Why do we use filters in telescopes for astronomical imaging? I have read that if we image without a filter we get no information about the color or SED of objects.  Can anyone elaborate the reasons for using filters for imaging/photometry?  What happens if we image without a filter?

In general, the CCDs used to capture images do not register the energy (therefore, colour) of the incident photons on them - they just count the number of photons observed by each pixel (or a value proportional to the number of photons, as the are not 100% efficient). So, they essentially just show overall brightness variations across the image.

If you want to capture colour information you therefore have to use filters. I.e., if you want to get information on the number of red photons (e.g., the intensity of red light incident on each pixel of your image), you use a filter to block out all other light. You can do this with multiple different filters to build up a full colour image.

In most commercial digital colour cameras the CCDs have a filter mask over four pixel patches: two filtering for green light, one for blue light, and one for red light. The outputs from these pixel is used to build the full colour image.


Is there an official list of objects in the sky? I would like to know how many known quasars there are in the Universe as well as x-ray binary black holes (not Neutron stars) but I don't seem to be able to.

I also want to know how many black holes have been found via gravitational wave detections but I think the answer to this question is 3.

Do you guys know any credible sources that list the objects in the sky via category and gives detailed info of how we found them (x-ray binary or gravitational waves or quasars).

Lists of astronomical objects are called "catalogues" and other answers have already mentioned the NGC catalogue of galaxies and nebulae. However, you asked about quasars and black hole x-ray binaries.

For quasars, there are several catalogues. None claim to be complete, as new discoveries happen all the time. The MILLIQUAS catalogue is a compendium of 460,222 type-I QSOs and AGN. It also contains about a million "Quasar candidates". (sources that may be quasars, 80% confidence)

For Black hole binaries, the difficulty is proving that the object is indeed a black hole. This usually requires calculating the mass of the accreting object, which is not always easy. The X-ray catalogues are classified by the mass of the normal stellar component. The High Mass x-ray binary catalogue notes objects that are "black hole candidates". It is in no way a definitive list, since no such list exists.


Types of Multiverses Can anyone recommend me books or other online resources on the concept of the Multiverse, it types, and about higher mathematical dimensions of space-time. I have been interested in such topics for a while but have not found any accessible book that could explain them in a way that's comprehensible to me. (I'm new to the subject of astrophysics, just getting started).

What is the numerical value of $c$, the SR does not say that. It is a global parameter of our Universe, we need to measure it and substitute it into the SR formulas.

There are also more trivial global parameters. Like the dimension of the spacetime. With (space) dimensions more or less than 3, the formulas of the classical mechanics would be the same.

All the essays talking about "multiverses" or "alternate Universes", are playing with the consequences of the current theories, but substituted with different global parameters than we have now. We could imagine, they are playing with alternative Universes.

However, here we have two major problems.

The logical problem. What does "Universe" mean? The word has latin root, "uni versum" means "to contain everything". The "Universe" is the thing what contains everything. How can multiple be existing then?

For example, the sci-fi topos on which a parallel 3D spacetime referring as "another Universe". No, it is not "another Universe". If there would be proof that such a thing exists, then "our Universe" and this "alternate Universe" would be the "Universe" on its original meaning. We could refer the alternative as an "other spacetime" or "parallel space" or so.

We have also a major problem, what is science. If we can access the parallel 3D space, i.e. if we can interact with it, then it is not "another Universe", it is just a newly found region of ours. But what is no interaction is possible between them and us? Then they are not existing. They might be existing like that our thoughts exist. In any case, it is not physics/astronomy any more, it is phylosophy in the best case.

Professional scientists generally get itchy if they need to answer not science questions. It is because you are assuming rhat they know it, while they know that it is not even science. For example, how hot is a 10 solar masses supernova, that is their field. Thus, your question will be likely closed as unclear, because it is.

What you can find:

You can find popular science texts about playing the current theories, but with different global constants. For example, "what if speed of light would be different" will give you likely useful google hits.
General relativity has some interesting topological extensions. For example, the Kerr-solution looks as if rotating black holes would bind different Minkowski-spacetimes together and so.
There might be interesting hits for googling for "string theory landscape".

In any case, make it more clear, what do you understand on "multiverse" or so, use science terminologies for that. You can find these terminologies by googling forums.

Probably the most popular google keywords will change some decade later.


White Dwarf/Degenerate Gas Behaviour just want to try to clarify a few things about the nature of White Dwarfs and degenerate gases.

So if a white dwarf is help up by an electron degenerate gas, the electrons are unable to lose energy by emitting photons, correct? They basically are trapped at whatever energy level they occupy and that's that. The nuclei that make up the majority of the mass of the star are still constituents of a "normal" gas however, correct? This means that these nuclei and emit photons to lose kinetic energy? How does that work exactly? I understand that electron can emit photos inside an atom to drop to lower energy levels and that they can emit photons of any energy level if they are free (particularly if they're in a magnetic field, synchroton radiation, yeah?). To protons and neutrons do something similar? I ask because I never read anything about protons/nuclei emitting photons. If this basic picture is correct, is this the mechanism by which a white dwarf cools down over time?

Final bit, since most white dwarfs are quite hot, the peak wavelength is going to be in the UV. If I wanted to determine the percentage of such a star's luminosity that is given off in the optical, do I just need to integrate Planck's blackbody equation over the range of wavelengths I'm interested in and then compare that to the total luminosity given by $L = 4\pi R^2 \sigma T^4$?

Thanks and Cheers!

The main method by which heat is transferred in a white dwarf is thermal conduction. Degenerate electrons have very long mean free paths and can conduct heat excellently.

White dwarfs lose energy by emitting photons from their outer, non-degenerate surface layers. They may also emit neutrinos from their interiors when they are really young and hot.

Gamma rays and X-rays are of course emitted by nuclei. This works just the same as atomic emission - charged particles (protons) change energy levels within the nucleus.

Most white dwarfs are not hot, they are similar to the Sun or cooler. Hot ones are easier to see, but only young white dwarfs are hot. But yes, your scheme is approximately (in that white dwarfs are not blackbodies) correct.


Why is space-based VLBI scattering sub-structure "Hopefully, a new promising tool to reconstruct the true image of observed background target(s)"? @HDE226868 mentioned in the observatory that the GBT@20 – The Celebration; Twenty Years of Innovation and Discovery was going on. There are plenty of YouTube videos and slide sets available as links. One of which is Yuri Kovalev (LPI and MIPT Moscow)'s Space VLBI and the GBT.
Below is a screenshot that mentions

Discovery of the scattering sub-structure

A tool to probe turbulent interstellar medium.
Must be taken into account by high resolution VLBI experiments.
Hopefully, a new promising tool to reconstruct the true image of observed background target. Critical for SgrA*.

and links to (these are my best-guess links)

Gwinn et al. (2014): Discovery of Substructure in the Scatter-Broadened Image of Sgr A*
Popov et al. (2017):  PSR B0329+54: Substructure in the scatter-broadened image discovered with RadioAstron on baselines up to 330,000 km

The caption from the image from Gwinn et al. (2014) reads:

Figure 2. Correlated flux density of Sgr A∗ at λ = 1.3 cm plotted against baseline length. Squares show the sensitive GBT-VLBA baselines, circles VLBA-VLBA baselines. Each point represents a 15-min vector average for one baseline after self-calibration; error bars show statistical ±1σ. The green curve shows the correlated flux density of the average scattering disk for an East-West baseline, as described in Sec. 2. The gray curves show quantiles of the predicted distribution of correlated flux density from substructure, with source and scattering parameters from Bower et al. (2006, 2014b), as described in Sec. 3 below.

Excess power at long baselines implies excess fine spatial structure. From what I understand this is belied to be a result of "turbulence" within the intervening interstellar medium, rather than from the original source being observed, and without being characterized and measured presents a serious limit to the resolution of VLBI, both EHT-scale and space-based, and apparently "Critical for SgrA*."
Question: This all seems to be somewhat analagous to astronomical seeing in that it needs to be understood and even measured before it can be compensated for, in order to obtain diffraction (baseline) limited resolution. But why is it "Hopefully, a new promising tool to reconstruct the true image of observed background target."?

It is well known that the study of pulsars is hampered by the scattering effect - the distortion of radiation during the passage of the interstellar medium. Even very compact pulsars look like lanterns in fog. It is clear that this effect should be observed when studying not only pulsars, but also other radiation sources. And here Radioastron plays its role. The pulsar, like a lighthouse, illuminates the Galaxy, and the interstellar medium, like fog, blurs its image. "Radioastron", in turn, "filters" this radiation. It turned out that against the background of blurry images of objects in the Universe, there are ultracompact specks - a scattering substructure. Accordingly, the effect of the fine structure of scattering discovered thanks to the space interferometer will help to understand the nature of the turbulence of the interstellar medium, to determine the distance to the clouds, causing scattering of radio waves. This is a significant contribution to astrophysics and is interesting in itself. Today we already have a better understanding of the physics of turbulent flows in the interstellar medium. And most importantly, if we do not correctly interpret the results of studying the Universe, we will be able to correctly interpret the results of observations.
At the center of our Galaxy is a massive black hole - about 10 million
times heavier than the Sun. It is simply necessary to study it, but
scattering of radiation interferes with it. Note that the study of
black holes in modern astrophysics is one of the fundamental,
fundamental problems. Scientists solving them now know about the
effect of scattering substructure discovered by us and take it into
account in their works. Moreover, having obtained a true picture of
the spots of the scattering substructure, it is possible to restore
the true image of the object hidden from us behind the scattering
screen. This opens up tremendous opportunities for astrophysicists.
There are already results: a significant number of publications have
appeared in leading scientific journals describing observational data
on the scattering substructure and mechanisms for reconstructing true
images of space objects.

Lighthouses in the sky. It is difficult to study our Galaxy without “illumination”.
https://poisknews.ru/magazine/20721/


Stellar life cycle flow chart with mass conditions and time scales I remember that in my nuclear astrophysics lecture a decade ago, our lecturer drew a large flow chart like diagram of stellar evolution in dependence of the mass of the star (in solar units) on the black-board. I found one from Wikipedia which is close to the one I remember, but less detailed than the one I remember. The content of the linked chart is summarized in my following sketch:

Sadly, neither me nor my former classmates recall which text book or publication the diagram came from, we suspect it was a composition of many sources. The diagram came with a brief explanation of each mass threshold, and it also outlined the governing equations in the respective stellar state. I also remember that there were rough approximates of the time scales of the respective state given.
Maybe somebody can help me out with alternative search terms, references, or a textual answer briefly summarizing the conditions and the state-of-the-art error bars for the mass thresholds and timescales?
Related

The life course for a massive star from birth to death using the HR Diagram and it comments are helpful, but only slightly related
What are the equations governing stellar evolution (Luminosity, Mass, Temperature, Radius) touches one aspect of what I am after
A flowchart from Stephan et al. 2019 outlining the outcomes of binary stellar evolution in the galactic center comes with the right amount of details, but is not for general star evolution.

