Chemical reactions in white dwarf and carbon allotropes White dwarfs consists mostly of carbon and oxygen. In my opinion, they are too hot to contain these elements in molecular form and hence chemical reactions does not happen (I think resulting CO2 will also decompose under such high temperature and pressure). I have two questions related to that:

1) After a few billion years, can the WD cool down enough to sustain chemical reactions which will result in production of CO2.

2) Given, there is very high pressure and temperature, after a few billion years what will be the dominant allotrope of carbon (Do we have, in future, diamonds in the sky ;)

I strongly doubt given the extreme density of white dwarfs( I think density of white dwarfs wont change over time), if any of the scenario is possible.

White dwarfs are objects the size of the Earth, but with a mass more similar to the Sun. Typical internal densities are $10^{9}$ to $10^{11}$ kg/m$^{3}$.

White dwarfs are born as the contracting core of asymptotic giant branch stars that do not quite get hot enough to initiate carbon fusion. They have initial central temperatures of $\sim 10^{8}$K, that swiftly (millions of years) drops to a few $10^{7}$K due to neutrino emission.

An important point to make is that after that, the interior of a white dwarf is almost isothermal. This is because the degenerate electrons that provide the pressure support also have extremely long mean free paths for scattering interactions and thus the thermal conductivity is extremely high.

The exterior of the white dwarf is a factor of 100 cooler than the interior. The temperature drop happens over a very thin shell (perhaps 1% of the outer part of the white dwarf), where the degenerate gas transitions to becoming non-degenerate at the surface.

This outer layer acts like an insulating blanket and makes the cooling timescales of white dwarfs very long. From interior temperatures of say $3\times 10^{7}$ K it takes a billion years or so to cool to $5\times 10^{6}$ K and then another 10 billion years to cool to around $10^{6}$K, and such white dwarfs, which must have arisen from the first stars that were born with progenitor masses of 5 to 8 solar masses, will be the coolest white dwarfs in the Galaxy.

At these temperatures there is no possibility of the carbon undergoing chemical reactions, it is completely ionised; the carbon and oxygen nuclei are in a crystalline lattice at these densities, surrounded by a degenerate electron gas. There is evidence that crystallisation does take place, via asteroseismology of some pulsating massive white dwarfs.

The details of the crystalline structure in these objects is unknown, and the subject of theoretical investigation. However, diamond is pure carbon and white dwarfs are expected to be a carbon/oxygen mixture. A further complication is that the process of crystallisation may be accompanied by gravitational separation of the carbon and oxygen, so that the inner core is more oxygen-rich than the outer core.

Original ideas were that the crystalline form would be body-centred cubic (bcc), but other more complex possibilites are opened up by the mixture of carbon and oxygen. bcc Carbon would be a new allotrope of carbon and not like diamond - it is a denser way of arranging the nuclei.

EDIT: To answer a point in the comments. Even if you were to wait trillions of years and allow white dwarfs to cool to the thousands or even hundreds of degrees that you might think would allow electrons to recombine and for chemistry to occur, that is not how it works. In the degenerate electron gas, the typical Fermi energy of the electrons is an MeV or so, compared with the eV-keV of bound electron states, and this is completely independent of the temperature. So the high electron number densities ensure that they will never recombine with the carbon nuclei (a theory first developed by Kothari 1938).


Why would the merger of spinning black holes within the accretion disk of a supermassive black hole cause them to "shoot straight up" out of the disk? I just tried to read the new New York Times article Two Black Holes Colliding Not Enough? Make It Three which links to the new 25-June-2020 Physical Review Letter Graham et al. Candidate Electromagnetic Counterpart to the Binary Black Hole Merger Gravitational-Wave Event S190521g
The event described is the merger of two black holes that were embedded in the accretion disk of a supermassive black hole in the center of a galaxy; i.e. in a quasar (loosely speaking).
The NYTimes article describes the following

In the story that Dr. Graham and his team patched together, the black holes were spinning, which caused a recoil that shot the merged result almost straight up and eventually out of the accretion disk at 120 miles per second, at which point the flare stopped. If the explanation is accurate, the black hole should fall back into the accretion disk at the same speed in a few months or a year, generating another flare. “We’ll be looking for that,” Dr. Graham said.

Question: Why would the merger of spinning black holes within the accretion disk of a supermassive black hole cause them to "shoot straight up" out of the disk? How was momentum conserved, i.e. did something else recoil and "shoot straight down"?

Let me see if I can try answering both parts of your question. The key is a combination of two things: 1) Most of the binary BHs in an accretion disk will have their binary orbits in the same plane as the accretion disk, so that "perpendicular to the binary plane" = "perpendicular to the accretion disk"; 2) The most effective form of binary recoil -- in which, as Steve Linton noted, excess linear momentum is carried away by gravitational waves (GWs) -- causes the merged remnant to be kicked in a direction perpendicular to the binary orbital plane.
For the first part, the combination of a gaseous accretion disk and compact, massive objects orbiting within the disk (stars, white dwarfs, neutron stars, black holes) is thought to accelerate the formation of binary objects (including binary black holes) -- but only for objects that are orbiting in the plane of the accretion disk. Since such objects are orbiting about the central SMBH in the same plane, when two such objects form a binary, they will generally end up orbiting each other in the same plane. So you should imagine a population of binaries within this disk, mostly with their own orbital planes aligned with the plane of the accretion disk. (Note that there is no reason to expect the spins of the individual BHs to be aligned with each other, which will be important later.)
Now we consider the gravitational recoil effect. The simplest scenario -- one that's not actually relevant here! -- for binary-BH merger involves two non-spinning BHs. If they are equal in mass, then everything is symmetric and you get no "kick". If they are unequal in mass, then the lower-mass BH has a higher orbital velocity, and will (due to relativistic beaming) give off momentum in the form of GWs in its orbital direction more effectively than is the case for the more massive BH. So the whole system gives off an excess of (linear) momentum in one direction, and the binary recoils to conserve momentum. There would be no net effect if the orbits were circular and unchanging (because then the "jet" of excess momentum in the GWs would just sweep through $360^{\circ}$ as the BHs orbited each other); but because the orbits are decaying due to the emission of GWs, you end up with a net asymmetric emission of momentum in the GWs, matched by a recoil of the merging binary. But this is in the plane of the binary, so it wouldn't produced the perpendicular kick we're interested in. (See this Astrobites article for a nice picture, taken from this article.)
Although the recoil in the case of non-spinning BHs had been studied theoretically for some time, recent simulations (starting around 2007) showed that if the BHs were spinning (as almost all black holes almost certainly are), and had their spins mis-aligned, then there was an additional, much stronger recoil effect -- a "superkick". The key is that this recoil effect causes the merging binary to be kicked in a direction perpendicular to the binary's orbital plane (see this Astrobites article for a discussion, although it doesn't really offer an nice, simple explanation of the underlying cause). Since, as noted above, the orbital plane of the binary is generally aligned with the plane of the accretion disk, the end result is the merger remnant being kicked approximately perpendicular to the accretion disk.


What could a cloud of mini radio dishes see? Suppose an astronomer gave a 1 m radio dish to 500 people scattered over the face of the Earth and connected them to the internet. The people are directed to set their radio antennae up in their backyard per given instructions. The astronomer would then control the array of radio antennae remotely via the internet.
Could such an array of radio dishes be used to do any kind of meaningful astronomical interferometry? If so, what kind? If not, why not?
I fully expect that the radio dishes wouldn’t hold a candle to the sensitivity other radio dish networks (EVT, VLA, EVLA, etc.) due to the tiny dish size. I expect time synchronization would be an issue, but perhaps they could find a clever way to synchronize. I imagine this array would have very high resolution due to the large number of large and small baselines — extended further by the Earth’s rotation. Between low dish sensitivity, large numbers of dishes and a huge baseline count, I’m curious what such a network could be capable of.

Partial answer:

I imagine this array would have very high resolution due to the large number of large and small baselines

and

I expect time synchronization would be an issue, but perhaps they could find a clever way to synchronize.

The 1 meter dishes are small and so for the dish to have any relevance to the project the wavelength has to be a lot smaller.
To do interferometry your synchronization will have to be extremely good; a 3 cm wavelength would be 10 GHz and the period is 0.1 nanosecond. You'll need to have all kinds of expensive and sophisticated electronics to maintain a stable and drift-free timebase at that level of accuracy.
You might be able to leverage your long baselines but fairly wide fields of view for these small dishes by using timing somehow. If you had ten distant dishes each pointed in 50 different directions and there was a sudden event, you might be able to localize it by arrival time differences, a bit like how lightning detectors work.
I don't know anything about it, but there is also something called intensity interferometry but it's not clear if that's actually helpful at all. See Hanbury Brown and Twiss effect and (paywalled) R. Hanbury Brown, R.Q. Twiss (1954) LXXIV. A new type of interferometer for use in radio astronomy

A new type of interferometer for measuring the diameter of discrete radio sources is described and its mathematical theory is given. The principle of the instrument is based upon the correlation between the rectified outputs of two independent receivers at each end of a baseline, and it is shown that the cross-correlation coefficient between these outputs is proportional to the square of the amplitude of the Fourier transform of the intensity distribution across the source. The analysis shows that it should be possible to operate the new instrument with extremely long baselines and that it should be almost unaffected by ionospheric irregularities.

See Boffin : a personal story of the early days of radar, radio astronomy and quantum optics
See also Dainis Dravens' 2010 BOSON INTERFEROMETRY From astronomy to particle physics, and back especially the radio part.
Only somewhat related, with some discussion of the effect before going on to discuss the optical implementation (but using electronic correlation): Intensity interferometry: Optical imaging with kilometer baselines
I wrote a related answer here: https://astronomy.stackexchange.com/a/42131/7982


What are some dark features on (or qualities to) the moon? I know there are mare, oceani, and lacunae, but anything else that I'm overlooking from lists of lunar features?

Does the "dark"/far side have a particular, maybe Latin scientific name (tho it gets lit as much as the near side, of course), or the unlit area aside from 'night'?

