d) =  \frac{1}{k_c + k_l \cdot d + k_q \cdot d^2}$$
In this formula, $d$ is the distance between the light source and the observers, so in our case it would be the height of Earth's atmosphere (if we assume homogenous distribution of soot/ BC over the air column).
$k_c$ is called the constant attenuation factor, $k_l$ is the linear attenuation factor, and $k_q$ stands for quadratic attenuation factor.
Such an expression would already be very helpful, but if there is science on the overall effect of particulates, that would be much cooler. I am dreaming of having a formula for visible magnitude of stars with naked eye in dependence of particulate concentration (for ideal, pitch-black nights without clouds).
Background
My question is inspired by What effect do aircraft have on night-time visibility? which does not have an answer yet.
I am aware that in the physical oceanography and in marine biology, there are extensive studies on light attenuation due to BC or other dissolved substances in water, and most search results are on that.
References and related questions

C.Liousse, H.Cachier, S.G.Jennings: Optical and thermal measurements of black carbon aerosol content in different environments: Variation of the specific attenuation cross-section (1993)
H. Horovath: Atmospheric light absorption — A review
(1993)
ESA on Desert dust plume over the Atlantic observed by Aeolus and Sentinel-5P

Aerosol physics is actually an old topic - yet still a very vibrant one.
One main factor on the extinction (which sums up scattering and absorption) other than the distance (which is about constant when looking at a a particular zenith distance) is the wavelength $\lambda$ you look at compared to the typical aerosol particle radius $r$ as its ratio affects the scattering regime we look at ($\lambda \gg r, \lambda \approx r, \lambda \ll r$). There's multiple papers on the aerosol influence on observations, like this (Stubbs et al) for PanSTARRS or Patat et al (2018) on extinction over Paranal which also looks at the temporal variability.
Looking at the typical aerosol concentrations (e.g. here by DWD), it seems that the smaller particles are more abundant. Thus we are in the Rayleigh or Mie scattering regimes where the extinction is due to scattering and does not depend much on the actual particle properties other than their size. The actual absorption only becomes important for the larger particles which are the least abundant.
All in all this might only be half an answer, but too long for a comment and maybe gives you leads to follow-up on. If you get a hand on Friedlander's "smoke, dust and haze", it might be a good read, too (especially chapters 5 and 13).


What's the useful magnification for a scope with 70mm objective lens? Some of the spotting scopes which are e.g. 70mm claim to magnify terrestrial objects at 75x. But yet they cannot produce sharp and clear images at even 30x-35x.

While 90mm scopes can do a better job than them but still they stop  producing clear view at higher magnification.

So, I want to know is there a formula from which I can estimate (if not accurately tell) their useful magnification at which they are able to focus objects?

I came across this:

Magnification = Focal length / Eye piece focal length

But not sure if that works for what I am asking.

I make telescopes and telescope optics (mirrors). The question is actually pretty complex. I get it a lot, because most people believe magnification is the end-all-be-all of telescopes, but reality is a little different. Here's what I mean by that:

You are correct that the actual magnification of the instrument (how much bigger the image is, compared to the object) is given by the formula:

M = F / f

M = magnification
F = focal length of the objective
f = focal length of the ocular (eyepiece)

The formula seems to suggest that you could get near-infinite magnification from any telescope by just reducing f, the focal length of the ocular. And it's technically true.

But the question is - how much can I increase magnification before it's too much to be useful?

When you zoom into an image with Photoshop, at some point you start seeing the pixels; any zooming past that point does not reveal more details. Telescopes do not have pixels, but the image the objective provides does not have an arbitrary amount of details - just like with Photoshop, arbitrarily increasing magnification with a telescope past some point will only give you a messy blur (not pixelated, just blurry).

So how much is too much? It depends.

If you had perfect optics, used in perfect conditions, then the maximum useful magnification would be:

Mu = 2 * D

Mu = maximum useful magnification
D = diameter of the objective in mm (a.k.a. aperture)

So for an instrument with a 75 mm aperture, the maximum useful magnification is around 150x. If you increase magnification past this point, the image is bigger, but it also gets progressively more blurry, so no additional details are revealed. But those are ideal numbers.

In practice, optical elements (lenses, mirrors) are not perfect. Only the most expensive instruments have optics that are anywhere near perfection. Imperfections add to the blur.

Even with perfect optics, if the telescope is not perfectly collimated (the lenses and mirrors are not perfectly aligned), that reduces performance too. Most refractors are collimated from factory and need no adjustment, but some cheap refractors may not be perfectly collimated. Other instruments such as dobsonian reflectors need periodic tuning to redo their collimation. When the instrument is miscollimated, the maximum useful magnification is reduced.