Edit from 2021-01-22
To make answering easier, I quickly layed out the existing diagram as BPMN which is editable e.g. with this freely available editor. The source code is available via https://pastebin.com/mrafLWa0 - please feel free to reuse, expand and distribution my diagram, just make you comment here where and what for.

My slapdash (and perhaps messy because I rendered it in PowerPoint) attempt at addressing stellar evolution. Open the flowchart in a new tab in fullscreen to see a better rendering of it.


Why do we have so few neutrinos to study? According to Wikipedia most neutrinos passing through the Earth emanate from the Sun. About 65 billion (6.5^10) solar neutrinos per second pass through every square centimeter perpendicular to the direction of the Sun in the region of the Earth.

However the IceCube Neutrino Observatory or other neutrino detectors only detect a few neutrinos every year. Why?

It's not that the neutrino detectors don't have a chance of detecting a neutrino, but that the neutrinos do not interact with matter (well hardly ever).

If there are $6.5\times 10^{10}$ neutrinos per second per $cm^2$ and if the detector has an area of $10^{6} cm^2$, and it detects 1 neutrino per year, then 1 out of $2.05\times 10^{24}$ neutrinos interacts with the detector. $2.05\times 10^{24}$ pass through the detector unnoticed.


What size objects can be created from collision of two asteroids? Does collision of two asteroids necessarily create only very small objects? (and how big is it "very small"?)

I am especially interested about collision of two asteroids with similar orbits (and thus relatively small difference of the speed vectors in the moment of collision).

Can such a collision create big (say 1km, or maybe 100m) objects?

Is it in this case likely that a created object would have something like a comet tail consisting of dust and small objects (probably gravitionally attracted to the resulting asteroid remnant)?

Since planets are forming from asteroid collisions, I would say those collisions won't necessarily create small objects.

Imagine throwing a snowball against a window.

If you throw it lightly, some snow will stick to the window and increase the mass of it (at least until it melts away) and the rest of the snowball will be scattered everywhere around but won't stick to the window.

If you throw the snowball too hard, the window will break, resulting in even more tiny pieces. But still, some snow will stick to some pieces of glass.

That's exactly what happens on asteroid impacts and depends on the speed, mass and structure of each asteroid.

The size of those debris should reach from microscopic dust up to gigantic rocks with several kilometer wide diameter.


Does any iron fuse in stars before they go supernova? I understand that iron and all heavier elements consume more energy to produce than they make, and that is what eventually leads to a supernova. I also understand that a lot of the heavier elements are produced during that supernova. However, what I'm wondering is, before the star goes supernova, does any of the iron fuse with other elements? Yes, there would be a net energy loss, but if there is only small amount of iron in the star, it would probably be able to handle that.

The "iron core" in a supernova is actually the end product of a nuclear statistical equilibrium that begins when the silicon core begins to fuse with alpha particles (helium nuclei). Exothermic reactions are possible right up to Nickel-62 (which is actually the nucleus with the highest binding energy per nucleon). In fact, successive, rapid alpha captures produce nuclei with the same number of protons and neutrons, but at the same time, the competing processes of photodisintegration and radioactive decay work in the other direction. The process is thought to mostly stop at Nickel-56 which, because heavier nuclei are more stable with $n/p>1$, then undergoes a couple of $\beta^{+}$ decays via Cobalt-56 to Iron-56. However, the core of a supernova just before it explodes is likely to contain a bit of a mixture of iron-peak isotopes.

Before all this happens it is possible for iron and nickel to undergo nuclear reactions if there is an appropriate source of free neutrons. The elements beyond iron in our universe are predominantly created by neutron-capture in either the r-process or the s-process.

The r-process is thought to occur after a core-collapse supernova (or a type Ia supernova) has been initiated. The neutron flux is created by the neutronisation of protons by a dense, degenerate electron gas in the collapsing core.

However, the s-process can occur outside the core of a massive star before it explodes. It is a secondary process because it needs iron nuclei to be present already - that is, the iron that is used for the seed nuclei is not produced inside the star, it was already present in the gas from which the star formed. The s-process in massive stars uses free neutrons produced during neon burning (so at advanced nuclear burning stages beyond helium, carbon and oxygen burning) and results in the addition of neutrons to iron nuclei. This builds up heavy isotopes, which may either be stable or undergo $\beta$ decay and/or further neutron captures to build up a chain of "s-process elements" (e.g.  Sr, Y, Ba) all the way up to lead. The overall process is endothermic, but the yields and reaction rates are so small that it has no major influence on the overall energetics of the star. The newly-minted s-process elements are easily blasted into the interstellar medium shortly afterwards when the supernova explodes.


What would be the best inertial reference frame to describe objects in the solar system (other than the solar system's barycenter)? For the purpose of a project, I require the components that describe the inertial frame to be as independent as possible from solar system bodies itself. Since the solar system barycenter is itself determined by the positions and masses of the solar system bodies, it will not be a good choice.

I was thinking that the frame that I want could be described by some nearby star that does not have planets of its own or else the galactic centre itself.

If such an inertial frame exists, I would like the same to be expressed in heliocentric cartesian coordinates.

Any relevant thoughts would be much appreciated!

Use the initial Center Of Mass (COM) as an origin.

It's trivial to show that the COM of a closed N-body system (in Newtonian physics) moves with a constant velocity.  This is a relatively easy system to work with and transforming to other coordinate systems, such as the heliocentric one you mention, is relatively trivial compared to computing the 3D N-body system with good accuracy.  A word of advice : if the math of converting between coordinate systems is a problem for you, the mathematics and algorithms of an accurate simulation will be likely much worse.  I'd suggest trying out such conversion mathematics as a warm up exercise and if you're comfortable with that then dive into the simulation.

COM can be shown to move at constant velocity because it's a closed system (or can be treated as such over the timescale you mention), and so net force on the COM is zero, meaning it has to move with constant velocity.

Adding a small body like an extrasolar asteroid makes next to no difference to the system as a whole.  It's a minute change.  You can treat such objects as special cases, ignoring their effect on the system and just calculating the effect of the system on them.  If you're planning to add very large bodies and see what happens then you need to include them in the general system all the way through the simulation.

Note there are a lot of existing simulations out there that implement a wide range of approaches.  This is a problem that has been extensively studied and a search online would yield a myriad of approaches.

The motion of the Barycenter is complex.  I'd avoid this complication in what is already gong to be a messy mathematical system to simulate.

Your main priorities should be gathering accurate data for initial starting values (non trivial) and choice of algorithms.  Choice or ordinate system is comparatively minor.


How should one rationally deal with the issue of space travelling alien civilizations? What kind of reasoning is appropriate to understand the as of today unanswered question of whether there are (other) interstellar space travelling civilizations in the Milky Way?

We have already sent probes towards the border of the Solar system. And even landed human beings on another celestial body and brought them home alive and well. If we extrapolate the 50 years of space travel, the 100 years of electronics (radio), the 400 years of physical science, to just a fraction of the biological age of humankind into the future (like a few thousands of years), interstellar travel is not out of the question for us or at least our artefacts. So I imagine two possible alternatives:

1) The Milky Way is cluttered by lots of space travelling civilizations like us and our future. Once one of them/us gets going, they'll soon be everywhere. The Sun orbits the Milky Way every 250 million years, about 2% of the age of the galaxy. Going to the nearest stars is enough to soon be everywhere. But if they are everywhere since almost always, they should be here, we should be their seed.

2) We are the only space travelling civilization in the entire galaxy, ever. But then what makes us unique? We consist of the most common elements and volatiles of the universe and our planet and star and galactic location all seem to be very typical. There's no known trace of any uniqueness here. Whatever could it be?

Are there more alternatives?

While we cannot say today which alternative is true, we should be able to at least specify the possible alternatives. But to me they all seem to be absurd! What would be a rational logical scientific approach to this apparent paradox?

That's basically the Fermi paradox. It seems likely that there are numerous civilizations in the galaxy, and yet we see no trace of them anywhere.

The Drake equation is often invoked to calculate the probability of existence of other civilizations, by compounding several other, more simple probabilities: the probability that a star has planets, that the planets are habitable, that life could arise on such a planet, etc. The proposed values so far indicate that there should be many civilizations in our galaxy.

Astronomy, just plain observations, might be one way to solve the paradox by detecting alien societies.

Broadcasting messages via radio has also been attempted, in the hope that someone might reply.

However, the only constant in this search has been the negative answer. There are some hypotheses trying to explain that:

the rare Earth hypothesis: maybe there's something very special about Earth that allows the evolution of intelligent life
the destruction hypotheses: maybe life tends to self-destruct, or maybe there are space marauders out there killing everyone
disasters: maybe things like gamma ray bursts tend to wipe out life periodically
the young Universe hypothesis: if there are very, very many universes in the Multiverse, then the vast majority of them are young. If so, in the vast majority of them there will be 0, or at most 1 civilizations. It's simply a matter of probability that we are in such a universe.
religious arguments such as the creation of Man was a special, unique act by a higher entity

Others argue that many civilizations might exist, but we don't see them. Possible reasons:

we're just too far from each other, in space, in time, or both
it's just too hard to travel between stars
we haven't been searching long enough, or we're looking or listening to the wrong things
they're not interested in us
they all go into a Technological Singularity pretty quickly and become god-like, undetectable and incomprehensible to us
they are too busy browsing their internets to be interested in the outer space (yes, that has been proposed seriously)
they are too alien for us to detect or comprehend
they are like the Na'vi in Avatar - a non-technological society
they deliberately avoid us (the Earth-as-a-zoo hypothesis)
they do try to talk to us, but evidence is being supressed by the government (yes, that is also a serious proposal)
they are here already, hiding amongst us (proposed by Iosif Shklovskii and Carl Sagan, so I guess it qualifies as serious - albeit highly hypothetical at best)

Truth is, we really don't know. All we can do now is speculate.