(I'm trying to name a kitten who was born July 19th during the Apollo XI 50th anniversary. All girls, the one white one with soft black extremities is Luna, one all black but the barest white dusting about her neck is Mare, then there's Mare's essentially identical twin, who needs the name.)

not sure about the dark side, but there are numerous other lunar features:

Wikipedia Lunar Features

Maybe Catena (a chain of craters) or Rima (lunar rilles)?


Could the accelerating expansion of universe inidicate we are surrounded by "denser" space? I'm quite novice about all this, only have read some vulgarization books and sites, but I wonder: could all the observable universe be falling outwards on some non-observable, more compact kind of barrier - I'll call it "edge"? Could this be a part of the explanation of cosmological accelerated expansion?

Like say we would be in a kind of bubble, or egg-shape thing? This thing could be inflating, and the closer matter would be from the edges, the quicker it would fall?

Good question "why would we be at the center of it?"... I know already there could be a kind of "flow" of matter at large scales which is already kind of violation of universe isotropy. Could we remove this issue with some kind gravity propagation itself?

Any sense to anybody it makes?

It's a pretty clever idea and a solid question that unseen mass might cause gravity outside the observable that tugs on the universe and might be the cause of dark energy as opposed to some unknown energy that seems to permeate all space.  It's not a new idea, but it's worth answering.

There's 2, possibly 3 (or 4) pretty big problems to that approach.   The first is that Dark Energy expansion is very uniform.  For it to be explained by gravity, that gravity would need to be close to uniform around the observable universe, with Earth close to the center.    A giant "edge" as you called it, or maybe we're in a low mass bubble surrounded by higher mass.    Not impossible, but somewhat improbable that the Earth is in the center of this theoretical gravity field pulling the observable universe apart.

The 2nd problem is time.  Gravity travels at the speed of light, so this gravity field would need to have existed prior to the big bang or be very close to the edge of the observable universe.  Still not impossible, but it's a difficult theory to work with as it requires a few leaps of faith.

A possible 3rd problem is that measurements show that the universe is very flat.  I don't know if a gravity field of the sort you suggest would be flat or not, I suspect not, unless it was enormously large.

Given enough time, maybe they can find a test to detect gravitons (if they exist) or, find dark energy and answer this one with better certainty.  The question is asked here on Quora if interested.

I'll add a 4th problem.   The velocity of observed young galaxies at the edge of the observable universe is close to the speed of light away from us in all directions.  Relative velocity that high is difficult to explain by high mass outside of the observable universe cause that mass would need to be enormous, measurably greater than the mass of the observable universe.   All the factors taken into account, it's an improbable explanation

If I've missed anything obvious or broken any laws of physics with the answer above, corrections are always welcome.


Can the Moon provide momentum to an object in Earth's orbit? Gravity Assisted Boost Can a satellite maintain an equatorial orbit around Earth near the Moon's orbit to receive partial gravitational boosts by gaining momentum as the satellite passes though the Moon's gravity well?

I'm not asking about hanging out in the Lagrange points, but passing between the L2 point and the Moon or the Moon and Earth aligned, in a Moon assisted orbit.

The Moon has a "Roche lobe", where the Moon's gravity dominates, and all you need for the situation you describe is that the Roche lobe lies physically outside the Moon.  That will be true, because the Lagrange points are on the outside edge of that Roche lobe, and it's outside the Moon.  The Moon's Roche lobe is shown here: http://hyperphysics.phy-astr.gsu.edu/hbase/Mechanics/lagpt.html


How is it possible that we haven't discovered anything in the Oort cloud yet? The Oort cloud is a hypothetical cloud of small icy bodies surrounding the Sun at more than 1000 AU. It is thought to be a vast reservoir of comets that occasionally get disrupted, sending comets towards the inner solar system.
I would guess that there are stellar occultations where a body from the Oort cloud passes in front of a background star, obscuring its light. This makes it surprising to me that we haven't detected any objects in the Oort cloud. Perhaps the sensitivity of our instruments is not high enough? Perhaps that in spite of the large number of objects that should exist in the Oort cloud, these occultations are actually infrequent, and you would have to wait tens of years before seeing one? Perhaps nobody has combed the data to find such signals?
Would any of our existing telescopes have been able to pick up the occultation if it was pointed in the right direction at the right time?

It's not very surprising that no Oort Cloud objects have been detected via occultations. They're extremely rare, even for our most advanced space-based observatories.
According to Ofek & Nakar 2010, published about one year after the launch of Kepler, the telescope would be able to detect between $0$ and $\sim100$ occultation events of Oort Cloud objects, assuming that it monitored $\sim10^5$ stars for a time $\tau=3$ years - both of which ended up being underestimates.$^{\dagger}$ Part of the reason for the uncertainty is that the parameters of the Oort Cloud are not tightly constrained. Varying the inner radius $r_{\text{min}}$ from 1000 to 5000 AU, or varying the index $\alpha $ of the number density distribution $n\propto r^{\alpha}$ from $-4$ to $-3$, can produce changes on the expected occultations by a couple orders of magnitude.
The problem becomes apparent when you consider that 1) occultations are intrinsically short and infrequent and 2) they're one-off events. It's not the same as detecting a transiting exoplanet, or observing an expected occultation from a minor planet whose orbit is already known. Therefore, it's hard to tell whether an event is a true occultation or simply a glitch of some sort. The second point of the paper was to present methods for validating possible events.
Ground-based surveys of Kuiper Belt Objects (KBOs) have been performed, but they suffer from atmospheric effects, as you might expect. Given that occultations would last on the order of 1 second, atmospheric scintillation becomes a problem for the high-cadence searches required (Alcock et al.). Given that some ground-based surveys just for KBOs a few kilometers across, like the Taiwanese-American Occultation Survey, have returned no detections (Zhang et al. 2008), it might not be surprising that ground-based Oort Cloud searches are also extremely difficult.

${\dagger}$ By comparison, the CoRoT mission, assuming it monitored $\sim10^4$ stars over the same period, was expected to detect essentially no occultations, according to the same analysis.


Recent origin of Saturn's rings Saturn was formed around billion years ago along with Jupiter. After which it had enough time to form moon which can break into rings.

Then why is it said that Saturn's ring are of recent origin - about few million years ago?

Wikipedia claims that Saturn's rings may be as old as Saturn itself; NASA says that they could be only a few hundred million years old.

The 4.something-billion-years-old fits in with one of the two theories of the rings' formation - that Saturn swept up material from the protoplanetary disk. The date makes sense, because there's no way the disk would last for billions of years.

The newer date does fit in with the moon-torn-apart hypothesis, which is perhaps the dominant hypothesis. The hypothetical progenitor moon could have been torn apart at any point between Saturn's birth and the present day. "A few hundred million years ago" fits in reasonably well.

I should add that all the theories surrounding the rings' formation have reasonable evidence in their favor, though there's quite a lot for the 4.something-billion-years theory (see here, here and here). Thank you, Cassini!


What is Earth's average temperature as a whole? I would like to know, what is the average temperature of Earth as a whole planet, not just its surface temperature. Hope the following scenario will make it more clear what I mean:

Imagine, that we isolated/sealed Earth (up to 30 km above sea level) as it is right now, so that there is no mass or energy being added or escaping and waited several millions of years till the temperature is everywhere the same.

What would that temperature be? When I tried to google it, I only could find references to surface temperature, and not to this.

EDIT: Based on remarks in comment section:

I am not an expert, that is why asked this question, however IMHO of a layman in this area, I presumed that radioactive decay wont play that much of a role (like a +-1%). And the role of a gravity is well beyond me. But the most important thing is, I pretty much wanted to know the average temperature by heat capacity of the Earth at this moment (so basically, all decays and all tidal forces excluded, even if they play significant role).

I would just like to have (a fairly confident) idea about the potential temperature, and no need for highly exact numbers and calculations, +-15% is absolutely fine.

What is Earth's average temperature as a whole?

Sealing the planet or using $E = mc^2$ isn't a practical means to measure the temperature. Here's how the average is usually determined:

Source: Tufts University - Global Warming - Heating by the greenhouse effect

  "... the average surface temperature of the planet is 288 degrees kelvin (15 degrees Celsius or 59 degrees Fahrenheit).".

Source: Space.com - What Is Earth's Average Temperature?

  "According to GISS, the global mean surface air temperature for that period was estimated to be 57 F (14 C). That would put the planet's average surface temperature in 2017 at 58.62 F (14.9 C).".


How does dark matter interact with black holes? Is dark matter trapped near / in black holes, like regular matter is, or does something different happen with dark matter?

It interacts gravitationally.That's all there is to it.

There is a big difference though to the way that normal and dark matter interact with black holes - dark matter is far less likely to be captured.

Given two lumps of matter, one normal one dark, with the same mass and angular momentum with respect to the black hole; only the normal matter is capable of shedding its angular momentum (normally an accretion disc is involved), which allows its orbit to shrink enough to be captured (within three Schwarzschild radii). Dark matter is dissipationless; if it has too much angular momentum it won't get captured.


How much more massive would Pluto have to be to clear its neighborhood? How much more massive would Pluto have to be to clear its neighborhood?

This Wikipedia page does a decent job of describing the orbit-clearing criterion, based on the original paper by Stern & Levison (2002), which can be found here (PDF).

In order to have cleared its orbit over a period of billions of years, an object needs a "Stern-Levison parameter" $\Lambda$ which is $> 1$; Pluto has $\Lambda \approx 3$-$4 \times 10^{-3}$. Since $\Lambda$ depends on the square of the object's mass (see the Wikipedia page, or Equation 4 of Stern & Levison), Pluto would need to be about fifteen to twenty times its current mass in order to cross the threshold.


Will the Heat Death scenario ever occur or just the Big Freeze? According to the Wikipedia page regarding the ultimate fate of the universe, heat death is a different scenario than the big freeze.

It states that in the big freeze, temperature will asymptotically approach absolute zero. However, the following section about heat death states that heat death may only occur if temperature reaches an eventual temperature minimum.