The atmosphere is turbulent. This is called "seeing" in astronomy. When seeing is bad, extra blur is added to the image, and you have to dial down the magnification.

Finally, magnification is not the ultimate parameter of a telescope. When you buy a car you don't ask the dealer "what's the fastest car you have?" You adjust the speed of the car to the conditions: drive at one speed on the freeway, at a different speed in a residential neighborhood, and at yet another speed in the parking lot at the local grocery store. Same with telescopes.

Some objects require a lot of magnification - things like Mars at opposition, or certain double stars, or small craters on the Moon, these all benefit from the highest magnification you could get. But objects such as Jupiter are better seen at more like a medium or medium-high magnification (Jupiter is low contrast so too much magnification washes out detail in most cases - but not always). And then there are nebulae and clusters which are better seen at low magnification. Horses for courses.

So what's the actual answer?

Well, you have the upper limit set in stone: 2 * D, or 150x for a 75 mm scope. But that's more like a theoretical bound. In practice the actual value will rarely be equal to that, usually will be a little below it, and sometimes much below it.

And then you have to choose your magnification based on the actual object you're observing: high, medium, or low.

This is why for experienced astronomers magnification is just a number. They've learned to adjust the instrument to the local conditions, to the object being observed, and even to the characteristics of the observer (younger folks tolerate very high magnification better than older ones - something to do with the narrow size of the beam of light exiting the scope at high magnification, then hitting floaters and other defects that accumulate with age).

This is why most people end up with a handful of eyepieces, and choose the best one depending on each case.


Is there an optimum orbit for a Hipparcos/Gaia-like parallax observatory The Gaia spacecraft is currently mapping the stars in the Milky Way by parallax. It orbits in the Earth-Sun L2 Lagrange point. The parallax observed from the spacecraft orbiting around the sun allows us to derive the distances to many stars in the milky way, as well as calculate their velocities.
Assuming no cost to placing such an observatory wherever we want around the sun, what would be the optimal orbit for optimal rate of science gains?
I apologise if this seems a big vague, let me elaborate.
If you imagine placing it in a smaller orbit, the parallax of distant stars (already very small) will be made even smaller. But information about the parallax will be returned quicker.
If you place the observatory in the orbit of Uranus, you'll have a much greater parallax and will be able to measure the distances of distant stars with much greater precision. However, it will take 80 years to complete an orbit, and the rate of distance measuring would be intolerably slow.
Intuitively, I feel like there is some obvious sweet spot that would produce measurements at an acceptable rate, though I also feel like my objective function here is ill-defined.
Would a highly eccentric orbit be preferred to a circular one? Would a particular orbit optimise the 'science gains' for one particular aspect at the expense of others (such as accuracy of nearby stars traded for accuracy of distant stars, or accuracy of stellar proper motion)? If there are multiple objectives to optimise, what would  the tradeoffs be?

There is several reasons why the L2 point is an optimum point for the satellite.

First you have a stable orbit with low gravitational perturbation and which is not too far away. So, you won't have a huge travel time. And the parallax is good for what the satellite is supposed to do.
Secondly in a Lagrange point you can easily point out the same point. In orbit, you will have some grey area.
Eventually, I think you are doing a mistake when you say : "Assuming no cost". In my opinion, if we are not sending a satellite like Gaia around another planet, it is because we want to avoid risks and sending the satellite, around Mars for example, adds a ton of risk.

That's why we use the L2 point to place a satellite in regards to all the parameters. But, you can choose another location if you want to maximize something in particular.


How far from the Sun can an Earth TCO (temporarily captured orbiter) go after leaving? Objects which are temporarily captured into an Earth-Moon system orbit (temporarily captured orbiters or TCOs) by definition will leave this orbit. While the term suggest an Earth orbit, it's really an n-body orbit involving the gravitational effects of the Moon and Sun as well.

They are also sometimes referred to as "mini-moons".

Upon exiting, can TCOs go as far out as the asteroid belt, or do they then just orbit the Sun in a heliocentric orbit very similar to the Earth's 1 AU?

Has anyone tracked or calculated the orbit of a known TCO to this extent?

As a side question (it can be split this off separately if necessary) have any of these been larger than say 10 meters in diameter?