Has the existence of earth-sized rogue planets been confirmed? The list of known rogue planets provided by Wikipedia is fairly short. Some are "candidates"; others "may be" red dwarfs. The two confirmed rogue planets are several times the size of Jupiter.

I understand that rogue planets can only be detected if they glow (which immediately brings the phrase "red dwarf" to mind) or if they cross the path of light emanating from a star. This makes detection extremely difficult.

The reason I ask is I'm not quite certain whether objects the size of Jupiter or smaller can even exist in interstellar space. Maybe dark matter grinds them to dust, or maybe they implode in the absence of any meaningful inertial frame - I don't know. Has anything Earth-sized been detected out there?

The list on Wikipedia looks current. It doesn't list planets that have only been detected as one-off microlensing events. This can detect Jupiter sized objects (with considerable uncertainty in mass). But observing a microlensing doesn't allow for study, so a microlensing is not considered to be enough evidence for naming or cataloging a known planet. So the short answer is no Earth sized rogue planets are known.

This is entirely due to the difficulty in detection. An Earth sized rogue planet would cool on its surface to a few degrees above absolute zero, and be undetectable with current telescopes. Microlensing events with small rogue planets would also be smaller and harder to detect above noise.

An Earth sized rogue planet has a perfectly good inertial frame, (it is the bound planets that are in a rotating, non-inertial frame), nor does a frame of reference have anything to do with implosions. Dark matter is doesn't grind anything since the main property of Dark matter is that it doesn't interact with regular matter.

In all likelihood, there are billions of rogue planets in the galaxy, and some are smaller, and some are Earth-sized and "rocky". They could have been formed around a star and then ejected.


How can "Geysers" on Europa reach heights of 100km? After seeing today's Washington Post's article NASA just saw Europa’s geysers erupting into space. Again., and especially the phrase "geysers erupting into space" I checked and Europa has a surface gravity of about 1.3 $m/s^2$, which is substantial - almost the same as the Moon's at about 1.6 $m/s^2$!

The BBC's website article Europa moon 'spewing water jets' says

  "Further evidence has been obtained to show that Jupiter's icy moon Europa throws jets of water out into space."

and

  "The suggestion is that the jets reach several hundred kilometres in height before then falling back on to Europa."

On earth, pressure can come from geological hydrostatic forces and by steam produced by geothermal heating, but they rarely rise beyond a few dozen meters. That would be perhaps 100 meters in Europa's gravity.

The images shown of the water above Europa show heights of 100 kilometers and more.

Unlike Earth rock being heavier than water, Europa's ice crust floats. What could be sources of hydrostatic pressure or pressurized steam near the surface that can create such "geysers" or "jets" that can project water to such heights in this gravity?

above:  The geyser Strokkur in Iceland, Earth, from here.

Technically those aren't "geysers" on Europa, they're cryovolcanos.    Though that definition may be a bit imprecise as well, but the 100 km eruptions on Europa probably have more in common with volcanic eruptions on Earth than geysers.   Europa undergoes significant tidal flexing which, in combination to it's solid icy crust, could be compared to explosive volcanic eruptions on Earth and not the cyclic geysers, which are far smaller.

Volcanoes on Earth can shoot material high into the atmosphere on Earth but volcanic ash isn't a good example because it's not so much "shot" into the air as it rides rising hot air to altitudes 30 and 40 KM high.

Magma fountains on Earth can reach a few hundred meters and Tephra (anything larger than Ash, from pebbles to large rocks/boulders) can be found as far as a mile from a volcano and can exit the volcano at several hundred meters per second.   From the Tephra article:

  Blocks and bombs as large as 8-30 tons have fallen as far away as 1 km
  from their source (Bryant, 1991). Small blocks and bombs have been
  known to travel as far away as 20-80 km (Scott, 1989)! Some of these
  blocks and bombs can have velocities of 75-200 m/s (Bryant, 1991)

Europa's Tidal-flexing can cause significant pressures within it's icy crust.  See here.  and here.

So, if we draw an Earth-Europa comparison, and assuming my math is right, using 1/2 at^2, all you'd need for material to reach 100 km on Europa is an exit velocity from the ground of a bit over 500 meters per second.   Or, some 20%-30% faster than a 22 caliber bullet.   That's faster than material usually leaves an Earth Volcano, but not all that much faster.   When frozen, Europa's outer icy crust is probably very hard and brittle.  In pressure spots where the pressure exceeds the structural integrity of the icy crust, it's not all that surprising that the results are pretty explosive, leaving the surface at 500 meters per second.

Ice is very strong and brittle and when cold enough, where the Earth's crust, while rocks are quite brittle and strong, the crust overall can also be quite malleable, especially a few miles down in hotter temperatures under an erupting volcano.

Very cold ice is resistant to cracking, but under enough pressure, when it finally does crack it can release explosive energy.   Ice cracks in water due to differential expansion, (Silicate rock doesn't expand with temperature nearly as much as frozen ice does), and while the cracking of an ice cube dropped into water might not seem like much, a big enough ice cube dealing with enough tidal heating can have explosive results.

While this fun little video might seem unrelated (turn down your volume before watching), it demonstrates the explosive force that rapid warming can have on a hard block of ice.  This isn't steam pressure, as the steam is all escaping.  This is the inside of the block expanding faster than the outside can and the result is impressive.   https://www.youtube.com/watch?v=epkRd-w3TGw   So, with that in mind, explosive energy from a hard, cold 10 mile thick icy surface under the influence of tidal flexing shouldn't be all that surprising.


visualisation of the universe's expansion in physics we were learning about the universe and the big bang. we were being told that everything came from one point and 'exploded' outwards, and is continuing to expand. I felt that this explanation was unclear, so I said that it was more like everything being one the surface of an expanding 'balloon', but with an extra dimension, a space which is the hyper surface of the expanding universe. my teacher disagreed and said it was just like a big explosion. I think this is wrong-it implies that there is an edge of some description, or that the universe is just an expanding sphere of stuff inside infinite space. what is the best way to visualise the universe?

You were correct, you teacher was incorrect.

It is the space that expands - much as a surface of a balloon does.

An explosion is a poor analogy in contrast because - as you suggest - it implies something to expand into.

For another way, not as accurate as the balloon analogy but maybe helpful - imagine being trapped inside an expanding loaf of bread in an oven. You have no knowledge of the world outside the loaf - and you may assume the loaf is infinite in extent. But as the dough rises the gaps inside the loaf get larger without the mass of the loaf itself increasing. You could compare our position in the universe to being inside such an infinite loaf.


How to calculate the expected surface temperature of a planet I'm writing a program to generate solar systems but I'm having trouble calculating the expected temperature of a planet. I have found a formula to calculate this, but I haven't been able to get a remotely correct answer out of it as it doesn't clearly state what units your supposed to use.

This formula I found:

$$4 \pi R ^ 2 ơ T ^ 4 = \frac{\pi  R ^ 2  L_{\odot}(1 - a)}{(4  \pi  d ^ 2)}$$

where $R$ is the planet's radius (not sure what units), $d$ is the distance from the Sun (it mentions AU), $a$ is the albedo, $L_{\odot}$ is the luminosity of the Sun (which I assume can be interchanged with the luminosity of any star), $T$ is the temperature of the planet (kelvin, this is what I'm trying to get), and $ơ$ is the Stefan-Boltzmann constant.

The site I found it on is notes for an astronomy college course. Here is the link:

http://www.astronomynotes.com/solarsys/s3c.htm#

Any help would be very much appreciated.

The formula

$$4 \pi R ^ 2 ơ T ^ 4 = \frac{\pi  R ^ 2  L_{\odot}(1 - a)}{4  \pi  d ^ 2}$$

is correct, if you want to calculate the radiative equilibrium temperature. You only need to use the right units. We can further simplify the formula to

$$T ^ 4 = \frac{ L_{\odot}(1 - a)}{16  \pi  d ^ 2 ơ}\;.$$

You should input the luminosity in watts, the distance to the star in meters and the Stefan-Boltzmann constant as

$$σ = 5.670373 × 10^{−8} \;\mathrm{W}\; \mathrm{m}^{−2}\; \mathrm{K}^{−4}.$$

The albedo is dimensionless. The resulting temperature will be in Kelvins. Let me make an example for Earth:

$d = 149,000,000,000 \;\mathrm{m}$

$L = 3.846×10^{26} \;\mathrm{W}$

Albedo of Earth is 0.29. (The Bond albedo should be used.) You will get

$$
T ^ 4 = \frac{ 3.846×10^{26}(1 - 0.29)}{16  \pi \times  (149,000,000,000) ^ 2 \times (5.670373 × 10^{−8})}=4,315,325,985 \;\mathrm{K}^4\;.
$$

After powering this number to 1/4, we obtain temperature 256 K, which is -17° C. This looks reasonable. The real average temperature on Earth is closer to 15° C, but the greenhouse effect is responsible for the difference.


How is the Hubble constant determined from gravitational waves? We know there is a discrepancy between measurements of the Hubble constant, $H_0$. On one side there is the method of the Planck mission, where they use the CMB and the $\Lambda$CDM model to determine the Hubble constant. On the other side, they use standard candles, like Cepheid variables (for example: Riess et al., 2019) and red giants (for example: Freedman et al., 2020). The very precise measurements don’t overlap and thus we need a ‘solution’.

One way to improve our knowledge about $H_0$ is by using gravitational waves to determine the Hubble constant (Feeney et al., Arxiv version). They compare the measured ‘strength’ of the gravitational wave with the ‘strength’ of the wave when it was created to determine the distance to the source object. But how do they calculate/determine the ‘strength’ of the wave at the source? I know they also use an EMS observation for calculating the redshift (and that redshift converted to radial velocity divided by the distance gives $H_0$), but from where do they calculate the source-‘strength’ of the gravitational wave? How can they otherwise determine how much energy the wave has lost? I can’t seem to find the answer.

If you measure the gravitational waveform from an inspiralling binary, you can at any point measure the amplitude, instantaneous frequency and the rate of change of frequency. The last two give you the "chirp mass", which is related to the product and sum of the binary component masses.

The amplitude of the gravitational wave then depends on the chirp mass and the distance to the source. Thus we can estimate the (luminosity) distance.

If we then know the redshift of the source, or at least the redshift of the galaxy it is in, then an independent estimate of the Hubble parameter is possible. Therefore this can only be done for gravitational wave sources with identified host galaxies (so far this can only be done for GW sources that show an electromagnetic counterpart - merging neutron stars).