If temperature never truly reaches absolute zero (the temperature minimum), can heat death even occur in the first place?

The Wikipedia article is somewhat confused on this point (at least one of the references does not support the sentence it is supposed to support).

The classic concept of the heat death of the universe was that eventually it will run out of free energy and there will hence be no possibility of any work. A more modern view of heat death is an approach to maximum entropy where the system will be in equilibrium. This does not rule out random fluctuations bringing it away from equilibrium, but they are so rare it does not matter.

Note that it is entirely possible to have a heat death at any temperature.

The big freeze would happen if expansion of the universe led to temperatures decreasing towards absolute zero. However, if there is accelerating expansion this generates a constant finite temperature horizon radiation, about $10^{-29}$ K. This means that the big freeze will never happen.

Had there not been horizon radiation it would in principle have been possible to avoid heat death by storing some background heat in a perfectly reflective box, waiting until the universe become colder, and use it to perform work. This can be repeated indefinitely. (Not really: the timescales quickly become long enough that random fluctuations destroy your device, but that it a separate issue).


How to calculate the integrated source flux density of a continuous source? I'm working with data from radio surveys. My images are .fits files with BUNIT Jy/beam. I want to calculate the net flux from a continuous source (not point source). When I simply added all the intensity values and converted it into Jy using the beam parameters, my professor said that it is not enough. He said, "You have to account for noise by taking 3 or 4-sigma level depending on the distribution of the noise". The distribution of noise (at places other than the source) is nearly Gaussian as I can see from a fits viewer.

By "take 3-sigma level", does he mean that I should fit a Gaussian to the noise, find sigma and then choose the pixels with intensity values greater than 3*sigma? If I do this, obviously I lose a lot of pixels. Also, if this is correct, should I subtract the mean of intensity of "noise" pixels from my source intensity?

An RMS noise is given in the header, with a caution that it varies highly around some sources. Is this RMS noise value in anyway related to the sigma that I calculated earlier?

Maybe too late but probably he is referring to only take into account pixels with an intensity greater than 3*sigma. If not, you will be counting random noise as continuum. Obviously you will lose a lot of pixels, but those pixels doesn't carry any information at all.


Why do stars twinkle but planets don't Many people say that stars twinkle because of Earth's atmosphere. However, the atmosphere is still there when we look at planets and don't see they twinkling.

Is there a clear scientific consensual explanation for the twinkling of some objects but not for others?

Note: just as a metter of distinction, my question is different from this one. In fact, the moon is obviously different from stars, while my question is based on the fact that the only actual difference between stars and planets is that planets move across the sky, differently from stars (since they are the sky's reference). Proof of my words is the late distinction of Uranus as a planet due its extremely slow motion, although it also doesn't twinkle.

The answer to this is that stars are effectively point sources, but planets have disks.  Borrowing the link to Sky and Telescope's explanation from @N. Steinle's comment.

The way I envision it is to think of the light from a celestial object as a fiber optic cable, continually being snipped at with scissors (which stand in for atmospheric turbulence changing the path of light traveling through it).  The cable from a star is a single glass fiber with a very bright beam shining down it.  The cable from a planet is a thick bundle of fibers with a much dimmer beam shining down each fiber in the bundle (compare the surface brightness of Mars with the surface brightness of the Sun)-- the total amount of light coming through the cables is comparable.  The scissors can cut through the single-fiber cable completely, but they can only take out a few strands of the thick bundle.  So atmospheric turbulence can cause the light from a star to be drastically interrupted, while the change for a planet is much less.  Therefore, there is visible twinkling from stars, but not from planets.


Second Telescope: Maksutov Cassegrain vs Refractor I bought my first telescope, a 3-inch refractor a few weeks ago. It has been a wonderful experience seeing the Moon, Jupiter, Saturn magnified and Andromeda, M33, M81 and much more for the first time.

Now I wish to get my second telescope. I was considering :
Explore Scientific ED APO 102mm F/7
Bresser Maksutov Cassegrain 152mm F/12.5

I live in a high altitude region with Bortle 3 skies. I hope to see basic details on Jupiter, Saturn and Mars at high contrast and clarity and perhaps the disc of Neptune. Also occasionally deep sky objects. Portability and fast setup time are also a consideration. The MC is about 60% of the price of the APO.

In addition to my 26 & 12 mm Plossl eyepieces I plan on getting a 50mm Erfle (Field Stop 40mm) and a 6mm Plossl.

For mount, I was contemplating between Bresser Exos 2 and iOptron CEM25P.

I wish to know your experience with refractors and MCs, in general, or with these specific models. Since the MC is cheaper and has a bigger aperture, will the APO have any advantages eg.contrast? Also, what are your thoughts on the mounts?

Thank you for your answer!

I have viewed through many telescopes over the years and in doing so have met many people like yourself. To suggest a F/12.5 Maksutov will satisfy you would be the equivalent of telling someone to dive into the deep end of a swimming pool without knowing if that person can swim.

If it's bang for buck that motivated your question I think you would get the most if you just explore the capabilities of the 3 inch refractor for a while. There is a lot to be seen with that telescope and you have barely scratched the surface.

Some prefer electronic eyepieces. Have you seen what they are capable of doing? I viewed with eyepieces for fifty years before trying a Mallincam ten years ago. I haven't used an eyepieces for serious viewing since.

Take a look at the NightSkiesNetwork. It is a group that regularly broadcasts their observing sessions.

https://www.nightskiesnetwork.ca/about

I cannot think of any better advice for someone like yourself than to stick your toe in the water for a few seconds before jumping in.


Can you provide a visual description of the sun rise? The Sun rises in the east only two times a year. The other days, the Sun rises in the east, but a little bit to the north, or to the south. How can you know the degree of how much north or south it is? Does it vary in different hemispheres?

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

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

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

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


Does the gravitational attraction near the surface of dense celestial objects diverge from inverse square? Does the gravitational attraction near the surface of dense celestial objects (neutron star, white dwarf itc) diverge (to infinity) from inverse square?

This question is inspired by the similarity between EM and gravity (inverse square force.). The paper by John Lekner here (doi:10.1098/rspa.2012.0133) shows that there is an electrostatic attraction between charged spheres no matter the polarities of the charges and that it diverges at close separation until electrical short for almost all charge ratios. I am wondering if there is a similar kind of gravitational inverse square divergence for anything other than a black hole.

Actually make this for a black hole as well, although I know that a black hole is not thought to have a normal surface.

Leckner's paper deals with the effect of induced polarization on the spheres. Electrons are redistributed, making the force different from what one would expect. The gravitational counterpart is tidal distortion: since the gravitational field is non-radial when you have two heavy masses close to each other, matter will move to make the surface an equipotential surface. This means that the gravitational acceleration at the surfaces will not be constant at all locations.

Doing an analytic solution of how two ellipsoids attract each other and deform seems to be tractable (e.g. see this question) but algebraically very tedious and likely involves lots of special functions. See the addendum below for an approximate numeric model.

Black holes produce another complication: since spacetime nearby is curved and expanded the meaning of the distance in the inverse square law becomes problematic. The Paczyński–Wiita potential is an approximation of the potential, and it deviates from the $U=-GM/r$ as $U_{PW}=-GM/(r-R_S)$ (where $R_S$  is the Schwarzschild radius). It makes the force increase faster than the classical potential as we approach $r=R_S$.

Addendum: I did a numeric exploration of the force between two ellipsoidal, self-gravitating masses with centres of mass separated by a given distance. To find the shape I started with spheres and adjusted the semi-major axis (while preserving volume) so the potential along the surface became more equal at the poles. After a few iterations this gives a self-consistent shape. Then I calculated the force (the derivative of the potential) due to this shape on the other mass.

The result is indeed that the force increases faster than $1/r^2$
as the bodies approach each other, since they elongate and eventually merge (a bit before this they will deviate from my ellipsoidal assumption). If one multiplies the force by the squared distance the product should be constant for pure $1/r^2$ forces, but it starts to increase as they approach enough. Note that this is a non-rotating model: with rotation the numbers will change and the ellipsoids will become tri-axial, but I suspect the qualitative behaviour stays the same.


Is there a distinction between NEOs and near-Earth asteroids? Is there a difference? My "real question" is in Space Exploration Meta (neo (near-earth-object) and near-earth-asteroid tags, do we need both?), but I think that astronomers will be able to help understand the situation and terminology.

Question: Is there a distinction between NEOs and near-Earth asteroids? Is there a difference? Or is it a distinction without a difference?

I noticed for example that here in Astronomy SE there is just near-earth-object.

Yes, Near Earth Objects (NEOs) include asteroids (Near Earth Asteroids (NEAs)) and a few percent of comets (Near Earth Comets (NECs)). As shown by the Update to Determine the Feasibility of Enhancing the Search and Characterization of NEOs (NEO SDT Report) in Section 2.3, the risk from Near Earth Comets is about 1% that of asteroids. So "NEO" tends to be assumed to mean "asteroid" unless otherwise stated.


Are there non thermal filaments outside our Galaxy? The question says it all: are there observations of something like non thermal filaments in other Galaxies other than our?

You have two questions here, which are somewhat mutually exclusive. I'll start with the second.
Are there observations of something like non thermal filaments in other galaxies other than our?
The paper you linked pretty plainly spells out that only a handful of nonthermal filaments (NTF) have been found, and all of them have been found near our galactic core. To quote the paper:

Observationally, the NTFs are unique to the GC

Specifically they mention that there were only 9 known NTFs previous to this paper:

nine previously identified isolated NTFs—the most recent NTF discovery is described by Reich (2003)

and this paper reports that they found another 20:

New high-resolution, wide-field 90 cm VLA observations of the Galactic Center (GC) region by Nord and coworkers have revealed 20 nonthermal filament (NTF) candidates.