No answers after a week, so I will put a toe into the water an make a rough estimate. I think a thorough answer would require both a significant 3D orbital dynamics simulation with a million particles, and a way to define dynamically a threshold for what passing bodies would be considered "just barely temporarily captured" and which *not quite temporarily captured".
There should in fact be some good papers to cite that report this kind of simulation. I'll try to look around; I think there is a good opportunity for one or two more answers here.
Also see Have there been any documented mini-moons since 2006 RH120? and all of the sources linked there!
An example of an object in a medium-term stable orbit in the Earth-Moon system might be found in a halo orbit around the Earth-Moon L1, or L2 points. A spherical envelope-back calculation using spherical cow milk for ink suggests that in the rotating frame, it cycles in a 100,000 km orbit every two weeks around the Lagrange point, or about 0.5 km/sec in that frame. In a non-rotating Earth-centered frame, the Moon itself is moving at about 1 km/sec.
Since these temporary orbits are weakly bound, I'm going to say that an object entering the Earth-Moon system at roughly 1.5 km/sec can hang around for a while, then exit with roughly that velocity.
Looking inthe Sun's inertial frame now, The Earth-Moon system is moving at about 30 km/sec. That 1.5 km/sec is now looking fairly small, only about 5%. So if it exited prograde or retrograde, it would end up in a heliocentric orbit with a semi-major axis no more than 10% different than the Earth's. TCO's are NEOs that take the time to flirt with the Earth-Moon system, then continue on as NEOs to perhaps flirt again another day.

EDIT: I just visited my mini-moon question, found Gravnik 2012 and realized the plots are there! Here are some of them. It turns out everything I said is right. That's not a surprise, because I believe I read this paper a few years ago.
1 billion initial conditions, 10 million particles actually integrated.
The population of natural Earth satellites Mikael Granvik, Jeremie Vaubaillon, and Robert Jedicke, Icarus, Volume 218, Issue 1, March 2012, Pages 262-277  https://doi.org/10.1016/j.icarus.2011.12.003

Abstract
We have for the first time calculated the population characteristics of the Earth’s irregular natural satellites (NESs) that are temporarily captured from the near-Earth-object (NEO) population. The steady-state NES size–frequency and residence-time distributions were determined under the dynamical influence of all the massive bodies in the Solar System (but mainly the Sun, Earth, and Moon) for NEOs of negligible mass. To this end, we compute the NES capture probability from the NEO population as a function of the latter’s heliocentric orbital elements and combine those results with the current best estimates for the NEO size–frequency and orbital distribution. At any given time there should be at least one NES of 1-m diameter orbiting the Earth. The average temporarily-captured orbiter (TCO; an object that makes at least one revolution around the Earth in a co-rotating coordinate system) completes (2.88 ± 0.82) rev around the Earth during a capture event that lasts (286 ± 18) d. We find a small preference for capture events starting in either January or July. Our results are consistent with the single known natural TCO, 2006 RH120, a few-meter diameter object that was captured for about a year starting in June 2006. We estimate that about 0.1% of all meteors impacting the Earth were TCOs.


RA and Dec of the Sun at J2000 I'm in the somewhat awkward position of asking something that, despite me thinking there should be hundreds and hundreds of web pages and datasets showing this particular value, I've been unable to find anything about it.

What, indeed, are the Right Ascension and Declination angles of the Sun for J2000 ? Wikipedia shows some values (RA = 286.13, Dec = 63.87), but it is not clear whether they are J2000 or not...

At noon on 1 Jan 2000 (UT) the sun's RA & dec were (J2000):

18hrs 45min 9.36s

-23deg, 2 m, 8.2s

This is per Guide9 software.


Is there a theoretical maximum size limit for a star? Some stars are simply huge. Eventually, though, wouldn't there be simply too much pressure or mass for the star to sustain itself? Wouldn't it eventually collapse into a black hole?

Is there a theoretical upper limit on the size of a star, and what is it based on?

According to current knowledge, yes. If the gas cloud is too massive, the pressure of the radiation prevents the collapse and the star formation.

The article Stars Have a Size Limit by Michael Schirber, it's about 150 Solar Masses. However, there's the Pistol Star, which is speculated to be 200 SM.

In the article 'Das wechselhafte Leben der Sterne' by Ralf Launhard (Spektrum 8/2013) there's a diagram with information that when the mass is over 100 SM, the star can't form because of radiation pressure. The exact value of the limit is not speculated in the article.


How many photons does it take to determine the existence of a distant object? This applies to any object, but I see the recent discovery of the oldest, most distant galaxy and it started me wondering what the limits are. Presumably you can do better with a bigger telescope and longer exposure, but I'm curious about literally how many photons you need to collect to decide whether or not something is actually there. And then presumably you need a bunch more to determine its red-shift and angular size and rotation etc.
I have absolutely no feel for these numbers.  Is it tens of photons or a millions of photons?