It is a bit more complicated than this, because the amplitude also depends on the inclination of the binary orbit to the line of sight. Fortunately, this can be constrained by measuring the relative amplitudes of the two possible gravitational wave polarisations (+ and ×) and this can be done if you have two interferometers with different orientation of their arms (like LIGO and VIRGO for example), which also helps to locate the GW source in the sky to look for electromagnetic counterparts.


Where does energy at the beginning of a star's lifecycle (before any nuclear reactions) come from? David Christian's Maps of Time has this to say about the period during which the first stars started to form from the diffuse clouds of hydrogen and helium that then made up the universe:

  Under the pull of their own gravity, the clouds of hydrogen and helium began to collapse in on themselves...As gravity packed each cloud into ever smaller spaces, pressure built up in the center. Increasing pressure means increasing temperatures, and so, as they shrank, each gas cloud began to heat up (~Kindle location 1469).

The part I'm working on understanding is "Increasing pressure means increasing temperatures." Does the pressure itself cause the molecules to heat up? Or is it the kinetic energy that the particles pick up as gravity pulls them towards the center of the cloud? Or is the phenomenon here something other than increased energy per unit mass?

Of course, once the nuclear chain reaction gets started in the star it's not hard to figure out where the energy comes from. I'm talking about before that.

This is basic thermodynamics.

When you compress a gas, you inject energy into it. Think of the pump you use to inflate the tires on your bike. It takes some force to move the piston, right? That effort is not wasted, but goes directly into the air in the pump. Now the air has more energy.

But what happens to a gas when you put energy into it? It's molecules jiggle around faster. Well, faster jiggling is basically the definition of higher temperature. By putting more energy into the gas, you raise its temperature.

You can actually tell that the bike pump is getting warmer if you pump quickly and forcefully - this is something you can experience yourself.

Same with stars - the whole star is the "bike pump", and gravity is the one who's pushing the piston. Due to compression (shrinking) under gravity, the gas gets hotter and hotter. It turns out a star has A LOT of gravitational energy, so the gas can get VERY hot.

In your terms, yes, it's the acceleration that molecules experience falling into the gravity well of the star that makes them move faster. Faster moving molecules = higher temperature. Pretty straightforward phenomenon, really.

Historically, gravitational compression was thought to be the main source of energy for the stars, before the discovery of nuclear physics. Helmholtz and Lord Kelvin proposed this hypothesis in the 1800s.

The pressure-temperature relation of any gas was originally known as the Gay-Lussac law. Now we know it's just a particular case of more general phenomena (ideal gas law) tying together pressure, temperature, volume, and various kinds of energy.

A spectacular application of the p-T relation is the so-called "fire piston" or "fire syringe", which can ignite small pieces of cotton or paper by just hitting a piston really hard (extremely strong compression = big temperature rise). Search Youtube for some videos like this one:

https://www.youtube.com/watch?v=4qe1Ueifekg


If getting shined on by a gamma ray is an extinction level event how do we detect them? Cosmic gamma ray bursts are seriously powerful and do a lot of damage. So how did we measure gamma ray frequencies without getting fried?

I have two possible explanations:

We never detected them directly but calculated their frequency from secondary radiation from the surrounding gas that was ionized.
We get a direct hit but the source is so far that beam divergence will be large enough for only a few photons per square meter to hit us, so there's not enough energy to be destructive.

The danger from gamma-ray bursts (GRBs) is the potential to reduce atmospheric ozone drastically, allowing harmful ultraviolet radiation to pass through the atmosphere and make life on Earth much more difficult. However, you have to be fairly close for this to happen. Gehrels et al. (2003) calculated that a typical GRB from a supernova would have to occur less than 8 parsecs from Earth to cause serious damage to the ozone layer - or, as they put it, to cause levels to drop low enough that they would

  double the “biologically active” UV ﬂux received at the surface

Obviously, though, we can detect gamma-rays from distances farther than 8 parsecs! The closest GRB detected so far, GRB 980425, is likely the result of SN 1998bw, and thus originated 140 million light-years away - over 42 million parsecs away. The radiation from it was nowhere near high enough to harm life on Earth.

Your second explanation, therefore, is the correct one. The intensity of the signal decreases according to the inverse-square law - that is, $I(r)\propto r^{-2}$ - so at large distances, the flux is small enough that there is negligible impact on Earth's atmosphere.

Phiteros also raised a good point: To cause severe problems from Earth, the gamma-ray emission from a GRB - which is often along an axis from both poles of the progenitor - would have to be pointed nearly directly at Earth. For example, there was once a scare that a supernova by either of two stars in the WR 104 system could cause problems for Earth, but the odds that the beam would hit Earth full-on are slim.


What would the effects be on Earth if Jupiter was turned into a star? In Clarke's book 2010, the monolith and its brethren turned Jupiter into the small star nicknamed Lucifer.  Ignoring the reality that we won't have any magical monoliths appearing in our future, what would the effects be on Earth if Jupiter was turned into a star?

At it's closest and furthest:

How bright would the "back-side" of the earth be with light from Lucifer?

How much heat would the small star generate on earth?

How many days or months would we actually have night when we circled away behind the sun?

How much brighter would the sun-side of earth be when Lucifer and the sun both shine on the same side of the planet?

Before I start, I'll admit that I've criticized the question based on its improbability; however, I've been persuaded otherwise. I'm going to try to do the calculations based on completely different formulas than I think have been used; I hope you'll stay with me as I work it out.

Let's imagine that Lucifer becomes a main-sequence star - in fact, let's call it a low-mass red dwarf. Main-sequence stars follow the mass-luminosity relation:

$$\frac{L}{L_\odot} = \left(\frac{M}{M_\odot}\right)^a$$

Where $L$ and $M$ are the star's luminosity and mass, and $L_\odot$ and $M_\odot$ and the luminosity and mass of the Sun. For stars with $M < 0.43M_\odot$, $a$ takes the value of 2.3. Now we can plug in Jupiter's mass ($1.8986 \times 10 ^{27}$ kg) into the formula, as well as the Sun's mass ($1.98855 \times 10 ^ {30}$ kg) and luminosity ($3.846 \times 10 ^ {26}$ watts), and we get

$$\frac{L}{3.846 \times 10 ^ {26}} = \left(\frac{1.8986 \times 10 ^ {27}}{1.98855 \times 10 ^ {30}}\right)^{2.3}$$

This becomes $$L = \left(\frac{1.8986 \times 10 ^ {27}}{1.98855 \times 10 ^ {30}}\right)^{2.3} \times 3.846 \times 10 ^ {26}$$

which then becomes

$$L = 4.35 \times 10 ^ {19}$$ watts.

Now we can work out the apparent brightness of Lucifer, as seen from Earth. For that, we need the formula

$$m = m_\odot - 2.5 \log \left(\frac {L}{L_\odot}\left(\frac {d_\odot}{d}\right) ^ 2\right)$$

where $m$ is the apparent magnitude of the star, $m_\odot$ is the apparent magnitude of the Sun, $d_\odot$ is the distance to the Sun, and $d$ is the distance to the star. Now, $m = -26.73$ and $d(s)$ is 1 (in astronomical units). $d$ varies. Jupiter is about 5.2 AU from the Sun, so at its closest distance to Earth, it would be ~4.2 AU away. We plug these numbers into the formula, and find

$$m = -6.25$$

which is a lot less brighter than the Sun. Now, when Jupiter is farthest away from the Sun, it is ~6.2 AU away. We plug that into the formula, and find

$$m = -5.40$$

which is dimmer still - although, of course, Jupiter would be completely blocked by the Sun. Still, for finding the apparent magnitude of Jupiter at some distance from Earth, we can change the above formula to

$$m = -26.73 - 2.5 \log \left(\frac {4.35 \times 10 ^ {19}}{3.846 \times 10 6 {26}}\left(\frac {1}{d}\right) ^ 2\right)$$

By comparison, the Moon can have an average apparent magnitude of -12.74 at full moon - much brighter than Lucifer. The apparent magnitude of both bodies can, of course, change - Jupiter by transits of its moon, for example - but these are the optimal values.

While the above calculations really don't answer most parts of your question, I hope it helps a bit. And please, correct me if I made a mistake somewhere. LaTeX is by no means my native language, and I could have gotten something wrong.

I hope this helps.

Edit

The combined brightness of Lucifer and the Sun would depend on the angle of the Sun's rays and Lucifer's rays. Remember how we have different seasons because of the tilt of the Earth's axis? Well, the added heat would have to do with the tilt of Earth's and Lucifer's axes relative to one another. I can't give you a numerical result, but I can add that I hope it wouldn't be too much hotter than it is now, as I'm writing this!

Second Edit

Like I said in a comment somewhere on this page, the mass-luminosity relation really only works for main-sequence stars. If Lucifer was not on the main sequence. . . Well, then none of my calculations would be right.


Are there any estimates for the combined mass of all near earth objects (NEOs)? I tried looking around but couldn't find any sources attempting to answer this particular question.  It's obviously impossible to know the exact answer, but all I'm looking for is an estimate.  My best guess would be that it's about a tenth (or less) of the mass of the asteroid belt, with a composition roughly 10-15% metallic and 85-90% carbonaceous or silicate.  Is anyone aware of any sources that attempt to give an estimate?  If not, what would you say is a reasonable estimate?

Thanks in advance.

Short answer: The total mass of the known NEOS is about $8 \times 10^{17}\rm\,kg$, if we assume average density of $2\rm\,g/cm^3$.
Long Answer: I used the NASA JPL data query website. Instead of mass, they have diameter for equivalent spherical shape in kilometers for NEOs.
Here are the top 10 results from the query:
1036 Ganymed (A924 UB)      37.675
109P/Swift-Tuttle           26
3552 Don Quixote (1983 SA)  19.0
433 Eros (A898 PA)          16.84
C/2001 OG108 (LONEOS)       13.6
C/1991 L3 (Levy)            11.6
1P/Halley                   11.0
4954 Eric (1990 SQ)         10.8
1627 Ivar (1929 SH)         9.12
1866 Sisyphus (1972 XA)     8.48

Out of 24877 entries for NEOs, there are only 1268 NEOs with an entry for the diameter.  If we convert these to volumes using $V= \frac{4}{3}\pi r^3$ and sum, we get $4.21 \times 10^{5}\rm\, km^3$ of total NEO volume.
If we assume an average asteroid density of $2\rm\, g/cm^3$ = $2000\rm\, kg/m^3$ = $2 \times 10^{12}\rm\,kg/km^3$ , then the total mass is $8.42 \times 10^{17}\rm\, kg$.
As a comparison, the total mass of the asteroid belt is estimated at a whopping $2.39 \times 10^{21}\rm\, kg$. So, even with comets, the NEOs are much less massive than the asteroid belt.
This estimate of total mass may be somewhat conservative, because it leaves out the mass of most of the known NEOs. However, I don't think it is too conservative since about 90% of the mass estimate is accounted for in the largest 10 asteroids/comets listed above.  The smaller stuff doesn't contribute much to the overall estimate.