So it seems the answer to your question, at least according to this paper, is that no, there are no observations of NTFs in other galaxies at this point in time.
Are there non thermal filaments outside our Galaxy?
I interpret this to mean, could there exist NTFs in other galaxies, and not so much have we actually seen NTFs in other galaxies. I think the answer to this, despite being a total guess with no proof to back it up, is a resounding yes. There's no reason to think our galaxy and it's structure/evolution is unique. Very likely NTFs exist in many other galaxies which formed under similar processes.
This article by the NRAO details an attempt to explain where these NTFs originate from and the proposed solution is that they are a result of starburst activity. To quote from the article:

Yusef-Zadeh determined that a number of the non-thermal filaments seemed to connect to concentrated areas of thermal emission, which identify pockets of star formation. Galatic Center Combined radio image from the Very Large Array and Green Bank Telescope. The linear filaments near the top are some of the nonthermal radio filaments (NRFs) studied by the researchers. Other features, such as supernova remnants (SNRs) and the area surrounding our Galaxy's supermassive black hole (Sgr A) are shown.
"What this showed us is that two seemingly disparate processes, thermal and non-thermal radio emission, can be created by the very same phenomenon," said Yusef-Zadeh. "In this case, that phenomenon is pockets of starburst activity."

If this is indeed the source of NTFs, then any galaxy which has or did have starburst formation would likely have NTFs as well.


constructing a transit light curve What type of equipment would I need to construct a transit light curve of a planet. It doesn't have to be an exoplanet. If somebody could chuck a list that would be really helpful.

Basically you need to have the following equipment list
1.- A mechanical system track objects: you can build with 3d printer like:this prototype and/or python tracking mechanical controller
2.- A cmos/ccd sensor camera with over exposed function and color filters ,like this usb computer or use dslr camera telescope or like
3.- Automatic comparative histogram  (Software), you can use python OpenCV
enter link description here
I hope this links can help you, basically you need compare all histograms of same object


What is the orbit of Earth around the Sun? According to wikipedia and other sources, a planet and a star always move in a circular orbit around the common center of mass of the both bodies, in case  of Earth and sun, this center of mass lies inside the sun. Therefore, Earth should orbit the sun in a circular orbit. So, my question is, why does the Earth orbits the sun in an Elliptical orbit?

And, if it does, and one focus is the sun, then what is the other focus?

According to wikipedia and other sources, a planet and a star always move in a circular orbit around the common center of mass of the both bodies ...

This is not true. In the absence of other gravitational sources, a planet and a star move in elliptical orbits about the common center of mass. Ancient scientists assumed circular orbits, but only because circles were somehow most pure. There were problems with this: It didn't match what those ancient scientists saw. So they modified these circular orbits by tacking on other circles. It was a mess.

Kepler cleaned this mess up by noting that the orbits of the planets are very close to elliptical rather than circular. This empirical result was later strengthened by Newton, who derived that elliptical orbits (not circular orbits) are a natural consequence of a force that follows an inverse square law. Gravitation is such a force.

The orbits of the planets about the solar system are not quite elliptical because planets are attracted to one another as well as to the Sun. Nonetheless, the orbits are still very close to elliptical. The Sun is more than 1000 times more massive than Jupiter. Jupiter, Saturn, and the other planets represent tiny perturbations on the nearly elliptical orbits of the planets about the Sun.

What about other perturbing effects, such as nearby stars and the Milky Way as a whole? These are immeasurably small. A passing star will perturb planetary orbits over the course of the tens of millions of years or more. The Milky Way as a whole has essentially no effect whatsoever. Theoretically, the effect most certainly does exist, but it is so small that it can be ignored, even over the lifespan of the solar system.


Are nebulae only visible from a distance? There's plenty of data about the density of nebulae, and obviously they're pretty thin.  This is space, after all, and these things are really, really big.

But I'm wondering about the practical experience of that density.  Are the colorful patterns of gas only visible when you're far away and looking through them, or if you were inside a nebula would you see colors all around you instead of the black of space?  Are any dense enough to obscure the view between planets of the same solar system?  Between a planet and its moon?

The colorful patterns of gas are caused by excitation of gas atoms, e.g. oxygen, from radiation of a nearby star. You would see the same colors as from a distance, as the light is emitted in almost random directions. But you wouldn't see the same region of nebula shining at the same time as seen from Earth, when excitation is caused by a flash of radiation from a central star, because the radiation travels concentrical away from the star, and the distances the light has to travel from the excited part of the nebula to the observer is dependent of the location of the observer.

Light may be scattered non-uniformly by dust-particles, so looking different from differnt positions, e.g. forming a halo around stars, depending of the composition and shape of the particles. The interplanetary dust in our solar system causes zodiacal light.

Our solar system is in an interstellar dust cloud right now, but it's not very dense.

Very dense clouds can form, but would become instable and collapse to a star or a protoplanetary disk, compare this site about dark nebula.


Pekin Chang'e 4 Mission, Shadow? I'm wondering why there is a shadow under the Yutu 2 rover, of the Chang'e 4 mission, since it is on the far/dark side of the moon.

The Chang'e 4 lander, as seen by the Yutu 2 rover. (Image credit: CNSA)

See the Far Side of the Moon in This Amazing Panorama from China's Chang'e 4 Lander

Oh sorry, I just understand what is going on. Being the far side of the moon, did not mean that it's never exposed to the sun.
So the Expression "Dark Side Of The Moon" is not so true.

Has  written in Wikidepia :

Tidal forces from Earth have slowed down the Moon's rotation to the point where the same side is always facing the Earth—a phenomenon called tidal locking. The other face, most of which is never visible from the Earth, is therefore called the "far side of the Moon". Over time, some parts of the far side can be seen due to libration.[7] In total, 59 percent of the Moon's surface is visible from Earth at one time or another. Useful observation of the parts of the far side of the Moon occasionally visible from Earth is difficult because of the low viewing angle from Earth (they cannot be observed "full on").

The phrase "dark side of the Moon" does not refer to "dark" as in the absence of light, but rather "dark" as in unknown: until humans were able to send spacecraft around the Moon, this area had never been seen.[1][2][3][4] While many misconstrue this to think that the "dark side" receives little to no sunlight, in reality, both the near and far sides receive (on average) almost equal amounts of light directly from the Sun. However, the near side also receives sunlight reflected from the Earth, known as earthshine. Earthshine does not reach the area of the far side that cannot be seen from Earth. Only during a full Moon (as viewed from Earth) is the whole far side of the Moon dark. The word "dark" has expanded to refer also to the fact that communication with spacecraft can be blocked while the spacecraft is on the far side of the Moon, during Apollo space missions for example.


What is the current theory for the formation of the Earth's Moon? Given that the Earth and Moon are often said in the literature to have very similar geochemistry, what is the current theory as to how the Earth's Moon formed?

The current accepted theory is known as the Giant Impact Hypothesis, where according to this NASA webpage "Origin of the Earth and Moon" (Taylor) a Mars sized object collided into the early Earth.

This theory allows explanations of (from the link above):

The chemical makeup:

  The giant impact hypothesis is consistent with our ideas for how planets were assembled and explains some important features of the Earth-Moon system, such as why the Moon has only a tiny metallic core.

and in terms of orbits:

  To account for the amount of angular momentum in the Earth-Moon system, Cameron estimated that the object would need to be about 10% the mass of Earth, about the size of Mars. (Angular momentum is the measure of motion of objects in curved paths, including both rotation and orbital motion. For the Earth and Moon this means the spin of each planet plus the orbital motion of the Moon around the Earth.)

Image source

However, recently, there has been some revisions to this theory.

According to NASA's page "NASA Lunar Scientists Develop New Theory on Earth and Moon Formation", based on concerns that the Mars-sized would likely to have had a different composition to the Earth (inconsistent with the current similarities in geochemistry).

The new hypothesis is (from the link above):

  After colliding, the two similar-sized bodies then re-collided, forming an early Earth surrounded by a disk of material that combined to form the moon. The re-collision and subsequent merger left the two bodies with the similar chemical compositions seen today.

This is also discussed in "Huge Moon-Forming Collision Theory Gets New Spin" (Wall, 2012), basing the revised hypothesis on the rotation rate, which is theorised to have been very fast, from the article:

  Earth's day had been just two to three hours long at the time of the impact, Cuk and Stewart calculate, the planet could well have thrown off enough material to form the moon (which is 1.2 percent as massive as Earth).

Further discussion, based on the geochemical makeup is discussed in "Geochemical Constraints on the
Origin of the Earth and Moon" (Jones and Palme) conclude that:

  Although none of these observations actually disproves
  the giant impact hypothesis, we find it disquieting that the
  obvious consequences expected of a giant impact are not
  observed to be fulfilled.

So, there is still some question as to precisely how the Moon was formed.


Does Absolute Velocity Exist? It seems everyone is on the same page about there being no "absolute velocity" due to the fact that everything is relativistic. However, this leaves me confused. This seems to be disconnected with the concept that the speed of light is in fact an unchanging constant. But to state that all speed is relative to an object is to essentially say, "speed does not exist, but is only relative to an observer". But yet, the speed of light is a calculable constant that never changes. How then, can we say, there is no such thing as absolute speed?

If I throw a flashlight in space that is turned on 50mph, does the speed of the emitted light from the flashlight increase by 50 miles per hour in the direction I threw it? Of course not. We know this. But if this is truly the case, how then can there not be an absolute velocity which could be calculated by determining the speed of light in relevance to the object we're measuring (since the speed of light is constant)?

Please help because this is bothering me.

Chris, you are actually on the verge of understanding how special relativity works. You're very close. You only need to take one extra step.

  to state that all speed is relative to an object is to essentially say, "speed does not exist, but is only relative to an observer"

This is correct. Absolute space, as in newtonian physics, does not exist. Same for absolute time. Both are just relations between objects, and between events.

As long as you hang on to the notions of absolute space and absolute time, relativity cannot be understood. If you let go of them, relativity actually seems natural.

Einstein simply noticed this:

A) All previous experiments (like Michelson-Morley) have shown that whenever you measure the speed of light, you always get the same result, no matter what.

B) Maxwell's equations for electromagnetic fields (such as light) include the speed of light, and the value of that speed is frame-independent. Which means, they always show the speed of light to be the same, no matter how the observer is moving.