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


Hole through the earth I have a problem: If I would dig a hole through the earth, and jump right in, disregarding the rotation of the earth, I should come back out on the other side.

However, orbital mechanics say otherwise: Specific Orbital Energy tells me:

v^2/2 - µ/r = -µ/2a

Now, on the planet, my relative velocity is 0 m/s, so v^2/2 = 0. The term simplifies to:

-µ/r = -µ/2a | / (-µ)
1/r = 1/2a | ^(-1)
r = 2a | /2
a = r/2

My apoapsis is 3000 km by this equation. But it should be 6000 km, since I come out of the other side.

Why is this, and did I do a wrong assumption?

The formula you are using is for a point mass, but a hole through the Earth experiences less gravity than a point mass, as the comment points out.  One can then ask, what if the Earth's mass were all at the center, why wouldn't the mass fly through and go to a distance r on the other side?  It's because a mass like that is a singularity, one that doesn't matter unless you fall right through it.  If the Earth's mass was really concentrated at a point, then falling from r would still tend to just miss that point, since it's just a point anyway, and the huge forces would turn the orbit around, such that the object fell from r, went to zero, turned around, and flew back out to r.  That's a very squeezed ellipse, with a=r/2, as desired.  The motion that goes straight through and out to the opposite r, on the other hand, is not an ellipse with the mass at one focus, so it violates Kepler's laws.  Again, the violation of Kepler's laws is that the mass is singular in a way that matters if you fall through a hole in it.  Interestingly, it doesn't matter how concentrated the mass is toward the center-- if there's a hole in it that the object falls through, then the object does not obey Kepler's laws.  It will only obey Kepler's laws if the orbiting object does not enter the mass.


Is the eccentricity of a planets orbit relevant to finding whether the distance between said planet and its host star determines seasons? The tilt of the Earth is the main factor that determines the seasons. It's also true that the eccentricity of Earth's orbit is very close to zero, implying that aphelion and perihelion distances are relatively close. If the eccentricity of Earth's orbit were close to but less than one, then would the distance between the Earth and the Sun become a dominant factor in determining the seasons?

Yes it would. If the sun was much further away the season on earth would certainly be more winter-like. The reason distance is not the deciding factor for earth is the variation in distance is fairly small.


Why are some universal entities round and others are flat? I've been wondering for a while now why some objects in the universe (such as stars, planets, etc.) are round, while others are flat (galaxies, planetary rings and protoplanetary disks).

I know these shapes come to be because of gravity. In planets, the gravity from the core pulls everything to the centre, thus, rounding the planet (same with stars). But with Saturn, for example, why have its rings formed as rings and not as a spherical layer of ice surrounding the planet?

The same with the Kuiper belt and asteroid belt, I would presume the mechanics there are the same as with a planetary orbit, hence, why they are belts and not clouds. But then what about the oort cloud? Why is that a huge cloud in the outermost solar system and not a huge ring of ice.

As for the solar system itself, I've read that the solar system is kind of egg-shaped, but shouldn't it be just like the galaxy (or the other way round)?

I've tried looking into this and wondering about what different laws of physics may apply in each case, but it's all really confusing.

Preliminary note: there is a continuity of more or less flattened shapes. No macroscopic objects are really spherical, or really flat. Take Jupiter: it has a rotation period around its axis of 10 hours, and you can clearly see that it is an ellipsoid. It's the centrifugal force, combined with the gravitational attraction, that shapes these objects.

Now to your question: Planets and stars are dense bodies, and rotate as single objects. On the other hand, the Oort cloud, Saturns rings, stellar clusters and galaxies are made of independent particles, free-floating in lots of empty space. Depending on the formation history and gravitational environment (e.g., Saturn and its moons) of these entities, the orbits of these individual particles can be more or less co-aligned, and form a more or less spherical clouds.


What is the current accepted theory for the fate of hot Jupiters? It is well established that one main feature of many hot jupiters is their close proximity to their parent star, usually the equivalent of being within the orbit of Mercury.  So, these planets are gas giants and are very hot (hence their category).

However, a few discoveries have lead to questioning about what is the fate of these planets.

example 1: HD 209458b  a.k.a. "Osiris"

According to the NASA page "Dying Planet Leaks Carbon-Oxygen", Osiris is doing more than 'evaporating', it is leaking carbon, oxygen alongside hydrogen in an envelope behind the planet that has been detected from Earth.  The significance of carbon and oxygen is stated from the article:

  Although carbon and oxygen have been observed on Jupiter and Saturn, it is always in combined form as methane and water deep in the atmosphere. In HD 209458b the chemicals are broken down into the basic elements. But on Jupiter or Saturn, even as elements, they would still remain invisible low in the atmosphere. The fact that they are visible in the upper atmosphere of HD 209458b confirms that atmospheric 'blow off' is occurring.