Scientifically important discoveries with the help of amateurs I am looking for examples where amateur astronomers in these days contributed sigificantly to important scientifically discoveries.

I am aware of one example: Victor Buso from Argentina was lucky enough to observe a Supernova just when it formed, which has led to follow-up observations by large scientific observatories, and in the end resulted in a Nature article which has Buso as a co-author.

What might some other similar examples be?

below: "Astronomer Victor Buso poses in front of the telescope with which he witnessed the birth of the supernova 2016gkg. (Victor Buso)" From here, borrowed from Does this amateur observation hold a record even including robotic searches for supernovae?. Click/open for full size.

I guess it depends on what you consider scientifically important, but users at Planet Hunters have helped with the discovery of several exoplanets by analyzing data from the Kepler Space Telescope, as well as the famous Tabby's star (KIC 8462852) which has led to a paper by Tabetha Boyajian co-written by the users who participated in the analysis of the light curve of the star.

The star is currently still the subject of numerous studies and publications.


Why carbon is considered volatile? Carbon is considered volatile by planetary science, eg Moon lacks volatiles and thus lacks carbon.
However volatiles are defined as "elements or substances with low boiling point", but Carbon boiling point is very high! Its sublimation point is 3900K, so it should be refractory and not volatile.
I am confused why it is volatile with such high sublimation point.
Regards,
Alex

Oxygen is the third most common element in the universe after hydrogen and helium. Elemental oxygen is very volatile, but it's also highly reactive and will combine with almost anything it encounters.
The resulting oxides can be volatile in the case of CO (boiling point 81.6 K) or carbon dioxide (sublimates at 194.7 K), or extremely refractory: silicon dioxide has a boiling point of 3220 K, making it less volatile than iron (boiling point 3134 K). Magnesium oxide doesn't even melt until it reaches 3125 K, just a little short of the boiling point of iron. Rocky materials are almost entirely composed of oxides and silicates, and overall are about half oxygen by mass.
Carbon will readily react at high temperatures with the oxygen in many oxides to produce CO or CO2 (this being the basis of many smelting processes on Earth). Since oxygen is so common, most of the elemental carbon that ends up accreting with rocky materials can be expected to end up getting oxidized.
The moon not only appears to have formed by a process that allowed most of its volatiles to escape, that process also enriched it with light elements. As a result, it has a small iron core, but lots of silicates and metal oxides for carbon to react with, forming volatile CO or CO2 in the process.


Is everything quantum entangled? So, I am learning about quantum entanglement in my quantum course and had a thought. If the Big Bang is the creation of everything - time included, and this came from a single "source". Would it not follow that all particles in the universe are entangled to each other? And perhaps, we cannot observe this entanglement because we are "inside the system" sort to speak. Are we not all connected?

I know the last part is cheesy, but the point is serious.

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

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

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

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


How fast will the sun grow when it becomes a red giant? We all know that our sun will become a red giant in a couple of billion years? But how fast will the process be when it starts? Would it be theoretically possible (disregarding the deadly effects of this event) for a human to experience the sun as its present size until it swallows earth?

Yes, the time is very long, but exactly how long depends on when you "start the clock" for the expansion to a red giant.  If you take the end of the "main sequence" (when the core hydrogen runs out) as the start of the expansion phase, it's more like a billion years to complete the process of becoming a maximally expanded red giant.  But the answer that gets 5 million years focuses only on the phase of most rapid expansion (which begins after the star is already quite a bit larger than the Sun).  Either way, our lives a the smallest blink of an eye to a star!


Why does Earth have more gravitational force than the Moon if I stand on it? Perhaps this is a silly question but I am trying to understand Einstein's general theory of relativity.

I would like to explain my question in more detail.

The greater the mass of an object, the greater the curvature of spacetime.

If I imagine the curvature of spacetime as a hole with a steep slope around it, I can picture that the closer I get to the center of the hole, the steeper the slope is and therefore the more I would be pushed into the center of the hole. I am not completely sure if that's a very good analogy but in general I think I get the picture of timespace curvature as depicted here:

Now, my question is why is it that the size of the hole matters if I am not standing on the edge (side) of the hole or in the hole but rather on an object that 'covers' the hole (Earth).

Let's say if I stand on a bridge that covers a 20-meter-wide hole and on a bridge the covers 500-meter-wide hole. Wouldn't the gravitational force experienced be the same since the surface on both bridges is flat?

The Earth doesn't cover the hole, the Earth is the hole. Gravity is an attractive force, so were you not standing on the Earth, the Earth's gravity would cause you to accelerate towards it. As you are standing on the Earth, you feel this acceleration as your weight, just as you would feel pressure if you were to push against a wall.

To take a more complex view, we would say that you are being attracted to the local center of gravity, which is located within the Earth. Take your bridges example, it doesn't matter that something obstructs you from reaching the center of gravity. You will feel the attraction of gravity regardless. If you jump while standing on a bridge, you would land just the same as if you jumped while standing on the ground (though please try not to miss the bridge). The graphic is meant to illustrate the strength of the force based on the mass of the attractive object and distance.

If we were to put you on a more massive object, which would have a steeper curvature, you would experience this as increased weight, which is why our lunar explorers bounced around on the moon instead of walking.


How is the term "solar system" defined? Could confirmation of a new planet lead to a change in this definition? Voyager 1 (https://en.wikipedia.org/wiki/Voyager_1) is approximately 140 AU away, and is considered to have left the solar system. Now, the layman's definition of solar system provided by Google is:

  ...the collection of eight planets and their moons in orbit around the
  sun, together with smaller bodies in the form of asteroids,
  meteoroids, and comets.

And from Wikipedia:

  The Solar System[a] is the gravitationally bound system of the Sun and
  the objects that orbit it, either directly or indirectly,[b] including
  the eight planets and five dwarf planets as defined by the
  International Astronomical Union (IAU).

At this point, Voyager 1 is well beyond the aphelion of all known planets, so according to the definitions above, so at least colloquially, saying that it has left the Solar System seems to make sense. However, according to Wikipedia, it was only officially considered to have left the Solar System once it left the heliopause. This is not really in conflict with the above definitions, as, to my knowledge, the heliopause extends beyond the orbit of all the original 8/9 planets at all times. The Encyclopedia Britannica (https://www.britannica.com/science/heliopause) states that the heliopause
extends for 123AU.

However, the hypothesised Planet 9/X is conjectured to have an aphelion of approximately 1,200 AU and a Perihelion of 200 AU (https://en.wikipedia.org/wiki/Planet_Nine). This would seem to bring the two competing definitions of "solar system" discussed here into sharp conflict.

Which definition would likely win out? Could the confirmation of such a planet force a reckoning on whether Voyager 1 has truly left the solar system?

Edit:
I wanted to add a more authoritative source that says Voyager 1 has indeed left the solar system.

  Voyager 1 is the first man-made object to leave our solar system and pass into interstellar space. Scientists confirmed this finding a year later after studying Voyager’s data, which showed clear changes in the plasma or ionized gas right outside of the solar bubble.

https://www.nasa.gov/directorates/heo/scan/images/history/August2012_2.html

The definition of the edge of the Solar System is not one of particular scientific importance, so it is not particularly well-defined. There are a number of plausible definitions, all reasonable.  I've listed one which occur to me below in order of increasing size:

The pancake-shaped volume which contains the orbits of the major planets.
The sphere big enough to contain the major planets.
The volume inside the solar bubble.
The thickish disk which contains the Kupier Belt.
The sphere which contains the Oort Cloud.
The Sun's Hill sphere.
The irregularly-shaped volume of space where the Sun is the closest star.

As you note, the discovery of a Planet X would change some of them, but not others.

How choose to define the edge of the Solar System depends a lot on what you intend to use it for.  I very much doubt that any one will win out, at least not until the definition of the Solar System makes an important difference somewhere: government, commerce, science.

So I think there's no clear answer to your question possible.


How do scientists know if an Earth-like planet is really Earth-like? I'm not sure how scientists determine that a planet is really an Earth-like planet. How do they do it?

The earth similarity index (ESI) is a weighted geometric mean of four similarities.

The formula documented on
http://phl.upr.edu/projects/earth-similarity-index-esi
(as of march 23, 2014) should be adjusted a bit, since
$n$ should be the weight sum instead of the number of planetary properties.
With the weights provided on the site we get

$$ESI =\left( \left(1-\left | \frac{r_E - r_P}{r_E + r_P} \right |\right)^{0.57}\cdot
 \left(1-\left | \frac{\rho_E - \rho_P}{\rho_E + \rho_P} \right |\right)^{1.07}\cdot\\
 \left(1-\left | \frac{v_E - v_P}{v_E + v_P} \right |\right)^{0.70}\cdot
 \left(1-\left | \frac{288\mbox{K} - \vartheta_P}{288\mbox{K} + \vartheta_P} \right |\right)^{5.58}\right)^{\frac{1}{7.92}},$$
with $r_E=6,371 \mbox{ km}~~$  Earth's radius, $r_P$  the radius of the planet,
$\rho_E=5.515\mbox{ g}/\mbox{cm}^3$ Earth's bulk density, $\rho_P$ the bulk density of the planet,
$v_E=11.2 \mbox{ km}/\mbox{s}~~$ the escape velocity on the surface of Earth, $v_P$ the escape velocity on the surface of  the planet, and
$\vartheta_P$ the surface temperature of the planet; the weight sum is $0.57+1.07+0.70+5.58=7.92$.

Mars as an example:
With $r_P=0.53 r_E$, $\rho_P=0.71 \rho_E$, $v_P=0.45v_E$, $\vartheta_P=227\mbox{ K}$ we get
$$ESI_s =\left( \left(1-\left | \frac{0.47}{1.53} \right |\right)^{0.57}\cdot
 \left(1-\left | \frac{0.29}{1.71} \right |\right)^{1.07}\cdot\\
 \left(1-\left | \frac{0.55}{1.45} \right |\right)^{0.70}\cdot
 \left(1-\left | \frac{61\mbox{ K}}{515\mbox{ K}} \right |\right)^{5.58}\right)^{\frac{1}{7.92}}=\\
(0.811241627\cdot 0.819676889\cdot 0.716163454\cdot 0.494865663)^{\frac{1}{7.92}}=
0.833189885$$ as surface similarity. (Some of the data used from here.)
Global similarity combines surface similarity with interior similarity.
Global similarity of Mars with Earth is about 0.7.