Both A and B tell the same thing: speed of light is always the same. But how can that be, when different observers are moving through absolute space at different speeds?

The answer is - absolute space is not a thing. Space is just a background for a special type of relation between objects, called "distance". Time is just a background for a special type of relation between events, called "duration". But neither space nor time are things, separate from objects and events. Only when you put together a whole lot of "distance" type of relations, space seems to appear.

The really important thing is speed of light. It's a fundamental constant of nature, like the Planck energy, etc. This universe is built in such a way that speed of light and other constants like it remain the same no matter what.

But space and time are derivative notions. They are not primordial realities like the speed of light. So they depend on the observer. For me, the bundle of distance relations we call "space" looks a certain way. For you, the same bundle looks a bit differently. This is okay, because space is not absolute; it's derived from a lot of other things.

When you throw the flashlight, the speed of light from it is the same for everyone. But how can that be, since different observers are moving in different ways? Simple: they all see the bundles of distance relations differently; those differences are such that speed of light always remains the same.

And that's special relativity in a nutshell.


Can Doppler shift make colors of two stars at same distance different? As you know, Doppler effect can change the wavelength of the light.
Suppose there are two stars at same distance and same temperature. If one star is receding and the other star is approaching, their colors would look different.
What I want to know is, how probable is it that two stars show different colors because of Doppler shift? In other words, if two stars at same distance have different colors, say blue and red, then is it safe to assume that two stars have different surface temperature?
Actually, this was the question from the final exam for Wave & Heat.

Q. Stars emit radiation whose spectrum is very similar to that of an ideal blackbody. Two stars, identical in size, are at the same distance from us. One of the stars appears reddish in color while the other one looks very blue. Choose the correct statement.
(a) The red star is hotter.
(b) The blue star is hotter.
(C) They could both have the same surface temperature.

I read that some stars near the galactic center moves very fast. http://en.wikipedia.org/wiki/S2_(star)
If we only consider Planck distribution, only (b) is possible. I'm wondering whether (c) is also a realistic case, or it is only conceptually possible.

As you correctly mention, there are rare stars which move particularly fast. Check also hypervelocity stars section here, where a stellar population moving at around $10^3 \textrm{km/s}$ is described (by the way, these velocities were most likely produced in the galactic center). Note also, that these stars are too fast to stay in the disk for long times. As another sidenote, galactic center in not accessible in visual due to dust extinction.

Redshifts are calculated particularly easy:
$$z=\dfrac{v}{c}$$,
where $v$ is the velocity along the line of sight, $c$ is the speed of light, $z=\dfrac{\delta\lambda}{\lambda}$ and $(v/c)^2\approx 0$ (which holds even for hypervelocity stars).

Because hypervelocity stars are rare, they most likely can be found only close to regular stars. Yet, being very optimistic, take $v=3\cdot 10^3 \textrm{km/s}$, and you obtain that $\dfrac{\delta\lambda}{\lambda}=10^{-2}$ under very favourable conditions. In general it is not small, but when it comes to colors, the change is negligible.

Say, green light at $\lambda=555 \textrm{nm}$ would now have $\lambda=550 \textrm{nm}$ or $560 \textrm{nm}$ in wavelegth, i.e. would still be very green. So Doppler effect has no bearing on your problem.

What has possible bearing, however, is dust reddening. One of the two stars might be obscured by the dust, unlike the other one, and look reddish. So, both b) and c) are possible answers.

P.S. And actually, if a star "appears reddish", it is most likely dust extinction. Black body temperature colors are either blue, bluish, white, yelowish, yellow, orange or red.


Where do new stars get their hydrogen from? When stars run out of hydrogen, they explode (though they also use heavier elements for some time) and form nebulae. In the nebulae, new stars are born which use hydrogen as their fuel.

So, my question is from where do these new stars get hydrogen for their fuel?

Stars only burn hydrogen in their core, where the temperature gets high enough for nuclear reactions to occur. They end their lives when they run out of fuel in the core, but lots of hydrogen still exists in their envelopes.

The Sun will burn 5-10% of its mass before exhausting its core. When this happens, the core will contract due to the radiation pressure disappearing. It then starts to burn hydrogen in a shell around the core. Eventually, when the Sun dies, it will have burned less than half of its hydrogen. Larger stars burn an even smaller fraction.

This means that, when stars die they still leave hydrogen behind for the next generation.

Galaxies can still run of of gas, though. Since after all, each $M_\odot$ of star formed burns of the order of $1\,M_\odot$, if a galaxy isn't fueled with new gas it will become depleted on a timescale of order $1$ over its specific star formation rate sSFR, which is its star formation rate SFR measured in Solar masses per year, divided by its stellar mass $M_*$ in Solar masses:
$$
t_\mathrm{depl} \sim \frac{1}{\mathrm{sSFR}}=\frac{M_*/M_\odot}{\mathrm{SFR}/M_\odot\,\mathrm{yr}^{-1}}.
$$
For instance, a $10^9\,M_\odot$ galaxy with a star formation rate of $10\,M_\odot\,\mathrm{yr}^{-1}$ will become depleted of gas in roughly $10^8$ years.

Gas-depleted galaxies do exist, and the more depleted they are, the lower the star formation rate is (e.g. Rose et al. 2010), but in general the timescale is longer than the above, since galaxies also accrete gas from the surrounding circumgalactic medium.


Where did this famous Planetary Precession Formula come from? The following equation (which I shall term the Planetary Precession Formula, PPF for short) famously appeared in a 1915  publication by Einstein where he indicated how it could be derived from his General Theory of Relativity (GTR).

$$\epsilon = \frac{24 \, \pi^3a^2}{c^2 T^2(1-e^2)}$$

where $\epsilon$ is the (anomalous, non-Newtonian) angular precession per orbit, $a$ is the orbit semi-major axis, $c$ is the speed of light, $T$ is the orbital period, $e$ is the orbit ellipticity.

The PPF formula accurately predicts the (anomalous, non-Newtonian) precession of Mercury and other Solar planets.

The formula was known in scientific circles well before 1915.  For example Gerber (1898) derived it from his own (widely-derided) model of gravity.  In the internet article Gerber's Gravity it is written that

  It became a fairly popular activity in the 1890s for physicists to propose various gravitational potentials based on finite propagation speed in order to account for some or all of Mercury's orbital precession.  Oppenheim published a review of these proposals in 1895.  The typical result of such proposals is a predicted non-Newtonian advance of orbital perihelia per revolution of... >

$$k\,\frac {\pi\,m}{L \,c^2} =  k \frac{4 \, \pi^3a^2}{c^2 T^2(1-e^2)}.$$

where $L = a(1 - e^2)$  is the semi-latus rectum of an ellipse, $m$ is a function of the angular speed $\omega$ of an orbiting planet: $m = a^3 \omega^2$ with $\omega = 2\pi/T$ and $k$ is a constant which might be derived from theory.

Clearly with $k = 6$ we get the PPF formula given above.

I wish to know where the $k\pi m/Lc^2$ expression comes from. From the article it would appear to come from the 28-page review paper by Oppenheim, 1895 which is scanned here. I have been through the scans of this paper but without finding that equation explicitly (the paper is in German which I know very poorly, Google Translate helps a bit but leaves a lot of ambiguity). It might be that the anonymous author of the article extracted the expression from a review of Oppenheim's paper or even the original (French & German) papers themselves, but he is not contactable.  Maybe someone here is familiar with this era of astrophysical history and can point me in the right direction?

I do not know where that formula was first published in full, but Oppenheim at least does something very close to it. First, keep in mind some of the relevant symbols in Oppenheim, though they're rather standard:
$$\def\anode{☊}\begin{eqnarray*}
k &=& \small\sqrt{G} = \small\text{Gaussian gravitational constant}\\
\anode &=& \small\text{longitude of the ascending node}\\
\omega &=& \small\text{argument of perihelion}\\
\varpi &=& \omega+\anode = \small\text{longitude of perihelion}\\
n &=& \small{k\sqrt{(m_0+m_1)/a^3}} = \small\text{mean motion}
\end{eqnarray*}$$
We can see that if we massage the notation, the proportionality we're looking for is equivalently stated as:
$$\frac{\pi m}{Lc^2} \leadsto \frac{\pi GM}{a(1-e^2)c^2} \leadsto \frac{\pi n^2a^2}{(1-e^2)c^2}\text{.}$$
Since the mean motion is $n\equiv 2\pi/T$, where $T$ is the orbital period,
$$\frac{n^3a^2}{c^2} = 2\pi\frac{n^2a^2}{c^2}\frac{1}{T}\text{,}$$
which means that if we want to talk about a perihelion advance of $\delta\varpi \propto \frac{\pi n^2a^2}{(1-e^2)c^2}$ per orbit, it is equivalent to talk about a term in the form
$$\frac{\mathrm{d}\varpi}{\mathrm{d}t} \propto \frac{n^3a^2}{(1-e^2)c^2} = \frac{n^3a^2}{c^2}\left(1+e^2+\mathcal{O}(e^4)\right)\text{.}$$
I cannot find where, if indeed anywhere, Oppenheim considers the missing factor of $(1-e^2)$ as belonging to the anomalous precession, but otherwise the formula is definitely there. I suspect that he only uses a circular orbit approximation, $e^2\approx 0$, because it's nowhere to be found in Sec. IV (Weber's 1846 theory) and performing his calculation (sans $1-e^2$) gives me $\delta\varpi = 13.72''$ per century for Mercury, in good agreement with his stated result of $\delta\varpi = 13.65''$, whereas putting in a factor of $(1-e^2)$ by hand gives $\delta\varpi = 14.32''$ instead.

Perhaps Oppenheim did not consider it explicitly and the author of the MathPages took it as obvious that the eccentricity factor should be there. Or perhaps there's a side comment in the text that I don't see; unfortunately, I'm nowhere fluent enough in German to understand much of what's going on.


How do we define "up and down" in space? Every model of the solar system seems to follow a specific orientation in which the upper side of the system is Earth's north half and the lower side of the system is Earth's southern half. So is North and South in space, or up and down based on Earth's up and down?