It is stated in the article that Osiris is likely to become a hypothesised class of exoplanet known as a Chthonian, which is defined in "Evaporation rate of hot Jupiters and formation of Chthonian planets" (Hebard et al. 2003) as

  new class of planets made of the residual central core of former hot
  Jupiters

These would be similar in size to Earth, but considerably denser.

example 2: CoRoT-7b

According to the NASA article "Most Earthlike Exoplanet Started out as Gas Giant", CoRoT-7b is an Earth-sized planet where a hot Jupiter usually is found, they describe it as

  is almost 60 times closer to its star than Earth, so the star appears almost 360 times larger than the sun does in our sky," Jackson said. As a consequence, the planet's surface experiences extreme heating that may reach 3,600 degrees Fahrenheit on the daylight side. CoRoT-7b's size (70 percent larger than Earth) and mass (4.8 times Earth's) indicate that the world is probably made of rocky materials.

The high day time temperature means that the star-facing side of the planet is likely to be molten, any tenuous atmosphere is also blasted away.  Scientists estimate that many Earth-masses may have been boiled off.  It also seems that the decreasing mass is causing the planet to be drawn closer to the star  -cause more material to be boiled off, hence the mass to decrease.

To summarise on of the scientists in the article:

  You could say that, one way or the other, this planet is disappearing before our eyes,"

The question

As these are only 2 examples of a possible process, the question is, what is the current accepted theory as to the fate of hot Jupiter exoplanets?

Could this also be the reason a hot Jupiter does not exist in our solar system?

This is a fairly loaded question in that it depends heavily on what a "hot Jupiter" actually is defined to be. What is "hot"? What is a "Jupiter"? In reality, there's a continuum of planetary masses and distances from their parent star, and in the literature you'll commonly see references to "hot Neptunes", "hot Saturns," etc.

The predominant theory as to how giant planets form is that they first coalesce from rock and ice beyond the ice line, the distance from the parent star at which water becomes solid. This distance is approximately where Mars lies today in our solar system. What's surprising about "hot gas planets" is that they are found within this ice line, significantly within. This implies that after they formed their cores, they migrated closer to their host stars via some currently undetermined process (for which there are several good candidates, but for now let's assume that the existence of hot planets shows that at least one of these processes operates quite regularly).

And what about the word "hot"? Well, for the planets that are closest to their parent stars, there is known to be a radius anomaly: The radii of these planets are significantly larger than models of giant planet structure irradiated by their host stars would predict. So I would define "hot" planets as gas giants whose radii are larger than what would be predicted by the standard models.

Now that we got some of the definitions out of the way, there's the question of survival. When giant planets are close to their parent stars, they become tidally locked. As a consequence, there is very little energy tidally dissipated on the surface of the giant planet, the shape of the planet is fixed and there are little internal motions. However, the giant planet also raises a tide on its host star as well, and because it takes a lot of angular momentum to change the spin of an object with 1,000 times more mass, the host stars are almost never going to be tidally locked to their closest planet.

The rate at which energy is dissipated within the star is highly uncertain, and this uncertainty is typically swept into a fudge parameter "Q," the quality factor, with lower quality factors reflecting more dissipation. "Q" is measured for certain bodies in our own solar system (i.e. Earth and Jupiter) and in some stellar binaries, but is highly variable from body to body, ranging from about 10 for the Earth to 10^8 for some stars.

Whether a planet survives to be observed today depends on how long the orbital decay time, which is determined by Q, compares to the age of the system. For some systems, such as WASP-12b and WASP-19b, which feature highly inflated hot Jupiters, Q is estimated to be small enough to cause them to fall into their host stars in a surprisingly short time (< 10^7 years).

One other possibility is that the gas surrounding the rock/ice core is blasted away by the tremendous amount of heat deposited into the planet. This leaves you with a relatively low-density planet that's somewhat devoid of iron, as the cores of giant planets form further from their host stars than the rocky planets. There are a few candidate close-in, Neptune-mass objects that may have been produced as a result of them losing the bulk of their atmospheres in this way (Example: GJ3470b).

As for our own solar system, the formation of a hot Jupiter would have likely destroyed the inner solar system as it migrated close to the Sun, owing to the fact that it would violently perturb the inner planets' orbits. Additionally, the Sun would likely be enhanced in metals owing to the accretion of metal-rich material from this giant planet. While it's potentially possible that there was a hot Jupiter in our solar system before the other planets formed, it currently seems unlikely.