Surface gravity can be calculated from radius and bulk density of a planet.
The radius of an expoplanet can be estimated by the transit method, relating the estimated diameter of the star to the brightness change of the
star during planet transit. The mass of the planet can be estimated by the wobble of the radial velocity of the star (using Doppler shift).
By mass estimate of the star and the orbital period of the planet the distance of the planet to the star can be estimated. An estimate of the
absolute brightness of the star can then be used to estimate the surface temperture of the planet. There exist more methods.
The accuracy of these estimates vary with the quality of the observations.

ESI values between 0.8 and 1.0 are considered as Earth-like.

Details of the planet's atmosphere, amount of surface water, and other details are not considered in the formula. So it's just a very rough prioritization. With future spectroscopic data further refinement could be possible.


Beneficial astronomical uses of satellite constellations like Starlink? Has any thought been given to software that would use predictable occultations by Earth-orbiting satellites to improve (rather than degrade) astronomical observations?

Potential challenges I could imagine might include:

The satellites’ high angular velocity compared with typical occulters.
Knowing enough about the outline and orientation of each satellite.
Diffraction effects.

Perhaps there are satellite design changes that could make them more useful, akin to star shades. Perhaps some camera or other sensor design changes could help capitalize on what seems an inevitable reality. And perhaps there are types of observations for which no amount of cleverness and computational resources could turn Starlink into a net win for astronomy.

Partial answer for now.

tl;dr: It will be an "exercise in futility".

This is a great question! With tens of thousands of very similar satellites in LEO the number of occultations that will occur from a given observatory will be much larger and better characterizable than with the current diverse population, so "if it were useful it would have been tried already" is not a valid dismissal.

However, in LEO they are moving quite fast, thousands of arcseconds per second of time, so any measurement may be limited by shot noise for small apertures, and since the satellites themselves are of the order of a few meters, for large aperture telescopes the occultation is partial and complex; while the satellite may block light for some part of the telescopes mirror, other parts will still be receiving light.

The speed of a satellite in LEO is about 7800 m/s, so if you have an 8 meter telescope the entire event will last about 1 millisecond. There may be some really interesting way to do some kind of very complicated reconstruction for a very bright object when a spacecraft occultation is observed by a large aperture telescope, since what you have is a complex aperture that moves through a variety of states over that one millisecond.

During that time the Earth's atmosphere will be in a fairly well "frozen" state, so one solution for astronomical seeing will probably work for the entire event

But my hunch is that there will be so many small uncertainties and ambiguities that this will end up as an exercise in futility, and events will be so infrequent and non-repeating that there will be no way to check the method.

Here are some plots of apparent angular velocity of satellites in LEO from 300 to 1000 km altitude. The new, large internet-delivering constellations will be closer to 300 than 1000 km, taken from and explained in depth in this answer.

One degree per second is of course 3600 arcseconds per second of time.


Eta Cassiopeiae star type; G0V? F9V? Both? For a story I am looking into the star system of Eta Cassiopeiae, which is a binary with a main sequence star called Achird (very similar to the sun) and a K7V star.
The issue I am running into is the discrepancies of what the star type is; some say it is a G0V and some say it is a F9V.
I've looked at a dozen sources so I won't link anything, but I am wondering if anyone might know the most reliable and or most current source to go off of.

$\eta$ Cassiopeiae A has an estimated mass of 0.972 M$_\odot$, an estimated temperature of 5973 K, and a B-V color index of ~0.58[1]. In addition to the spectrum of the star, we look at these and other properties when attempting to classify main sequence stars. You can see a table [here] which shows the bulk properties of each spectral type that we can use in addition to spectrum to help classify the star. The mass estimate places it into the G5V category, while the temperature and B-V place it definitively into the G0V category[2]. Hence, I would go with G0V. The mass is a little on the light side for the G0V category, but the temperature and color index and right on for the characteristic values for G0V.


If two stars collide, what is the probability that they merge to form a single star? After looking at What are the odds that the Sun hits another star? and answering it (crudely), now I'd like to ask the following:
What is the probability that if two stars collide, their cores merge to form one larger, more massive star?

Fairly good.
Two stars of mass $M$ falling from infinity straight towards each other until they merge at distance $2R$ will get kinetic energy $GM^2/R$. This is a lot, for two suns it is $1.8978\times 10^{41}$ J. However, compared to the binding energy of even a single star, $\approx 3GM^2/5R$ this is less(the sun has binding energy $2.2774\times 10^{41}$ J, and a double mass $2^{1/3}R$ radius same density merged star $7.2302\times 10^{41}$ J, 3.17 times more). So there is not enough energy released to blow up a star, but is is about a quarter of it: a lot of matter is going to get ejected or end up in orbits through a heated envelope that will take a while to simmer down.
The key issue is whether the cores get slowed down enough by the encounter to remain bound, becoming a common envelope binary. A direct hit clearly would work, but glancing collisions may allow the cores to miss each other: now the question is whether the envelope can absorb enough kinetic energy. A rough estimate may be that there is significant slowing if the mass scooped up/pushed aside $\pi r_{core}^2 \rho_{envelope} R$ becomes comparable to $m_{core}$. For two sun-like stars with $r_{core}=0.2R_\odot$ this seem to happen, but much hydrodynamics may occur complicating things.
Glebbek's dissertation on stellar mergers estimates a rough condition for the orbital angular momentum to exceed the maximum spin angular momentum of the merged star as $$\frac{r_p}{R_1+R_2} > k^4\frac{(1+q)^{\xi+4}}{2q^2}$$ where $k^2\approx 0.05$, $\xi\approx 0.6$, $r_p$ the periastron distance, and $q=M_2/M_1$. This typically is exceeded: there is a lot of angular momentum that needs to be shed (for example by blowing off a lot of heated gas). For example, two sun-like stars having $r_p=R_\odot/2$ has a LHS of 1/4 and a RHS of 0.0303.
That dissertation also contains numerical simulations of various merger cases.


Plotting Equatorial coordinates to X,Y plane (simulating telescope/camera view) I would like to simulate a telescope camera view, projecting (RA, Dec) coordinate sets to a 2D plane. So effectively the need is to convert Equatorial coordinates to (x, y) coordinates, when we can know how many degrees a single pixel spans.
For convenience I'm using degrees for RA and Dec here, and the code is just pseudocode.
The following base information is provided as an example:
Imaging equipment:

Telescope focal length in mm (flen): 420
Image resolution (widthPx, heightPx): 4000 * 3000
Pixel size in microns (pxSize): 8
Binning (bin): 1

So from my understanding we can calculate a few things from this for starters.
The field of view in degrees should be:
fovW = radToDeg( 2 * atan((pxSize * bin * widthPx / 1000) / (2 * flen)) )
fovH = radToDeg( 2 * atan((pxSize * bin * heightPx / 1000) / (2 * flen)) )

And from this we could derive:
pixelsPerDegWidth = widthPx / fovW;
pixelsPerDegHeight = heightPx / fovH;

For locations closer to the equator I can get some valid looking results using this formula:
deltaRa = starRa - imageCenterRa
deltaRa = deltaRa > 180 ? 360 + deltaRa : deltaRa
deltaDec = -1 * (starDec - imageCenterDec)

posX = deltaRa * cosdeg(starDec) * pixelsPerDegWidth + 0.5 * widthPx
posY = deltaDec * pixelsPerDegHeight + 0.5 * heightPx

With some catalog star positions projected using this formula, I can even get Astrometry.net solver to correctly identify the center position.
However when we move to the poles this comes crashing down.
So let's go ahead and use a target center point near the pole:

RA 30 degrees (imageCenterRa)
Dec 89 degrees (imageCenterDec)

Let's also say that we have a hypothetical star at RA, Dec (220, 89.2) - this is close to the pole, and both the star and the pole gets included in the image.
What kind of formula should we use to get the correct location and to draw the area correctly even in the polar regions?
I've been looking for an example for a while but haven't come up with any. I've seen some formulas that may be just the thing I'm looking for but my thick skull loves examples - hence why I included some exact numbers here, as I would love to see how this would be calculated in practice.
So to summarize: how would we calculate the (X, Y) coordinate in the image for the star at RA, Dec (220, 89.2)?

This isn’t code for a full implementation, but you could do something like this:

Transform your RA, Dec coordinates (which are spherical polar coordinates, just all with the same $r$) to Cartesian $x, y, z$ coordinates.   (The $r$ value you would use here is arbitrary, for example $r = 1$, and should be the same for all stars, not their real distances; you're just projecting them onto a sphere so you can do the transformation.)

Apply a coordinate frame rotation using your field-center RA, Dec coordinates (or perhaps their complements) as the rotation angles, so that the field center direction is the new $z$ axis.

Plot the resulting new $x, y$ coordinates, ignoring $z$.

You might have to do a further rotation of those $x, y$ coordinates to get the plot oriented the way you want, or maybe there’s a clever way to do that with the proper choice of angles in step 2.


Eyepiece::Astromania or Meoptex I am looking into getting a zoom eyepiece for myself, and not sure on what basis to evaluate and compare various options.
For some background on the type of observing I plan to do see my earlier question: Good eyepieces for Saturn and Jupiter
Here are some examples of zoom eyepieces that I've looked at so far

Astromania 1.25" 7-21mm Zoom Eyepiece Apparent field of view is 30- degree at 21mm and 43- degree at 7mm
Meoptex 1.25"-7-21mm Zoom Eyepiece No further information

On these pages there are links to several more options.
It would be a very great help if anyone could help me figure out on what basis I might choose one of these or quickly rule some out, or recommend how I could search for more options.

7-21mm is a pretty common zoom eyepiece option, available from the companies you identified as well as others like SVBONY and Orion. (8-24mm is another popular option).
Doing a quick search for the Meoptex, it appears to have the same apparent FOV's as the Astromania, as well as the same eye relief.  It's not impossible that they are the same design, repackaged for another company.
Another feature to look for in telescope eyepieces is "fully-multi coated" which will minimize reflections and stray light (Astromania, Orion, and SVBONY list this feature, but Meoptex does not).
IN my limited experience with telescopes, it generally holds true that the more expensive options have better performance, but your total performance will be limited by your poorest component.  (A \$200 eyepiece won't make a $25 telescope any better)
You can also look for used eyepieces, advice, and reviews on cloudynights.com.