For humans up and down is related to gravity. So you would need to choose a reference first. If you are in outer space and your feet point to the sun and your head to Jupiter, you'd be rightside up (if the sun is the center) and upside down (if Jupiter is the reference).

North and south do not have any physical relationship with up and down. Neither in our planet, nor in space.

In space you have to choose a coordinate system. In those systems you have a reference plane, latitude and longitude. For latitude (+90 to -90) the positives numbers are those that would correspond to the north of the equatorial plane.


What if a primordial black hole went through the Sun? How much mass would the Sun lose to a primordial black hole that (initially) has 10 Earth masses, passing through or close to the center of the Sun at solar escape velocity? How massive would the Sun end up and how massive would the black hole become?

The answer is that there is a limited rate at which matter that can be crammed into a black hole of this size. That rate is small enough that the black hole will traverse through the Sun hardly picking up any mass (when expressed as a fraction of its original mass).
Details:
You could use the Bondi-Hoyle accretion rate:
$$\dot{M} = \frac{dM}{dt} = \pi \rho \frac{G^2 M^2}{c_s^3}\ ,$$
where $\rho$ is the density of the solar material and $c_s$ is its sound speed.
The biggest values will occur at the centre because the density is greatest at the centre and increases more rapidly than the cube of the sound speed. Using rough values of say, $c_s \sim 500$ km/s and a density of $10^5$ kg/m$^3$, we get, for $M=6\times 10^{25}$ kg,
$$ \dot{M} = 4\times 10^{19} \left(\frac{\rho}{10^3\ {\rm kg\ m}^{-3}}\right)\left( \frac{c_s}{300\ {\rm km\ s}^{-1}}\right)^{-3}\ {\rm kg\ s}^{-1} $$
The black hole would be travelling at around 600 km/s, so would traverse the central 10% of the Sun in around 100s (the accretion rate gets orders of magnitude lower in the outer parts of the Sun, so can be neglected), so might increase its mass by $4\times 10^{21}$ kg (or 0.007%). This would be insufficient to slow it appreciably and it would shoot out of the other side and disappear into space again. Note that a larger black hole would pick up more mass as a fraction of its original mass because $\dot{M} \propto M^2$.
All this assumes that the accretion can occur at the Bondi-Hoyle rate. This is very unlikely for two reasons - the first is turbulence which reduces the accretion efficiency. The second is radiation pressure.  There will be some sort of super-heated region immediately around the black hole (just imagine trying to compress $4 \times 10^{19}$ kg of material into a 20 cm diameter hole every, the event horizon diameter for a 10 Earth-mass black hole, every second) and the radiation from this region will exert an outward pressure on the surrounding gas. The limiting Eddington accretion rate, where the radiation pressure from the hot gas immediately around the black hole, exerts enough pressure to halt further infalling gas, is given by
$$ \dot{M} \leq \frac{4 \pi G M m_p}{\epsilon c \sigma_T}\ ,$$
where $m_p$ is the proton mass and $\sigma_T = 6.6\times 10^{-29}$ m$^2$ is the Thomson scattering cross-section for free electrons in the solar plasma (i.e. the effective area an electron presents to radiation). The $\epsilon$ parameter is the radiative efficiency of the infalling gas (i.e. what fraction of its kinetic energy is radiated away) and is usually assumed to be of order 0.1. Taken together, this gives a maximum accretion rate (for a ten Earth-mass black hole) of
$$ \dot{M} = 4 \times 10^{10} \left(\frac{\epsilon}{0.1}\right)^{-1}\ {\rm kg\ s}^{-1}$$
Whether the Eddington limit scales down to micro black holes and dense environments like this, I am unsure. But it seems clear that the accretion rate will be way lower than the Bondi-Hoyle rate. I conclude that the Bondi-Hoyle accretion rate is probably a vast over-estimate, and that the black hole would shoot through the Sun and leave on the other side, having hardly accreted any mass (as a fraction of its own mass).


At what distance from Earth would our Sun be the same apparent magnitude as the next brightest star in the sky? When I stand outside looking at the night sky, to my untrained eye, everything except the moon looks like a star.  I know intellectually that some are planets circling our sun, and some are entire galaxies far away, but they all look basically the same.

How far out do you have be be from our sun for it to appear the same as any other 'star' in the sky?

Edit to clarify

As we migrate out through the solar system, when we look up to the sky the sun will get dimmer the farther away we get. On earth there is no doubt which star is our sun.

As we occupy bodies in the solar system farther from the sun, where will we be when the sun appears to be the same brightness as any other star in the sky?

One way to answer would be to consider the brightest star in our sky (other than the Sun), which is Sirius.  Then determine how far you would have to be from our Sun for it to be as bright as Sirius is from here.

That turns out to be 1.8 light years.  That's not even halfway to the nearest star, so if you're in any other star system, then our Sun is just another star.  If you're anywhere in our solar system, even way out in the Oort cloud, then our Sun is way brighter than anything else.


How does water exist within the frost line of a star? My supposition is that within the distance of the so called frost line around a star, water in empty space sublimates, evaporates. I further suppose that this means that vaporized water is pushed to outside the frost line by solar wind or some other mechanism. (Please correct me if this is wrong)

But there are some ways for water (not necessarily liquid) to exist within the frost line. I wonder under what conditions that is possible. What the origin or history such water can have. And how common or likely it is.

I suggest some possibilities to be extended, elaborated or refuted:

Water under enough atmospheric pressure to keep it from escaping into space.
Water in cold shadow, as in some craters on the lunar poles.
Water inside rock, as in the mantle of the Earth (and other planets?)
Water ice transported by comets across the frost line.
Water vapor hanging around in empty inner inter-planetary space, for example after a planetary collision which melts most matter into a cloud. Or maybe permanently.

When you talk about the frost line you are talking about a protostar, and its protoplanetary disk, not an actual star. So, there is no empty space nor strong solar winds.

Inside the frost line, water does not condense into ice grains, but the individual molecules could as well become attracted by the planets-to-be.


Could dark energy be negative gravity? Main question: Could dark energy (the mysterious accelerating expansion of the universe) be explained by "negative gravity"?

"Spin off" questions:

Does antimatter have negative gravity?
If antimatter has negative gravity, and is present in the universe with equal amounts of matter and antimatter, would it produce the observed amount of dark energy in the universe?

My theory that I am trying to validate:

Gravity is the flow of space into matter, and it flows out of antimatter.  If this is true, antimatter should have "negative gravity", I would expect it to have a repelling effect, while matter has an attracting effect.  For example, if antimatter were on earth (somehow without annihilating), it would fall to the planet.  However, if there were an antimatter planet, objects would "fall up" off it.  In fact, I believe the antimatter planet would break apart because of this quality.  Furthermore, I don't believe antimatter planets or stars would form because of this quality (e.g. it would remain anti-hydrogen).  I also theorize that it would be in parts of the universe where matter isn't (e.g. in the voids outside the galactic clusters), and would be very diffuse (because it would repel itself).  I believe the theorized repelling effect of antimatter could be what we are observing as dark energy (hence why I am asking these questions).  I just searched for this online to see if anyone else has this theory, and yes, someone has something very similar here: http://www.universetoday.com/84934/antigravity-could-replace-dark-energy-as-cause-of-universes-expansion/

** Edit **

It seems others have caught on and simulated the idea of negative mass as dark energy in the universe.  They have run simulations with equal mass and negative mass and ended up with a result that resembles the current universe!  They say negative mass might also explain dark matter!  Check it out here: https://youtu.be/MZtS7cBMIc4

This video also shows how negative mass might react: https://youtu.be/uAJlg8MDAlU

Could dark energy (the mysterious accelerating expansion of the universe) be explained by "negative gravity"?

But it already is "negative gravity". In general relativity, the stress-energy tensor $T_{\mu\nu}$ describes the energy, momentum, and stress of matter in spacetime. Through the Einstein field equation, it is connected with Ricci curvature $R_{\mu\nu}$, which consists of half of the twenty independent degrees of freedom of spacetime curvature.

The intuitive geometrical meaning of Ricci curvature is as follows: suppose you have a small ball of test particles, initially all at the same velocity with respect to one another, and allow them to fall freely under gravity. Then the Ricci curvature (contracted with their intitial velocity) measures the initial acceleration of the volume of the ball of test particles divided by the initial volume. Thus, this is a good way to tell if something is locally "attractive" or "repulsive", based on whether a ball of test particles around it starts shrinking or expanding.

Through the Einstein field equation, stress-energy is connected to Ricci curvature. As a result, the volume of the small ball of test particles shrinks proportionally to $\rho+3p$, where $\rho$ is the energy density and $p$ is the average of the principal stresses--for a perfect fluid, the pressure. Naturally, if $\rho+3p < 0$, this is "negative gravity" in the sense that the test ball will start expanding instead. This quantity is also what's seen in the cosmological Friedmann equations. Since the Friedmann-Robertson-Walker family of solutions assume that the universe is homogeneous and isotropic on the large scale, it's sensible that treat the large-scale the same as the above local condition.

Dark energy has $\rho+3p<0$, and in particular (for cosmological constant), $\rho = -p > 0$.

To recap: (1) in general relativity, gravity depends on the stress-energy tensor, which has more than just the mass-energy density, and numerically pressure is more important because there are three spatial dimensions while only one time dimension, (2) we can characterize locally "negative gravity" as $\rho+3p < 0$, or, more generally, as any the violation of the strong energy condition, because this characterizes whether a small ball of test particles is attracted or repulsed by the local stress-energy content.

  Does antimatter have negative gravity?

Not in general relativity. Energy is the gravitational charge, and antimatter has positive energy; thus it should gravitate in the same way.

  Gravity is the flow of space into matter, and it flows out of antimatter.

If you're thinking of some analogy with electric charge, with electric field lines flowing out of positive charge and into negative charge, antimatter has positive mass, which is required by quantum field theory.

  ... I just searched for this online to see if anyone else has this theory, and yes, someone has something very similar here: ...

Some particles, such as photons, are their own anti-particle. This has immediate implication for Villata's claim that particles and antiparticles behave oppositely in a (normal) matter-produced gravitational field: it predicts that gravitational lensing cannot happen, since if light is bent one way by a gravitational field, Villata's stance predicts it's also bent in the the reverse direction.

Your stance is different from Villata's, but the corresponding statement for your theory is that light cannot produce a gravitational field. That's a lot harder to specifically test for, but it's clear that it's massively inconsistent with general relativity.


Can a neutron star ever be less than about 1.44 solar masses (the Chandrasekhar limit)? Why not? I learned about the Chandrasekhar limit as being the UPPER limit, in terms of mass, for a white dwarf...
But, I have never heard of a neutron star being BELOW that mass, so I have wondered, recently, if that is also a LOWER limit for neutron stars...
I suppose that the squished-together protons and electrons that the neutron star is made of might lose their 'degeneracy', if the gravitational potential is (or becomes) low enough, and the star might 'poof-out' (or proof back out) to a white dwarf...
Has this ever been known to happen?
Can a neutron star experience Hawking radiation and actually lose mass, like a black hole?
If black holes can, theoretically at least, exist at very small masses as long as they are small enough voluminously (compacted within their Swarzschild or Kerr radii), then why can't neutron stars?

A number of neutron stars in binary systems have measured masses below $1.44 M_\odot$ (e.g. a pulsar of mass $1.251 \pm 0.021 M_\odot$, McKee et al. 2020). I think the current lowest mass contender is $1.174 \pm 0.004 M_\odot$ (Martinez et al. (2015). See the plot below with a pictorial representation of the current state of neutron star mass measurements from Horvath et al. (2020). Conveniently, the vertical line is I think at about 1.4 solar masses.

The Chandrasekhar mass for a ball of iron at the centre of a core collapse supernova is more like $1.15 M_\odot$ because ionised iron has 2.15 mass units per electron, rather than the 2 of carbon or oxygen, and because the electron energy required to neutronise iron is much lower than for carbon and oxygen.
The theoretical lower limit to the neutron star mass is about $0.1-0.2M_\odot$, but none have been observed below $1M_\odot$ and there isn't any known astrophysical mechanism to produce them.
A much fuller answer to this last part, which I won't cut and paste here can be found at Physics SE. In brief, the lower limit arises because the adiabatic index of material in the interior falls below 4/3 (because of inverse beta decay,  the formation of neutron-rich heavy nuclei and the disappearance of some free neutrons) and the material is too compressible to form a stable star with a negative binding energy.


Can red dwarf stars have a spectral type of L? We have the "normal" main sequence stars, OBAFGKM. Below K and M are the brown dwarfs, L,T, and Y. Zooming into the area at the edges of M and L, we have spectral types M9, M9.5, L0, and L1. So I am wondering, is there any red dwarf that has a spectral type of L?

Yes. Stars (those objects that are supported by hydrogen fusion) can be as cool as spectral type L2.
Brown dwarfs can be as warm as M4/5 when they are young.
i.e. there isn't a clear spectral type Vs mass relationship. It depends on age too.


Cosmological redshift - How do we know it's not caused by the observer's time dilation? Instead of cosmological redshift being caused by the metric expansion of space, why couldn't cosmological redshift be caused by our perspective moving faster in time?  Said another way, if the observer moved through time faster than the emission source, couldn't this time dilation be an explanation of cosmological redshift?  Instead of the metric expansion of space, why not "cosmic time dilation"?

As far as I know, there is no physical mechanism in physics to explain how the metric expansion can cause red shift.  It appears that we assume metric expansion can cause red shift with no underlying physics or testing.

Alternatively, red shift caused by time dilation is well tested, proven, and understood.  It is something we can play with here on Earth.

Wouldn't assuming cosmological redshift is caused by the observer moving faster in time than the emission source in the past makes more sense than the metric expansion of space causing cosmological redshift?

Of course when measuring redshift from distant sources, the Doppler and gravitational effects must be accounted for.  That's not what I'm asking about.

Edit:  I probably need some math help, or a pointer to a better example to play with.

With a Hubble Constant of 71.9, a lightyear expands about $7.3483\times10^{-11}\ \%$ per year.  The Pinwheel Galaxy is 20.87 million light years light years away.  If I continuously redshift the Pinwheel Galaxy by the expansion of space, I get a redshift of .00153 using this equation:

$z = 1 - e^{(7.3483\times10^{-11})(2.087\times10^{7})}\ $

The pinwheel galaxy has a measured Redshift of .000804.

Is there an equation that gives a better result without magic values for $z$ or a real world example with known values that works?

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

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

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

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


A simple question about Saha Equation The Saha Equation:

$$\frac{N_{i+1}}{N_{i}} = \frac{g_{i+1}}{g_{i}}\frac{g_e}{N_e h^3}(2 \pi m_{e} kT)^{3/2} \exp(- \chi_i / kT)$$

where $\chi_i$ is ionization potential. My problem is about the sign ($+$ or $-$) of $\chi_i$. I know that its formula is something like:

$$\chi_i = (13.6\ \mathrm{eV}) Z^2 \left(\frac{1}{n_i^2} - \frac{1}{n_j^2}\right)$$

and in this case $n_j$ is $\infty$. But I get confused about it's sign ($+$ or $-$). Can someone simply say what is the sign of the formula for ionization potential and the sign of the formula for the Energy we need to go from $n_i$ to $n_j$.

I mean: What is the complete formula for $E_m - E_n$ and also $\chi_{n}$? (with the right sign so they can be used in the first Saha Equation above)

Just to provide a source, the Ionization potential for a hydrogen-like atom is given by:

$$\chi_n = \frac{(13.6\:\mathrm{eV})Z^2}{n^2}$$

This is the energy necessary to ionize an electron in a hydrogen-like atom that "begins" at the $n^\mathrm{th}$ energy level. This equation is positive because you need to input energy to the electron to break it out of its potential well. Note that you may see a nearly identical equation for the energy of the electrons which will be negative. That implies the electrons themselves have negative energy because they're in the electric potential well and you need a positive amount of energy to ionize them.

Now, when you look at the Saha equation, you see a negative sign in the exponential term: $\exp(-\chi_i/kT)$. It was already established that $\chi_i$ is positive and so also are $k$ and $T$ simply by their nature. The negative sign comes from the derivation and meaning of this exponential term. Describing the full derivation would involve an entire lecture, but to point you along the right track, the exponential term is known as the Boltzmann Factor and ultimately comes from the fact that particle energies are very often distributed according to the Boltzmann distribution. That negative sign is simply an artifact of the Boltzmann distribution concerning how particles like to probabilistically arrange their energies. If that sign was a positive, the universe would just blow up so be careful with your signs!


Double Planet Orbiting Wide Binary Star? Right now I'm writing a sci-fi story that develops in a double planet. But that's not all, the double planet (along with a couple more planets) orbit a big star and the whole system is orbited by another star (about 100 AU from the central star.)

I need to know how implausible this is. I don't want my story to drift from speculation into impossible worlds. I'm guessing that the occurrence of such a planetary arrangement is very low likely, but is such a system even possible at all?

Now, if it's possible, how messed up the days and seasons would be? I'm bending my brain to figure out what would be the length of day and night in such a system (the planets are very close to each other.)

Any help with this will be really appreciated.

Generally speaking, there's 2 types of planet systems in a binary - see pretty picture.   Source.  The writer is a sci-fi writer so the entire article might be of interest to you.

Due to binary star tidal forces, there are some setups that are unlikely but in your example with a 100 AU distance between the stars, an orbit around the larger star is reasonable.

A double planet system is less likely.   It's unlikely to form on it's own during planet formation as that requires too much planetary angular momentum during formation.

It's possible, but also unlikely to form by giant impact, as that's more likely to leave 1 planet and 1 moon.  I've read (but can't find an article right now) that there's a giant impact size ratio and it's in the planet-moon range, not planet planet.   Much less than 1 to 1.  Pluto-Charon is 9-1 and Earth-Moon 81-1.   A giant impact is also unlikely to create an Iron rich core for both objects.   It's not a good way to create planet-planet.

That leaves a 3rd possibility, also unlikely, but perhaps the most likely of the bunch is planet capture.    Planets can form in Trojan points in the same orbit (Theia).  The difficulty with orbital capture is that the velocity needs to be just right and capture's are likely to be significantly elongated orbits, which, maybe, over time, perhaps with a 3rd asteroid, could even out to slightly more circular.   This is very improbable, but it might be the most likely way to form a double planet.

My 2 cents, as a layman.


Stellarium simulation I'm interested lately in Stellarium and the functionalities which it provides especially scripting. I want to make a simulation that shows the celestial sphere with some planets like Venus, Mars, and the Pole-star.

I would make a kind of animation that starts from a departure point (France) and arrives in (Canada), but in this animation, I couldn't figure out how to synchronise between time and position (Supposing that I'm travelling to Canada at a fixed speed like in a vessel).

I've already looked for some examples in the scripts folder and I read the documentation and I found that I can use setObserverLocation() and maybe setDate() but I don't know to synchronise between both of them to make the animation realistic.

This doesn't really answer your question, but may help you get started. If you run https://github.com/barrycarter/bcapps/blob/master/STACK/bc-stell-travel.pl and follow the instructions at https://github.com/barrycarter/bcapps/blob/master/STACK/README you can create a video similar to https://www.youtube.com/watch?v=_4ej5rfYm4A but higher quality (youtube lowers the quality of uploaded videos).

With the default settings, it's a ~6 hour Mach 1 flight from Menton, France to Rampart House, YT Canada starting at 2017-02-10 at 00:00:00 GMT at approximately 120x regular speed.

The only vaguely interesting things about the default flight:

Since the flight goes fairly far north, you can see Polaris rising in the Northern sky for most of the flight.
Since you traveling westwards fairly quickly, once you get far north enough, the stars go "backwards": they rise in the west and set in the east.