Is the Apophis asteroid a concern for planet Earth? I was watching the TED lecture How to defend Earth from asteroids , and the speaker Phil Plait spoke of the possibility of this asteroid hitting Earth. Is this really a concern?

Apophis is of no real concern to us as far as we can tell.  That said, it doesn't mean another asteroid doesn't have our name on it...

According to the most reliable data we have regarding this particular asteroid it only has a 1 chance in about 250,000 of hitting the earth.  University of Hawaii states:

  “Our new orbit solution shows that Apophis will miss Earth’s surface in 2036 by a scant 20,270 miles, give or take 125 miles,” Tholen said. “That's slightly closer to Earth than most of our communications and weather satellites.” He credits the large telescopes and superb atmospheric conditions on Mauna Kea for being able to make these determinations.

Considering how big space is, this is pretty much a hit in cosmic terms.  Keep in mind that this information has been readily available since 2009 at the NASA website as well.


How to calculate declination of Moon? Is there any way to calculate declination of Moon for any day?I'll subtract inclination of Earth on the ecliptic from the result. I'll get Moon's the shortest angular distance from Sun.
Thanks for all answers!

The complexity of the formula depends on the precision you need.
We can make a crude approximation with two sine waves: one for the Moon's travel eastward around the ecliptic over a
tropical month

$$\delta_1 = 23.4^\circ \sin {2 \pi (u - 10.75) \over 27.32158}$$

and one for the Moon's travel north or south of the ecliptic over a
draconic month

$$\delta_2 = 5.1^\circ \sin {2 \pi (u - 20.15) \over 27.21222}$$

both modified as the Moon goes faster or slower over an anomalistic month

$$u = t - 0.48 \sin {2 \pi (t - 3.45) \over 27.55455}$$

where t is the number of days since 2000-01-01 12:00 TT (J2000.0).
Then

$$\delta = \delta_1 + \delta_2$$

is within 1° of the Moon's geocentric declination most of the time.
For a topocentric declination, subtract
1° sin (observer's geographic latitude).

If you need higher precision, see
van Flandern and Pulkkinen 1979
or Meeus 1999
for more complicated formulas.

Note: the u and t offsets above are near a 0° longitude crossing, an ascending node, and an apogee in January 2000 but are tweaked to minimize RMS differences from JPL HORIZONS lunar positions for 2000-2018.


What is the longest natural bound orbit chain observed? Define a bound orbit chain as a list of successively less massive bodies, each in a bound orbit with the bodies preceding it in the list.
Then an example of a bound orbit chain would be: <Sun, Earth, Moon>, since the Earth orbits the Sun, and the Moon orbits the Earth and the Sun (and they are listed from most to least massive).  This is a chain of length 3.
Question: What is the longest natural bound orbit chain we've ever observed?
Related: Do moons have moons?

Certainly 4 (perhaps 5):
The longest certainly-known chain is "3":  Sun-Earth-Moon.
Rhea (a moon of Saturn) has been observed to have a ring of orbiting material: That gives Sun-Saturn-Rhea-Ring for four levels (including unresolved dust)
Looking further afield:
DH Tauri and DI Tauri are a binary pair of similarly sized T-Tauri stars (both small red dwarfs); they are in a mutual orbit.
DH Tauri has a large sub-stellar companion.
The companion has an accretion disc of orbiting matter.
So we have DI-Tauri > DH-Tauri > DH-Tauri-b > matter in disc, for four levels of orbit.
Even this isn't very good, because DH Tauri is bigger than DI Tauri, so it is a bit of a stretch to say that DH Tauri orbits DI.  If anything it is the other way round.  There are very few planets found in orbit around the secondary star in a system and none of those have likely moons. Moreover the disc of matter around the sub-stellar object is not known to have actual "moons" embedded in it.
So "4" seems to be the best we can do...
But if we can add in Galaxy - Star - planet -moon - dust/ring? we get 5 levels.


Why do the upper-atmosphere clouds of Venus appear to have that V shape? IF I understand correctly, the atmosphere moves in the same direction as the rotational spin, but about 60 times faster. It is driven from the hot side of Venus to the cold side (the difference being great due to the planet's slow rotation allowing time for it to heat up and cool down). Is my understanding correct?

But why does the visible flow in the cloud tops describe a sideways V shape, with the flow appearing to move diagonally from the equator toward the poles? Is that also because the equator is hotter than the poles, driving the flow in that hotter-to-colder direction?

This is supplementary to antispinward's excellent answer and provides additional sources and a visualization from the JAXA spacecraft Atasuki. It has been shamelessly borrowed from Would it be possible to "ride the wave" on Venus?

The recently published paper in Nature Geoscience Atmospheric mountain wave generation on Venus and its influence on the solid planet’s rotation rate has open-access links in Science News, Motherboard, and Science.

I haven't found a preprint in arXiv yet.

There is a stationary gravity wave in Venus's very dense atmosphere. It is intermittent, but has been detected several times. See the NYTimes article Venus Smiled, With a Mysterious Wave Across Its Atmosphere for a discussion of recent observations by JAXA. See Akatsuki and Happy Birthday, Akatsuki!, celebrating it's first Venusian year at Venus.

above: "A sequence of images showing the stationary nature of the bow-shape wave above Venus when it was observed in December 2015. Planet-C" from NYTimes. credit: Planet-C/JAXA

above: "An illustration of how gravity waves travel up mountains and into Venus’s atmosphere. Credit ESA" From How Mountains Obscured by Venus’s Clouds Reveal Themselves.


How fast do we travel through space? Could someone give the rough estimates of speeds of our travel through space?

I can easily find the two nearby ones:

Earth's own rotation - depending on latitude, up to 40,075 km/24h = 463 m/s
Earth's orbital speed around the Sun - 940 000 000 km/year = 29,786 m/s

Now how do these fare?

Motion of Sun within the arm of the galaxy, around the center of the galaxy?
Motion of the galaxy within the local supercluster?
(if possible to measure at all) movement of the supercluster within the universe?

According to the Astronomical Society of the Pacific's website "How Fast Are You Moving When You Are Sitting Still?", we are moving:

Rotation of the Earth: 1600 km/h

Orbital rotation of the Earth around the sun: 107,000 km/h

The sun around the centre of the galaxy: 792,000 km/h

Going further, according to the article "Milky Way Galaxy: Facts About Our Galactic Home" (Redd, 2013) quotes the sun's speed around the centre of the galaxy as 828,000 km/h

The Andromeda Galaxy and the Milky Way are headed for a collision at 112 km/sec

According to "Milky Way moving away from void" (2007), the Milky Way galaxy is moving from the local void at 600,000 mph (960,000km/h)

According to the article "The Universe is expanding at 74.2 km/sec/Mpc" (Plait, 2009), the Universe is expanding at a rate of 74.2km/sec per megaparsec (3.26 million light years).

There seems to be some divergence of values in the articles, but these give a good estimate.


Which of Jupiter's moons did I most likely see? I was gazing through my binoculars at Jupiter last night and noticed 3 (maybe 4) tiny dots in line with it. A first I thought they might have been background stars but quickly realised their colour was similar to that of Jupiter and their brightness was larger than expected considering their size. As a result, I'm pretty certain I was looking at Jupiter's moons.

Presumably, I was seeing the largest of its moons (Io, Europa, Ganymede, Callisto). They were far too tiny to see any details, obviously, but would it have been only those moons I could see? I assume the rest are too small. And is seeing those 4 moons as tiny bright dots through my 7x50s feasible?

The galilean moons are visible in binoculars, indeed. Yesterday evening (assuming you're in the western hemisphere), 3 of them were visible as separate dots. See the chart here:

http://www.wolframalpha.com/input/?i=jupiter+galilean+moons+on+march+6+2016+at+9pm+pst

You can adjust the query to reflect your time zone, and the actual date/time of the observation.

There are also smartphone apps that allow you to determine the position of the moons at any given time. On Android I use the Moons of Jupiter app.


What are the exact physical parameters used to calculate Mercury precession with Einstein theory? NASA measured 43,13 arc seconds per century. General relativity predicts 42,98 arc seconds per century.

I try to find out what the parameters' values such as $G$, $M_{sun}$, $\omega_{min}$ at aphelion radius, etc. used to get this exact 42,98.

With a new model developed by R. Plamondon, we get exactly 43,13 and also it matched perfectly with Mercury's period of revolution around Sun.

My problem is the following. When I put my parameters into Einstein metric, I got something unexpected.

In very a general post-Newtonian metric for a two-body system with the first body oblate, where $M\equiv m_1+m_2$ is the total mass, $\mu\equiv m_1m_2/M$ is the reduced mass, and $p\equiv a(1-e^2)$ is the semi-latus rectum of the orbit, the perihelion advance per orbit is
$$\small\delta\varpi = 6\pi\frac{GM}{pc^2}\left[\underbrace{\frac{2-\beta+2\gamma}{3}}_{\text{GTR}=1} + \underbrace{\frac{2\alpha_1-\alpha_2+\alpha_3+2\zeta_2}{6}}_{\text{GTR}=0}\frac{\mu}{M} + \frac{J_2R^2c^2}{2GMp}\right]\text{.}$$
According to GTR, $\beta=\gamma=1$ exactly. The second term contains various parameters relating to preferred-frame effects and energy-momentum non-conservation. In GTR, they are all identically zero, while experimentally, $\alpha_1\lesssim10^{-4}$ and the rest are many orders of magnitude less than that. Finally, in the third term, $J_2$ is the quadrupole moment of the first body; $J_2R^2 = (C-A)/m_1$ where $C$ and $A$ are moments of inertia about rotational and equatorial axes, respectively. A derivation can be found Will §7.3.

  I try to find out what the parameters' values such as $G$, $M_{sun}$, $\omega_{min}$ at aphelion radius, etc. used to get this exact 42,98.

This is just the front coefficient in the above: if $T$ is the orbital period of Mercury, then
$$6\pi\frac{GM/c^2}{a(1-e^2)}\frac{1}{T} = \frac{42.98''}{\mathrm{century}}\text{,}$$
and we can restate the GTR prediction of Mercury as
$$\dot\varpi_{\text{☿}}^{\scriptsize\text{GTR}} = \left.\frac{42.98''}{\mathrm{century}}\right.\left[1+2958J_2\right]\text{,}$$
where $J_2$ is the gravitational quadrupole moment of the Sun.