Was comet ISON in a Kepler orbit prior to disintegrating? Was comet ISON in a Kepler orbit prior to disintegrating? If so, it is possible over time that Kepler orbits can be altered by increases or decreases in the mass of of celestial bodies?
For instance if ISON was in a Kepler orbit and it lost mass did it maintain its orbit and disintegrate in its orbital track or does it alter its track due to the loss of mass?

To your first question, yes, it was in a Kepler orbit, (this interactive model shows just how it appeared). ISON never completed a full circuit around the Sun since this was it's first time coming out of the Oort Cloud and it disintegrated.

Further answering, the orbit of a body around another in space is not affected by it's mass. Remember the experiment that proves two object fall at the same speed regardless of it's mass (here the experiment is made on the moon with a feather and a hammer) and remember that orbiting is just falling at a certain speed and angle in respect to another body.


Does the Sun turn around a big star? The Moon orbits the Earth.
The Earth orbits the Sun.

Does the Sun orbit another bigger star?
If so, does this star orbit, in turn, a very big star?
... etc ...

What are all the intermediate subsystems up to motion around the center of the Milky Way?

The Sun is not within the gravitational sphere of influence of any other star. The centre of mass of the solar system (which is very close to the Sun) instead orbits in the general Galactic gravitational potential. Because this has a roughly cylindrical symmetry (the Galaxy is basically a disk with a bulge in the middle), this means that it executes a (roughly) circular orbit around the centre of the Galaxy, taking about 230 million years to do so. At the same time it is wobbling in the vertical plane of the Galactic disc, up and down with a cycle of about 70 million years (see How far is the Earth/Sun above/below the galactic plane, and is it heading toward/away from it? )

There is really nothing intermediate because the stars in the milky way form a "collisionless system" they don't really interact gravitationally on an individual basis. Stars can be influenced by perturbations in the smooth Galactic potential caused by massive star clusters, giant molecular clouds and spiral arms. This is thought to be why the velocity dispersions of stars around regular circular orbits increases with age.


Virial Shock Heating In the review paper on protoclusters, Roderik Overzier (pg 25) mentions the concept of virial shock heating which seems to be a process involved with galaxy cluster formation.

I'm not fully certain what a virialized galaxy cluster is apart from the fact that it's in hydrostatic equilibrium, and is one of the key distinguishing features between a protocluster and a fully formed galaxy cluster.

I vaguely understand that shock heating relates to the heating that occurs when something is travelling at supersonic speed within a medium and produces a shock.

Virial shocks appear when gas falls in toward a galaxy, and the shocked gas stays hot so the pressure in the post-shock gas can support the shock and keep it away from the galactic disk.
 The alternative is if the post-shock gas cools and loses pressure support, and the (non-virial) shock collapses right down to the galactic disk.  The significance is that if gas falls directly onto the high-density disk, it has different effects than if it meets a low-density shock well away from the galaxy.  There's a lot of information at https://academic.oup.com/mnras/article/345/1/349/984798


If a black hole has a mass similar to a star, why does it have a different gravitational pull? Say there is a black hole with 1 solar mass, having the sun's mass, it would have the same gravity as the sun, right? But it still has an event horizon, so why does it have such a strong gravitational pull despite its mass?

Actually, it doesn't. If a black hole and a star have the same mass, their gravity is the same. As the formula for the attraction (gravity) between two objects is given as $F=\dfrac{GMm}{r^2}$, where $G$ is the gravitational constant, $M$ is the mass of the black hole/star, $m$ is the mass of the smaller object, and $r$ is the distance between the two. Even though a black hole is tiny and a star is much, much bigger than a black hole, the radius of each object does not matter. It only depends on mass and distance.
The reason why it seems like the gravitational pull is stronger is because the black hole indeed is much smaller than the star with the same mass. If you had a $1 M_\odot$ star and a black hole of the same mass, at distances less than $1 R_\odot$ would mean that you would be inside the star,  and still outside the black hole. Therefore, you would feel gravity from the black hole and accelerate toward it, while (assuming you weren't vaporized by the Sun's internal temperatures, you would experience drag from the hydrogen gas, as well as (even if the gas was essentially dragless) more mass being concentrated in the other direction (away from your vector of falling) as a star is not a point-like particle, its mass is distributed throughout the star.


Do Pluto and Charon have unusual Lagrange points? The usual examples of Lagrange points one most commonly encounters, Sun-Earth  and Earth-Moon Lagrange points, are examples of 3-body problems where $M_1\gg M_2\gg M_3$. The Pluto-Charon system, however, are much closer in their relative masses, so much so that their barycenter is outside Pluto's surface. From Wikipedia:

  Pluto and Charon are sometimes considered a binary system because the barycenter of their orbits does not lie within either body. The IAU has not formalized a definition for binary dwarf planets, and Charon is officially classified as a moon of Pluto.

How does this affect the orbital stability of the five Pluto-Charon Lagrange points?

The L1, L2, and L3 points are unstable in any orbital system. (source)

The L4 and L5 points of a pair of bodies are only stable if the larger of the bodies is at least 25 times as massive than the smaller (source).  The ratio of the Pluto/Charon system is only 8.7.  Because of this, none of the Lagrange points are stable, and an object orbiting at any of them will require active station-keeping to compensate for perturbations in the orbit.


How can I clean my lenses without causing scratching or damages to the lens surface? Sometimes you can be trying to observe very faint or very small objects in the sky and it wouldn't pay to have these obscured or marred by a scratch or dust on the lenses.

I keep to good practices and make sure to replace the cap on the lenses when finished in order to stop dust buildup but inevitably it will get a little dusty and dirty, taking this into consideration:

What steps do I need to take to ensure that my lenses are kept in top condition?

Can I simply use a damp cloth or sponge and dab the lenses or should I use a softer material, like someone with glasses might have a softer lens cleaner?

Do I require a special cleaning fluid or can a good quality clean be done without any additional cleaning agents?

Clean your eyepieces in the same way you would clean a camera lens. An air blower, soft (i.e. camel hair) brush, lens tissue or microfibre cloth, and a solvent or alcohol will do the trick.

When I was an undergraduate I did some work with lasers and very expensive dielectric coated mirrors. The techniques here will show you how to carefully clean optics. Never completely wet your lens tissue or cloth. One drop on a lens tissue is usually enough, or a couple of drops on a microfibre cloth.

Essentially, you need to:

Use a blower to blow off as much dust as you can.
Then use a soft brush.
Dampen a small square of lens tissue with pure alcohol or acetone (see warning below). Gently draw or wipe the tissue across the surface once, and then discard. Repeat if necessary.
You may need to fog the lens with your breath to help remove water-soluble gunk.

Be insanely careful if you are using a solvent such as acetone. Although a good for cleaning optics, it's not very nice to paint and plastic so it should be a last resort for very stubborn marks well away from the edges of the lens (so you'll never touch the paint with it). Methyl or isopropyl alcohol are safer but less effective, or a generic lens cleaner from a camera store.


Is there any evidence that the Gas Giant planets in our solar system are experiencing orbital migration? Planetary migration is defined by Lubow and Ida (2010) in their article Planet Migration as

  the process by which a planet’s orbital radius changes in time. The
  main agent for causing gas giant planet migration is the gravitational interaction of the young
  planet with the gaseous disk from which it forms.

Relevant, it is believed by some that the gas giant planets in our solar system experienced orbital migration.  According to the article Wandering Gas Giants and Lunar Bombardment (Taylor, 2006), as an example,

  Outward migration of Saturn might have triggered a dramatic increase in the bombardment rate on the Moon 3.9 billion years ago, an idea testable with lunar samples.

So my question is is there any evidence (not opinions) to suggest any form of migration occurring with our solar system's gas giants?

There is no direct evidence of change of orbits of the gaseous planets of the Solar System. But one could wonder about the stability of the Solar System, and if such events could happen again.

Theory:

The Solar System is a chaotic system, as most of gravitational systems involving $N$ bodies. The KAM theorems (for Kolmogorov, Arnold and Moser, the three authors of the theory) show that, for non-degenerate Hamiltonian systems, quasiperiodic trajectories exist, that could ensure the stability of the system. But unfortunatly the KAM theorems cannot be applied to the Solar System, because the planets are too massive.

Simulations:

It leaves us with numerical simulations, that show that the Solar System is indeed chaotic, and that planetary collisions, ejection and migration can happen. But if it is true for the telluric planets, it is not the case for the gaseous planets: one reason could be that they are not much perturbated by the inner planets.

Sources:

Laskar (1994)
Laskar (2012)


Exo Planets and Planets difference? After hearing that it has been confirmed that we found seven similar exoplanets like earth. I thought exoplanets were dead stars. After doing a little bit of research I found out that exoplanets are planets that are orbiting a star, NOT the sun. But isn't a star like the Sun? Can someone please explain this to me?

Not all planets are exoplanets, but all exoplanets are planets.

When we differentiate the two, we use "planets" to refer to the eight planets revolving around our Sun. The Greek prefix "exo" (outside) should indicate that exoplanets orbit other stars.

In reality, you can call exoplanets "planets", and you'll often hear "Astronomers have found x planets around star y". The word "exoplanet" is used interchangeably with planets from other planetary systems.


app to locate star by coordinates I've named a star and would love to be able to find it in the sky - is there an app I can use to locate stars given a specific coordinate?

There are apps which will help guide you to more significant stars (Google skymap comes to mind if you're looking for a simple mobile system - Stellarium may also be a good option), but it's highly unlikely that any app will include the location and names of so-called 'personal' stars, as the bodies who sell such things don't have any the authority to do so and are not generally recognized in the actual world of Astronomy.

Note that the IAU (International Astronomy Union) is the body generally recognized to have the responsibility of naming and classification of stars and other bodies. They also come out strongly against such practices and services:

  As an international scientific organization, the IAU dissociates itself entirely from the commercial practice of "selling" fictitious star names or "real estate" on other planets or moons in the Solar System. Accordingly, the IAU maintains no list of the (several competing) enterprises in this business in individual countries of the world.  Readers wanting to contact such enterprises despite the explanations given below should search commercial directories in their country of origin.

Link to IAU page on buying of star names