  NASA measured 43,13 arc seconds per century. General relativity predicts 42,98 arc seconds per century.

If you're referring to one of Anderson et al.'s analyses, the value is $43.13\pm0.14$. Under GTR, this value is compatible with any $J_2$ up to about
$$\begin{eqnarray*}\frac{J_2R^2c^2}{2GMp}\lesssim 0.0067 &\Longleftrightarrow & J_2\lesssim 2.3\times10^{-6}\text{.}\end{eqnarray*}$$
But what is the Sun's quadrupole moment? This is a somewhat sticky question, but the results of lunar laser ranging experiments are incompatible with $J_2$ of more than about $3\times10^{-6}$, while using helioseismology, Pijpers (1998) estimates $J_2 = (2.18\pm0.06)\times 10^{-7}$. Using Pijper's value yields a GTR prediction of $43.01''$ per century, which is compatible with Anderson et al.'s analysis.

  With a new model developed by R. Plamondon, we get exactly 43,13 and also it matched perfectly with Mercury's period of revolution around Sun.

I have no idea what Plamondon's model is, but I would be rather suspicious of any model that gives "exactly" $43.13$ when that's just the middle of an interval of statistical uncertainty according to one of the many prior analyses of Mercury's orbit.

  When I put my parameters into Einstein metric, I got something unexpected.

If we ignore the post-Newtonian parameters other than $\beta$ and $\gamma$, the static post-Newtonian metric is
$$\mathrm{d}s^2 = -\left(1+2\Phi+2\beta\Phi^2\right)\mathrm{d}t^2 + \left(1-2\gamma\Phi\right)\mathrm{d}S_\text{Euclid}^2\text{,}$$
where the Newtonian potential of the Sun is modeled up to the quadrupole term
$$\Phi = -\frac{M_\odot}{r}\left[1-J_2\frac{R_\odot^2}{r^2}\frac{3\cos^2-1}{2}\right]\text{.}$$
Deriving the perihelion advance of Mercury with this metric is Exercise 40.5 in MTW, and it matches the general result referenced above with $M\equiv M_\odot+M_{☿} \approx M_\odot$ and sans the middle term. For the Sun-Mercury system, $\mu/M\sim 10^{-7}$, so the middle term is even more irrelevant than bounds on the PPN parameters would suggest.

References:

Will, C. M., Theory and experiment in gravitational physics, (Cambridge University Press, Cambridge, 1993), 2nd edition.
Anderson, J. D., Campbell, J. K., Jurgens, R. F., Lau, E. L., Newhall, X. X., Slade, M. A., Standish Jr, E. M., in Proceedings of the Sixth Marcel Grossmann Meeting on General Relativity, ed. by Sato, H. and Nakamura, T., (World Scientific, Singapore, 1992).
Pijpers, F. P., Mon. Not. R. Astron. Soc. 297, L76 (1998). [arXiv:astro-ph/9804258]
Misner, C. W., Thorne, K. S., and Wheeler, J. A., Gravitation, (Freeman,
San Francisco, 1973).


Why does the 70-solar-mass black hole in the LB-1 system challenge current astrophysics? Recently scientists discovered a large stellar-mass black hole, which (previously) they believed that it shouldn't be possible in our galaxy!

Source

  An international team of scientists say they have discovered a
  stellar-mass black hole with a mass 70 times greater than the sun — so
  large it defies current theories on how black holes of its kind form.

Why shouldn't it be present in our galaxy system? Are there any physical or mathematical calculations that defy its existence?

References:

Chinese Academy of Sciences Headquarters Press Release
Nature Paper - A wide star–black-hole binary system from radial-velocity measurements
Nature EPDF - A wide star–black-hole binary system from radial-velocity measurements

There're several pieces of information one needs to understand this.

Although stars more massive than 70 solar masses exist, when they become black holes, they usually lose mass in the process. The exact amount of mass lost depends on the metallicity (which is a technical term that describes how much "metals" - the astronomer's definition of metals is anything heavier than hydrogen & helium - is in the star). The companion star that we see today is at solar metallicity, so it's probable that the original star (the one that became the black hole) was also at solar metallicity. Unfortunately, that means it shouldn't leave a 70-mass black hole remnant. From the paper:

  This [70 solar mass black hole] would strongly challenge current stellar evolution models, which only allow for the formation of black holes up to $25 M_{sun}$ at solar metallicity.

Where, then, did this 70-solar mass black hole come from? The paper discusses a few alternatives. The obvious one is that two smaller black holes simply merged to form this one. Problem with that is, you still need two 35-solar mass black holes, and 35 is clearly > 25. (In principle you could also have a 25-solar mass black hole merge with a 45-solar mass black hole, but that still leaves the question of where the 45-solar mass black hole came from in the first place.) Note also that this black hole probably didn't arise from neutron stars merging, since neutron stars have a mass limit of about 2 solar masses. Finally there is three black holes merging into one, but this is unlikely: mergers are already rare events, and having two mergers must be even rarer.

Here're a few more unlikely explanations I can think of:

Perhaps it's a primordial black hole. Problem with this is that primordial black holes are something of a unicorn - a last resort explanation - because they work whenever you need a missing mass. If you can explain an observation without invoking primordial black holes, that's much preferable. See also the next bullet point.
Perhaps the black hole formed elsewhere, in an environment that isn't at solar metallicity, then travelled to this place and captured its companion star. Problem with this is that stellar capture is not a likely process. In the same way we can say that the Earth probably formed around the Sun; it didn't form elsewhere and was captured by the Sun later. There's another problem, which is that the star is observed to have an eccentricity of almost zero (this means its orbit is roughly circular). Newton's laws predict an elliptical orbit. There are processes that will drive the orbits towards circular, but the time taken is long. A (rare) stellar capture event that leads to almost exactly a circular orbit is even more unlikely.

The paper discusses a couple of more credible alternatives:

Perhaps this was a triple system where one star became a black hole, and then this black hole "fell into" one of the other stars and ate it from the inside.
Perhaps there was a "fallback supernova". This is when a star goes supernova, but the ejected material somehow falls back onto the stellar remnant. This has never been directly observed, and this might be the first.
Perhaps something is wrong with the measurement, e.g. an unaccounted for systematic effect. This is the most mundane explanation.

In any case, the system is now an attractive target for telescopes.

Update: there are now several articles claiming an error in the analysis, and that there is no 70-solar mass black hole in this system.


Black hole, escape velocity, going up? I do know nothing can escape a black hole, and I understand popsci explanation of geodesics of space-time around and inside the event horizon.

I however have some trouble with following thought experiment, given there are many references that crossing event horizon of SMBH should be totally unnoticeable for the subject falling through:

I am at gym - a regular one, on Earth. Escape velocity of Earth is around 11.2 km/s. Yet I can lift weights up say one meter even at slower speed, say one meter per second.

Next, consider me hovering one meter above event horizon. As far as I know, for non-rotating BH this is totally possible. Similar to earth, I could lift weight up one meter at speed less than escape velocity (which at this point is speed of light minus x).

Now, I’m hovering one meter below the event horizon. Would I not be able to lift the weight up one meter at speed less than escape velocity (which at this point is speed of light plus y; not sure if x and y are equal), or would I not be able to hover at this position?

Assuming answer is no to either or both, how this would appear to me as the subject?

Nothing can be stationary inside the event horizon and everything, even light must travel inwards. Therefore there is no way that your analogy can work and is why using the idea of escape velocity is a really bad idea in General Relativity.
i.e. You cannot be at a fixed $r$ coordinate to lift a weight. You must be falling inwards, as must any weight you lift, as indeed must a light beam you direct radially outward.


Computing the average hours of sunshine at given location&orientation I made it my geeky aim to see which of two flats that we are considering moving into is sunnier. The first flat faces S and the second faces NNW; both are at the same height above ground and have similar terrain (buildings/trees) around.

Conventional wisdom would suggest the S-facing one is (much) sunnier, but I am not sure if this is really so, since I don't think it is the case that the Sun, when it's above the horizon, spends most of its time in the Southern-hemispace/hemifield.

Furthermore, I'd like to get a more quantitative estimate of how much direct sunlight each flat gets per day (averaged across seasons). How can I use tools available on the Internet (probably GoogleMaps-based) to find this out?

My only solution to date was to generate Sun trajectories (elliptics) for each flat, using the SunCalc.net website; these suggest the NNW-facing flat hardly ever sees the sun, except around the summer solstice. Is this a reliable estimate, or can anyone suggest another way to solve this riddle?

Between September and March (ie winter, assuming a Northern Hemisphere location) The sun rises in the South East, is Due south at midday and sets in the South west. Your South facing window would get sun for as long as the sun is up. The North West facing window would only get a glimpse of the sun around sunset (and it would then be low and weak)

Between March and September (Northern hemisphere summer) The sun rises in the North East, it is in the south between roughly 6am and 6pm, then sets in the North West. Your Southern window would get 12 hours of sun each day. The North West window would get some sun in the evening, and this sun would be low (weak but shining into the room)

Overall the South facing window would get a lot more sun than the North West window. The reason is that the sun is always above a point on the Earth that is between 23.5 degrees North or South of the equator. For anyone in the Northern Hemisphere, the sun is usually South of them.

A quantatative estimate of this could be done, but the actual values you got would depend on the Latitude and precise aspect of the two windows. However the result from SunCalc looks to be robust. The North West facing window will only get a little sun on summer evenings, and little or nothing during the winter.


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

I was trying to search for Mars from Michigan (at elevation of 30 deg and azimuth of 220 deg) around 11:45 PM with binoculars of 20x60 but didn't get successful.

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

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


How would the pocket cellular clock work? In a museum in Lviv, I saw a pocket cellular clock. I don't have a photo, but it was a small disc that had 2 or 3 filaments in it which were pointed at the stars (one of them was Andromeda, I think). Through analyzing the position of the stars it was possible to determine the time of the night. I'm not sure if knowing the day of the year was promised.

I couldn't find the information about the way the clock was used. Does anyone know how such device was used?

The device that you saw is called a nocturnal. It calculates the local time based on the month of the year and the position of Polaris and one or more other stars.

Since stars change position throughout the night, they can be used to determine time; but the positions of stars change throughout the year, so the input of the month of the year is needed. Polaris is used as a relatively fixed reference point.

More information:

Nocturnal (instrument) - Wikipedia